GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 13 Oct 2019, 18:42

### GMAT Club Daily Prep

#### 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

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

# 62 candies were equally distributed to a group of children. If the num

Author Message
TAGS:

### Hide Tags

Manager
Joined: 13 Jun 2012
Posts: 199
Location: United States
WE: Supply Chain Management (Computer Hardware)
62 candies were equally distributed to a group of children. If the num  [#permalink]

### Show Tags

16 Oct 2018, 13:07
5
00:00

Difficulty:

95% (hard)

Question Stats:

48% (03:00) correct 52% (02:41) wrong based on 50 sessions

### HideShow timer Statistics

62 candies were equally distributed to a group of children. If the number of candies left were two less than the number of children then which of the following CANNOT be the number of candies received by each child?

a.1
b.3
c.7
d.9
e.15
Director
Joined: 19 Oct 2013
Posts: 520
Location: Kuwait
GPA: 3.2
WE: Engineering (Real Estate)
62 candies were equally distributed to a group of children. If the num  [#permalink]

### Show Tags

16 Oct 2018, 14:31
Turkish wrote:
62 candies were equally distributed to a group of children. If the number of candies left were two less than the number of children then which of the following CANNOT be the number of candies received by each child?

a.1
b.3
c.7
d.9
e.15

Let the number of children be x
the number of candies be c

the remainder equation can be set up as $$\frac{62}{x} = c + (\frac{x-2}{x})$$

Multiply by x

the equation becomes $$62 = c*x + x - 2$$

take x as a common factor it becomes $$62 + 2 = x (c+1)$$

You can plug in number of candies in the equation all of them will give us an integer for x (the number of children) EXCEPT FOR answer choice D the logic here is that we cannot have half a child?

if we substitute the number candies to be 1, we get number of children 64/2 = 32

if we substitute the number candies to be 3, we get the number of children 64/4 = 16

If we substitute the number candies to be 7, we get the number of children 64/8 = 8

If we substitute the number of candies to be 9, we get the number of children 64/10 = 6.40 not possible!

if we substitute the number of candies to be 15, we get the number of children to be 64/16 = 4

I hope this helps!
CEO
Joined: 12 Sep 2015
Posts: 3990
Re: 62 candies were equally distributed to a group of children. If the num  [#permalink]

### Show Tags

16 Oct 2018, 14:46
1
Top Contributor
Turkish wrote:
62 candies were equally distributed to a group of children. If the number of candies left were two less than the number of children then which of the following CANNOT be the number of candies received by each child?

a.1
b.3
c.7
d.9
e.15

There's a nice rule that says, If N divided by D equals Q with remainder R, then N = DQ + R
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3

Let n = number of children
Let c = number of candies that each child received.

So, we can rephrase the given information as: When 62 candies are divided among n children, each child receives c candies and there are (n-2) candies remaining.
In other words, 62 divided by n equals c with remainder (n-2)

Applying the above rule, we get: 62 = cn + n-2
Add 2 to both sides to get: 64 = cn + n
Factor out the n to get: 64 = n(c + 1)
Divide both sides by n to get: 64/n = c + 1
Subtract 1 from both sides to get: c = 64/n - 1

This tells us that c is 1 less than some divisor of 64 (64/n must be a divisor of 64)

a. 1. Since 2 is a divisor of 64, we can see that 1 is 1 less than some divisor of 64. ELIMINATE A
b. 3. Since 4 is a divisor of 64, we can see that 3 is 1 less than some divisor of 64. ELIMINATE B
c. 7. Since 8 is a divisor of 64, we can see that 7 is 1 less than some divisor of 64. ELIMINATE C
d. 9. 9 is NOT is 1 less than some divisor of 64. KEEP D
e. 15. Since 16 is a divisor of 64, we can see that 15 is 1 less than some divisor of 64. ELIMINATE E

RELATED VIDEO FROM OUR COURSE

_________________
Test confidently with gmatprepnow.com
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 935
Re: 62 candies were equally distributed to a group of children. If the num  [#permalink]

### Show Tags

16 Oct 2018, 15:27
Turkish wrote:
62 candies were equally distributed to a group of children. If the number of candies left were two less than the number of children then which of the following CANNOT be the number of candies received by each child?

a.1
b.3
c.7
d.9
e.15

$$\left. \matrix{ {\rm{\# children}}\,\,\,{\rm{:}}\,\,\,n \ge 2\,\,{\mathop{\rm int}} \hfill \cr {\rm{\# candies/child}}\,\,\,{\rm{:}}\,\,\,c \ge 1\,\,{\mathop{\rm int}} \,\, \hfill \cr} \right\}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,nc\,\,\,{\rm{candies}}\,\,{\rm{distributed}}$$

$$?\,\,\,:\,\,\,c\,\,\,\underline {{\text{CANNOT}}}$$

$$62 - nc = n - 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,n\left( {c + 1} \right) = 64\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{FOCUS}}\,!} \,\,\,\,c + 1\,\,{\text{is}}\,\,{\text{a}}\,\,{\text{positive}}\,\,{\text{divisor}}\,\,{\text{of}}\,\,64\,\,\,\,\left( * \right)$$

$$?\,\,\mathop = \limits^{\left( * \right)} \,\,\left( D \right)\,\,\,\,\,\,\,\left[ {\,10\,\,{\text{is}}\,\,{\text{not}}\,\,{\text{a}}\,\,{\text{positive}}\,\,{\text{divisor}}\,\,{\text{of}}\,\,64\,} \right]\,\,\,\,\,$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
VP
Joined: 07 Dec 2014
Posts: 1222
62 candies were equally distributed to a group of children. If the num  [#permalink]

### Show Tags

16 Oct 2018, 18:49
Turkish wrote:
62 candies were equally distributed to a group of children. If the number of candies left were two less than the number of children then which of the following CANNOT be the number of candies received by each child?

a.1
b.3
c.7
d.9
e.15

let c=number of children
x=number of candies received by each child
x=[62-(c-2)]/c→
x=(64-c)/c
plugging in all answer choices for x,
only 9 makes c a non-integer
D
Math Expert
Joined: 02 Aug 2009
Posts: 7941
Re: 62 candies were equally distributed to a group of children. If the num  [#permalink]

### Show Tags

16 Oct 2018, 20:00
1
Turkish wrote:
62 candies were equally distributed to a group of children. If the number of candies left were two less than the number of children then which of the following CANNOT be the number of candies received by each child?

a.1
b.3
c.7
d.9
e.15

Hi..

A good method would be to make use of choices here
First let us make the equation..
Let x be the number of candies given to each of n child..
So left is n-2..
Therefore n*x+n-2=62......n (x+1)=64
Now the choices..
a.1......n(1+1)=64....n=32
b.3.....n(3+1)=64...n=16
c.7.....n(7+1)=64...n=8
d.9.....n(9+1)=64..10n=64...n=6.4 Not possible as number of children cannot be a fraction
e.15....n(15+1)=64...n=4

D
_________________
Re: 62 candies were equally distributed to a group of children. If the num   [#permalink] 16 Oct 2018, 20:00
Display posts from previous: Sort by