Min had some $2-notes and some $10-notes.
712 of the-notes were $2-notes and the rest were $10-notes. After Min had spent $2800 worth of $10-notes and
47 of $2-notes, she had
37 of the notes left. Find the number of $2-notes Min had left.
| |
$2-notes |
$10-notes |
Total |
Comparing $2-notes
and $10-notes
at first |
7x7 =
49 u |
5x7 =
35 u |
12x7 =
84 u |
| Before |
7x7 =
49 u |
|
7x12 =
84 u |
| Change |
- 4x7 =
- 28 u |
- 280 |
- 4x12 =
- 48 u |
| After |
3x7=
21 u |
35 u - 280 |
3x12 =
36 u |
The total number of notes at first is the repeated. Make the total number of notes the same. LCM of 12 and 7 is 84.
The number of $2-notes is repeated.
Number of $10-notes
= 2800 ÷ 10
= 280
Total number of notes spent = 28 u + 280
28 u + 280 = 48 u
48 u - 28 u = 280
20 u = 280
1 u = 280 ÷ 20 = 14
Number of $2-notes that Min had left
= 21 u
= 21 x 14
= 294
Answer(s): 294
