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30 May 2020, 16:15
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
30% (02:23) correct
70%
(02:12)
wrong
based on 76
sessions
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Not Attempted Yet
S= {1,2,3,4,5,6,7,8,9,10,11,12}
At each step, we select any two numbers from the list and replace them with either their sum or their absolute difference. We repeat the step until there is only 1 number left in the list S. Which of the following can be that last number?
A. 43
B. 50
C. 65
D. 77
E. 82
At each step, we select any two numbers from the list and replace them with either their sum or their absolute difference. We repeat the step until there is only 1 number left in the list S. Which of the following can be that last number?
A. 43
B. 50
C. 65
D. 77
E. 82
yashikaaggarwal
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30 May 2020, 18:21
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IMO B
Let say none of the no. Is subtracted and is added onto one another.
The absolute sum is 78
So option E can never be the choice.
Option D = 77 (78-1) But we need a pair of two to determine one. Let say 2-1=1 but the changes in 78 will be like 78-2-1+1=76. Option D not possible.
Option C = 65 (78-13)
Let say 12+1=13 is subtracted from whole. But, the operation won't be just 78-13. The no. 1&12 won't be added to the sum as well.
So (2+3)+(4+5)+(6+7)+(8+9)+(10+11)-(12+1)
=52 ((option C out))
(Moreover the pair
Option A = 43 (78-35)
Just like option C just 35 can't be minus. (It can be (1+3)+(4+5)+(6+7)+(8+9)-(11+12)+(10+2)
It will be 78-35-35 that is 8. Not possible
(We can see a multiple of two us always deducted from the total to drive value and C And A were lacking that)
Option B = 50 (78-28)
28 is an even no. (-14-14)
The operation can be
(1+3)+(4+5)+(6+7)+(8+9)+{(10+11)-(12+2)}
= 50 (answer)
Posted from my mobile device
Let say none of the no. Is subtracted and is added onto one another.
The absolute sum is 78
So option E can never be the choice.
Option D = 77 (78-1) But we need a pair of two to determine one. Let say 2-1=1 but the changes in 78 will be like 78-2-1+1=76. Option D not possible.
Option C = 65 (78-13)
Let say 12+1=13 is subtracted from whole. But, the operation won't be just 78-13. The no. 1&12 won't be added to the sum as well.
So (2+3)+(4+5)+(6+7)+(8+9)+(10+11)-(12+1)
=52 ((option C out))
(Moreover the pair
Option A = 43 (78-35)
Just like option C just 35 can't be minus. (It can be (1+3)+(4+5)+(6+7)+(8+9)-(11+12)+(10+2)
It will be 78-35-35 that is 8. Not possible
(We can see a multiple of two us always deducted from the total to drive value and C And A were lacking that)
Option B = 50 (78-28)
28 is an even no. (-14-14)
The operation can be
(1+3)+(4+5)+(6+7)+(8+9)+{(10+11)-(12+2)}
= 50 (answer)
Posted from my mobile device
31 May 2020, 17:43
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Firstly, we have the numbers 1, 2, …, 12. The sum of these numbers is 78 (Eliminate E), call it X. We select two numbers, says, a and b and replace them with either their sum a + b or their difference a - b(a>=b).
If we replace these two numbers by their sum, then the sum of all numbers remains the same.
If we replace these two numbers by their difference, then the sum of all numbers on paper is X - (a + b) + (a - b) = X - 2b, still an even number.
So, for each step, the sum of the numbers still remains as an even number. We can eliminate A, C and D.
B
If we replace these two numbers by their sum, then the sum of all numbers remains the same.
If we replace these two numbers by their difference, then the sum of all numbers on paper is X - (a + b) + (a - b) = X - 2b, still an even number.
So, for each step, the sum of the numbers still remains as an even number. We can eliminate A, C and D.
B
nick1816
