practice and homework lesson 9.4 answer key

9.1 Verifying Trigonometric Identities and Using Trigonometric Identities to Simplify Trigonometric Expressions

sin 2 θ − 1 tan θ sin θ − tan θ = ( sin θ + 1 ) ( sin θ − 1 ) tan θ ( sin θ − 1 ) = sin θ + 1 tan θ sin 2 θ − 1 tan θ sin θ − tan θ = ( sin θ + 1 ) ( sin θ − 1 ) tan θ ( sin θ − 1 ) = sin θ + 1 tan θ

This is a difference of squares formula: 25 − 9 sin 2 θ = ( 5 − 3 sin θ ) ( 5 + 3 sin θ ) . 25 − 9 sin 2 θ = ( 5 − 3 sin θ ) ( 5 + 3 sin θ ) .

9.2 Sum and Difference Identities

2 + 6 4 2 + 6 4

2 − 6 4 2 − 6 4

1 − 3 1 + 3 1 − 3 1 + 3

cos ( 5 π 14 ) cos ( 5 π 14 )

9.3 Double-Angle, Half-Angle, and Reduction Formulas

cos ( 2 α ) = 7 32 cos ( 2 α ) = 7 32

cos 4 θ − sin 4 θ = ( cos 2 θ + sin 2 θ ) ( cos 2 θ − sin 2 θ ) = cos ( 2 θ ) cos 4 θ − sin 4 θ = ( cos 2 θ + sin 2 θ ) ( cos 2 θ − sin 2 θ ) = cos ( 2 θ )

cos ( 2 θ ) cos θ = ( cos 2 θ − sin 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ cos ( 2 θ ) cos θ = ( cos 2 θ − sin 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ

10 cos 4 x = 10 ( cos 2 x ) 2 = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x . = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ] = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x . = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x ) = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x ) = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x ) 10 cos 4 x = 10 ( cos 2 x ) 2 = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x . = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ] = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x . = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x ) = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x ) = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x )

− 2 5 − 2 5

9.4 Sum-to-Product and Product-to-Sum Formulas

1 2 ( cos 6 θ + cos 2 θ ) 1 2 ( cos 6 θ + cos 2 θ )

1 2 ( sin 2 x + sin 2 y ) 1 2 ( sin 2 x + sin 2 y )

− 2 − 3 4 − 2 − 3 4

2 sin ( 2 θ ) cos ( θ ) 2 sin ( 2 θ ) cos ( θ )

tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ

9.5 Solving Trigonometric Equations

x = 7 π 6 , 11 π 6 x = 7 π 6 , 11 π 6

π 3 ± π k π 3 ± π k

θ ≈ 1.7722 ± 2 π k θ ≈ 1.7722 ± 2 π k and θ ≈ 4.5110 ± 2 π k θ ≈ 4.5110 ± 2 π k

cos θ = − 1 , θ = π cos θ = − 1 , θ = π

π 2 , 2 π 3 , 4 π 3 , 3 π 2 π 2 , 2 π 3 , 4 π 3 , 3 π 2

9.1 Section Exercises

All three functions, F F , G G , and H H , are even.

This is because F ( − x ) = sin ( − x ) sin ( − x ) = ( − sin x ) ( − sin x ) = sin 2 x = F ( x ) F ( − x ) = sin ( − x ) sin ( − x ) = ( − sin x ) ( − sin x ) = sin 2 x = F ( x ) , G ( − x ) = cos ( − x ) cos ( − x ) = cos x cos x = cos 2 x = G ( x ) G ( − x ) = cos ( − x ) cos ( − x ) = cos x cos x = cos 2 x = G ( x ) and H ( − x ) = tan ( − x ) tan ( − x ) = ( − tan x ) ( − tan x ) = tan 2 x = H ( x ) . H ( − x ) = tan ( − x ) tan ( − x ) = ( − tan x ) ( − tan x ) = tan 2 x = H ( x ) .

When cos t = 0 , cos t = 0 , then sec t = 1 0 , sec t = 1 0 , which is undefined.

sin x sin x

sec x sec x

csc t csc t

sec 2 x sec 2 x

sin 2 x + 1 sin 2 x + 1

1 sin x 1 sin x

1 cot x 1 cot x

tan x tan x

− 4 sec x tan x − 4 sec x tan x

± 1 cot 2 x + 1 ± 1 cot 2 x + 1

± 1 − sin 2 x sin x ± 1 − sin 2 x sin x

Answers will vary. Sample proof:

cos x − cos 3 x = cos x ( 1 − cos 2 x ) = cos x sin 2 x cos x − cos 3 x = cos x ( 1 − cos 2 x ) = cos x sin 2 x

Answers will vary. Sample proof: 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = sec 2 x + tan 2 x = tan 2 x + 1 + tan 2 x = 1 + 2 tan 2 x 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = sec 2 x + tan 2 x = tan 2 x + 1 + tan 2 x = 1 + 2 tan 2 x

Answers will vary. Sample proof: cos 2 x − tan 2 x = 1 − sin 2 x − ( sec 2 x − 1 ) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x cos 2 x − tan 2 x = 1 − sin 2 x − ( sec 2 x − 1 ) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x

Proved with negative and Pythagorean identities

True 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3 ( sin 2 θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3 ( sin 2 θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ

9.2 Section Exercises

The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures x , x , the second angle measures π 2 − x . π 2 − x . Then sin x = cos ( π 2 − x ) .   sin x = cos ( π 2 − x ) .   The same holds for the other cofunction identities. The key is that the angles are complementary.

sin ( − x ) = − sin x , sin ( − x ) = − sin x , so sin x sin x is odd. cos ( − x ) = cos ( 0 − x ) = cos x , cos ( − x ) = cos ( 0 − x ) = cos x , so cos x cos x is even.

6 − 2 4 6 − 2 4

− 2 − 3 − 2 − 3

− 2 2 sin x − 2 2 cos x − 2 2 sin x − 2 2 cos x

− 1 2 cos x − 3 2 sin x − 1 2 cos x − 3 2 sin x

csc θ csc θ

cot x cot x

tan ( x 10 ) tan ( x 10 )

sin ( a − b ) = ( 4 5 ) ( 1 3 ) − ( 3 5 ) ( 2 2 3 ) = 4 − 6 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) − ( 4 5 ) ( 2 2 3 ) = 3 − 8 2 15 sin ( a − b ) = ( 4 5 ) ( 1 3 ) − ( 3 5 ) ( 2 2 3 ) = 4 − 6 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) − ( 4 5 ) ( 2 2 3 ) = 3 − 8 2 15

cot ( π 6 − x ) cot ( π 6 − x )

cot ( π 4 + x ) cot ( π 4 + x )

sin x 2 + cos x 2 sin x 2 + cos x 2

They are the same.

They are the different, try g ( x ) = sin ( 9 x ) − cos ( 3 x ) sin ( 6 x ) . g ( x ) = sin ( 9 x ) − cos ( 3 x ) sin ( 6 x ) .

They are the different, try g ( θ ) = 2 tan θ 1 − tan 2 θ . g ( θ ) = 2 tan θ 1 − tan 2 θ .

They are different, try g ( x ) = tan x − tan ( 2 x ) 1 + tan x tan ( 2 x ) . g ( x ) = tan x − tan ( 2 x ) 1 + tan x tan ( 2 x ) .

− 3 − 1 2 2 , or  − 0.2588 − 3 − 1 2 2 , or  − 0.2588

1 + 3 2 2 , 1 + 3 2 2 , or 0.9659

tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 − tan x tan ( π 4 ) = tan x + 1 1 − tan x ( 1 ) = tan x + 1 1 − tan x tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 − tan x tan ( π 4 ) = tan x + 1 1 − tan x ( 1 ) = tan x + 1 1 − tan x

cos ( a + b ) cos a cos b = cos a cos b cos a cos b − sin a sin b cos a cos b = 1 − tan a tan b cos ( a + b ) cos a cos b = cos a cos b cos a cos b − sin a sin b cos a cos b = 1 − tan a tan b

cos ( x + h ) − cos x h = cos x cosh − sin x sinh − cos x h = cos x ( cosh − 1 ) − sin x sinh h = cos x cos h − 1 h − sin x sin h h cos ( x + h ) − cos x h = cos x cosh − sin x sinh − cos x h = cos x ( cosh − 1 ) − sin x sinh h = cos x cos h − 1 h − sin x sin h h

True. Note that   sin ( α + β ) = sin ( π − γ )     sin ( α + β ) = sin ( π − γ )   and expand the right hand side.

9.3 Section Exercises

Use the Pythagorean identities and isolate the squared term.

1 − cos x sin x , sin x 1 + cos x , 1 − cos x sin x , sin x 1 + cos x , multiplying the top and bottom by 1 − cos x 1 − cos x and 1 + cos x , 1 + cos x , respectively.

a) 3 7 32 3 7 32 b) 31 32 31 32 c) 3 7 31 3 7 31

a) 3 2 3 2 b) − 1 2 − 1 2 c) − 3 − 3

cos θ = − 2 5 5 , sin θ = 5 5 , tan θ = − 1 2 , csc θ = 5 , sec θ = − 5 2 , cot θ = − 2 cos θ = − 2 5 5 , sin θ = 5 5 , tan θ = − 1 2 , csc θ = 5 , sec θ = − 5 2 , cot θ = − 2

2 sin ( π 2 ) 2 sin ( π 2 )

2 − 2 2 2 − 2 2

2 − 3 2 2 − 3 2

2 + 3 2 + 3

− 1 − 2 − 1 − 2

a) 3 13 13 3 13 13 b) − 2 13 13 − 2 13 13 c) − 3 2 − 3 2

a) 10 4 10 4 b) 6 4 6 4 c) 15 3 15 3

120 169 , – 119 169 , – 120 119 120 169 , – 119 169 , – 120 119

2 13 13 , 3 13 13 , 2 3 2 13 13 , 3 13 13 , 2 3

cos ( 74° ) cos ( 74° )

cos ( 18 x ) cos ( 18 x )

3 sin ( 10 x ) 3 sin ( 10 x )

− 2 sin ( − x ) cos ( − x ) = − 2 ( − sin ( x ) cos ( x ) ) = sin ( 2 x ) − 2 sin ( − x ) cos ( − x ) = − 2 ( − sin ( x ) cos ( x ) ) = sin ( 2 x )

sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos 2 θ − sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan 2 θ = sin ( θ ) cos θ tan 2 θ = cot ( θ ) tan 2 θ = tan 3 θ sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos 2 θ − sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan 2 θ = sin ( θ ) cos θ tan 2 θ = cot ( θ ) tan 2 θ = tan 3 θ

1 + cos ( 12 x ) 2 1 + cos ( 12 x ) 2

3 + cos ( 12 x ) − 4 cos ( 6 x ) 8 3 + cos ( 12 x ) − 4 cos ( 6 x ) 8

2 + cos ( 2 x ) − 2 cos ( 4 x ) − cos ( 6 x ) 32 2 + cos ( 2 x ) − 2 cos ( 4 x ) − cos ( 6 x ) 32

3 + cos ( 4 x ) − 4 cos ( 2 x ) 3 + cos ( 4 x ) + 4 cos ( 2 x ) 3 + cos ( 4 x ) − 4 cos ( 2 x ) 3 + cos ( 4 x ) + 4 cos ( 2 x )

1 − cos ( 4 x ) 8 1 − cos ( 4 x ) 8

3 + cos ( 4 x ) − 4 cos ( 2 x ) 4 ( cos ( 2 x ) + 1 ) 3 + cos ( 4 x ) − 4 cos ( 2 x ) 4 ( cos ( 2 x ) + 1 )

( 1 + cos ( 4 x ) ) sin x 2 ( 1 + cos ( 4 x ) ) sin x 2

4 sin x cos x ( cos 2 x − sin 2 x ) 4 sin x cos x ( cos 2 x − sin 2 x )

2 tan x 1 + tan 2 x = 2 sin x cos x 1 + sin 2 x cos 2 x = 2 sin x cos x cos 2 x + sin 2 x cos 2 x = 2 sin x cos x . cos 2 x 1 = 2 sin x cos x = sin ( 2 x ) 2 tan x 1 + tan 2 x = 2 sin x cos x 1 + sin 2 x cos 2 x = 2 sin x cos x cos 2 x + sin 2 x cos 2 x = 2 sin x cos x . cos 2 x 1 = 2 sin x cos x = sin ( 2 x )

2 sin x cos x 2 cos 2 x − 1 = sin ( 2 x ) cos ( 2 x ) = tan ( 2 x ) 2 sin x cos x 2 cos 2 x − 1 = sin ( 2 x ) cos ( 2 x ) = tan ( 2 x )

sin ( x + 2 x ) = sin x cos ( 2 x ) + sin ( 2 x ) cos x = sin x ( cos 2 x − sin 2 x ) + 2 sin x cos x cos x = sin x cos 2 x − sin 3 x + 2 sin x cos 2 x = 3 sin x cos 2 x − sin 3 x sin ( x + 2 x ) = sin x cos ( 2 x ) + sin ( 2 x ) cos x = sin x ( cos 2 x − sin 2 x ) + 2 sin x cos x cos x = sin x cos 2 x − sin 3 x + 2 sin x cos 2 x = 3 sin x cos 2 x − sin 3 x

1 + cos ( 2 t ) sin ( 2 t ) − cos t = 1 + 2 cos 2 t − 1 2 sin t cos t − cos t = 2 cos 2 t cos t ( 2 sin t − 1 ) = 2 cos t 2 sin t − 1 1 + cos ( 2 t ) sin ( 2 t ) − cos t = 1 + 2 cos 2 t − 1 2 sin t cos t − cos t = 2 cos 2 t cos t ( 2 sin t − 1 ) = 2 cos t 2 sin t − 1

( cos 2 ( 4 x ) − sin 2 ( 4 x ) − sin ( 8 x ) ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( 8 x ) − sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 x ) ) = cos 2 ( 8 x ) − sin 2 ( 8 x ) = cos ( 16 x ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) − sin ( 8 x ) ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( 8 x ) − sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 x ) ) = cos 2 ( 8 x ) − sin 2 ( 8 x ) = cos ( 16 x )

9.4 Section Exercises

Substitute   α     α   into cosine and   β     β   into sine and evaluate.

Answers will vary. There are some equations that involve a sum of two trig expressions where when converted to a product are easier to solve. For example: sin ( 3 x ) + sin x cos x = 1.   sin ( 3 x ) + sin x cos x = 1.   When converting the numerator to a product the equation becomes: 2 sin ( 2 x ) cos x cos x = 1 2 sin ( 2 x ) cos x cos x = 1

8 ( cos ( 5 x ) − cos ( 27 x ) ) 8 ( cos ( 5 x ) − cos ( 27 x ) )

sin ( 2 x ) + sin ( 8 x ) sin ( 2 x ) + sin ( 8 x )

1 2 ( cos ( 6 x ) − cos ( 4 x ) ) 1 2 ( cos ( 6 x ) − cos ( 4 x ) )

2 cos ( 5 t ) cos t 2 cos ( 5 t ) cos t

2 cos ( 7 x ) 2 cos ( 7 x )

2 cos ( 6 x ) cos ( 3 x ) 2 cos ( 6 x ) cos ( 3 x )

1 4 ( 1 + 3 ) 1 4 ( 1 + 3 )

1 4 ( 3 − 2 ) 1 4 ( 3 − 2 )

1 4 ( 3 − 1 ) 1 4 ( 3 − 1 )

cos ( 80° ) − cos ( 120° ) cos ( 80° ) − cos ( 120° )

1 2 ( sin ( 221° ) + sin ( 205° ) ) 1 2 ( sin ( 221° ) + sin ( 205° ) )

2 cos ( 31° ) 2 cos ( 31° )

2 cos ( 66.5° ) sin ( 34.5° ) 2 cos ( 66.5° ) sin ( 34.5° )

2 sin ( −1.5° ) cos ( 0.5° ) 2 sin ( −1.5° ) cos ( 0.5° )

2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) − 2 ( sin ( 4 x ) cos ( 3 x ) − sin ( 3 x ) cos ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) = 4 sin ( 3 x ) cos ( 4 x ) 2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) − 2 ( sin ( 4 x ) cos ( 3 x ) − sin ( 3 x ) cos ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) = 4 sin ( 3 x ) cos ( 4 x )

sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 4 sin x cos 2 x sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 4 sin x cos 2 x

2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( .5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) ) 2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( .5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) )

2 cos ( 35° ) cos ( 23° ) , 1.5081 2 cos ( 35° ) cos ( 23° ) , 1.5081

− 2 sin ( 33° ) sin ( 11° ) , − 0.2078 − 2 sin ( 33° ) sin ( 11° ) , − 0.2078

1 2 ( cos ( 99° ) − cos ( 71° ) ) , −0.2410 1 2 ( cos ( 99° ) − cos ( 71° ) ) , −0.2410

It is an identity.

It is not an identity, but 2 cos 3 x 2 cos 3 x is.

tan ( 3 t ) tan ( 3 t )

2 cos ( 2 x ) 2 cos ( 2 x )

− sin ( 14 x ) − sin ( 14 x )

Start with cos x + cos y . cos x + cos y . Make a substitution and let x = α + β x = α + β and let y = α − β , y = α − β , so cos x + cos y cos x + cos y becomes cos ( α + β ) + cos ( α − β ) = cos α cos β − sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β cos ( α + β ) + cos ( α − β ) = cos α cos β − sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β

Since x = α + β x = α + β and y = α − β , y = α − β , we can solve for α α and β β in terms of x and y and substitute in for 2 cos α cos β 2 cos α cos β and get 2 cos ( x + y 2 ) cos ( x − y 2 ) . 2 cos ( x + y 2 ) cos ( x − y 2 ) .

cos ( 3 x ) + cos x cos ( 3 x ) − cos x = 2 cos ( 2 x ) cos x − 2 sin ( 2 x ) sin x = − cot ( 2 x ) cot x cos ( 3 x ) + cos x cos ( 3 x ) − cos x = 2 cos ( 2 x ) cos x − 2 sin ( 2 x ) sin x = − cot ( 2 x ) cot x

cos ( 2 y ) − cos ( 4 y ) sin ( 2 y ) + sin ( 4 y ) = − 2 sin ( 3 y ) sin ( − y ) 2 sin ( 3 y ) cos y = 2 sin ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y cos ( 2 y ) − cos ( 4 y ) sin ( 2 y ) + sin ( 4 y ) = − 2 sin ( 3 y ) sin ( − y ) 2 sin ( 3 y ) cos y = 2 sin ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y

cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x

tan ( π 4 − t ) = tan ( π 4 ) − tan t 1 + tan ( π 4 ) tan ( t ) = 1 − tan t 1 + tan t tan ( π 4 − t ) = tan ( π 4 ) − tan t 1 + tan ( π 4 ) tan ( t ) = 1 − tan t 1 + tan t

9.5 Section Exercises

There will not always be solutions to trigonometric function equations. For a basic example, cos ( x ) = −5. cos ( x ) = −5.

If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution.

