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Tough overlapping sets problem

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Tough overlapping sets problem

by Mo2men » Sat Jun 17, 2017 2:05 pm
There are 500 cars on a sales lot, all of which have either two doors or four doors. There are 165 two-door cars on the lot. There are 120 four-door cars that have a back-up camera. Eighteen percent of all the cars with back-up cameras have standard transmission. If 40% of all the cars with both back-up cameras and standard transmission are two-door cars, how many four-door cars have both back-up cameras and standard transmission?

(A) 18
(B) 27
(C) 36
(D) 45
(E) 54

Source: Magoosh

OA: B

Any easy way to solve this problem?
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by GMATGuruNY » Sat Jun 17, 2017 7:11 pm
Mo2men wrote:There are 500 cars on a sales lot, all of which have either two doors or four doors. There are 165 two-door cars on the lot. There are 120 four-door cars that have a back-up camera. Eighteen percent of all the cars with back-up cameras have standard transmission. If 40% of all the cars with both back-up cameras and standard transmission are two-door cars, how many four-door cars have both back-up cameras and standard transmission?

(A) 18
(B) 27
(C) 36
(D) 45
(E) 54
Let BS = cars with both back-up cameras and standard transmission and B = cars with back-up transmission.
We can PLUG IN THE ANSWERS, which represent the number of 4-door cars with both back-up cameras and standard transmission.

D: 45
Since 40% of the cars with both back-up cameras and standard transmission have two doors, the 45 four-door cars with both back-up cameras and standard transmission constitute 60% of BS:
45 = (60/100)(BS)
BS = (100/60)45 = (5/3)45.

Since 18% of the cars with back-up cameras also have standard transmission, the 45(5/3) cars with both features constitute 18% of B:
45(5/3) = (18/100)B
B = (100/18)(5/3) (45) = (50/9)(5/3)(45) = (250/27)(45) =1250/3 ≈ 416.66.
Since the resulting value for B is a non-integer, eliminate D.

Since the number of 4-door cars with back-up cameras = 120, B≈416 implies that the number of 2-door cars with back-up cameras ≈ 416-120 = 296.
The value in red is too great, given that there are only 165 2-door cars on the lot.
Implication:
The correct answer choice must be SMALLER.
Eliminate E.

Notice the following:
The portion in blue implies that B = (250/27)(answer choice).
For the portion in blue to yield an integer value for B, the correct answer choice must be a MULTIPLE OF 27.
Eliminate A and C.

The correct answer is B.
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by DavidG@VeritasPrep » Sun Jun 18, 2017 9:47 am
Mo2men wrote:There are 500 cars on a sales lot, all of which have either two doors or four doors. There are 165 two-door cars on the lot. There are 120 four-door cars that have a back-up camera. Eighteen percent of all the cars with back-up cameras have standard transmission. If 40% of all the cars with both back-up cameras and standard transmission are two-door cars, how many four-door cars have both back-up cameras and standard transmission?

(A) 18
(B) 27
(C) 36
(D) 45
(E) 54

Source: Magoosh

OA: B

Any easy way to solve this problem?
I thought about it similarly:

Call the number of two-door cars with a back-up camera 'c.'
We know 120 four-door cars have back-up cameras, so the total number of cars with a back-up camera: c + 120

18%, or 9/50 of those have standard transmission, so (9/50)(c + 120) have both features
If 40% of the cars with both features are two-door cars, then 60%, or 3/5 are four-door cars. Four-door cars with both features =
(3/5)(9/50)(c + 120) = (27/250)(c + 120) = 27 * (1/250) * (c + 120) This is the value we want.

We know it's a multiple of 27, so that leaves B or E. If we set 27 * (1/250) * (c + 120) to 54, we get c = 380. Clearly too big, as there are 165 two-door cars total. That leaves B
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by Matt@VeritasPrep » Thu Jun 22, 2017 9:54 pm
Let's call the answer x.

We know that x = 60% of all backup+standard.

We know that backup+standard = 18% of (120 + y), where y is the number of two door cars with backup cameras.

So x = 60% of 18% of (120 + y), or 10.8% of (120 + y), or 12.96 + .108y.

y is at most 165, so our answer must be A or B.

If we try A, we get .108y = 5.04, not an integer.

If we try B, we get .108y = 14.04, an integer!

So B's our pick, and we're set.
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