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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

C(d+1) is even : Insufficient 3 cases: Case 1: C is even, d+1 is even => C is even, D is Odd. Case 2: C is even, d+1 is Odd => C is even, D is Even. Case 3: C is Odd, d+1 is even => C is Odd, D is Odd.

Statement 2: (C+2)(D+4) is even: Insufficient 3 Cases: Case 1:(C+2)is even, (D+4) is even => C and D : Both even Case 2: (C+2) is even, (D+4) is Odd => C is even, D is Odd Case 3: (C+2) is Odd, (D+4) is even => C is Odd, D is Odd.

Together: Insufficient: As multiple cases exist for both the statements.

(1) C*(D+1) is even. In order the product to be even C must be even or D+1 must be even, or both. Hence it's not necessary C to be even. Not sufficient.

(2) (C+2)(D+4) is even. Notice that this is the same as C*(D+4) is even. Again, in order the product to be even C must be even or D+4 must be even, or both. Hence it's not necessary C to be even. Not sufficient.

(1)+(2) Both D+1 and D+4 cannot be even, thus C must be even. Sufficient.

Intially I went to do it all in head. But it works better if you do it on the paper--substitution with number.

C even?

1. say NO and check: 5 *(7+1) = even, say YES and check: 8*(7+1) 0 even -> not sufficient, clearly not A

2. say NO and check: (5+2)*(2+4) = even say YES and check: (6+2)*(2+4) = even -> not sufficient, clearly not B

3. say No and check: 5*(7+1) = even; (5+2)*(7+4) = odd -> restricted 5*(8+1) = odd; (5+2)*(8+4) = even ->restricted say YES and check: just one look and you know (1) and (2) both are satisfied

Maybe it is not that faster. With numbers there are too many things to assume if I do it in head.

S1: c(d+1) is even means either c is even or d+1 is even..... not suff

S2: (c+2)(d+4) is even means either (c+2) is even or (d+4) is even... not suff

Both together: c(c+2)(d+1+3) = c(d+1) + 3c + 2(d+1+3) which implies Even + __ + Even = Even

Therefore 3c is even... Hence c is even...

Answer C... _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

Statement 1 is insufficient since c could be even or odd and (d+1) could also be even or odd and still c * (d+1) could be even.

Refer the below table.

c ***** d+1 ***** c (d+1) E ***** E ***** E E ***** O ***** E O ***** E ***** E

Statement 2 is insufficient. (c+2) (d+4) = cd + 2d + 4c + 8. (Here 2d, 4c and 8 are even). Hence c * d should even, however we cannot say whether c is even.

Refer the below table.

c ***** d ***** c * d E ***** E ***** E E ***** O ***** E O ***** E ***** E

Combining two statements.

c(d+1) and cd is even. Also if d is even then (d+1) is odd and vice versa.

Now set up a table.

c ***** d ***** (d+1) **** cd *********** c(d+1) O ***** E ***** O **** Even *********** Odd --- This combination does not work E ***** O ***** E **** Even *********** Even --- This works and hence c is even.

Hence combining both statements would answer the question. Answer C. _________________

Support GMAT Club by putting a GMAT Club badge on your blog

Re: If c and d are integers, is c even? [#permalink]

Show Tags

22 Feb 2011, 01:58

I don't know the best way; but here's how I would solve it;

1. c(d+1) is even c even and d+1 even i.e. d odd OR c odd and d+1 even i.e. d odd OR c even and d+1 odd i.e. d even

We can see that c can be even or odd. Not sufficient.

2. (c+2)(d+4) is even

c+2 even and d+4 even i.e. c even and d even OR c+2 odd and d+4 even i.e. c odd and d even OR c+2 even and d+4 odd i.e. c even and d odd

C can be even or odd. Not sufficient.

If you see the odd case for c in both statements; c odd and d+1 even i.e. d odd c+2 odd and d+4 even i.e. c odd and d even You see that for c=odd; 1 statement says d=odd; 2nd statement says d=even; Conflict; D can't be odd and even at the same time.

If you consider c=even; c even and d+1 even i.e. d odd c+2 even and d+4 odd i.e. c even and d odd Both statements match.

c even and d+1 odd i.e. d even c+2 even and d+4 even i.e. c even and d even Both statements match.

Re: If c and d are integers, is c even? [#permalink]

Show Tags

22 Feb 2011, 02:00

I tried this approach:

Statement 1:

c(d+1) is even

Therefore, we have three possibilities. a) c - E & (d+1) - E => c is even and d is odd b) c - E & (d+1) - O => c is even and d is even c) c - O & (d+1) - E => c is odd and d is odd

So c and be even or odd!