π 3 , 2 π 3 π 3 , 2 π 3

3 π 4 , 5 π 4 3 π 4 , 5 π 4

π 4 , 5 π 4 π 4 , 5 π 4

π 4 , 3 π 4 , 5 π 4 , 7 π 4 π 4 , 3 π 4 , 5 π 4 , 7 π 4

π 4 , 7 π 4 π 4 , 7 π 4

7 π 6 , 11 π 6 7 π 6 , 11 π 6

π 18 , 5 π 18 , 13 π 18 , 17 π 18 , 25 π 18 , 29 π 18 π 18 , 5 π 18 , 13 π 18 , 17 π 18 , 25 π 18 , 29 π 18

3 π 12 , 5 π 12 , 11 π 12 , 13 π 12 , 19 π 12 , 21 π 12 3 π 12 , 5 π 12 , 11 π 12 , 13 π 12 , 19 π 12 , 21 π 12

1 6 , 5 6 , 13 6 , 17 6 , 25 6 , 29 6 , 37 6 1 6 , 5 6 , 13 6 , 17 6 , 25 6 , 29 6 , 37 6

0 , π 3 , π , 5 π 3 0 , π 3 , π , 5 π 3

π 3 , π , 5 π 3 π 3 , π , 5 π 3

π 3 , 3 π 2 , 5 π 3 π 3 , 3 π 2 , 5 π 3

0 , π 0 , π

π − sin − 1 ( − 1 4 ) , 7 π 6 , 11 π 6 , 2 π + sin − 1 ( − 1 4 ) π − sin − 1 ( − 1 4 ) , 7 π 6 , 11 π 6 , 2 π + sin − 1 ( − 1 4 )

1 3 ( sin − 1 ( 9 10 ) ) 1 3 ( sin − 1 ( 9 10 ) ) , π 3 − 1 3 ( sin − 1 ( 9 10 ) ) π 3 − 1 3 ( sin − 1 ( 9 10 ) ) , 2 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) 2 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , π − 1 3 ( sin − 1 ( 9 10 ) ) π − 1 3 ( sin − 1 ( 9 10 ) ) , 4 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) 4 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , 5 π 3 − 1 3 ( sin − 1 ( 9 10 ) ) 5 π 3 − 1 3 ( sin − 1 ( 9 10 ) )

π 6 , 5 π 6 , 7 π 6 , 11 π 6 π 6 , 5 π 6 , 7 π 6 , 11 π 6

3 π 2 , π 6 , 5 π 6 3 π 2 , π 6 , 5 π 6

0 , π 3 , π , 4 π 3 0 , π 3 , π , 4 π 3

There are no solutions.

cos − 1 ( 1 3 ( 1 − 7 ) ) cos − 1 ( 1 3 ( 1 − 7 ) ) , 2 π − cos − 1 ( 1 3 ( 1 − 7 ) ) 2 π − cos − 1 ( 1 3 ( 1 − 7 ) )

tan − 1 ( 1 2 ( 29 − 5 ) ) tan − 1 ( 1 2 ( 29 − 5 ) ) , π + tan − 1 ( 1 2 ( − 29 − 5 ) ) π + tan − 1 ( 1 2 ( − 29 − 5 ) ) , π + tan − 1 ( 1 2 ( 29 − 5 ) ) π + tan − 1 ( 1 2 ( 29 − 5 ) ) , 2 π + tan − 1 ( 1 2 ( − 29 − 5 ) ) 2 π + tan − 1 ( 1 2 ( − 29 − 5 ) )

0 , 2 π 3 , 4 π 3 0 , 2 π 3 , 4 π 3

sin − 1 ( 3 5 ) , π 2 , π − sin − 1 ( 3 5 ) , 3 π 2 sin − 1 ( 3 5 ) , π 2 , π − sin − 1 ( 3 5 ) , 3 π 2

cos − 1 ( − 1 4 ) , 2 π − cos − 1 ( − 1 4 ) cos − 1 ( − 1 4 ) , 2 π − cos − 1 ( − 1 4 )

π 3 π 3 , cos − 1 ( − 3 4 ) cos − 1 ( − 3 4 ) , 2 π − cos − 1 ( − 3 4 ) 2 π − cos − 1 ( − 3 4 ) , 5 π 3 5 π 3

cos − 1 ( 3 4 ) cos − 1 ( 3 4 ) , cos − 1 ( − 2 3 ) cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( − 2 3 ) 2 π − cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( 3 4 ) 2 π − cos − 1 ( 3 4 )

0 , π 2 , π , 3 π 2 0 , π 2 , π , 3 π 2

π 3 π 3 , cos −1 ( − 1 4 ) cos −1 ( − 1 4 ) , 2 π − cos −1 ( − 1 4 ) 2 π − cos −1 ( − 1 4 ) , 5 π 3 5 π 3

π + tan −1 ( −2 ) π + tan −1 ( −2 ) , π + tan −1 ( − 3 2 ) π + tan −1 ( − 3 2 ) , 2 π + tan −1 ( −2 ) 2 π + tan −1 ( −2 ) , 2 π + tan −1 ( − 3 2 ) 2 π + tan −1 ( − 3 2 )

2 π k + 0.2734 , 2 π k + 2.8682 2 π k + 0.2734 , 2 π k + 2.8682

π k − 0.3277 π k − 0.3277

0.6694 , 1.8287 , 3.8110 , 4.9703 0.6694 , 1.8287 , 3.8110 , 4.9703

1.0472 , 3.1416 , 5.2360 1.0472 , 3.1416 , 5.2360

0.5326 , 1.7648 , 3.6742 , 4.9064 0.5326 , 1.7648 , 3.6742 , 4.9064

sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) , 3 π 2 sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) , 3 π 2

π 2 , 3 π 2 π 2 , 3 π 2

7.2 ∘ 7.2 ∘

5.7 ∘ 5.7 ∘

82.4 ∘ 82.4 ∘

31.0 ∘ 31.0 ∘

88.7 ∘ 88.7 ∘

59.0 ∘ 59.0 ∘

36.9 ∘ 36.9 ∘

Review Exercises

sin − 1 ( 3 3 ) sin − 1 ( 3 3 ) , π − sin − 1 ( 3 3 ) π − sin − 1 ( 3 3 ) , π + sin − 1 ( 3 3 ) π + sin − 1 ( 3 3 ) , 2 π − sin − 1 ( 3 3 ) 2 π − sin − 1 ( 3 3 )

sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 )

cos ( 4 x ) − cos ( 3 x ) cos x = cos ( 2 x + 2 x ) − cos ( x + 2 x ) cos x = cos ( 2 x ) cos ( 2 x ) − sin ( 2 x ) sin ( 2 x ) − cos x cos ( 2 x ) cos x + sin x sin ( 2 x ) cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + sin x ( 2 ) sin x cos x cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + 2 sin 2 x cos 2 x = cos 4 x − 2 cos 2 x sin 2 x + sin 4 x − 4 cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x = sin 2 x ( sin 2 x + cos 2 x ) − 4 cos 2 x sin 2 x = sin 2 x − 4 cos 2 x sin 2 x cos ( 4 x ) − cos ( 3 x ) cos x = cos ( 2 x + 2 x ) − cos ( x + 2 x ) cos x = cos ( 2 x ) cos ( 2 x ) − sin ( 2 x ) sin ( 2 x ) − cos x cos ( 2 x ) cos x + sin x sin ( 2 x ) cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + sin x ( 2 ) sin x cos x cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + 2 sin 2 x cos 2 x = cos 4 x − 2 cos 2 x sin 2 x + sin 4 x − 4 cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x = sin 2 x ( sin 2 x + cos 2 x ) − 4 cos 2 x sin 2 x = sin 2 x − 4 cos 2 x sin 2 x

tan ( 5 8 x ) tan ( 5 8 x )

− 24 25 , − 7 25 , 24 7 − 24 25 , − 7 25 , 24 7

2 ( 2 + 2 ) 2 ( 2 + 2 )

2 10 , 7 2 10 , 1 7 , 3 5 , 4 5 , 3 4 2 10 , 7 2 10 , 1 7 , 3 5 , 4 5 , 3 4

cot x cos ( 2 x ) = cot x ( 1 − 2 sin 2 x ) = cot x − cos x sin x ( 2 ) sin 2 x = − 2 sin x cos x + cot x = − sin ( 2 x ) + cot x cot x cos ( 2 x ) = cot x ( 1 − 2 sin 2 x ) = cot x − cos x sin x ( 2 ) sin 2 x = − 2 sin x cos x + cot x = − sin ( 2 x ) + cot x

10 sin x − 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + 1 ) 10 sin x − 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + 1 )

− 2 2 − 2 2

1 2 ( sin ( 6 x ) + sin ( 12 x ) ) 1 2 ( sin ( 6 x ) + sin ( 12 x ) )

2 sin ( 13 2 x ) cos ( 9 2 x ) 2 sin ( 13 2 x ) cos ( 9 2 x )

3 π 4 , 7 π 4 3 π 4 , 7 π 4

0 , π 6 , 5 π 6 , π 0 , π 6 , 5 π 6 , π

3 π 2 3 π 2

No solution

0.2527 , 2.8889 , 4.7124 0.2527 , 2.8889 , 4.7124

1.3694 , 1.9106 , 4.3726 , 4.9137 1.3694 , 1.9106 , 4.3726 , 4.9137

Practice Test

sec ( θ ) sec ( θ )

− 1 2 cos θ + 3 2 sin θ − 1 2 cos θ + 3 2 sin θ

1 − cos ( 64 ∘ ) 2 1 − cos ( 64 ∘ ) 2

2 cos ( 3 x ) cos ( 5 x ) 2 cos ( 3 x ) cos ( 5 x )

4 sin ( 2 θ ) cos ( 6 θ ) 4 sin ( 2 θ ) cos ( 6 θ )

x = cos –1 ( 1 5 ) x = cos –1 ( 1 5 )

3 5 , – 4 5 , – 3 4 3 5 , – 4 5 , – 3 4

tan 3 x – tan x sec 2 x = tan x ( tan 2 x – sec 2 x ) = tan x ( tan 2 x – ( 1 + tan 2 x ) ) = tan x ( tan 2 x – 1 – tan 2 x ) = – tan x = tan ( – x ) = tan ( – x ) tan 3 x – tan x sec 2 x = tan x ( tan 2 x – sec 2 x ) = tan x ( tan 2 x – ( 1 + tan 2 x ) ) = tan x ( tan 2 x – 1 – tan 2 x ) = – tan x = tan ( – x ) = tan ( – x )

sin ( 2 x ) sin x – cos ( 2 x ) cos x = 2 sin x cos x sin x – 2 cos 2 x – 1 cos x = 2 cos x – 2 cos x + 1 cos x = 1 cos x = sec x = sec x sin ( 2 x ) sin x – cos ( 2 x ) cos x = 2 sin x cos x sin x – 2 cos 2 x – 1 cos x = 2 cos x – 2 cos x + 1 cos x = 1 cos x = sec x = sec x

Amplitude: 1 4 1 4 , period: 1 60 1 60 , frequency: 60 Hz

Amplitude: 8, fast period: 1 500 1 500 , fast frequency: 500 Hz, slow period: 1 10 1 10 , slow frequency: 10 Hz

D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) , 31 second

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• Authors: Jay Abramson
• Publisher/website: OpenStax
• Book title: Algebra and Trigonometry
• Publication date: Feb 13, 2015
• Location: Houston, Texas
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• Section URL: https://openstax.org/books/algebra-and-trigonometry/pages/chapter-9

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Into Math Grade 7 Module 9 Lesson 4 Answer Key Draw and Analyze Shapes to Solve Problems

We included  HMH Into Math Grade 7 Answer Key PDF   Module 9 Lesson 4 Draw and Analyze Shapes to Solve Problems to make students experts in learning maths.

HMH Into Math Grade 7 Module 9 Lesson 4 Answer Key Draw and Analyze Shapes to Solve Problems

I Can use tools to draw or construct figures that meet given criteria, and I can analyze the resulting figures.

Step It Out

C. How many triangles can be formed? Answer: Only one triangle can be formed for the given details. Explanation: From the given information Lucas has the first two strips are 4 feet long and 7 feet long. When Lucas nails them together, the two pieces of wood form a 50° angle. So, he can form only one triangle.

D. How many triangles did you draw in Parts A-C given two sides and the measure of an angle that is not between them? Answer: four triangles can be drawn as shown in above pictures.

Turn and Talk How are the situations in Step It Out 1 and Step It Out 2 different? How are they the same? Answer: As, the angle between the two lines and radius increases, the number of cuts will decreases. we observed, for larger angle and foe smaller angle and smaller units length and larger units length.

Check Understanding

On Your Own

B. Can she draw a triangle with the same side lengths as Travis’s triangle but with different angle measures? Explain. Answer: Its an equilateral triangle with same side lengths and same angle 60° as shown above. Explanation: An equilateral triangle is a triangle with all three sides of equal length , corresponding to what could also be known as a “regular” triangle.

Lesson 9.4 More Practice/Homework

Question 8. It has been raining for hours, and 0.75 inch of rain has fallen. If there are 1.5 inches of rain in a single storm, Carmen knows that the basement of the building she is in will flood. The rain begins falling at a rate of 0.175 inch per hour! Write an inequality to represent how long it can continue to rain at this rate before the basement floods. Answer: Rate of rainfall is 4.28 cm need 4hrs 28 min to fill the basement. Explanation: Inches of rain the basement will flood in = 1.5 inch Inches of rain which has already fallen     = 0.75 inch Remaining Inches of rain which will fall for the basement to flood in = 1.5 – 0.75 = 0.75 inches of rain Rate of rain fall = 0.175 per hour Time taken for the rain to fill the basement = Remaining Inches of rain ÷ Rate of rain fall = 0.75 ÷ 0.175 = 4.28 Time taken for the rain to fill the basement = 4 hours 28 minutes

Question 9. Two complementary angles have measures of (4x + 5)° and (2x + 7)°. What is the value of x? What are the measures of the angles? Answer: 13.5° Explanation: Two complementary angles have measures of (4x + 5)° and (2x + 7)° (4x + 5)° + (2x + 7)° = 90° 6x + 9 = 90 6x = 90 – 9 6x = 81 x = 13.5°

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Texas Go Math Grade 4 Lesson 9.1 Answer Key Remainders

Refer to our Texas Go Math Grade 4 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 4 Lesson 9.1 Answer Key Remainders.

Essential Question

Investigate

Materials: square tiles

Erica and 2 friends are playing a game of dominoes. There are 19 dominoes in the set. Erica wants each player to receive the same number of dominoes. Can she divide them equally among the 3 players? Why or why not? Answer: No, she divide them equally among the 3 players. Explanation: 19 dominoes can not be divided in to 3 equal parts, when we divide 1 is left over.

You can use an array to find the number of dominoes each player will receive.

D. Find the number of rows of three tiles and the number of tiles left over. Record your answer. __________ rows of 3 tiles __________ tile left over So, each player gets __________ dominoes with __________ domino left over. Answer: 6 rows of 3 tiles 1 tile left over So, each player gets 6 dominoes with 1 domino left over. Explanation: Total dominoes are 19 6 rows of 3 tiles 18 tiles 1 tile left over So, each player gets 6 dominoes with 1 domino left over.

Mathematical Processes Explain how the model helped you find the number of dominoes each player receives Why is 1 tile left over. Answer: The mode is easy to apply operation on different  numbers of elements of an array. First, we provide values like the number of elements and array. Then a number  to divide all elements of an array using a for a loop. Then we are going  to display the final numbers.

Make Connections

When a number cannot be divided evenly, the amount left over is called the remainder.

Use square tiles to find 29 ÷ 5.

• Use 29 tiles.
• Use the tiles to make rows of 5. The number of tiles left the remainder.

Mathematical Processes How do you know when there will be a remainder in a division problem? Answer: the given number is not perfectly divisible and some remainder is left

Share and Show

Use tiles or draw an array to find the quotient and remainder.

Problem Solving

H.O.T. Multi-Step What’s the Error?

Daily Assessment Task

Fill in the bubble completely to show your answer.

Question 11. Jordyn explains to her friends how to play checkers. She has 29 minutes to explain to 3 teams. She wants to take the same number of whole minutes to explain to each team. What is the greatest number of whole minutes she can take to explain the rules to each team? (A) 8 minutes (B) 4 minutes (C) 9 minutes (D) 10 minutes Answer: Option(C) Explanation: Jordyn has 29 minutes to explain to 3 teams. She wants to take the same number of whole minutes to explain to each team. 29 ÷ 3 = 9 and 5 minutes is left over. The greatest number of whole minutes she can take to explain the rules to each team is 9

Question 13. Multi-Step Barb puts 60 games into boxes to donate. Each box can hold 7 games. How many more games does Barb need so that all the boxes can have 7 games? (A) 3 games (B) 8 games (C) 2 games (D) 9 games Answer: Option(A) Explanation: Barb puts 60 games into boxes to donate. Each box can hold 7 games. Total boxes 60 ÷ 7 = 8 is quotient and 4 is remainder. Number of games Barb needs so that all the boxes can have 7 games 7 – 4 = 3

TEXAS Test Prep

Go Math Lesson 9.1 4th Grade Reminder Question 14. Rena has 23 DVDs that she wants to sort into rows of 6 each. How many DVDs will Rena have left over? (A) 3 (B) 5 (C) 17 (D) 4 Answer: Option(B) Explanation: Rena has 23 DVDs that she wants to sort into rows of 6 each. 23 ÷ 6 = 3 quotient and 5 is remainder. 5 DVDs are left over with Rena.

Texas Go Math Grade 4 Lesson 9.1 Homework and Practice Answer Key

Question 17. Miguel has 48 pieces of quartz. He wants to divide the pieces equally into 9 bags. How many pieces of quartz will be in each bag? How many pieces will be left over? Use tiles or draw an array to help you find the answer.

Question 18. Lacy has 65 tickets for the school carnival. She and 6 friends will share the tickets equally. How many tickets will each person receive? How many tickets will be left over? Use tiles or draw an array to help you find the answer.

Lesson Check

Question 19. Rebecca made 36 cookies for the school, bake sale. If she puts 5 cookies in a bag, how many bags will she use? (A) 6 (B) 5 (C) 8 (D) 7 Answer: Option(D) Explanation: Rebecca made 36 cookies for the school, bake sale. If she puts 5 cookies in a bag, Total bags she use 36 ÷ 5 = 7

Question 20. Felix has 60 towels to stack on a store shelf. He puts 8 towels in each stack. He puts any remaining towels in a drawer below the shelf. How many towels does Felix put in the drawer? (A) 7 (B) 4 (C) 6 (D) 5 Answer: Option(A) Explanation: Felix has 60 towels to stack on a store shelf. He puts 8 towels in each stack. Total stacks she arranges 60 ÷ 8 = 7 is quotient and 3 is remainder He puts 3 remaining towels in a drawer below the shelf. 7 towels Felix put in the drawer.

Question 21. Lynn is packing 52 books for her family’s move. Each box can hold 9 books. What is the least number of boxes Lynn will need to pack all of the books? (A) 7 (B) 8 (C) 6 (D) 5 Answer: Option(C) Explanation: Lynn is packing 52 books for her family’s move. Each box can hold 9 books. 52 ÷ 9 = 5 is quotient and 7 is remainder The least number of boxes Lynn will need to pack all of the books, as 7 is left , he can pack 6 boxes.

