Three packets, U, V and W contained a certain number of erasers. Packet W contained
27 as many erasers as U and V combined. There were 61 more erasers in Packet U than in Packet W. Packet V contained 137 more erasers than Packet W. How many erasers were in Packet V?
Packet U |
Packet V |
Packet W |
Total |
7 u |
2 u |
9 u |
2 u + 61 |
2 u + 137 |
|
|
The total number of erasers in Packet U and Packet V is repeated.
2 u + 61 + 2 u + 137 = 7 u
4 u + 198 = 7 u
7 u - 4 u = 198
3 u = 198
1 u = 198 ÷ 3 = 66
Number of erasers in Packet V
= 2 u + 137
= 2 x 66 + 137
= 269
Answer(s): 269