There were some grapefruits in 3 bags, L, M and N. 25% of the number of grapefruits in Bag L was equal to 10% of the number of grapefruits in Bag M. The number of grapefruits in Bag N was 70% of the total number of grapefruits. After Pierre removed 20% of the grapefruits in Bag N, there were 1210 more grapefruits in Bag N than in Bag M. In the end, how many grapefruits should be transferred from Bag M to Bag N so that the number of grapefruits in Bag L would be the same as Bag M?
Bag L |
Bag M |
Bag N |
2x3 |
5x3 |
|
3x7 |
7x7 |
6 u |
15 u |
49 u |
|
Bag L |
Bag M |
Bag N |
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 L =
110 Bag M
Bag L : Bag M
4 : 10
2 : 5
70% =
70100 =
710The total number of grapefruits in Bag L and Bag M is the combined repeated identity. Make the total number of grapefruits in Box L and Bag M the same. LCM of 7 and 3 = 21
Number of grapefruits removed from Bag N
=
20100 x 49 u
= 9.8 u
Number of grapefruits left in Bag N
= 49 u - 9.8 u
= 39.2 u
Number of more grapefruits in Bag N than Bag M
= 39.2 u - 15 u
= 24.2 u
24.2 u = 1210
1 u = 1210 ÷ 24.2 = 50
|
Bag L |
Bag M |
Bag N |
Before |
6 u |
15 u |
39.2 u |
Change |
|
- 9 u |
+ 9 u |
After |
6 u |
6 u |
48.2 u |
Number of grapefruits to be transferred from Bag M to Bag N so that the number of grapefruits in Bag L would be the same as Bag M.
= 15 u - 6 u
= 9 u
= 9 x 50
= 450
Answer(s): 450