There were some guavas in 3 bags, V, W and X. 25% of the number of guavas in Bag V was equal to 10% of the number of guavas in Bag W. The number of guavas in Bag X was 70% of the total number of guavas. After Ken removed 20% of the guavas in Bag X, there were 968 more guavas in Bag X than in Bag W. In the end, how many guavas should be transferred from Bag W to Bag X so that the number of guavas in Bag V would be the same as Bag W?
Bag V |
Bag W |
Bag X |
2x3 |
5x3 |
|
3x7 |
7x7 |
6 u |
15 u |
49 u |
|
Bag V |
Bag W |
Bag X |
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 Bag V =
110 Bag W
Bag V : Bag W
4 : 10
2 : 5
70% =
70100 =
710The total number of guavas in Bag V and Bag W is the combined repeated identity. Make the total number of guavas in Box V and Bag W the same. LCM of 7 and 3 = 21
Number of guavas removed from Bag X
=
20100 x 49 u
= 9.8 u
Number of guavas left in Bag X
= 49 u - 9.8 u
= 39.2 u
Number of more guavas in Bag X than Bag W
= 39.2 u - 15 u
= 24.2 u
24.2 u = 968
1 u = 968 ÷ 24.2 = 40
|
Bag V |
Bag W |
Bag X |
Before |
6 u |
15 u |
39.2 u |
Change |
|
- 9 u |
+ 9 u |
After |
6 u |
6 u |
48.2 u |
Number of guavas to be transferred from Bag W to Bag X so that the number of guavas in Bag V would be the same as Bag W.
= 15 u - 6 u
= 9 u
= 9 x 40
= 360
Answer(s): 360