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Schools: IIM (A) ISB '24
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If y > x, is y > 64 ?

(Statement1): x has 7 factors (all even but one) and is divisible by only one prime number.

\(a^{6}\) —> (6+1)= 7 factors
—> a should be prime number and even number at the same time—> 2 could be prime and even
\(x= 2^{6} = 64\) —> y >64 (yes)
Sufficient

(Statement2): |y - 64| > |x - 64|
—> square both sides
\((y—64)^{2} > (x—64)^{2}\)

\(y^{2} —128y+ 64^{2} > x^{2} —128x+64^{2} \)
\(y^{2}—128y > x^{2} —128x\)
(y—x)( y+x) —128(y—x) > 0
(y—x)(y+x—128) >0

Given that y—x >0, then —> y+x—128 >0
—> y+x >128

Now, we have two inequalities:
y—x > 0
y+x >128
By adding together:
—> 2y > 128 —> y > 64 (Yes)
Sufficient

The answer is D

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Bunuel
If y > x, is y > 64

(1) x has 7 factors (all even but one) and is divisible by only one prime number.
(2) |y - 64| > |x - 64|


Are You Up For the Challenge: 700 Level Questions
From 1
X=2^6=total 7 factors ( 6+ 1)
X=64
since Y>X
Y>64
Sufficient
From 2
Y>X
so Y-64>X-64( we can add or subtract in inequality)
and lY-64l>lX-64l
for this to be true Y>64
Sufficient
D:)
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I didn't understand how Statement 2 is sufficient. Could someone post step by step answer
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ayushsaxena04
I didn't understand how Statement 2 is sufficient. Could someone post step by step answer

if using step process , here how it works

consider what is given in the question , y > x

consider both mods give negative value ie , . suppose y < 64 and X < 64, you get

-(y-64)>-(x-64)
-y+64 > -x+ 64

ie X > Y , but this contradicts with condition given in the question iteself, so in statement 2 consider Y>64 and solve you again get Y > X . so you can say Y>64.
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