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15 Jul 2018, 21:41
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C
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Difficulty:
25%
(medium)
Question Stats:
78% (01:58) correct
22%
(01:48)
wrong
based on 225
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Set S consist of the values 11, -9, 100, 0, x, and y. if x not equal to y, what is the median of the set S?
(1) The average (arithmetic mean) of X and Y is 308.
(2) The mode of Set S is 0.
(1) The average (arithmetic mean) of X and Y is 308.
(2) The mode of Set S is 0.
vaibhav1221
Joined: 19 Nov 2017
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Schools: IIMA '25 (A) Columbia CBS'26 (D) Fuqua '26 (D) ISB '25 (WL) Darden '26 (D) Kenan-Flagler '26 (WL)
GMAT 1: 710 Q49 V38
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Schools: IIMA '25 (A) Columbia CBS'26 (D) Fuqua '26 (D) ISB '25 (WL) Darden '26 (D) Kenan-Flagler '26 (WL)
GMAT 1: 710 Q49 V38
Posts: 292
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Updated on: 15 Jul 2018, 23:33
Originally posted by vaibhav1221 on 15 Jul 2018, 21:56.
Last edited by vaibhav1221 on 15 Jul 2018, 23:33, edited 1 time in total.
Last edited by vaibhav1221 on 15 Jul 2018, 23:33, edited 1 time in total.
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Statement 1:
mean of x and y is 308
x= 1 and y = 307
-9,0,1,11,100,307 Median = 6
x = -2 y = 310
-9,-2,0,11,100,310 Median = 5.5
NS
Statement 2:
Mode is 0
x = 0 y =101
-9,0,0,11,100,101 Median = 5.5
x = 0 y = 3
-9,0,0,3,11,100 Median = 1.5
NS
Statement 1&2:
x = 0/308 y = 308/0
-9,0,0,11,100,308 Median = 5.5
Sufficient.
Therefore C
mean of x and y is 308
x= 1 and y = 307
-9,0,1,11,100,307 Median = 6
x = -2 y = 310
-9,-2,0,11,100,310 Median = 5.5
NS
Statement 2:
Mode is 0
x = 0 y =101
-9,0,0,11,100,101 Median = 5.5
x = 0 y = 3
-9,0,0,3,11,100 Median = 1.5
NS
Statement 1&2:
x = 0/308 y = 308/0
-9,0,0,11,100,308 Median = 5.5
Sufficient.
Therefore C
15 Jul 2018, 22:29
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Bunuel
St1:- Given, A.M(x,y)=308 or x+y=616 ;with \(x\neq{y}\),
There are many possibilities of (x,y) which yield x+y=616
So, We have more than one value of median.
Insufficient.
St2:- Mode of set is 0; implies 0 is appears more than one time in the set S.
Since all other members are fixed; hence, either x or y can be 0 not both.
So, when x=0, y may be any number except 0. So, we can't determine a unique median value.
Similarly, when y=0, x may be any number except 0. So, we can't determine a unique median value.
Insufficient.
(1)+(2), we need to fix x+y=626 and x=0 or y=0.
Two cases:
When x=0; then y=626, set S={-9,0,0,11,100,616} So, Median(S)=\(\frac{(0+11)}{2}\)=5.5
When y=0; then x=626, set S={-9,0,0,11,100,616} So, Median(S)=\(\frac{(0+11)}{2}\)=5.5
We have obtained an unique value of median in both the cases.
Sufficient.
Ans. (C)
