10 | 1 | days |
2 |
(B + C)'s 1 day's work = | ❨ | 1 | + | 1 | ❩ | = | 7 | . |
9 | 12 | 36 |
Work done by B and C in 3 days = | ❨ | 7 | x 3 | ❩ | = | 7 | . |
36 | 12 |
Remaining work = | ❨ | 1 - | 7 | ❩ | = | 5 | . |
12 | 12 |
Now, | 1 | work is done by A in 1 day. |
24 |
So, | 5 | work is done by A in | ❨ | 24 x | 5 | ❩ | = 10 days. |
12 | 12 |
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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