There were some passion fruits in 3 boxes, N, P and Q. 25% of the number of passion fruits in Box N was equal to 10% of the number of passion fruits in Box P. The number of passion fruits in Box Q was 70% of the total number of passion fruits. After Sean removed 20% of the passion fruits in Box Q, there were 484 more passion fruits in Box Q than in Box P. In the end, how many passion fruits should be transferred from Box P to Box Q so that the number of passion fruits in Box N would be the same as Box P?
Box N |
Box P |
Box Q |
2x3 |
5x3 |
|
3x7 |
7x7 |
6 u |
15 u |
49 u |
|
Box N |
Box P |
Box Q |
Before |
6 u |
15 u |
49 u |
Change |
|
|
- 9.8 u |
After |
6 u |
15 u |
39.2 u |
25% =
25100 =
14 10% =
10100 =
110 14 Box N =
110 Box P
Box N : Box P
4 : 10
2 : 5
70% =
70100 =
710The total number of passion fruits in Box N and Box P is the combined repeated identity. Make the total number of passion fruits in Box N and Box P the same. LCM of 7 and 3 = 21
Number of passion fruits removed from Box Q
=
20100 x 49 u
= 9.8 u
Number of passion fruits left in Box Q
= 49 u - 9.8 u
= 39.2 u
Number of more passion fruits in Box Q than Box P
= 39.2 u - 15 u
= 24.2 u
24.2 u = 484
1 u = 484 ÷ 24.2 = 20
|
Box N |
Box P |
Box Q |
Before |
6 u |
15 u |
39.2 u |
Change |
|
- 9 u |
+ 9 u |
After |
6 u |
6 u |
48.2 u |
Number of passion fruits to be transferred from Box P to Box Q so that the number of passion fruits in Box N would be the same as Box P.
= 15 u - 6 u
= 9 u
= 9 x 20
= 180
Answer(s): 180