If the average of 5 positive integers is 40 and the

Question

If the average of 5 positive integers is 40 and the difference between the largest and the smallest of these 5 numbers is 10, what is the maximum value possible for the largest of these 5 integers?

A. 50
B. 52
C. 49
D. 48
E. 44

Bunuel · Accepted Answer

The average of 5 positive integers is 40 --> the sum of these 5 integers is 5*40=200;
The difference between the largest and the smallest of these 5 numbers is 10 --> let the smallest integer be x and the largest x+10;

Now, we want to maximize x+10.

General rule for such kind of problems:  
to maximize one quantity, minimize the others; 
to minimize one quantity, maximize the others.

Hence, in order to maximize x+10 we should minimize all other terms, thus make them equal to the smallest integer x. We would have: x+x+x+x+(x+10)=200 --> x=38 --> x+10=48.

Answer: D.

Hope it's clear.

PareshGmat · Answer

Let the 5 integers be as follows:

To have maximum value of a+10, let all others be "a"

a+10 ............ a ............... a .............. a .............. a

Addition = 200

5a + 10 = 200

a = \frac{190}{5} = 38

a+10 = 48 (Maximum Value)

Answer = D

sunny12 · Answer

My method was,

x + (x+1) + (x+2) + (x+3) + (x+4) / 5 = 40

Thus 5x + 10 = 200 
and x = 38

Since we know the range is 10, that is ( x + 4 ) - x = 10

so, ( x + 4) - 38 = 10

(x + 4) = 48

** Take (x + 4) as a whole number.

GMATinsight · Answer

Sum of 5 Integer (a, b, c, d, e) = 5*40 = 200

e - a = 10
i.e. e = a+10

For e to be maximum remaining 4 MUST be as small as possible
Since smallest of 5 numbers is a so to minimize other numbers we can take them equal to the smallest of 5 numbers

i.e. a+a+a+a+(a+10) = 200
i.e. 5a = 190
i.e. a = 38

i.e. Largest e = 38+10 = 48

Answer: option D

rahulkashyap · Answer

why cant we solve taking a=e-10

GMATinsight · Answer

We can solve it like that too

in that case the equation will look like

(e-10)+(e-10)+(e-10)+(e-10)+e = 200
i.e. 5e - 40 = 200
i.e. 5e = 240
i.e. e = 48

I hope this helps

mvictor · Answer

I solved it the same way bunuel explained.
sum is 200.
x is the smallest
x+10 is the greatest
to maximize x+10, suppose all numbers are the same
x+x+x+x+x+10=200
5x=190
x=190/5
x=38
x+10=48.

TheGSK · Answer

It average of 5 integers is 40, then their sum is  40 *5=200  
Let max value be  x  .
Therefore smallest value is  x -10 
for one value to be maximum (i.e. x ) , all others have to be minimum (i.e. x- 10 ) . 
So,  4(x -10) + x = 200 
 x =48 
Ans. D

stonecold · Answer

Excellent Question
Here is what i did in this Question =>

Using =>

Mean=\frac{Sum}{#}

Sum(5)=40*5=200

Now to maximise the largest quantity,we must minimise all other quantities.
Let a be the smallest element
The series would be -> a,a,a,a,a+10

Hence 5a+10=200
a=38

Helen the largest element = 38+10 = 48

Hence D

GyMrAT · Answer

Let a, b, c, d, e be the 5 positive integers, where a is smallest & e is largest.

Given average = 40, hence a+b+c+d+e = 5*40 = 200......(i)

& e - a = 10......(ii)

for e to be maximum, we need all other numbers to be as small as possible, hence a=b=c=d

We get from (i) 4a + e = 200.......(iii)

Solving (ii) & (iii), we get a = 38 & hence Maximum value of e = 48

Answer D.

Thanks,
GyM

JeffTargetTestPrep · Answer

We are given that we have a sum of 40 x 5 = 200 and a range of 10.

Let’s strategically check our answer choices:

If the largest integer is 50, then the 4 other numbers could be 40. However, 50 + 40 x 4 = 210, which is greater than 200. Thus, we see 50 (and also 52) cannot be correct.

Let’s check 49:

49 + 4 x 39 = 49 + 156 = 205, which is also greater than 200.

Let’s check 48:

48 + 38 x 4 = 48 + 152 = 200, so 48 is the largest number in the set.
Alternate Solution:

Let us denote the greatest element in this set by M and the smallest element in this set by m.

To maximize the greatest integer in the set, we should keep the remaining integers as small as possible. For that purpose, let’s assume all the elements in the set besides M are equal to m. Then, according to the information given in the question, we have

(4m + M)/5 = 40

4m + M = 200

and

M - m = 10

Let’s multiply the last equation by -1 and add to the preceding equation:

5m = 190

m = 38

Substituting m = 38 in M - m = 10, we see that the greatest value of M is 48.

Answer: D

almaskz · Answer

In a questions like this one, I like to assign variables to the value the question asks. 
So, for example in this question we are looking for "the maximum value possible for the largest of these 5 integers". 
Let it be "x".
Then, to maximize "x", we need to minimize all other unknowns. Since the question stem does not say anything about integers being different (which is some times a case), we can minimize 4 other numbers to be (x-10) (Inferred from difference the largest and the smallest being "10").

the  first 4  numbers =   (x-10)  
the  fifth  number =   x   
Now, from this formula:  total number of items * avg = sum of items 
5*40 = 200 = (x-10)*4 + x
200 = 4x-40+x
5x = 240
 x = 48 
 Hence, the answer is D

The main take a way , at least for me, is assigning the variable ("x") to the thing we need to solve for at the end. This way, we can avoid mistakes solving for something different than the question actually asks. 
On some of the trickier questions there would be an answer choice 38, that would be intended to punish those who assign x to the smallest number in the set and by the time he/she finishes the calculations, forgets to add extra 10 to find the largest number is the set.

Kinshook · Answer

Given: 
1. The average of 5 positive integers is 40.
2. The difference between the largest and the smallest of these 5 numbers is 10.

Asked: What is the maximum value possible for the largest of these 5 integers?

Let the numbers be {s, s, s, s, s+10} since to maximise largest number, all other numbers need to be minimised
4s + s + 10 = 5*40 = 200
5s = 200 - 10 = 190
s = 190/5 = 38
Largest number = s + 10 = 48

IMO D

Archit3110 · Answer

sum of nos ; 5*40 ; 200
range os no ; 10
let smallest be x so largest no ; 10+x
5x+10=200
x=38
largest ; 48
IMO D

pierjoejoe · Answer

one way i like to solve this kind of problem is just imagine there are 5 numbers and among them there is a number called S that is the smallest, and a number that is L that is the largest. so no number can be greater than L or smaller than S.

now when we want to find the maximum of the largest (as we are doing in this case) we want to minimize all the other values in one of the equations that the problem statement offers. in this case the equation is the average:

S + X1+ X2 + X3 + X4 + L = 200
if we want to find the maximum of L we need to minimize all the other values. what is the minimum of any value? we said it before! it is S!

so we can just write S + S + S +S +S + L = 200
thus the condition mecomes
4S + L = 200

i like to call this condition "average condition" because it comes from the average. the problem always put some other limitations in this types of problems. otherwise we can just set S to zero and the maximum value for L would be 200.
the other condition is just that the largest L minus the smallest S is 10

put the 2 equations in a linear system:
4S + L = 200
L-S = 10

there are 2 equations and 2 variables, just find the results that are:
S = 38 and L = 48

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