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9 consecutive integers that start with m + 2

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by sanju09 » Wed Feb 18, 2009 6:14 am
The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with m + 2?

A. m + 4
B. n + 6
C. n + 3
D. m + 5
E. n + 4
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Re: 9 consecutive integers that start with m + 2

by x2suresh » Wed Feb 18, 2009 6:25 am
sanju09 wrote:The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with m + 2?

A. m + 4
B. n + 6
C. n + 3
D. m + 5
E. n + 4
m,m+1,m+2,m+3,m+4

n=m-1+15/5= m+3

m+2,.. m+10

m+1+1, m+1+2,.... m+1+9

avg = m+1+ (9*10)/2*9
= m+6
=n+3

C
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by billzhao » Wed Feb 18, 2009 6:52 am
My answer is n+4 ,

What is OA?
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n+4

by kanha81 » Wed Feb 18, 2009 11:48 am
This is how I solved; however a shorter and a quicker version is appreciated.

A(5)=n=(m+..+m+4)/5
=> n=5(m+2)/5
=> n=m+2

A(9)=(m+2..+m+10)/9
=> A(9)=9(m+6)/9
=> A(9)=m+6=m+2+4=n+4

Hence, [E]

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Re: n+4

by sanju09 » Thu Feb 19, 2009 3:06 am
kanha81 wrote:This is how I solved; however a shorter and a quicker version is appreciated.

A(5)=n=(m+..+m+4)/5
=> n=5(m+2)/5
=> n=m+2

A(9)=(m+2..+m+10)/9
=> A(9)=9(m+6)/9
=> A(9)=m+6=m+2+4=n+4

Hence, [E]

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by Mr2Bits » Thu Feb 19, 2009 7:21 am
Answer is E.

Easiest way is to plug in values for M

M=1

(1+2+3+4+5)/5 = 3=N

M+2 = 3
(3+4+5+6+7+8+9+10+11)/9 = 7 = N+4
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