The ratio of the ages of A and B is 3 : 5.
The ratio of the ages of B and C is 3 : 5.
B's age is the common link to both these ratio. Therefore, if we make the numerical value of the ratio of B's age in both the ratios same, then we can compare the ages of all 3 in a single ratio.
The can be done by getting the value of B in both ratios to be the LCM of 3 and 5 i.e., 15.
The first ratio between A and B will therefore be 9 : 15 and
the second ratio between B and C will be 15 : 25.
Now combining the two ratios, we get A : B : C = 9 : 15 : 25.
Let their ages be 9x, 15x and 25x.
Then, the sum of their ages will be 9x + 15x + 25x = 49x
The question states that the sum of their ages is 147.
i.e., 49x = 147 or x = 3.
Therefore, B's age = 15x = 15*3 = 45
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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