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
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
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..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
