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w, x and y are positive integers...

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w, x and y are positive integers...

by The Jock » Mon Sep 06, 2010 9:23 am
w, x and y are positive integers such that . If the average (arithmetic mean) of w, x and y is 20, is w > 15 ?

(1) y = 28

(2) One of the three numbers is 17

With option 1 and 2, I could find that x= 15 but in that case I believe conditions should not suffice because we are looking for x>15.
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by The Jock » Mon Sep 06, 2010 9:25 am
Oh...sorry, for a moment i forgot that in DS answer can be NO also. mea culpa...
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by selango » Mon Sep 06, 2010 9:52 am
w+x+y=60

w>15?

stmt1,

y=28

w+x=32

W=17,x=15,w>15

w=14,x=18,w<15

Insuff

stmt2,

One of 3 numbers is 17

We dont know which one is 17

Insuff

Combining 1 and 2,

w+x=32

We dont know whether W or X is 17.

Insuff

Pick E
Last edited by selango on Mon Sep 06, 2010 10:01 am, edited 1 time in total.
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by The Jock » Mon Sep 06, 2010 9:57 am
Hey,

the answer is C, with options 1 and 2 you come to know that w=15, but w>15, so that answer is a definite no and we got our answer, that is C.
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by selango » Mon Sep 06, 2010 10:03 am
How come you arrived at W=15?
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by The Jock » Mon Sep 06, 2010 5:22 pm
w+x+y = 60

from option 1:
a. w=16, x=16,y=28: Is w>15- yes
b. w=15, x=17,y=28; Is w>15- no

from option 2:
a. w=17, x=18, y=25; Is w>15- yes
b. w=15, x =17, y=28; Is w>15- no

from option 1 and option 2:
w=15, x=17, y=28

so we know that w can not be greater than 15. So C
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by selango » Mon Sep 06, 2010 7:34 pm
Dude,

you are assuming values so that to make w!>15.But you need to consider all possibilites.

Now from ur example I ll rephrase the first stmt

w+x+y = 60

from option 1:

y=28

a. w=17, x=15,y=28: Is w>15- yes
b. w=15, x=17,y=28; Is w>15- no

from option 2:

one of 3 numbers is 17

a. w=17, x=15, y=28; Is w>15- yes
b. w=15, x =17, y=28; Is w>15- no

Now 2 options are available.W=17 or w=15.Still we don't know the value of W.
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by The Jock » Tue Sep 07, 2010 12:06 am
Oh, how can i miss that, really sorry, the answer should be E.
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