ISEE Mathematics Achievement Question 50: Answer and Explanation

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Question: 50

8. A car dealer sold three different makes of cars. The price of the first make was $4200, the second $4800, and the third $5400. The total sales were $360,000. If three times as many of the third car were sold as the first, and twice as many of the second make were sold than the first, how many cars of the third make were sold

  • A. 15
  • B. 24
  • C. 36
  • D. It cannot be determined by the information given.

Correct Answer: C

Explanation:

The correct answer is (C). Solve this problem as you would any mixture-value problem. The numbers of cars sold are all related to the number of those sold for $4200. Call the number of $4200 cars sold x. Then, the number of $5400 cars sold is 3x, and the number of $4800 cars is 2x.

The value of $4200 cars sold is $4200 o x.

The value of $4800 cars sold is $4800 o 2x.

The value of $5400 cars sold is $5400 o 3x.

The sum of these values equals the total sales.

($4200 o x) + ($4800 o 2x) + (5400 o 3x) = $360,000

$4200x + $9600x + $16,200x = $360,000

$30,000x = $360,000

x = $360,000 ÷ $30,000

x = 12

Since x = 12 of the $4200 cars, 3x, or 36, of the $5400 model were sold.

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