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Arithmetic from 11th Edition

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by truplayer256 » Mon Feb 23, 2009 4:29 pm
If m>0 and x is m percent of y, then, in terms of m, y is what percent of x?
x=my/100
y=?/100*x
We want the answer in terms of m. So, from x=my/100, we can solve for m. After we do this, we get:
m=100x/y
We want ?/100 in terms of m. From y=?x/100, ?=100y/x.
100y/x= 10000/m E
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by welcome » Mon Feb 23, 2009 7:20 pm
IMO E.
Shubham.
590 >> 630 >> 640 >> 610 >> 600 >> 640 >> 590 >> 640 >> 590 >> 590
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Re: Arithmetic from 11th Edition

by x2suresh » Mon Feb 23, 2009 7:26 pm
jfranco23 wrote:If m>0 and x is m percent of y, then, in terms of m, y is what percent of x?

A. 100m
B. (1/100m)
C. (1/m)
D. (10/m)
E. (10000/m)
x= m/100*y

y= 100/m *x

y is (100/m)/100 % of x
--> 1/m %

C
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Re: Arithmetic from 11th Edition

by sureshbala » Mon Feb 23, 2009 9:13 pm
jfranco23 wrote:If m>0 and x is m percent of y, then, in terms of m, y is what percent of x?

A. 100m
B. (1/100m)
C. (1/m)
D. (10/m)
E. (10000/m)
Given x/y = m%=m/100

Obviously y/x = 100/m = 10000/m %
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Re: Arithmetic from 11th Edition

by x2suresh » Mon Feb 23, 2009 9:21 pm
Agree with E..

Silly mistake.
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