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06 Apr 2022, 03:30
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Set A consists of five consecutive positive integers. Set B consists of every integer that is in set A as well as the sum of every pair of two distinct integers from set A, discarding any numbers that already appear in set B. What is the greatest integer in set B ?
(1) The range of set B is 13.
(2) There are exactly 12 integers in set B.
Are You Up For the Challenge: 700 Level Questions
(1) The range of set B is 13.
(2) There are exactly 12 integers in set B.
Are You Up For the Challenge: 700 Level Questions
06 Apr 2022, 03:56
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We have five consecutive positive integers in set A:
a, a+1, a+2, a+3, a+4
To make set B, we use these five integers, along with any possible sum of two of them. When we add two positive integers, the sum will become larger, so the smallest thing in set B will be 'a'. The largest thing in set B will be the sum of the two largest things in A, which are a+3 and a+4, so the largest value in B will be 2a + 7. The range of B is thus 2a + 7 - a = a + 7, and if, as Statement 1 tells us, the range is 13, then a = 6, and the largest value in set B must be 19. So Statement 1 is sufficient.
Statement 2 really doesn't tell us much, because it will be true for almost any set you can make, unless you use very small numbers. If, say, set A is the set 100, 101, 102, 103, 104, then set B will contain these five values, along with all of the values between 100+101 = 201 (the smallest sum you can make) and 103+104 = 207 (the largest sum you can make). So Set B will contain the five values between 100 and 104, and the seven values between 201 and 207, and will contain 12 values in total. As long as the values in A don't overlap with the sums you can make, you'll have 12 values in set B (though you'll have less than 12 if you use a set of small numbers, like 1, 2, 3, 4, 5, because then some of the sums overlap with the original set A).
So the answer is A.
a, a+1, a+2, a+3, a+4
To make set B, we use these five integers, along with any possible sum of two of them. When we add two positive integers, the sum will become larger, so the smallest thing in set B will be 'a'. The largest thing in set B will be the sum of the two largest things in A, which are a+3 and a+4, so the largest value in B will be 2a + 7. The range of B is thus 2a + 7 - a = a + 7, and if, as Statement 1 tells us, the range is 13, then a = 6, and the largest value in set B must be 19. So Statement 1 is sufficient.
Statement 2 really doesn't tell us much, because it will be true for almost any set you can make, unless you use very small numbers. If, say, set A is the set 100, 101, 102, 103, 104, then set B will contain these five values, along with all of the values between 100+101 = 201 (the smallest sum you can make) and 103+104 = 207 (the largest sum you can make). So Set B will contain the five values between 100 and 104, and the seven values between 201 and 207, and will contain 12 values in total. As long as the values in A don't overlap with the sums you can make, you'll have 12 values in set B (though you'll have less than 12 if you use a set of small numbers, like 1, 2, 3, 4, 5, because then some of the sums overlap with the original set A).
So the answer is A.
General Discussion
06 Oct 2023, 09:01
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IanStewart
Bunuel chetan2u @IanStewart- In statement 2, will we not use 5C2 =10 additional numbers in Set B? So in total 15 numbers? So it will mean 3 numbers are repeated and therefore discarded from Set B but overall it won't give us much so the statement 2 is insufficient. Just trying to understand whether there will be 12 values or 15 values based on calc.
