Jeremy has a carton containing some green pears and yellow pears. If he adds in 26 green pears, 0.7 of the pears in the carton will be yellow pears. If he adds in 54 yellow pears, ³₄ of the pears in the carton will be green pears. How many pears are there in the carton?

	Yellow pears	Green pears	Yellow pears	Green pears
Before	7 u	3 u - 26	3 p - 54	1 p
Change		+ 26	+ 54
After	7 u	3 u	3 p	1 p

0.7 = ⁷₁₀ = ⁷₁₀

If he adds 26 green pears,
Number of green pears in the end
= 10 u - 7 u
= 3 u

If he adds 54 yellow pears,
Number of green pears in the end
= 4 p - 3 p
= 1 p

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

7 u = 3 p - 54 --- (1)

3 u - 26 = 1 p
3 u = 1 p + 26 --- (2)

Make u the same.
(1)_x3
21 u = 9 p - 162 --- (3)

(2)_x7
21 u = 7 p + 182 --- (4)

(3) = (4)
9 p - 162 = 7 p + 182
9 p - 7 p = 162 + 182
2 p = 344

1 p = 344 ÷ 2 = 172

Number of pears at first
= 1 p + 3 p - 54
= 4 p - 54
= 4 x 172 - 54
= 688 - 54
= 634

Answer(s): 634