Do I need to find the value of separate variables?

[acho510]

Question: An animal shelter uses 21 cups of food to feed 12 dogs and 15 cats one meal each. How many cups of food are needed to feed one meal to 16 dogs and 20 cats.

I set this up as a proportion: , and I got x = 28.

Answer choices are:
a) 20
b) 24
c) 28
d) not enough information given

My question is: Do I have to find the separate value of how many cups for dogs or cats to find the correct answer? Or is the way I solved correct?

 

[MarkFL]

Hello, and welcome to FMH!

Since the ratio of cats to dogs remained the same, the way you worked it is fine.

 

[Dr.Peterson]

My question is: Do I have to find the separate value of how many cups for dogs or cats to find the correct answer? Or is the way I solved correct?

Your answer is correct; but the way you solved it lacks explicit mention of the fact that the ratio is the same, as MarkFL pointed out, so it is not a complete explanation.

An alternative approach that makes the constant ratio visible, and would therefore catch the error if it were not, is like this:

An animal shelter uses 21 cups of food to feed 12 dogs and 15 cats one meal each. --> 7 cups feed 4 dogs and 5 cats.​

16 dogs and 20 cats are 4 times as much as 4 dogs and 5 cats, so it takes 4*7 = 28 cups to feed them​

An algebraic method might use separate variables for the number of dogs and cats, and would turn out to be equivalent to this.

But if you were aware not only that the amount per dog or cat might be different, but that the constant ratio of dogs to cats makes that moot, then your thinking was good.

 

[prerit]

Hi,
Given the problem, my approach would be like:

12d+15c=21-----eq1
16d+20c=x--------eq2

d=dogs, c=cat, x=unknown qty

multilply eq1 by 16 and eq2 by 12 we get
192d+240c=336-----eq1a
192d+240c=12x-----eq2a

crossing out d and c term we get -12x+336=0
-12x=-336
x=336/12
x=28.

 

[Steven G]

Hi,
Given the problem, my approach would be like:

12d+15c=21-----eq1
16d+20c=x--------eq2

d=dogs, c=cat, x=unknown qty

multilply eq1 by 16 and eq2 by 12 we get
192d+240c=336-----eq1a
192d+240c=12x-----eq2a

crossing out d and c term we get -12x+336=0
-12x=-336
x=336/12
x=28.

Since the LCM(12,16)=48 not 192, you could have multiplied eq1 by 4 and eq3 by 3 which will involve smaller numbers than you used. Again, your method only works sice the ratios remained the same.