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# Tom drove from X to Y and then from Y to X and took different routes b

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Manager
Joined: 17 May 2017
Posts: 130
GPA: 3
Tom drove from X to Y and then from Y to X and took different routes b  [#permalink]

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03 Jul 2017, 02:31
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45% (medium)

Question Stats:

65% (01:40) correct 35% (01:48) wrong based on 140 sessions

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Tom drove from X to Y and then from Y to X and took different routes back and forth. It took him 40 miles per hour to move from X to Y. Meanwhile, it took him 50 miles per hour to come from Y to X. What was the average speed for his entire travel?

1) The distance from the return trip was 4/5 of the entire distance.
2) The distance from the return trip was 160 miles.
Manager
Joined: 17 May 2017
Posts: 130
GPA: 3
Re: Tom drove from X to Y and then from Y to X and took different routes b  [#permalink]

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03 Jul 2017, 02:41
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 52130
Re: Tom drove from X to Y and then from Y to X and took different routes b  [#permalink]

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03 Jul 2017, 02:59
2
Tom drove from X to Y and then from Y to X and took different routes back and forth. It took him 40 miles per hour to move from X to Y. Meanwhile, it took him 50 miles per hour to come from Y to X. What was the average speed for his entire travel?

$$Average \ speed = \frac{(Total \ distance)}{(Total \ time)}$$

(1) The distance from the return trip was 4/5 of the entire distance. Say the entire distance is d, then the distance from Y to X is 4d/5 and the distance from X to Y is d/5.

$$Average \ speed = \frac{(Total \ distance)}{(Total \ time)}=\frac{d}{\frac{(\frac{d}{5})}{40}+\frac{(\frac{4d}{5})}{50}}=\frac{1}{\frac{(\frac{1}{5})}{40}+\frac{(\frac{4}{5})}{50}}$$ (by reducing d). Sufficient.

(2) The distance from the return trip was 160 miles. Not sufficient.

Hope it's clear.
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Director
Joined: 27 May 2012
Posts: 665
Re: Tom drove from X to Y and then from Y to X and took different routes b  [#permalink]

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01 Aug 2018, 04:36
1
haardiksharma wrote:
Tom drove from X to Y and then from Y to X and took different routes back and forth. It took him 40 miles per hour to move from X to Y. Meanwhile, it took him 50 miles per hour to come from Y to X. What was the average speed for his entire travel?

1) The distance from the return trip was 4/5 of the entire distance.
2) The distance from the return trip was 160 miles.

Dear Moderator,
Found this DS question while going through the PS problems in Rate/Distance , Hope you will do the needful. Thank you.
_________________

- Stne

Math Expert
Joined: 02 Sep 2009
Posts: 52130
Re: Tom drove from X to Y and then from Y to X and took different routes b  [#permalink]

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01 Aug 2018, 05:09
stne wrote:
haardiksharma wrote:
Tom drove from X to Y and then from Y to X and took different routes back and forth. It took him 40 miles per hour to move from X to Y. Meanwhile, it took him 50 miles per hour to come from Y to X. What was the average speed for his entire travel?

1) The distance from the return trip was 4/5 of the entire distance.
2) The distance from the return trip was 160 miles.

Dear Moderator,
Found this DS question while going through the PS problems in Rate/Distance , Hope you will do the needful. Thank you.

Moved to DS forum. Thank you.
_________________
Intern
Joined: 03 Sep 2017
Posts: 10
Location: India
GMAT 1: 700 Q44 V42
GPA: 3.4
Re: Tom drove from X to Y and then from Y to X and took different routes b  [#permalink]

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01 Aug 2018, 05:54
Bunuel wrote:
Tom drove from X to Y and then from Y to X and took different routes back and forth. It took him 40 miles per hour to move from X to Y. Meanwhile, it took him 50 miles per hour to come from Y to X. What was the average speed for his entire travel?

$$Average \ speed = \frac{(Total \ distance)}{(Total \ time)}$$

(1) The distance from the return trip was 4/5 of the entire distance. Say the entire distance is d, then the distance from Y to X is 4d/5 and the distance from X to Y is d/5.

$$Average \ speed = \frac{(Total \ distance)}{(Total \ time)}=\frac{d}{\frac{(\frac{d}{5})}{40}+\frac{(\frac{4d}{5})}{50}}=\frac{1}{\frac{(\frac{1}{5})}{40}+\frac{(\frac{4}{5})}{50}}$$ (by reducing d). Sufficient.

(2) The distance from the return trip was 160 miles. Not sufficient.

Hope it's clear.

Hello, I was wondering about this part - "the distance from Y to X is 4d/5 and the distance from X to Y is d/5."

I understood why the distance from Y to X is 4d/5, but how did we conclude that the distance from X to Y is d/5?

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 52130
Re: Tom drove from X to Y and then from Y to X and took different routes b  [#permalink]

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01 Aug 2018, 06:05
vedanths wrote:
Bunuel wrote:
Tom drove from X to Y and then from Y to X and took different routes back and forth. It took him 40 miles per hour to move from X to Y. Meanwhile, it took him 50 miles per hour to come from Y to X. What was the average speed for his entire travel?

$$Average \ speed = \frac{(Total \ distance)}{(Total \ time)}$$

(1) The distance from the return trip was 4/5 of the entire distance. Say the entire distance is d, then the distance from Y to X is 4d/5 and the distance from X to Y is d/5.

$$Average \ speed = \frac{(Total \ distance)}{(Total \ time)}=\frac{d}{\frac{(\frac{d}{5})}{40}+\frac{(\frac{4d}{5})}{50}}=\frac{1}{\frac{(\frac{1}{5})}{40}+\frac{(\frac{4}{5})}{50}}$$ (by reducing d). Sufficient.

(2) The distance from the return trip was 160 miles. Not sufficient.

Hope it's clear.

Hello, I was wondering about this part - "the distance from Y to X is 4d/5 and the distance from X to Y is d/5."

I understood why the distance from Y to X is 4d/5, but how did we conclude that the distance from X to Y is d/5?

Posted from my mobile device

(1) The distance from the return trip was 4/5 of the entire distance.

So, if the total distance is d, then the distance from Y yo X is 4/5*d and the distance from X to Y is d - 4/5*d = d/5.
_________________
Re: Tom drove from X to Y and then from Y to X and took different routes b &nbs [#permalink] 01 Aug 2018, 06:05
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