RATIO AND PROPORTION WORD PROBLEMS

About "Ratio and proportion word problems"

On this web page, we are going to see ratio and proportion word problems.

Before we practice solving problems on ratio and proportion, first we have to know the shortcuts which are required.

Now, Let us look at some  on ratio and proportion word problems.

Ratio and proportion word problems

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)

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, then, find the speed of the first train.

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.

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Problem 11 :

Two numbers are respectively 20% and 50% are more than a third number, Find the the ratio of the two numbers.

Solution :

Let "x" be the third number.

Then,

the first number = (100+20)% of x  =  120% of x  =  1.2x

the first number = (100+50)% of x  =  150% of x  =  1.5x

First no. : second no. = 1.2x = 1.5x

1.2x : 1.5x---------------> 12x : 15x

Dividing by (3x), we get 4 : 5

Hence, the ratio of two numbers is 4:5

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Problem 12 :

The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively.In what ratio, the liquids in both the vessels be mixed to obtain a new mixture in vessel C consisting half milk and half water?

Solution :

Vessel A :

[ 4:3 ------> 4+3 = 7,       M----> 4/7,        W----> 3/7 ]

Let "x" be the quantity of mixture taken from vessel A to obtain a new mixture in vessel C.

Quantity of milk in "x" = (4/7)x = 4x/7

Quantity of water in "x" = (3/7)x = 3x/7

Vessel B :

[ 2:3 ------> 2+3 = 5,       M----> 2/5,        W----> 3/5 ]

Let "y" be the quantity of mixture taken from vessel B to obtain a new mixture in vessel C.

Quantity of milk in "y" = (2/5)y = 2y/5

Quantity of water in "y" = (3/5)y = 3y/5

Vessel A and B :

Quantity of milk from A and B = (4x/7) + (2y/5) = (20x+14y) / 35

Quantity of water from A and B = (3x/7) + (3y/5) = (15x+21y) / 35

According to the question, vessel C must consist half of the milk and half of the water.

That is, in vessel C, quantity of milk and water must be same.

There fore,

Quantity of milk from (A+B)  =  Quantity of water from (A+B)

(20x+14y) / 35 =  (15x+21y) / 35

20x + 14y  =  15x+21y

5x  =  7y

x/y  =  7/5

x : y  =  7 : 5

Hence, the required ratio is 7:5

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Problem 13 :

A vessel contains 20 liters of a mixture of milk and water in the ratio 3:2. From the vessel, 10 liters of the mixture is removed  and replaced with an equal quantity of pure milk. Find the ratio of milk and water in the final mixture obtained.

Solution :

[ 3 : 2 ------> 3+2 = 5,       M----> 3/5,        W----> 2/5 ]

In 20 liters of mixture,

no. of liters of milk = 20x3/5 = 12 liters

no. of liters of water = 20x2/5 = 8 liters

Now, 10 liters of mixture removed.

In this 10 liters of mixture, milk and water will be in the ratio 3:2.

No. of liters of milk in this 10 liters = 10x3/5 = 6 liters

No. of liters of water in this 10 liters = 10x2/5 = 4 liters

After removing 10 liters (1st time),

No. of liters of milk in the vessel = 12 - 6 = 6 liters

No. of liters of water in the vbessel  8 - 4 = 4 liters

Now,  we add 10 liters of pure milk in the vessel,

After adding 10 liters of pure milk in the vessel,

No. of liters of milk in the vessel = 6 + 10 = 16 liters

No. of liters of water in the vessel  = 4 + 0 = 4 liters

After removing 10 liters of mixture and adding 10 liters of pure milk,

the ratio of milk and water = 16 : 4 -----> 4 : 1

Hence, the required ratio is 4 : 1

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Problem 14 :

If \$782 is divided among three persons A, B and C in the ratio            1/2 : 2/3 : 3/4, then   find the share of A.

Solution :

Given ratio ---> 1/2 : 2/3 : 3/4

First let us convert the terms of the ratio into integers.

L.C.M of denominators (2, 3, 4) = 12

When we multiply each term of the ratio by 12, we get

12x1/2 : 12x2/3 : 12x3/4 ------> 6 : 8 : 9

From the ratio 6 : 8 : 9,

Share of A = 6x

Share of B = 8x

Share of C = 9x

We know that ( A + B + C ) = 782

6x + 8x + 9x = 782 --------> 23x = 782 ------> x = 34

Share of A = 6x = 6(34) = 204

Hence, the share of A = \$ 204.

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Problem 15 :

An amount of money is to be divided among P, Q and R in the ratio  3 : 7 : 12. The difference between the shares of P and Q is \$2400. What will be the difference between the shares of Q and R?

Solution :

From the given ratio 3 : 7 : 12,

Share of P = 3x

Share of B = 7x

Share of C = 12x

Difference between the shares of P and Q is \$ 2400

That is,

Q - P = 2400 -------> 7x - 3x = 2400 -------> 4x = 2400 -------> x = 600

R - Q = 12x - 7x  =  5x  =  5(600)  =  3000

Hence, the difference between the shares of Q and R is \$3000.

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We hope the the students would have understood the example problems explained "Ratio and proportion word problems"

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