Question 22. Paul has 39 vocabulary cards to study. If he puts the cards in groups of 6, how many groups will he have, and how many cards will he have left over, if any? (A) 6 groups with 3 cards left over (B) 6 groups with 4 cards left over (C) 6 groups with no cards left over (D) 5 groups with 4 cards left over Answer: Option(A) Explanation: Paul has 39 vocabulary cards to study. If he puts the cards in groups of 6, 39 ÷ 6 = 6 is quotient and 3 is remainder He have 6 groups and 3 cards left over.

Question 23. Multi-Step Alyson has 46 beads to make bracelets. Each bracelet has 5 beads. How many more beads does Alyson need so that all the beads she has are used? (A) 9 (B) 4 (C) 1 (D) 5 Answer: Option(B) Explanation: Alyson has 46 beads to make bracelets. Each bracelet has 5 beads. 46 ÷ 5 = 9 is quotient and 1 is remainder. 4 more beads Alyson need so that all the beads she has are used, as each bracelet has 5 beads.

Question 24. Multi-Step Brad has saved $23 to spend on posters for his room. If each poster costs $4, how much more money does Brad need to save to buy 6 posters? (A) $6 (B) $4 (C) $3 (D) $1 Answer: Option(D) Explanation: Brad has saved $23 to spend on posters for his room. If each poster costs $4, he need 6 posters for his room, So, 6 x 4 = $24 Total money Brad need to save to buy 6 posters 24 – 23 = $1

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Go Math! 4 Common Core, Grade: 4 Publisher: Houghton Mifflin Harcourt

Go math 4 common core, title : go math 4 common core, publisher : houghton mifflin harcourt, isbn : 054758783x, isbn-13 : 9780547587837, use the table below to find videos, mobile apps, worksheets and lessons that supplement go math 4 common core., textbook resources.

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Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions

Big Ideas Math Answers Grade 4 Chapter 9 Multiply Whole Numbers and Fractions is an essential study material for students. Basic concepts of Multiplication of Whole Numbers and Fractions are clearly explained to score good marks for students. Also, in-depth knowledge of Chapter 9 Multiplies Whole Numbers and Fractions Big Ideas Grade 4 Math Answer Key mentioned with a clear explanation. Furthermore, BIM Grade 4 Multiply Whole Numbers and Fractions Answer Key is very convenient for students who are struggling to learn that concept. We have also given extra practice in small groups for the best practice of the students.

Big Ideas 4th Grade Chapter 9 Multiply Whole Numbers and Fractions Math Book Answer Key

Every problem is solved and given with a detailed explanation to help the students while their preparation. We have included the BIM Grade 4 Chapter 9 Solution Key in all the ways such as quiz, formative assessment, homework, or extra practice, etc. Practice in all the ways to get a complete grip on the concept. Provide a bright future for students, by giving the Chapter 9 Multiplies Whole Numbers and Fractions Big Ideas Grade 4 Math Answers as a reference.

Lesson: 1 Understand Multiples of Unit Fractions

Lesson 9.1 Understand Multiples of Unit Fractions

Understand multiples of unit fractions homework & practice 9.1.

Lesson: 2 Understand Multiples of Fractions

Lesson 9.2 Understand Multiples of Fractions

Understand multiples of fractions homework & practice 9.2.

Lesson: 3 Multiply Whole Numbers and Fractions

Lesson 9.3 Multiply Whole Numbers and Fractions

Multiply whole numbers and fractions homework & practice 9.3.

Lesson: 4 Multiply Whole Numbers and Mixed Numbers

Lesson 9.4 Multiply Whole Numbers and Mixed Numbers

Multiply whole numbers and mixed numbers homework & practice 9.4.

Lesson: 5 Problem Solving: Fraction Operations

Lesson 9.5 Problem Solving: Fraction Operations

Problem solving: fraction operations homework & practice 9.5.

Performance Task

Multiply Whole Numbers and Fractions Performance Task

Multiply whole numbers and fractions activity, multiply whole numbers and fractions chapter practice.

Explore and Grow

Think and Grow: Multiples of Unit Fractions

Show and Grow

Write the fraction as a multiple of a unit fraction.

Explanation: We write \(\frac{2}{3}\) in addition equation as \(\frac{2}{3}\) =\(\frac{1}{3}\) + \(\frac{1}{3}\) and multiplication as fraction \(\frac{2}{3}\) represents 2 parts that are each \(\frac{1}{3}\) of the whole, \(\frac{2}{3}\) =2 X \(\frac{1}{3}\).

Question 3. \(\frac{6}{5}\) Answer: \(\frac{6}{5}\) = \(\frac{1}{5}\)+\(\frac{1}{5}\) +\(\frac{1}{5}\)+ \(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\) = 6 X \(\frac{1}{5}\).

Explanation: We write \(\frac{6}{5}\) in addition equation as \(\frac{6}{5}\)= \(\frac{1}{5}\)+ \(\frac{1}{5}\) +\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+ \(\frac{1}{5}\) and multiplication as fraction \(\frac{6}{5}\) represents 6 parts that are \(\frac{1}{5}\) of the whole \(\frac{6}{5}\) = 6 X \(\frac{1}{5}\).

Question 4. \(\frac{7}{100}\) Answer: \(\frac{7}{100}\) = \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\)+ \(\frac{1}{100}\)+ \(\frac{1}{100}\)+ \(\frac{1}{100}\) = 7 X \(\frac{1}{100}\).

Explanation: We write \(\frac{7}{100}\) in addition equation as \(\frac{7}{100}\)= \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\) + \(\frac{1}{100}\)+ \(\frac{1}{100}\)+ \(\frac{1}{100}\)+ \(\frac{1}{100}\) and multiplication as fraction \(\frac{7}{100}\) represents 7 parts that are \(\frac{1}{100}\) of the whole, \(\frac{7}{100}\)= 7 X \(\frac{1}{100}\).

Apply and Grow: Practice

Question 5. \(\frac{2}{6}\) Answer: \(\frac{2}{6}\) = \(\frac{1}{6}\) + \(\frac{1}{6}\) = 2 X \(\frac{1}{6}\).

Explanation: We write \(\frac{2}{6}\) in addition equation as \(\frac{2}{6}\)= \(\frac{1}{6}\) + \(\frac{1}{6}\) and multiplication as fraction \(\frac{2}{6}\) represents 2 parts that are \(\frac{1}{6}\) of the whole,\(\frac{2}{6}\) = 2 X \(\frac{1}{6}\).

Question 6. \(\frac{6}{8}\) Answer: \(\frac{6}{8}\) = \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\)+ \(\frac{1}{8}\)+ \(\frac{1}{8}\)+ \(\frac{1}{8}\) = 6 X \(\frac{1}{8}\).

Explanation: We write \(\frac{6}{8}\) in addition equation as \(\frac{6}{8}\) = \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\)+ \(\frac{1}{8}\)+ \(\frac{1}{8}\)+ \(\frac{1}{8}\) and multiplication as fraction \(\frac{6}{8}\) represents 6 parts that are \(\frac{1}{8}\) of the whole, \(\frac{6}{8}\) =6 X \(\frac{1}{8}\).

Question 7. \(\frac{5}{4}\) Answer: \(\frac{5}{4}\) = \(\frac{1}{4}\) + \(\frac{1}{4}\) +\(\frac{1}{4}\) + \(\frac{1}{4}\) +\(\frac{1}{4}\) = 5 X \(\frac{1}{4}\).

Explanation: We write \(\frac{5}{4}\) in addition equation as \(\frac{5}{4}\)=\(\frac{1}{4}\) + \(\frac{1}{4}\) +\(\frac{1}{4}\) +\(\frac{1}{4}\) + \(\frac{1}{4}\) and multiplication as fraction \(\frac{5}{4}\) represents 5 parts that are \(\frac{1}{4}\) of the whole, \(\frac{5}{4}\) = 5 X \(\frac{1}{4}\).

Question 8. \(\frac{3}{12}\) Answer: \(\frac{3}{12}\)= \(\frac{1}{12}\) + \(\frac{1}{12}\) + \(\frac{1}{12}\) = 3 X \(\frac{1}{12}\)

Explanation : We write \(\frac{3}{12}\) in addition equation as \(\frac{3}{12}\)= \(\frac{1}{12}\) + \(\frac{1}{12}\) + \(\frac{1}{12}\) and multiplication as fraction \(\frac{3}{12}\) represents 3 parts that are \(\frac{1}{12}\) of the whole, \(\frac{3}{12}\) = 3 X \(\frac{1}{12}\).

Question 9. \(\frac{8}{100}\) Answer: \(\frac{8}{100}\) = \(\frac{1}{100}\) + \(\frac{1}{100}\)+\(\frac{1}{100}\)+ \(\frac{1}{100}\)+\(\frac{1}{100}\)+\(\frac{1}{100}\)+\(\frac{1}{100}\)+ \(\frac{1}{100}\) = 8 X \(\frac{1}{100}\).

Explanation : We write \(\frac{8}{100}\) in addition equation as \(\frac{8}{100}\) = \(\frac{1}{100}\) + \(\frac{1}{100}\)+\(\frac{1}{100}\)+ \(\frac{1}{100}\)+\(\frac{1}{100}\)+ \(\frac{1}{100}\)+ \(\frac{1}{100}\)+\(\frac{1}{100}\) and multiplication as fraction \(\frac{8}{100}\) represents 8 parts that are \(\frac{1}{100}\) of the whole, \(\frac{8}{100}\)= 8 X \(\frac{1}{100}\).

Question 10. \(\frac{7}{10}\) Answer: \(\frac{7}{10}\) = \(\frac{1}{10}\) +\(\frac{1}{10}\) +\(\frac{1}{10}\) + \(\frac{1}{10}\) +\(\frac{1}{10}\) +\(\frac{1}{10}\) +\(\frac{1}{10}\) = 7 X \(\frac{1}{10}\).

Explanation: We write \(\frac{7}{10}\) in addition equation as \(\frac{7}{10}\) = \(\frac{1}{10}\) + \(\frac{1}{10}\) +\(\frac{1}{10}\) +\(\frac{1}{10}\) +\(\frac{1}{10}\) + \(\frac{1}{10}\) +\(\frac{1}{10}\) and multiplication as fraction \(\frac{7}{10}\) represents 7 parts that are \(\frac{1}{10}\) of the whole,\(\frac{7}{10}\) = 7 X \(\frac{1}{10}\).

Question 11. Structure When a fraction is written as a multiple of a unit fraction, what is the relationship between the numerator of the fraction and the number that is multiplied by the unit fraction? Answer: When a fraction is written as a multiple of a unit fraction, the relationship between the numerator of the fraction and the number that is multiplied by the unit fraction, is the whole number times the unit fraction.

Explanation : We know that the numerator of a unit fraction is always one. So When a fraction is written as a multiple of a unit fraction the number will be always the whole number (whole number of times the unit fraction) and is multiplied by unit fraction. For example \(\frac{3}{8}\) we write it as 3 X \(\frac{1}{8}\) here it is 3 and this 3 is multiplied by unit fraction.

DIG DEEPER! Write the mixed number as a multiple of a unit fraction.

Question 12. 1\(\frac{1}{8}\) Answer: 1\(\frac{1}{8}\) = \(\frac{9}{8}\) = \(\frac{1}{8}\)+\(\frac{1}{8}\)+ \(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+ \(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)= 9 X \(\frac{1}{8}\) .

Explanation: First we convert a mixed number to an improper fraction, It is to multiply the bottom number 8 by the “regular” number 1, add in the top number 1, and then put this on top of the original bottom number as 9 by 8. 1\(\frac{1}{8}\) = 1+\(\frac{1}{8}\) = \(\frac{9}{8}\) We write \(\frac{9}{8}\) in addition equation as \(\frac{9}{8}\)= \(\frac{1}{8}\)+ \(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+ \(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\) +\(\frac{1}{8}\) and multiplication as fraction \(\frac{9}{8}\) represents 9 parts that are \(\frac{1}{10}\) of the whole, \(\frac{9}{8}\) = 9 X \(\frac{1}{8}\).

Question 13. 3\(\frac{4}{5}\) Answer: 3\(\frac{4}{5}\) = \(\frac{19}{5}\)=\(\frac{1}{5}\)+\(\frac{1}{5}\)+ \(\frac{1}{5}\) +\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+ \(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+ \(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+ \(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+ \(\frac{1}{5}\)= 19 X \(\frac{1}{5}\).

Explanation: First we convert a mixed number to an improper fraction, It is to multiply the bottom number 5 by the “regular” number 3, add in the top number 4, and then put this on top of the original bottom number as 19 by 5. 3\(\frac{4}{5}\)=3+\(\frac{4}{5}\) = \(\frac{19}{5}\) We write \(\frac{19}{5}\) in addition equation as \(\frac{19}{5}\) = \(\frac{1}{5}\)+ \(\frac{1}{5}\)+\(\frac{1}{5}\) +\(\frac{1}{5}\)+\(\frac{1}{5}\)+ \(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+ \(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+ \(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\)+ \(\frac{1}{5}\)+\(\frac{1}{5}\) and multiplication as fraction \(\frac{19}{5}\) represents 19 parts that are \(\frac{1}{5}\) of the whole,\(\frac{19}{5}\)= 19 X \(\frac{1}{5}\).

Question 13.

Think and Grow: Modeling Real Life

Explanation: Given a juice stand worker uses \(\frac{1}{2}\) of an orange to garnish each drink. The worker has \(\frac{7}{12}\) of an orange left. So the worker can garnish more \(\frac{7}{12}\) minus \(\frac{1}{2}\) means \(\frac{7}{12}\) – \(\frac{1}{2}\) = \(\frac{1}{12}\), Given \(\frac{7}{12}\) of an orange is left , to write orange left as a multiple of \(\frac{1}{12}\) we have orange left as \(\frac{7}{12}\) so write it as 7 parts that are \(\frac{1}{12}\) of the whole, \(\frac{7}{12}\)= 7 X \(\frac{1}{12}\).

Question 14. A piece of rope is \(\frac{8}{5}\) meters long. You cut the rope into \(\frac{1}{5}\) meter long pieces. How many pieces do you cut? Answer: 8 pieces I have cut.

Explanation: Given a piece of rope is \(\frac{8}{5}\) meters long and it has been cut into \(\frac{1}{5}\) meter long pieces, therefore pieces of rope cut is \(\frac{8}{5}\) divided by \(\frac{1}{5}\) = \(\frac{8}{5}\) X latex]\frac{5}{1}[/latex] = 8.

Question 15. A restaurant serves \(\frac{4}{10}\) of a meatloaf to 4 customers. Each customer receives the same amount of meatloaf. What fraction of the meatloaf does each customer receive? Answer: The fraction of the meatloaf each customer receives is \(\frac{1}{10}\)

Explanation: Given a restaurant serves \(\frac{4}{10}\) of a meatloaf to 4 customers and each customer receives the same amount of meatloaf so fraction of the meatloaf each customer receives is  \(\frac{4}{10}\)  divided by 4 = \(\frac{4}{10}\) x \(\frac{1}{4}\) = \(\frac{1}{10}\).

Explanation: I use 3\(\frac{3}{4}\) pounds of trail mix to make treat bags, We convert a mixed number to an improper fraction,It is to multiply the bottom number 4 by the “regular” number 3, add in the top number 3, and then put this on top of the original bottom number as 15 by 4, it becomes \(\frac{15}{4}\),Each bag i kept \(\frac{1}{4}\) pound of trail mix so total number of treat bags required to make are \(\frac{15}{4}\) divided by \(\frac{1}{4}\)=\(\frac{15}{4}\) X \(\frac{4}{1}\)=15 treat bags.

Question 17. DIG DEEPER! You walk from home to school and then back home again each day for 5 days. Altogether, you walk \(\frac{10}{8}\) miles. What is the distance from your home to school? Explain. Answer: The distance from home to school is \(\frac{1}{8}\) miles

Explanation: Total number of miles walked is \(\frac{10}{8}\) in 5 days that to from home to school and then back home again means it is twice a day , Let us take X as the distance from home to school is, Given 2X x 5=\(\frac{10}{8}\) miles, X = Then each day number of miles walked is  \(\frac{10}{8}\) divided by 10 which is  equal to  \(\frac{10}{8}\) x \(\frac{1}{10}\) =\(\frac{1}{8}\), Therefore X = \(\frac{1}{8}\) miles.

Explanation: We write \(\frac{2}{12}\) in addition equation as \(\frac{2}{12}\)= \(\frac{1}{12}\) + \(\frac{1}{12}\) and multiplication as fraction \(\frac{2}{12}\) represents 2 parts that are \(\frac{1}{12}\) of the whole, \(\frac{2}{12}\)= 2 X \(\frac{1}{12}\).

Explanation: We write \(\frac{5}{100}\) in addition equation as \(\frac{5}{100}\)= \(\frac{1}{100}\)+ \(\frac{1}{100}\)+\(\frac{1}{100}\)+\(\frac{1}{100}\)+\(\frac{1}{100}\) and multiplication as fraction \(\frac{5}{100}\) represents 5 parts that are \(\frac{1}{100}\) of the whole, \(\frac{5}{100}\) = 5 X \(\frac{1}{100}\).

Question 3. \(\frac{3}{6}\) Answer: \(\frac{3}{6}\)=\(\frac{1}{6}\)+\(\frac{1}{6}\)+\(\frac{1}{6}\) = 3 X \(\frac{1}{6}\).

Explanation: We write \(\frac{3}{6}\) in addition equation as \(\frac{3}{6}\)=\(\frac{1}{6}\)+ \(\frac{1}{6}\)+\(\frac{1}{6}\) and multiplication as fraction \(\frac{3}{6}\) represents 3 parts that are \(\frac{1}{6}\) of the whole,\(\frac{3}{6}\) = 3 X \(\frac{1}{6}\).

Question 4. \(\frac{7}{10}\) Answer: \(\frac{7}{10}\) =\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+ \(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\) = 7 X \(\frac{1}{10}\)

Explanation: We write \(\frac{7}{10}\) in addition equation as \(\frac{7}{10}\)= \(\frac{1}{10}\)+ \(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+\(\frac{1}{10}\)+ \(\frac{1}{10}\)+\(\frac{1}{10}\) and multiplication as \(\frac{7}{10}\) represents 7 parts that are \(\frac{1}{10}\) of the whole,\(\frac{7}{10}\) = 7 X \(\frac{1}{10}\).

Question 5. \(\frac{4}{5}\) Answer: \(\frac{4}{5}\) = \(\frac{1}{5}\)+\(\frac{1}{5}\)+ \(\frac{1}{5}\)+\(\frac{1}{5}\) = 4 X \(\frac{1}{5}\).

Explanation: We write \(\frac{4}{5}\) in addition equation as \(\frac{4}{5}\)=\(\frac{1}{5}\)+ \(\frac{1}{5}\)+\(\frac{1}{5}\)+\(\frac{1}{5}\) and multiplication as \(\frac{4}{5}\) represents 4 parts that are \(\frac{1}{5}\) of the whole,\(\frac{4}{5}\)= 4 X \(\frac{1}{5}\).

Question 6. \(\frac{9}{2}\) Answer: \(\frac{9}{2}\) = \(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+ \(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+ \(\frac{1}{2}\)+\(\frac{1}{2}\) = 9 X \(\frac{1}{2}\).

