SOLVING PROPORTIONS WITH VARIABLES

Solve each Proportion :

Problem 1 :

2/r = 12/18

Solution :

2/r = 12/18

By using cross multiplication, we get

2 × 18 = 12 × r

36 = 12r

Dividing both parts by 12, we get

36/12 = r

r = 3

Problem 2 :

50/20 = 5/h

Solution :

50/20 = 5/h

By using cross multiplication, we get

50 × h = 5 × 20

50h = 100

Dividing both parts by 50, we get

h = 100/50

h = 2

Problem 3 :

16/6 = 3d/9

Solution :

16/6 = 3d/9

By using cross multiplication, we get

16 × 9 = 3d × 6

144 = 18d

Dividing both parts by 18, we get

144/18 = d

 d = 8

Problem 4 :

45/27 = x/3

Solution :

45/27 = x/3

By using cross multiplication, we get

45 × 3 = x × 27

135 = 27x

Dividing both parts by 27, we get

135/27 = x

x = 5

Problem 5 :

v/84 = 2/21

Solution :

v/84 = 2/21

By using cross multiplication, we get

v × 21 = 2 × 84

21v = 168

Dividing both parts by 21, we get

v = 168/21

v = 8

Problem 6 :

b/8 = 8/64

Solution :

b/8 = 8/64

By using cross multiplication, we get

b × 64 = 8 × 8

64b = 64

Dividing both parts by 64, we get

b = 64/64

b = 1

Problem 7 :

13/3 = 26/c

Solution :

13/3 = 26/c

By using cross multiplication, we get

13 × c = 26 × 3

13c = 78

Dividing both parts by 13, we get

c = 78/13

c = 6

Problem 8 :

27/15 = 72/a

Solution :

27/15 = 72/a

By using cross multiplication, we get

27 × a = 72 × 15

27a = 1680

Dividing both parts by 27, we get

a = 1680/27

a = 40

Problem 9 :

15/n = 5/12

Solution :

15/n = 5/12

By using cross multiplication, we get

15 × 12 = 5 × n

180 = 5n

Dividing both parts by 5, we get

180/5 = n

n = 36

Problem 10 :

20/8 = 4k/32

Solution :

20/8 = 4k/32

By using cross multiplication, we get

20 × 32 = 4k × 8

640 = 32k

Dividing both parts by 32, we get

640/32 = k

k = 20

Solve, using a proportion.

Problem 11 :

If 8 cases of merchandise cost $60, what would a dozen cases cost?

Solution :

Cost of 8 merchandise = 60

1 dozen = 12

Let x be the cost of 12 cases.

8 : 60 = 12 : x

Writing into fraction

8/60 = 12/x

8x = 12(60)

x = 12(60)/8

x = 90

So, the reqired cost is $90.

Problem 12 :

Benito saves $6 out of every $20 he earns. He earned $90 one month. How much did he save?

Solution :

Let x be the amount he is saving for one month when he earns $90.

8 : 20 = x : 90

8/20 = x/90

8(90) = 20x

x = 8(90) / 20

x = 36

So, he will save $36.

Problem 13 :

A man who is 6 feet tall casts a shadow that is 11 feet long. At the same time, a tree casts a shadow that is 33 feet long. What is the height of the tree?

Solution :

Let x be the height of the tree.

6 : 11 = x : 33

6/11 = x/33

6(33) = 11x

x = 6(33) / 11

x = 6(3)

= 18 ft

So, the required height of the tree is 18 ft.

Problem 14 :

Mr. Savage used 3 gallons of paint to cover 1,350 square feet of wall space. At this rate, how much paint will be needed to cover 1,800 square feet?

Solution :

To cover 1350 square feet of space, he needs 3 gallons of paint.

Let x be the amount of paint needed to cover 1800 square feet.

3 : 1350 = x : 1800

3/1350 = x/1800

3(1800) = 1350x

x = 3(1800) / 1350

x = 4

So, 4 gallons of paint is needed to cover 1800 sqaure feet.

Problem 15 :

In his last game, the Rams’ quarterback completed 10 out of 18 passes he threw. At this rate how many passes will he attempt if he completes 15 passes?

Solution :

Let x be the number persons who will pass.

10/18 = x/15

10(15) = 18x

x = 150/18

x = 8.3

Approximately 8.

Problem 16 :

Serena paid a tax of $288 on a house assessed at $48,000. Using the same tax rate, find the tax on a house assessed at $59,000.

Solution :

Let x be the required amount.

288 : 48000 = x : 59000

288/48000 = x/59000

288(59000) = 48000x 

x = 288(59000)/48000

x = 354