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by govind_raj_76 » Wed Jun 02, 2010 7:56 pm
Variables x, y and z represent positive numbers. x is directly proportional to y and y is inversely proportional to z.

What is the vale of x if z = 25 ?

1) y=2 whenever z = 50.


2) x=2 whenever z = 100.

Please help how to approach this problem.
Govind
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Source: — Data Sufficiency |
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by liferocks » Wed Jun 02, 2010 8:11 pm
x=my and y=n/z where m and n are constants

so x=mn/z
question is what is mn/25..so we need to know the values of both m and n
From 1
n=50*2..but no information about m.not sufficient

From 2
mn=100*2.......so mn/25=100*2/25 =8....sufficient

Ans option B
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by selango » Wed Jun 02, 2010 8:17 pm
x is directly proportional to y

-->x=ay (a is constant)

y is inversly proportional to z

-->y=b/z (b is constant)

When z =25,y=b/25 and x=ab/25

From stmt1,

z=50,y=b/50

2=b/50

b=2.

We dont know value of a ,so insufficient.

From stmt2,

z=100

y=b/100

x=ab/100

2=ab/100

ab=200

x=200/25=8

So B insufficient.

Option B

Please correct me if I am wrong.
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by eloka » Wed Jun 02, 2010 8:18 pm
Hallo..

I am not sure if I did it right...but this is what I got so far.

Alert: K = constant------> because its proportional so I add K

x,y,z = positive, number

X = K1*Y
Y = K2*Z

Question: what is X if z=25?

1) y=2, z=50

X=K1*Y
Y=K2*Z

.............................................................
Y =2
X=K1*Y
X=K1*2

meanwhile,

X=K1.Y
Y=K2*Z
2=K2*Z

substitute

X=K1*(K2*Z)
X=K1*K2*Z
X=K1*K2*25

we still cannot calculate K1 &K2
NOT UNIQUE

2) X =2, Z=100

X=K1*Y

2=K1*Y
Y=K2*Z
Y=K2*100

2=K1*Y
Y=K2*Z

X=K1*K2*Z
2=K1*K2*100
K1*K2= 2/100
K1*K2=1/50

.......................
the question: what is X if z=25?

X=K1*K2*Z
= 1/50*25
=1/2



UNIQUE

B
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by selango » Wed Jun 02, 2010 8:21 pm
eloka

Y is inversly proportional to z

So y=k/z, not y=kz
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by eloka » Wed Jun 02, 2010 8:25 pm
OOPPS..sorry I didnt read the question carefully. Forgive me..

Math formula (according to my brain)


X= K1*(Y)
Y=K2*(1/Z)


from there, we can solve the problems..
Warm wishes to all of you who are striving to conquer the GMAT

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