TEST ON RATIO AND PROPORTION

Problem 1 :

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

Solution :

Let 3x, 5x and 7x be the ages of three boys

Average age of 3 boys = 25

Sum of three ages/3 = 25

(3x + 5x + 7x)/3 = 25

15x / 3 = 25

5x = 25

x = 25/5

x = 5

3x = 3(5) ==> 15

5x = 5(5) ==> 25

7x = 7(5) ==> 35

So, the ages of three boys are 15, 25 and 35.

Problem 2 :

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

Solution :

The angles be 2x, 7x and 11x.

Sum of angles of triangle = 180

2x + 7x + 11x = 180

20x = 180

x = 180/20

x = 9

2(9) ==> 18

7(9) ==> 63

11x = 11(9) ==> 99

So, the interior angles are 18, 63 and 99.

Problem 3 :

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

Solution :

Let x be the third number.

The first number = 20% more than x

= 120% of x

= 1.2x

Second number = 50% more than x

= 150% of x

= 1.5x

Ratio of two numbers = 1.2 x : 1.5 x

= 1.2/1.5

= 12/15

= 4/5

So, the required ratio is 4 : 5.

Problem 4 :

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 :

Share of A : Share of B : Share of C = 1/2 : 2/3 : 3/4

x/2 + 2x/3 + 3x/4 = 782

(6x + 8x + 9x)/12 = 782

23x / 12 = 782

23x = 782(12)

23x = 9384

x = 9384/23

= 408

A's share = 1/2 of 408

= 204

= 247

Problem 5 :

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 :

Let 3x, 7x and 12x be the shares of P, Q and R.

7x - 3x = 2400

4x = 2400

x = 2400/4

x = 600

Difference between Q and R = 12x - 7x

= 5x

= 5(600)

= 3000

So, the difference between the amounts Q and R is $3000

Problem 6 :

Carter's SUV requires 8 gallons of gasoline to travel 148 miles. How many gasoline, to the nearest gallon, will he need for 500 mile trip ?

Solution :

To cover the distance of 148 miles, we need 8 gallons of gasoline. To cover the distance of 500 miles trip, we need to find the quantity of gasoline required.

Let x be the required amount of gasoline.

148 : 8 = 500 : x

148/8 = 500/x

148x = 500(8)

x = 500(8) / 148

x = 27.02

So, 27 gallons of gasoline is needed.

Problem 7 :

If 5x : 3 = (x + 14) : 2, what is the value of x ?

Solution :

5x : 3 = (x + 14) : 2

5x/3 = (x + 14)/2

2(5x) = 3(x + 14)

10x = 3x + 42

10x - 3x = 42

7x = 42

x = 42/7

x = 6

So, the value of x is 6.

Problem 8 :

If 15 people can repair a road of length 150 meters, at the same rate, how many people are needed to repair a road of length 420 meters.

Solution :

Let x be the required number of people.

15 : 150 = x : 420

15/150 = x/420

15(420) = 150x

x = 420(15)/150

x = 42

So, the required number of people is 42.

Problem 9 :

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

Solution :

Weight of John = 56.7 kg

Let x be the new weight.

7 : 6 = 56.7 : x

7/6 = 56.7/x

7x = 56.7(6)

x = 56.7(6) / 7

x = 48.6

So, the new weight is 48.6 kg.

Problem 10 :

If a : b = c : d = 2.5 : 1.5, what are the values of ad : bc and a + c  :  b + d ?

Solution : 

a : b = c : d = 2.5 : 1.5

a/b = c/d = 2.5/1.5

a = 2.5 k, b = 1.5 k, c = 2.5 m, d = 1.5 m

ad : bd = {2.5k)(1.5m) : (1.5k)(2.5m)

ad : bd = 3.75 km : 3.75 km

ad : bd = 1 : 1

a + c : b + d = (2.5k + 2.5m) : (1.5k + 1.5m)

= 2.5(k + m) : 1.5(k + m)

= 2.5 : 1.5

= 25/15

a + c : b + d = 5 : 3