Jeremy has a carton containing some yellow pears and green pears. If he adds in 17 yellow pears, 0.6 of the pears in the carton will be green pears. If he adds in 45 green pears, ³₄ of the pears in the carton will be green pears. How many pears are there in the carton?

	Green pears	Yellow pears	Green pears	Yellow pears
Before	3 u	2 u - 17	3 p - 45	1 p
Change		+ 17	+ 45
After	3 u	2 u	3 p	1 p

0.6 = ⁶₁₀ = ³₅

If he adds 17 yellow pears,
Number of yellow pears in the end
= 5 u - 3 u
= 2 u

If he adds 45 green pears,
Number of yellow pears in the end
= 4 p - 3 p
= 1 p

The number of each type in both scenarios remain unchanged at first.

3 u = 3 p - 45 --- (1)

2 u - 17 = 1 p
2 u = 1 p + 17 --- (2)

Make u the same.
(1)_x2
6 u = 6 p - 90 --- (3)

(2)_x3
6 u = 3 p + 51 --- (4)

(3) = (4)
6 p - 90 = 3 p + 51
6 p - 3 p = 90 + 51
3 p = 141

1 p = 141 ÷ 3 = 47

Number of pears at first
= 1 p + 3 p - 45
= 4 p - 45
= 4 x 47 - 45
= 188 - 45
= 143

Answer(s): 143