Ratio proportion and variation problems with solutions

This article explained some examples with a solution of ratio, proportion and variation chapter

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Ratio and Proportion Questions with Solutions | Quantitative Aptitude

Example-1: The ratio between two numbers is 5 : 6 and the sum of their squares is 244. Then find the numbers

Solution: Let the two numbers be 5a and 6a respectively.

⇒(5a)² + (6a)² = 244

⇒25a² + 36a² = 244

⇒61 a² = 244 ⇒ a² = 4 ⇒ a=2

So the numbers are 5 x 2, 6 x 2

i.e 10 & 12

Example-2: Find the numbers if the ratio between two numbers is 3 : 7. and their LCM is 210

Solution: Let the numbers be 3a and 7a

LCM is 3 × 7 × a = 210

21a = 210

a = 10

The numbers are 30 and 70.

Example-3: Find the fourth proportional to 5, 8, 20

Solution: Take the fourth proportion as ‘a’

Then 5 : 8 :: 20 : a

According to the property of proportions

Product of extremes = Product of means

5a = 160

a = 160/5 = 32

Example-4: Find the third proportional to 36 & 48

Solution: According to the property of proportions, the three quantities x, y & z are in continued proportion

i.e x : y :: y : z are in proportion then y² = xz

So take the third proportion is ‘a” then

36 : 48 :: 48 : a

a = 48 x 48 /36 = 64

Example-5 : If p : q = 5 : 9, q : r = 6 : 8, find p : q : r

Solution: Find the LCM of 9 & 6 is 18

Now the ratio p : q = 5 : 9 = 10 : 18 ( Multiplying with 2 (18/9=2))

Ratio q : r = 6 : 8 = 18 : 24 ( Multiplying with 2 (18/6=3))

Therefore ratio of p : q : r = 10 : 18 : 24

Example – 6 : If a/b = 3/4, then find the value of the expression ( 5a – 3b)/(7a – 2b).

Solution: Here assume the value as a = 3 and b = 4, then

( 5a – 3b)/(7a – 2b) = (15 – 12 ) / (21 – 8) = 3/13

Example – 7 : If 4a = 5b = 3c then find value of a : b : c

Solution: 4a = 5b = 3c then $\frac{a}{4} = \frac{b}{5} = \frac{c}{3}$

Now a : b : c = 1/4 : 1/5 : 1/3 = 15/60 : 12/60 : 20/60

a : b : c = 15 : 12 : 20

Example -8 : Find the mean proportion of 27 and 3

Solution: We know that

a : b :: b : c are in proportion

b² = ac

Mean proportion of 27 and 3

= $\sqrt{27 \times 3}$ = 9

Example -9: What must be added to each number 25, 19, 10, and 7 so that resultant numbers are in proportion?

Solution: Let ‘a’ be added to each number, and then they are in proportion

i.e 25 + a : 19 + a = 10 + a : 7 + a

Now, according to the property of proportion

Product of extremes = Product of means

⇒ ( 25 + a ) ( 7 + a ) = ( 19 + a ) ( 10+ a )

⇒ 175 + 32a + a² = 190 + 29a + a²

⇒ 3a = 15

⇒ a = 5

Example -10 : If x : y = y : z, then x⁴ : y⁴is equal to

Solution: Here x : y = y : z

Now, according to the property of proportion

Product of extremes = Product of means

y² = xz

⇒ $\frac{x^4}{y^4} = \frac{x^4}{y^{2^2}} = \frac{x^4}{(xz)^2} = \frac{x^2}{z^2}$

Therefore x⁴ : y⁴= x² : z²

Example-11 : How much can be added to the term ratio 5 : 8 to make it equal to 1 : 2

Solution: Let ‘a’ be added to a ratio of 5 : 8 to make 1 : 2

Then

$\frac{5+ a}{8 + a} = \frac{1}{2}$

a = -2

Example – 12 : If a : b = c : d = e : f = 5 : 6 then find the value of $\frac{ap + cq + er}{bp + dq + fr}$

Solution: According to the property of equal ratios

$\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{5}{6}$

then $\frac{a + c + e}{b + d + f} = \frac{5}{6}$

Now, according to the property of multiplication or divide of ratio

$\frac{ap}{bp} = \frac{cq}{dq} = \frac{er}{fr} = \frac{5}{6}$ then

$\frac{ap + cq + er}{bp + dq + fr} = \frac{5}{6}$

Example-13: Two numbers are in the ratio 5 : 6 and if 4 is subtracted from each, they are reduced to a ratio 4 : 5. Find a bigger number

Solution: The ratio of two numbers is 5 : 6

Let these numbers 5a & 6a

Subtract 4 from each, then a new ratio 4 : 5

i.e 5a – 4 : 6a – 4 = 4 : 5

⇒ 25a – 20 = 24a – 16

⇒ a = 4

Now these numbers are 20 & 24

So the bigger number is 24

Example-14: A bag contains one rupee coin, two rupee coins, and five rupee coins in the ratio of 3 : 4 : 5. If there are all Rs. 288 in the bag, are many coins of one rupee there?

