n a group of 10 students, the mean of the lowest 9 scores is 42

Question

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

A 5
B 3
C 4
D 6
E 7

Show Answer

Official Answer

C

A 5
B 3
C 4
D 6
E 7

DashingNightmare2 · Answer

Sum of lowest 9 scores = 42*9 = 378 
Sum of Highest 9 scores = 47*9 = 423

For Finding  Maximum mean ,

Let's suppose all initial 9 scores =42 , lets find last score
Sum of 8 lowest scores excluding the least score =42*8 = 336 
highest score =423 - 336 =87
Total score = 42 * 9 + 87 = 465
Mean =  \frac{465}{10}  = 46.5

For Finding  Minimum mean ,

Lets take
least score =0, => sum of 8 lowest scores except least scores= 378 => highest score = 423-378 = 45(not possible, because highest number >= mean of last 9 numbers)

least score =1 =>sum of 8 lowest scores except least scores= 378 -1 =>=> highest score = 423-377 =46(not possible)

least score =2 =>sum of 8 lowest scores except least scores= 378 -2 =>=> highest score = 423-376 =47(possible)

Hence Total = 423 +2 =425 => median of all 10 numbers =  \frac{425}{10}  = 42.5

Difference = 46.5 - 42.5 = 4

Hence C.

Kinshook · Answer

Given: 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. 
Asked: For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by

Let the scores in increasing order be x1, x2, .... x10 where x1 is lowest score and x10 is highest score.
x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 = 42*9 = 378 
x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 = 47*9 = 423

x10 - x1 = 423 - 378 = 45

(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10)max is when x1 = 42; and x10 = 42 + 45 = 87
(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10)max = 378 + 87 = 465
Mean_max = 465/10 = 46.5

(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10)min is when x10 = 47
(x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10)min = 378 + 47 = 425
Mean_min = 425/10 = 42.5

Difference = Mean_max - Mean_min = 46.5 - 42.5 = 4

IMO C

huongguyen · Answer

1.
Top 9 lowest score (a1 -> a9): Mean = 42 => Sum = 42*9 = 378
<=> a1 + a2+ ...+ a9 = 378
<=> [a2 + ...+a9] = 378 - a1

2.
Top 9 highest score (a2->a10): Mean = 47 => Sum = 47 * 9 = 465
<=> a2 + a3 + ...+a9 + a10 = 465
<=> 378 - a1 + a10 = 465
<=> a10 - a1 = 45

The List reaches its Max value when a1: Max <=> a1 = 42 => a10 = 42 + 45 = 87
Sum total list = 42*9 + 87 = 465
=> Mean = 465/10 = 46,5

The list reaches its min value when a10 = min <=> a10 = 47 => a1 = 47 - 45 = 2
Sum total list = 2 + 47*9 = 425
=> Mean = 425/10 = 42,5

The diference = 46,5 - 42,5 = 4
IMO: C

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n a group of 10 students, the mean of the lowest 9 scores is 42

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