The sum of 5 different positive 2-digit integers is 130. What is the highest possible value of the largest of these integers?
A) 88
B) 84
C) 78
D) 74
E) 68
Please help me solve it. I can't figure it out.
Pls let me know how you'd solve this problem
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Let the 5 different integers be a, b, c, d and e, such that a<b<c<d<e.somberblue wrote:The sum of 5 different positive 2-digit integers is 130. What is the highest possible value of the largest of these integers?
A) 88
B) 84
C) 78
D) 74
E) 68
It is given that a+b+c+d+e = 130.
To MAXIMIZE the value of e, we must MINIMIZE the value of a+b+c+d.
Since a, b, c and d must be 4 different positive 2-digit integers, the smallest possible value for a+b+c+d is as follows:
a+b+c+d = 10+11+12+13 = 46.
Since a+b+c+d = 46, and a+b+c+d+e = 130, we get:
46 + e = 130
e = 84.
The correct answer is B.
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Hi somberblue,
This is an example of a "limit" question (a prompt that asks you to solve for the "lowest" or "highest" possible value) and Mitch'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 Mitch 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: B
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
This is an example of a "limit" question (a prompt that asks you to solve for the "lowest" or "highest" possible value) and Mitch'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 Mitch 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: B
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
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Any 2 digit number AB (=N) comprises of 10A + B where A and B are single digit integers and A > 0.
Let's test the largest given value of A first:
Let A = 8
So N = 80 + B
Subtract this from 130 to get 50-B
The smallest 2-digit numbers are 10, 11 and 12 so subtract these all from 50-B to get 17-B
The smallest remaining number, 17-B > 12 (because no 2 numbers are the same)
So -B > -5
B < 5
Hence maximum B = 4
Therefore N = 80 + B = 84
Answer B
Let's test the largest given value of A first:
Let A = 8
So N = 80 + B
Subtract this from 130 to get 50-B
The smallest 2-digit numbers are 10, 11 and 12 so subtract these all from 50-B to get 17-B
The smallest remaining number, 17-B > 12 (because no 2 numbers are the same)
So -B > -5
B < 5
Hence maximum B = 4
Therefore N = 80 + B = 84
Answer B
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