The arithmetic mean of a sequence of 15 integers is 78. The largest

Question

The arithmetic mean of a sequence of 15 integers is 78. The largest value of the set is five times the smallest value of the set. If the mean of the set is equal to the median, then what is the maximum possible value for the range of the set?

A.

B.

C.

D.

E.

A.  52
 B.  104
 C.  112
 D.  208
 E.  260

EgmatQuantExpert · Accepted Answer

Solution 
  Given:  
 • The arithmetic mean of a sequence of 15 integers is 78
• The largest value of the set is five times the smallest value of the set
• The mean of the set is equal to the median

To find:  
 • The maximum possible value for the range of the set

Approach and Working:   
 • We know that the range of a set = the largest value – the smallest value

So, for the range to be maximum, we need to maximise the value of largest element of the set and minimise the value of smallest element of the set.
 • We know that the mean = the median = the 8th element of the sequence = 78
• So, to minimise the first element of the sequence, all the elements before the median element must be same. Let us assume it be “a”
• And, to maximise the last element of the sequence, all the elements after the median except the last element must be minimum = the median = 78
• The last element of the sequence will be = 5 * a

We are also given that the mean of the set = median = 78
 • Implies, sum of all 15 elements in the sequence = 78 * 15
• Thus, a + a + a + a + a + a + a + 78 + 78 + 78 + 78 + 78 + 78 + 78 + 5a = 78 * 15
 o 12a = 78 * 15 – 78 * 7
o  a = 78 * \frac{8}{12} = 52

• Therefore, range of the set = 5a – a = 4a = 4 * 52 = 208

Hence the correct answer is Option D.

Answer: D

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KomalSg · Answer

If we want maximum range, we want 5x to be as large as possible and x to be as small as possible
x - - - - - - 78 - - - - - - 5x

This is only possible if all the other numbers are 78 
13*78 = 1014

Total sum of 15 numbers = 1170
Subtracting sum of 13 no.s from sum of 15 no.s = 1170-1014
= 156

which is equal to x+5x = 6x=156
x=26

Range = 5x-x=4x=4*26=104

Vikas1211 · Answer

My approach for this problem was

Sum of nos. 78*15=1170
Sum of nos of a sequence= n/2*(a+an)
Where n total nos of digit, a = first digit and an = last digit 
Therefore, 15/2*(a+an)= 1170

a+an=156
an = 5a (given in question)
6a=156
a=26, an=130
Range =130-26, 104 (B)

Feel free for correction ?

Posted from my mobile device

Archit3110 · Answer

sequence
XXXXXXX 787878787878 5X = 78* 15
12X+ 78*7 = 1170
X= 52
5X= 260
RANGE = 5X-X = 260-52 = 208
IMO D

nitesh50 · Answer

We have to minimise X and maximize 5x.

So
X X X X X X X 78 78 78 78 78 78 78 5X

Is the correct sequence.

Hence 
7*(78-X)=5X-78
8*78=12X

Hence X=52
4X=208

Option D is correct.

gracie · Answer

(x+5x)/2=78 mean/median
x=26
5x=130
130-26=104
B

zapak · Answer

We have to maximise the value of 5x and minimise the value of x such that the sum of the sequence remains 15*78=1170.

Hence, the sequence can be where mean is equal to median

x,x,x,x,x,x,x, 78, 78,78,78,78,78,78, 5x

The sum of the above sequence is 7x+ 78*7 + 5x = 15*78-----> 12x= 8*78-----> x=52

Hence the maximum range of the value of the set is 5x-x= 4x= 4* 52= 208

Hence answer is D.,

puneetb · Answer

My approach was:
l=5a

Median=(a+l)/2=Mean=78

-->>:::Range=l-a=104

I think I missed the point of maximizing the Range but accidentally did so...!!?

Billallkhan · Answer

My question is why is this being assumed
"Thus, a + a + a + a + a + a + a + 78 + 78 + 78 + 78 + 78 + 78 + 78 + 5a = 78 * 15"

Rashed12 · Answer

Dear  Bunuel ,

What mistake I have made in the math. In my sense, the answer is B. I would be glad if you help me to find out the mistake.

(x+5x)/2=78 ( Mean and Median is the same)
x=26
5x=130
130-26=104

teja8341 · Answer

EgmatQuantExpert

" So, to minimise the first element of the sequence, all the elements before the median element must be same. Let us assume it be a "

Is'nt it that all the elements before the median element must be equal to the median to minimise the first element?

Please correct me if I'm wrong, having tough time understanding these type of questions.

adityaganjoo · Answer

In order to maximise the range, while keeping the median equal to mean, terminal numbers of the series must be as far as possible. That is, smallest term is as small as possible, while largest term is as large as possible, and setting up the intermediate terms in such a way that median = 78, and sum of all terms is 78*15

Thus, ascending order must be: t, t, t, t, t, t, t, 78, 78, 78, 78, 78, 78, 78, 5t (78 is the smallest value a term after median can have)
Therefore, t(7) + 78(7) + 5t = 78(15)
or, 12t = 78(8)
or, t = (26)(2)

Range = 5t - t = 4t = 26*2*4 = 26*8 = 208

Therefore, (D)

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The arithmetic mean of a sequence of 15 integers is 78. The largest

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The arithmetic mean of a sequence of 15 integers is 78. The largest

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