80% of the caramel apples in Packet B were vanilla caramel apples and the rest were banana caramel apples. Packet C had 30% more vanilla caramel apples than Packet B and twice as many caramel apples than the total number of caramel apples in Packet B. Find the percentage of the banana caramel apples in Packet C that would need to be transferred into Packet B, so that there were an equal number of vanilla and banana caramel apples in Packet B.
|
Packet B |
Packet C |
|
Vanilla |
Banana |
Vanilla |
Banana |
|
5 u |
10 u |
Before |
4 u |
1 u |
5.2 u |
4.8 u |
Change |
|
+ 3 u |
|
- 3 u |
After |
4 u |
4 u |
5.2 u |
1.8 u |
80% =
80100 =
45 100 %+ 30% = 130%
Total number of caramel apples in Packet B
= 4 u + 1 u
= 5 u
Total number of caramel apples in Packet C
= 2 x 5 u
= 10 u
Number of vanilla caramel apples in Packet C
= 130% x 4 u
=
130100 x 4 u
= 5.2 u
Number of vanilla caramel apples in Packet C
= 10 u - 5.2 u
= 4.8 u
Number of banana caramel apples to be transferred from Packet C to Packet B
= 4 u - 1 u
= 3 u
Percentage of caramel apples to be transferred from Packet C
=
34.8 x 100%
= 62.5%
Answer(s): 62.5%