Refer to Worksheet on Ratios while preparing for the exam. You can perform great by practicing various problems from Ratio Worksheets. Multiple models of questions along with the detailed solutions and explanations are given here. The Ratios Practice Worksheet is designed by the experts as per the latest syllabus.

So, students who wish to learn the ratio concepts can refer to these Word Problems on Ratios and score well in the examinations. All the concepts of ratios along with the basic concept of ratio, the ratio in the simplest form, conversion of ratios, comparison of ratios, and finding the equivalent ratios are given in this article.

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## Simplifying Ratios Worksheets

1. Express the following ratios in the simplest form.

(a) 60 : 90

(b) 500 : 200

(c) 3 : 15.3

(d) 4.2 : 14

(e) 16a : 24a

(f) 2 \(\frac { 1 }{ 6 } \) âˆ¶ 1 \(\frac { 1 }{ 3 } \)

(g) \(\frac { 2 }{ 8 } \) âˆ¶ \(\frac { 2 }{ 10 } \)

(h) \(\frac { 4 }{ 3 } \) âˆ¶ \(\frac { 10 }{ 7 } \) : 2

(i) 4 \(\frac { 2 }{ 5 } \) âˆ¶ 2 \(\frac { 1 }{ 2 } \) : 2 \(\frac { 2 }{ 5 } \)

(j) \(\frac { 4 }{ 5 } \) âˆ¶ \(\frac { 6 }{ 10 } \)

(k) 1 âˆ¶ \(\frac { 1 }{ 2 } \) âˆ¶ \(\frac { 1 }{ 3 } \)

(l) 8mÂ²n : 20mnÂ²

(m) a dozen to a score

(n) 18 months : 5 years

(o) 10 m : 1200 cm

## Solution:

(a) Given that 60: 90

List the factors of 60

The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.

List the factors of 90.

The factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.

Find the greatest common factor of 60 and 90, GCF(60, 90)

GCF of 60 and 90 is 30.

Divide 60 and 90 each by the GCF.

\(\frac { 60 }{ 30 } \) = 2

\(\frac { 90 }{ 30 } \) = 3

Use the whole number results to rewrite the ratio in the simplest form.

Therefore, the final answer is 2:3

(b) Given that 500 : 200

List the factors of 500

The factors of 500 are 1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, and 500.

List the factors of 200.

The factors of 200 are 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, and 200.

Find the greatest common factor of 500 and 200, GCF(500, 200)

GCF of 500 and 200 is 100.

Divide 500 and 200 each by the GCF.

\(\frac { 500 }{ 100 } \) = 5

\(\frac { 200 }{ 100 } \) = 2

Use the whole number results to rewrite the ratio in the simplest form.

Therefore, the final answer is 5: 2

(c) Given that 3: 15.3

Find the greatest common factor of 3 and 15.3, GCF(3, 15.3)

GCF of 3 and 15.3 is 3.

Divide 3 and 15.3 each by the GCF.

\(\frac { 3 }{ 3 } \) = 1

\(\frac { 15.3 }{ 3 } \) = 3.1

Use the whole number results to rewrite the ratio in the simplest form.

Therefore, the final answer is 1: 3.1

(d) Given that 4.2: 14

Divide 4.2 and 14 each by the 14.

\(\frac { 4.2 }{ 14 } \) = 0.3

\(\frac { 14 }{ 14 } \) = 1

Use the whole number results to rewrite the ratio in the simplest form.

Therefore, the final answer is 0.3: 1

(e) Given that 16a : 24a

Find the greatest common factor of 16a and 24a, GCF(16a, 24a)

GCF of 16a and 24a is 8a.

Divide 16a and 24a each by the GCF.

\(\frac { 16a }{ 8a } \) = 2

\(\frac { 24a }{ 8a } \) = 3

Use the whole number results to rewrite the ratio in the simplest form.

Therefore, the final answer is 2 : 3

(f) Given that 2 \(\frac { 1 }{ 6 } \) âˆ¶ 1 \(\frac { 1 }{ 3 } \)

Find the greatest common factor of 2 \(\frac { 1 }{ 6 } \) and 1 \(\frac { 1 }{ 3 } \), GCF(2 \(\frac { 1 }{ 6 } \), 1 \(\frac { 1 }{ 3 } \))

GCF of 2 \(\frac { 1 }{ 6 } \) and 1 \(\frac { 1 }{ 3 } \) is \(\frac { 1 }{ 6 } \).