Explanation: We write \(\frac{9}{2}\) in addition equation as \(\frac{9}{2}\) = \(\frac{1}{2}\)+ \(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+ \(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\) and multiplication as \(\frac{9}{2}\) represents 9 parts that are \(\frac{1}{2}\) of the whole, \(\frac{9}{2}\)= 9 X \(\frac{1}{2}\).

Question 7. \(\frac{8}{8}\) Answer: \(\frac{8}{8}\)=1=\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+ \(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+ \(\frac{1}{8}\) = \(\frac{8}{8}\) = 8 X \(\frac{1}{8}\).

Explanation: \(\frac{8}{8}\)= 1 ,but we write the fraction as a addition of a unit fraction and multiple of a unit fraction, We write \(\frac{8}{8}\) in addition equation as \(\frac{8}{8}\) = \(\frac{1}{8}\)+ \(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\)+ \(\frac{1}{8}\)+\(\frac{1}{8}\)+\(\frac{1}{8}\) and multiplication as \(\frac{8}{8}\) represents 8 parts that are \(\frac{1}{8}\) of the whole,\(\frac{8}{8}\)= 8 X \(\frac{1}{8}\).

Question 8. \(\frac{6}{3}\) Answer: \(\frac{6}{3}\)=2,\(\frac{6}{3}\)= \(\frac{1}{3}\)+\(\frac{1}{3}\)+ \(\frac{1}{3}\)+\(\frac{1}{3}\)+\(\frac{1}{3}\)+\(\frac{1}{3}\) = 6 X \(\frac{1}{3}\).

Explanantion: \(\frac{6}{3}\)= 2 ,but we write the fraction as a addition of a unit fraction and multiple of a unit fraction, We write \(\frac{6}{3}\) in addition equation as \(\frac{6}{3}\)= \(\frac{1}{3}\)+ \(\frac{1}{3}\)+\(\frac{1}{3}\)+\(\frac{1}{3}\)+\(\frac{1}{3}\)+ \(\frac{1}{3}\) and multiplication as \(\frac{6}{3}\) represents 6 parts that are \(\frac{1}{3}\) of the whole,\(\frac{6}{3}\)= 6 X \(\frac{1}{3}\).

Explanation: Given that Newton’s fraction as sum of 9 unit fractions and each unit fraction is one twelfth of a whole. So Newton’s fraction is \(\frac{9}{12}\) we write \(\frac{9}{12}\) in addition equation as \(\frac{9}{12}\)=  \(\frac{1}{12}\) +\(\frac{1}{12}\) +\(\frac{1}{12}\) + \(\frac{1}{12}\) +\(\frac{1}{12}\) +\(\frac{1}{12}\) +\(\frac{1}{12}\) + \(\frac{1}{12}\) +\(\frac{1}{12}\) and multiplication as \(\frac{9}{12}\) represents 9 parts that are \(\frac{1}{12}\) of the whole,\(\frac{9}{12}\)= 9 X \(\frac{1}{12}\).

Explanation: Given 6 number of grapefruit halves so there are total grapefruit as \(\frac{6}{2}\) we write \(\frac{6}{2}\) in addition equation as \(\frac{6}{2}\)=\(\frac{1}{2}\)+ \(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+\(\frac{1}{2}\)+ \(\frac{1}{2}\),so the multiplication equation is \(\frac{6}{2}\) represents 6 parts that are \(\frac{1}{2}\) of the whole,\(\frac{6}{2}\)= 6 X \(\frac{1}{2}\).

Explanation: 1. 5 X \(\frac{1}{4}\) = \(\frac{5}{4}\) 2. \(\frac{1}{4}\)+\(\frac{1}{4}\)+\(\frac{1}{4}\)+\(\frac{1}{4}\)= 5 X \(\frac{1}{4}\)= \(\frac{5}{4}\) 3. \(\frac{5}{4}\) =\(\frac{5}{4}\) 4. 4 X \(\frac{1}{5}\)= \(\frac{4}{5}\)≠  \(\frac{5}{4}\) So in the above expressions only 4 X \(\frac{1}{5}\)  does not belong with the other three expressions.

Explanation: I have \(\frac{6}{8}\) cup of blueberries and i can put \(\frac{1}{8}\) cup of blueberries in each pan cake so total number of pancakes made are \(\frac{6}{8}\) divided by \(\frac{1}{8}\) = \(\frac{6}{8}\) X \(\frac{8}{1}\)= 6 pancakes.

Question 13. DIG DEEPER! You cut a loaf of zucchini bread into 20 equal slices. You and your friends eat \(\frac{3}{10}\) of the slices. You want to put each leftover slice into its own bag. How many bags do you need? Answer: 14 bags are needed.

Explanation : Given a loaf of zucchini bread is cut  into 20 equal slices and i ,friend eat \(\frac{3}{10}\) of the slices, Total number of slices of zucchini bread =20, Number of slices ate = \(\frac{3}{10}\) means it says out of 10 slices 3 were ate, so out of 20 slices  2 x 3= 6 slices were eaten. Number of slices left = 20 – 6 =14 slices and one bag contains only one slice. Therefore number of bags needed are 14.

Question 14. Is 46 a multiple of 45. Answer: No 46 is not a multiple of 45.

Explanation: A multiple is a number that can be divided by another number a certain number of times without a remainder. 45 when divided by 46 a certain times will leave a remainder, So 46 is not a multiple of 45 or a factor is one of two or more numbers that divides a given number without a remainder. As factors of 45 are 1,3,5,9,15 and factors of 46 are 1,2,23, both do not have common factors so 46 is not a multiple of 45.

Question 15. Is 3 a factor of 75? Answer: Yes 3 is a factor of 75 Explanation: We know that a factor is one that divides a given number without a remainder and 75 when divided by 3 leaves no remainder, So 3 is a factor of 75.

The multiplication expression is \(\frac{2}{6}\) +\(\frac{2}{6}\) +\(\frac{2}{6}\) + \(\frac{2}{6}\) = 4 X \(\frac{2}{6}\) = \(\frac{1}{6}\) +\(\frac{1}{6}\)+ \(\frac{1}{6}\) +\(\frac{1}{6}\)+ \(\frac{1}{6}\) +\(\frac{1}{6}\)+ \(\frac{1}{6}\) +\(\frac{1}{6}\) = 8 X \(\frac{1}{6}\).

Explanation: \(\frac{2}{6}\) +\(\frac{2}{6}\) +\(\frac{2}{6}\) +\(\frac{2}{6}\) the multiplication equation is  4 parts that are \(\frac{2}{6}\) of the whole as  4 X \(\frac{2}{6}\) \(\frac{1}{6}\) +\(\frac{1}{6}\)+ \(\frac{1}{6}\) +\(\frac{1}{6}\)+ \(\frac{1}{6}\) +\(\frac{1}{6}\)+ \(\frac{1}{6}\) +\(\frac{1}{6}\) = 8 X \(\frac{1}{6}\) the multiplication equation is 8 parts that are \(\frac{1}{6}\) of the whole as 8 X \(\frac{1}{6}\).

Structure Compare your expressions. How are they the same? How are they different? Same means the result of both expressions is same = \(\frac{8}{6}\) Different means in one expression we take value of \(\frac{2}{6}\) 4 times and in another expression we take \(\frac{1}{6}\) 8 times.

Explanation: Same means \(\frac{2}{6}\) +\(\frac{2}{6}\) +\(\frac{2}{6}\) +\(\frac{2}{6}\) = 4 X \(\frac{2}{6}\) = \(\frac{8}{6}\) and \(\frac{1}{6}\) +\(\frac{1}{6}\)+ \(\frac{1}{6}\) +\(\frac{1}{6}\)+\(\frac{1}{6}\) +\(\frac{1}{6}\)+ \(\frac{1}{6}\) +\(\frac{1}{6}\) = 8 X \(\frac{1}{6}\) =\(\frac{8}{6}\) , both have results similar as \(\frac{8}{6}\), Different means here we are taking expressions as addition values of  \(\frac{2}{6}\) by 4 times and in other expression we are taking addition values of \(\frac{1}{6}\) by 8 times.

Think and Grow: Multiples of Fractions

Explanation: To write 2 X \(\frac{3}{5}\) multiple of unit fraction first we write \(\frac{3}{5}\) as multiple of 3 X \(\frac{1}{5}\) then we multiply it by 2 as 2 X 3 X \(\frac{1}{5}\)=6 X \(\frac{1}{5}\). Show and Grow

Write the product as a multiple of a unit fraction. Then find the product.

Explanation: First we write \(\frac{4}{5}\)  as  \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\) = 4 X \(\frac{1}{5}\) then multiply it by 2 which is  2 X 4 X \(\frac{1}{5}\) = 8 X \(\frac{1}{5}\) = \(\frac{8}{5}\).

Explanation: First we write \(\frac{2}{10}\) as \(\frac{1}{10}\) +\(\frac{1}{10}\) = 2 X \(\frac{1}{10}\) then multiply it by 3 which is 3 X 2 X \(\frac{1}{10}\) = 6 X \(\frac{1}{10}\) =\(\frac{6}{10}\).

Explanation: First we write \(\frac{3}{2}\) as \(\frac{1}{2}\) +\(\frac{1}{2}\) + \(\frac{1}{2}\) = 3 X \(\frac{1}{2}\) then we multiply it by 4 which is 4 X 3 X \(\frac{1}{2}\)= 12 X \(\frac{1}{2}\)= \(\frac{12}{2}\)=6.

Write the product as a multiple of a unit fraction. Then find the product.

Explanation: First we write \(\frac{2}{3}\) as \(\frac{1}{3}\)+\(\frac{1}{3}\)= 2 X \(\frac{1}{3}\) then multiply it by 5 which is 5 X 2 X \(\frac{1}{3}\)= 10 X \(\frac{1}{3}\)= \(\frac{10}{3}\).

Explanation: First we write \(\frac{5}{8}\) as \(\frac{1}{8}\)+ \(\frac{1}{8}\)+ \(\frac{1}{8}\) + \(\frac{1}{8}\)+ \(\frac{1}{8}\) = 5 X \(\frac{1}{8}\) then multiply it by 6 which is 6 X 5 X \(\frac{1}{8}\)=30 X \(\frac{1}{8}\)= \(\frac{30}{8}\).

Explanation: First we write \(\frac{7}{4}\) as \(\frac{1}{4}\)+\(\frac{1}{4}\)+ \(\frac{1}{4}\)+ \(\frac{1}{4}\)+\(\frac{1}{4}\)+\(\frac{1}{4}\)+ \(\frac{1}{4}\) = 7 x \(\frac{1}{4}\) then we multiply it by 9 which is 9 X 7 X \(\frac{1}{4}\)= 63 X \(\frac{1}{4}\) = \(\frac{63}{4}\).

Explanation: First we write \(\frac{4}{12}\) as \(\frac{1}{12}\)+ \(\frac{1}{12}\)+ \(\frac{1}{12}\)+\(\frac{1}{12}\)= 4 X \(\frac{1}{12}\) then we multiply it by 7 which is 7 X 4 X \(\frac{1}{12}\)= 28 X \(\frac{1}{12}\)=\(\frac{28}{12}\) we can also simplify 7 X \(\frac{4}{12}\) as 4,12 can be divided by 4 we can also write 7 X \(\frac{4}{12}\) as 7 X  \(\frac{1}{3}\)= \(\frac{7}{3}\).

Explanation: First we write \(\frac{9}{6}\) as \(\frac{1}{6}\) +\(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) +\(\frac{1}{6}\) +\(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) +\(\frac{1}{6}\) =9 X \(\frac{1}{6}\) then we multiply it by 8 which is 9 X 8 X \(\frac{1}{6}\) = 72 X \(\frac{1}{6}\) = \(\frac{72}{6}\)  further as 6 can divide 72 and leaving no remainder we write \(\frac{72}{6}\) =12.

Explanation: First we write \(\frac{20}{100}\) = \(\frac{2}{10}\) as \(\frac{1}{10}\) + \(\frac{1}{10}\) = 2 X \(\frac{1}{10}\) then multiply it by 10 which is 2 X 10 X \(\frac{1}{10}\)= 20 X \(\frac{1}{10}\)=\(\frac{20}{10}\) and 10 can divide 20 leaving no remainder further we can write \(\frac{20}{10}\) as 2.

Number Sense Find the unknown number.

Explanation: ___ X \(\frac{8}{10}\) = \(\frac{16}{10}\)= to get the unknown number we take it as P, P X \(\frac{8}{10}\) = \(\frac{16}{10}\) means P = \(\frac{16}{10}\) X \(\frac{10}{8}\)  we get 2 , so P=2.

Explanation: Let us take unknown letter as P so 4 X \(\frac{P}{2}\) = \(\frac{20}{2}\) , 4 X P= 2 X \(\frac{20}{2}\) to get value of P we take 4 to other side , Therefore P=20 X \(\frac{1}{4}\) = 5, so the unknown number is 5.

Explanation: Let us take unknown number as P given 3 X \(\frac{9}{P}\) = \(\frac{27}{100}\) , 27 X[ latex]\frac{1}{P}[/latex] = \(\frac{27}{100}\), Therefore P= 27 X \(\frac{100}{27}\)=100.

Explanation: 5 X \(\frac{3}{6}\)= 5 X  \(\frac{1}{2}\) since the denominator of each given fraction is 2 divide the space between every pairs of two consecutive integers (on the number line) in 2 equal parts each part so obtained will represent the fraction \(\frac{1}{2}\) , so on the number line we move 5 parts on the right-side of zero as shown in the figure.

Explanation : Given A bird keeper uses a \(\frac{1}{3}\) cup scoop to feed 3 birds each bird he feeds \(\frac{2}{3}\) cup of birdseed, Three birds each need \(\frac{2}{3}\) cup of birdseed so the bird keeper fills the scoop as 3 X \(\frac{2}{3}\) = 3 X _____P______ X \(\frac{1}{3}\) here P is the  number of times the bird keeper fills the cup scoop. 3 X 2 X \(\frac{1}{3}\) =  3 X P X \(\frac{1}{3}\) , when we equate both sides, we get 3 X 2 = 3 X P , therefore P = 2. Show and Grow

Question 15. A chef makes 4 servings of honey1oatmeal. She uses a \(\frac{1}{2}\) tablespoon measuring spoon to measure \(\frac{3}{2}\) tablespoons of honey for each serving. How many times does she fill the measuring spoon? Answer: 3 times the chef fill the measuring spoon.

Explanation: Chef uses a \(\frac{1}{2}\) table spoon as measuring spoon, To measure honey for each serving she needed for each serving is \(\frac{3}{2}\) , so 4 X \(\frac{3}{2}\) , Let us take P as number of times of measuring spoon , Given 4 X \(\frac{3}{2}\) = 4 X _P_ X \(\frac{1}{2}\), 4 X ___3___ X \(\frac{1}{2}\)= 4 X _P_ X \(\frac{1}{2}\) , we equate both sides  we get value of P as 3 , therefore 3 times the chef fill the measuring spoon.

Question 16. You have 7 magnetic blocks that are each \(\frac{9}{100}\) meter long. You connect the ends of the blocks to make a snake. What fraction of a meter is the block snake? Answer: \(\frac{63}{100}\) meter is the block snake.

Explanation: Given 7 magnetic blocks and each are \(\frac{9}{100}\) meter long. After connecting the ends of the blocks to make a snake is 7 X \(\frac{9}{100}\) = 7 X 9 X \(\frac{1}{100}\) = 63 X \(\frac{1}{100}\) = \(\frac{63}{100}\) meter.

Explanation: Given friend roller-skates \(\frac{15}{10}\) miles each day means \(\frac{3}{2}\) miles each day so for 5 days it is 5 X \(\frac{3}{2}\) = 5 X 3 X \(\frac{1}{2}\) = 15 X \(\frac{1}{2}\) = \(\frac{15}{2}\) miles.

Explanation: To write the product as a multiple of a unit fraction we write it \(\frac{2}{3}\) as 2 X \(\frac{1}{3}\) and multiply by 2 = 2 X 2 X \(\frac{1}{3}\) = 4 X \(\frac{1}{3}\) = \(\frac{4}{3}\)

Explanation: To write the product as a multiple of a unit fraction we write it \(\frac{5}{8}\) as 5 X \(\frac{1}{8}\) and multiply by 3 = 3 X 5 X \(\frac{1}{8}\) = 15 X \(\frac{1}{8}\) = \(\frac{15}{8}\)

Explanation: To write the product as a multiple of a unit fraction we write it \(\frac{8}{2}\) as 8 X \(\frac{1}{2}\) and multiply by 4 = 4 X 8 X \(\frac{1}{2}\) = 32 X \(\frac{1}{2}\) = \(\frac{32}{2}\) = 16.

Explanation: To write the product as a multiple of a unit fraction we write it \(\frac{9}{10}\) as 9 X \(\frac{1}{10}\) and multiply by 5 = 5 X 9 X \(\frac{1}{10}\) = 45 X \(\frac{1}{10}\) = \(\frac{45}{10}\) = \(\frac{9}{2}\).

Explanation: To write the product as a multiple of a unit fraction we write it \(\frac{6}{5}\) as 6 X \(\frac{1}{5}\) and multiply by 8 = 8 X 6 X \(\frac{1}{5}\) = 48 X \(\frac{1}{5}\) = \(\frac{48}{5}\).

Explanation: To write the product as a multiple of a unit fraction we write it \(\frac{2}{4}\) as 2 X \(\frac{1}{4}\) and multiply by 10 = 2 X 10 X \(\frac{1}{4}\) = 20 X \(\frac{1}{4}\) = \(\frac{20}{4}\) = \(\frac{10}{2}\).

Explanation: We have the value of expression of 4 X \(\frac{7}{8}\) as \(\frac{28}{8}\) So 1.(4 X 7) x \(\frac{1}{8}\)= \(\frac{28}{8}\) 2. \(\frac{28}{8}\)=\(\frac{28}{8}\) 3. 4 X 7 ≠ \(\frac{28}{8}\) 4. \(\frac{32}{7}\) ≠ \(\frac{28}{8}\) 5. \(\frac{7}{8}\)+\(\frac{7}{8}\)+\(\frac{7}{8}\)+\(\frac{7}{8}\) = \(\frac{28}{8}\), Therefore expressions 1,2,5 are equivalent to \(\frac{28}{8}\). The expressions equivalent to 4 X \(\frac{7}{8}\) are  (4 X 7) x \(\frac{1}{8}\) , \(\frac{28}{8}\),\(\frac{7}{8}\)+\(\frac{7}{8}\)+\(\frac{7}{8}\)+\(\frac{7}{8}\)

Explanation: 6 X \(\frac{6}{2}\)= 6 X 6 X \(\frac{1}{2}\)=36 X \(\frac{1}{2}\)= \(\frac{36}{2}\) and 5 X \(\frac{7}{2}\) =5 X 7 X \(\frac{1}{2}\)= 35 X \(\frac{1}{2}\), Now comparing both sides 36 X \(\frac{1}{2}\) and 35 X \(\frac{1}{2}\) we see 36 X \(\frac{1}{2}\) > 35 X \(\frac{1}{2}\) so 6 X \(\frac{6}{2}\) is greater.

Question 9. Structure Your friend fills a \(\frac{3}{4}\) – cup measuring cup with rice 2 times. Write an equation to show how much rice she uses.

Answer: Friend uses \(\frac{3}{2}\) cups.

