Three packets, N, P and Q contained a certain number of erasers. Packet Q contained
14 as many erasers as N and P combined. There were 98 more erasers in Packet N than in Packet Q. Packet P contained 134 more erasers than Packet Q. How many erasers were in Packet P?
Packet N |
Packet P |
Packet Q |
Total |
4 u |
1 u |
5 u |
1 u + 98 |
1 u + 134 |
|
|
The total number of erasers in Packet N and Packet P is repeated.
1 u + 98 + 1 u + 134 = 4 u
2 u + 232 = 4 u
4 u - 2 u = 232
2 u = 232
1 u = 232 ÷ 2 = 116
Number of erasers in Packet P
= 1 u + 134
= 1 x 116 + 134
= 250
Answer(s): 250