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.
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
Each previous number is multiplied by 2.
? 8 m shadow means original height = 12 m
? 1 m shadow means original height = 12/8 m
? 100 m shadow means original height = (12/8) x 100 m
= (6/4) x 100 = 6 x 25 = 150 m
Let 8% of 96 = y of 1/25
? (8 x 96)/100 = y/25
? y = (8 x 96 x 25)/100 = 192
Since the principal is not given, so data is inadequate.
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