5 | 1 |
2 |
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 |
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
Subtract 20, 25, 30, 35, 40, 45 from successive numbers. So 0 is wrong.
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