Angela's family drank

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Angela's family drank

by sanju09 » Wed Mar 09, 2011 4:55 am
One morning each member of Angela's family drank a B-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
A. 3
B. 4
C. 5
D. 6
E. 7



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by Anurag@Gurome » Wed Mar 09, 2011 5:14 am
sanju09 wrote:One morning each member of Angela's family drank a B-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
A. 3
B. 4
C. 5
D. 6
E. 7

[spoiler]https://www.gmatmaths.com[/spoiler]
Can you please check that again, is it B-ounce mixture of coffee with milk? That should be a number instead.
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by sanju09 » Wed Mar 09, 2011 5:29 am
Anurag@Gurome wrote:
sanju09 wrote:One morning each member of Angela's family drank a B-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
A. 3
B. 4
C. 5
D. 6
E. 7

[spoiler]https://www.gmatmaths.com[/spoiler]
Can you please check that again, is it B-ounce mixture of coffee with milk? That should be a number instead.
If each member of Angela's family drank a B-ounce mixture of coffee with milk, then it means that each of them have drunk an equal amount of mixture.
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by rros0770 » Thu Mar 10, 2011 6:55 am
I'm coming up with [spoiler]5 (inluding Angela)[/spoiler]

I solved it by selecting/plugging in values for the coffee and milk:
6 ounces coffee
4 ounces milk
(10 ounces of milk+coffee total)


Using the above selected values:
1/6 coffee + 1/4 milk = B-ounces
1 ounce coffee + 1 ounce milk = 2 ounce mixture

Therefore you can make FIVE 2 ounce mixtures of coffee+milk


Curious to see if my above logic is correct, also, I would like to see an algebraic solution to this.
I wasn't able to solve using variables fast enough...
[/spoiler]

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by anshumishra » Thu Mar 10, 2011 5:07 pm
sanju09 wrote:One morning each member of Angela's family drank a B-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
A. 3
B. 4
C. 5
D. 6
E. 7



[spoiler]https://www.gmatmaths.com[/spoiler]
Algebraically :
n-> people in the family
c -> no. of cups of coffee
m -> no. of cups of milk

As, everybody consumes equal (B-ounce) of the mixture (milk/coffee)
=> c/6+m/4 = (c+m)/n
=> 4cn + 6 mn = 24c + 24m
=> 2cn + 3mn = 12c + 12m
=> 3m(n-4) = 2c(6-n)

Since, m > 0, c> 0, the only integral value of "n" for which both sides have same sign is 5.
Therefore, the answer is 5. C
Thanks
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by saurabh_maths » Thu Mar 10, 2011 6:01 pm
Is there any other way to solve this ?

nothing comes up in my mind as of now ?

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by Night reader » Thu Mar 10, 2011 6:34 pm
@sanju, this is very tricky problem. I first deducted the amounts Angela drank of milk and coffee, and I got 3/4 of milk and 5/6 of coffee left with her family. So the remaining family members' number should be derived from (3M/4 + 5C/6)/(M/4 + C/6) where M is milk and C is coffee. The resulting number should be an integer!

(3M/4 + 5C/6)/(M/4 + C/6) = i {i is integer}
(9M+10C)/(3M+2C) = i OR (9M+10C)=i*(3M+2C).

Now let's try the answer choices. Start with A) (9M+10C)=(3-1)(3M+2C), we plug in (3-1) because our ratio does not include Angela. So 9M+10C=6M+4C, and 3M=-6C this cannot be correct;

B) (9M+10C)=(4-1)(3M+2C), 9M+10C=9M+6C, AND 4C=0 this cannot be correct;

C) (9M+10C)=(5-1)(3M+2C), 9M+10C=12M+8C, AND 3M=2C this is correct;

D) (9M+10C)=(6-1)(3M+2C), 9M+10C=15M+10C, AND 6M=0 this cannot be correct;

E) (9M+10C)=(7-1)(3M+2C), 9M+10C=18M+12C, AND 9M=-2C this cannot be correct;


answer is C
sanju09 wrote:One morning each member of Angela's family drank a B-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
A. 3
B. 4
C. 5
D. 6
E. 7



[spoiler]https://www.gmatmaths.com[/spoiler]
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com

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by Night reader » Thu Mar 10, 2011 7:07 pm
congrats Anshu, nice work done!
anshumishra wrote:
sanju09 wrote:One morning each member of Angela's family drank a B-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?
A. 3
B. 4
C. 5
D. 6
E. 7



[spoiler]https://www.gmatmaths.com[/spoiler]
Algebraically :
n-> people in the family
c -> no. of cups of coffee
m -> no. of cups of milk

As, everybody consumes equal (B-ounce) of the mixture (milk/coffee)
=> c/6+m/4 = (c+m)/n
=> 4cn + 6 mn = 24c + 24m
=> 2cn + 3mn = 12c + 12m
=> 3m(n-4) = 2c(6-n)

Since, m > 0, c> 0, the only integral value of "n" for which both sides have same sign is 5.
Therefore, the answer is 5. C
My knowledge frontiers came to evolve the GMATPill's methods - the credited study means to boost the Verbal competence. I really like their videos, especially for RC, CR and SC. You do check their study methods at https://www.gmatpill.com

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by rros0770 » Thu Mar 10, 2011 7:31 pm
Thanks Anshu, you made that look too easy. I see where I was getting hung up now