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Re: If C+D=11 and C & D are positive integers, which of the foll [#permalink]
26 Apr 2012, 00:10

2

This post received KUDOS

Expert's post

sugu86 wrote:

Hi guys,

Way to solve the below problem ?

If C+D=11 and C & D are positive integers, which of the following is a possible value of 5C+8D ?

(A) 55 (B) 61 (C)69 (D) 83 (E) 88

If C + D = 11 and C and D are positive integers, which of the following is a possible value for 5C + 8D?

A. 55 B. 61 C. 69 D. 83 E. 88

C+D=11 --> C=11-D --> 5C + 8D=5(11-D)+8D=55+3D, so the answer must be 55 plus some multiple of 3.

Only B and E satisfy this: 61=55+3*2 and 88=55+3*11, but the second case is not possible because in this case D=11 and C=0, and we are told that C is a positive integers.

Answer: B.

P.S. Do not shorten or reword the questions you post. _________________

Re: If C + D = 11 and C and D are positive integers, which of [#permalink]
26 Apr 2012, 22:37

Expert's post

sugu86 wrote:

If C + D = 11 and C and D are positive integers, which of the following is a possible value for 5C + 8D?

A. 55 B. 61 C. 69 D. 83 E. 88

You can also use simple brute force here if nothing works. Since C+D = 11, C and D would take values from one of the pairs {1, 10}, {2, 9}, {3, 8}, {4, 7}, {5, 6}

5C + 8D could be 5*1 + 8*10 = 85 or 5*10 + 8*1 = 58 5C + 8D could be 5*2 + 8*9 = 82 or 5*9 + 8*2 = 61 (that's one of the options)

Mind you, brute force works only for some low level questions (but it is an option). _________________

Re: Extra Equation Strategie [#permalink]
20 Mar 2013, 19:58

3

This post received KUDOS

c + d = 11 5c + 8d = 5c + 5d + 3d = 5(c+d) + 3d = 55 + 3d. Now 3d is a multiple of 3, so the right answer should be 55 + multiple of 3. Or we can subtract 55 from the options and see which one results in a multiple of 3. 61 - 55 = 2*3 88 - 55 = 33 = 3*11. But if d = 11 then c = 0. So, 88 cannot be the option. Hence, B is the right answer.

Please give a kudo if you like my explanation.

Last edited by Abhii46 on 20 Mar 2013, 20:58, edited 1 time in total.

Re: Extra Equation Strategie [#permalink]
20 Mar 2013, 20:29

1

This post received KUDOS

Expert's post

Abhii46 wrote:

c + d = 11 5c + 8d = 5c + 5d + 3d = 5(c+d) + 3d = 55 + 3d. Now 3d is a multiple of 3, so the right answer should be 55 + multiple of 3. Or we can subtract 55 from the options and see which one results in a multiple of 3. 61 - 55 = 2*3 Hence B is the right answer.

Please give a kudo if you like my explanation.

Hi Abhii46, your explanation makes sense, but it still leaves open the possibility of 88, since that is 55 + 33. However, 88 cannot be an answer to this question, so the answer is indeed B. Can anyone explain why E isn't a valid option?

(Hint: Think of what would have happened had the answer choice been 85 instead.)

Re: Extra Equation Strategie [#permalink]
20 Mar 2013, 20:55

1

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88 cannot be the option because then for that d has to be 11, which makes c = 0. Sorry, I missed that. 85 can be the answer, which makes d = 10 and c = 1.

Re: Extra Equation Strategie [#permalink]
20 Mar 2013, 21:08

1

This post received KUDOS

Expert's post

marcovg4 wrote:

This problem is presented in one of the MGMAT strategy books (Algebra):

If c+d=11 and c and d are positive integers, wich of the followings is a possible value for 5c+8d?

A) 55 B) 61 C) 69 D) 83 E) 88

SO c+d = 11. Now 5c+8d = 5(c+d)+3d = 55+3d. Thus as d is an integer, when 55 is subtracted from the given options should be divisible by 3. Taking the options, we see that A=55 gives d=0. Not possible as it states d is a positive integer.Again E=88 gives d=11, but this will make c=0 which is again not possible. Thus the only answer is B=61.

Re: Extra Equation Strategie [#permalink]
20 Mar 2013, 21:28

2

This post received KUDOS

Expert's post

marcovg4 wrote:

Ok, so 0 must be considered an Even number, not a positive nor negative integer. Sorry, I confuse those unique properties of the zero.

Thanks everyone, very helpful explanations

Hi marcovg4, you're completely right, zero is even and neither positive nor negative (the only number with this property.

Zero is one of those numbers that comes up over and over on the GMAT (I may have to write a ROn Point blog about that soon!). Aside from being even even, and neither positive nor negative, it also has the properties of not being a prime number, being a multiple of every number, 0!=1, x^0 = 1, and being neither red nor black (this applies to the roulette wheel only ).

Re: If C + D = 11 and C and D are positive integers, which of [#permalink]
21 Mar 2013, 09:44

Expert's post

Rock750 wrote:

VeritasPrepKarishma wrote:

C and D would take values from one of the pairs {1, 10}, {2, 9}, {3, 8}, {4, 7}, {5, 6}.

Actually we don't know if C>D or C<D , hence there will be more pairs to consider : {10,1} , {9,2} , {8,3} and so on ...

Correct if i am wrong but i think that the brutal force here would take more time to be attempted

If you read the explanation further, you will notice that I have considered both possibilities for each pair

Quote from my post above: "5C + 8D could be 5*1 + 8*10 = 85 or 5*10 + 8*1 = 58 5C + 8D could be 5*2 + 8*9 = 82 or 5*9 + 8*2 = 61 (that's one of the options)"

C can be 1 and D can be 10 or C can be 10 or D can be 1 etc.

And also, as I mentioned before - brute force works only for some low level questions. I don't particularly endorse it but it is an option and if nothing else comes to mind, just try some values - things might fall in place. Point is, it's better than staring at the question wondering what to do. _________________

Re: If C + D = 11 and C and D are positive integers, which of [#permalink]
28 Mar 2013, 03:23

2

This post received KUDOS

5C+8D can't be 55 or 88. So, eliminate options A and E.

5C + 8D = 5 * (C + D) + 3D = 5 * 11 + 3D = 55 + 3D

Subtracting 55 from options B,C and D,we get 6,14 and 28 respectively. Here only 6 is divisible by 3. Hence, Option B is the answer. ------------------------------------------------ Please press KUDOS if you like my post. _________________

Re: If C + D = 11 and C and D are positive integers, which of [#permalink]
10 Dec 2014, 09:00

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