SATMathematics Practice Test 1

The range of the 21 fish is $24-8=16$ inches, and the range of the $20$ fish after the $24$-inch measurement is removed is $16-8=8$ inches. The change in range, 8 inches, is much greater than the change in the mean or median.Choice A is incorrect. Let m be the mean of the lengths, in inches, of the 21 fish. Then the sum of the lengths, in inches, of the 21 fish is $21m$. After the $24-$inch measurement is removed, the sum of the lengths, in inches, of the remaining 20 fish is $21m-24$, and the mean length, in inches, of these 20 fish is $\frac{21m-24}{20}$,, which is a change of $\frac{24-m}{20}$ inches Since m must be between the smallest and largest measurements of the 21 fish, it follows that $8 < m < 24$, from which it can be seen that the change in the mean, in inches, is between $\frac{24-24}{20}=0$,which is a change of $\frac{24-m}{20}=\frac{4}{5}$ and so must be less than the changein the range, 8 inches. Choice B is incorrect because the median length of the 21 fish is the length of the 11th fish, 12 inches. After removing the 24-inch measurement, the median of the remaining 20 lengths is the average of the 10th and 11th fish, which would be unchanged at 12 inches. ChoiceD is incorrect because the changes in the mean, median, and range of the measurements are different.