The sum of 5 different positive 2-digit integers is 130. What is the

We need the other 4 numbers to be different and smallest possible.

10, 11, 12, 13 + x =130
x=84

Answer B.

Answer = B. 84

Highest number = 130 - (Addition of all 4 smallest numbers)

= 130 - (10+11+12+13)

= 130 - 46 = 84

Let the 5 different positive integers be a,b,c,d and e.
a+b+c+d+e = 130

It has been said that these numbers are all different and also 2-digit integers.
The smallest 2 digit integer is 10.
The other smallest integers are 11,12,13. The sum of the smallest 4 integers are 46.

The last integer is 120 - 46 = 84(Option B)

Since the 5 numbers are all different, let's let A, B, C, D and E represent the numbers.
Furthermore, let's say A < B < C , D < E, which means E is the biggest value.

We're told that A + B + C + D + E = 130
In order to MAXIMIZE the value of E, we must MINIMIZE the other 4 values.
In other words, we want to MINIMIZE the value of A + B + C + D
Since the numbers must be 2-digit integers, the smallest values for A, B, C and D are 10, 11, 12 and 13 respectively.

So, plugging those values into our equation, we get: 10 + 11 + 12 + 13 + E = 130
Simplify: 46 + E = 130
E = 84

Answer:

Cheers,
Brent

Hi All,

This is an example of a "limit" question (a prompt that asks you to solve for the "lowest" or "highest" possible value) and Brent's solution is perfect (it's exactly how I would approach this question). Sometimes Quant questions are more about "playing" with a series of restrictions and saying "what if…?" than doing lots of algebra.

There is an algebra approach to this prompt, but it still involves aspects of what Brent showed in his approach (TESTing VALUES, minimizing values so you can maximize one of them):

Since we're dealing with 5 DIFFERENT positive 2-digit integers, to maximize 1 of them, we have to minimize all of the others.

I'll call the smallest 2-digit integer….X
So the next 3 smallest would be….(X+1), (X+2) and (X+3)

Those 4 values sum to 4X + 6

Since the sum of all 5 terms = 130, the 5th value (the biggest one) is….130 - (4X+6)

If you make X the smallest positive 2-digit number possible, you'd have X=10. Plug THAT value into the calculation and you get…

130 - (40+6) = 84

Final Answer:

While I would NOT recommend approaching the question in this way (since it requires so much more work and time), it does get you to the correct answer.

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

Since we want to maximize the largest integer, we must minimize the four smallest integers. The smallest 2-digit integers are 10, 11, 12, and 13,which have a sum of 46. Thus, the largest integer would be 130 - 46 = 84.

Answer: B