Divide 2 \(\frac { 1 }{ 6 } \) and 1 \(\frac { 1 }{ 3 } \) each by the GCF.

2 \(\frac { 1 }{ 6 } \) : \(\frac { 1 }{ 6 } \) = 13

1 \(\frac { 1 }{ 3 } \) :Â \(\frac { 1 }{ 6 } \) = 8

Use the whole number results to rewrite the ratio in the simplest form.

Therefore, the final answer is 13: 8

(g) Given that \(\frac { 2 }{ 8 } \) âˆ¶ \(\frac { 2 }{ 10 } \)

\(\frac { 2 }{ 8 } \) âˆ¶ \(\frac { 2 }{ 10 } \) = \(\frac { 1 }{ 4 } \) âˆ¶ \(\frac { 1 }{ 5 } \)

\(\frac { 1 }{ 4 } \) âˆ¶ \(\frac { 1 }{ 5 } \) = 5 : 4

Use the whole number results to rewrite the ratio in the simplest form.

Therefore, the final answer is 5: 4

(h) Given that \(\frac { 4 }{ 3 } \) âˆ¶ \(\frac { 10 }{ 7 } \) : 2

\(\frac { 4 }{ 3 } \) âˆ¶ \(\frac { 10 }{ 7 } \) : 2

Divide the given numbers with 2.

\(\frac { 2 }{ 3 } \) âˆ¶ \(\frac { 2 }{ 7 } \) : 1

Use the whole number results to rewrite the ratio in the simplest form.

Therefore, the final answer is 14: 15: 21

(i) Given that 4 \(\frac { 2 }{ 5 } \) âˆ¶ 2 \(\frac { 1 }{ 2 } \) : 2 \(\frac { 2 }{ 5 } \)

Divide the given numbers with 2.

2 \(\frac { 1 }{ 5 } \) âˆ¶ 1 \(\frac { 1 }{ 4 } \) : 1 \(\frac { 1 }{ 5 } \)

Use the whole number results to rewrite the ratio in the simplest form.

Therefore, the final answer is 44: 25: 24

(j) Given that \(\frac { 4 }{ 5 } \) âˆ¶ \(\frac { 6 }{ 10 } \)

Divide the given numbers with 2.

\(\frac { 2 }{ 5 } \) âˆ¶ \(\frac { 3 }{ 10 } \)

\(\frac { 20 }{ 15 } \)

Divide the above fraction with 5.

Therefore, the final answer is 4 : 3

(k) Given that 1 âˆ¶ \(\frac { 1 }{ 2 } \) âˆ¶ \(\frac { 1 }{ 3 } \)

1 âˆ¶ \(\frac { 1 }{ 2 } \) âˆ¶ \(\frac { 1 }{ 3 } \) = 6 : 3 : 2

Therefore, the final answer is 6 : 3 : 2

(l) Given that 8mÂ²n : 20mnÂ²

Find the greatest common factor of 8mÂ²n and 20mnÂ², GCF(8mÂ²n, 20mnÂ²)

GCF of 8mÂ²n and 20mnÂ² is 4mn.

Divide 8mÂ²n and 20mnÂ² each by the GCF.

8mÂ²n : 4mn = 2m

20mnÂ² :Â 4mn = 5n

Use the whole number results to rewrite the ratio in the simplest form.

Therefore, the final answer is 2m: 5n

(m) a dozen to a score

a dozen = 12

a score = 20

12: 20

Find the greatest common factor of 12 and 20, GCF(12, 20)

GCF of 12 and 20 is 4.

Divide 12 and 20 each by the GCF.

12 : 4 = 3

20 :Â 4 = 5

Use the whole number results to rewrite the ratio in the simplest form.

The answer for a dozen to a score in a simple ratio is 3: 5

(n) 18 months: 5 years

1 year = 12 months

5 years = 5 * 12 = 60

18 months: 60 months

Find the greatest common factor of 18 months and 60 months, GCF(18 months, 60 months)

GCF of 18 months and 60 months is 6 months.

