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# the average of 5 consecutive numbers is 87. if each of the

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VP
Joined: 30 Sep 2004
Posts: 1480
Location: Germany
the average of 5 consecutive numbers is 87. if each of the [#permalink]

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15 Feb 2005, 03:53
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the average of 5 consecutive numbers is 87. if each of the original numbers is reduced by 2 and then divided by 5, the average of the final set of numbers is

a 14,4
b 15,2
c 16,4
d 17,0
e 18,5
VP
Joined: 13 Jun 2004
Posts: 1115
Location: London, UK
Schools: Tuck'08

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15 Feb 2005, 04:35
Same...D

87*5 = 435
435 - 2*10 = 425
425/5 = 85
85/5 = 17

Manager
Joined: 11 Jan 2005
Posts: 57
Location: Mexico City

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15 Feb 2005, 09:50
I like the (87-2)/5 approach as a time saver.

I did it by multiplying out 87*5 then subtracting 10 and dividing by two bringing me to 85 which i divided by 5.

(87-2)/5 is money.
Manager
Joined: 13 Feb 2005
Posts: 63
Location: Lahore, Pakistan

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15 Feb 2005, 22:42
dont we have to divide all the numbers by 5? thus dividing the sum with 25?
VP
Joined: 13 Jun 2004
Posts: 1115
Location: London, UK
Schools: Tuck'08

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15 Feb 2005, 22:54
faizaniftikhar83 wrote:
dont we have to divide all the numbers by 5? thus dividing the sum with 25?

The exact sentence from the problem is :

if each of the original numbers is reduced by 2 and then divided by 5, the average of the final set of numbers is ...

But you thought it was the following one :

if each of the original numbers is reduced by 2 , the average of the final set of numbers is
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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15 Feb 2005, 23:18
(87-2)/5 = 17

Try out some sets to see that this is true =)
Manager
Joined: 13 Feb 2005
Posts: 63
Location: Lahore, Pakistan

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16 Feb 2005, 01:44
Antmavel wrote:
faizaniftikhar83 wrote:
dont we have to divide all the numbers by 5? thus dividing the sum with 25?

The exact sentence from the problem is :

if each of the original numbers is reduced by 2 and then divided by 5, the average of the final set of numbers is ...

But you thought it was the following one :

if each of the original numbers is reduced by 2 , the average of the final set of numbers is

but still....the clause in blue means that we have to divide each number by 5...not divide the sum with 5
VP
Joined: 13 Jun 2004
Posts: 1115
Location: London, UK
Schools: Tuck'08

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16 Feb 2005, 02:44
divide each number by five and do the sum is the same than do the sum and divide by five...try with few exemples :

(5+6+7)/3 = 5/3+6/3+7/3 = 6

(4+5+6+7)/2 = 4/2+5/2+6/2+7/2 = 11
Manager
Joined: 13 Feb 2005
Posts: 63
Location: Lahore, Pakistan

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16 Feb 2005, 02:54
Antmavel wrote:
divide each number by five and do the sum is the same than do the sum and divide by five...try with few exemples :

(5+6+7)/3 = 5/3+6/3+7/3 = 6

(4+5+6+7)/2 = 4/2+5/2+6/2+7/2 = 11

.... ....how silly of me.....thanx for helping me out in this scenario antmavel
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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16 Feb 2005, 07:48
D is is...

X/5=87

x=87*5=435

(X1-2)+...(x5-2)= 435-10=425!

425/25=17!
16 Feb 2005, 07:48
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