The average (arithmetic mean) of 3 different positive integers is 100

(a+b+c)/3 = 100
a+b+120 = 300
=> a+b = 180 =>a = 180 - b
b<120 => a>60
smallest a is 61 => C

Average=100={120+x+y}/3
X+y=180
Assume x to be smallest and start plugging in the answers to get y.
Now neither x nor y can be larger than 120. So a and b makes y greater than 120. Therefore are incorrect. From the rest 61 is the smallest.

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Avg of 3 nos will be -
(x+ 2nd largest + largest)/3
=> (x+ 119 +120) / 3 =100
=> x= 61 (C)

What if the other 2 numbers are 109 and 71 or 100 and 80? It is no where mentioned that the 2nd number has to be immediate smaller number than 120. So how we can eliminate 109/100? Please explain.

Hello Tannu,

We need to find the smallest number out of the 3 and we are given that largest number is 120.
We also know that we have 3 different positive integers, i.e, the remaining 2 numbers are not equal.

Consider a parallel example: x + y = 5. (where x and y are positive integers)

If we want to find the smallest out of x and y, the OTHER number HAS TO BE AS LARGE AS POSSIBLE, that is 4.
Why?
Consider y = 3:
x+3=5. Therefore x = 2. But is this the smallest POSSIBLE value that 'x' can take? 'x' will be as small as possible only when y will be as large as possible. In this case, if we take x + 4 = 5, then x will be equal to 1 (this can be the smallest possible positive integer. Therefore y has to be 4 (or 2nd number immediately smaller to 5, as you have pointed out).

With this understanding, you can re-read the solutions provided above and hopefully understand it better.
Hope this helped.

Thank you..

The average (arithmetic mean) of 3 different positive integers is 100 and the largest of these 3 integers is 120. What is the least possible value of the smallest of these 3 integers?

(A) 1
(B) 10
(C) 61
(D) 71
(E) 80

Ans= The 2nd largest no can be 119 only otherwise is be > than 120 and violate the statement. Hence 3rd no will be Total sum - sum of 2 nos = 3rd no, i.e.... 300-(120+119) = 61.

KUDOS!!

[quote="HKD1710"]The average (arithmetic mean) of 3 different positive integers is 100 and the largest of these 3 integers is 120. What is the least possible value of the smallest of these 3 integers?

(A) 1
(B) 10
(C) 61
(D) 71
(E) 80


The sum will be equal to 300. As in the question it is given that the numbers are different.
We need to maximise the one of the two unknown numbers. 120,119. Therefore the last number will be 61.
C

Posted from my mobile device

You can use the "Deviations from the Mean" Approach to Solve pretty quickly.

Rule: For any Given Set, the Total +Positive Deviations ABOVE the Mean must = the Total -Negative Deviations BELOW the Mean



We are given that 120 is the Largest of the 3 Integers. Also, the Mean = 100.

In order to get the Least Possible Value for the Smallest Integer, we should Maximize the +Positive Deviations above the Mean.

Since each Integer has to be different, the next largest Integer = 119.

120 - 100 = +20 Deviation Above the Mean

119 - 100 = +19 Deviation Above the Mean

-------------------
100
-------------------

the 3rd Integer must have a -39 Deviation BELOW the Mean to match the Total Deviations above the Mean.

100 - 39 = 61

61 is the Smallest Number we can get.

-C-

Average: \(\frac{Sum }{ Total }\)

=> 100 = \(\frac{Sum }{ 3}\)

=> Sum = 100 * 3 = 300

The largest number is 120.

=> Therefore the sum of the remaining two integers is 300 - 120 = 180.

=> If both integers are equal: values will be 90 and 90.

As we have been asked possible least integer then One of the 90 can become 120 (90+30 - as that is the maximum limit).

But as they are different positive integers, the other integer can maximum have a value as 119.(90+29)

=> And least integer will be 90-29 = 61.

So, the numbers are 61, 119, 120.

Answer C

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