Fred and Ian had 90 pencils together. Ian had 2 less pencils than Fred. Both of them sold some of their pencils. Ian sold 20% more pencils than what Fred sold. In the end, Ian had half as many pencils left as Fred. How many pencils had Ian left?
| |
Fred
(1) |
Ian
(2) |
Make p the same
(2)x2 = (3) |
Comparing Fred
and Ian at first |
2 more |
|
|
| Before |
46 |
44 |
88 |
| Change |
- 5 u |
- 6 u |
- 12 u |
| After |
2 p |
1 p |
2 p |
Number of pencils that Ian had at first
= (90 - 2) ÷ 2
= 88 ÷ 2
= 44
Number of pencils that Fred had at first
= 44 + 2
= 46
Number of pencils that Ian sold in percent
= 100% + 20%
= 120%
120% =
120100 =
65 Make p the same.
(1) = (3)
46 - 5 u = 88 - 12 u
12 u - 5 u = 88 - 46
7 u = 42
1 u = 42 ÷ 7 = 6
Number of pencils that Ian had left
= 44 - 6 u
= 44 - 6 x 6
= 44 - 36
= 8
Answer(s): 8
