Risa and Xuan have some markers in the ratio 3 : 13. Risa has 50 less markers than Xuan. How many markers must Xuan give to Risa so that they will have the same number of markers?
| |
Risa |
Xuan |
Total |
| Before |
3 u |
13 u |
16 u |
| Change |
+ 5 u |
- 5 u |
|
After |
1x8 =
8 u |
1x8 =
8 u |
2x8 =
16 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 16 is 16.
Difference in the number of markers at first
= 13 u - 3 u
= 10 u
10 u = 50
1 u = 50 ÷ 10 = 5
Number of markers that each of them had in the end
= 16 u ÷ 2
= 8 u
Number of markers that Xuan must give to Risa
= 8 u - 3 u
= 5 u
= 5 x 5
= 25
Answer(s): 25
