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

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

B
_________________

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

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

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.
_________________

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

Let the number of two-door (2D) cars with back-up cameras (BUC) be 'x'. We already know that the number of 4D cars with BUC is 120. Therefore, the total number of cars with BUC = (x+120) of which 18% (9/50th) have standard transmission (ST).
Therefore, the number of cars with both BUC and ST = (9/50)(x+120) of which 40% (2/5th) are 2D.
So, obviously, the number of 4D cars with both BUC and ST = (3/5)*{9/50)(x+120)} = 27(x/250+120/250) = 27{(x+120)/250}. Obviously, this number cannot be a fraction which means (x+120)/250 cannot be a fraction.
Therefore, (x+120) muct be divisible by 250 and the only value of 'x' which makes this possible is 130 (the next possible value is 130+250=380 which is not possible since the total number of 2D cars is 165). So:
The number of 4D cars with both BUC and ST = 27{(130+120)/250} = 27(250/250) = 27.
ANS:B

a+b+c+d+e+f+g+h=500
a+b+c+d=165 Therefore e+f+g+h=335
e+g=120 Therefore f+h=215
18%(a+e+c+g)=(a+e) Therefore 82%(a+e)=18%(c+g)
40%(a+e)=a Therefore 40%e=60%a Therefore a/e = 2/3
e=??

From all this now micro equations are
1. a+b+c+d=165
2. e+g=120
3. f+h=215
4. 82%(a+e)=18%(c+g)
5. a/e = 2/3
since in equation 4, a can be replaced by e, g can be replaced by e from equation 2, only left is c which cannot be replaced. Therefore this has to be solved by multiple cases method

The equation 4 will become 82%(2e/3+e)=18%*(c+120-e)
or (c+120)/e=232/27
Therefore e=27k and c=232k-120
k=1 therefore e=27 c=110 therefore possible solution
k=2 therefore e=53 c=344 which is not possible. Therefore answer is e=27

Let the number of cars with back up cameras = X

18% of X have standard transmission

40% of these 18% of X are two door, which means that 60% of 18% of X are four door.

The question wants to know what 60% of 18% of X is.

60% of 18% = .6*.18 = .108.

.108 = 108/1000=54/500=

27/250.

X*27/250 is the answer and has to be an integer.

If X=250 then the answer could be 27 and if X=500 the answer could be 54, both among the answer choices.

But if X=500, that would mean that all cars have back up cameras. Since we know there are 165 two door cars, there are 335 four door cars. But only 120 of these have cameras, so 215 have no camera, which violates the premise above, so X can't be 500 but instead 250 and the answer 27.

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Hello mikemcgarry!

Can I solve this one using a double matrix?

­I think x will be 130 because if we consider x=13 then 13*0.108=1.404 and 12.96+1.404 is not equal to 27. However, if we take x=130 i.e. 2-door cars with cameras, it satisfies because anyways total 2-door cars are 165 and 130 is less than 165. Please correct me if I am wrong.