(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.
Number of pages typed by Ravi in 1 hour = | 32 | = | 16 | . |
6 | 3 |
Number of pages typed by Kumar in 1 hour = | 40 | = 8. |
5 |
Number of pages typed by both in 1 hour = | ❨ | 16 | + 8 | ❩ | = | 40 | . |
3 | 3 |
∴ Time taken by both to type 110 pages = | ❨ | 110 x | 3 | ❩ | hours |
40 |
= 8 | 1 | hours (or) 8 hours 15 minutes. |
4 |
5 | 5 |
11 |
5 | 6 |
11 |
6 | 5 |
11 |
6 | 6 |
11 |
5 | 5 |
11 |
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 |
Then, 4x + 6y = | 1 | and 3x + 7y = | 1 | . |
8 | 10 |
Solving the two equations, we get: x = | 11 | , y = | 1 |
400 | 400 |
∴ 1 woman's 1 day's work = | 1 | . |
400 |
⟹ 10 women's 1 day's work = | ❨ | 1 | x 10 | ❩ | = | 1 | . |
400 | 40 |
Hence, 10 women will complete the work in 40 days.
2(A + B + C)'s 1 day's work = | ❨ | 1 | + | 1 | + | 1 | ❩ | = | 15 | = | 1 | . |
30 | 24 | 20 | 120 | 8 |
Therefore, (A + B + C)'s 1 day's work = | 1 | = | 1 | . |
2 x 8 | 16 |
Work done by A, B, C in 10 days = | 10 | = | 5 | . |
16 | 8 |
Remaining work = | ❨ | 1 - | 5 | ❩ | = | 3 | . |
8 | 8 |
A's 1 day's work = | ❨ | 1 | - | 1 | ❩ | = | 1 | . |
16 | 24 | 48 |
Now, | 1 | work is done by A in 1 day. |
48 |
So, | 3 | work will be done by A in | ❨ | 48 x | 3 | ❩ | = 18 days. |
8 | 8 |
A's 2 day's work = | ❨ | 1 | x 2 | ❩ | = | 1 | . |
20 | 10 |
(A + B + C)'s 1 day's work = | ❨ | 1 | + | 1 | + | 1 | ❩ | = | 6 | = | 1 | . |
20 | 30 | 60 | 60 | 10 |
Work done in 3 days = | ❨ | 1 | + | 1 | ❩ | = | 1 | . |
10 | 10 | 5 |
Now, | 1 | work is done in 3 days. |
5 |
∴ Whole work will be done in (3 x 5) = 15 days.
∴ 1 woman's 1 day's work = | 1 | . |
320 |
(16 x 15) men can complete the work in 1 day.
∴ 1 man's 1 day's work = | 1 |
240 |
So, required ratio |
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= 4 : 3 (cross multiplied) |
Work done by X in 4 days = | ❨ | 1 | x 4 | ❩ | = | 1 | . |
20 | 5 |
Remaining work = | ❨ | 1 - | 1 | ❩ | = | 4 | . |
5 | 5 |
(X + Y)'s 1 day's work = | ❨ | 1 | + | 1 | ❩ | = | 8 | = | 2 | . |
20 | 12 | 60 | 15 |
Now, | 2 | work is done by X and Y in 1 day. |
15 |
So, | 4 | work will be done by X and Y in | ❨ | 15 | x | 4 | ❩ | = 6 days. |
5 | 2 | 5 |
Hence, total time taken = (6 + 4) days = 10 days.
1 |
4 |
1 |
10 |
7 |
15 |
8 |
15 |
8 |
15 |
Explanation:
A's 1 day's work = | 1 | ; |
15 |
B's 1 day's work = | 1 | ; |
20 |
(A + B)'s 1 day's work = | ❨ | 1 | + | 1 | ❩ | = | 7 | . |
15 | 20 | 60 |
(A + B)'s 4 day's work = | ❨ | 7 | x 4 | ❩ | = | 7 | . |
60 | 15 |
Therefore, Remaining work = | ❨ | 1 - | 7 | ❩ | = | 8 | . |
15 | 15 |
C's 1 day's work = | 1 | - | ❨ | 1 | + | 1 | ❩ | = | 1 | - | 7 | = | 1 | . |
3 | 6 | 8 | 3 | 24 | 24 |
A's wages : B's wages : C's wages = | 1 | : | 1 | : | 1 | = 4 : 3 : 1. |
6 | 8 | 24 |
∴C's share (for 3 days) = Rs. | ❨ | 3 x | 1 | x 3200 | ❩ | = Rs. 400. |
24 |
22 | 1 | days |
2 |
22 | 1 | days |
2 |
The time difference is (3 - 1) 2 days while B take 3 days and A takes 1 day.
If difference of time is 2 days, B takes 3 days.
If difference of time is 60 days, B takes | ❨ | 3 | x 60 | ❩ | = 90 days. |
2 |
So, A takes 30 days to do the work.
A's 1 day's work = | 1 |
30 |
B's 1 day's work = | 1 |
90 |
(A + B)'s 1 day's work = | ❨ | 1 | + | 1 | ❩ | = | 4 | = | 2 |
30 | 90 | 90 | 45 |
∴ A and B together can do the work in | 45 | = 22 | 1 | days. |
2 | 2 |
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