Jeremy has a carton containing some yellow pears and green pears. If he adds in 41 yellow pears, 0.7 of the pears in the carton will be green pears. If he adds in 49 green pears, ⁷₈ 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	7 u	3 u - 41	7 p - 49	1 p
Change		+ 41	+ 49
After	7 u	3 u	7 p	1 p

0.7 = ⁷₁₀ = ⁷₁₀

If he adds 41 yellow pears,
Number of yellow pears in the end
= 10 u - 7 u
= 3 u

If he adds 49 green pears,
Number of yellow pears in the end
= 8 p - 7 p
= 1 p

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

7 u = 7 p - 49 --- (1)

3 u - 41 = 1 p
3 u = 1 p + 41 --- (2)

Make u the same.
(1)_x3
21 u = 21 p - 147 --- (3)

(2)_x7
21 u = 7 p + 287 --- (4)

(3) = (4)
21 p - 147 = 7 p + 287
21 p - 7 p = 147 + 287
14 p = 434

1 p = 434 ÷ 14 = 31

Number of pears at first
= 1 p + 7 p - 49
= 8 p - 49
= 8 x 31 - 49
= 248 - 49
= 199

Answer(s): 199