Divide 18 months and 60 months each by the GCF.

18 months : 6 months = 3

60 months :Â 6 months = 10

Use the whole number results to rewrite the ratio in the simplest form.

The answer for a dozen to a score in a simple ratio is 3: 10

(o) 10 m: 1200 cm

1 m = 100 cm

10 m = 10 * 100 = 1000

1000 cm : 1200 cm

Find the greatest common factor of 1000 cm and 1200 cm, GCF(1000 cm, 1200 cm)

GCF of 1000 cm and 1200 cm is 200 cm.

Divide 1000 cm and 1200 cm each by the GCF.

1000 cm : 200 cm = 5

1200 cm :Â 200 cm = 6

Use the whole number results to rewrite the ratio in the simplest form.

The answer for a dozen to a score in a simple ratio is 5: 6

2. In a school library, the ratio of English books to that of Hindi books is 6: 7. If there are 156 English books, find the number of Hindi books in the library?

## Solution:

Given that in a school library, the ratio of English books to that of Hindi books is 6: 7.

English : Hindi = 6 : 7

Substitute 156 in the place of English

156 : Hindi = 6 : 7

We know that product of extreme terms = product of mean terms

156 * 7 = Hindi * 6

Hindi = \(\frac { 1092 }{ 6 } \) = 182

Therefore, the number of Hindi books in the library are 182.

3. The ratio of the number of male and female teachers in a school is 4: 9. If the number of male teachers is 72, find the number of female teachers?

## Solution:

Given that the ratio of the number of male and female teachers in a school is 4: 9.

male teachers: female teachers = 4: 9

Let the female teachers = x

Substitute 72 in the place of male teachers

72 : female teachers = 4 : 9

We know that product of extreme terms = product of mean terms

72 * 9 = Female teachers * 4

Female teachers = \(\frac { 648 }{ 4 } \) = 162

Therefore, the number of Female teachers are 162.

4. An alloy contains zinc and copper in the ratio 14: 18. Find the weight of copper if it has 63 kg of zinc?

## Solution:

Given that an alloy contains zinc and copper in the ratio 14 : 18.

Let the weight of zinc be 14x

Let the weight of copper = 18x

As per the given information, 14x = 63 kg

x = \(\frac { 63 }{ 14 } \) = 4.5 kg

Find the weight of the copper by multiplying 4.5 kg with 18.

Substitute the x value in 18x.

The weight of the copper = 18 (4.5 kg) = 81 kg

Therefore, the weight of the copper is 81 kg.

5. Compare the following ratios.

(a) 4: 6 and 8: 10

(b) 22: 38 and 38: 42

(c ) 1 âˆ¶ \(\frac { 2 }{ 3 } \) and \(\frac { 2 }{ 3 } \) âˆ¶ \(\frac { 1 }{ 2 } \)

(d) \(\frac { 4 }{ 3 } \) : \(\frac { 5 }{ 9 } \) and \(\frac { 2 }{ 7 } \) âˆ¶ \(\frac { 3 }{ 5 } \)

## Solution:

(a) Given that 4 : 6 and 8 : 10.

4 : 6 can write as \(\frac { 4 }{ 6 } \)

8 : 10 can write as \(\frac { 8 }{ 10 } \)

Firstly, find the LCM of the 6Â and 10.

The LCM of the 6 and 10 is 30.

\(\frac { 4 Ã— 5 }{ 6 Ã— 5 } \) = \(\frac { 20 }{ 30 } \)

\(\frac { 8 Ã— 3 }{ 10 Ã— 3 } \) = \(\frac { 24 }{ 30 } \)

Now, we can compare \(\frac { 20 }{ 30 } \) and \(\frac { 24 }{ 30 } \)

20 is less than 24.

\(\frac { 20 }{ 30 } \) < \(\frac { 24 }{ 30 } \)

Therefore, 4: 6 < 8 : 10

(b) Given that 22 : 38 and 38 : 42.

22 : 38 can write as \(\frac { 22 }{ 38 } \)

38 : 42 can write as \(\frac { 38 }{ 42 } \)

Firstly, find the LCM of the 38 and 42.

