A can complete a certain work in 30 days. B is 25% more efficient than A and C is 20% more efficient than B. They all worked together for 3 days. B alone can complete the remaining work in:

A. 12 days

B. 20 days

C. 18 days

D. 15 days

SSC CGL 2019 Tier 1 4 March 2020 Shift 1

Solution:

In this question from Time and Work,

Let us see the shortcut method first:

Let us assume the efficiency of A be 4 (it means A can do 4 units of work in a day)

Therefore, the efficiency of B will be = 125% of the efficiency of A = 5 (it means B can do 5 units of work in a day)

Therefore, the efficiency of C will be = 120% of the efficiency of B = 6 (it means B can do 6 units of work in a day)

Total units of work = 30 x 4 = 120 units

Total work done by A, B and C in 3 days = (4 + 5 + 6) x 3 = 45 units

Remaining work = 120 - 45 = 75 units

The number of days required by B to complete the remaining work = 75/5 = 15 days.

Hence, (D) is the correct answer.

Let us see the basic method:

The efficiency of A = One day work of A = 1/30

The Efficiency of B = One day work of B = 1/30 + (25/100) x (1/30) = 1/30 + 1/120 = 1/24

The Efficiency of C = One day work of C = 1/24 + (20/100) x (1/24) = 1/24 + 1/120 = 1/20

Let the remaining work completed by B in x days

As per the question, we get

(1/30 + 1/24 + 1/20) x 3 + (1/24)x = 1

x/24 = 1 - 3/30 - 3/24 - 3/20

x/24 = (120 - 12 - 15 - 18)/120

x/24 = 75/120

x = (75 x 24)/120

x = 15

Hence, (D) is the correct answer.