Jeremy has a carton containing some bunches of green grapes and bunches of red grapes. If he adds in 15 bunches of green grapes, 0.6 of the bunches of grapes in the carton will be bunches of red grapes. If he adds in 54 bunches of red grapes,
67 of the bunches of grapes in the carton will be green bunches of grapes. How many bunches of grapes are there in the carton?
|
Bunches of red grapes |
Bunches of green grapes |
Bunches of red grapes |
Bunches of green grapes |
Before |
3 u |
2 u - 15 |
6 p - 54 |
1 p |
Change |
|
+ 15 |
+ 54 |
|
After |
3 u |
2 u |
6 p |
1 p |
0.6 =
610 =
35If he adds 15 bunches of green grapes,
Number of bunches of green grapes in the end
= 5 u - 3 u
= 2 u
If he adds 54 bunches of red grapes,
Number of bunches of green grapes 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 - 54 --- (1)
2 u - 15 = 1 p
2 u = 1 p + 15 --- (2)
Make u the same.
(1)
x2 6 u = 12 p - 108 --- (3)
(2)
x3 6 u = 3 p + 45 --- (4)
(3) = (4)
12 p - 108 = 3 p + 45
12 p - 3 p = 108 + 45
9 p = 153
1 p = 153 ÷ 9 = 17
Number of bunches of grapes at first
= 1 p + 6 p - 54
= 7 p - 54
= 7 x 17 - 54
= 119 - 54
= 65
Answer(s): 65