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Re: M18-26 [#permalink]
so the question asks to find an APPROXIMATE value? You could have mentioned that in question
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Re: M18-26 [#permalink]
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]
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]
Hi Bunuel

can you please help me in understanding why it it out of 150 not 200?
what percent of the cars without air conditioners had traction control systems?

AC AC'
TC 6 57 63
TC' 44 93 137
50 150 200

why it is not (57/200)*100?
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Re: M18-26 [#permalink]
Expert Reply
ManoramaS 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



Hi Bunuel

can you please help me in understanding why it it out of 150 not 200?
what percent of the cars without air conditioners had traction control systems?

AC AC'
TC 6 57 63
TC' 44 93 137
50 150 200

why it is not (57/200)*100?


The question asks: what percent of the cars without air conditioners had traction control systems?

So, we need (cars without air conditioners but with air conditioners)/(cars without air conditioners) = 57/150.
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Re: M18-26 [#permalink]
Expert Reply
I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M18-26 [#permalink]
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