Three packets, H, J and K contained a certain number of pencils. Packet K contained
411 as many pencils as H and J combined. There were 89 more pencils in Packet H than in Packet K. Packet J contained 199 more pencils than Packet K. How many pencils were in Packet J?
Packet H |
Packet J |
Packet K |
Total |
11 u |
4 u |
15 u |
4 u + 89 |
4 u + 199 |
|
|
The total number of pencils in Packet H and Packet J is repeated.
4 u + 89 + 4 u + 199 = 11 u
8 u + 288 = 11 u
11 u - 8 u = 288
3 u = 288
1 u = 288 ÷ 3 = 96
Number of pencils in Packet J
= 4 u + 199
= 4 x 96 + 199
= 583
Answer(s): 583