Model 7 Working with individual wages Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 7 EXERCISES

Top 10,000+ Aptitude Memory Based Exercises

The following question based on time & work topic of quantitative aptitude

Questions : Some carpenters promised to do a job in 9 days but 5 of them were absent and remaining men did the job in 12 days. The original number of carpenters was

(a) 18

(b) 16

(c) 24

(d) 20

The correct answers to the above question in:

Answer: (d)

Using Rule 1,

Let the original number of carpenters be x.

$M_1D_1 = M_2D_2$

x × 9 = (x - 5) × 12

9x = 12x - 60

3x = 60 ⇒ x = 20

Practice time & work (Model 7 Working with individual wages) Online Quiz

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Read more working with individual wages Based Quantitative Aptitude Questions and Answers

Question : 1

If x men can do a piece of work in x days, then the number of days in which y men can do the same work is

a) $x^2y$ days

b) $x^2/y$ days

c) xy days

d) $y^2/x$ days

Answer: (b)

Using Rule 1,

$M_1D_1 = M_2D_2$

$x . x = y . D_2$

$D_2 = x^2/y$ days

Question : 2

If 7 men working 7 hrs a day for each of 7 days produce 7 units of work, then the units of work produced by 5 men working 5 hrs a day for each of 5 days is

a) $343/25$

b) $49/125$

c) $25/343$

d) $125/49$

Answer: (d)

Using Rule 1,

${M_1D_1T_1}/W_1 = {M_2D_2T_2}/W_2$

${7 × 7 × 7}/7 = {5 × 5 × 5}/W_2$

49 × $W_2$ = 125

$W_2 = 125/49$

Question : 3

Seventy-five men are employed to lay down a railway line in 3 months. Due to certain emergency conditions, the work was to be finished in 18 days. How many more men should be employed to complete the work in the desired time ?

a) 375

b) 350

c) 300

d) 325

Answer: (c)

Using Rule 1,

$M_1D_1 = M_2D_2$

75 × 90 = $M_2$ × 18

$M_2 = {75 × 90}/18 = 375$

Number of additional men

= 375 - 75 = 300

Question : 4

‘x’ number of men can finish a piece of work in 30 days. If there were 6 men more, the work could be finished in 10 days less. The original number of men is

a) 15

b) 12

c) 6

d) 10

Answer: (b)

MenDays
x30
x + 620

x + 6 : x : : 30 : 20

${x + 6}/x = 30/20 = 3/2$

2x + 12 = 3x

3x - 2x = 12 ⇒ x = 12

Using Rule 1,

Here, $M_1 = x, D_1 = 30, M_2 = x + 6, D_2$ = 20

$M_1D_1 = M_2D_2$

x × 30 = (x + 6) × 20

$3x = 2x + 12$ ⇒ x = 12

Question : 5

If the work done by (x –1) men in (x + 1) days is to the work done by (x + 2) men in (x - 1) days are in the ratio 9 : 10, then the value of x is equal to :

a) 8

b) 7

c) 5

d) 6

Answer: (a)

Using Rule 1,

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

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

$9/10 = {(x - 1)(x + 1)}/{(x + 2)(x - 1)} = {x + 1}/{x + 2}$

10x + 10 = 9x + 18

x = 18 - 10 = 8

Question : 6

10 men working 6 hours a day can complete a work in 18 days. How many hours a day must 15 men work to complete the same work in 12 days ?

a) 15 days

b) 12 days

c) 6 days

d) 10 days

Answer: (c)

Using Rule 1,

MenDaysWorking hours
10186
1512x

where, x is working hrs/days

∴ ${15 : 10}/{12 : 18}]$ : : 6 : x

15 × 12 × x = 10 × 18 × 6

$x = {10 × 18 × 6}/{15 × 12} = 6$ hours

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