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**Problem 1 :**

The average age of three boys is 25 years and their ages are in the proportion 3:5:7. The age of the youngest boy is

**Solution :**

From the ratio 3 : 5 : 7, the ages of three boys are 3x, 5x and 7x.

Average age of three boys = 25

(3x+5x+7x)/3 = 25 ----------> 15x = 75 -----------> x = 5

Age of the first boy = 3x = 3(5) = 15

Age of the first boy = 5x = 5(5) = 25

Age of the first boy = 7x = 7(5) = 105

**Hence the age of the youngest boy is 15 years**

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**Problem 2 :**

John weighs 56.7 kilograms. If he is going to reduce his weight in the ratio 7:6, find his new weight.

**Solution :**

Original weight of John = 56.7 kg (given)

He is going to reduce his weight in the ratio 7:6

His new weight = (6x56.7)/7 = 6x8.1 = 48.6 kg.

**Hence his new weight = 48.6 kg**

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**Problem 3 :**

The ratio of the no. of boys to the no. of girls in a school of 720 students is 3:5. If 18 new girls are admitted in the school, find how many new boys may be admitted so that the ratio of the no. of boys to the no. of girls may change to 2:3.

**Solution :**

Sum of the terms in the given ratio = 3+5 = 8

So, no. of boys in the school = 720x(3/8)= 270

No. of girls in the school = 720x(5/8)= 450

Let "x" be the no. of new boys admitted in the school.

No. of new girls admitted = 18 (given)

**After the above new admissions,
**

no. of boys in the school = 270+x

no. of girls in the school = 450+18 = 468

The ratio after the new admission is 2 : 3 (given)

So, (270+x) : 468 = 2 : 3

3(270+x) = 468x2 (using cross product rule in proportion)

810 + 3x = 936

3x = 126

x = 42

**Hence the no. of new boys admitted in the school is 42
**

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**Problem 4 :**

The monthly incomes of two persons are in the ratio 4:5 and their monthly expenditures are in the ratio 7:9. If each saves $50 per month, find the monthly income of the second person.

**Solution :**

**From the given ratio of incomes ( 4 : 5 ), **

Income of the 1st person = 4x

Income of the 2nd person = 5x

**(Expenditure = Income - Savings)
**

Then, expenditure of the 1st person = 4x - 50

Expenditure of the 2nd person = 5x - 50

Expenditure ratio = 7 : 9 (given)

So, (4x - 50) : (5x - 50) = 7 : 9

9(4x - 50) = 7(5x - 50) (using cross product rule in proportion)

36x - 450 = 35x - 350

x = 100

Then, income of the second person = 5x = 5(100) = 500.

**Hence, income of the second person is $500**

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**Problem 5 :**

The ratio of the prices of two houses was 16:23. Two years later when the price of the first has increased by 10% and that of the second by $477, the ratio of the prices becomes 11:20. Find the original price of the first house.

**Solution :**

From the given ratio 16:23,

original price of the 1st house = 16x

original price of the 2nd house = 23x

**After increment in prices,
**

price of the 1st house = 16x + 10% of 16x = 16x + 1.6x = 17.6x

price of the 2nd house = 23x+477

**After increment in prices, the ratio of prices becomes 11:20
**

Then we have, 17.6x : (23x + 477) = 11 : 20

20(17.6x) = 11(23x+477) (using cross product rule)

352x = 253x + 5247

99x = 5247

x = 53

Then, original price of the first house = 16x = 16(53) = 848

**Hence, original price of the first house is $848**

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**Problem 6 :**

Find in what ratio will the total wages of the workers of a factory be increased or decreased, if there be a reduction in the number of workers in the ratio 15:11 and an increment in their wages in the ratio 22:25.

**Solution :**

Let us assume,

x = No. of workers, y = Average wages per worker

Then, total wages = (no. of workers) x (wages per worker)

Total wages = xy or 1xy ------------ (1)

After reduction in workers in the ratio15 : 11,

no. of workers = 11x / 15

After increment in wages in the ratio 22 : 25,

wages per worker = 25y / 22

Now, the total wages = (11x / 15)(25y / 15) = 5xy / 6

Therefore, total wages after changes = 5xy / 6 ------------ (2)

**From (1) and (2) we get that, **

the total wages get decreased from xy to 5xy / 6.

So, decrement ratio = xy : 5xy/6 -----> 1 : 5/6 ------> 6 : 5

**Hence, the total wages will be decreased in the ratio 6 : 5**

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**Problem 7 :**

If the angles of a triangle are in the ratio 2:7:11, then find the angles.

**Solution :**

From the ratio 2 : 7 : 11,

the three angles are 2x, 7x, 11x

In any triangle, sum of the angles = 180

So, 2x + 7x + 11x = 180°

20x = 180 -------> x = 9

Then, the first angle = 2x = 2(9) = 18°

The second angle = 7x = 7(9) = 63°

The third angle = 11x = 11(9) 99°

**Hence the angles of the triangle are (18°, 63°, 99°)**

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**Problem 8 :**

The ratio of two numbers is 7:10. Their difference is 105. Find the numbers.

**Solution :**

From the ratio 7 : 10,

the numbers are 7x, 10x.

Their difference = 105

10x - 7x = 105 ------> 3x = 105 --------> x = 35

Then the first number = 7x = 7(35) = 245

The second number = 10x = 10(35) = 350

**Hence the numbers are 245 and 350. **

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**Problem 9 :**

A , B and C are three cities. The ratio of average temperature between A and B is 11:12 and that between A and C is 9:8. The ratio between the average temperature of B and C is

**Solution :**

From A : B = 11 : 12 and A : C = 9 : 8, we find A in common.

The values corresponding to A in both the ratios are different.

**First we have to make them to be same. **

Value corresponding to A in the 1st ratio = 11

Value corresponding to A in the 2nd ratio = 9

L.C.M of (11, 9 ) = 99

First ratio, A : B = 11 : 12 = (11x9) : (12x9) = 99 : 108

Second ratio, A : C = 9 : 8 = (9x11) : (8x11) = 99 : 88

Clearly,

A : B = 99 : 108 ----------- (1)

A : C = 99 : 88 ---------------(2)

The values corresponding to A in both the ratios are same.

From (1) and (2), we get B : C = 108 : 88

By simplification, we get B : C = 27 : 22

**Hence, the ratio between the average temperature of B and C is 27 : 22 **

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**Problem 10 :**

The ratio between the speeds of two trains is 7:8. If the second train runs 400 kms. in 5 hours, the speed of the first train is

**Solution :**

From the given ratio 7 : 8,

Speed of the first train = 7x

Speed of the second train = 8x ----------(1)

Second train runs 400 kms in 5 hours (given)

**[Hint : Speed = Distance / Time]**

So, speed of the second train = 400/5 = 80 kmph -------(2)

From (1) and (2), we get

8x = 80 -------> x = 10

So, speed of the first train = 7x = 7(10) = 70 kmph.

**Hence, the speed of the second train is 70 kmph.**

**More Quantitative Aptitude Topics**

**10. Problems on Boats and Streams **

**11. Problems on Ratio and Proportion**

**16. Compound Interest Problems**

**17. Permutation and Combination Problems**

**22. Word Problems on Simple Equations**

**23. Word Problems on Simultaneous Equations **

**24. Problems on Quadratic Equations**

**25. Word Problems on Quadratic equations**

**26. Pipes and Cisterns Shortcuts**

Apart from the problems explained above, if you want to know more about ratio and proportion, please click here.

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