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algebraic expressions

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by rahul.s » Sun Feb 07, 2010 6:17 am
in a used car lot, there are 3 times as many red cars as green cars. if tomorrow 12 green cars are sold and 3 red cars are added, then there will be 6 times as many red cars as green cars. how many green cars are currently in the lot?

there are no options for this problem.

OA: 25
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by papgust » Sun Feb 07, 2010 6:27 am
R = 3G
R-3G = 0 ....... (1)

(R+3)=6*(G-12)

R+3 = 6G - 72

R-6G = -75 ..... (2)

Using (1) and (2),
G = 25
R = 75
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by bhumika.k.shah » Sun Feb 07, 2010 6:34 am
But in the real gmat there would be options and so by back solving u r getting the answer.

G = 25 . Therefore , R = 75

Now 25-12 ( green cars sold) = 13
75+3 = 78 (Red cars added) = 78

Therefore, red cars are six times the green cars.

Hence Answer is 25

Why bother much about the equations and blah when back solving takes less time. ?

rahul.s wrote:
in a used car lot, there are 3 times as many red cars as green cars. if tomorrow 12 green cars are sold and 3 red cars are added, then there will be 6 times as many red cars as green cars. how many green cars are currently in the lot?

there are no options for this problem.

OA: 25
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by shashank.ism » Sun Feb 07, 2010 6:35 am
rahul.s wrote:in a used car lot, there are 3 times as many red cars as green cars. if tomorrow 12 green cars are sold and 3 red cars are added, then there will be 6 times as many red cars as green cars. how many green cars are currently in the lot?

there are no options for this problem.

OA: 25


Let red cars be denoted by R and Green cars by G.
R= 3G = 0 ....... (i)

(R+3)=6(G-12) --> putting r from (i) we get , 3G+3 = 6(G-12) --> G+1 = 2(G-12) --> G=25. Its quite simple
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by bhumika.k.shah » Sun Feb 07, 2010 6:46 am
Papgust ,
i did it the other way. I thought its 3R = G . But still landed up with the same answer . Whats the flaw in the way i solved the sum ?

papgust wrote:R = 3G
R-3G = 0 ....... (1)

(R+3)=6*(G-12)

R+3 = 6G - 72

R-6G = -75 ..... (2)

Using (1) and (2),
G = 25
R = 75
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by shashank.ism » Sun Feb 07, 2010 6:52 am
bhumika.k.shah wrote:Papgust ,
i did it the other way. I thought its 3R = G . But still landed up with the same answer . Whats the flaw in the way i solved the sum ?

papgust wrote:R = 3G
R-3G = 0 ....... (1)

(R+3)=6*(G-12)

R+3 = 6G - 72

R-6G = -75 ..... (2)

Using (1) and (2),
G = 25
R = 75
Bhumika I think there is no problem in solving back way ... but if for a simple problem u go on putting values and unluckily you didn't struck correct one in first 2 attempts , you already lost your precious minutes. better to go by exact process if solution seems to be small..

If just by seeing answer you feel that this could be the answer then you can give a try for sure...

Its just matter of your own choice and how fats are you at doing things..
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