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QUANTITATIVE SECTION 6 :- ques10 of 20 ID: Q01-86 GGREClub
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22 Sep 2017, 11:37
22 Sep 2017, 11:37
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Question Stats:
75% (01:48) correct
25% (01:11) wrong based on 8 sessions
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It takes Joey the postman 1 hours to run a 3 mile long route every day. He delivers packages and then returns to the post office along the same path. If the average speed of the round trip is 5 mile/hour, what is the speed with which Joey returns?
a 11
b 12
c 13
d 14
e 15
Answer given is 12
But it should be 15 and here is my reasoning-
let his speed for first half of the journey be 3 miles an hour
let the return journey (2nd half) be x miles an hour
now,
avg speed =5 mile an hour
we know if the Av speed = \(\frac{2*x1 *x2}{x1+x2}\) (Where X1 = speed m/hr in first half and X2 = speed in 2nd half)
(Plz Note - This formula only hold valid if both forward and return journey covered equal distance as in this case)
therefore \(\frac{2*3*x}{3+x}\)=5
Solving for X we get x=15 m/hr
a 11
b 12
c 13
d 14
e 15
Answer given is 12
But it should be 15 and here is my reasoning-
let his speed for first half of the journey be 3 miles an hour
let the return journey (2nd half) be x miles an hour
now,
avg speed =5 mile an hour
we know if the Av speed = \(\frac{2*x1 *x2}{x1+x2}\) (Where X1 = speed m/hr in first half and X2 = speed in 2nd half)
(Plz Note - This formula only hold valid if both forward and return journey covered equal distance as in this case)
therefore \(\frac{2*3*x}{3+x}\)=5
Solving for X we get x=15 m/hr
Re: QUANTITATIVE SECTION 6 :- ques10 of 20 ID: Q01-86 GGREClub
[#permalink]
22 Sep 2017, 22:08
22 Sep 2017, 22:08
Hi everybody plz check my reasoning so that I know Iam in the right track.
Many thanks in Advance
Many thanks in Advance
Re: QUANTITATIVE SECTION 6 :- ques10 of 20 ID: Q01-86 GGREClub
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26 Sep 2017, 05:51
26 Sep 2017, 05:51
pranab01 wrote:
Hi everybody plz check my reasoning so that I know Iam in the right track.
Many thanks in Advance
Many thanks in Advance
You are definitely right. I've used a similar approach:
\(5\frac{m}{h}=\frac{3m+3m}{1h+Xh}\)
using the definition of average speed. Then, I find Xh, i.e. the time spent in doing the return trip
\(Xh=\frac{1}{5}h=12min\)
That's where 12 appears. But it is the time spent not the speed.
Finally, to get the average speed of the return trip I just have to compute
\(\frac{3m}{\frac{1}{5}h}=3m*5h=15\frac{m}{h}\)
Answer E!
