bobdylan wrote:At a pet center 5 cans of food are used in 6 days to feed 10 dogs. Each dog is provided with the same amount of food. How many cans of food would be needed for 12 days if 3 of the dogs were sold?
a.84
b. 70
c. 49
d. 25
e. 24

Are you sure that the answer choices provided are correct?

Because I am getting 7 cans as answer. below is my explanation:

Cans of food consumed by 1 dog in 1 day = 5/(10*6) = 1/12
Therefore, Cans of food consumed by 7 dogs in 12 days = 1/12 * 12 * 7 = 7

Please cross-check the answer choices from the source.

Sorry, it was 35 cans of food!
At a pet center 35 cans of food are used in 6 days to feed 10 dogs. Each dog is provided with the same amount of food. How many cans of food would be needed for 12 days if 3 of the dogs were sold?
a.84
b. 70
c. 49
d. 25
e. 24

Sorry, it was 35 cans of food!
At a pet center 35 cans of food are used in 6 days to feed 10 dogs. Each dog is provided with the same amount of food. How many cans of food would be needed for 12 days if 3 of the dogs were sold?
a.84
b. 70
c. 49
d. 25
e. 24

Then the answer would be 49.

Let me know if my solution is not clear to you.

IMO C:

Solution is,

Number of cans required for 12 days = 35 x (7/10) x (12/6) = 49

bobdylan wrote:At a pet center 35 cans of food are used in 6 days to feed 10 dogs. Each dog is provided with the same amount of food. How many cans of food would be needed for 12 days if 3 of the dogs were sold?
a.84
b. 70
c. 49
d. 25
e. 24

The amount of food needed is VARIABLE: it depends on the number of dogs and the number of days.
Food = dog*days.

But the amount of food PER CAN is CONSTANT: it remains the same value, regardless of the number of dogs or the number of days.
(dogs*days)/cans = k.

Since the amount of food per can must always be the same value, we get the following:
(dogs*days)/cans = (dogs*days)/cans.

Thus, given that 10 dogs over 6 days require 35 cans and that 7 dogs over 12 days require x cans, we get:
(10*6)/35 = (7*12)/x
x = 49.

The correct answer is C.

I tried to attempt the Question using this approach but lets says food per dog has to be constant So,


(cans x days)/no. of dogs = k
So using this approach we have

(6x35)/10 = n x 12/7


and hence the answer is in fractions. Please let me know where am i getting wrong.

hey_thr67 wrote:I tried to attempt the Question using this approach but lets says food per dog has to be constant So,


(cans x days)/no. of dogs = k
So using this approach we have

(6x35)/10 = n x 12/7


and hence the answer is in fractions. Please let me know where am i getting wrong.

@hey_thr67

You are multiplying 'cans of food' with the 'no. of the days'. It is wrong.
We have to divide the cans by no. of days and by no. of dogs that will give us the avg food consumption by 1 dog in 1 day.

Hope it helps!

hey_thr67 wrote:I tried to attempt the Question using this approach but lets says food per dog has to be constant So,


(cans x days)/no. of dogs = k
So using this approach we have

(6x35)/10 = n x 12/7


and hence the answer is in fractions. Please let me know where am i getting wrong.

x/y = k implies the following: if x increases by a factor, then y increases by the same factor.

In my solution, (dogs*days)/cans = k.
This relationship holds true because:
If the number of dogs doubles, then the number of cans must double -- even if the number of days is unchanged.
If the number of days doubles, then the number of cans must double -- even if the number of dogs is unchanged.

In your solution, (cans * days)/no. of dogs = k.
This relationship is not valid: the number of days is NOT directly proportional to the number of dogs.
Quite the opposite.
If the number of cans remains constant, then, as the number of dogs INCREASES, the number of days must DECREASE.
TWICE AS MANY dogs can be fed for HALF the number of days.