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## Comment on

Testing Possible Cases## Hi Brent,

Need your help: https://gmatclub.com/forum/if-m-p-s-and-v-are-positive-and-m-p-s-v-which-of-the-following-160298.html

I tried solving it by input-output method and I don't think that's the best approach for it. I chose the following values: m = 3; p = 5; s = 7; v = 2

## Question link: https:/

Question link: https://gmatclub.com/forum/if-m-p-s-and-v-are-positive-and-m-p-s-v-which...

Yeah, the input-output method MAY help you eliminate some answer choices, but it won't take you all the way to the correct answer.

Here's my full solution: https://gmatclub.com/forum/if-m-p-s-and-v-are-positive-and-m-p-s-v-which...

Cheers,

Brent

## Hey Brent :) Can you give me

## My solution to question 390:

My solution to question 390: https://gmatclub.com/forum/a-school-administrator-will-assign-each-stude...

My solution to question 176: https://gmatclub.com/forum/if-n-3-8-2-8-which-of-the-following-is-not-a-...

Aside: Here are links to solutions to ALL official guide questions: https://gmatclub.com/forum/gmac-official-guides-the-master-directory-lin...

Cheers,

Brent

## https://gmatclub.com/forum/if

Hi Brent. I just wanted to say thank you for all these videos and resources. If I had looked at the above linked questions before taking your course, I would have been in shock. But now I am having much much greater success.

## Good to hear!!

Good to hear!!

## Hi Brent,

Please review my solution.

statement 1: xy+y=ODD ; y(x+1)=Odd . This implies y= odd and x+1 = odd. If x+1=odd then x= even.(sufficient)

statement 2: 6x-3y=ODD; 3(2x-y)=odd. We know 3 is odd.(2x-y)=Odd ; x can be even or x can be odd. (Not sufficient)

So, opt A is the correct ans.

## That's a perfectly-reasoned

That's a perfectly-reasoned solution. In fact, it's exactly how I'd solve such a question.

That said, if students aren't sure how to proceed with similar questions, they can always just test all possible cases as we have done above.

## If r and s are integers and

A. r

B. s

C. r + s

D. rs - r

E. r^2 + s

I drew a table.there are two scenarios

When rs+r is odd...

1/r=odd,s =even,rs+r odd

2/ r=even, S= odd, rs+r odd..

Confused what to do!

## That's a good approach. Let's

That's a good approach. Let's start it from the beginning by testing all four possible cases:

1) r is EVEN and s is EVEN: Here, rs + r is EVEN

2) r is EVEN and s is ODD: Here, rs + r is EVEN

3) r is ODD and s is EVEN: Here, rs + r is ODD

4) r is ODD and s is ODD: Here, rs + r is EVEN

Aha! In your solution, you say that, if r=even and s=odd, then rs+r is odd, but this is not the case.

For example, if r = 0 and s = 1, then rs + r = (0)(1) + 0 = 0, which is even.

As we can see above, case #3 is the only scenario that yields an ODD value of rs + r.

So, it must be the case that r is ODD and s is EVEN

Does that help?

## https://gmatclub.com/forum/if

Basically we test four cases;but in here two cases is being tasted in your solution. Would you please explain?

## Solution link: https:/

Solution link: https://gmatclub.com/forum/if-x-and-y-are-integers-is-x-even-226190.html...

Let's compare this question to the question I answered in the above comments (If r and s are integers and rs + r is odd, which of the following must be even?)

For the rs+r question, we're given the expression rs + r, and our job is to determine the circumstances in which rs + r is odd.

Notice that rs + r is an algebraic EXPRESSION (not an EQUATION).

As such, we can test all 4 possible cases (e.g., r is odd and s is even) to see which case(s) yields an odd value for rs + r

In the other question (If x and y are integers, is x even?), statement 1 tells us that x + y = y⁵

Notice that x + y = y⁵ is an EQUATION.

So, while we COULD try testing all four cases, it's more convenient to rewrite x + y = y⁵ as x = y⁵ - y, and then see what happens to x when y is ODD and when y is EVEN.

That said, we could have also tested all four cases. Let's do that.

1) x is EVEN and y is EVEN:

Here, x + y = y⁵ becomes EVEN + EVEN = EVEN⁵

Simplify: EVEN + EVEN = EVEN

This works.

2) x is EVEN and y is ODD:

Here, x + y = y⁵ becomes EVEN + ODD = ODD⁵

Simplify: EVEN + ODD = ODD

This works.

3) x is ODD and y is EVEN:

Here, x + y = y⁵ becomes ODD + EVEN = EVEN⁵

Simplify: ODD + EVEN = EVEN

Does NOT work.

4) x is ODD and y is ODD:

Here, x + y = y⁵ becomes ODD + ODD = ODD⁵

Simplify: ODD + ODD = ODD

Does NOT work.

Cases 1 and 2 are the only ones that work.

In both cases, x is EVEN, so we can answer the target question with certainty.

Does that help?

Cheers,

Brent

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