Jeremy has a carton containing some green pears and yellow pears. If he adds in 12 green pears, 0.7 of the pears in the carton will be yellow pears. If he adds in 58 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 - 12 |
3 p - 58 |
1 p |
Change |
|
+ 12 |
+ 58 |
|
After |
7 u |
3 u |
3 p |
1 p |
0.7 =
710 =
710If he adds 12 green pears,
Number of green pears in the end
= 10 u - 7 u
= 3 u
If he adds 58 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 - 58 --- (1)
3 u - 12 = 1 p
3 u = 1 p + 12 --- (2)
Make u the same.
(1)
x3 21 u = 9 p - 174 --- (3)
(2)
x7 21 u = 7 p + 84 --- (4)
(3) = (4)
9 p - 174 = 7 p + 84
9 p - 7 p = 174 + 84
2 p = 258
1 p = 258 ÷ 2 = 129
Number of pears at first
= 1 p + 3 p - 58
= 4 p - 58
= 4 x 129 - 58
= 516 - 58
= 458
Answer(s): 458