(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.
4th observation i.e., 8 is the median.
47/10000 = .0047
.02 =(2/100) x 100% =2%
Let N / 11 = 233
Then, N = 233 x 11 = 2563
? Missing digit is 5.
121012 = 12 x 10084 + 4
? remainder = 4
We have the important relation, More work, More time (days)
? A piece of work can be done in 6 days.
? Three times of work of same type can be done in 6 x 3
= 18 days
? = 750.0003 ÷ 19.999
? ? ? 750 ÷ 20
? ? ? 375 ? 38
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