Explanation: Friend fills a \(\frac{3}{4}\) – cup with rice 2 times, so she uses 2 X \(\frac{3}{4}\) = 2 X 3 x \(\frac{1}{4}\) = 6 X \(\frac{1}{4}\)= \(\frac{6}{4}\)= \(\frac{3}{2}\).

Explanation: Each bottle needs \(\frac{20}{4}\) cups of water and have a \(\frac{1}{4}\) – cup measuring cup, Now we take P as number of times for filling so \(\frac{20}{4}\) = P X \(\frac{1}{4}\), 20 X \(\frac{1}{4}\)=P X \(\frac{1}{4}\), Equating both sides we get  P = 20,  so it is 20 times to fill the measuring cups of water for one bottle, Given that to make a tornado we require 2 bottles, 2 X 20 cups of water = 40 cups of water.

Question 11. DIG DEEPER! You and your friend are each selling 12 coupon books. So far, you have sold \(\frac{2}{6}\) of your books. Your friend has sold 3 times as many as you. What fraction of your friend’s coupon books has she sold?

Answer: We will get the fraction as whole 1 so friend has sold all the 12 coupon books.

Explanation: So far i sold \(\frac{2}{6}\) books out of 12 coupon books. Lets us take P number of books, i have sold P out of 12 coupon books, P= 12 X \(\frac{2}{6}\) =12 X 2 X \(\frac{1}{6}\) = 24 X \(\frac{1}{6}\) = \(\frac{24}{6}\) =4 , So i have sold 4 coupon books, Given friend has sold 3 times as many as I, Means 3 X 4 =12 means friend has sold all the coupon books

Review & Refresh

Find the product.

Question 12. 12 × 47 = ___ Answer: 12 X 47 = 564

Explanation: When 12 is multiplied by 47 we get the product as 564.

Question 13. 35 × 31 = ___ Answer: 35 X 31 =1,085

Explanation: When 35 is multiplied by 31 we get the product as 1,085.

Question 14. 58 × 49 = ___ Answer: 58 X 49 = 2,842

Explanation: When 58 is multiplied by 49 we get the product as 2,842.

We notice that each expression can be written as the product as a multiple of a unit fraction, The product of a number and a counting number is a multiple of the number. We can find multiples of unit fractions. To find the product of a whole number and a fraction without using models, First we multiply the numerator by the whole number, then rewrite the expression as a multiple of a unit fraction and denominator stays the same we get the product results.

Think and Grow: Multiply Whole Numbers and Fractions

Explanation: To find 4 X \(\frac{2}{3}\) Method 1:  Multiply the numerator by the whole number as 4 X 2 and divide by 3= \(\frac{8}{3}\) Method 2 : Then writing the expression as a multiple of unit fraction \(\frac{2}{3}\)= 4 X ( 2 X \(\frac{1}{3}\) ) applying associative property of multiplication (4 X 2) X \(\frac{1}{3}\)= 8 X \(\frac{1}{3}\)=\(\frac{8}{3}\)

Explanation: 3 X \(\frac{2}{4}\) can be simplified as 3 X \(\frac{1}{2}\) now the expression is in multiple of unit fraction so  3 X \(\frac{1}{2}\) =\(\frac{3}{2}\).

Explanation: 7 X \(\frac{3}{2}\)  first we write the expression as multiple of unit fraction so 7 X ( 3 X \(\frac{1}{2}\) ) applying associative property of multiplication ( 7 X  3) X \(\frac{1}{2}\) ) = 21 X \(\frac{1}{2}\) = \(\frac{21}{2}\)

Explanation : Already the expression 2 X \(\frac{1}{5}\) is a multiple of unit fraction so we multiply the whole with the numerator and numerator is being same, 2 X \(\frac{1}{5}\) = \(\frac{2}{5}\).

Explanation: 5 X \(\frac{3}{10}\), we write the expression as multiple of unit fraction so 5 X 3 X \(\frac{1}{10}\)= 15 X \(\frac{1}{10}\)= \(\frac{15}{10}\) further it can be simplified, as both numerator and denominator can be divided by 5 so \(\frac{15}{10}\)=\(\frac{3}{2}\).

Explanation: 6 X  \(\frac{7}{8}\), we write the expression as multiple of unit fraction so 6 X 7 X \(\frac{1}{8}\) applying associative property of multiplication 6 X ( 7 X \(\frac{1}{8}\) )= ( 6 X 7 ) X \(\frac{1}{8}\)= 42 X \(\frac{1}{8}\)= \(\frac{42}{8}\) and further also the fraction can be simplified, as both numerator and denominator can be divided by 2, making \(\frac{42}{8}\) as \(\frac{21}{4}\).

Explanation: 8 X  \(\frac{9}{10}\), we write the expression as multiple of unit fraction so 8 X 9 X \(\frac{1}{10}\) applying associative property of multiplication 8 X ( 9 X \(\frac{1}{10}\) )= (8 X 9 ) X \(\frac{1}{10}\)= 72 X \(\frac{1}{10}\)= \(\frac{72}{10}\) and further also the fraction can be simplified, as both numerator and denominator can be divided by 2, making \(\frac{72}{10}\) as \(\frac{36}{5}\).

Explanation: 3 X  \(\frac{60}{100}\),  we write the expression as multiple of unit fraction, So 3 X 60 X \(\frac{1}{100}\) applying associative property of multiplication 3 X (60 X \(\frac{1}{100}\) )= (3 X 60 ) X \(\frac{1}{100}\)= 180 X \(\frac{1}{100}\)= \(\frac{180}{100}\) and further also the fraction can be simplified, as both numerator and denominator can be divided by 10, making \(\frac{180}{100}\) as \(\frac{18}{10}\).

Explanation : \(\frac{4}{2}\) X 10 , we write the expression as multiple of unit fraction, So 4 X \(\frac{1}{2}\) X 10= 40 X \(\frac{1}{2}\)=\(\frac{40}{2}\) as the fraction can be further simplified as both numerator and denominator can be divided by 2, making it as 20.

Explanation: 7 X \(\frac{2}{4}\), First we write expression as a multiple of unit fraction 7 X 2 X \(\frac{1}{4}\) =14 X \(\frac{1}{4}\) = \(\frac{14}{4}\) Now 5 X \(\frac{3}{4}\) is also written as a multiple of unit fraction 5 X 3 X \(\frac{1}{4}\) = 15 X \(\frac{1}{4}\) =\(\frac{15}{4}\) comparing \(\frac{14}{4}\) and \(\frac{15}{4}\) as the denominators are same same 4 and in numerators 15 is great than 14, So  \(\frac{14}{4}\)  is less than or smaller or < \(\frac{15}{4}\) , 7 X \(\frac{2}{4}\) is smaller < 5 X \(\frac{3}{4}\)

Answer: 4 X \(\frac{1}{8}\) is greater than > 4 X \(\frac{1}{10}\).

Explanation: 4 X \(\frac{1}{8}\) already the expression is in multiple of unit fraction, we write as \(\frac{4}{8}\) and even the expression 4 X \(\frac{1}{10}\) is in multiple of unit fraction we write as \(\frac{4}{10}\) because in order to compare the fractions we must first turn their different denominators into the same denominators to make denominators same we multiply 4 X \(\frac{1}{8}\) by 10 /10 the value will not change = 4 X 10 X \(\frac{1}{80}\)=40 X \(\frac{1}{80}\) =  \(\frac{40}{80}\) now other expression 4 X \(\frac{1}{10}\) by 8/8 so that the value = 4 X 8 X \(\frac{1}{80}\) = 32 X \(\frac{1}{80}\) now as denominator are same we can compare only numerators \(\frac{40}{80}\) with \(\frac{32}{80}\) as 40 is greater than 32, So  \(\frac{4}{8}\) is greater than > \(\frac{4}{10}\) , 4 X \(\frac{1}{8}\) is greater than > 4 X \(\frac{1}{10}\).

Explanation: First we write expressions as a multiple of unit fraction \(\frac{4}{3}\) X 9 = 9 X 4 X \(\frac{1}{3}\) =36 X \(\frac{1}{3}\) =\(\frac{36}{3}\) now other expression  6 X \(\frac{6}{3}\) as = 6 X 6 X \(\frac{1}{3}\) = 36 X \(\frac{1}{3}\) = \(\frac{36}{3}\) , As denominator is same in the both expression we compare numerators and even the numerators are same in both the expression so given both products are equal . Therefore \(\frac{4}{3}\) X 9 =, is equal to 6 X \(\frac{6}{3}\).

Question 13. YOU BE THE TEACHER Your friend says the product of 6 and \(\frac{5}{8}\) is \(\frac{5}{48}\). Is your friend correct? Explain. Answer: No she is not correct as the product of  6 X \(\frac{5}{8}\) ≠ \(\frac{5}{48}\).

Explanation: We write the expression as a product  6 and \(\frac{5}{8}\) = 6 X \(\frac{5}{8}\) to find the product of a whole number and a fraction we multiply the numerator by the whole number. The denominator stays the same. 6 X 5 X  \(\frac{1}{8}\) =30 X \(\frac{1}{8}\) = \(\frac{30}{8}\) now comparing with \(\frac{5}{48}\) as denominators are different in order to compare the fractions we must first turn their different denominators into the same denominators to make denominators same we multiply \(\frac{30}{8}\) with numerator and denominator as 6= \(\frac{30}{8}\) X \(\frac{6}{6}\) we get \(\frac{180}{48}\), now comparing with \(\frac{5}{48}\) being denominators same, now numerators 180,5 both are not same so friend is not correct as the product of  6 X \(\frac{5}{8}\) ≠ \(\frac{5}{48}\).

Question 14. Open-Ended The product of a whole number and a fraction is \(\frac{24}{10}\). What could the two factors be? Answer: The product is given as \(\frac{24}{10}\) and the two factors can be (1,24) or (2,12) or (3,8) or (4,6) with multiple of \(\frac{1}{10}\).

Explanation: Given the product of a whole number and a fraction is \(\frac{24}{10}\) to find the two factors, we first write the product as whole with multiple of unit fraction,\(\frac{24}{10}\), we write as 24 X \(\frac{1}{10}\), now the whole part is 24 we find the factors for 24, two more numbers that multiply to make 24 are 1 x 24 = 2 x 12 = 3 x 8 = 4 x 6 = 24. This means the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 so the two factors can be (1,24) X \(\frac{1}{10}\) or  (2,12) X  \(\frac{1}{10}\) or (3,8) X \(\frac{1}{10}\) or (4,6) X \(\frac{1}{10}\) = 24 X \(\frac{1}{10}\) = \(\frac{24}{10}\).

Explanation: Given a short roller-coaster track is \(\frac{3}{10}\) mile long and longer roller-coaster track is about 4 times as long as the short track to find how long is the longer roller-coaster track we multiply short roller-coaster track  \(\frac{3}{10}\) mile by 4, 4 X \(\frac{3}{10}\) now we write expression as a multiple of unit fraction so 4 X 3 X \(\frac{1}{10}\) = 12 X \(\frac{1}{10}\) =\(\frac{12}{10}\)  as numerator and denominator can be further simplified as both can be divided by 2 making it as \(\frac{6}{5}\).

Explanation: Given the Renaissance Tower in Dallas, Texas is \(\frac{27}{100}\) kilometer tall and The Burj Khalifa is about 3 times as tall as the Renaissance Tower, To find how tall is the Burj Khalifa we multiply the Renaissance Tower in Dallas, Texas length by 3, 3 X \(\frac{27}{100}\) we write expression as a multiple of unit fraction 3 X 27 X \(\frac{1}{100}\) = 81 X \(\frac{1}{100}\) =\(\frac{81}{100}\), So the Burj Khalifa is \(\frac{81}{100}\) kilometers tall.

Question 16. You water 6 plants using \(\frac{3}{5}\) liter of water for each. How many liters of water do you use? Between which two whole numbers does your answer lie? Answer: \(\frac{18}{5}\) liters of water is used, The answer will lie in between 3 and 4.

Explanation: Given \(\frac{3}{5}\) liter of water for each is watered, for 6 plants total number of liters of water used is 6 X \(\frac{3}{5},now we write as multiple of unit fraction as 6 X 3 X [latex]\frac{1}{5}\) = 18 X \(\frac{1}{5}\)= \(\frac{18}{5}\) liters of water is used.

Now we divide \(\frac{18}{5}\) we get the value as 3.6,means between 3 and 4 whole numbers our answer will lie.

Question 17. DIG DEEPER! You have 6 cups of strawberries. You want to make 4 strawberry-banana smoothies and 4 strawberry-kiwi smoothies. Each smoothie needs \(\frac{2}{3}\) cup of strawberries. Do you have enough? If not, how many more cups of strawberries do you need? Answer: Yes we have enough cups of strawberries, we dont require more cups of strawberries.

Explanation: Given to make each smoothie we need \(\frac{2}{3}\) cup of strawberries first to make 4 strawberry-banana smoothies we need 4 X \(\frac{2}{3}\) cup of strawberries = 4 X 2 X \(\frac{1}{3}\)= 8 X \(\frac{1}{3}\) =\(\frac{8}{3}\), Now similarly for making 4 strawberry-kiwi smoothies we need 4 X \(\frac{2}{3}\) cup of strawberries = 4 X 2 X \(\frac{1}{3}\)= 8 X \(\frac{1}{3}\) = \(\frac{8}{3}\),Now total cups of strawberries requried to make smoothie is \(\frac{8}{3}\) + \(\frac{8}{3}\) = 2 X \(\frac{8}{3}\)= 2 X 8 X \(\frac{1}{3}\)=16 X \(\frac{1}{3}\)= \(\frac{16}{3}\) totals cups of strawberries required to make smoothies. Now we compare \(\frac{16}{3}\) with 6, To compare first we make both denominators same so we multiply 6 with \(\frac{3}{3}\)= 6 X \(\frac{3}{3}\)= 6 X 3 X \(\frac{1}{3}\)= 18 X \(\frac{1}{3}\)= \(\frac{18}{3}\) as both the denominators are same we compare \(\frac{16}{3}\) and \(\frac{18}{3}\) by seeing numerator 16 is less than 18 means we have enough cups of strawberries.

Explanation: 3 X \(\frac{3}{5}\) can be written as multiple of unit fraction as 3 X (3 X \(\frac{1}{5}\)) applying associative property of multiplication ( 3 X 3 ) X \(\frac{1}{5}\)= 9 X \(\frac{1}{5}\) = \(\frac{9}{5}\).

Explanation: 1 X \(\frac{6}{8}\) can be written as multiple of unit fraction as 1 X ( 6 X  \(\frac{1}{8}\)) applying associative property of multiplication ( 1 X 6 ) X \(\frac{1}{8}\) = 6 X \(\frac{1}{8}\)= \(\frac{6}{8}\) as numerator and denominator can be further simplified as both can be divided by 2 making as \(\frac{6}{8}\)= \(\frac{3}{4}\)X \(\frac{2}{2}\)=\(\frac{3}{4}\).

Explanation: 4 X \(\frac{10}{12}\) can be written as multiple of unit fraction as 4 X ( 10 X  \(\frac{1}{12}\)) applying associative property of multiplication ( 4 X 10 ) X \(\frac{1}{12}\) = 40 X \(\frac{1}{12}\)= \(\frac{40}{12}\) as numerator and denominator can be further simplified as both can be divided by 4 as \(\frac{40}{12}\)= \(\frac{10}{3}\) X \(\frac{4}{4}\)=\(\frac{10}{3}\) X 1 = \(\frac{10}{3}\).

Explanation: 7 X \(\frac{6}{10}\) can be written as multiple of unit fraction as 7 X ( 6 X  \(\frac{1}{10}\)) applying associative property of multiplication ( 7 X 6 ) X \(\frac{1}{10}\) = 42 X \(\frac{1}{10}\)= \(\frac{42}{10}\) as numerator and denominator can be further simplified as both can be divided by 2 as \(\frac{42}{10}\)= \(\frac{21}{5}\) X \(\frac{2}{2}\)=\(\frac{21}{5}\) X 1 = \(\frac{21}{5}\).

Explanation: \(\frac{4}{6}\) X 5  can be written as multiple of unit fraction as 5 X ( 4 X  \(\frac{1}{6}\)) applying associative property of multiplication ( 5 X 4 ) X \(\frac{1}{6}\) = 20 X \(\frac{1}{6}\)= \(\frac{20}{6}\) as numerator and denominator can be further simplified as both can be divided by 2 as \(\frac{20}{6}\) = \(\frac{10}{3}\) X \(\frac{2}{2}\)=\(\frac{10}{3}\) X 1 =\(\frac{10}{3}\).

Explanation: 8 X \(\frac{5}{2}\) can be written as multiple of unit fraction as 8 X ( 5 X  \(\frac{1}{2}\)) applying associative property of multiplication ( 8 X 5 ) X \(\frac{1}{2}\) = 40 X \(\frac{1}{2}\) = \(\frac{40}{2}\) further it can be simplified as \(\frac{40}{2}\)=20

Explanation: \(\frac{70}{100}\) X 6  can be written as multiple of unit fraction as 6 X ( 70 X  \(\frac{1}{100}\)) applying associative property of multiplication ( 6 X 70 ) X \(\frac{1}{100}\) = 420 X \(\frac{1}{100}\)= \(\frac{420}{100}\) as numerator and denominator can be further simplified as both can be divided by 10 as \(\frac{420}{100}\)= \(\frac{42}{10}\) X \(\frac{10}{10}\)=\(\frac{42}{10}\) X 1 =\(\frac{42}{10}\).

Explanation: 10 X \(\frac{9}{3}\) can be written as multiple of unit fraction as 10 X ( 9 X  \(\frac{1}{3}\)) applying associative property of multiplication ( 10 X 9 ) X \(\frac{1}{3}\) = 90 X \(\frac{1}{3}\)= \(\frac{90}{3}\) further it can be simplified  because 90 can be divided by 3 without leaving remainder making \(\frac{90}{3}\) as 30.

Explanation: 2 X \(\frac{4}{2}\) we write as multiple of unit fraction 2 X 4 X \(\frac{1}{2}\) = 8 X \(\frac{1}{2}\) =  \(\frac{8}{2}\) and 8 X \(\frac{1}{2}\) is already multiple of unit fraction = latex]\frac{8}{2}[/latex] now we can compare both sides  as the denominators are same and numerators are same both the expressions are equal 2 X \(\frac{4}{2}\) = 8 X \(\frac{1}{2}\).

Explanation: 9 X \(\frac{1}{5}\) already it is multiple of unit fraction we write it as \(\frac{9}{5}\) and 9 X \(\frac{1}{12}\) is also a multiple of unit fraction we write it as \(\frac{9}{12}\) now comparing first we make both denominators same so the multiples of 5 are 5,10,15,20,25,30,35,40,45,50,55,60,65 and multiples of 12 are 12,24,36,48,60,72 so the least common multiple of both 5 and 12 is 60 , So to get denominator 60 we multiply top and bottom by 12 for \(\frac{9}{5}\) = \(\frac{9}{5}\) X \(\frac{12}{12}\)= \(\frac{108}{60}\) , So to get denominator 60 we multiply top and bottom by 5 for \(\frac{9}{12}\)= \(\frac{9}{12}\) X \(\frac{5}{5}\)= \(\frac{45}{60}\) Now comparing \(\frac{108}{60}\) with \(\frac{45}{60}\)  as both denominator are same, As we see numerator it is clear that 108 is more than 45 so \(\frac{108}{60}\) > \(\frac{45}{60}\) , 9 X \(\frac{1}{5}\) > \(\frac{9}{12}\).

