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

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M7MBA: Moderator; Posts: 2058; Joined: Sun Oct 29, 2017 4:24 amFollowed by: 5 members; Thanked: 1 times

by M7MBA » Wed May 20, 2020 7:24 am

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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?

A. 15
B. 20
C. 25
D. 50
E. 75

[spoiler]OA=C[/spoiler]

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Source: Beat The GMAT — Problem Solving | Wed May 20, 2020 7:24 am

anonylfc: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Wed May 20, 2020 9:34 pm

Re: 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 ca

by anonylfc » Fri May 22, 2020 8:21 am

Lets assume green cars to be 'x'.
Lets assume red cars to be 'y'

According to first statement => y=3x

Now if 3 red cars are added => y+3
If 12 green cars are sold => x-12
Now the relationship between them ?
=> y+3 = 6 (x-12)

Now we know that y=3x. Hence,
3x+3 = 6(x-12)
3x+3=6x-72
3+72=6x-3x
75=3x
25=x

Which is the answer. Hope this was helpful ! Please upvote if it was !

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8246; Joined: Sat Apr 25, 2015 10:56 amFollowed by: 29 members; Location: Los Angeles, CA; Thanked: 43 times

Re: 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 ca

by Scott@TargetTestPrep » Mon May 25, 2020 11:11 am

M7MBA wrote: ↑
Wed May 20, 2020 7:24 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?

A. 15
B. 20
C. 25
D. 50
E. 75

[spoiler]OA=C[/spoiler]

Solution:

Let r = the number of red cars and g = the number of green cars currently in the lot. We can create the equations:

r = 3g

and

r + 3 = 6(g - 12)

Substituting 3g for r in the second equation, we have:

3g + 3 = 6(g - 12)

3g + 3 = 6g - 72

75 = 3g

25 = g

Answer: C

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