There are 3 different positive integers.

Question

There are 3 different positive integers. If their average (the arithmetic mean) is 8, what are their values?
1) The largest integer is twice the smallest integer.
2) One of them is 9.

chetan2u · Accepted Answer

Hi,
 This is a typical so called C trap .. 
you can form an equation with the 1st statement and the 2nd statement gives you a value of variable..

But whenever you see some restrictions in the values of variables, for example some max, min value or a range is given, check that statement. 
In most probability , yopu will find the answer in it..

lets see the statements..
 1) The largest integer is twice the smallest integer. 
let the smaller one be x and the largest be 2x..
the middle one will be between x and 2x..
we can find ways as engr12 has correctly given..

But say the sum of number is big, we may have to work on lots of value, and in this case best will be to find the limits of x..

lets find the  limits of value of x 
say the middle number is x+1, minmum possible..
so x+x+1+2x=24..
4x+1=24..
or x=5.6..
therefore x <6..

second case let it be just less than the largest number..
x+2x-1+2x=24..
5x=25..
x=5..

In both cases only x=5 fits in..
lets check for x=5..
numbers are x,2x-1,2x... 5,9,10..

we can check no other will fit in..

Suff

2) One of them is 9.
clearly insuff..

ENGRTOMBA2018 · Answer

The language of the question is not good. Mention the complete question without rewording or shortening the original question. Check the question again and do mention the official answer.

As for the question, OA should be A as shown below:

You are given that the 3 integers are positive and distinct. Their average = 8 ---> sum of the 3 distinct positive integers = 24.

Per statement 1, let a,b,c be the 3 distinct positive integers with  a>b>c . Thus a=2c and a+b+c = 24 ---> 3c+b = 24 --> possible cases

2,21,1 or 4, 18, 2 or 6,15,3 or 8,12,4 or 10,9,5 or 14,3,7 for (a,b,c). Out of these 4 only possible or allowed case (based on a>b>c) is that of (10,9,5) and hence 1 unique set possible. Sufficient.

Per statement 2, 1 of the 3 integers = 9. Clearly not sufficient.

Hence A should be the correct answer.

Hope this helps.

DmitryFarberMPrep · Answer

Cuhmoon, did you miss that the third integer had to be in between the other two? If we overlook that, there are plenty of possible combinations. It's an easy restriction to miss!

cuhmoon · Answer

Thanks Dmitry. But is that implied or should that be part of the question?

DmitryFarberMPrep · Answer

It is actually stated, but it's subtle. If the  largest  is twice the  smallest , then the remaining number must by definition be in between the two. The problem is that if we just represent this as, say, 3x + y = 24, we end up removing that information from the setup. It might be helpful to start by making three blanks in a row and labeling them while this information is fresh.

cuhmoon · Answer

Thanks a lot for the explanation!

cuhmoon · Answer

Great explanation! Thanks a lot.

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

There are 3 different positive integers. If their average (the arithmetic mean) is 8, what are their values?
1) The largest integer is twice the smallest integer.
2) One of them is 9.

In the original condition, there are 3 variables(a<b<c) and 1 equation(a+b+c=24), which should match with the number of equations. So you need 2 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make C the answer.
When 1) & 2), 5, 9, and 10 are valid, which is unique. So, the answer is C. However, this is an integer and statistics question, one of the key questions, and apply the mistake type 4(A). For 1), only 5, 9, 10 are valid as well, because they are different positive integers. Thus, 6,6,12 -> impossible and 4,8,12 -> impossible. Therefore, the answer is A.

 For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

cuhmoon · Answer

Great explanation! Thanks v much!!

There are 3 different positive integers.

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There are 3 different positive integers.

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