When each of the children in a room was given the same number of

Question

When each of the children in a room was given the same number of crayons from a box containing 45 crayons, the number of crayons left in the box was 3 less than the number of children in the room. Which of the following could NOT have been the number of children in the room?

A. 3
B. 8
C. 12
D. 14
E. 16

Show Spoiler

GMAT official practice test 5.

A. 3
B. 8
C. 12
D. 14
E. 16

GMAT official practice test 5.

chetan2u · Accepted Answer

Let the number of children and crayons given to each be x and y. 
Thus, number of crayons distributed = x*y, and crayons left over = x-3

Therefore,  x*y +  (x-3)=45.....xy+x=48......x(y+1)=48

Hence, x is a factor of 48.
In the given options, 14 is NOT a factor of 48.

Hence D

MrMusaev · Answer

The fastest way is to test each answer:

A) 45 % 3 = 0 (remainder is 3 less)
B) 45 % 8 = 5 (remainder is 3 less)
C) 45 % 12 = 9 (remainder is 3 less)
D) 45 % 14 = 17 (remainder is 3 more) -> answer
E) 45 % 16 = 13 (remainder is 3 less)

Bunuel · Answer

Can you please specify from which exact Official Guide this question is? I could not find it in any. Or do you mean it's from the GMAT Prep Focus Exams? Thank you!

Bunuel · Answer

Similar question from GMAT Prep Focus https://gmatclub.com/forum/when-each-of ... 21605.html Hope it helps.

andrewjohn8 · Answer

Yes, sorry, this is from a GMAT Focus Practice exam.

Bunuel · Answer

___________________________________________________
Thank you!

GmatKnightTutor · Answer

When each of the children in a room was given the same number of crayons from a box containing 45 crayons, the number of crayons left in the box was 3 less than the number of children in the room. Which of the following could NOT have been the number of children in the room?

A. 3
B. 8
C. 12
D. 14
E. 16

Show Spoiler

GMAT official practice test 5.

A. 3
B. 8
C. 12
D. 14
E. 16

Finding four that work may be easier than the one that doesn't!

3 children - You can tell that 45 crayons can be divided perfectly among 3 children, so it's possible to be left with 0 crayons.... which is 3 less than the number of children in this scenario. This works.

8 children - Finding your target could be helpful. What is 3 less than 8? It's 5. So we are basically checking to see whether there is a multiple of 8 that is equal to (45-5) = 40. 8 times 5 is 40 so this options works, too!

12 children - Use the same process. What is 3 less than 12? 9. What is 45-9? 36. Is 36 a multiple of 12? Yes! This works, too.

14 children - What is 3 less than 14? 11. What is 45-11? 34. Is 34 a multiple of 14? No.

16 children - What is 3 less than 16? 13. What is 45-13? 32. Is 32 a multiple of 16? Yes.

Once you get the "process" you could probably get through some of the options more quickly.

(D) is your answer.

ADHDDyslexic · Answer

It's from the GMAT official practice test 5.

Mathclub1205 · Answer

The only accurate way (without guessing values) that I found:

Let x be the number of kids and k be the whole part of 45/x.

We can rewrite the problem as:

(i) 45-kx = x-3
(ii) x + kx = 48
(iii) x(k+1) = 48.

Which means that 48 must divisible by x in order to be solvable.

Only 14 doesn't divide 48.

D it is.

GMATinsight · Answer

i.e. 45+3 = 48 crayons should be evenly divisible among the children

i.e. Number of children = Factor of 48

Optly 14 from the options is not a factor hence

Answer: Option D

cryuss · Answer

can u explain how this works

Krunaal · Answer

If there are 3 children, then there should be 0 crayons left, so we need to check whether 45 crayons are divisible by 3 children. Yes, 3 * 15.

If there are 8 ch, then 5 cr left, remaining 40 divisible by 8? Yes, 8 * 5

12 ch, 9 cr left, 36 divisible by 12? Yes, 12 * 3

14 ch, 11 cr left, 34 divisible by 14? Nope.

16 ch, 13 cr left, 32 divisible by 16? Yes, 16 * 2

bumpbot · Answer

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When each of the children in a room was given the same number of

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