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The average of 8 numbers is 21 .I f each of the numbers is

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The average of 8 numbers is 21 .I f each of the numbers is [#permalink] New post 14 Oct 2003, 09:16
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The average of 8 numbers is 21 .I f each of the numbers is multiplied by 8, the average of the new set of numbers is?
A. 8
B. 21
C. 29
D. 168
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shubhangi

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 [#permalink] New post 14 Oct 2003, 09:38
168

(n + (n + 1) + (n + 2) + ... + (n + 7)) / 8 = 21 ...... (1)

(8n + 8(n + 1) + ... + 8(n + 7))/8 = ?
= 8 * (n + (n + 1) + ... + (n + 7))/8
= 8 * 21 [ using (1) ]
= 168

Of course,
The other way doing this is by balancing out the original equation by multiplying by 8 on both sides.
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question [#permalink] New post 14 Oct 2003, 17:04
wonder_gmat wrote:
168

(n + (n + 1) + (n + 2) + ... + (n + 7)) / 8 = 21 ...... (1)

(8n + 8(n + 1) + ... + 8(n + 7))/8 = ?
= 8 * (n + (n + 1) + ... + (n + 7))/8
= 8 * 21 [ using (1) ]
= 168

Of course,
The other way doing this is by balancing out the original equation by multiplying by 8 on both sides.


Wonder,
Does your solution assume that the nums are consecutive?

Thanks,
exy18
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Re: question [#permalink] New post 14 Oct 2003, 17:14
exy18 wrote:
wonder_gmat wrote:
168

(n + (n + 1) + (n + 2) + ... + (n + 7)) / 8 = 21 ...... (1)

(8n + 8(n + 1) + ... + 8(n + 7))/8 = ?
= 8 * (n + (n + 1) + ... + (n + 7))/8
= 8 * 21 [ using (1) ]
= 168

Of course,
The other way doing this is by balancing out the original equation by multiplying by 8 on both sides.


Wonder,
Does your solution assume that the nums are consecutive?

Thanks,
exy18

Not necessarily. I just used consecutive integers here for sake of ease in explaining. The bottom line is you have to multiply the original average by the same number that you multiply the original numbers.
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 [#permalink] New post 14 Oct 2003, 20:14
To write it siimply:
(x1 + x2 +... + x8)/8 = 21 (given) ............ a

Multiply both sides of equation a by 8

(8*x1 + 8*x2 +.... + 8*x8)/8 = 21*8 = 168

thanks
  [#permalink] 14 Oct 2003, 20:14
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The average of 8 numbers is 21 .I f each of the numbers is

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