Question

The girls' soccer team held a fundraising dog & cat wash. They charged $6

Answer 1

Answer:

The team washed 27 cats and 41 dogs.

Step-by-step explanation:

Given:

Total number of animals washed = 68

Total money earned = (487+85) = 572$

Charged for each cat = 6$

Charged for each dog = 10$

Let the number of cats be 'x' and number of dogs be 'y'.

According to the question:

$x+y =68$          ...equation (i)

$6x+10y=572$  ...equation (ii)

Let arrange the first equation.

$x=68-y$         ...equation (iii)

Plugging the value of x from (iii) to (i).

We have,

⇒ $6(68-y)+10y=572$

⇒ $408-6y+10y=572$

⇒ $408+4y=572$

⇒ $4y=572-408$

⇒ $4y=164$

⇒ $y=\frac{164}{4}$

⇒ $y=41$

Plugging the value of y=41 in equation (iii) we have .

⇒ $x=68-41=27$

So the number of dogs washed = 41 and the number of cats washed = 27

by the girls soccer team for the fundraising.