Ratio proportion and variation problems with solutions

In this article explained about some examples with solution of ratio proportion and variation chapter

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Ratio and Proportion Questions with Solutions | Quantitative Aptitude

Example-1 : Ratio between two numbers is 5 : 6 and 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

Numbers are 30 and 70.

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

Solution: Take fourth proportion is ‘a’

Then 5 : 8 :: 20 : a

According to property of proportions

Product of extreams = Product of means

5a = 160

a = 160/5 = 32

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

Solution: According to 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 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 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 much 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 then they are in proportion

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

Now according to property of proportion

Product of extreams = 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 property of proportion

Product of extreams = 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 : What much be added to the term ratio 5 : 8 so as to make it equal to 1 : 2

Solution: Let ‘a’ be added to 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 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 property of Multiply 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: Find the value of $\frac{a+b-c}{a} = \frac{b+c-a}{b} = \frac{c+a-b}{c} = ?$

Solution: Here all ratios are equal

$ratio proportion and variation problems with solutions and formulas$

Now by the property of equal ratio

$ratio proportion and variation problems with solutions and formulas$

Example-14 : Two numbers are in the ratio 5 : 6 and if 4 is subtracted from each, they are reduced to ratio 4 : 5. Find bigger number

Solution: Ratio of two numbers is 5 : 6

Let these numbers 5a & 6a

Subtract 4 from each then 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 bigger number is 24

Example-15: A bag contain one rupee coins, two rupee coins and five rupee coins in the ratio of 3 : 4 : 5. If there are in all Rs. 288 in the bag, has many coins of one rupee are there?

Solution: Let ‘a’ be added to 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-16: One milk boy adds 2 liter of water to 12 liter of milk and another 2 liter of water to 10 liter of milk. What is the ratio of 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 ratio of their strengths 12/14 : 10/12

= 12 x 12 : 10 x 14

= 36 : 35

Example-17: One milk boy adds equal quantity of 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 do 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 first ratio out of 14 liters having 9L milk and 5L of water

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

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

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

So ratio of new mixture is 17 : 11

Example-18: Two vessels contain quantity of 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 first ratio out of 45 liters having 15L water and 30L of milk

In 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 ratio of new mixture is 21 : 39

i.e 7 : 13

Example-19: 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-20 : A person traveling with a constant speed, he took 8 minutes 40 seconds to reach his office and 9 minutes to return, using a different route. Find 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 ratio of length is proportional to time

i.e 8 min 40 sec : 9 min

520 : 540 ( converted into seconds)

26 : 27

Example -21: 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 – 22 : The ration of P’s salary to Q’s salary is 2 : 3. The ration of Q,s salary to R’s salary is 4 : 5. What is the ration of P’s salary to R’s salary?

Solution: Find LCM of 3 and 4 (Both values are representing to ‘ 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 -23: 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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