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Two sets of 4 consecutive positive integers have exactly one integer in common. The sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set?
- a)4
- b)7
- c)8
- d)12
- e)it cannot be determined from the information given.
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Two sets of 4 consecutive positive integers have exactly one integer i...
A = { ( a ) , ( a + 1 ) , ( a + 2 ) , ( a + 3 )}
B = { ( a + 3 ) , ( a + 4 ) , ( a + 5 ) , ( a + 6 )}
Sum of A = 4a + 6
Sum of A = 4a + 18
Difference = (4a + 18) - (4a + 6) => 12
The correct answer answer is (D)
Most Upvoted Answer
Two sets of 4 consecutive positive integers have exactly one integer i...
To solve this problem, let's assume the first set of consecutive positive integers as {a, a+1, a+2, a+3} and the second set as {b, b+1, b+2, b+3}, where a and b are positive integers.
Since the two sets have exactly one integer in common, we can deduce that either (a+3) = b or a = (b+3).
Case 1: (a+3) = b
In this case, the sum of the first set is a + (a+1) + (a+2) + (a+3) = 4a + 6.
The sum of the second set is b + (b+1) + (b+2) + (b+3) = 4b + 6.
The difference between the two sums is (4b + 6) - (4a + 6) = 4(b - a).
Since b - a = 3, the difference between the two sums is 4 * 3 = 12.
Case 2: a = (b+3)
In this case, the sum of the first set is (b+3) + (b+4) + (b+5) + (b+6) = 4b + 18.
The sum of the second set is b + (b+1) + (b+2) + (b+3) = 4b + 6.
The difference between the two sums is (4b + 18) - (4b + 6) = 12.
Therefore, in both cases, the sum of the integers in the set with greater numbers is 12 greater than the sum of the integers in the other set. Hence, the correct answer is option (D) 12.
Since the two sets have exactly one integer in common, we can deduce that either (a+3) = b or a = (b+3).
Case 1: (a+3) = b
In this case, the sum of the first set is a + (a+1) + (a+2) + (a+3) = 4a + 6.
The sum of the second set is b + (b+1) + (b+2) + (b+3) = 4b + 6.
The difference between the two sums is (4b + 6) - (4a + 6) = 4(b - a).
Since b - a = 3, the difference between the two sums is 4 * 3 = 12.
Case 2: a = (b+3)
In this case, the sum of the first set is (b+3) + (b+4) + (b+5) + (b+6) = 4b + 18.
The sum of the second set is b + (b+1) + (b+2) + (b+3) = 4b + 6.
The difference between the two sums is (4b + 18) - (4b + 6) = 12.
Therefore, in both cases, the sum of the integers in the set with greater numbers is 12 greater than the sum of the integers in the other set. Hence, the correct answer is option (D) 12.
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Two sets of 4 consecutive positive integers have exactly one integer in common. The sum of the integers in the set with greater numbers is how much greater than the sum of the integers in the other set?a)4b)7c)8d)12e)it cannot be determined from the information given.Correct answer is option 'B'. Can you explain this answer?
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