Three positive integers a, b, and c are such that their average is 20

a+b+c=60 & a<b<c

By hit and trial, c=25 gives b=23 and a = 12

Hence C=25.

Hi All,

As 'thick' as this question might appear, it can be solved with a bit of 'brute force' and a combination of basic arithmetic skills. We can use the answer choices and TEST THE ANSWERS, but we'll also need to 'play around' a bit with the information that we're given (so we can determine what answers are possible and what answers are not).

We're given a number of facts about the variables A, B and C:
1) They're POSITIVE INTEGERS
2) Their average is 20
3) A ≤ B ≤ C
4) The median is (A + 11)

We're asked for the LEAST possible value of C.

To start, since there are 3 variables, we know that B is the median. So we can think of the three variables as A, (A+11) and C. By extension, we know that A is NOT 'close' to C; this means that A is probably a lot less than the average and C is well ABOVE the average.

The average of the variables is 20, so their sum is 60.

Let's TEST 23...

IF....
C = 23
A+A+11 = 37
2A = 26
A = 13
The numbers would be 13, 24 and 23...
This is NOT possible (B ≤ C is not true in this case)
ELIMINATE 23.

Let's TEST 24...

IF....
C = 24
A+A+11 = 36
2A = 25
A = 12.5
This is NOT possible (the variables have to be INTEGERS)
ELIMINATE 24.

Let's TEST 25...

IF....
C = 25
A+A+11 = 35
2A = 24
A = 12
The numbers would be 12, 23 and 25...
This MATCHES everything we were told, so this MUST be the answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

According to the condition, a, b and c can be equal to each other so let them be. They will form an equidistant set whose average = median = b = 20.

a + 11 = 20
a = 9

Then, form a grid and list down possible values.

a b c
9 20 31
10 21 29
11 22 27
12 23 25 -> bingo!
13 24 23 -> rejected because it will violate main condition (a <= b <= c). We can stop here.

Answer (C).

to minimize c, we have to maximize a
we can suppose that we have: a+a+11+a+11=20*3
3a+22=60
3a=38
38/3 is not an integer, thus, minimum value of c must be greater than a+11
a+12 - doesn't work (37/3)
a+13 - works (36/3=12)
a=12, b=23, c=25.
minimum value of c is 25.

Good logic.

Assume the numbers are a, a+11 and a+11 (to minimize the greatest value c)
You get a = 38/3 = 12.67
Here itself, you see that a cannot be 13 since that will add up to a higher value than 60. So a must be 12 - the highest integer value it can take.
This gives c = 60 - 12 - 12 - 11 = 25

This means the median is b=a+11. Since the sum is 60,a=(60-11-c)/2. The only value of c that gives us an integer value for a and makes the median a+11 is 25. Answer is C

Responding to a pm:



We need to minimise c.

Since median is 'a + 11', and since b is the middle value as per a ≤ b ≤ c, we know that 'b' is the median. So
b = a + 11
What this means is that 'b' is 11 more than 'a'. So in terms of 'a', we can write 'b = a + 11'.

The values will be a, a+11, c.
Note that 'a' cannot be 'a + 10'. What you are saying here is
a = a + 10
How is that possible? This gives 0 = 10 which is not true.

Now since we want to minimise c, we can give it the same value as b i.e. 'a + 11' because b ≤ c.
This doesn't give us an integer so we take the smallest value of c that does give us an integer.

When a becomes 38/3, it means a must be less than 13.

If a = 12, then b = 23

This is 8 below and 3 above the average, so we need 5 more above the average = 25.

 

­The solution looks correct! However, I get a different answer if I try to do this the other way around - by substituting "a" instead of "c". Can someone help me find out where I am going wrong? 

So,
b ≤ c,

a+11 ≤ c

and we have that 2a + c = 49 => a = \(\frac{49-c}{2}\)

Putting the value of a to the above inequality : 

\(\frac{49-c}{2}\) + 11 ≤ c

49-c+22 \(\leq\) 2c

71 ≤ 3c

23 ≤ c



 ­

odd_major

A couple of adjustments:

1) 71/3 = 23.666, so the lowest integer value we can get from your work is 24, not 23.

2) We still need to make sure that our values work for the whole system. Since 2a+11 = even + odd, the sum of a and b must be odd. Therefore, to make 60, c must also be odd. That makes 25 the lowest allowable value. From there, we can verify that 2a + 11 = 35 works (a=12, b=23).

(If the even/odd trick didn't jump out, we'd still want to check our value. If we use c=24, we get 2a+11 = 36, 2a = 25, a = 12.5, so the integer issue pops up and grabs us.)