Required number of hours is the number of terms of the series 40 + 45 + 50 +... as speed increases every hour.
Given, sum of the series is 385
? a = 40, d=5, s = 385 and n= ?
Using S = n/2[2a + (n - 1)d]
? 385 = n/2[80 + 5n - 5]
? 770 = 5n2 + 75n
? n2 + 15n - 154 = 0
? n2 + 22n - 7n - 154 = 0
? n(n + 22) - 7(n + 22) = 0
? (n + 22) (n - 7) = 0
? n = 7
Remaining part = 1/12 - 1/20 = (5 - 3)/60 = 2/60 = 1/30 i.e., 1/30 th part is filled by B in 1 min
Hence, required time to fill the whole tank = (165 + 1 + 1) min = 167 min
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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