CAT 1 problem 8

[gmatwork]

Is x > y?

(1) > y

(2) x3 > y

This problem is from CAT 1 , answer is (C)

I have trouble solving these types of DS questions quickly. I took too long to solve this problem. I know that testing numbers is more efficient than algebra on these types of DS problems but coming up with the right set of numbers was a challenge for me. It took me too long to solve and got it wrong.

Another type of problems I have issue with is the number line DS problems.

Can you give some advise as to how to improve accuracy and timing on these types of DS questions.

[tim]

We need the full problem written. Statement 1 is obviously incomplete..

[gmatwork]

Sorry it had a typo....

Is x > y?

(1) under-root(x) > y

(2) x^3 > y

ans:(C)

[stud.jatt]

Is x > y?

(1) > y

(2) x3 > y

This problem is from CAT 1 , answer is (C)

I have trouble solving these types of DS questions quickly. I took too long to solve this problem. I know that testing numbers is more efficient than algebra on these types of DS problems but coming up with the right set of numbers was a challenge for me. It took me too long to solve and got it wrong.

Another type of problems I have issue with is the number line DS problems.

Can you give some advise as to how to improve accuracy and timing on these types of DS questions.

Unfortunately there is no shortcut or formula to solve these types of questions. To answer them quickly and accurately you have to be thoroughly well versed with the properties of numbers.

Question Statement: is x > y

(1) sq.rt(x) > y OR x > y^2

this tells us that x is > 0 (as square of a number is always positive) and gives us the following pieces of information
a) for x, y > 1; x is always > y as in this range y^2 > y and hence x > y^2 > y

b) for 0 < x, y < 1; we don't get a definite answer, x may or may not be greater than y
If x lies between y and y^2 it will be smaller than y but still be greater than y^2 (as the square of a number lying between 0 and 1 is smaller than the number itself) and values of x > y will ofcourse satisfy the equation in statement 1.

Thus Statement 1 is insufficient

(2) x^3 > y

This can be analyzed as
a) For x, y > 1; we don’t get a definite answer, x may or may not be greater than y
If x lies between y and cube root(y), it will be smaller than y, but will still satisfy this equation, as do values of x > y

b) For 0 < x, y < 1; in this case x > y,
the reason is that for this range x^3 is smaller than x and hence x > x^3 > y

Thus Statement 2 is also insufficient

Taking both the statements together i.e. x > y^2 and x^3 > y we get our answer that x > y irrespective of the range that x or y may lie in.

This may seem like a very long and time consuming explanation but when you are comfortable with number properties and have had enough practice on these types of questions, you will be doing half of all these calculations subconsciously and the time to solve this question would be much less.

[tim]

thanks!

[gmatwork]

for stad.jatt:

In statement 1 I don't think you can square both sides and say x> square of y; since we don't know the sign of y

Thanks for the reply.

[stud.jatt]

for stad.jatt:

In statement 1 I don't think you can square both sides and say x> square of y; since we don't know the sign of y

Thanks for the reply.

for erpriyankabishnoi:

The sign of y doesn't matter here as statement 1 mentions

(1) sq.rt(x) > y

the fact that the square root of x is mentioned implies that x is a positive number as the GMAT wouldn't ask for the sq. rt of a negative number. (Though square roots of negative numbers do exist in the realm of complex numbers, you can not compare them to a real number in an inequality equation)

so if y is negative, x is automatically greater than y because x is a positive number. As this is a trivial case, I didn't state it in the solution.

However when x and y are positive, we get the two subcases as mentioned above and hence statement 1 is insufficient.

I hope this clarifies the solution.

[jnelson0612]

Great, thanks. Yes, we will never take the square root of a negative number on the GMAT, and whenever we take the square root of a number the result will always be positive.

[Jhenna12]

It was simple harder than i thought. Thanks for the information

[tim]

:)

[sachin.w]

Taking both the statements together i.e. x > y^2 and x^3 > y we get our answer that x > y irrespective of the range that x or y may lie in.

I don't quite understand how this has been derived.

Please elaborate

[RonPurewal]

here's another approach, which involves less work once a certain preliminary fact is on the ground.

preliminary fact:
x is always between √x and x^3. (the only exception to this statement occurs if x is 0 or 1, in which case all three expressions have the same value.)
if you don't see why this is true, consider the different types of numbers:
* if x > 1, then √x is less than x and x^3 is greater than x.
* if 0 < x < 1, then √x is greater than x and x^3 is less than x.
* if x is negative, then √x doesn't exist, so we needn't consider negative values.


(1) isn't sufficient because √x can be bigger than x. for instance, if x = 1/4, then √x = 1/2.
if y is, say, 1/3, then that's less than √x but not less than x (giving a "no" to the question).
it's much easier to get a "yes" to this question -- almost any values you pick at random will do that. (for instance, if x = 4, √x = 2, y = anything less than 2, you get "yes".)
insufficient.

(2) isn't sufficient because x^3 is (usually) bigger than x. for instance, if x = 2, then x^3 = 8.
so... if x = 2, x^3 = 8, y = 5, you get "no" to the question.
if x = 2, x^3 = 8, y = 1, you get "no" to the question.
insufficient.

together:
from the above, x is "sandwiched" by √x and x^3. so, if both √x and x^3 are greater than y, then the whole "sandwich" -- including x -- is greater than y.
(if you have one of the two exceptional cases, x = 0 or x = 1, then the statements are immediately sufficient because they are all saying the same thing.)

so (c)