The value of each of 5 numbers is at least 8. The average (arithmetic

Question

The value of each of 5 numbers is at least 8. The average (arithmetic mean) of these 5 numbers is 24. If the average of two of the numbers is 18, what is the largest possible value that any of these 5 numbers can have?

A. 28
B. 50
C. 68
D. 76
E. 84

A. 28
B. 50
C. 68
D. 76
E. 84

Skywalker18 · Answer

Number of terms , n= 5
 Average = 24
 a1+a2+a3+a4+a5 = 24*5=120
 
 average of two of the numbers is 18
 Sum of these 2 numbers , a3+a4 = 36
 The least possible value of any number is 8
 For the largest possible value (let's say a5) , the other 2 numbers a1 and a2 need to be 8 each . Therefore , a1+a2 = 16
 a1+a2+a3+a4 = 36+16 = 52
 Therefore a5 = 120 - 52
 = 68
 
Answer C

Abhishek009 · Answer

a + b + c + d + e = 120

a + b  + c + d + e = 120

36  + c + d + e = 120

c + d + e = 84

Replace the value of 2 numbers as 8

c + d  + e = 84

8+8  + e = 84

So , the maximum possible value of e = 68 (84-16)

Hence , answer is C. 68

keats · Answer

The value of each of 5 numbers is at least 8. The average (arithmetic mean) of these 5 numbers is 24. If the average of two of the numbers is 18, what is the largest possible value that any of these 5 numbers can have?

A. 28
B. 50
C. 68
D. 76
E. 84

_________________________
Let the five numbers be a,b,c,d, and e (each being at least 8)
a + b + c + d + e = 120
Let the two numbers whose average is 18 be a, and b. Therefore, a+b =36
Now,
c + d + e = 120 - 36 = 84
Largest possible value can be achieved by giving c and d minimum possible value i.e. 8
SO, E ( maximum value ) can be 84 - (8+8) = 84-16 = 68
So answer should be  C

nipunjain14 · Answer

Average => (a1+a2+a3+a4+a5)/5 =24 
Thus => a1+a2+a3+a4+a5 = 120
 now least value is 8, so let a1 = 8

average of 2 numbers is 18 
thus , let (a2+a3) /2 =18
a2+a3 = 36

This tells that sum of last 2 numbers should be :
120 -8-36 =76

Then by elimination , eliminate 84 and 76
highest possible value of a5 should be 68 from the answer choice

Answer : C

elegantm · Answer

Sum of 5 integers = 5 x 24 = 120
Sum of 2 integers = 2 x 16 = 36
So, Sum of remaining 3 integers = 120 - 36 = 84

In order to maximize one integer, we need to minimize remaining 2 integers.

As per condition, lowest value that any integer could take is 8. Thus, the sum of these 2 integers would be 16
So remaining integer would be = 84 - 16 = 68

Hence Answer C

JeffTargetTestPrep · Answer

We are given that 5 numbers have an average of 24 and that each number is at least 8.
 
Using the formula average = sum/number, we see that the sum of the 5 numbers is 5 x 24 = 120.
 
Since the average of 2 of the numbers is 18, the sum of those 2 numbers is 36. If we minimize the value of 2 more numbers, each of those numbers should be 8, and thus the sum of those 2 numbers is 16. Thus, the largest possible number in the set could be:
 
120 - (36 + 16) = 120 - 52 = 68
 
Answer: C

arvind910619 · Answer

68
a+b+c+d+e=120
a+b=36
c+d+e=84
now according to the condition that each number is at least 8 keep c and d as 8 then we have e=68

bumpbot · Answer

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The value of each of 5 numbers is at least 8. The average (arithmetic

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