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CAT 2020 Slot 2 — QA Question 24

Change in AverageMedium

In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by

Answer & solution

  • 4

  • B

    5

  • C

    3

  • D

    6

Solution

Medium

Let hh be the highest score, ll the lowest, and xx the sum of the middle 8. The two given means fix x+lx+l and x+hx+h; subtracting gives a fixed gap h−lh-l. The group mean depends only on x+h+lx+h+l, so push the scores to the extremes allowed to get the max and min possible group means.

1

Translate the two means. Lowest 9 scores exclude hh; highest 9 exclude ll.

x+l=9×42=378(mean of lowest 9)  ⋯(1)x+h=9×47=423(mean of highest 9)  ⋯(2)\begin{aligned} &x+l=9\times 42=378 \quad\text{(mean of lowest 9)}\ \ \cdots(1)\\ &x+h=9\times 47=423 \quad\text{(mean of highest 9)}\ \ \cdots(2) \end{aligned}
2

Find the fixed gap. Subtract (1) from (2):

h−l=423−378[(2)-(1)]⇒ h−l=45\begin{aligned} &h-l=423-378 \quad\text{[(2)-(1)]}\\ &\Rightarrow\ h-l=45 \end{aligned}
3

Minimum mean. The group total is x+h+lx+h+l. From (1), x+l=378x+l=378, so total =378+h=378+h, minimised by the smallest valid hh. Since hh is the highest of all 10, it cannot be below the high-9 mean 4747, so hmin⁡=47h_{\min}=47 (then l=47−45=2l=47-45=2).

Min mean=378+4710=42510⇒ Min mean=42.5\begin{aligned} &\text{Min mean}=\frac{378+47}{10}=\frac{425}{10}\\ &\Rightarrow\ \text{Min mean}=42.5 \end{aligned}
4

Maximum mean. From (2), x+h=423x+h=423, so total =423+l=423+l, maximised by the largest valid ll. Since ll is the lowest of all 10, it cannot exceed the low-9 mean 4242, so lmax⁡=42l_{\max}=42 (then h=42+45=87h=42+45=87).

Max mean=423+4210=46510⇒ Max mean=46.5\begin{aligned} &\text{Max mean}=\frac{423+42}{10}=\frac{465}{10}\\ &\Rightarrow\ \text{Max mean}=46.5 \end{aligned}
5

Take the difference. Subtract the results of steps 3 and 4:

Max mean−Min mean=46.5−42.5⇒ =4\begin{aligned} &\text{Max mean}-\text{Min mean}=46.5-42.5\\ &\Rightarrow\ =4 \end{aligned}
44
CAT 2020 Slot 2 QA Q24: In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. — Solution | TheCATExam