Thus statement is not sufficient.

Statment 2: (c+2) (d+4) is even Therefore, there are three possibilities. a) (c+2) is even and (d+4) is even => c is even and d is even b) (c+2) is even and (d+4) is odd => c is even and d is odd c) (c+2) is odd and (d+4) is even => c is odd and d is even So c can be even or odd! Thus statement is not sufficient.

Both Statements together: For the cases c even and d odd or even, both the statements will always be true! Thus c is even! Ans: 'C' _________________

"Wherever you go, go with all your heart" - Confucius

Re: If c and d are integers, is c even? [#permalink]

Show Tags

06 Jan 2014, 09:43

If c and d are integers, is c even?

(1) c(d+1) is even (2) (c+2)(d+4) is even

Here we have a number properties question. We will be dealing with evens/odds and multiplication/addition. A couple of things to keep in mind before we look at the statements: 1. If c is even it will have at least one factor of 2 2. The rules for addition for evens/odds: E+E = E; O+O = E; E+0 = O; O+E = O 3. The rules for multiplication for evens/odds: We will arrive at an even product EXCEPT when we have O*O which will produce an O result

1. Statement 1 is true if : a. C is Even and D is Even E(E+1) = E*O = E b. C is Even and D is odd E(O+1) = E*E = E c. C is Odd and D is Odd O(O+1) = O*E = E a and b give us an answer of yes; c gives us an answer of no - INSUFFICIENT

2. Statement 2 is true if: a. C is Even and D is Even (E+2)(E+4) = E*E = E b. C is Even and D is Odd (E+2)(O+4) = E*O= E c. C is Odd and D is Even (O+2)(E+4) = O*E=E a and b give us an answer of yes; c gives us an answer of no - INSUFFICIENT

When we take the statements together we notice that the two possible scenarios they have in common are a and b (C is Even and D is Even OR C is Even and D is Odd). In both scenarios C is Even, so we have one definitive answer to the question. SUFFICIENT the answer is C

This may seem like a lengthy explanation; I provided every step for clarity. Make sure to memorize rules for number properties, and you can solve a question similar to this one in two minutes or less.

Re: If c and d are integers, is c even? [#permalink]

Show Tags

08 Mar 2015, 12:22

Expert's post

Hi All,

It appears that all of the posters used Number Property rules to answer this question. You can also TEST VALUES; in that way, you can physically "see" how those same Number Property rules "work" in real life.

We're told that C and D are integers. We're asked if C is even. This is a YES/NO question.

Fact 1: (C)(D+1) is even

This means that one or the other (or both) of the "terms" must be even....

IF.... C = 2 D = 1 (2)(2) = 4 The answer to the question is YES

IF.... C = 1 D = 1 (1)(2) = 2 The answer to the question is NO Fact 1 is INSUFFICIENT

Fact 2: (C+2)(D+4) = even

Just as in Fact 1, this means that one or the other (or both) of the "terms" must be even....

IF.... C = 2 D = 1 (4)(5) = 20 The answer to the question is YES

IF... C = 1 D = 2 (3)(6) = 18 The answer to the question is NO Fact 2 is INSUFFICIENT

Combined, we know.... (C)(D+1) is even (C+2)(D+4) is even

Since D is an integer, ONLY ONE of the two terms - (D+1) and (D+4) - will be even; the other will be odd. As such, the other term in each of the products (the one with the C in it) MUST be even....

Re: If c and d are integers, is C even? 1) c(d+1) is even 2) (c+2)(d+4) [#permalink]

Show Tags

28 Mar 2016, 13:46

1) c(d+1) is even: c is even and d is odd OR c is even and d is even OR c is odd and d is odd - Not sufficient 2) (c+2)(d+4) is even: c even and d even OR c odd and d even OR c even and d odd - Not sufficient Combine: Only common part: c is even and d is odd OR c even and d even => c is even - Sufficient Answer C

Part 2 of the GMAT: How I tackled the GMAT and improved a disappointing score Apologies for the month gap. I went on vacation and had to finish up a...

Cal Newport is a computer science professor at GeorgeTown University, author, blogger and is obsessed with productivity. He writes on this topic in his popular Study Hacks blog. I was...

So the last couple of weeks have seen a flurry of discussion in our MBA class Whatsapp group around Brexit, the referendum and currency exchange. Most of us believed...