Model 7 Working with individual wages Practice Questions Answers Test with Solutions & More Shortcuts

Question : 26 [SSC CGL Tier-I 2013]

A man undertakes to do a certain work in 150 days. He employs 200 men. He finds that only a quarter of the work is done in 50 days. The number of additional men that should be appointed so that the whole work will be finished in time is :

a) 50

b) 125

c) 75

d) 100

Answer: (d)

Using Rule 1,

200 men do $1/4$ work in 50 days.

${M_1D_1}/W_1 = {M_2D_2}/W_2$

${200 × 50}/{1/4} = {M_2 × 100}/{3/4}$

$M_2 × 100 = 200 × 50 × 3$

$M_2$ = 300

Additional men = 100

Question : 27 [SSC MTS 2013]

A contractor undertook to finish a work in 92 days and employed 110 men. After 48 days, he found that he had already done $3/5$ part of the work, the number of men he can withdraw so that the work may still be finished in time is :

a) 30

b) 35

c) 45

d) 40

Answer: (a)

Using Rule 1,

${M_1D_1}/W_1 = {M_2D_2}/W_2$

${110 × 48}/{3/5} = {M_2 × 44}/{2/5}$

$M_2$ × 44 × 3 = 110 × 48 × 2

$M_2 = {110 × 48 × 2}/{44 × 3}$ = 80

Number of men can be withdrawn

= 110 - 80 = 30

Question : 28 [SSC CGL Tier-I 2013]

One man, 3 women and 4 boys can do a piece of work in 96 hours, 2 men and 8 boys can do it in 80 hours, 2 men and 3 women can do it in 120 hours. 5 men and 12 boys can do it in

a) 44 hours

b) 43$7/11$ hours

c) 39$1/11$ hours

d) 42$7/11$ hours

Answer: (b)

Using Rule 1,

1 hour’s work of 1 man and 4 boys = $1/160$

[Since,2 men and 8 boys can do the work in 80 hrs.]

1 hour’s work of 1 man 3 women and 4 boys = $1/96$

1 hour’s work of 3 women

= $1/96 - 1/160 = {10 - 6}/960 = 1/240$

1 hour’s work of 2 men

= $1/120 - 1/240 = 1/240$

1 hour’s work of 4 boys

= $1/160 - 1/480$

= ${3 - 1}/480 = 1/240$

2 men = 3 women = 4 boys

2 men + 8 boys = 12 boys

5 men + 12 boys = 22 boys

By $M_1D_1 = M_2D_2$

12 × 80 = 22 × $D_2$

$D_2 = {12 × 80}/22 = 480/11 = 43{7}/11$ hours

Question : 29 [SSC CHSL 2012]

30 men can repair a road in 18 days. They are joined by 6 more workers. Now the road can be repaired in

a) 17 days

b) 16 days

c) 14 days

d) 15 days

Answer: (d)

Using Rule 1,

$M_1D_1 = M_2D_2$

30 × 18 = 36 × $D_2$

$D_2 = {30 × 18}/36$ = 15 days

Question : 30 [SSC CGL Prelim 2008]

If p men working p hours per day for p days produce p units of work, then the units of work produced by n men working n hours a day for n days is

a) $n^3/p^2$

b) $n^2/p^2$

c) $p^2/n^2$

d) $p^3/n^2$

Answer: (a)

Since, P men working P hours/ day for P days produce P units of work.

1 man working 1 hour/day for 1 day produce

$P/P^3 =1/P^2$ units of work

n men working n hours a day for n day’s produce $n^3/P^2$ units of work

Using Rule 1,

Here, $M_1 = p, D_1 = p, T_1 =p,W_1$ = p

$M_2 = n, D_2 = n, T_2 = n, W_2$ = ?

$M_1D_1T_1W_2 = M_2D_2T_2W_1$

p × p × p × $w_2$ = n × n × n × p

$w_2 = n^3/p^2$

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