Quant Basics - How am I wrong?

[user2031]

I took the test on Quant basics (Chapter 1). I have 3 questions, that not only did I fail, but I do not know why I was wrong!

[Leaderboard]

First: explain why you got the answer you did.

[user2031]

Well I will:
Q1:
Quantity simplifies to: \sqrt{(a-3)^2} = \pm {(a-3)} which further simplifies to. (a-3); because GRE only cares about positive roots. Hence C

Q2:
\sqrt[3]{1000} = a.b = +10
pairs = [(-2,-5), (-1,-10),(1,10),(2,5) ] ; all integers
n(pairs) =4 (Hence 6>4).

Q3:
(1+\frac{1}{3}) * a = \frac{4}{3}*a \simeq 1.33 * a (Hence C)

[ganesh]

@user2031

Quantity simplifies to: \sqrt{(a-3)^2} = \pm {(a-3)}√(a−3)2=±(a−3)\sqrt{(a-3)^2} = \pm {(a-3)} which further simplifies to. (a-3)(a−3)(a-3); because GRE only cares about positive roots. Hence C

How do you know a-3 itself is positive? LHS is |a-3|, RHS is a-3, do they need to be the same?

Btw have you seen the solution videos for these?

@user2031

(1+13)∗a=43∗a≃1.33∗a(1+\frac{1}{3}) * a = \frac{4}{3}*a \simeq 1.33 * a (Hence C)

You’re approximating, you can’t do that in a QC question.

[Leaderboard]

@user2031

√1000=a.b=+10\sqrt[3]{1000} = a.b = +10
pairs = [(-2,-5), (-1,-10),(1,10),(2,5) ] ; all integers
n(pairs) =4 (Hence 6>4).

What about (5, 2)?

[user2031]

The question asked about distinct pairs. Hence, (5,2) === (2,5), am I not getting that right?

[Leaderboard]

No - they are not the same pair. If the “distinct” word was removed, then maybe.