Answer: \(\frac{6}{10}\)X 4 < 9 X \(\frac{3}{10}\).

Explanation: \(\frac{6}{10}\)X 4 we write as multiple of unit fraction 4 X 6 X \(\frac{1}{10}\)= 24 X \(\frac{1}{10}\) =\(\frac{24}{10}\) and 9 X \(\frac{3}{10}\) as 9 X 3 X \(\frac{1}{10}\)= 27 X \(\frac{1}{10}\)=\(\frac{27}{10}\), We can compare both as denominators of both are 10 we compare numerators as 24 with 27 as 27 is great,\(\frac{24}{10}\) < \(\frac{27}{10}\), \(\frac{6}{10}\) X 4 < 9 X \(\frac{3}{10}\).

Question 13. Writing Explain one way to multiply a whole number by a fraction. Answer: First we write fraction as multiple of unit fraction then applying associative property of multiplication we multiply the whole numbers first then with the unit fraction,later writing product.

Explanation: Let us take the whole number as 2 and fraction as \(\frac{3}{5}\),Here first we write fraction as multiple of unit fraction as 2 X 3 X \(\frac{1}{5}\) now applying associative property of multiplication, we multiply the whole numbers first then with the unit fraction, 6 X \(\frac{1}{5}\) as \(\frac{6}{5}\).

Question 14. Number Sense Between which two whole numbers does the product of 10 and \(\frac{5}{6}\) lie ? Answer: In between 8 and 9 whole numbers the product of 10 and \(\frac{5}{6}\) lies.

Explanation: 10 X \(\frac{5}{6}\) we first write it as multiple of unit fraction as 10 X ( 5 X \(\frac{1}{6}\)) applying associative property of multiplication, We write (10 X 5) X \(\frac{1}{6}\)= 50 X \(\frac{1}{6}\) = \(\frac{50}{6}\) if we divide \(\frac{50}{6}\))  we get 8.333 which is in between 8 and 9 whole numbers.

Question 15. Modeling Real Life A seamstress makes fabric dog collars. An extra small dog collar is \(\frac{1}{3}\) yard long. A large dog collar is 2 times as long as the extra small dog collar. How long is the large dog collar? Answer: The large dog collar is \(\frac{2}{3}\) yards long.

Explanation: Given the extra small dog collar is \(\frac{1}{3}\) yard long and a large dog collar is 2 times as long as the extra small dog collar, so the large dog collar = 2 X \(\frac{1}{3}\) as the fraction is multiple of unit fraction we write as \(\frac{2}{3}\) , therefore the large dog collar is \(\frac{2}{3}\) yards long.

Answer: Newton has 16 baseball cards, Descartes has 4 baseball cards and my friend has 2 baseball cards, over all there are total 22 baseball cards.

Explanation: Given Newton has 16 baseball cards and Descartes has \(\frac{1}{4}\) as many cards as Newton so\(\frac{1}{4}\) of 16 = 16 X \(\frac{1}{4}\) cards  it is already as multiple of unit fraction we can write as \(\frac{16}{4}\) further can be simplified because 16 can be divide by 4 so we get as 4 means Descartes has 4 cards.Now my friend has \(\frac{1}{2}\) as many cards as Descartes So Descartes has 4 cards means my friend has \(\frac{1}{2}\) out of 4= 4 X \(\frac{1}{2}\)  as it is already a multiple of unit fracrtion, we take it as \(\frac{4}{2}\) further we can simplify as 4 is divisiblle by 2 we get 2 cards. Newton has 16 baseball cards,Descartes has 4 baseball cards and my friend has 2 baseball cards, over all there are total 22 baseball cards.

Structure How does the Distributive Property relate to your models? Explain.

Explanation: According to the distributive property, multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together. We use distribute property while relating the models first we draw each multiplying addend individually and then add the products together to give the result. Example : 2 X 1 \(\frac{3}{8}\) using the distributive property we write 2 X 1 \(\frac{3}{8}\) as 2 X ( 1 + \(\frac{3}{8}\) )here first we draw model of 2 X 1 then we draw model of 2 X \(\frac{3}{8}\) then we add the products together as 2 + \(\frac{6}{8}\)= 2 + \(\frac{3}{4}\)=\(\frac{11}{4}\). Think and Grow: Multiply Whole Numbers and Mixed Numbers

Explanation: First we write the mixed number as a fraction then multiply, so 1 \(\frac{7}{12}\)= 1 + \(\frac{7}{12}\)= (1 X 12 + 7 ) by 12 = \(\frac{19}{12}\) now we multiply it with the whole, 2 X \(\frac{19}{12}\) now we write in unit fraction and multiply  2 X 19 X \(\frac{1}{12}\)= 38 X \(\frac{1}{12}\) = \(\frac{38}{12}\), we can write the product as mixed fraction as 12 goes in 3 times. 3 will be our whole number 12 X 3 is 36 and we have 2 left over (38 −36 ). 2 will be our numerator and our denominator will be the same 12 = \(\frac{38}{12}\)=  3 \(\frac{2}{12}\).

Explanation: First we write the mixed number as a fraction then multiply, so 2 \(\frac{1}{2}\)= 2 + \(\frac{1}{2}\)= (2 X 2 + 1) by 2 = \(\frac{5}{2}\) now we multiply it with the whole, 5 X \(\frac{5}{2}\) . We write in unit fraction and multiply  5 X 5 X \(\frac{1}{2}\)= 25 X \(\frac{1}{2}\) = \(\frac{25}{2}\), we can write the product as mixed fraction as 2 goes in 12 times. 12 will be our whole number 2 X 12 is 24 and we have 1 left over (25 −24 ). 1 will be our numerator and our denominator will be the same 2= \(\frac{25}{2}\)= 12 \(\frac{1}{2}\)

Explanation: First we write the mixed number as a fraction then multiply, so 7 \(\frac{4}{6}\)= 7 + \(\frac{4}{6}\)=(7 X 6 + 4) by 6 = \(\frac{46}{6}\) now we multiply it with the whole, 4 X \(\frac{46}{6}\) now we write in unit fraction and multiply  4 X 46 X \(\frac{1}{6}\)= 184 X \(\frac{1}{6}\) = \(\frac{184}{6}\), we can write the product as mixed fraction as 6 goes in 30 times. 30 will be our whole number 6 X 30 is 180 and we have 4 left over (184 −180 ). 4 will be our numerator and our denominator will be the same 6= \(\frac{184}{6}\)= 30 \(\frac{4}{6}\) and \(\frac{184}{6}\) can also written as \(\frac{92}{3}\) because both 184 and 6 can also be further divided by 2.

Explanation: First we write the mixed number as a fraction then multiply, so 3 \(\frac{9}{10}\)= 3 + \(\frac{9}{10}\)= (3 X 10 + 9) by 10 = \(\frac{39}{10}\) now we multiply it with the whole, 8 X \(\frac{39}{10}\) now we write in unit fraction and multiply  8 X 39 X \(\frac{1}{10}\)= 312 X \(\frac{1}{10}\) = \(\frac{312}{10}\), we can write the product as mixed fraction as 10 goes in 31 times. 31 will be our whole number 10 X 31 is 310 and we have 2 left over (312 −310 ). 2 will be our numerator and our denominator will be the same 10= \(\frac{312}{10}\)= 31 \(\frac{2}{10}\).

Question 10. Reasoning How can you check whether your answer in Exercise 9 is reasonable? Answer: Exercise 9 is reasonable.

Explanation: In the above problem 6 X 9 \(\frac{11}{100}\) we got product as \(\frac{5466}{100}\) =54 \(\frac{66}{100}\)  we can write the product as mixed fraction as 100 goes in 54 times. 54 will be our whole number 100 X 54 is 5400 and 5466-5400, leaving 66 as remainder so we consider the whole part 54,As 54 is reasonable close to 100.

Explanation: First we write the mixed number as a fraction then multiply, so 6 \(\frac{1}{3}\)= 6 + \(\frac{1}{3}\)= (6 X 3 + 1) by 3 = \(\frac{19}{3}\) now we multiply it with the whole, 9 X \(\frac{19}{3}\) now we write in unit fraction and multiply  9 X 19 X \(\frac{1}{3}\)= 171 X \(\frac{1}{3}\) = \(\frac{171}{3}\), As 3 goes 57 times leaving no remainder so we get 9 X 6 \(\frac{1}{3}\)=57 which matches with the friend’s product so friend is correct.

Explanation:

Explanation: First we write mixed fraction as fraction so 2 \(\frac{1}{2}\) =( 2 X 2 + 1 ) by 2 = \(\frac{5}{2}\) , now we multiply with whole number 2 = 2 X \(\frac{5}{2}\) = 2 X 5 X \(\frac{1}{2}\) = 10 X \(\frac{1}{2}\) = \(\frac{10}{2}\) further it can be simplified as both 10,2 can be divided by 2 so \(\frac{10}{2}\) = 5 hours that day.

Question 13. A small bag of popcorn has 1\(\frac{7}{8}\) cups of popcorn. A large bag has 7 times as many cups of popcorn as the small bag. How many cups of popcorn are in the large bag? Answer: The large bag has \(\frac{105}{8}\) cups of popcorn or 13 \(\frac{1}{8}\) cups of popcorn.

Explanation: Given A small bag of popcorn has 1\(\frac{7}{8}\) cups of popcorn so first we write mixed fraction as fraction as 1 \(\frac{7}{8}\)= ( 1 X 8 + 7 ) by 8 =\(\frac{15}{8}\) ,as a large bag has 7 times as many cups of popcorn as the small bag so 7 times \(\frac{15}{8}\)= 7 X 15 X \(\frac{1}{8}\)=105 X \(\frac{1}{8}\)=\(\frac{105}{8}\) cups of popcorn. We can write the product as mixed fraction as 8 goes in 13 times. 13 will be our whole number 8 X 13 is 104 and we have 1 left over (105 −104 ). 1 will be our numerator and our denominator will be the same 8= \(\frac{105}{8}\)= 13 \(\frac{1}{8}\) cups of popcorn.

Question 14. DIG DEEPER! A deli worker prepares 4 packages of meat. Each package contains 3\(\frac{1}{3}\) pounds of meat. Your friend says the deli worker uses between 12 and 13 pounds of meat. Is your friend correct? Explain. Answer: No, friend is not correct as deli worker uses between 13 and 14 pounds of meat, not between 12 and 13 pounds of meat.

Explanation: Given each package contains 3\(\frac{1}{3}\) pounds of meat. First we write mixed fraction as fraction as ( 3 X 3 +1 ) by 3= \(\frac{10}{3}\) now deli worker prepares 4 packages of meat so 4 times  \(\frac{10}{3}\)= 4 X 10 X \(\frac{1}{3}\)= 40 X \(\frac{1}{3}\)=\(\frac{40}{3}\) , we will write the product as mixed fraction as 3 goes in 13 times. 13 will be our whole number 3  X 13 is 39 and we have 1 left over (40 −39 ). 1 will be our numerator and our denominator will be the same 3= \(\frac{105}{8}\)= 13 \(\frac{1}{3}\), so the whole number is greater than 13 means deli worker uses between 13 and 14 pounds of meat not between 12 and 13 pounds of meat.

Explanation: Given an athlete runs 5\(\frac{3}{4}\) miles 2 times each day, First we write mixed fraction as fraction as ( 5 X 4 +3) by 4 =\(\frac{23}{4}\) now we multiply by 2 , 2 X \(\frac{23}{4}\)= 2 X 23 X \(\frac{1}{4}\)=46 x \(\frac{1}{4}\)= \(\frac{46}{4}\), Now for 1 week it is 7 X \(\frac{46}{4}\)= 7 X 46 X \(\frac{1}{4}\)= 322 X \(\frac{1}{4}\)= \(\frac{322}{4}\) now we will write the product as mixed fraction as 4 goes in 80 times. 80 will be our whole number 4 X 80 is 320 and we have 2 left over (322 −320 ). 2 will be our numerator and our denominator will be the same 4= \(\frac{322}{4}\)= 80 \(\frac{2}{4}\) , As 80 \(\frac{2}{4}\) is greater than 80,  athlete runs more than 80 miles in 1 week he meets his goal.

Explanation: First we write the mixed number as a fraction then multiply, so 1 \(\frac{1}{12}\) = 1 + \(\frac{1}{12}\) = ( 1 X 12 + 1 )by 12 = \(\frac{13}{12}\) now we multiply it with the whole, 2 X \(\frac{13}{12}\) now we write in unit fraction and multiply  2 X 13 X \(\frac{1}{12}\)= 26 X \(\frac{1}{12}\) = \(\frac{26}{6}\), Now the fraction can be further divided by 2 as 26,6 both goes in 2 after dividing it becomes as \(\frac{13}{6}\) now we can write the product as mixed fraction as 6 goes in 2 times. 2 will be our whole number 6 X 2 is 12 and we have 1 left over (13 −12 ). 1 will be our numerator and our denominator will be the same 6= \(\frac{13}{6}\)= 2 \(\frac{1}{6}\).

Explanation: First we write the mixed number as a fraction then multiply, so 3 \(\frac{5}{6}\) = 3 + \(\frac{5}{6}\) =(3 X 6 + 5) by 6 = \(\frac{23}{6}\) now we multiply it with the whole, 2 X \(\frac{23}{6}\) now we write in unit fraction and multiply  2 X 23 X \(\frac{1}{6}\)= 46 X \(\frac{1}{6}\) = \(\frac{46}{6}\), Now the fraction can be further divided by 2 as 46,6 both goes in 2 after dividing it becomes as 23,3 = \(\frac{23}{3}\) now we can write the product as mixed fraction as 3 goes in 7 times. 3 X 7  is 21 , 7 will be our whole number and  (23 −21 ) we have 2 left over. 2 will be our numerator and our denominator will be the same 3= \(\frac{23}{3}\)= 7 \(\frac{2}{3}\).

Explanation: First we write the mixed number as a fraction then multiply, so 3 \(\frac{6}{10}\) = 3 + \(\frac{6}{10}\) =( 3 X 10 + 6 ) by 10 = \(\frac{36}{10}\) now we multiply it with the whole, 4 X \(\frac{36}{10}\) now we write in unit fraction and multiply  4 X 36 X \(\frac{1}{10}\)= 144 X \(\frac{1}{10}\) = \(\frac{144}{10}\), Now the fraction can be further divided by 2 as 144,10 both goes in 2 after dividing it becomes as 72,5 = \(\frac{72}{5}\) now we can write the product as mixed fraction as 5 goes in 14 times. 5 X 14  is 70 ,14 will be our whole number and  (72 −70 ) we have 2 left over. 2 will be our numerator and our denominator will be the same 5 = \(\frac{72}{5}\)= 14 \(\frac{2}{5}\).

Explanation: First we write the mixed number as a fraction then multiply, so 2 \(\frac{3}{8}\) = 2 + \(\frac{3}{8}\) =(2 X 8 + 3) by 8 = \(\frac{19}{8}\) now we multiply it with the whole, 5 X \(\frac{19}{8}\) now we write in unit fraction and multiply  5 X 19 X \(\frac{1}{8}\)= 95 X \(\frac{1}{8}\) = \(\frac{95}{8}\), Now we can write the product as mixed fraction as 8 goes in 11 times. 8 X 11  is 88 ,11 will be our whole number and  (95 −88 ) we have 7 left over. 7 will be our numerator and our denominator will be the same 8= \(\frac{95}{8}\)= 11 \(\frac{7}{8}\).

Explanation: First we write the mixed number as a fraction then multiply, so 8 \(\frac{20}{100}\) = 8 + \(\frac{20}{100}\) = ( 8 X 100 + 20 ) by 100 = \(\frac{820}{100}\) now we multiply it with the whole, 100 X \(\frac{820}{100}\) now we write in unit fraction and multiply  10 X 820 X \(\frac{1}{100}\) = 8200 X \(\frac{1}{100}\) = \(\frac{8200}{100}\), Now the fraction can be further divided by 100 as 8200,100 both goes in 100 after dividing by 100  it becomes as 82.

Question 7. Reasoning Without calculating, is the product of 7 and 5\(\frac{3}{4}\) greater than or less than 35? Explain. Answer: 7 X 5 \(\frac{3}{4}\) is greater than ( > ) 35.

Explanation: Given to compare between 7 X 5 \(\frac{3}{4}\) and 35, first we take 7 X 5 \(\frac{3}{4}\) to this expression if we apply distributive property means if multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together. example a X b \(\frac{c}{d}\) = a X b + a X \(\frac{c}{d}\) So 7 X 5 \(\frac{3}{4}\) = 7 X 5 + 7 X \(\frac{3}{4}\) means 35 + 7 X  \(\frac{3}{4}\) So by comparing left side it is addition of 35 plus some value ,it will be more than 35 only and right hand side it it 35 so obiviously 7 X 5 \(\frac{3}{4}\) is greater than ( > ) 35.

Explanation: To caluculate the value of 4 X 2 \(\frac{8}{10}\) , First we write the mixed number as a fraction then multiply, so 2 \(\frac{8}{10}\) = 2 + \(\frac{8}{10}\) = ( 2 X 10 + 8 ) by 10 = \(\frac{28}{10}\) now we multiply it with the whole, 4 X \(\frac{28}{10}\) now we write in unit fraction and multiply  4 X 28 X \(\frac{1}{10}\)= 112 X \(\frac{1}{10}\) = \(\frac{112}{10}\), now we can write the product as mixed fraction as 10 goes in 11 times. 10 X 11  is 110 , 11 will be our whole number and  (112 −110 ) we have 2 left over. 2 will be our numerator and our denominator will be the same 10= \(\frac{112}{10}\)= 11 \(\frac{2}{10}\) is same as friend, So friend is correct.

Question 9. Number Sense Between which two whole numbers does the product of 9 and 7 \(\frac{1}{8}\) lie? Answer: The product of 9 and 7 \(\frac{1}{8}\) lies in between 64 and 65.

Explanation: First we write the mixed number 7 \(\frac{1}{8}\) as a fraction then multiply, so 7 \(\frac{1}{8}\) = 7 + \(\frac{1}{8}\) = 7 X 8 + 1 by 8 = \(\frac{57}{8}\) now we multiply it with the whole, 9 X \(\frac{57}{8}\) now we write in unit fraction and multiply  9 X 57 X \(\frac{1}{8}\) = 513 X \(\frac{1}{8}\) = \(\frac{513}{8}\), now we can write the product as mixed fraction as 8 goes in 64 times. 8 X 64  is 512 , 64 will be our whole number and  (513 −512 ) we have 1 left over. 1 will be our numerator and our denominator will be the same 8 = \(\frac{513}{8}\)= 64 \(\frac{1}{8}\). Any how we got 64 whole number and next whole number is 65 means the product of 9 and 7 \(\frac{1}{8}\) lies in between 64 and 65.

