Jeremy has a carton containing some green pears and yellow pears. If he adds in 41 green pears, 0.6 of the pears in the carton will be yellow pears. If he adds in 42 yellow pears,
67 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 |
3 u |
2 u - 41 |
6 p - 42 |
1 p |
Change |
|
+ 41 |
+ 42 |
|
After |
3 u |
2 u |
6 p |
1 p |
0.6 =
610 =
35If he adds 41 green pears,
Number of green pears in the end
= 5 u - 3 u
= 2 u
If he adds 42 yellow pears,
Number of green pears in the end
= 7 p - 6 p
= 1 p
The number of each type in both scenarios remain unchanged at first.
3 u = 6 p - 42 --- (1)
2 u - 41 = 1 p
2 u = 1 p + 41 --- (2)
Make u the same.
(1)
x2 6 u = 12 p - 84 --- (3)
(2)
x3 6 u = 3 p + 123 --- (4)
(3) = (4)
12 p - 84 = 3 p + 123
12 p - 3 p = 84 + 123
9 p = 207
1 p = 207 ÷ 9 = 23
Number of pears at first
= 1 p + 6 p - 42
= 7 p - 42
= 7 x 23 - 42
= 161 - 42
= 119
Answer(s): 119