There were some mangosteens in 3 bags, P, Q and R. 25% of the number of mangosteens in Bag P was equal to 10% of the number of mangosteens in Bag Q. The number of mangosteens in Bag R was 60% of the total number of mangosteens. After Archie removed 20% of the mangosteens in Bag R, there were 136 more mangosteens in Bag R than in Bag Q. In the end, how many mangosteens should be transferred from Bag Q to Bag R so that the number of mangosteens in Bag P would be the same as Bag Q?
Bag P |
Bag Q |
Bag R |
2x2 |
5x2 |
|
2x7 |
3x7 |
4 u |
10 u |
21 u |
|
Bag P |
Bag Q |
Bag R |
Before |
4 u |
10 u |
21 u |
Change |
|
|
- 4.2 u |
After |
4 u |
10 u |
16.8 u |
25% =
25100 =
14 10% =
10100 =
110 14 Bag P =
110 Bag Q
Bag P : Bag Q
4 : 10
2 : 5
60% =
60100 =
35The total number of mangosteens in Bag P and Bag Q is the combined repeated identity. Make the total number of mangosteens in Box P and Bag Q the same. LCM of 7 and 2 = 14
Number of mangosteens removed from Bag R
=
20100 x 21 u
= 4.2 u
Number of mangosteens left in Bag R
= 21 u - 4.2 u
= 16.8 u
Number of more mangosteens in Bag R than Bag Q
= 16.8 u - 10 u
= 6.8 u
6.8 u = 136
1 u = 136 ÷ 6.8 = 20
|
Bag P |
Bag Q |
Bag R |
Before |
4 u |
10 u |
16.8 u |
Change |
|
- 6 u |
+ 6 u |
After |
4 u |
4 u |
22.8 u |
Number of mangosteens to be transferred from Bag Q to Bag R so that the number of mangosteens in Bag P would be the same as Bag Q.
= 10 u - 4 u
= 6 u
= 6 x 20
= 120
Answer(s): 120