Jane and Xuan have some markers in the ratio 3 : 7. Jane has 60 less markers than Xuan. How many markers must Xuan give to Jane so that they will have the same number of markers?
| |
Jane |
Xuan |
Total |
| Before |
3 u |
7 u |
10 u |
| Change |
+ 2 u |
- 2 u |
|
After |
1x5 =
5 u |
1x5 =
5 u |
2x5 =
10 u |
The total number of markers remains unchanged. Make the total number of markers at first and in the end the same. LCM of 2 and 10 is 10.
Difference in the number of markers at first
= 7 u - 3 u
= 4 u
4 u = 60
1 u = 60 ÷ 4 = 15
Number of markers that each of them had in the end
= 10 u ÷ 2
= 5 u
Number of markers that Xuan must give to Jane
= 5 u - 3 u
= 2 u
= 2 x 15
= 30
Answer(s): 30
