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

 It is currently 25 Mar 2019, 11:12 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # M06-12

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

2
15 00:00

Difficulty:   95% (hard)

Question Stats: 65% (02:56) correct 35% (02:56) wrong based on 186 sessions

### HideShow timer Statistics

A farm has chickens, cows and sheep. The number of chickens and cows combined is 3 times the number of sheep. If there are more cows than chickens or sheep, and together, cows and chickens have a total of 100 feet and heads, how many sheep live at the farm?

A. 5
B. 8
C. 10
D. 14
E. 17

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

2
4
Official Solution:

A farm has chickens, cows and sheep. The number of chickens and cows combined is 3 times the number of sheep. If there are more cows than chickens or sheep, and together, cows and chickens have a total of 100 feet and heads, how many sheep live at the farm?

A. 5
B. 8
C. 10
D. 14
E. 17

Let $$C$$ be a number of chickens, $$W$$ - cows, and $$S$$ - sheep.

Chickens and cows together have 100 feet and heads: $$100=1C+1W+4W+2C$$, assuming that 1 is the number of heads per either unit and 4 and 2 are the number of legs per $$C$$ and $$W$$ accordingly. Therefore: $$100 = 3C + 5W$$.

It is given that $$W \gt C$$, and we also see that $$C$$ should be a multiple of 5 for $$3C + 5W$$ to add up to 100. Picking a few numbers, we get two pairs: (5, 17) and (10, 14).

$$S = \frac{(C+W)}{3}$$; $$5 + 17 = 22$$ which is not divisible by 3, but $$10 + 14 = 24$$ which is divisible by 3, therefore only the second pair will work. So, $$S = 8$$.

_________________
Intern  Joined: 11 Oct 2013
Posts: 6
Location: India
Concentration: International Business, Strategy
Schools: Smeal '17 (A\$)
GPA: 3.01
WE: Supply Chain Management (Manufacturing)

### Show Tags

Could you please explain why C should be a multiple of 5?
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

1
7
mat88 wrote:
Could you please explain why C should be a multiple of 5?

3C + 5W = 3C + (a multiple of 5) = 100 = (a multiple of 5), thus 3C must also be a multiple of 5, which means that C must be a multiple of 5.

WHEN THE SUM OR THE DIFFERENCE OF NUMBERS IS A MULTIPLE OF AN INTEGER
1. If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.

2. If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.

3. If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.

For more check here: divisibility-multiples-factors-tips-and-hints-174998.html
_________________
Current Student B
Joined: 02 Sep 2014
Posts: 88
Location: United States
GMAT 1: 770 Q50 V44 GPA: 3.97

### Show Tags

1
1
I did this slightly differently.
We have 100 = 3C + 5W = 3 (C+W) + 2W = 9S + 2W.
Now we know W>S. If S = 14, 17, W will be negative which is not possible. So D, E are eliminated.
If S = 10, W=2 not possible since W>S. Not C.
If S = 5, then W = 55/2 which is not integer. Not possible.
So S has to be 8, which gives W = 14. Looks good.
Intern  B
Joined: 28 Mar 2014
Posts: 18

### Show Tags

1
There are three times the number of chickens and cows than sheep ----> 3*(w+c) = S why not this ? how comes this eqn s=(c+w)/3? kindly response pls
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

suyashtcs wrote:
There are three times the number of chickens and cows than sheep ----> 3*(w+c) = S why not this ? how comes this eqn s=(c+w)/3? kindly response pls

Edited to remove ambiguity. Hope it's better now.
_________________
Intern  Joined: 04 Mar 2015
Posts: 6

### Show Tags

I didn't understand this
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

I didn't understand this

Please elaborate what didn't you understand? Thank you.
_________________
Intern  Joined: 04 Mar 2015
Posts: 6

### Show Tags

first of all, I would like to thank you for your help
Actually, I didn't understand how 100 equal 1C+1W+4W+2C as well as why C should be a multiple of 5 for 3C+5W to add up to 100. and How we get these two pairs: (5, 17) and (10, 14).

Highly appreciate your help
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

first of all, I would like to thank you for your help
Actually, I didn't understand how 100 equal 1C+1W+4W+2C as well as why C should be a multiple of 5 for 3C+5W to add up to 100. and How we get these two pairs: (5, 17) and (10, 14).

Highly appreciate your help

1. We are given that together, cows and chickens have a total of 100 feet and heads. A cow has 1 head and 4 feet and a chicken has 1 head and 2 feet, so 100=1C+1W+4W+2C. This is also answered in the solution. Check highlighted parts below.

Chickens and cows together have 100 feet and heads: $$100=1C+1W+4W+2C$$, assuming that 1 is the number of heads per either unit and 4 and 2 are the number of legs per $$C$$ and $$W$$ accordingly. Therefore: $$100 = 3C + 5W$$.

