Let the ages of mother, father and boys be M, F, B1 and B2, respectively.
The total age of four member
= 19 x 4
= 76 yr.
Given (B1 + B1)/2 = 11/2
? B1 + B2 = 11
M + F + B1 + B2 = 76 - 11
? M + F = 65 .......(i)
According to the question,
(B1 + B2 + F) / 3 = (B1 + B2 + M) / 3 + 3
? B1 + B2 + F) = B1 + B2 + M + 9
? F = M + 9
? F - M = 9 ............. (ii)
From Eqs (i) and (ii), we get
F = 37 yr and M = 28 yr
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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