70% of the candies in Packet B were vanilla candies and the rest were cranberry candies. Packet C had 25% more vanilla candies than Packet B and four times as many candies than the total number of candies in Packet B. Find the percentage of the cranberry candies in Packet C that would need to be transferred into Packet B, so that there were an equal number of vanilla and cranberry candies in Packet B.
|
Packet B |
Packet C |
|
Vanilla |
Cranberry |
Vanilla |
Cranberry |
|
10 u |
40 u |
Before |
7 u |
3 u |
8.75 u |
31.25 u |
Change |
|
+ 4 u |
|
- 4 u |
After |
7 u |
7 u |
8.75 u |
27.25 u |
70% =
70100 =
710 100 %+ 25% = 125%
Total number of candies in Packet B
= 7 u + 3 u
= 10 u
Total number of candies in Packet C
= 4 x 10 u
= 40 u
Number of vanilla candies in Packet C
= 125% x 7 u
=
125100 x 7 u
= 8.75 u
Number of vanilla candies in Packet C
= 40 u - 8.75 u
= 31.25 u
Number of cranberry candies to be transferred from Packet C to Packet B
= 7 u - 3 u
= 4 u
Percentage of candies to be transferred from Packet C
=
431.25 x 100%
= 12.8%
Answer(s): 12.8%