Let a, b and c be the amount invested in schemes P, Q and R, respectively.
Then, according to the question,
[(a x 10 x 1)/100] + [(b x 12 x 1)/100] + [(c x 15 x 1)/100] = 3200
? 10a + 12b + 15x = 320000 .....(i)
Now, c = 240% of b = 12b/5 ....(ii)
and c = 150% of a = 3a/2
? a = 2c/3 = (2/3 x 12/5) b = 8b/5 .....(iii)
From Eqs. (i), (ii) and (iii), we get
16b + 12b + 36b = 320000
? 64b = 320000
? b = 5000
? Sum invested in scheme Q = ? 5000
.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.
NA
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