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 |
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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