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There are 500 cars on a sales lot, all of which have either two doors

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There are 500 cars on a sales lot, all of which have either two doors [#permalink]

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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


This is one from a set of 15 challenging problems. To see all 15, as well as the OE for this particular question, see:
Challenging GMAT Math Practice Questions

Mike :-)
[Reveal] Spoiler: OA

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Re: There are 500 cars on a sales lot, all of which have either two doors [#permalink]

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New post 29 Mar 2017, 20:23
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mikemcgarry 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


This is one from a set of 15 challenging problems. To see all 15, as well as the OE for this particular question, see:
Challenging GMAT Math Practice Questions

Mike :-)


Hi,

One method I can suggest is ...

First let's find the max these can be
Say cars with camera is C..
So 18% of C have both C & S...
Now 120 of 4-door cars have C..
Say all 2-door cars have C, that is 165..
So C has to be lesser than or equal to 120+165 or 285

What is the number of 4-door cars with C&S
60% of 18% of C or \(\frac{60*18}{100*100}*C= 0.108*C\)

But 0.108*C has to be an integer
There are three DECIMALS, so it has to multiplied with 3 TENS or 3*2s and 3*5s..
108 is a multiple of 4 so 2*2s are already there so C has to be multiple of (3-2)*2s and 3*5s..
That is 2*5*5*5=250.. so 250 or 250*2=500
Also C cannot be MORE than 285, so ONLY possible value is 250..

Now 0.108*C= 0.108*250=27

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Re: There are 500 cars on a sales lot, all of which have either two doors [#permalink]

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New post 01 Jun 2017, 18:34
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

This is one from a set of 15 challenging problems. To see all 15, as well as the OE for this particular question, see:
Challenging GMAT Math Practice Questions

@Bunuel..Can you pls provide a solution. Not able to understand. Thanks

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Re: There are 500 cars on a sales lot, all of which have either two doors [#permalink]

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New post 16 Jun 2017, 20:00
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Chets25 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

This is one from a set of 15 challenging problems. To see all 15, as well as the OE for this particular question, see:
Challenging GMAT Math Practice Questions

@Bunuel..Can you pls provide a solution. Not able to understand. Thanks


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?
Hi,
Here, In this question he is stating that
18 percent of all cars with Back-up cams have ST .
18%(BackCams) = Back cams + Std. Trans
Now, Of these above 40 % are 2 doors . But, We need for 4-Doors.
hence, 60% of those with Back Cams + Std.Trans are 4-doors.

Now what we need is 4-doors with Back cams + std.Trans
= 60% of (Back Cams + std. Trans)
= 60 % of (18%(BackCams))
= 0.108 * ( Back Cams ) ...........Equation 1

Now In the question we are given that 500 are 4-doors and 165 are 2-door cars.
4- door cars = 500 - 165 = 335 .
Number of 4 - doors with Back Cams + Standard transmission can never be in decimals. Never seen a half built or 0.23 of a car.

Hence,
In equation 1.
0.108 * ( Back Cams ) --- which represents the number of 4-doors + back cams + Std. Trans can never be in decimals.
It is only possible when Back Cams are in the multiples of 250
i.e. Back Cams are 250 , 500, 750 ....
But The maximum back Cams can take should be less than 500 .
Because it is given that. - There are 120 four-door cars that have a back-up camera. that is there are some cars which do not have back up cameras.
hence number of cars with Back Cams = 250

Therefore, Our answer = 0.108 (Back Cams) = 0.108 * 250 .= 27

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There are 500 cars on a sales lot, all of which have either two doors [#permalink]

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New post 17 Jun 2017, 06:42
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mikemcgarry 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



1. Let 120 four-door cars and x two-door cars have a back-up camera
2. 18% of (120+x) have standard transmission
3. 60% of the above are four-door cars with both. i.e., 60/100*18/100* (120+x).
4. Simplifying we get 12.96+0.108x . This number has to be an integer. 0.108*x has to end in 0.04 or x has to end in 3.
5. If we try 13 for x , we get an integer value which is 27.
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Re: There are 500 cars on a sales lot, all of which have either two doors [#permalink]

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New post 06 Aug 2017, 03:31
mikemcgarry 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


This is one from a set of 15 challenging problems. To see all 15, as well as the OE for this particular question, see:
Challenging GMAT Math Practice Questions

Mike :-)


Has to be a max of 285 cars (165 two doors and 120 4 door cars with back up transmission)
Now use options
Option b)
27*100/60=45
45*100/18=250.
Rest all exceeds 285 using similar process.
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There are 500 cars on a sales lot, all of which have either two doors [#permalink]

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New post 25 Aug 2017, 21:09
guhabhishek wrote:

Has to be a max of 285 cars (165 two doors and 120 4 door cars with back up transmission)
Now use options
Option b)
27*100/60=45
45*100/18=250.
Rest all exceeds 285 using similar process.

Agree I think it's the best solution here

max 285 cars 18% of 285 makes 50, 60% of 50 makes 30, so C D E are out, top to bottom analysis
now bottom top, lets take 27 and plug the numbers, 27 is 60% then 18 is 40% together 45 cars which makes 18% of total with back-up camera, 250 cars
if we take 18, then 18 is 60% and 12 is 40% summing comes to 30, 30 is 18% of something...ooops not an integer

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Re: There are 500 cars on a sales lot, all of which have either two doors [#permalink]

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New post 16 Sep 2017, 05:46
Let A be number 0f 2 door cars with back up camera.
120 - number of 4 door cars with back up camera.

Now 0.18 * (A + 120) is number of cars with backup camera and standard transmission.
of this 40% - 2 doors, so remaining 60% - 4 doors

so our answer is 0.6 * 0.18 ( A + 120), Let our answer be x.
=> (3/5)*(9/50) * (A + 120) = x => (27/250) * (A+120) = x
rearranging , A = (250/27) * x - 120,

Now note prime factorization of 250 is ( 2 * 5 ^ 3) and prime factorization of 27 is (3 ^ 3), so GCD of both 250 and 27 is 1,
so to make A an integer, x has to be multiple of 27

Let x = 27, then A = 130
Let x = 54, then A = 380 , but this is not possible, as total of 2 - door cars is 165.
so x has to be 27, which is B

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Re: There are 500 cars on a sales lot, all of which have either two doors   [#permalink] 16 Sep 2017, 05:46
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