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Statistics and Sets Problems

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Statistics and Sets Problems

by swerve » Thu Feb 13, 2020 7:33 am
Quick Sell Outlet sold a total of \(40\) televisions, each of which was either a Model \(P\) television or a Model \(Q\) television. Each Model \(P\) television sold for \(\$p\) and each Model \(Q\) sold for \(\$q\). The average (arithmetic mean) selling price of the \(40\) televisions was \(\$141\). How many of the \(40\) televisions were Model \(P\) televisions?

1) The Model \(P\) televisions sold for \(\$30\) less than the Model \(Q\) televisions.
2) Either \(p=120\) or \(q=120\)

The OA is C

Source: GMAT Prep
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Source: — Data Sufficiency |

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Re: Statistics and Sets Problems

by Jay@ManhattanReview » Thu Feb 13, 2020 10:33 pm
swerve wrote: ↑
Thu Feb 13, 2020 7:33 am
 Quick Sell Outlet sold a total of \(40\) televisions, each of which was either a Model \(P\) television or a Model \(Q\) television. Each Model \(P\) television sold for \(\$p\) and each Model \(Q\) sold for \(\$q\). The average (arithmetic mean) selling price of the \(40\) televisions was \(\$141\). How many of the \(40\) televisions were Model \(P\) televisions?

1) The Model \(P\) televisions sold for \(\$30\) less than the Model \(Q\) televisions.
2) Either \(p=120\) or \(q=120\)

The OA is C

Source: GMAT Prep
Say x no. of P models were sold; thus, (40 – x) no. of Q models were sold.

=> px + q(40 – x) = 141*40

x(p – q) + 40q = 141*40

Since we have three variables, none of the statements alone can work.

(1) and (2) together

From (1), we know that q > p, and it is given that the average price of the 40 TVs = $141; thus, p must be less than $141 and q must be greater than $141.

=> only p can be equal to $120 so that q = 120 + 30 = $150.

Again from x(p – q) + 40q = 141*40, we have

x(p – q) + 40 q = 141*40 => x(120 – 150) + 40*150 = 141*40 => x = 12. Sufficient.

The correct answer: C

Hope this helps!

-Jay
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