ACT 2020 July(C01) Math Question 19 Explanation - ACT Helper

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2020 July(C01) Math Question 19
Statistics
- Mean, Median, Mode Combinations
Choice A
Let the 9th number be 91 which is greater than 90.
$T h e m e a n o f S e t A = (62 + 76 + 76 + 80 + 82 + 87 + 94 + 96) \div 8 = \frac{653}{8} = 81.625$ $T h e m e d i a n o f S e t A = (80 + 82) \div 2 = 81$ $T h e m e a n o f S e t B = \frac{653 + 91}{9} \approx 82.66$ $T h e m e d i a n o f S e t B = 82$ $\Rightarrow 82.66 > 81.625$ $\Rightarrow 82 > 81$
The mean and the median of Set B will each be greater than the mean and the median of Set A.
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