The LCM of the 38 and 42 is 798.

\(\frac { 22 Ã— 21 }{ 38 Ã— 21 } \) = \(\frac { 462 }{ 798 } \)

\(\frac { 38 Ã— 19 }{ 42 Ã— 19 } \) = \(\frac { 722 }{ 798 } \)

Now, we can compare \(\frac { 462 }{ 798 } \) and \(\frac { 722 }{ 798 } \)

462 is less than 722.

\(\frac { 462 }{ 798 } \) < \(\frac { 722 }{ 798 } \)

Therefore, 22 : 38 < 38 : 42.

(c) Given that 1 âˆ¶ \(\frac { 2 }{ 3 } \) and \(\frac { 2 }{ 3 } \) âˆ¶ \(\frac { 1 }{ 2 } \).

1 âˆ¶ \(\frac { 2 }{ 3 } \) can write as \(\frac { 3 }{ 2 } \)

\(\frac { 2 }{ 3 } \) âˆ¶ \(\frac { 1 }{ 2 } \) can write as \(\frac { 4 }{ 3 } \)

Firstly, find the LCM of the 2 and 3.

The LCM of the 2 and 3 is 6.

\(\frac { 3 Ã— 3 }{ 2 Ã— 3 } \) = \(\frac { 9 }{ 6 } \)

\(\frac { 4 Ã— 2 }{ 3 Ã— 2 } \) = \(\frac { 8 }{ 6 } \)

Now, we can compare \(\frac { 9 }{ 6 } \) and \(\frac { 8 }{ 6 } \)

9 is greater than 8.

\(\frac { 3 }{ 2 } \) > \(\frac { 4 }{ 3 } \)

Therefore, 1 âˆ¶ \(\frac { 2 }{ 3 } \) > \(\frac { 2 }{ 3 } \) âˆ¶ \(\frac { 1 }{ 2 } \).

(d) Given that \(\frac { 4 }{ 3 } \) : \(\frac { 5 }{ 9 } \) and \(\frac { 2 }{ 7 } \) âˆ¶ \(\frac { 3 }{ 5 } \)

\(\frac { 4 }{ 3 } \) : \(\frac { 5 }{ 9 } \) can write as \(\frac { 36 }{ 15 } \)

\(\frac { 2 }{ 7 } \) âˆ¶ \(\frac { 3 }{ 5 } \) can write as \(\frac { 10 }{ 21 } \)

Firstly, find the LCM of the 15 and 3.

The LCM of the 2 and 3 is 6.

\(\frac { 3 Ã— 3 }{ 2 Ã— 3 } \) = \(\frac { 9 }{ 6 } \)

\(\frac { 4 Ã— 2 }{ 3 Ã— 2 } \) = \(\frac { 8 }{ 6 } \)

Now, we can compare \(\frac { 9 }{ 6 } \) and \(\frac { 8 }{ 6 } \)

9 is greater than 8.

\(\frac { 3 }{ 2 } \) > \(\frac { 4 }{ 3 } \)

Therefore, 1 âˆ¶ \(\frac { 2 }{ 3 } \) > \(\frac { 2 }{ 3 } \) âˆ¶ \(\frac { 1 }{ 2 } \).

6. (a) If a : b = 12 : 10 and b : c = 20 : 18, find a : c.

(b) If x : y = \(\frac { 1 }{ 3 } \) âˆ¶ \(\frac { 1 }{ 4 } \) and y : z = \(\frac { 1 }{ 4 } \) âˆ¶ \(\frac { 1 }{ 5 } \), find x : z

(c) If p : q = 6 : 10 and q : r = 12 : 14 find p : q : r

(d) If a : b = 12 : 14 and b : c = 6 : 10, find a : b : c

## Solution:

(a) Given that a : b = 12 : 10 and b : c = 20 : 18.

12 : 10 can write as \(\frac { 12 }{ 10 } \)

20 : 18 can write as \(\frac { 20 }{ 18 } \)

To get a : c, multiply a/b and b/c

a/b Ã— b/c = a/c

\(\frac { 12 }{ 10 } \) Ã— \(\frac { 20 }{ 18 } \)

a/c = \(\frac { 4 }{ 3 } \)

Therefore, the final answer is a : c = \(\frac { 4 }{ 3 } \).

