Three packets, U, V and W contained a certain number of erasers. Packet W contained
211 as many erasers as U and V combined. There were 56 more erasers in Packet U than in Packet W. Packet V contained 161 more erasers than Packet W. How many erasers were in Packet V?
Packet U |
Packet V |
Packet W |
Total |
11 u |
2 u |
13 u |
2 u + 56 |
2 u + 161 |
|
|
The total number of erasers in Packet U and Packet V is repeated.
2 u + 56 + 2 u + 161 = 11 u
4 u + 217 = 11 u
11 u - 4 u = 217
7 u = 217
1 u = 217 ÷ 7 = 31
Number of erasers in Packet V
= 2 u + 161
= 2 x 31 + 161
= 223
Answer(s): 223