Three packets, F, G and H contained a certain number of pens. Packet H contained
27 as many pens as F and G combined. There were 57 more pens in Packet F than in Packet H. Packet G contained 156 more pens than Packet H. How many pens were in Packet G?
Packet F |
Packet G |
Packet H |
Total |
7 u |
2 u |
9 u |
2 u + 57 |
2 u + 156 |
|
|
The total number of pens in Packet F and Packet G is repeated.
2 u + 57 + 2 u + 156 = 7 u
4 u + 213 = 7 u
7 u - 4 u = 213
3 u = 213
1 u = 213 ÷ 3 = 71
Number of pens in Packet G
= 2 u + 156
= 2 x 71 + 156
= 298
Answer(s): 298