(b) Given that x : y = \(\frac { 1 }{ 3 } \) âˆ¶ \(\frac { 1 }{ 4 } \) and y : z = \(\frac { 1 }{ 4 } \) âˆ¶ \(\frac { 1 }{ 5 } \).

x : y = \(\frac { 1 }{ 3 } \) âˆ¶ \(\frac { 1 }{ 4 } \) can write as \(\frac { 4 }{ 3 } \)

y : z = \(\frac { 1 }{ 4 } \) âˆ¶ \(\frac { 1 }{ 5 } \) can write as \(\frac { 5 }{ 4 } \)

To get x : z, multiply x/y and y/z

x/y Ã— y/z = x/z

\(\frac { 4 }{ 3 } \) Ã— \(\frac { 5 }{ 4 } \)

x/z = \(\frac { 5 }{ 3 } \)

Therefore, the final answer is x : z = \(\frac { 5 }{ 3 } \).

(c) Given that p : q = 6 : 10 and q : r = 12 : 14.

p : q = 6 : 10 can write as \(\frac { 6 }{ 10 } \) : 1

q : r = 12 : 14 can write as 1 : \(\frac { 14 }{ 12 } \)

Therefore, p : q : r = \(\frac { 6 }{ 10 } \) : 1 : \(\frac { 14 }{ 12 } \)

Taking the L.C.M. of 10 and 12

The L.C.M. of 10 and 12 is 60.

= \(\frac { 6 }{ 10 } \) Ã— 60 : 1 Ã— 60 : \(\frac { 14 }{ 12 } \) Ã— 60

= 36 : 60 : 70

Therefore, the final answer is p: q: r = 36: 60: 70.

(d) Given that a : b = 12 : 14 and b : c = 6 : 10.

a : b = 12 : 14 can write as \(\frac { 12 }{ 14 } \) : 1

b : c = 6 : 10 can write as 1 : \(\frac { 10 }{ 6 } \)

Therefore, p : q : r = \(\frac { 12 }{ 14 } \) : 1 : \(\frac { 10 }{ 6 } \)

Taking the L.C.M. of 14 and 6

The L.C.M. of 14 and 6 is 42.

= \(\frac { 12 }{ 14 } \) Ã— 42 : 1 Ã— 42 : \(\frac { 10 }{ 6 } \) Ã— 42

= 36 : 42 : 70 = 18 : 21 : 35

Therefore, the final answer is a: b: c = 18: 21: 35.

7. Are the following ratios equivalent?

(a) 6: 10 and 30: 50

(b) 2: 8 and 4 : 6

(c) 16: 6 and 48: 60

(d) 5: 7 and 10: 14

## Solution:

(a) Given that 6: 10 and 30: 50.

From the given data, extreme terms are 6 and 50, mean terms are 10 and 30.

Find the Product of extreme terms and mean terms.

Product of extreme terms = 6 Ã— 50 = 300

Product of mean terms = 10 Ã— 30 = 300.

Compare the Product of extreme terms and the Product of mean terms.

If the product of means = product of extremes, the given ratios are equivalent.

300 = 300

Therefore, 6 : 10 and 30 : 50 ratios are equivalent.

(b) Given that 2: 8 and 4: 6.

From the given data, extreme terms are 2 and 6, mean terms are 8 and 4.

Find the Product of extreme terms and mean terms.

Product of extreme terms = 2 Ã— 6 = 12

Product of mean terms = 8 Ã— 4 = 32.

Compare the Product of extreme terms and the Product of mean terms.

If the product of means = product of extremes, the given ratios are equivalent.

12 is not equal to 32

Therefore, 2: 8 and 4: 6 ratios are not equivalent.

(c) Given that 16: 6 and 48: 60.

From the given data, extreme terms are 16 and 60, mean terms are 6 and 48.

Find the Product of extreme terms and mean terms.

Product of extreme terms = 16 Ã— 60 = 960

Product of mean terms = 6 Ã— 48 = 288.

Compare the Product of extreme terms and the Product of mean terms.