Explanation: Given Athlete A hold a 2 \(\frac{1}{2}\) kilogram plate while doing squats, Athlete B holds a plate that is 4 times heavier than Athlete A’s, So Athlete B holds 4 X 2 \(\frac{1}{2}\), First we write the mixed number as a fraction then multiply, so 2 \(\frac{1}{2}\) = 2 + \(\frac{1}{2}\) = ( 2 X 2 + 1 ) by 2 = \(\frac{5}{2}\) now we multiply it with the whole, 4 X \(\frac{5}{2}\) now we write in unit fraction and multiply  4 X 5 X \(\frac{1}{2}\) = 20 X \(\frac{1}{2}\) = \(\frac{20}{2}\) as 20,2 goes by 2 we further divide both by 2 we get 10,1 = \(\frac{20}{2}\) = 10, therefore Athlete B holds 10 kilograms of plate while doing squats.

Question 11. DIG DEEPER! A zoo nutritionist orders 5 \(\frac{1}{4}\) tons of apples and 7\(\frac{2}{4}\)tons of bananas each year to feed the animals. She orders 6 times as many tons of herbivore pellets than tons of fruit. How many tons of herbivore pellets does the nutritionist order? Answer: Explanation:

First we write the mixed numbers into fractions and subtract, So 9 \(\frac{1}{4}\) = 9 +\(\frac{1}{4}\) = (9 x 4 + 1) by 4 = \(\frac{37}{4}\) and  6 \(\frac{1}{4}\) = 6 + \(\frac{1}{4}\) = 6 X 4 + 1 by 4 = \(\frac{25}{4}\) now subtracting \(\frac{37}{4}\) – \(\frac{25}{4}\) to subtract a fraction from another we need to first make sure both fractions have the same denominator as both denominators have same value 4 now we can subtract one numerator from the other to make one fraction as 37 – 25 by 4 = \(\frac{12}{4}\) as 12, 4 goes by 4 we further can divide both by 4 we get 12,1 = \(\frac{12}{4}\) = 3 .

Explanation: First we write the mixed numbers into fractions and subtract, So 8 \(\frac{4}{12}\) = 8 +\(\frac{4}{12}\) = ( 8 x 12 + 4 ) by 12 = \(\frac{100}{12}\) and  1 \(\frac{10}{12}\) = 1 + \(\frac{10}{12}\) = 1 X 12 + 10 by 12 = \(\frac{22}{12}\) now subtracting \(\frac{100}{12}\) – \(\frac{22}{12}\) to subtract a fraction from another we need to first make sure both fractions have the same denominator as both denominators have same value 3 now we can subtract one numerator from the other to make one fraction as 100 – 22 by 12 = \(\frac{78}{12}\) since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as 12 goes in 6 times. 12 X 6  is 72 , 6 will be our whole number and  (78 −12 ) we have 6 left over. 6 will be our numerator and our denominator will be the same 12, 6 \(\frac{6}{12}\) , Further \(\frac{78}{12}\) can be simplified as 78,12 both goes by  2 making them as 39, 6=\(\frac{39}{6}\), even further \(\frac{39}{6}\) can be simplified as 39,6 both goes by 3 making them as 13,2 =\(\frac{13}{2}\), Since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as 2 goes in 6 times. 6 X 2  is 12 , 6 will be our whole number and  (13 −12 ) we have 1 left over. 1 will be our numerator and our denominator will be the same 3 = \(\frac{78}{12}\) = 6 \(\frac{6}{12}\) or \(\frac{78}{12}\) = \(\frac{39}{6}\) = \(\frac{13}{2}\) = 6 \(\frac{1}{2}\).

Reasoning Explain how you can tell whether you need more than or less than 6 cups of blueberries without calculating Answer: To make 3 batches of recipe we require \(\frac{21}{4}\) pounds of strawberries or 5 \(\frac{1}{4}\) pounds of strawberries required, \(\frac{3}{2}\) cups of orange juice or 1 \(\frac{1}{2}\) cups of orange juice, 12 tablespoons of lemon juice,12 tablespoons of honey and \(\frac{27}{4}\) cups of blue berries are required or 6 \(\frac{3}{4}\) cups of blueberries are required. We require more than 6 cups of blueberries why because if we see product required is 3 X 2 \(\frac{1}{4}\) cups of blue berries without calculating , Applying distributive property we get (3 X 2 ) + ( 3 X \(\frac{1}{4}\) ) = 6 + ( 3 X \(\frac{1}{4}\) ) it clearly shows that we require more than 6 cups of blueberries.

Explanation: Given  to make 1 batch of Recipe Ladybug Ice pops we require  1 \(\frac{3}{4}\) pounds of strawberries , \(\frac{1}{2}\) cup of orange juice , 4 tablespoons of lemon juice, 4 tablespoons of honey, 2 \(\frac{1}{4}\) cups of blue berries. To find out how much of each ingredient needed to make 3 batches of recipe we multiply each ingredient with 3. Strawberries needed are 3 X 1 \(\frac{3}{4}\), First we write the mixed number as a fraction then multiply so 1 \(\frac{3}{4}\) = ( 1 X 4 + 3 ) by 4 = \(\frac{7}{4}\) , so 3  X  \(\frac{7}{4}\)  now we write in unit fraction and multiply  3 X 7 X \(\frac{1}{4}\) = 21 X \(\frac{1}{4}\) = \(\frac{21}{4}\) pounds of strawberries required, Since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as 4 goes in 5 times. 5 X 4  is 20 , 5 will be our whole number and  (21 −20 ) we have 1 left over. 1 will be our numerator and our denominator will be the same 4 =5 \(\frac{1}{4}\) pounds of strawberries required, Orange juice required 3 X \(\frac{1}{2}\) already it is in unit fraction so we require \(\frac{3}{2}\) cups of orange juice Since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as 2 goes in 1 times. 1 X 2 is 2, 1 will be our whole number and (3-2), we have 1 left over. 1 will be our numerator and our denominator will be the same 2 = 1 \(\frac{1}{2}\) cups of orange juice. Lemon juice required is 3 X 4 = 12 tablespoons of lemon juice, 3 X 4 = 12 tablespoons of honey, now blue berries required are 3 X 2 \(\frac{1}{4}\) cups, First we write the mixed number as a fraction then multiply so 2 \(\frac{1}{4}\) = ( 2 X 4 + 1 ) by 4 = \(\frac{9}{4}\) , now  3  X  \(\frac{9}{4}\)  now we write in unit fraction and multiply  3 X 9 X \(\frac{1}{4}\) = 27 X \(\frac{1}{4}\) = \(\frac{27}{4}\) cups of blue berries are required, Since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as 4 goes in 6 times. 6 X 4 is 24, 6 will be our whole number and (27-24) we have 3 left over. 3 will be our numerator and our denominator will be the same 4 = 6 \(\frac{3}{4}\) cups of blueberries are required. Now comparing blueberries we require more than 6 cups of blueberries why because we use is 3 X 2 \(\frac{1}{4}\) cups of blue berries are required without calculating, Applying distributive property  we get (3 X 2 ) +( 3 X \(\frac{1}{4}\) )= 6 +( 3 X \(\frac{1}{4}\)), it clearly shows that we require more than 6 cups of blueberries.

Think and Grow: Problem Solving: Fraction Operations

Explanation: We know to convert a temperature from degrees Celsius to degrees Fahrenheit is Fahrenheit = Celsius temperature X \(\frac{9}{5}\) + 32 . So multiply the Celsius temperature by \(\frac{9}{5}\), then add 32, Given the Celsius temperature as 10 0 in the figure above, First we write \(\frac{9}{5}\) in unit fraction and multiply by Celsius temperature so \(\frac{9}{5}\)=9 X \(\frac{1}{5}\) now we multiply by 10 as 10 X 9 X \(\frac{1}{5}\) = 90 X \(\frac{1}{5}\) = \(\frac{90}{5}\) this value can be further simplified as 90,5 both goes by  5 making them as 18,1, Now we got result as 18 , We add 18 + 32  we will get 50 0 Fahrenheit.

Understand the Problem

What do you know?

• To convert a temperature from degrees Celsius to degrees Fahrenheit, multiply the Celsius temperature by \(\frac{9}{5}\), then add 32.
• The thermometer shows 10 degrees Celsius. What do you need to find?
• You need to find the temperature shown by the thermometer in degrees Fahrenheit.

Make a Plan

How will you solve?

• First, multiply the Celsius temperature, 10 degrees, by \(\frac{9}{5}\).
• Then add 32 to the product.

Explanation : To convert a temperature from degrees Celsius to degrees Fahrenheit, Let us take  p as multiply of Celsius temperature by \(\frac{9}{5}\), Now we take f as p+32, First we write \(\frac{9}{5}\) in unit fraction and multiply by Celsius temperature, so \(\frac{9}{5}\)=9 X \(\frac{1}{5}\) now multiply by 10 = 10 X 9 X \(\frac{1}{5}\) = 90 X \(\frac{1}{5}\) = \(\frac{90}{5}\) this value we take as p this can be further simplified as 90,5 both goes by  5 making them as 18, Now we got p as 18 we add 32 to p , we get f, So f = 18 + 32 = 50 0 So the temperature shown by the thermometer is 50 0 Fahrenheit.

Question 1. Show how to solve the example above using one equation. Answer: F= C X \(\frac{9}{5}\) + 32 degrees.

Explanation: To solve the example above using one equation, Let us take Fahrenheit as F , Celsius as C so to convert a temperature from degrees Celsius to degrees Fahrenheit we multiply C with \(\frac{9}{5}\) and add 32.

Understand the problem. What do you know? What do you need to find? Explain.

Answer: We know Celsius temperature, We need to find the temperature shown by the thermometer in degrees Fahrenheit.

Explanation: Given the temperature in Celsius and we know the formula to convert from Celsius temperature to Fahrenheit as F= C X \(\frac{9}{5}\) + 32 degrees so we substitute the values and find the temperature.

Question 2. You make a friendship bracelet with 3 pink strings and 2 blue strings. Each string is 3\(\frac{3}{4}\) feet long. How many feet of string do you use?

Answer: \(\frac{75}{4}\) feet long strings is used or 18 \(\frac{3}{4}\) feet long strings is used.

Explanation: Given to make a friendship bracelet we use 3 pink strings , 2 blue strings means total 5 strings are needed and each string is 3\(\frac{3}{4}\) feet long, First we convert mixed fraction into fraction as ( 3 X 4 + 3 ) by 4 = \(\frac{15}{4}\) ,Now we will convert into unit fraction as 15 X \(\frac{1}{4}\) now we multiply with 5 , 5 X 15 X \(\frac{1}{4}\) =\(\frac{75}{4}\), since the numerator is greater than the denominator , we can further simplify it into a mixed fraction as 4 goes in 18 times. 18 X 4 is 72, 18 will be our whole number and (75-72) we have 3 left over. 3 will be our numerator and our denominator will be the same 4 = 18 \(\frac{3}{4}\), So \(\frac{75}{4}\) feet long strings is used or 18 \(\frac{3}{4}\) feet long strings is used.

Question 3. A smoothie store worker makes 4 peanut butter banana smoothies and 2 fruit smoothies. The worker uses \(\frac{2}{3}\) cup of bananas in each smoothie. How many cups of bananas does the worker need?

Answer: The worker needed  \(\frac{12}{3}\) cups of bananas or 4 cups of bananas are needed to make smoothie.

Explanation: Given a smoothie store worker makes 4 peanut butter banana smoothies and 2 fruit smoothies. The worker uses \(\frac{2}{3}\) cup of bananas in each smoothie so the number of cups of bananas the worker needed are 4 X latex]\frac{2}{3}[/latex] plus 2 X \(\frac{2}{3}\) first we convert both into unit fractions and multiply 4 X 2 X \(\frac{1}{3}\) = 8 X \(\frac{1}{3}\) = \(\frac{8}{3}\)  of peanut butter banana smoothies and 2 X 2 X \(\frac{1}{3}\) = 4 X \(\frac{1}{3}\) = \(\frac{4}{3}\) fruit smoothies, now we add both \(\frac{8}{3}\) + \(\frac{4}{3}\) as both denominators are same we add numerators as 8 + 4 =12 as numerator and denominator as 3 = \(\frac{12}{3}\) as numerator is greater than denominator further can be simplified and 12 divides by 3 gives 4 with 0 remainder so the worker needed  \(\frac{12}{3}\) cups of bananas or 4 cups of bananas are needed to make smoothie.

Understand the problem. Then make a plan. How will you solve? Explain.

Question 4. Your friend walks her dog for \(\frac{1}{4}\) mile each day. She then runs 2\(\frac{3}{4}\) miles each day. How many total miles does she walk her dog and run in 1 week?

Answer: Total 21 miles my friend walks her dog and run in 1 week.

Explanation: Given my friend walks her dog for \(\frac{1}{4}\) mile each day , She then runs 2\(\frac{3}{4}\) miles each day. So in each day she walk her dog and run  is \(\frac{1}{4}\) mile + 2\(\frac{3}{4}\) miles, Then we multiply with 7 because we have to calculate for 1 week. First we convert mixed fraction into fraction 2 \(\frac{3}{4}\) , ( 2 X 4 + 3 ) by 4 = \(\frac{11}{4}\), Now we add with \(\frac{1}{4}\) + \(\frac{11}{4}\) = \(\frac{12}{4}\) as numerator is greater than denominator further can be simplified and 12 divides by 4 gives 3 with 0 remainder, so 3 miles for each day, Now for 1 week it is 7 X 3  = 21 miles. My friend walks 21 miles her dog and run in 1 week.

Question 5. Hair donations must be 12 inches longer. Your friend’s hair is 7 inches long. Her hair grows about \(\frac{1}{2}\) inch each month. Can she donate her hair in 8 months?

No friend cannot donate her hair in 8 months as in 8 months it grows to 11 inches long and for hair donations it must be 12 inches longer.

Given friend’s hair is 7 inches long. Her hair grows about \(\frac{1}{2}\) inch each month. In 8 months it grows 8  X \(\frac{1}{2}\) = \(\frac{8}{2}\) as numerator is greater than denominator further can be simplified and 8 divides by 2 gives 4 with 0 remainder, in 8 months it will grow 4 inches , So after 8 months it will be 7 + 4 =11 inches long as 11 is less than 12 and hair donations must be 12 inches longer so friend cannot donate her hair in 8 months.

Question 6. Today you walk \(\frac{6}{10}\) mile from the Martin Luther King Jr. Memorial to the Washington Monument. Tomorrow you will walk about 4 times as far from the Washington Monument to the White House. About how much farther will you walk tomorrow?

Tomorrow i will walk father \(\frac{18}{10}\) miles  or \(\frac{9}{5}\) miles or 1 \(\frac{4}{5}\) .

Given today i walked \(\frac{6}{10}\) mile from the Martin Luther King Jr. Memorial to the Washington Monument, Tomorrow i will walk about 4 times as far from the Washington Monument to the White House , So number of miles walked tomorrow will be 4 X \(\frac{6}{10}\) first we write in unit fraction 4 X 6 X \(\frac{1}{10}\) = 24 X \(\frac{1}{10}\)  = \(\frac{24}{10}\) as numerator is greater than denominator further can be simplified as 24,10 both can be divided by 2 gives 12,5 so \(\frac{24}{10}\) = \(\frac{12}{5}\) now farther i will walk tomorrow will be \(\frac{24}{10}\) – \(\frac{6}{10}\) as denominator of both are same we substract numerators , 24 – 6 =18 and denominator will remain same 10 = \(\frac{18}{10}\) miles or further can be simplified and 18,10 both can be divided by by 2 gives 9, 5 = \(\frac{18}{10}\) miles = \(\frac{9}{5}\) miles as numerator is greater than denominator we can write it in mixed fraction as 5 goes in 1 times. 5 X 1 is 5 , 1 will be our whole number and (9 −5 ) we have 4 left over. 4 will be our numerator and our denominator will be the same 5 = 1 \(\frac{4}{5}\) .

The 60-pound dog eats 10 \(\frac{1}{2}\)  more cups of food than the 20-pound dog in 1 week.

Given A 20-pound dog eats 1 \(\frac{1}{2}\) cups  and a 60-pound dog eats 3 cups, the recommended amount of food each day. Now we need to calculate how much more food does the 60-pound dog eat in 1 week than the 20-pound dog, So Step 1 : First we find how much food the 20-pound dog eats in 1 week, let us take it as b = 7 X 1 \(\frac{1}{2}\) we write mixed fraction as fraction 1 \(\frac{1}{2}\) = ( 1 X 2 + 1 )by 2 = \(\frac{3}{2}\) now multiply by 7 = 7 X \(\frac{3}{2}\) we write into unit fraction 7 X 3 X \(\frac{1}{2}\) = 21 X \(\frac{1}{2}\) = \(\frac{21}{2}\) cups, Step 2: Now we find how much food the 60-pound dog eats in 1 week let us take it as c = 7 X 3 = 21 cups, Step 3: Subtract b from c to find how much more food the 60-pound dog eats in 1 week. c- b = 21 – \(\frac{21}{2}\) to subtract first we make both denominators same, we multiply and divide 21 by 2 making as (21 x 2) by 2 = \(\frac{42}{2}\) –\(\frac{21}{2}\) as denominators are same we subtarct numerators 42-21=21 will be numerator and denominator is same 2 =\(\frac{21}{2}\) as numerator is greater than denominator and 2 goes 10 times, 2 X 10 is 20 , 10 will be our whole number and  (21 −20 ) we have 1 left over. 1 will be our numerator and our denominator will be the same 2 = 10 \(\frac{1}{2}\) . Therefore the 60-pound dog eats 10 \(\frac{1}{2}\) more cups of food than the 20-pound dog in 1 week.

Answer: To feed the food for both dogs for 1 week we require \(\frac{182}{4}\) cups or \(\frac{91}{2}\) or 45 \(\frac{1}{2}\) cups.

Explanation: From the table 40-pound dog eats 2 \(\frac{1}{4}\) and 100-pound dog eats 4 \(\frac{1}{4}\) the recommended amounts of dog food each day, to find how much food is needed to feed both dogs for 1 week first we calculate for 40-pound dog= 7 X 2 \(\frac{1}{4}\) we write the mixed fraction as a fraction 2 x 4 + 1 by 4 = \(\frac{9}{4}\) and multiply by 7 , 7 X \(\frac{9}{4}\) now we write in unit fraction 7 X 9 X \(\frac{1}{4}\) = 63 X \(\frac{1}{4}\) = \(\frac{63}{4}\) now we calculate for 100-pound dog eats 4 \(\frac{1}{4}\) in 1 week is 7 X 4 \(\frac{1}{4}\) we write the mixed fraction as a fraction ( 4 X 4 + 1 )by 4 = \(\frac{17}{4}\) and multiply by 7 , 7 X \(\frac{17}{4}\) now we write in unit fraction 7 X 17 X \(\frac{1}{4}\) = 119 X \(\frac{1}{4}\) = \(\frac{119}{4}\). Now to feed the food for both dogs for 1 week is \(\frac{63}{4}\) + \(\frac{119}{4}\) as denominators are same we add numerators we get 63 + 119 = 182 as numerator, \(\frac{182}{4}\) further can be simplifed as 182 and 4 goes by 2 we write as \(\frac{91}{2}\) numerator is greater we can write in mixed fraction as 2 goes 45 times, 2 X 45 is 90 , 45 will be our whole number and (91 −90 ) we have 1 left over. 1 will be our numerator and our denominator will be the same 2, making as 45 \(\frac{1}{2}\) cups.

