Cody has 321 mochi balls and Pierre has 799 mochi balls. How many mochi balls must Pierre give to Cody so that Cody will have nine times as many mochi balls as Pierre?
|
Cody |
Pierre |
Total |
Before |
321 |
799 |
1120 |
Change |
+ ? |
- ? |
|
After |
9 u |
1 u |
10 u |
Total number of mochi balls
= 321 + 799
= 1120
Total number of mochi balls
= 9 u + 1 u
= 10 u
10 u = 1120
1 u = 1120 ÷ 10 = 112
Number of mochi balls that Pierre must give Cody
= 799 - 112
= 687
Answer(s): 687