problems

1) Train A starts from X at 12:00 p.m & reach at Y on 2:30 p.m & train B starts from Y at 12:15 and reach at X on 2:15 so,when both trains crosses each other?

2) Two trains are travelling from point A to point B such that the speed of first train is 65kmps and the speed of 2train is 29km/hr.Where is the distance b/w A & B such that the slower train reached 5hrs late compared to the faster?

3) A man buys spirit at Rs 60 per litre, adds water to it and then sells it at Rs 75 per liter.What is the ratio of spirit to water if his profit in the sell is 37.5%?

4) A certain quantity of petrol is found to be adulterated to the extent of 10%. What proportion of the adulterated petrol should be replaced with the pure petrol to take the purity level to 98%?

5) Boat goes downstream from P to Q in 2hrs,upstrem in 6hrs and if speed of stream was 1/2 of boat in still water.Find distance PQ? ans:- 6 / 4 / 10 / none.

Try to give the explanations of every sum.

4 Answers

11
SANDIPAN CHAKRABORTY ·

EDITED

1)The answer is 1:15 pm.

Let the total distance b/w thestations be D

In the 15 mins till B starts A travells D/10

so eff dist b/w A and B now is (9/10)D

Let after time t the trains meet

then for train A

distnc travelled L in time t

L = D2.5 x t ..........(1)

for train B to meet A , B must have travelled (9/10)D - L in time t

(9/10)D - L = D2 x t .........(2)

PLugging in the value of L in (2) from (1) we get t = 1 hr

So they meet after 1 hr 15 mins wrt to train A (the 15 mins A travelled till B started)

so they meet at 12:00 + 1 hr 15 mins

====>1:15 pm

1
1.618 ·

Shud b under 9-10th forum...:)

11
SANDIPAN CHAKRABORTY ·

Q3) ans is spirit:water = 10:1

The cost of 1L spirit is Rs 60

Let xL water be added per litre of spirit

Now SP of 1 L mixture = Rs 75

so (x+1) L costs = Rs (x+1)75

% profit = 75(x+1) - 6060 x 100 = 37.5

solving we get x = 110

so 1L spirit : 110 L water

or 10 L spirit : 1L water

so ratio spirit:water = 10 : 1

11
SANDIPAN CHAKRABORTY ·

Q4)

10 % adultration means 90 parts petrol has 10 parts impurity

Let amount of petrol added be x

so 90 + x100 + x = 98100

==> x = 400 parts...

now 400 parts of petrol to be added per 100 parts of adultrated

petrol...

so 400% of adultrated petrol to be replaced with pure petrol

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