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Source: — Problem Solving |
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by shankar.ashwin » Wed Oct 19, 2011 9:49 pm
Let x=1.
5 consec int - (1,2,3,4,5)
Avg = a =3.

Now if first number is x+3 = 4
We have (4,5,6,7,8,9,10,11,12)
And Avg = 8 (or) a+5 B
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by rijul007 » Wed Oct 19, 2011 9:54 pm
x, x+1, x+2, x+3, x+4

5x+10 = 5a
5x = 5a-10
x = a-2

Sum of 9 consecutive integers = 9x+3+4+5+6+7+8+9+10+11 =9x+9(3+11)/2 [sum of AP]
=>9x + 63

average = (9x+63)/9 = x+7 = a-2 +7 [because x=a-2]
avg = a+5


B is the correct option
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by GMATGuruNY » Wed Oct 19, 2011 9:57 pm
silverflamein wrote:The average of 5 consecutive integers starting with x is a. What is the average of 9 consecutive integers that start with x+3 ?

A - x+a
B - a+5
C - 2x
D - 2a
E - 2x+a
With evenly spaced integers, the average = the median.

Let x=1.
The integers are 1,2,3,4,5.
a = 3.

Given x+3 = 1+3 = 4:
The integers are 4,5,6,7,8,9,10,11,12.
Average = 8. This is our target.

Now we plug x=1 and a=3 into the answers to see which yields our target of 8.

Only answer choice B works:
a+5 = 3+5 = 8.

The correct answer is B.
Last edited by GMATGuruNY on Thu Oct 20, 2011 7:15 am, edited 2 times in total.
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by Anurag@Gurome » Wed Oct 19, 2011 11:57 pm
silverflamein wrote:The average of 5 consecutive integers starting with x is a. What is the average of 9 consecutive integers that start with x+3 ?
Average of 5 consecutive integers starting with x will be the 3rd integer = (x + 2)
Hence, (x + 2) = a

Average of 9 consecutive integers starting with (x + 3) will be the 5th integer = (x + 3) + 4 = (x + 7) = (a + 5)

The correct answer is B
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by rohit_gmat » Thu Oct 20, 2011 3:28 am
used mitch's method as shown above..

but while finding the avg did - 4+12/2 = 8

(but i wasted a lil time at first coz i did 4+13/2 then i was stuck :P so be careful :P)
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