A walks 1 km in 5 min.
Then, distance covered by A in 1 min = 1000/5 = 200 m
B walks 1 km in 8 min.
Then, distance covered by B in 1 min = 1000/8 = 125 m
C walks 1 km in 10 min.
Then, distance covered by C in 1 min = 1000/101 = 100 m
Let A meets B and C in x and y min, respectively.
Then, according to the question,
Distance covered by C in(x + 2) min = Distance covered by A in x min
? 100(x + 2) = 200 x
?100 x + 200 = 200 x
? 200 = 100 x
? x = 200/100 = 2 min
Now, for A and B
Distance covered by B in (y + 1) min = Distance covered by A in y min
? 125(y + 1) = 200y
? 125y + 125 = 200y
? 125 = 200y - 125y
? 125 = 75y
? y = 125/75 = 5/3 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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