There were some peaches in 3 boxes, F, G and H. 25% of the number of peaches in Box F was equal to 10% of the number of peaches in Box G. The number of peaches in Box H was 70% of the total number of peaches. After Japheth removed 25% of the peaches in Box H, there were 435 more peaches in Box H than in Box G. In the end, how many peaches should be transferred from Box G to Box H so that the number of peaches in Box F would be the same as Box G?
Box F |
Box G |
Box H |
2x3 |
5x3 |
|
3x7 |
7x7 |
6 u |
15 u |
49 u |
|
Box F |
Box G |
Box H |
Before |
6 u |
15 u |
49 u |
Change |
|
|
- 12.25 u |
After |
6 u |
15 u |
36.75 u |
25% =
25100 =
14 10% =
10100 =
110 14 Box F =
110 Box G
Box F : Box G
4 : 10
2 : 5
70% =
70100 =
710The total number of peaches in Box F and Box G is the combined repeated identity. Make the total number of peaches in Box F and Box G the same. LCM of 7 and 3 = 21
Number of peaches removed from Box H
=
25100 x 49 u
= 12.25 u
Number of peaches left in Box H
= 49 u - 12.25 u
= 36.75 u
Number of more peaches in Box H than Box G
= 36.75 u - 15 u
= 21.75 u
21.75 u = 435
1 u = 435 ÷ 21.75 = 20
|
Box F |
Box G |
Box H |
Before |
6 u |
15 u |
36.75 u |
Change |
|
- 9 u |
+ 9 u |
After |
6 u |
6 u |
45.75 u |
Number of peaches to be transferred from Box G to Box H so that the number of peaches in Box F would be the same as Box G.
= 15 u - 6 u
= 9 u
= 9 x 20
= 180
Answer(s): 180