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20 May 2017, 03:46
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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In a sequence of 10 consecutive integers, what is the average of all numbers in sequence?
(1) The sum of the first 5 integers is 410
(2) The sum of the last 5 integers is 435
(1) The sum of the first 5 integers is 410
(2) The sum of the last 5 integers is 435
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20 May 2017, 03:57
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In a sequence of 10 consecutive integers, what is the average of all numbers in sequence?
(1) The sum of the first 5 integers is 410
(2) The sum of the last 5 integers is 435
(1) The sum of the first 5 integers is 410
(2) The sum of the last 5 integers is 435
Hi,
We know there are 10 CONSECUTIVE integers..
We are looking at average of these 10 numbers or middle of 5&6 number.
Let's see the statements..
1) The sum of the first 5 integers is 410
So average of these numbers or 3rd number =410/5=82
So 5 & 6 number will be 82+2 & 82+3...
Average=\(\frac{84+85}{2}\)=84.5
Sufficient
(2) The sum of the last 5 integers is 435
same as above 8 number =435/5=87..
So 5&6 number are 87-3 & 87-2..
Average =84+85)/2=84.5
Sufficient
D
20 May 2017, 04:11
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Another way of looking at it: Let the 10 consecutive integers be
a, a+1, a+2, .... a+9
If we know a, we know the answer
Statement 1. Sum of first 5 integers = 410
a + (a+1) +... + (a+4) = 410.
We can find a using this. Sufficient
Statement 2. Sum of last 5 integers = 435
(a+5) + (a+6) +... + (a+9) = 435
We can find a using this. Sufficient
Hence D
(OA is given to be C. Either its wrong or I am missing something)
a, a+1, a+2, .... a+9
If we know a, we know the answer
Statement 1. Sum of first 5 integers = 410
a + (a+1) +... + (a+4) = 410.
We can find a using this. Sufficient
Statement 2. Sum of last 5 integers = 435
(a+5) + (a+6) +... + (a+9) = 435
We can find a using this. Sufficient
Hence D
(OA is given to be C. Either its wrong or I am missing something)
