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M18-26

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M18-26  [#permalink]

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New post 16 Sep 2014, 01:04
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  45% (medium)

Question Stats:

70% (02:12) correct 30% (02:33) wrong based on 240 sessions

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Among the 200 cars displayed at an auto show, 50 had built-in air conditioners and 63 had traction control systems. If 12% of cars with air conditioners had traction control systems, what percent of the cars without air conditioners had traction control systems?

A. 35
B. 36
C. 38
D. 40
E. 45

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Re M18-26  [#permalink]

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New post 16 Sep 2014, 01:04
Official Solution:

Among the 200 cars displayed at an auto show, 50 had built-in air conditioners and 63 had traction control systems. If 12% of cars with air conditioners had traction control systems, what percent of the cars without air conditioners had traction control systems?

A. 35
B. 36
C. 38
D. 40
E. 45

Of the 50 cars with air conditioners, 6 (\(12\%\) of \(50 = \frac{50}{100}*12 = 6\)) had traction control systems. This means that \(63 - 6 = 57\) cars had traction control systems but did not have air conditioners. As there were \(200 - 50 = 150\) cars without air conditioners, the answer to the question is \(\frac{57}{150}*100\% = 38\%\).

Answer: C
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Re: M18-26  [#permalink]

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New post 13 Mar 2016, 22:47
Bunuel wrote:
Official Solution:

Among the 200 cars displayed at an auto show, 50 had built-in air conditioners and 63 had traction control systems. If 12% of cars with air conditioners had traction control systems, what percent of the cars without air conditioners had traction control systems?

A. 35
B. 36
C. 38
D. 40
E. 45

Of the 50 cars with air conditioners, 6 (\(12\%\) of \(50 = \frac{50}{100}*12 = 6\)) had traction control systems. This means that \(63 - 6 = 57\) cars had traction control systems but did not have air conditioners. As there were \(200 - 50 = 150\) cars without air conditioners, the answer to the question is \(\frac{57}{150}*100\% = 38\%\).

Answer: C


Shouldnt the question ask "what percent of the cars without air conditioners only had traction control systems"?
I did 63/150 as a result.
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Re: M18-26  [#permalink]

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New post 13 Mar 2016, 22:53
sinhap07 wrote:
Bunuel wrote:
Official Solution:

Among the 200 cars displayed at an auto show, 50 had built-in air conditioners and 63 had traction control systems. If 12% of cars with air conditioners had traction control systems, what percent of the cars without air conditioners had traction control systems?

A. 35
B. 36
C. 38
D. 40
E. 45

Of the 50 cars with air conditioners, 6 (\(12\%\) of \(50 = \frac{50}{100}*12 = 6\)) had traction control systems. This means that \(63 - 6 = 57\) cars had traction control systems but did not have air conditioners. As there were \(200 - 50 = 150\) cars without air conditioners, the answer to the question is \(\frac{57}{150}*100\% = 38\%\).

Answer: C


Shouldnt the question ask "what percent of the cars without air conditioners only had traction control systems"?
I did 63/150 as a result.


Hi sinhap07,

Its very important not to miss out on important terms..
Here it says
what percent of the cars withoutair conditioners had traction control systems?
the usage of without means only traction control systems..

we are only talking of only two thing , so without one of those, would mean only ONE ..
Hope it helps
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Re: M18-26  [#permalink]

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New post 06 Jul 2017, 06:11
Bunuel wrote:
Among the 200 cars displayed at an auto show, 50 had built-in air conditioners and 63 had traction control systems. If 12% of cars with air conditioners had traction control systems, what percent of the cars without air conditioners had traction control systems?

A. 35
B. 36
C. 38
D. 40
E. 45


Total number of cars \(= 200\)

Number of Cars with built-in air conditioners \(= 50\)

Total number of cars without built-in air conditioners \(= 200 - 50 = 150\)

Number of Cars that had traction control systems \(= 63\)

\(12\)% of cars with air conditioners had traction control systems \(=12\)% of \(50 = \frac{12}{100} * 50 = 6\)

Therefore, \(6\) cars with built-in air conditioners had traction control systems.

Number of cars with traction control systems but without built-in air conditioners \(= 63 - 6 = 57\)

Required Percentage = (Number of cars with traction control systems but without built-in air conditioners / Total number of cars without built-in air conditioners) \(* 100\)

Required Percentage \(= \frac{57}{150}*100 = 38\)%

Answer (C)...
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Re: M18-26  [#permalink]

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New post 06 Jul 2017, 09:03
so the question asks to find an APPROXIMATE value? You could have mentioned that in question
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Re: M18-26  [#permalink]

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New post 06 Jul 2017, 09:25
rma26 wrote:
so the question asks to find an APPROXIMATE value? You could have mentioned that in question


It is not approximate value.

\(\frac{57}{150}*100\) is exactly \(= 38\)%.

Please check the calculation.
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Re: M18-26  [#permalink]

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New post 06 Jul 2017, 13:48
sashiim20 wrote:
rma26 wrote:
so the question asks to find an APPROXIMATE value? You could have mentioned that in question


It is not approximate value.

\(\frac{57}{150}*100\) is exactly \(= 38\)%.

Please check the calculation.


Do you have a quick way of getting the percentage from 57/150? I got that part relatively quickly but I had trouble getting 38.

Posted from my mobile device
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Re: M18-26  [#permalink]

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New post 06 Jul 2017, 13:58
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gmathopeful19 wrote:
sashiim20 wrote:
rma26 wrote:
so the question asks to find an APPROXIMATE value? You could have mentioned that in question


It is not approximate value.

\(\frac{57}{150}*100\) is exactly \(= 38\)%.

Please check the calculation.


Do you have a quick way of getting the percentage from 57/150? I got that part relatively quickly but I had trouble getting 38.

Posted from my mobile device


Reduce by 3 to get 19/50, which is the same as 38/100.
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Re: M18-26  [#permalink]

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New post 08 Jul 2017, 11:37
This is a beginner level question, but how did you calculate 57/150?
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Re: M18-26  [#permalink]

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New post 08 Jul 2017, 11:39
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M18-26  [#permalink]

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New post 09 Jul 2017, 02:13
bdb12 wrote:
This is a beginner level question, but how did you calculate 57/150?



There are multiple ways to do that:

\(\frac{57}{150}\)= \(\frac{57}{(50*3)}\)=\(\frac{19}{50}\).........multiply by 2/2 to convert denominator to 100

Then the result would be \(\frac{38}{100}\)


Another way

\(\frac{57}{150}\) = \(\frac{57}{(10 *15)}\)........divide by 3

\(\frac{19}{(10*5)}\) 19 is almost equal to 20

\(\frac{20}{(10*5)}\) = \(\frac{2}{5}\)= 0.4 So it should be slightly less than 0.4 the 38 is our target.

I hope it helps
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M18-26   [#permalink] 09 Jul 2017, 02:13
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