Solution: Let ‘a’ be added to a ratio of 3 : 4 : 5 to make 288 rupees

i.e one rupee coins = 3a , two rupee coins = 4a & five rupee coins = 5a

Now equal the value of all coins

3a + 2 (4a) + 5(5a) = 288

Simplifying the above equation, we get

a = 8

Number of one rupee coins = 3 x 8 = 24

Example-15: One milk boy adds 2 liters of water to 12 liters of milk and another 2 liters of water to 10 liters of milk. What is the ratio of the strength of milk in the two mixtures?

Solution

Strength of milk in the first mixture

$\frac{12}{12+2} = \frac{12}{14}$

Strength of milk in the second mixture

$\frac{10}{10+2} = \frac{10}{12}$

Therefore, the ratio of their strengths 12/14 : 10/12

= 12 x 12 : 10 x 14

= 36 : 35

Example-16: One milk boy adds equal quantities of a mixture of milk and water in the ratio 9 : 5 and 4 : 3 respectively. Both the mixtures are now mixed thoroughly. Find the ratio of milk to water in the new mixture obtained?

Solution: Here two mixtures

one is 9 : 5 and another is 4 : 3

take the LCM of 14 (9 + 5) , 7 (4 +3) is 14

Both mixtures are mixed in equal quantities

In the first ratio out of 14 liters having 9L of milk and 5L of water

Second ratio 4 : 3 = 8 : 6 ( multiplying with 2 for each)

In the second ratio out of 14 liters having 8L of milk and 6L of water

In the new mixture having 17L ( 9+8) milk and 11L (5 +6) of water

So the ratio of the new mixture is 17 : 11

Example-17: Two vessels contain a quantity of a mixture of water and milk in the ratio 1 :2 & 2 : 3 respectively. Both the mixtures are mixed in the ratio of 3 : 1, Find the ratio after mixing the two mixtures.

Solution: Here two mixtures

one is 1 : 2 and another is 2 : 3

take the LCM of 3 (1 +2) , 5 (2 + 3) is 15

Both mixtures are mixed in 3 : 1

So take quantities of first mixture is 45L ( 15 x3) and second mixture is 15 L( 15 x 1)

In the first ratio out of 45 liters having 15L water and 30L of milk

In the second ratio out of 15 liters having 6L water and 9L of milk

In the new mixture having 21L ( 15 + 6) water and 39L (30+9) of milk

So the ratio of the new mixture is 21 : 39

i.e 7 : 13

Example-18: If P : Q = 3 : 4 , Q : R = 5 : 9 and R : S = 16 : 15, then ratio between P and S is

Solution: Here First find the ratio of P : Q : R

P : Q = 3 : 4 & Q : R = 5 : 9

P : Q : R = 15 : 20 : 36 ( Multiplying with 5 for the ratio of P : Q & 4 for the ratio of Q : R )

Now equal to the ratios of P : Q : R & R : S ( LCM of both R values of 36 & 16 is 144 )

P : Q : R = 15 : 20 : 36 = 60 : 80 : 144 ( Multiplying with 4 for the ratio of P : Q : R )

R : S = 16 : 15 = 144 : 135 ( Multiplying with 9 for the ratio of R : S )

P : Q : R : S = 60 : 80 : 144 : 135

Ratio of P : S = 60 : 135 = 4 : 9

Example-19 : A person traveling with a constant speed, took 8 minutes 40 seconds to reach his office and 9 minutes to return, using a different route. Find the ratio of the lengths of the two routes.

Solution: We know that D = ST ( D = distance , S = Speed & T = time)

D ∝ S ( If time is constant )

D ∝ T ( If speed is constant )

Now in our case speed is constant so the ratio of length is proportional to the time

i.e 8 min 40 sec : 9 min

520 : 540 ( converted into seconds)

26 : 27

Example -20: The ratio of present ages of two sisters P, Q is 1 : 2 and 5 years back the ratio was 1 : 3 what will be the ratio of their ages after 5 years?

Solution: Ages of P and Q are 1a & 2b

Now 5 years back the ratio is 1 : 3 so

1a – 5 : 2b -5 = 1 : 3

Simplifying the above ratio3a -15 = 2a -5

a = 10

The ages of P and Q are 10, 20

After 5 years their age 10+5 , 20+5

So the ratio is 15 : 25

i.e 3 : 5

Example – 21 : The ratio of P’s salary to Q’s salary is 2 : 3. The ratio of Q,s salary to R’s salary is 4 : 5. What is the ratio of P’s salary to R’s salary?

Solution: Find LCM of 3 and 4 (Both values represent ‘ Q ‘)

The LCM is 12

Now covert ‘ Q ‘ values in each ratio to 12

Thus, Ration – 1 = 2 : 3 = 8 : 12

Ratio – 2 = 4 : 5 = 12 : 15

Thus, P : Q : R = 8 : 12 : 15

Hence P : Q = 8 : 15

Example -22: Ratio of the earning of P and Q is 4 : 7 of the earning of A increased by 50% and those of Q decreases by 25 % the new ratio of their earning becomes

Solution: Let the original earning of P and Q are 4x and 7x

New earning of = 150 % of 4x = 150 x 4x / 100 = 6x

New earning of Q = 75 % of of 7x = 75 x 7x / 100 = 21x/4

Ratio of P and Q after new earning = 6x : 21x/4

Now the above ratio can be written as 8 : 7

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