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Is the average (arithmetic mean) of x, y and z equal to 16? 19 Feb 2020, 02:23
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E
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35% (medium)

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67% (01:50) correct 33% (00:59) wrong based on 64 sessions

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Is the average (arithmetic mean) of x, y and z equal to 16?

(1) The sum of x and z is equal to 50.

(2) The sum of x and y is equal to 46
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E
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Is the average (arithmetic mean) of x, y and z equal to 16? 19 Feb 2020, 03:20
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Q. Is x+y+z=48 ?

Let's start with (1)+(2) : x+z = 50 and x+y = 46
If x=49, z=1, y=-3 then x+y+z=47 (No)
If x=48, z=2, y=-2 then x+y+z=48 (Yes)
Since we cannot even deduce from (1)+(2) whether x+y+z=48 or not, we can eliminate options (C),(A),(B),(D)

FINAL ANSWER IS (E)
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Is the average (arithmetic mean) of x, y and z equal to 16? 19 Feb 2020, 03:25
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Both answer possible as nopthing is mentioned about x which can also be negative value.
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Is the average (arithmetic mean) of x, y and z equal to 16? 19 Feb 2020, 04:33
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(a) x+z=50
(b) x+y=46

we have 2 equations with 3 unknowns
we don't have limits for the numers and because of this we can have x+y+z=50+z and z be any possible number and the equation can be adjusted to fit the set Z
Thus E, not enough info

I know the explanation is bad, but it somehow makes sense to me haha, I need to see the other answeres to know how to explain it better

Answer E
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Is the average (arithmetic mean) of x, y and z equal to 16? 19 Feb 2020, 05:40
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Is the average (arithmetic mean) of x, y and z equal to 16?

(1) The sum of x and z is equal to 50.

(2) The sum of x and y is equal to 46

We are given that, (x + y +z)/3 =16?
From statement (1), (x + z) = 50
(x + z) = 50, we do not know the value of y. Insufficient.

From statement (2), x + y =46 but z is unknown. Insufficient.

Combining both statements, we have x + z + x + z = 50 + 46 = 96
Insufficient.
Answer: E
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Is the average (arithmetic mean) of x, y and z equal to 16? 19 Feb 2020, 08:23
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Quote:
Is the average (arithmetic mean) of x, y and z equal to 16?

(1) The sum of x and z is equal to 50.

(2) The sum of x and y is equal to 46

if avg of xyz = 16, then sum xyz = 16*3 = 48

(1) insufic

x+z=50, y=?

(2) insufic

x+y=46, z=?

(1/2) insufic

if xyz=48, then x+x+y+z=96, x+48=96, x=48
if x=48, then 48+y=46, y=-2; 48+z=50, z=2

if xyz≠48, then z-y=4, z=4+y; x=46-y
x+y+z=y+(46-y)+(4+y)=y+50, y=?
since we don't know y, insufic

Ans (E)
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Is the average (arithmetic mean) of x, y and z equal to 16? Updated on: 20 Feb 2020, 02:36
Originally posted by Archit3110 on 19 Feb 2020, 09:18.
Last edited by Archit3110 on 20 Feb 2020, 02:36, edited 1 time in total.
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to determine whether x+y+z= 48
#1
The sum of x and z is equal to 50
answer is yes if value of y is -ve integer and yes if y is +ve integer; insufficient
#2
The sum of x and y is equal to 46
insufficient
from 1 &2
we dont get values
IMO E: sufficient

Is the average (arithmetic mean) of x, y and z equal to 16?

(1) The sum of x and z is equal to 50.

(2) The sum of x and y is equal to 46
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Is the average (arithmetic mean) of x, y and z equal to 16? 19 Feb 2020, 16:10
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(1) x + z = 50 but we can not calculate average since y value is not known. --> Not sufficient
(2) same as 1, z value is not known. so average can not be calculated --> Not sufficient

Let's combine both - all we know from both (1) and (2) is y is 4 units less than z.

So any of the following combinations is possible

z=0, y=-4, x=50 --> Avg = 46/3 = 15.*
z=3, y=-1, x=47 --> Avg =49/3 = 16.*

(1) and (2) Together do not help find out the avg of x, y, z.

The answer is E.
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