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The average of five integers is 63, and none of these integers is

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The average of five integers is 63, and none of these integers is [#permalink]

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The average of five integers is 63, and none of these integers is greater than 100. If the average of three of the integers is 65, what is the least possible value of one of the other two integers?

A. 5
B. 15
C. 20
D. 21
E. 30
[Reveal] Spoiler: OA

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Re: The average of five integers is 63, and none of these integers is [#permalink]

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New post 05 Nov 2015, 12:11
Bunuel wrote:
The average of five integers is 63, and none of these integers is greater than 100. If the average of three of the integers is 65, what is the least possible value of one of the other two integers?

A. 5
B. 15
C. 20
D. 21
E. 30


The sum of two other integers = 63*5-65*3 = 120
the least possible value of one integer when the other integer has the highest value of 100 =120 - 100 = 20

Ans C

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The average of five integers is 63, and none of these integers is [#permalink]

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New post 05 Nov 2015, 16:08
Bunuel wrote:
The average of five integers is 63, and none of these integers is greater than 100. If the average of three of the integers is 65, what is the least possible value of one of the other two integers?

A. 5
B. 15
C. 20
D. 21
E. 30


When it comes to averages, we know that average value = (sum of n values)/n
We can rewrite this into a useful formula: sum of n values = (average value)(n)

The average of five integers is 63
So, the sum of ALL 5 integers = (63)(5) = 315

The average of three of the integers is 65
So, the sum of the 3 integers = (65)(3) = 195

So, the sum of the 2 REMAINING integers = 315 - 195 = 120

If the sum of the 2 REMAINING integers = 120, and we want to minimize one value, we must MAXIMIZE the other value.
100 is the maximum value so let 1 integer = 100, which means the other must equal 20

Answer: C

Cheers,
Brent
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Re: The average of five integers is 63, and none of these integers is [#permalink]

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New post 06 Nov 2015, 11:52
Bunuel wrote:
The average of five integers is 63


Σ5 = 315

Bunuel wrote:
If the average of three of the integers is 65


Σ5 = 315

Bunuel wrote:
none of these integers is greater than 100, what is the least possible value of one of the other two integers?


Σ3 = 195
Σ2 = 120
x + y = 120

x & y < 100

Now the condition here is if we maximise one we will be minimuziing the other value.

Lets check the options now -

X + Y =>120

A. 5 - Impossible , the other value is 115
B. 15 Impossible , the other value is 105

C. 20 , the other value is 100
D. 21, the other value is 115
E. 30 , the other value is 90

Now check for the maximimum possible value of the one integer, to mazimize the other Integer

among the given options only (C) is the best.
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Re: The average of five integers is 63, and none of these integers is [#permalink]

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New post 18 Aug 2016, 13:44
I selected option (d) also noticed that 40% of the folks who tried this question opted for the same option. The questions say's "none of these integers is greater than 100" i.e. they can be equal to 100 or less. Is this a fair assumption ? This is the part I missed and opted for (d).

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Re: The average of five integers is 63, and none of these integers is [#permalink]

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New post 18 Aug 2016, 13:58
gary391 wrote:
I selected option (d) also noticed that 40% of the folks who tried this question opted for the same option. The questions say's "none of these integers is greater than 100" i.e. they can be equal to 100 or less. Is this a fair assumption ? This is the part I missed and opted for (d).


Hi! Gary,

IMO, unless it is not written in the question that any of the number is not equal to or greater than 100, we must assume that number can be = 100
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Re: The average of five integers is 63, and none of these integers is [#permalink]

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New post 18 Aug 2016, 14:04
Bunuel wrote:
The average of five integers is 63, and none of these integers is greater than 100. If the average of three of the integers is 65, what is the least possible value of one of the other two integers?

A. 5
B. 15
C. 20
D. 21
E. 30


This is how I solved it:-

Average of 5 integers is 63. That means five numbers are 63, 63, 63, 63, 63

Average of 3 integers is 65. That means three integers are 65, 65, 65 (if we notice that 65 is 2 greater than 63, and hence total 6 is more in this care compared to three integers in above scenario)

Hence, we need to subtract 6 from the remaining two integers to have same total. We can write the remaining two integers as 60, 60 (Subtract 3 from each)

Since out of these two numbers the largest number can be 100, the smallest can be 20 to get the same total.


B is the answer
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Re: The average of five integers is 63, and none of these integers is [#permalink]

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New post 18 Aug 2016, 20:58
Bunuel wrote:
The average of five integers is 63, and none of these integers is greater than 100. If the average of three of the integers is 65, what is the least possible value of one of the other two integers?

A. 5
B. 15
C. 20
D. 21
E. 30


You can use the method of deviations.
The average of all number is 63 but of three is 65. So these three numbers give a positive deviation of 3*2 = 6
If the greatest number is 100 (maximum allowed), the smallest number will be the least. 100 is 37 more than 63.
So the lest numbers can be 6 + 37 = 43 less than 63. The least number would be 20.

Answer (C)

For more, check: http://www.veritasprep.com/blog/2012/05 ... eviations/
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Re: The average of five integers is 63, and none of these integers is [#permalink]

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Re: The average of five integers is 63, and none of these integers is   [#permalink] 05 Jan 2018, 07:53
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