## Ratios and Proportions Increment/Decrement Problems

Below are four questions where ratios are involed.

Question 1
Section A of class VIII had boys and girls in the ratio of 2:1. The current number of boys in the class is 40. Few girl students where shifted from section B to Section A. This changed the ratio of boys and girls to 2:3. Find the number of girls who got shifted from section B.

a. 30
b. 40
c. 10
d. 5
e. 15

Let the original number of boys be B and that of girls be G

Then B/G = 2/1.
It is given that B = 40
Therefore, 40/G = 2, Or G = 20

Let number of girls who were shifted from Section B be x. It is given that the new ratio of boys B to the increased number of girls i.e G + x is 2/3.
Then B/(G+x) = 2/3
Or 40/20+x = 2/3
20+x = 60
Or x = 40

Question 2
Number of players in sports namely badminton, tennis and table tennis where in the ratio of 2:3:4. Originally there were 16 badminton players. If 25% of the badmitton players withdrew due to some reason, what is the new ratio of the players.

a. 3:6:8
b. 3:7:8
c. 3:6:4
d. 6:6:8
e. 1:6:8

Let the number of players in Badminton,tennis and table tennis be P,Q and R respectively.

It is given that P:Q:R=2:3:4

From the above ratio, we can easily infer that P/Q = 2/3...(1) and P/R = 2/4 = 1/2...(2)

It is given that P = 16
From equation 1, P/Q = 2/3 or 16/Q = 2/3. Therefore Q = 16 x 3/2 = 24

Similarly from equation 2, P/R = 1/2 or 16/R = 1/2. Therefore, R = 32

Let the number of badminton players present after 25% withdrew be P1

Then P1 = 75% of P or 75/100 x 16
Therefore P1 = 12

New ratio is P1 : Q : R or 12:24:32 or 3:6:8

Question 3
Let the ratio of water and milk be 1:4 in a filled can of capacity 100 litres. The mixture was added with more water so as to change the ratio of 3:8. Find the quantity of water present originally and the amount of water (in litres) added thereafter.

a. 15
b. 8
c. 12
d. 6
e. 10

Let the original quantities of water and milk be W and M respectively.

It is given that W/M = 1/4 ...(1)

Also it is given that, total capacity of the can = W + M = 100 ...(2)

If a fraction exists of the form a/b = c/d. Then we can very well write a/a+b = c/c+d

Applying similar logic to equation 1 we get, W/(W + M) = 1/(1 + 4) = 1/5

But we know W + M = 100

Therefore, W/100 = 1/5 or W = 20 litres.

Substituting W = 20 litres in equation 2, we get

M = 100 - 20 = 80 litres.

Let x quantity of water be added so that the ratio becomes 3/8

Therefore W + x / M = 3/8 ...(3)

But we know W = 20 and M = 80

Therefore, equation 3 becomes, 20 + x / 80 = 3/8

Or 20 + x = 30

x = 10 litres.

We have found W = 20 litres and x = 10 litres

Question 4
Assume the ratio of monthly wages to team leaders and team members in factory be 2:1. If salary of team leaders is hiked by 25% and that of team members is hiked by 30%, what would be the new ratio.

a. 25/13
b. 13/25
c. 25/26
d. 26/25
e. 25/12

Let the monthly wages of team leaders and team members be L and M respectively.

It is given that L/M = 2/1

Let the hiked salary (by 25%) of team leaders be L1. Therefore L1 = 125/100 X L = 1.25L

Let the hiked salary (by 30%) of team members be M1. Therefore M1 = 130/100 X M = 1.3M

New ratio L1/M1 = 1.25L/1.3M

But we already know L/M = 2/1

Therefore, L1/M1 = 1.25/1.3 x 2/1 = 2.5/1.3 = 25/13

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