If the product of means = product of extremes, the given ratios are equivalent.

960 is not equal to 288.

Therefore, 16: 6 and 48: 60 ratios are not equivalent.

(d) Given that 5: 7 and 10: 14.

From the given data, extreme terms are 5 and 14, mean terms are 7 and 10.

Find the Product of extreme terms and mean terms.

Product of extreme terms = 5 Ã— 14 = 70

Product of mean terms = 7 Ã— 10 = 70.

Compare the Product of extreme terms and the Product of mean terms.

If the product of means = product of extremes, the given ratios are equivalent.

70 is equal to 70.

Therefore, 5: 7 and 10: 14 ratios are equivalent.

8. Two numbers are in the ratio 8 : 3. If their difference is 75, find the numbers.

## Solution:

Given that two numbers are in the ratio 8 : 3 and the difference between them is 75.

From the given data, let the numbers are 8a and 3a.

Now, subtract the 3a from 8a.

8a – 3a = 75

5a = 75

a = \(\frac { 75 }{ 5 } \)

a = 15.

Now, substitute the a value to find the numbers.

The first number = 8a = 8 Ã— 15 = 120

The second number = 3a = 3 Ã— 15 = 45

Therefore, the two numbers are 120 and 45.

9. Two numbers are in the ratio 6: 4 and their sum is 120. Find the numbers?

## Solution:

Given that two numbers are in the ratio 6 : 4 and their sum is 120.

From the given data, let the numbers are 6a and 4a.

Now, add 6a and 4a.

6a + 4a = 120

10a = 120

a = \(\frac { 120 }{ 10 } \)

a = 12.

Now, substitute the a value to find the numbers.

The first number = 6a = 6 Ã— 12 = 72

The second number = 4a = 4 Ã— 12 = 48

Therefore, the two numbers are 72 and 48.

10. The three angles of a scalene triangle are in the ratio 4: 6: 10. Find the measures of the angles.

## Solution:

Given that the three angles of a scalene triangle are in the ratio 4 : 6 : 10.

From the given data, let the numbers are 4a, 6a, and 10a.

We know that the sum of three angles of a triangle is 180Â°.

Now, add 4a, 6a, and 10a.

4a + 6a + 10a = 180Â°

20a = 180Â°

a = \(\frac { 180Â° }{ 20 } \)

a = 9Â°.

Now, substitute the a value to find the numbers.

The first angle = 4a = 4 Ã— 9Â° = 36Â°

The second angle = 6a = 6 Ã— 9Â° = 54Â°

Also, the third angle = 10a = 10 Ã— 9Â° = 90Â°

Therefore, the three angles are 36Â°, 54Â°, and 90Â°.

11. Divide $369 in the ratio 4: 5.

## Solution:

Given that two numbers are in the ratio 4 : 5.

From the given data, let the numbers are 4a and 5a.

Now, add 4a and 5a.

4a + 5a = 9a

Given 9a = $369

9a = $369

a = \(\frac { $369 }{ 9 } \)

a = $41.

Now, substitute the a value to find the numbers.

The first number = 4a = 4 Ã— $41 = $164

The second number = 5a = 5 Ã— $41= $205

Therefore, the two numbers are $164 and $205.

12. Divide 72 chocolates in the ratio of 4: 6: 8.

## Solution:

Given that two numbers are in the ratio 4 : 6 : 8.

From the given data, let the numbers are 4a, 6a, and 8a.

Now, add 4a, 6a, and 8a.

4a + 6a + 8a = 18a

Given 18a = 72

18a = 72

a = \(\frac { 72 }{ 18 } \)

a = 4.

Now, substitute the a value to find the numbers.

The first number = 4a = 4 Ã— 4 = 16

The second number = 6a = 6 Ã— 4 = 24

Also, the second number = 8a = 8 Ã— 4 = 32

Therefore, the three numbers are 16, 24, and 32.

13. A-line segment of length 40 cm is divided into three parts in the ratio 3: 2: 5. Find the length of each part?

## Solution:

Given that a line segment of length 40 cm is divided into three parts in the ratio 3 : 2 : 5.

From the given data, let the numbers are 3a, 2a, and 5a.

Now, add 3a, 2a, and 5a.

