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# If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002,

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If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002,  [#permalink]

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Updated on: 02 Nov 2012, 02:37
5
00:00

Difficulty:

55% (hard)

Question Stats:

69% (02:01) correct 31% (02:12) wrong based on 221 sessions

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If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and x is 999, what is the value of x?

A. 999
B. 1001
C. 1003
D. 1004
E. 1005

Originally posted by ruturaj on 09 Jul 2011, 02:21.
Last edited by Bunuel on 02 Nov 2012, 02:37, edited 1 time in total.
Renamed the topic, edited the question and moved to PS forum.
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Re: Faster way to solve such questions  [#permalink]

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13 Jul 2011, 05:00
3
2
ruturaj wrote:
If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and x is 999, what is the value of x?
999
1001
1003
1004
1005

When I say the mean of the ages of a group is 25, it means that effectively every person's age can be replaced by 25 and the sum of ages will remain the same. The sum of deviation of all numbers from the mean is 0. Deviation means how much the number is more or less than the mean.

If mean is 999, deviation of 993 from mean is -6.
Deviation of 994 from mean is -5.
The sum of all deviations except that of x is:
- 6 - 5 - 3 - 2 - 1 + 2+ 2 + 3 + 5 = -5
But this sum should be 0. Hence deviation of x from 999 should be +5. Therefore, x should be 1004.
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Re: Faster way to solve such questions  [#permalink]

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09 Jul 2011, 03:42
7
4
ruturaj wrote:
If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and x is 999, what is the value of x?
999
1001
1003
1004
1005

I think the fastest way to do this is a simple understanding of mean .
Sum of the deviations of the numbers in the set from the mean is always zero

993, 994, 996, 997, 998, 1001, 1001, 1002, 1004

mean is 999

so the list is -6-5-3-2-1+2+2+3+5... this shud total to zero
but this is -5 , hence we need a number that is 5 more than the mean to get a +5 and make it zero
hence the answer is 999+ 5 1004

D
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Re: Faster way to solve such questions  [#permalink]

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09 Jul 2011, 03:20
1
993 = 1000 - 7

994 = 1000 - 6

996 = 1000 - 4

997 = 1000 - 3

998 = 1000 - 2

1001 = 1000 + 1

1001 = 1000 + 1

1002 = 1000 + 2

1004 = 1000 + 4

So Sum of terms = (1000 * 9 - 22 + 8) + x

Average = (9000 - 14 + x)/10 = 999

999 = 1000 - 1

So (9000 - 14 + x) = 10000 - 10

=> x = 1000 + 4 = 1004

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Re: Faster way to solve such questions  [#permalink]

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09 Jul 2011, 03:36
As avg is given we can divide them in 2 groups
1st is of elements below 999 and 2nd is ofelements abv 999
if we find now the diff that each element has with 999
that is (999-each element) for the 1st group (eg. 999-998=1,999-997=2,....)
and (each element-999) for the 2nd group(eg. 1001-999=2,1002-999,....)
now add the diff for each group separately

for 1st group sum is 17 and for 2nd group its 12

17-12=5

so we need an element which is in the 2nd group and 5 more than 999 thats 1004

This method relies on the fact that avg is the middle point of a set from Weightage point of view....
I think this way it cab be solved within a min...
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Re: Faster way to solve such questions  [#permalink]

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09 Jul 2011, 05:22
2
$$\frac{3 + 4 + 6 + 7 + 8 + 11 + 11 + 12 + 14 + x}{10} = 9$$

$$\frac{76 + x}{10} = 9$$

$$x = 90 - 76 = 14$$

$$990 + 14 = 1004$$
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Re: Faster way to solve such questions  [#permalink]

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01 Nov 2012, 22:59
ruturaj wrote:
If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and x is 999, what is the value of x?
999
1001
1003
1004
1005

This problem can be solved in 8 secs

When you have sequences like this with very little differences we may deviate with signs and differences.
To get rid of them and wasting time in adding big numbers, we can replace these big numbers with very small numbers.

Remember, every sec in GMAT is Invaluable

example:
If 993(the first term) is assumed as 1, 994 --> 993+1 is replaced by 1+1 ie 2, and so on so forth,then the sequence will be

Mean is 999 which is 993+6 ie 1+6=7

(1+2+4+5+6+9+9+10+12+x)/10=7
x=12, which is 1 + 11 = 993+11= 1004

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Re: If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002,  [#permalink]

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01 Oct 2014, 03:33
Mean = 999

Dividing the numbers in 2 groups; < 999 & > 999

Group I : < 999

999-6 = 993

999-5 = 994

999-3 = 996

999-2 = 997

999-1 = 998

Total -ve = -17

Group II: > 999

999+2 = 1001

999+2 = 1001

999+3 = 1002

999+5 = 1004

Total +ve = 12

For the mean (including x) to be 999, total +ve should be equal to total -ve

So, 999+5 = 1004

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If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002,  [#permalink]

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14 Feb 2015, 05:59
Great tips! I did it somehow different:

993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 , x | M = 999

1000*4...=.4000
900*5.....=.4500
90*5.......=..450
+28........=....28
+8..........=.....8

$$\frac{8986+x}{10}$$= 999

x = 1004
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Re: If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002,  [#permalink]

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12 Feb 2018, 17:36
ruturaj wrote:
If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and x is 999, what is the value of x?

A. 999
B. 1001
C. 1003
D. 1004
E. 1005

Since the numbers are “pretty” large and each is “pretty” close to the average, without using the conventional formula (i.e., average = sum/quantity), we can determine the value of x by using the difference between each known number and the average. If we subtract 999, the average, from each number, we have:

-6, -5, -3, -2, -1, 2, 2, 3, 5

We see that the sum of the differences is (-17) + 12 = -5. To counter the -5 (i.e., 5 less than the average), we need a value that is 5 more than the average. That is, we need x to be 999 + 5 = 1004 so that the average of all values (including x) will be 999.

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Re: If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002,  [#permalink]

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29 Aug 2019, 06:34
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Re: If the average of 993, 994, 996, 997, 998, 1001, 1001, 1002,   [#permalink] 29 Aug 2019, 06:34
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