2. Your second question is answered here: m06-183709.html#p1452616

3. This is also addressed in the solution - it's done by number picking.
_________________
Intern  Joined: 04 Mar 2015
Posts: 6

### Show Tags

Thank you very much,
Intern  Joined: 08 Dec 2015
Posts: 1

### Show Tags

I think this is a high-quality question and the explanation isn't clear enough, please elaborate. "4 and 2 are the number of legs per C and W accordingly" - wrong
Board of Directors P
Joined: 17 Jul 2014
Posts: 2561
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30 GPA: 3.92
WE: General Management (Transportation)

### Show Tags

a trully 700-lvl question...I incorrectly answered, but now I see why I did the mistake.

cows have 4 legs
chicks have 2 legs.
cow and chicks have 1 head each :D
suppose cows=K, chicks =C.
we thus have:
4K+2C+K+C=100
5K+3C=100
We also know that K+C=3S. we then must find values for K and C so that K+C would be divisible by 3.
C=1 - won't work
C=2 wont work
C=3 wont work
C=4 - won't work
C=5 -> K=17. K+C=22 - won't work.
C=8 - won't work
C=9 - won't work
C=10 -> K=14 -> C+K=24 - works.
C=11 no
C=12 no
C=15 -> K=11. won't work, since K must be > than C.

as we see, only K=14, and C=10 works.
K+C=3S
24=3S
S=8
Intern  Joined: 04 Jul 2015
Posts: 1

### Show Tags

I think this is a poor-quality question and I agree with explanation.
Current Student B
Joined: 19 Mar 2013
Posts: 3

### Show Tags

How do you know that you can use the same variable for feet and hands?
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

dkrunic wrote:
How do you know that you can use the same variable for feet and hands?

Please check here: m06-183709.html#p1594846
_________________
Intern  B
Joined: 07 Jan 2017
Posts: 19
Location: India
Concentration: Other, Other
Schools: UConn"19
WE: Medicine and Health (Pharmaceuticals and Biotech)

### Show Tags

A farm has chickens, cows and sheep. The number of chickens and cows combined is 3 times the number of sheep. If there are more cows than chickens or sheep, and together, cows and chickens have a total of 100 feet and heads, how many sheep live at the farm?

A. 5
B. 8
C. 10
D. 14
E. 17

Let $$C$$ be a number of chickens, $$W$$ - cows, and $$S$$ - sheep.

Chickens and cows together have 100 feet and heads: $$100=1C+1W+4W+2C$$, assuming that 1 is the number of heads per either unit and 4 and 2 are the number of legs per $$C$$ and $$W$$ accordingly. Therefore: $$100 = 3C + 5W$$.

It is given that $$W \gt C$$, and we also see that $$C$$ should be a multiple of 5 for $$3C + 5W$$ to add up to 100. Picking a few numbers, we get two pairs: (5, 17) and (10, 14).

$$S = \frac{(C+W)}{3}$$; $$5 + 17 = 22$$ which is not divisible by 3, but $$10 + 14 = 24$$ which is divisible by 3, therefore only the second pair will work. So, $$S = 8$$.

another pair of (5,25) is possible and as per this the answer can be C.10
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

Sachin07 wrote:
A farm has chickens, cows and sheep. The number of chickens and cows combined is 3 times the number of sheep. If there are more cows than chickens or sheep, and together, cows and chickens have a total of 100 feet and heads, how many sheep live at the farm?

A. 5
B. 8
C. 10
D. 14
E. 17

Let $$C$$ be a number of chickens, $$W$$ - cows, and $$S$$ - sheep.

Chickens and cows together have 100 feet and heads: $$100=1C+1W+4W+2C$$, assuming that 1 is the number of heads per either unit and 4 and 2 are the number of legs per $$C$$ and $$W$$ accordingly. Therefore: $$100 = 3C + 5W$$.

It is given that $$W \gt C$$, and we also see that $$C$$ should be a multiple of 5 for $$3C + 5W$$ to add up to 100. Picking a few numbers, we get two pairs: (5, 17) and (10, 14).

$$S = \frac{(C+W)}{3}$$; $$5 + 17 = 22$$ which is not divisible by 3, but $$10 + 14 = 24$$ which is divisible by 3, therefore only the second pair will work. So, $$S = 8$$.

another pair of (5,25) is possible and as per this the answer can be C.10

Dear Sachin07, please note that all questions here are correct, so if you are getting answer which differs from the OA it means that you've done something wrong.

For example, c = 5 and w = 25 doe not give a sum of 100 for 3c + 5w:

3*10 + 5*25 =155.
_________________
SVP  D
Joined: 26 Mar 2013
Posts: 2102

### Show Tags

Let $$K$$ be a number of chickens, $$C$$ - cows, and $$S$$ - sheep.

Translation the word question into formulas:

chickens and cows combined is 3 times the number of sheep: $$K$$ + $$C$$ = 3$$S$$.......1

Chickens and cows together have 100 feet and heads: $$3K + 5C=100$$. It is known that chicken has 1 head and 2 feet and cow has 1 head and 4 feet...........2

cows than chickens or sheep: C> K & S............3

From 1: K=3S - C

Apply into 2 , we get: 9S+2C=100..... Here I want to make one equation in sheep and cow to make use of the info in (3).

Let's analyze the equation above, 100 is even & 2C is even so 9S MUST be even too. Hence S is even number...........Eliminate choices A & E

Rearrange 2: C= 50- (9S)/2

Plug in choice B, S=8 .......C=14 ......C>S........our target answer.

You may want to examine choice C, S=10..........C=5......C<S... incorrect M06-12   [#permalink] 13 Feb 2017, 01:23

Go to page    1   2    Next  [ 25 posts ]

Display posts from previous: Sort by

# M06-12

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.  