3a + 2a + 5a = 10a

Given 10a = 40 cm

10a = 40 cm

a = \(\frac { 40 }{ 10 } \)

a = 4.

Now, substitute the a value to find the numbers.

The first number = 3a = 3 Ã— 4 = 12

The second number = 2a = 2 Ã— 4 = 8

Also, the second number = 5a = 5 Ã— 4 = 20

Therefore, the three numbers are 12, 8, and 20.

14. A certain amount is divided into two parts in the ratio of 8: 10. If the first part is 240, find the total?

## Solution:

Given that a certain amount is divided into two parts in the ratio 8: 10. The first part is equal to 240.

From the given data, let the numbers are 8a and 10a.

Now, calculate a from 8a = 240.

a = \(\frac { 240 }{ 8 } \)

a = 30

Now, substitute the a value to find the second part.

The second part = 10a = 10 Ã— 30 = 300.

Now, add the first part and second part to find out the total.

Total = 240 + 300 = 540

Therefore, the total amount is 540.

15. Divide $688 into three parts such that the first part is 4/3 of the second and the ratio between the second and third is 8: 10.

## Solution:

Given that $688 divided into three parts such that the first part is 4/3 of the second and the ratio between the second and third is 8: 10

From the given data, let the common multiple of these ratio = a.

Then, the second and third ratio = 8a and 10a

Given that the first part is 4/3 of the second part

The first part is \(\frac { 4 }{ 3 } \) Ã— 8a = \(\frac { 32a }{ 3 } \)

Given \(\frac { 32a }{ 3 } \) + 8a + 10a = $688

Given \(\frac { 86a }{ 3 } \) = $688

86a = $2064

Now, calculate a from 86a = $2064.

a = \(\frac { $2064 }{ 86 } \)

a = 24

Now, substitute the a value to find the answer.

The first part = \(\frac { 32a }{ 3 } \) = \(\frac { 32 }{ 3 } \) Ã— 24 = 256.

The second part = 8a = 8 Ã— 24 = 192.

The third part = 10a = 10 Ã— 24 = 240.

Therefore, the final answer is 256, 192, 240.

16. Divide 216 bottles of juice between A and B in the ratio \(\frac { 2 }{ 5 } \) : \(\frac { 2 }{ 7 } \).

## Solution:

Given that 216 bottles of juice divided between A and B in the ratio \(\frac { 2 }{ 5 } \) : \(\frac { 2 }{ 7 } \)

From the given data, let the common multiple of these ratio = a.

Then, the first and second numbers are = \(\frac { 2a }{ 5 } \) and \(\frac { 2a }{ 7 } \)

Now, add \(\frac { 2a }{ 5 } \) and \(\frac { 2a }{ 7 } \)

\(\frac { 2a }{ 5 } \) + \(\frac { 2a }{ 7 } \) = \(\frac { 24a }{ 35 } \)

Now, calculate a from \(\frac { 24a }{ 35 } \) = 216.

a = \(\frac { 7560 }{ 24 } \)

a = 315

Now, substitute the a value to find the answer.

The first part = \(\frac { 2 }{ 5 } \) a = \(\frac { 2 }{ 5 } \) Ã— 315 = 126.

The second part = \(\frac { 2 }{ 7 } \) a = \(\frac { 2 }{ 7 } \) Ã— 315 = 90.

Therefore, the final answer is 126, and 90.

17. The sides of the quadrilateral are in the ratio 2: 4: 6: 8 and the perimeter is 80 cm. Find the length of each side?

## Solution:

Given that the sides of the quadrilateral are in the ratio 2 : 4 : 6 : 8 and the perimeter is 80 cm.

From the given data, let the common multiple of these ratio = a.

Add 2a, 4a, 6a, and 8a.

2a + 4a + 6a + 8a = 80.

20a = 80

a = 80/20 = 4

Now, substitute the a value to find the answer.

The first part = 2a = 2 Ã— 4 = 8.

The second part = 4a = 4 Ã— 4 = 16.

Also, the third part = 6a = 6 Ã— 4 = 24.

The fourth part = 8a = 8 Ã— 4 = 32.

Therefore, the length of each side is 8, 16, 24, and 32.