60% of the caramel apples in Packet N were banana caramel apples and the rest were vanilla caramel apples. Packet P had 25% more banana caramel apples than Packet N and twice as many caramel apples than the total number of caramel apples in Packet N. Find the percentage of the vanilla caramel apples in Packet P that would need to be transferred into Packet N, so that there were an equal number of banana and vanilla caramel apples in Packet N.
|
Packet N |
Packet P |
|
Banana |
Vanilla |
Banana |
Vanilla |
|
5 u |
10 u |
Before |
3 u |
2 u |
3.75 u |
6.25 u |
Change |
|
+ 1 u |
|
- 1 u |
After |
3 u |
3 u |
3.75 u |
5.25 u |
60% =
60100 =
35 100 %+ 25% = 125%
Total number of caramel apples in Packet N
= 3 u + 2 u
= 5 u
Total number of caramel apples in Packet P
= 2 x 5 u
= 10 u
Number of banana caramel apples in Packet P
= 125% x 3 u
=
125100 x 3 u
= 3.75 u
Number of banana caramel apples in Packet P
= 10 u - 3.75 u
= 6.25 u
Number of vanilla caramel apples to be transferred from Packet P to Packet N
= 3 u - 2 u
= 1 u
Percentage of caramel apples to be transferred from Packet P
=
16.25 x 100%
= 16%
Answer(s): 16%