Explanation: Given friend makes strawberry jam to fill \(\frac{1}{2}\) of a jar and makes 5 times as much raspberry jam as strawberry jam means he makes 5 X \(\frac{1}{2}\) of jars of raspberry jam now we write as fraction \(\frac{5}{2}\) as numerator is greater, we can write in mixed fraction as 2 goes 2 times, 2 X 2 is 4 , 2 will be our whole number and (5 −4 ) we have 1 left over. 1 will be our numerator and our denominator will be the same 2, making as 2 \(\frac{1}{2}\) ,here 2 is whole there fore full jars of raspberry jam is 2. So friend makes \(\frac{5}{2}\) jars= 2 \(\frac{1}{2}\) jars in that full jars are 2, So 2 jars of raspberry jam he makes.

Question 2. You buy a pair of jeans that originally cost $22. The pair of jeans is \(\frac{1}{2}\) off the original price. You pay with a $20 bill. How much change do you receive? Answer: I receive $9 as change. Explanation: The pair of jeans is \(\frac{1}{2}\) off the original price and original cost is $ 22, So the cost to be paid is $22 X \(\frac{1}{2}\) = \(\frac{22}{2}\) further this can be simplified as both can be divided by 2 numerator becomes 11 and denominator as 1 = 11, therefore the original cost for the pair of jeans is $ 11, Now i have paid $ 20bill, i will receive a change that is $20 – $11= $9, i receive $9 as a change.

Explanation: Given the observation deck of the Space Needle in Seattle is 520 feet above ground and the tip of the Space Needle is 85 feet above the observation deck means the height of the entire Space Needle is 520 + 85 = 605 feet above the ground to tip, now an artist makes a replica that is \(\frac{1}{100}\) the height of the entire Space Needle. So the height of the replica is 605 X \(\frac{1}{100}\) = \(\frac{605}{100}\) further can be simplifed as 605 and 100 goes by 5 we get numerator as 121 and denominator as 20 = \(\frac{121}{20}\) as numerator is greater we can write in mixed fraction as 20 goes 6 times, 20 X 6 is 120 , 6 will be our whole number and  (121 −120 ) we have 1 left over. 1 will be our numerator and our denominator will be the same 20, making as 6 \(\frac{1}{20}\) feet tall.

Explanation: Given paper needed for Art Project orange is \(\frac{3}{8}\) and for 30 students it is 30 X \(\frac{3}{8}\) we write in unit fraction as 30 X 3 X \(\frac{1}{8}\) = 90 X \(\frac{1}{8}\) = \(\frac{90}{8}\) as numerator is greater we can write in mixed fraction as 8 goes 11 times, 8 X 11 is 88, 11 will be our whole number and  (90 −88 ) we have 2 left over. 2 will be our numerator and our denominator will be the same 8, making as 11 \(\frac{2}{8}\) sheets of orange is required, given black paper needed for Art Project is \(\frac{5}{8}\) and for 30 students it is 30 X \(\frac{5}{8}\) we write in unit fraction as 30 X 5 X \(\frac{1}{8}\) = 150 X \(\frac{1}{8}\) = \(\frac{150}{8}\) as numerator is greater we can write in mixed fraction as 8 goes 18 times, 8 X 18 is 144, 18 will be our whole number and  (150 −144 ) we have 6 left over. 6 will be our numerator and our denominator will be the same 8, making as 18 \(\frac{6}{8}\) black sheets are required. Therefore for 30 students to complete the Art Project orange sheets required are 11 \(\frac{2}{8}\) and 18 \(\frac{6}{8}\) black sheets of paper is required.

Question 5. Writing Write and solve a two-step word problem with mixed numbers that can be solved using multiplication.

Answer: A two-step problem is a word problem that requires two operations to solve it. We take two mixed numbers and slove using multiplication. In step one we use multiplication and add and in another step we use addition and again use multiplication as explained below with an example,

Explanation: For example: Percy walks 2 \(\frac{5}{4}\) kilometers for 2 days and 1 \(\frac{1}{4}\) kilometers for 5 days , how many kilometers does Percy walks in 1 week? Given in first 2 days Percy walks 2 \(\frac{5}{4}\) kilometers so In first step we change mixed fraction to fraction by using multiplication and then add 2 X 4 + 5 by 4 = \(\frac{13}{4}\) kilometers for 1 day for 2 days it is 2 X \(\frac{13}{4}\) = 2 X 13 X \(\frac{1}{4}\) = 26 X \(\frac{1}{4}\) = \(\frac{26}{4}\) kilometers for 2 days , Now for rest of 5 days we calculate. Here 1 \(\frac{1}{4}\) we change the mixed fraction to fraction as ( 1 X 4 + 1 ) by 4 = \(\frac{5}{4}\) for 1 day, For 5 days it will be 5 X \(\frac{5}{4}\), 5 X 5 X \(\frac{1}{4}\) = 25 X \(\frac{1}{4}\) = \(\frac{25}{4}\) kilometers for rest 5 days , Now total number of kilometers Percy walks is  \(\frac{26}{4}\) + \(\frac{25}{4}\) as denominators are same we add numerators we get 26 + 25 as 51 and denominator being same as 4 = \(\frac{51}{4}\) kilometers Percy walks in 1 week and further can be simplified as numerator is greater we can write in mixed fraction as 4 goes 12 times, 4 X 12 is 48, 12 will be our whole number and  (51 −48 ) we have 3 left over. 3 will be our numerator and our denominator will be the same 4, making as 12 \(\frac{3}{4}\) kilometers Percy walks in 1 week.

Explanation: In order to compare the fractions we must first turn their different denominators into the same denominators to make denominators same we multiply \(\frac{4}{5}\) with both numerator and denominator by 2 the value will not change = \(\frac{4}{5}\)  X \(\frac{2}{2}\)= \(\frac{8}{10}\) now as denominators are same 10 we can compare only numerators \(\frac{9}{10}\) with \(\frac{8}{10}\) as 9 is greater than 8, So \(\frac{9}{10}\) is greater than > \(\frac{8}{10}\) or \(\frac{9}{10}\) is greater than > \(\frac{4}{5}\)

Now convert each one of \(\frac{3}{8}\) and \(\frac{5}{6}\) into an equivalent fraction having  2 4 as denominator, \(\frac{3}{8}\) = \(\frac{3}{8}\) X \(\frac{3}{3}\) = \(\frac{9}{24}\) and \(\frac{5}{6}\) = \(\frac{5}{6}\) X \(\frac{4}{4}\) = \(\frac{20}{24}\) now as denominators are same 24 we can compare only numerators \(\frac{9}{24}\) with \(\frac{20}{24}\) as 9 is less than 20, so \(\frac{9}{24}\) < is less than \(\frac{20}{24}\) or \(\frac{3}{8}\) < is less than \(\frac{5}{6}\)

b. To calculate how much more water is in the Purple jar than the Green jar we subtract the quantity of purple jar minus quantity in green jar as \(\frac{15}{4}\) cups – \(\frac{3}{4}\) cup, as denominators are same for the both we will sustract numerators 15-3 = 12 and denominator remains the same as 4, \(\frac{12}{4}\) as both numerator and denominator can be divided by 4 we get numerator as 3 and denominator 1 so Purple Jar is 3 cups more than the Green jar.

c. Total number of cups of water are used in all is adding Green, Blue, Orange, Red, Purple – \(\frac{3}{4}\) + \(\frac{6}{4}\) + \(\frac{9}{4}\)  + \(\frac{12}{4}\) + \(\frac{15}{4}\) as denominators are same we add numerators as 3 + 6 + 9 + 12 + 15 = 45 and denominator being same as 4 = \(\frac{45}{4}\) as numerator is greater we can also write in mixed fraction as 4 goes 11 , 4  X 11 is 44, 11 will be our whole number and  (45 −44 ) we have 1 left over. 1 will be our numerator and our denominator will be the same 4, making as 11 \(\frac{1}{4}\), therfore total number of cups of water used in all is \(\frac{45}{4}\) cups or 11 \(\frac{1}{4}\) cups.

d. We got Purple jar contains \(\frac{15}{4}\) cups to this we will  add another \(\frac{3}{4}\) cup of water = \(\frac{15}{4}\) + \(\frac{3}{4}\) as denominator is same we add numerators as 15 + 3 = 18 and denominator is same 4 so now the Purple jar contains \(\frac{18}{4}\) cups , to find the whole value we write in mixed fraction as 4 goes , 4 X 4 is 16, 4 will be our whole number and  (18 −16 ) we have 2 left over. 2 will be our numerator and our denominator will be the same 4, making as 4 \(\frac{2}{4}\) = 4 + \(\frac{2}{4}\) and given that each jar can hold 4 cups of water but as we are getting more than 4 it will overflow, so it is not possible to add another \(\frac{3}{4}\) cup of water to the purple jar. e. Jars more than half full means each jars should be above 2 cups means to check we need denominator to be same that is 4 so we multiply denominator and numerator by 4, we get \(\frac{8}{4}\) to be more than half full means each jars numerators should be more than 8 and all jars denominators being same as 4, now we check with green jar, \(\frac{8}{4}\)  and \(\frac{3}{4}\) as 3 is not greater than 8 so not green jar, blue jar we have \(\frac{6}{4}\) cups here also numerator 6 is not more than 8 so even we will not conside blue jar, orange jar \(\frac{9}{4}\) now here numerator 8 is less than 9 so we will consider orange jar, Now red jar as red jar has 3 cups which is above 2 cups we consider red jar and we have purple jar as \(\frac{15}{4}\) here numerator 15 is more than numerator 8 so we consider purple jar, therefore jars with more than half full are Orange, Red and Purple.

f. Now we will add another \(\frac{3}{4}\) cup of water to the green jar means \(\frac{3}{4}\) + \(\frac{3}{4}\) as denominators as same we add numerators we get \(\frac{6}{4}\) which is equal to blue color jar, we know the jar that has the least amount of water makes the lowest sound before it was the green jar which had least amount of water, so it had lowest sound, so now after adding \(\frac{3}{4}\) cup of water to green jar it has become equal to blue jar so now both green and blue jar has least amount of water so they both make the lowest sound.

Three In a Row: Fraction Multiplication

Directions:

• Players take turns.
• On your turn, spin both spinners.
• Multiply the whole number and the fraction or mixed number. Cover the product.
• If the product is already covered, you lose your turn.
• The first player to get three in a row wins!

9.1 Understand Multiples of Unit Fractions

Explanation: We write \(\frac{3}{10}\) in addition equation as \(\frac{3}{100}\)= \(\frac{1}{10}\) + \(\frac{1}{10}\) + \(\frac{1}{10}\) and multiplication as unit fraction \(\frac{3}{10}\) represents 3 parts that are \(\frac{1}{10}\) of the whole, \(\frac{3}{10}\) = 3 X \(\frac{1}{10}\)

Explanation: We write \(\frac{4}{8}\) in addition equation as \(\frac{4}{8}\)= \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) + \(\frac{1}{8}\) and multiplication as unit fraction \(\frac{4}{8}\) represents 4 parts that are \(\frac{1}{8}\) of the whole, \(\frac{4}{8}\) = 4 X \(\frac{1}{8}\)

Question 3. \(\frac{7}{2}\) Answer: \(\frac{7}{2}\) = \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) = 7 X \(\frac{1}{2}\).

Explanation: we write \(\frac{7}{2}\) in addition equation as  \(\frac{7}{2}\) = \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) = 7 X \(\frac{1}{2}\) and multiplication as unit fraction \(\frac{7}{2}\) represents 7 parts that are \(\frac{1}{2}\) of the whole, \(\frac{7}{2}\) = 7 X \(\frac{1}{2}\).

Question 4. \(\frac{56}{100}\) Answer: \(\frac{56}{100}\)  = \(\frac{14}{25}\) = \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) = 14 X \(\frac{1}{25}\)

Explanation: \(\frac{56}{100}\) first we simplify as both numerator and denominator can be divided by 4 we get numerator as 14 and denominatior as 25 =  \(\frac{14}{25}\) now we write \(\frac{14}{25}\) in addition equation as \(\frac{14}{25}\) = \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) + \(\frac{1}{25}\) = 14 X \(\frac{1}{25}\) and multiplication as unit fraction \(\frac{14}{25}\) represents 14 parts that are \(\frac{1}{25}\) of the whole, \(\frac{14}{25}\) = 14 X \(\frac{1}{25}\) .

Explanation: Given unit fraction is one sixth of a whole is \(\frac{1}{6}\) and Newton’s fraction is 5 unit fractions means it is equal to \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) = 5 X \(\frac{1}{6}\) already the fraction is a multiple of a unit fraction = 5 X \(\frac{1}{6}\) , 5 is multiple and \(\frac{1}{6}\) is a unit fraction.

9.2 Understand Multiples of Fractions

Explanation: First we write \(\frac{2}{4}\)  as  \(\frac{1}{4}\) + \(\frac{1}{4}\) = 2 X \(\frac{1}{4}\) then multiply it by 2 which is  2 X 2 X \(\frac{1}{4}\) = 4 X \(\frac{1}{4}\) = \(\frac{4}{4}\) as numerator and denominator are same both get cancelled so we get 1 , \(\frac{4}{4}\) = 1.

Explanantion: As \(\frac{9}{12}\) can be simplified as both 9,12 can go with 3 leaving numerator as 3 and denominator as 4= \(\frac{3}{4}\), First we write \(\frac{3}{4}\)  as  \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) = 3 X \(\frac{1}{4}\) then multiply it by 3 which is  3 X 3 X \(\frac{1}{4}\) = 9 X \(\frac{1}{4}\) is multiple of unit fraction, The product is \(\frac{9}{4}\) .

Explanation: First we write \(\frac{3}{5}\)  as  \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\) = 3 X \(\frac{1}{5}\) then multiply it by 4 which is 4 X 3 X \(\frac{1}{5}\) = 12 X \(\frac{1}{5}\) = \(\frac{12}{5}\).

Explanantion: As \(\frac{8}{10}\) can be simplified as both 8,10 can go with 2, leaving numerator as 4 and denominator as 5 = \(\frac{4}{5}\), First we write \(\frac{4}{5}\)  as \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\) + \(\frac{1}{5}\) = 4 X \(\frac{1}{5}\) then multiply it by 7 which is  7 X 4 X \(\frac{1}{5}\) = 28 X \(\frac{1}{5}\) is multiple of unit fraction, The product is \(\frac{28}{5}\).

Explanation: As \(\frac{6}{3}\) can be simplified as both can be divided by 2, leaving numerator as 2 and denominator as 1 ,  \(\frac{6}{3}\) = 2  , now it is 8 X 2 = 16 as we got \(\frac{6}{3}\) =2, so we get results as whole not in fraction.

Explanation: As \(\frac{30}{8}\) can be simplified as both can be divided by 2 , we get numerator as 15 and denominator as 4, \(\frac{30}{8}\) = \(\frac{15}{4}\) = \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) + \(\frac{1}{4}\) = 15 X \(\frac{1}{4}\) now multiply with 10 X 15 X \(\frac{1}{4}\) = 150 X \(\frac{1}{4}\) = \(\frac{150}{4}\)  this can be further simplified as 150 and 4 both can be further divided by 2 making numerator as 75 and denominator as 2 = \(\frac{75}{2}\) .

9.3 Multiply Whole Numbers and Fractions

Explanation : First we write \(\frac{5}{8}\) in unit fraction as 5 X \(\frac{1}{8}\) now we multiply with whole 4 X 5 X \(\frac{1}{8}\) = \(\frac{20}{8}\) further can be simplifed as 20,8 both can be divided by 4 we get numerator as 5 and denominator 2, therefore 4 X \(\frac{5}{8}\) = \(\frac{5}{2}\) .

Explanantion: We can simplify \(\frac{9}{6}\) as both goes by 3 giving numerator as 3 and denominator as 2 so we get \(\frac{9}{6}\) = \(\frac{3}{2}\) now we write in unit fraction as 3 X \(\frac{1}{2}\) now we multiply by 3, 3 X 3 X \(\frac{1}{2}\) = 9 X \(\frac{1}{2}\) = \(\frac{9}{2}\).

Explanation: first we write \(\frac{7}{12}\) as unit fraction = 7 X \(\frac{1}{12}\), now we multiply with 5 as 5 X 7 X \(\frac{1}{12}\) = 35 X \(\frac{1}{12}\) = \(\frac{35}{12}\) .

Explanation: We can simplify \(\frac{30}{100}\) as both goes by 10 giving numerator as 3 and denominator as 10 so we get \(\frac{3}{10}\)  now we write in unit fraction as 3 X \(\frac{1}{10}\) and multiply by 7 , 7 X 3 X \(\frac{1}{10}\)  = 21 X \(\frac{1}{10}\) =  \(\frac{21}{10}\) .

Explanantion: We can simplify \(\frac{8}{4}\) as both goes by 4 giving numerator as 2 and denominator as 1 so we get 2 as whole, now we multiply with 9 we get 2 X 9 = 18.

9.4 Multiply Whole Numbers and Mixed Numbers

First we write the mixed number as a fraction then multiply, so 2 \(\frac{5}{8}\) = ( 2 X 8 + 5 ) by 8 = \(\frac{21}{8}\) now we multiply by 4 we get 4 X \(\frac{21}{8}\) = 4 X 21 X \(\frac{1}{8}\) = 84 X \(\frac{1}{8}\) =\(\frac{84}{8}\) further can be simplified as both 84,8 goes by 4 we get numerator as 21 and denominator as 2 =\(\frac{21}{2}\) as numerator is greater than denominator we can write in mixed fraction as 2 goes in 10 times. 10 will be our whole number 2 X 10 is 20 and we have 1 left over (21 −20 ), 1 will be our numerator and our denominator will stay the same 2, \(\frac{21}{2}\) = 10 \(\frac{1}{2}\).

9.5 Problem Solving: Fraction Operations

Question 24. Modeling Real Life Your friend rides her bike \(\frac{2}{6}\) mile to your house and \(\frac{2}{6}\) mile back home 4 times in 1 week. How far does she bike in all? Answer: My friend rides her bike by \(\frac{8}{3}\) miles or 2 \(\frac{2}{3}\) miles.

Explanation: Given friend rides her bike \(\frac{2}{6}\) mile to your house and \(\frac{2}{6}\) mile back home, first we calculate for 1 time, so first we will add \(\frac{2}{6}\) + \(\frac{2}{6}\) as both have same denominators we will add numerators as ( 2 + 2 = 4 ). So we get \(\frac{4}{6}\) miles for 1 time , now she rides 4 times means 4 X \(\frac{4}{6}\) = \(\frac{16}{6}\) this can be further simplified as 16,6 both goes by 2 we get numerator as 8 and denominator as 3 = \(\frac{8}{3}\) as numerator is greater than denominator we can write in mixed fraction as 3 goes in 2 times. 2 will be our whole number 3  X 2 is 6 and we have 2 left over (8 −6 ), 2 will be our numerator and our denominator will stay the same 3, we get mixed fraction as 2 \(\frac{2}{3}\) miles.

Conclusion:

Access the links given in the Big Ideas Math Book 4th Grade Answer Key Chapter 9 Multiply Whole Numbers and Fractions and practice well for the exams. We have solved all the questions in an easy and simple manner. We hope the information provided in this article has brought a smile to your face. Share Big Ideas Math Grade 4 Answer Key Chapter 9 Multiply Whole Numbers and Fractions.

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