Model 6  Efficiency of the worker Practice Questions Answers Test with Solutions & More Shortcuts

Question : 36 [SSC CHSL 2011]

A can do a work in 21 days. B is 40% more efficient than A. The number of days required for B to finish the same work alone is

a) 18 days

b) 10 days

c) 12 days

d) 15 days

Answer: (d)

Time taken by B

= ${21 × 100}/140$ = 15 days

Using Rule 17
If A can do a work in 'x' days and B is R% more efficient than A, then 'B' alone will do the same work in $x × 100/{100 + R}$ days

Here, x = 21, R = 40%

Time taken By B = $x × 100/{100 + R}$ days

= $21 × 100/140$ days = 15 days

Question : 37 [SSC DEO 2008]

A takes twice as much time as B and thrice as much as C to complete a piece of work. They together complete the work in1 day. In what time, will A alone complete the work.

a) 4 days

b) 9 days

c) 5 days

d) 6 days

Answer: (d)

Let time taken by C to complete the work = x days

Time taken by A to complete the work = 3x days

and time taken by B to complete the work = ${3x}/2$ days

According to the question,

$1/{3x} + 1/{{3x}/2} + 1/x = 1$

$1/{3x} + 2/{3x} + 1/x = 1$

${1 + 2 + 3}/{3x} = 1$

$6/{3x} = 1 ⇒ 2/x = 1 ⇒ x = 2$

Time taken by A

= 3x = 3 × 2 = 6 days

Question : 38 [SSC CGL Tier-I 2016]

If 10 people can do a job in 20 days, then 20 people with twice the efficiency can do the same job in

a) 40 days

b) 5 days

c) 10 days

d) 20 days

Answer: (b)

In second case, the efficiency of a man is twice to that in the first case.

$M_1D_1 = M_2D_2$

10 × 20 = 2 × 20 × $D_2$

$D_2 = {10 × 20}/{2 × 20}$ = 5 days.

IMPORTANT quantitative aptitude EXERCISES

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