500 ps

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dunkin77: Master | Next Rank: 500 Posts; Posts: 269; Joined: Sun Apr 01, 2007 5:41 am

by dunkin77 » Tue Apr 03, 2007 10:05 pm

Hi,

Somehow my answer was 110...and which is not even in the choices... The correct answer is D), can you help? thank you!

. If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160

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Source: Beat The GMAT — Problem Solving | Tue Apr 03, 2007 10:05 pm

ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

Re: 500 ps

by ajith » Tue Apr 03, 2007 10:13 pm

Let the number of hours the motorist drives be x
Let his speed be y

Now in normal course he travels xy miles

(x+1)(y+5)=xy+70

=> 5x+y+5 = 70
5x +y = 65

Now if he travels 2 hours longer and 10 miles faster

The distance travelled = (x+2)(y+10)

= xy + 10x+2y+20 = xy+130+20 =xy+150

ie he travels 150 miles extra

Always borrow money from a pessimist, he doesn't expect to be paid back.

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jayhawk2001: Community Manager; Posts: 789; Joined: Sun Jan 28, 2007 3:51 pmFollowed by: 1 member; Location: Silicon valley, California; Thanked: 30 times

by jayhawk2001 » Tue Apr 03, 2007 10:14 pm

Let x be the time and y be the speed

We have

(x+1)*(y+5) - xy = 70
So, 5x + y = 65

We are asked to find

(x+2)*(y+10) - xy
= 10x + 2y + 20
= 2 (5x + y) + 20
= 2 * 65 + 20
= 150

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dunkin77: Master | Next Rank: 500 Posts; Posts: 269; Joined: Sun Apr 01, 2007 5:41 am

by dunkin77 » Tue Apr 03, 2007 10:32 pm

I was right until 5x +y = 65 but did not think of assigning 65 into the next step.

Thank you for your help

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Cybermusings: Legendary Member; Posts: 559; Joined: Tue Mar 27, 2007 1:29 amFollowed by: 2 members; Thanked: 5 times

by Cybermusings » Wed Apr 04, 2007 3:05 am

Distance = Speed * Time. Let speed be x and time taken be y

70 = (x + 5) (y + 1) - xy

xy - x + 5y + 5 - xy = 70

5y + x = 65

New Distance = (x + 10) (y + 2)
= xy + 2x + 10y + 20

= xy + 2 (x + 5y) + 20

= xy + 2 (65) + 20

= xy + 130 + 20 = xy + 150

Hence 150 miles

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