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26 May 2016, 11:39
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D
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What is the average (arithmetic mean) of fifteen consecutive integers?
(1) The average of the first eleven integers is 7.
(2) The average of the last eleven integers is 9.
(1) The average of the first eleven integers is 7.
(2) The average of the last eleven integers is 9.
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26 May 2016, 22:55
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Since the integers are consecutive, knowing a particular term is sufficient to answer the question.
St1: The average of the first eleven integers is 7. --> 11/2 * (a + a + 10) = 77
a + 5 = 7
a = 2
We can find the average of 15 consecutive integers
Sufficient
St2: The average of the last eleven integers is 9. --> Similar to statement 1. We can find the 5th term and the subsequent average of the sequence.
Sufficient.
Answer: D
St1: The average of the first eleven integers is 7. --> 11/2 * (a + a + 10) = 77
a + 5 = 7
a = 2
We can find the average of 15 consecutive integers
Sufficient
St2: The average of the last eleven integers is 9. --> Similar to statement 1. We can find the 5th term and the subsequent average of the sequence.
Sufficient.
Answer: D
26 May 2016, 23:25
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Key to solving this problem are words: consecutive integers.
Statement 1
The average of the first eleven integers is 7. This means the sum of these eleven integers is 77.
This is nothing but an AP with a common difference of 1. Our job remains to find "a", the first term.
Once "a" is known, I can find sum of fifteen consecutive integers. Sum is known, then average is also known to me.
AP formula for sum is \(\frac{n}{2}\) [2a+(n-1)*d]
here n=11, d=1, a is the unknown.
2a+10 = 14; a is known.
Statement 1 is sufficient
Statement 2
Given: The average of the last eleven integers is 9
The first term of this AP is nothing but the 5th term of combined.
For an AP nth term = a+(n-1)d
Here 5th term = a+4
Applying AP formula for sum is \(\frac{n}{2}\) [2a+(n-1)*d]
here n=11, d=1, "a" here is (a+4)
2(a+4)+ 10 = 18; a is known.
Statement 2 is sufficient
So Answer would be D
Statement 1
The average of the first eleven integers is 7. This means the sum of these eleven integers is 77.
This is nothing but an AP with a common difference of 1. Our job remains to find "a", the first term.
Once "a" is known, I can find sum of fifteen consecutive integers. Sum is known, then average is also known to me.
AP formula for sum is \(\frac{n}{2}\) [2a+(n-1)*d]
here n=11, d=1, a is the unknown.
2a+10 = 14; a is known.
Statement 1 is sufficient
Statement 2
Given: The average of the last eleven integers is 9
The first term of this AP is nothing but the 5th term of combined.
For an AP nth term = a+(n-1)d
Here 5th term = a+4
Applying AP formula for sum is \(\frac{n}{2}\) [2a+(n-1)*d]
here n=11, d=1, "a" here is (a+4)
2(a+4)+ 10 = 18; a is known.
Statement 2 is sufficient
So Answer would be D
