Question #f8a322d9 - SAT Suite Question Bank

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Modded SAT Question Bank by Abdullah Mallik

Two data sets of $23$ integers each are summarized in the histograms shown. For each of the histograms, the first interval represents the frequency of integers greater than or equal to $10$ , but less than $20$ . The second interval represents the frequency of integers greater than or equal to $20$ , but less than $30$ , and so on. What is the smallest possible difference between the mean of data set A and the mean of data set B?

$0$
$1$
$10$
$23$

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Correct Answer: B

Rationale

Choice B is correct. The histograms shown have the same shape, but data set A contains values between $20$ and $60$ and data set B contains values between $10$ and $50$ . Thus, the mean of data set A is greater than the mean of data set B. Therefore, the smallest possible difference between the mean of data set A and the mean of data set B is the difference between the smallest possible mean of data set A and the greatest possible mean of data set B. In data set A, since there are $3$ integers in the interval greater than or equal to $20$ but less than $30$ , $4$ integers greater than or equal to $30$ but less than $40$ , $7$ integers greater than or equal to $40$ but less than $50$ , and $9$ integers greater than or equal to $50$ but less than $60$ , the smallest possible mean for data set A is $\frac{(3 \cdot 20) + (4 \cdot 30) + (7 \cdot 40) + (9 \cdot 50)}{23}$ . In data set B, since there are $3$ integers greater than or equal to $10$ but less than $20$ , $4$ integers greater than or equal to $20$ but less than $30$ , $7$ integers greater than or equal to $30$ but less than $40$ , and $9$ integers greater than or equal to $40$ but less than $50$ , the largest possible mean for data set B is $\frac{(3 \cdot 19) + (4 \cdot 29) + (7 \cdot 39) + (9 \cdot 49)}{23}$ . Therefore, the smallest possible difference between the mean of data set A and the mean of data set B is $\frac{(3 \cdot 20) + (4 \cdot 30) + (7 \cdot 40) + (9 \cdot 50)}{23} - \frac{(3 \cdot 19) + (4 \cdot 29) + (7 \cdot 39) + (9 \cdot 49)}{23}$ , which is equivalent to $\frac{(3 \cdot 20) - (3 \cdot 19) + (4 \cdot 30) - (4 \cdot 29) + (7 \cdot 40) - (7 \cdot 39) + (9 \cdot 50) - (9 \cdot 49)}{23}$ . This expression can be rewritten as $\frac{3 (20 - 19) + 4 (30 - 29) + 7 (40 - 39) + 9 (50 - 49)}{23}$ , or $StartFraction 23 Over 23 EndFraction$ , which is equal to $1$ . Therefore, the smallest possible difference between the mean of data set A and the mean of data set B is $1$ .

Choice A is incorrect. This is the smallest possible difference between the ranges, not the means, of the data sets.

Choice C is incorrect. This is the difference between the greatest possible mean, not the smallest possible mean, of data set A and the greatest possible mean of data set B.

Choice D is incorrect. This is the smallest possible difference between the sum of the values in data set A and the sum of the values in data set B, not the smallest possible difference between the means.

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