Practice Working with individual wages - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) 7 men can complete a piece of work in 12 days. How many additional men will be required to complete double the work in 8 days ?
(a)
(b)
(c)
(d)
| Work | Days | Men |
| 1 | 12 | 7 |
| ↓ | ↑ | ↓ |
| 2 | 8 | x |
∴ ${1 : 2}/{8 : 12}]$ : : 7 : x
where, x is no. of men
1 × 8 × x = 2 × 12 × 7
$x = {7 × 12 × 2}/8 = 21$
Number of additional men = 21 - 7 = 14
Metod 2 : Using Rule 1,
$M_1D_1W_2 = M_2D_2W_1$
7 × 12 × 2 = $M_2$ × 8 × 1
$M_2 = {7 × 12 × 2}/8 = 21$
No. of additional men = 21 - 7 = 14
Q-2) Two persons can complete a piece of work in 9 days. How many more persons are needed to complete double the work in 12 days?
(a)
(b)
(c)
(d)
| Work | Days | Persons |
| 1 | 9 | 2 |
| ↓ | ↑ | ↓ |
| 2 | 12 | x |
where x = number of persons
∴ ${1 : 2}/{12 : 9}]$ : : 2 : x
1×12 × x = 2 × 9 × 2
$x = {2 × 9 × 2}/12$ = 3
Using Rule 1,
Here, $M_1 = 2, W_1 = 1, D_1$ = 9
$M_2 = ?, W_2 = 2, D_2$ = 12
$M_1D_1W_2 = M_2D_2W_1$
2 × 9 × 2 = $M_2$ × 12 × 1
$M_2 = 36/12$ = 3
Q-3) 24 men can do a piece of work in 17 days. How many men will be able to do it in 51 days ?
(a)
(b)
(c)
(d)
$M_1D_1 = M_2D_2$
24 × 17 = $M_2$ × 51
$M_2 = {24 × 17}/51$ = 8 men
Q-4) 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)
(b)
(c)
(d)
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
Q-5) If 72 men can build a wall of 280 m length in 21 days, how many men could take 18 days to build a similar type of wall of length 100 m?
(a)
(b)
(c)
(d)
Using Rule 1,
We know that
$W_1/{M_1D_1} = W_2/{M_2D_2}$
$280/{72 × 21} = 100/{x × 18}$
Where x = number of men
x × 18 × 280 = 100 × 72 × 21
$x = {100 × 72 × 21}/{18 × 280}$ = 30
Q-6) ‘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)
(b)
(c)
(d)
| Men | Days |
| x | 30 |
| ↑ | ↓ |
| x + 6 | 20 |
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
Q-7) 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)
(b)
(c)
(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
Q-8) 4 mat-weavers can weave 4 mats in 4 days. At the same rate how many mats would be woven by 8 mat-weavers in 8 days ?
(a)
(b)
(c)
(d)
Using Rule 1,
| Weaver | Days | Mats |
| 4 | 4 | 4 |
| ↓ | ↓ | ↓ |
| 8 | 8 | x |
∴ ${4 : 8}/{4 : 8}]$ : : 4 : x
where, x is no. of mats
4 × 4 × x = 8 × 8 × 4
$x = {8 × 8 × 4}/{4 × 4}$ = 16
Q-9) A wall of 100 metres can be built by 7 men or 10 women in 10 days. How many days will 14 men and 20 women take to build a wall of 600 metres ?
(a)
(b)
(c)
(d)
Using Rule 1,
7 men ≡ 10 women
or 1 man = $10/7$ women
14 men + 20 women
= $({10 × 14}/7 + 20)$ women = 40 women
Now, more work, more days
More women, less days
∴ ${Work - {1 : 6}}/{Women - {40 : 10}}]$ : : 10 : x
Where x = number of days
1 × 40 × x = 6 × 10 × 10
or $x = 600/40$ = 15
Q-10) 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)
(b)
(c)
(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