General Questions
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times as fast as Sunil. How many days it will take for the three together to complete the work ?Time taken by Ramesh alone = (2/3) x 4 = 8/3 days
∴ Their 1 day's work = (1/4 + 1/6 + 3/8) = 19/24
So, together they can finish the work in 24 /19 days, i.e., 15/19 days.
| A's 1 hour's work = | 1 | ; |
| 4 |
| (B + C)'s 1 hour's work = | 1 | ; |
| 3 |
| (A + C)'s 1 hour's work = | 1 | . |
| 2 |
| (A + B + C)'s 1 hour's work = | ![]() |
1 | + | 1 | ![]() |
= | 7 | . |
| 4 | 3 | 12 |
| B's 1 hour's work = | ![]() |
7 | - | 1 | ![]() |
= | 1 | . |
| 12 | 2 | 12 |
B alone will take 12 hours to do the work.
| Whole work is done by A in | ![]() |
20 x | 5 | ![]() |
= 25 days. |
| 4 |
Now,
![]() |
1 - | 4 | ![]() |
i.e., | 1 | work is done by A and B in 3 days. | |
| 5 | 5 |
Whole work will be done by A and B in (3 x 5) = 15 days.
A's 1 day's work
| = | 1 | , (A + B)'s 1 day's work = | 1 | . |
| 25 | 15 |
B's 1 day's work = |
![]() |
1 | - | 1 | ![]() |
= | 4 | = | 2 | . |
| 15 | 25 | 150 | 75 |
| So, B alone would do the work in | 75 | = 37 | 1 | days. |
| 2 | 2 |
Let 1 man's 1 day's work = x and 1 boy's 1 day's work = y.
| Then, 6x + 8y = | 1 | and 26x + 48y = | 1 | . |
| 10 | 2 |
Solving these two equations, we get :
| x = | 1 | and y = | 1 | . |
| 100 | 200 |
(15 men + 20 boy)'s 1 day's work
| = | ![]() |
15 | + | 20 | ![]() |
= | 1 | . |
| 100 | 200 | 4 |
15 men and 20 boys can do the work in 4 days.
| (A + B)'s 1 day's work = | 1 |
| 10 |
| C's 1 day's work = | 1 |
| 50 |
(A + B + C)'s 1 day's work
| = | ![]() |
1 | + | 1 | ![]() |
= | 6 | = | 3 | . .... (i) |
| 10 | 50 | 50 | 25 |
A's 1 day's work = (B + C)'s 1 day's work .... (ii)
| From (i) and (ii), we get: 2 x (A's 1 day's work) = | 3 |
| 25 |
A's 1 day's work = |
3 | . |
| 50 |
B's 1 day's work |
![]() |
1 | - | 3 | ![]() |
= | 2 | = | 1 | . |
| 10 | 50 | 50 | 25 |
So, B alone could do the work in 25 days.
(P + Q + R)'s 1 hour's work
| = | ![]() |
1 | + | 1 | + | 1 | ![]() |
= | 37 | . |
| 8 | 10 | 12 | 120 |
Work done by P, Q and R in 2 hours
| = | ![]() |
37 | x 2 | ![]() |
= | 37 | . |
| 120 | 60 |
| Remaining work = | ![]() |
1 - | 37 | ![]() |
= | 23 | . |
| 60 | 60 |
| (Q + R)'s 1 hour's work = | ![]() |
1 | + | 1 | ![]() |
= | 11 | . |
| 10 | 12 | 60 |
| Now, | 11 | work is done by Q and R in 1 hour. |
| 60 |
So,
| 23 | work will be done by Q and R in | ![]() |
60 | x | 23 | ![]() |
= | 23 | hours | |
| 60 | 11 | 60 | 11 |
2 hours.
So, the work will be finished approximately 2 hours after 11 A.M., i.e., around 1 P.M.
| 1 woman's 1 day's work = | 1 |
| 70 |
| 1 child's 1 day's work = | 1 |
| 140 |
(5 women + 10 children)'s day's work
| = | ![]() |
5 | + | 10 | ![]() |
= | ![]() |
1 | + | 1 | ![]() |
= | 1 |
| 70 | 140 | 14 | 14 | 7 |
5 women and 10 children will complete the work in 7 days.
| (A + B)'s 20 day's work = | ![]() |
1 | x 20 | ![]() |
= | 2 | . |
| 30 | 3 |
| Remaining work = | ![]() |
1 - | 2 | ![]() |
= | 1 | . |
| 3 | 3 |
| Now, | 1 | work is done by A in 20 days. |
| 3 |
Therefore, the whole work will be done by A in (20 x 3) = 60 days.
| B's 10 day's work = | ![]() |
1 | x 10 | ![]() |
= | 2 | . |
| 15 | 3 |
| Remaining work = | ![]() |
1 - | 2 | ![]() |
= | 1 | . |
| 3 | 3 |
| Now, | 1 | work is done by A in 1 day. |
| 18 |
|
1 | work is done by A in | ![]() |
18 x | 1 | ![]() |
= 6 days. |
| 3 | 3 |
P can complete the work in (12 x 8) hrs. = 96 hrs.
Q can complete the work in (8 x 10) hrs. = 80 hrs.
P's1 hour's work = |
1 | and Q's 1 hour's work = | 1 | . |
| 96 | 80 |
| (P + Q)'s 1 hour's work = | ![]() |
1 | + | 1 | ![]() |
= | 11 | . |
| 96 | 80 | 480 |
| So, both P and Q will finish the work in | ![]() |
480 | ![]() |
hrs. |
| 11 |
Number of days of 8 hours each
= |
![]() |
480 | x | 1 | ![]() |
= | 60 | days = 5 | 5 | days. |
| 11 | 8 | 11 | 11 |


A's 1 day's work =