Oliver had three boxes, containing a total of 1400 plastic bowls. The number of plastic bowls in Box M to the total number of plastic bowls was 3 : 10. He sold 359 plastic bowls from Box N and sold
13 of the plastic bowls in Box P. The number of plastic bowls left in Box N to the number of plastic bowls left in Box P was 3 : 1. How many plastic bowls were there in Box N at first?
|
Box M |
Box N |
Box P |
Total |
Before |
3 u
|
7 u (980)
|
10 u (1400) |
|
|
6 p + 359 |
3 p |
|
Change |
|
- 359 |
- 1 p |
|
|
|
|
2 p |
|
After |
|
3x2 = 6 p |
1x2 = 2 p |
|
10 u = 1400
1 u = 1400 ÷ 10 = 140
Number of plastic bowls in Box N and Box P at first
= 7 u
= 7 x 140
= 980
The number of bowls left in P is the repeated identity.
LCM of 1 and 2 = 2
6 p + 359 + 3 p = 980
6 p + 3 p = 980 - 359
9 p = 621
1 p = 621 ÷ 9 = 69
Number of bowls in Box N at first
= 6 p + 359
= 6 x 69 + 359
= 414 + 359
= 773
Answer(s):773