Tough Speed, Rate, Time DS

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Tough Speed, Rate, Time DS

by knight247 » Sun Jun 30, 2013 12:34 am
Don't have an OA for this one. However, I get the answer C

Detailed explanations would be appreciated. Many thanks in advance.
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by luckypiscian » Mon Jul 01, 2013 10:23 pm
I would say no need to calculate

speed speed
jen 4 y/s 5 y/s
linda a y/s a+i y/s

60 yards 60 yards
|--------------------|---------------------|

We know Jen's speed for both first and last 60 yards so both, distance covered by him at any point in time / time taken by him to cover a particular distance can be calculated.

statement 1:
18 secs -> Distance covered by Jen = Distance covered by Linda + 21 yards
it also tells us that linda is still in first half

We should be able to equate and calculate linda's speed in first half with the above info, but we have no info about her speed in second half, Hence avg. can't be calculated.
INSUFFICIENT

statement 2:
Time taken by Linda = Time taken by Jen + 3

Forget if they ran at 2 different speeds, think of it as if they ran constantly at their avg speed.
We know the total distance covered, hence should be able to equate time taken by both to complete the race and get linda's avg. speed.
SUFFICIENT

If you still want to calculate
statement 1:
18 secs -> Distance covered by Jen = Distance covered by Linda + 21 yards
it also tells us that linda is still in first half

Jen would have covered for 60 yards in 60/4 = 15 secs
in next 3 (18 - 15) seconds he would have covered 3 * 5 = 15 yards
Hence distance covered by linda = Distance covered by Jen - 21
= 60 + 15 - 21
= 54 yards (still in first half)
Linda's speed in first half, a = 54/18 y/s = 3 y/s
but speed in the second half is a + i = 3 + i ( where i is the increase in speed )
We have no info about it. Hence
INSUFFICIENT

statement 2:
Time taken by Linda = Time taken by Jen + 3

Time taken by Jen = (60/4) + (60/5) = 15 + 12 = 27 secs
Hence
Time taken by Linda = 27 + 3 = 30 secs
Linda's avg speed = 120 yards/ 30 secs = 4 y/s
SUFFICIENT