Rate Problem 2

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phoenix9801: Master | Next Rank: 500 Posts; Posts: 190; Joined: Thu Jan 14, 2010 8:29 pm; Thanked: 2 times

by phoenix9801 » Thu May 31, 2012 4:29 pm

Hose 1 takes twice as long as hose 2 to fill an empty pool. Working together at their respective constant rates, the hoses can fill the pool in 6 hours. How many hours would it take the slower hose to fill the tank working alone?

A) 18
B) 15
C) 12
D) 9
E) 8

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Source: Beat The GMAT — Problem Solving | Thu May 31, 2012 4:29 pm

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 » Thu May 31, 2012 4:56 pm

phoenix9801 wrote:Hose 1 takes twice as long as hose 2 to fill an empty pool. Working together at their respective constant rates, the hoses can fill the pool in 6 hours. How many hours would it take the slower hose to fill the tank working alone?

A) 18
B) 15
C) 12
D) 9
E) 8

Time and rate are RECIPROCALS.
If H1 takes TWICE AS LONG as H2, then H1 works HALF AS FAST as H2.

Let the rate for H1 = 1 unit per hour and the rate for H2 = 2 units per hour.
Their combined rate = 1+2 = 3 units per hour.
When they work together for 6 hours, the amount of work produced = r*t = 3*6 = 18 units. This is the value of the pool.

Time for H1 alone to fill the pool = w/r = 18/1 = 18 hours.

The correct answer is A.

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GMAT Kolaveri: Master | Next Rank: 500 Posts; Posts: 287; Joined: Fri Mar 23, 2012 12:33 amFollowed by: 6 members; Location: Pune,India; Thanked: 60 times

by GMAT Kolaveri » Thu May 31, 2012 6:50 pm

Backsolving:

Assume slower one to be =18
Faster one = 9 (since slower take twice the time)

1/18 + 1/9
= 1/18 +2/18
=3/18

Total time taken(combined) = 18/3 = 6 (given in the question) Hence slower one =18

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