Is the average (arithmetic mean) of w, x, y, and z less than 28?

Question

Competition Mode Question

(1) The average (arithmetic mean) of w, x, and y is less than 28.
(2) The average (arithmetic mean) of x, y, and z is less than 28.

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unraveled · Answer

Is the average (arithmetic mean) of w, x, y, and z less than 28?
 Mean = \frac{w + x + y + z }{ 4}  < 28
So, Sum of w,x,y and z = w + x + y + z < 112

(1) The average (arithmetic mean) of w, x, and y is less than 28.
Sum w + x + y < 84
Nothing about z so both possibilities exist. Low z -  Mean  < 28. If z is relatively high then  Mean  > 28

INSUFFICIENT.

(2) The average (arithmetic mean) of x, y, and z is less than 28.
Sum x + y + z < 84
Similar to statement 1 both possibilities exist.

Together 1 and 2.
Adding both sums from the two statements
w + 2(x + y) + z < 168
Again nothing concrete about individual values is available.

Answer E.

exc4libur · Answer

average=sum/terms, sum=average(terms);

Is the sum(w,x,y,z)<28(4)?

(1) The average (arithmetic mean) of w, x, and y is less than 28.   insufic

sum(w,x,y)<28(3)

(2) The average (arithmetic mean) of x, y, and z is less than 28.   insufic

sum(z,x,y)<28(3)

(1 and 2)   insufic

sum(w,x,y)<28(3)
sum(z,x,y)<28(3)
w+z+2x+2y<28(6)

Ans (E)

Archit3110 · Answer

find
is sum w+x+y+z<112
#1
The average (arithmetic mean) of w, x, and y is less than 28
w+x+y<84
no info z
insufficient
#2The average (arithmetic mean) of x, y, and z is less than 28
x+y+z<84
no info about w ; insufficient
from 1 &2
we get both yes and no of 
w+x+y+z<112
IMO E

Is the average (arithmetic mean) of w, x, y, and z less than 28?

(1) The average (arithmetic mean) of w, x, and y is less than 28.
(2) The average (arithmetic mean) of x, y, and z is less than 28.

eakabuah · Answer

We are to determine if the arithmetic mean of w, x, y, and z is less than 28.

Statement 1: The average (arithmetic mean) of w, x, and y is less than 28.
Clearly statement 1 is not sufficient. The fact that w,x, and y have a mean less than 28 is not sufficient since the fourth number can be very large and push the mean above 28 or it could be small and the mean will remain below 28.

Statement 2: The average (arithmetic mean) of x, y, and z is less than 28.
Also not sufficient. The fact that x,y, and z, three numbers, have a mean less than 28 does not mean that adding the fourth number will not push the mean above 28 or maintain it below 28 since the fourth number could be very large or less than 28.

1+2
Sufficient.
Statement 1 says that the three numbers (i.e. w,x, and y) can each be replaced by a single number that is less than 28. At the same time, statement 2 says the three numbers (i.e. x,y, and z) can each be replaced by a number that is less than 28. Combining both statements we can infer that the four numbers (i.e. x,y, and z) can each be replaced by a number that is less than 28. Hence the mean of the four numbers is less than 28.

The answer is therefore C.

pandajee · Answer

IMO
Its C..
Example .. considering all positive numbers or one negative number

Posted from my mobile device

minustark · Answer

Is the average (arithmetic mean) of w, x, y, and z less than 28?

(1) The average (arithmetic mean) of w, x, and y is less than 28.
(2) The average (arithmetic mean) of x, y, and z is less than 28.

The question is asking whether w + x + y + z < 112.

(1) w + x + y < 78. still need to know the value of z. not sufficient.

(2) x + y + z < 78. same as 1. not sufficient.

Together, it may be the case that w, x, y, z all are less than 28. again, suppose each of x and y is equal to 1 and each of w and z is equal to 75. in that case the average of w, x, y and z will be greater than 28. so not sufficient.

E is the answer.

madgmat2019 · Answer

(1) The average (arithmetic mean) of w, x, and y is less than 28............here z can be anything high and can change the equation of the question......insufficient
(2) The average (arithmetic mean) of x, y, and z is less than 28.............here w can be anything high and can change the equation of the question......insufficient

Combining we can be < 28.....all of them <28 individually

Also can be > 28 if
In 1) w,x,y.....80,1,2... respectively....sum <84
In 2) x,y,z.....1,2,80... respectively....sum <84

So w,x,y,z....80,1,2,80....avg >28

OA:E

lacktutor · Answer

Is w+ x+ y+ z < 112 ???

(Statement1): w+ x+ y < 84 
No info about what z is. 
Clearly insufficient

(Statement2): x+ y + z < 84
No info about what w is.
Clearly insufficient.

Taken together 1&2, we’ll get that 
w+ x+ y+ z < 168 —(x +y) 
—> (x+y ) could be anything up to 84. Well, it is not sufficient

The answer is E

Posted from my mobile device

chetan2u · Answer

chondro48 We are looking for whether w+x+y+z<28*4....w+x+y+z<112 Clearly each is insufficient individually, so let us see when combined.. (I) \frac{w+x+y}{3}<28.......w+x+y<84 (II) \frac{z+x+y}{3}<28.......z+x+y<84 As all are dealing with Inequality sign '<', it is difficult to get an answer.. Say x and y are 0, w and z can be anything less than 84. If w and z are say 80, 0+0+80+80=160>112 If w and z are also 0, 0+0+0+0=0<112.. If you want to do a bit more of calculations Add the two w+x+y+z+x+y<84+84.......(w+x+y+z)+(x+y)<168 For example.. Say all are 1, then 1+1+1+1<112..YES x=40 and y=40, then w and z will be <84-40-40 or <4...w+z+x+y=3+3+40+40=86<112, so YES x=4 and y=6, then w and z will be <84-4-6 or <74......w+z+x+y=70+70+4+6=150 >112, so NO, not <112

GMATinsight · Answer

Answer: Option E

Video solution by  GMATinsight

https://www.youtube.com/watch?v=941TQxMmhTA

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Is the average (arithmetic mean) of w, x, y, and z less than 28?

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Is the average (arithmetic mean) of w, x, y, and z less than 28?

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