Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Question really means: Are there 2, 2, 2 and 3 in the factor box of product ABC? Answer is A

1. a, b, and c are consecutive even integers
SUFF. 3 consecutive even integers mean there are three 2s in the prime box of abc. Now we just need a 3. Every three consecutive even numbers will be a multiple of 3.
Now we have all primes required to make it to 24

2. a*b is divisible by 12
INSUFF. We can only establish that the product ABC has 2, 2 and 3 in its factor box. Can't tell if C is an even number.

Is A positive? B.

1. x2 - 2x + A is positive for all x (do you mean x^2?)
SUFF. When x = 1 or -1 we get:
x^2 - 2x + A
1-2=-1+A = positive
so A must be positive to yield a positive number

2. Ax2 + 1 is positive for all x[/quote]
INSUFF.
A could be 1, x could be 5:
1 + 25 + 1 = positive

but A could also be -0.5:
-0.5 + 25 + 1 = positive

I knew it was too easy... yeah, 0 is even. duh! Answer must be C.

From 1: BC are even, so we have 2, 2
From 2: AB gives we have 2, 2, 3

From both, we have 2, 2 from stat1, and 2, 2, 3 from stat2. We could have some overlapping, but we can break it down like this:
C gives us 2
AB gives us 2, 2, 3

The reason I didnt pick A for number 1 is because of 0.

Is 0 an even integer?

For number 2. If x=100 A can be a -#.

10000-200+(-1000)= 8800

What am I missing here?

1. The fact that zero is an even integer isn't a problem. Keep in mind that 0 is divisible by all numbers. 0/24, for instance, is 0. So, if one of a, b, c is equal to 0, a*b*c is divisible by 24. If not, by asdert's reasoning, a*b*c is still divisible by 24. Either way, I is sufficient. The answer is A.

2. It's important to see the logic of the question here. We're looking for a value of A that yields a positive value for x^2 - 2x + A for all x.

It's true that A can be negative and still make x^2 - 2x + A positive for some x. But let's suppose, as in the example here, that A = -1000. If x = 100, x^2 - 2x + A is positive. But if x = 1, it's not. If A = -1000, x^2 - 2x + A is not positive for all x.

In particular, x can always be 0. If x^2 - 2x + A > 0 when x = 0, then A > 0. Statement I is sufficient.

But for II, suppose A = 0. Then Ax^2 + 1 = 1 (> 0) for all x. So A isn't necessarily positive. II is insufficient. Answer's A.

The reason I didnt pick A for number 1 is because of 0.

Is 0 an even integer?

For number 2. If x=100 A can be a -#.

10000-200+(-1000)= 8800

What am I missing here?

1. The fact that zero is an even integer isn't a problem. Keep in mind that 0 is divisible by all numbers. 0/24, for instance, is 0. So, if one of a, b, c is equal to 0, a*b*c is divisible by 24. If not, by asdert's reasoning, a*b*c is still divisible by 24. Either way, I is sufficient. The answer is A.

2. It's important to see the logic of the question here. We're looking for a value of A that yields a positive value for x^2 - 2x + A for all x.

It's true that A can be negative and still make x^2 - 2x + A positive for some x. But let's suppose, as in the example here, that A = -1000. If x = 100, x^2 - 2x + A is positive. But if x = 1, it's not. If A = -1000, x^2 - 2x + A is not positive for all x.

In particular, x can always be 0. If x^2 - 2x + A > 0 when x = 0, then A > 0. Statement I is sufficient.

But for II, suppose A = 0. Then Ax^2 + 1 = 1 (> 0) for all x. So A isn't necessarily positive. II is insufficient. Answer's A.

I see the trick in question 1.

but i still dont see why S1 in question 2 is sufficient.

The reason I didnt pick A for number 1 is because of 0.

Is 0 an even integer?

For number 2. If x=100 A can be a -#.

10000-200+(-1000)= 8800

What am I missing here?

1. The fact that zero is an even integer isn't a problem. Keep in mind that 0 is divisible by all numbers. 0/24, for instance, is 0. So, if one of a, b, c is equal to 0, a*b*c is divisible by 24. If not, by asdert's reasoning, a*b*c is still divisible by 24. Either way, I is sufficient. The answer is A.

2. It's important to see the logic of the question here. We're looking for a value of A that yields a positive value for x^2 - 2x + A for all x.

It's true that A can be negative and still make x^2 - 2x + A positive for some x. But let's suppose, as in the example here, that A = -1000. If x = 100, x^2 - 2x + A is positive. But if x = 1, it's not. If A = -1000, x^2 - 2x + A is not positive for all x.

In particular, x can always be 0. If x^2 - 2x + A > 0 when x = 0, then A > 0. Statement I is sufficient.

But for II, suppose A = 0. Then Ax^2 + 1 = 1 (> 0) for all x. So A isn't necessarily positive. II is insufficient. Answer's A.

I see the trick in question 1.

but i still dont see why S1 in question 2 is sufficient.

How's this:

Yezz is right to say that A > x(2-x). But the point is that this has to hold for all values of x, not just for one particular value of x. So think about how this inequality behaves for different values of x.

If x is less than 0, then x (2 - x) is negative.
If x = 0 or 2, then x (2 - x) is zero.
If x> 0 and less than 2, then x (2 - x) is positive.
If x > 2, then x (2 - x) is negative.

A is greater than x (2 - x) whatever x is. So it must be greater than zero.

You could also think about it geometrically. The problem is equivalent to saying that the graph of y = A is always above the graph of y = x (2 - x). That means A must be more than the maximum value of x (2 - x), not just more than some particular value of x (2 - x). The maximum value of x (2 - x) is 1. So A must be greater than 1.

I'm still confused by the 1st one. If one of th integers is 0, then the combinations could be: -4,-2,0 -2,0,2 0,2,4. None of those are divisible by 24.

I'm still confused by the 1st one. If one of th integers is 0, then the combinations could be: -4,-2,0 -2,0,2 0,2,4. None of those are divisible by 24.

Yes they are as Johnrb pointed out.

0/24=0 This will be the case for any integer. So -4(-2)(0)=0/24=0 it is divisible.

Quick question what is 0/0? Is this still undefined or just 0?

I'm still confused by the 1st one. If one of th integers is 0, then the combinations could be: -4,-2,0 -2,0,2 0,2,4. None of those are divisible by 24.

Do we have to consider negative number when problem talks about consecutive even integers?

I'm still confused by the 1st one. If one of th integers is 0, then the combinations could be: -4,-2,0 -2,0,2 0,2,4. None of those are divisible by 24.

Do we have to consider negative number when problem talks about consecutive even integers?

I thought we only consider positive integers...

No, you definitely need to consider them. Integers, by definition, are all whole numbers, positive and negative.

0/0 is undefined. 0/(anything else) is 0, so it's all good for that one.

As for the second one, I'm on board with A. It's a hellish problem, but you see by plugging in some numbers for x, the equation is negative around the number +1. So A has to be a positive number to counter-act that. Under all other conditions, A could be negative and it wouldn't matter, but since we're told the equation is ALWAYS positive, A must be positive.

The second statement is A times some x^2. x^2 is always positive, but that doesn't mean A has to be. As long as Ax^2 > -1, this will work out, so A could be some negative number to make that happen.