w, x and y are positive integers such that w ≤ x ≤ y. 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
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Arithmetic mean
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sivaelectric
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Frankenstein
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Hi,
avg of w,x,y is 20. So w+x+y = 20*3 = 60
From(1): y=28 => w+x = 32
a)w=16,x=16. w>15
b)w=14,x=18. w<15
Insufficient
From(2):
a)w=17,y=20,z=23... w>15
b)w=14,y=17,z=29... w<15
Insufficient
Both (1)&(2): w+x = 32 and one of them is 17. So, w=15 and x=17 as w ≤ x
Sufficient
Hence C
avg of w,x,y is 20. So w+x+y = 20*3 = 60
From(1): y=28 => w+x = 32
a)w=16,x=16. w>15
b)w=14,x=18. w<15
Insufficient
From(2):
a)w=17,y=20,z=23... w>15
b)w=14,y=17,z=29... w<15
Insufficient
Both (1)&(2): w+x = 32 and one of them is 17. So, w=15 and x=17 as w ≤ x
Sufficient
Hence C
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cans
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w<=x<=y
w+x+y=60
a)y=28
w+x=32
both w,x can be 16. also w can be 14 and x can be 18. insufficient
b) y can't be 17 as its largest.
if x=17, we can arrange w and y any way we want. Thus insufficient
a&b together) y=28. w+x=32. if w=17,x=15 not possible
thus x=15 and w=15
IMO C
w+x+y=60
a)y=28
w+x=32
both w,x can be 16. also w can be 14 and x can be 18. insufficient
b) y can't be 17 as its largest.
if x=17, we can arrange w and y any way we want. Thus insufficient
a&b together) y=28. w+x=32. if w=17,x=15 not possible
thus x=15 and w=15
IMO C
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Cans!!
