Abi, Xylia and Anna have 378 buttons. Abi has
34 as many buttons as Xylia. Anna's buttons is
15 of the total number of Abi's and Xylia's buttons. How many buttons must Xylia give to Anna so that both Abi and Anna will have the same number of buttons?
Abi |
Xylia |
Anna |
Total |
5x7 = 35 u |
1x7 = 7 u |
|
3x5 = 15 u |
4x5 = 20 u |
|
|
15 u |
20 u |
7 u |
378 |
Total number of buttons
= 15 u + 20 u + 7 u
= 42 u
42 u = 378
1 u = 378 ÷ 42 = 9
Abi |
Xylia |
Anna |
Total |
15 u |
20 u |
7 u |
378 |
|
- 8 u |
+ 8 u |
|
15 u |
12 u |
15 u |
378 |
Number of buttons that Xylia must give to Anna
= 15 u - 7 u
= 8 u
= 8 x 9
= 72
Answer(s): 72