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inequality

Expert replies
Source: — Data Sufficiency |

by NikolayZ » Fri Oct 30, 2009 6:19 am
Hey Duongthang!

we need to know which is greater : 0.2n or 0.1n+0.05

From 1) n<0.1. let's plug in even 0.1, yo my mind it doesn't matter here : to plug in 0.1,0.095 or 0.01.

0.2*0.1=0.02, whereas 0.1*0.1+0.05=0.051. Hence 0.02<0.051 - Sufficient.
(you can do it yourself with another variables, such as 0.01 or whatever. The whole inequality won't change)

From 2) Here, we need to notice that n is positive, and if it is >0 but <0.1, the right part of inequality is greater.
when it is >0.1, we get the following:
0.2*0.2=0.04
and
0.1*0.2+0.05=0.09, still sufficient.
finally, plug in 0.9
0.2*0.9=0.18
and
0.1*0.9+0.05=0.09+0.05=0.14. left part becomes greater that the right.
Hence 2) - insufficient.
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by punitkaur » Fri Oct 30, 2009 6:47 am
According to me the answer should be C and not A.

Which is greater 0.2n or 0.1n+0.05

Let x=0.2n, y=0.1n+0.05

A) Given n<0.1

Therefore x<0.02 , y<0.06

Using the above inequalities

a)if x=0.01, y=0.01 ,y=x.
b)if x=0.01, y=0.02 , y>x

INSUFFICIENT to say which is greater

B) Given n>0.01

Therefore x >0.002, y>0.051
a)if x = 0.052, y=0.052 , y=x
b)if x=0.051, y=0.052, y>x

INSUFFICIENT

C) Considering A & B together
0.002 < x < 0.02
0.052 < y < 0.06

y > x SUFFICIENT

Hence C

Whats the OA?
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by NikolayZ » Fri Oct 30, 2009 7:28 am
Oops.
Sorry for disinformation. I don't why, but i even did not check a couple of possible numbers <0.1 =(
Thank you Punitkaur.
Indeed the answer is C.
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by ayashlaha » Fri Oct 30, 2009 10:11 am
I still think that the answer is A

In Punit Kaur's example:

if X=0.01 then n = 0.05
If n=0.05 then Y=0.055

Hence Y is greater.

In my mind I would rewrite the problem statement as:
is 0.2n = 0.1n + 0.05
=> Is 0.1n = 0.05

this is only true if n =0.5

However it is stated that n< 0.1 , hence statement 1 is sufficient.

Stetement 2 by itself is inconclusive if n is a large integer number.

Please post official answer or alternate explaination if you think I'm wrong.

Thanks.
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