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

Question

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

KarishmaB · Accepted Answer

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.

sudhir18n · Answer

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

subhashghosh · Answer

993 = 1000 - 7

994 = 1000 - 6

996 = 1000 - 4

997 = 1000 - 3

998 = 1000 - 2

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

Answer - D

Anonymous · Answer

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...

pike · Answer

\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

ChallengeAnything · Answer

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

PareshGmat · Answer

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

Answer = D

pacifist85 · Answer

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
Addition..=8986

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

x = 1004

JeffTargetTestPrep · Answer

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.

Answer: D

GMATGuruNY · Answer

Sum of the 10 integers = (count)(average) = 10*999 = 999  0  .
Implication of the blue digit above:
When the units digits of the 10 integers are added together, the resulting sum must have a units digit of 0.
Sum of the given units digits = 3+4+6+7+8+1+1+2+4 = 36.
When added to 36, the correct answer must yield a units digit of 0.
Only D works:
36 + 1004 = 1040

D

JoeAa · Answer

fab and efficient calculation

Imrane12 · Answer

I see this question differently from all the previously given ways of answering.

Taking all the last digits of the numerator numbers before x, it gives something even. Multiplying the other side of the equation by 10 to get rid of the denominator in the equation, we have 9990 and that is a even number. It simply means that x gotta be even so that the addition + x = 9990, in other words:

e+ ... = e, "..." got to be even. Looking at the answers, we are actually blessed to have only one given even number, so it has to be D. Yes, in the case we had two even numbers in the given solutions, it would have been harder, but here we can leverage the clearly indicative answers with assurance. Even if we had two even numbers as solution options, we could still compute the last digits of the numbers before x and find the exact number we are looking for as unit digit of the correct even number in the solution options.

I think this approach solves the answer without even having to worry about computing derivation, simply using logic of odd / even

bumpbot · Answer

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

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