a PS from Gmat

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diebeatsthegmat: Legendary Member; Posts: 1119; Joined: Fri May 07, 2010 8:50 amFollowed by: 3 members; Thanked: 29 times

by diebeatsthegmat » Sat Sep 03, 2011 7:52 pm

Anand starts from a point P towards point Q, where PQ = 90 km. After 1 hour, Ram starts from P and catches up with Anand after 2 more hours. After meeting they continue to travel towards Q. On reaching Q, Ram reverses his direction and meets Anand 6 hours after the first meeting. Find Anand's speed.

(A). (45/7) kmph
(B). (60/7) kmph
(C). (40/7) kmph
(D). (30/7) kmph
(E). (65/7) kmph

i dont know where i got wrong, but according to my logic. R goes 2 hours to catch A so it means 90/2=45 kms so A goes 45 kms in 3 hour becuase A went 1 hour earlier than R
but in 6 hours R met a again it means R went all the S in 4 hours and came back in 2 hours to meet A so A goes totally 2+1+6=9 hours thus his average is 90/9=10???
ii just odnt understand....but am i right or i was wrong?

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Source: Beat The GMAT — Problem Solving | Sat Sep 03, 2011 7:52 pm

jbivins: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Tue Aug 30, 2011 7:55 pm; Thanked: 1 times

by jbivins » Sat Sep 03, 2011 9:52 pm

This one took a bit so I def would have had to skip it on the GMAT or guess

but I solved it like this..

RR=rate of R
RA=rate of A

so we know that in 8 hours (2 hours to catch up plus 6 to meet back up)
R went 90 + (90-distance A already went)

distance A went = rate * time =RA*9 (he left an hour before R)

so

RR*8=90+(90-RA*9)

now we know from the first statement (R caught up to A in 2 hours when A left an hour before) that RR=(3/2)RA

final equation
[spoiler]
(3/2)RA*8=90+(90-RA*9)
12RA=180-9RA
21RA=180
RA=180/21=60/21 so B????[/spoiler]

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sun Sep 04, 2011 3:02 am

diebeatsthegmat wrote:Anand starts from a point P towards point Q, where PQ = 90 km. After 1 hour, Ram starts from P and catches up with Anand after 2 more hours. After meeting they continue to travel towards Q. On reaching Q, Ram reverses his direction and meets Anand 6 hours after the first meeting. Find Anand's speed.

(A). (45/7) kmph
(B). (60/7) kmph
(C). (40/7) kmph
(D). (30/7) kmph
(E). (65/7) kmph

i dont know where i got wrong, but according to my logic. R goes 2 hours to catch A so it means 90/2=45 kms so A goes 45 kms in 3 hour becuase A went 1 hour earlier than R
but in 6 hours R met a again it means R went all the S in 4 hours and came back in 2 hours to meet A so A goes totally 2+1+6=9 hours thus his average is 90/9=10???
ii just odnt understand....but am i right or i was wrong?

We can plug in the answers, which represent Anand's speed.
The distance is a multiple of 10.
The times given in the problem are 2,3, and 6.
The numerator of the correct answer likely is a multiple of 10, 2, 3 and 6.

Answer choice B: 60/7.
Distance traveled by Anand in 3 hours = 180/7 kilometers.
Since Ram catches up in 2 hours, rate for Anand = d/t = (180/7)/2 = 90/7 kilometers per hour.

6 hours later:
Distance traveled by Anand = r*t = (60/7)*6 = 360/7 kilometers.
Total distance traveled by Anand = 180/7 + 360/7 = 540/7 kilometers.
Remaining distance for Anand = 90 - 540/7 = 90/7 kilometers.

Distance traveled by Ram = r*t = (90/7)*6 = 540/7 kilometers.
Total distance traveled by Ram = 180/7 + 540/7 = 720/7 kilometers.
Distance traveled in the reverse direction = 720/7 - 90 = 90/7 kilometers.

Success!
Ram and Anand are the same distance from point Q.

The correct answer is B.

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DonPaw: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Sat May 01, 2010 7:57 pm

by DonPaw » Sun Sep 04, 2011 10:31 am

I solved it following way :

total distance traveled by both Anand and Ram = 180
9 A + 8 R = 180 -- Eq 1

relation between R & A,

3 A = 2 R -- Eq 2

After substitution, we will get A = 60/7 i.e. B

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mehrasa: Master | Next Rank: 500 Posts; Posts: 279; Joined: Fri Nov 05, 2010 5:43 pmFollowed by: 1 member; Thanked: 15 times

by mehrasa » Tue Sep 06, 2011 12:37 am

could someone please give a straightforward solution which can be underestand for this prob? thnx

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