Jeremy has a carton containing some bunches of red grapes and bunches of green grapes. If he adds in 40 bunches of red grapes, 0.7 of the bunches of grapes in the carton will be bunches of green grapes. If he adds in 48 bunches of green grapes,
34 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 green grapes |
Bunches of red grapes |
Bunches of green grapes |
Bunches of red grapes |
Before |
7 u |
3 u - 40 |
3 p - 48 |
1 p |
Change |
|
+ 40 |
+ 48 |
|
After |
7 u |
3 u |
3 p |
1 p |
0.7 =
710 =
710If he adds 40 bunches of red grapes,
Number of bunches of red grapes in the end
= 10 u - 7 u
= 3 u
If he adds 48 bunches of green grapes,
Number of bunches of red grapes 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 - 48 --- (1)
3 u - 40 = 1 p
3 u = 1 p + 40 --- (2)
Make u the same.
(1)
x3 21 u = 9 p - 144 --- (3)
(2)
x7 21 u = 7 p + 280 --- (4)
(3) = (4)
9 p - 144 = 7 p + 280
9 p - 7 p = 144 + 280
2 p = 424
1 p = 424 ÷ 2 = 212
Number of bunches of grapes at first
= 1 p + 3 p - 48
= 4 p - 48
= 4 x 212 - 48
= 848 - 48
= 800
Answer(s): 800