A cat keeps eating to gain weight while a dog keeps doing exercise. Later, the cat's weight increases by 20% and the dog's weight decrease by 10% such that their weights become the same. By what percentage is the cat weight less than the dog's original weight?

Question

Mathematics

Brendan Bush

12 November, 10:32

A cat keeps eating to gain weight while a dog keeps doing exercise. Later, the cat's weight increases by 20% and the dog's weight decrease by 10% such that their weights become the same. By what percentage is the cat weight less than the dog's original weight?

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Answers (1)

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King Mccormick · Answer 1

Step-by-step explanation:

We will call the dog's original weight d and the cat's c. according to the problem 1.2c=.9d

1.2c is the cat's weight increased by 20% and. 9d is the dog's weight decreased by 10% Now we just solve this with algebra for the dog's original weight.

1.2c=.9d

(1.2/.9) c = d

(4/3) c = d

Now, this tells us that the dog's original weight was 4/3 of the cat's. or about 33.33% larger

It kinda looks like the question is then asking what the cat's current weight is in comparison to the dog's original. If that's the case we need just one more step. We want the dog's original weight and the cat's current weight in terms of the same variable. Well, we have that.

Current cat = 1.2c or 6/5, since we have the other in fractional form.

original dog = (4/3) c

Now, what do we multiply the first to get to the second? Well, we need common denominators. 15 is the lcm of 3 and 5, so we'll use that.

6/5 = 18/15 and 4/3 = 20/15

Nowwe want to know what we have to multiply 18 by to get to 20. some algebra will find that.

18x=20

x = 20/18

x = 10/9

that's about 1.11, so the dog's original weight is about 11% higher than the cat's current weight.

Hopefully that's the answer it wanted, if not or if there is something you didn't understand let me know.