Question

In a toy shop, there were 32 more bears than rabbits. When 2/7 of the rabbits were sold, there were 80 rabbits fewer than bears. (a) How many bears were in the shop at first? (b) What fraction of the bears must be sold so that there would be the same number of bears and rabbits left? Give your answer as a fraction in its simplest form.

Answer 1

so, what do you know?
b = r+32
5/7 r = b-80
solve for r and b, and then finish it off

post your work if you get stuck.

Answer 2

b = number of bears

r = number rabbits

In a toy shop, there were 32 more bears than rabbits means:

b = r + 32

Subtract 32 to both sides

b - 32 = r

r = b - 32

When 2 / 7 of the rabbits were sold there are 5 / 7 rabbits left.

When 2 / 7 of the rabbits were sold, there were 80 rabbits fewer than bears means:

5 / 7 r = b - 80

Replace r by = b - 32 in this equation.

5 / 7 ( b - 32 ) = b - 80

Multiply both sides by 7.

5 ( b - 32 ) = 7 b - 560

5 b - 160 = 7 b - 560

Add 160 to both sides.

5 b = 7 b - 400

Subtract 7 b to both sides.

- 2 b = - 400

b = - 400 / - 2

b = 200

(a)

In the shop at first was 200 bears.

(b)

Use formula:

r = b - 32

r = 200 - 32 = 168

In the shop at first was 168 rabbits.

When 2 / 7 of the rabbits were sold:

2 ∙ 168 / 7 = 336 / 7 = 48 rabbits were sold.

168 - 48 = 120 rabbits left.

In order to have 120 bears, 200 - 120 = 80 bears must be sold.

Fraction of the bears must be sold = 80 / 200 = 40 ∙ 2 / 40 ∙ 5 = 2 / 5