Three packets, E, F and G contained a certain number of buttons. Packet G contained
16 as many buttons as E and F combined. There were 87 more buttons in Packet E than in Packet G. Packet F contained 129 more buttons than Packet G. How many buttons were in Packet F?
Packet E |
Packet F |
Packet G |
Total |
6 u |
1 u |
7 u |
1 u + 87 |
1 u + 129 |
|
|
The total number of buttons in Packet E and Packet F is repeated.
1 u + 87 + 1 u + 129 = 6 u
2 u + 216 = 6 u
6 u - 2 u = 216
4 u = 216
1 u = 216 ÷ 4 = 54
Number of buttons in Packet F
= 1 u + 129
= 1 x 54 + 129
= 183
Answer(s): 183