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A dealer originally bought 15 identical cars. If each car was sold for

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A dealer originally bought 15 identical cars. If each car was sold for  [#permalink]

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New post 05 Jan 2018, 01:18
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Question Stats:

74% (01:43) correct 26% (01:53) wrong based on 23 sessions

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A dealer originally bought 15 identical cars. If each car was sold for q dollars, which was 60 percent above the original cost per car, then in terms of q, what was the total cost of the cars, in dollars, to the dealer?

A. q/250
B. 5q/8
C. 75q/8
D. 9q/10
E. 24q

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A dealer originally bought 15 identical cars. If each car was sold for  [#permalink]

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New post 05 Jan 2018, 01:34
Bunuel wrote:
A dealer originally bought 15 identical cars. If each car was sold for q dollars, which was 60 percent above the original cost per car, then in terms of q, what was the total cost of the cars, in dollars, to the dealer?

A. q/250
B. 5q/8
C. 75q/8
D. 9q/10
E. 24q


Since the car was sold for q dollars which was 1.6 times the cost, the cost of the car would be \(\frac{q}{1.6}\)

Since he bought 15 identical cars, the cost of the cars would be \(15*\frac{q}{1.6} = \frac{150q}{16} = \frac{75q}{8}\)(Option C)
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Re: A dealer originally bought 15 identical cars. If each car was sold for  [#permalink]

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New post 05 Jan 2018, 09:48
Bunuel wrote:
A dealer originally bought 15 identical cars. If each car was sold for q dollars, which was 60 percent above the original cost per car, then in terms of q, what was the total cost of the cars, in dollars, to the dealer?

A. q/250
B. 5q/8
C. 75q/8
D. 9q/10
E. 24q

Another approach: assign values.
When you see fractions in answer choices, choose small values.
Almost always, if chosen values are small, the fractional answers can be assessed quickly.

Sell price is 60 percent above cost. SP = 1.6 * cost price
Use two integers. 5 * 16 = 80, so 5 * 1.6 = 8
You could also use cost = $10 and SP = $16

Let cost = $5 per car
Then Sell Price = $5 * 1.6 = $8

So q = $8
Total cost = ($5 * 15) = $75

With q = 8, find the answer that yields $75

Eliminate A, B, and D immediately.
Their numerators and denominators make their results much too small:
\(\frac{8}{250}\), \(\frac{40}{8}\), and \(\frac{72}{10}\), respectively

Eliminate E, too.
24 * 8 will not yield an answer whose units digit is 75
(And 24 * 8 > 160, more than twice what we need)

Check C) 75q/8

\(($75 * 8) = ($150 * 4) = $600\)
(double one factor and halve the other)

\(\frac{$600}{8}= \frac{$300}{4} = \frac{$150}{2}\)= $75. Correct.

Answer C
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Re: A dealer originally bought 15 identical cars. If each car was sold for &nbs [#permalink] 05 Jan 2018, 09:48
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