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Junior | Next Rank: 30 Posts
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by smashinonions » Fri Sep 26, 2008 4:24 am
Another one
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by nikhilagrawal » Fri Sep 26, 2008 4:36 am
ans is d,

a) since avg is close to 21, it tells more sweater were sold.

b) on solving d eqn I got 20 shirts and 12 sweaters (hit and trail).


Let me my ans is right or wrong .
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by 2009wish » Fri Sep 26, 2008 5:16 am
from A we get the equation ----

2 (no of shirts) = 3 (no of sweaters )

from this we know that number of shirts is more than number of sweaters.

from B we get the eq------

3x + 5y=84 (after simplification) , as x and y are objects and so have to be positive numbers.

both are individually sufficient......D

what is the OA?
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by manulath » Fri Sep 26, 2008 5:42 am
Let shirts be x and sweaters be y

I: 15x + 25y = 21(x+y)
or 4y = 6x
y = 1.5x

sufficient - more sweaters are sold.

II: 15x + 25y = 420
3x + 5y = 84
3x = 84 - 5y
x = 28 - 5y/3

As you see x and y have to be whole numbers
so y has to be multiple of 3
y=3, x=23
y=6, x=18
y=9, x=13
y=12, x=8
y=15, x=3

so II alone is not sufficient

Answer should be A
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by vishubn » Fri Sep 26, 2008 7:03 am
shirt - 15 $
sweater-25$

# of sweaters > # of shirts???

ad/bce

statement 1- avg= n(15)+m(25)/n+m=21
two unknows strike out option ad

statement 2- n(15) + m(25)=420

Statement 1 + statement 2--> simultaneous equation

OA -c ???

Vishu
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by niraj_a » Fri Sep 26, 2008 7:30 am
this question has been addressed earlier here -

https://www.beatthegmat.com/gmat-prep-sh ... 10447.html
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