Angela's family drank

[sanju09]

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]

[Anurag@Gurome]

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.

[sanju09]

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.

[rros0770]

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]

[anshumishra]

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

[saurabh_maths]

Is there any other way to solve this ?

nothing comes up in my mind as of now ?

[Night reader]

@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

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]

[Night reader]

congrats Anshu, nice work done!

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

[rros0770]

Thanks Anshu, you made that look too easy. I see where I was getting hung up now