To solve the problem, we need to determine the number of 1, 2, and 5 rupee notes that Abhimanyu has, given the total value of the notes and their ratios.
Let's denote:
- [tex]\( x \)[/tex] as the number of 5 rupee notes,
- [tex]\( 3x \)[/tex] as the number of 2 rupee notes (since the number of 2 rupee notes is three times the number of 5 rupee notes),
- [tex]\( y \)[/tex] as the number of 1 rupee notes.
We are given the following information:
1. The total value of the notes is 300 rupees.
2. The total number of notes is 160.
We can set up two equations based on this information:
1. Total number of notes equation:
[tex]\[ x + 3x + y = 160 \][/tex]
Simplifying, we get:
[tex]\[ 4x + y = 160 \][/tex]
2. Total value of the notes equation:
[tex]\[ 5x + 2 \times 3x + y = 300 \][/tex]
Simplifying, we get:
[tex]\[ 5x + 6x + y = 300 \][/tex]
[tex]\[ 11x + y = 300 \][/tex]
Now, we have a system of two linear equations:
[tex]\[ (1) \quad 4x + y = 160 \][/tex]
[tex]\[ (2) \quad 11x + y = 300 \][/tex]
To find [tex]\( x \)[/tex] (the number of 5 rupee notes), we subtract the first equation from the second equation:
[tex]\[ (11x + y) - (4x + y) = 300 - 160 \][/tex]
[tex]\[ 7x = 140 \][/tex]
[tex]\[ x = 20 \][/tex]
Now that we have [tex]\( x \)[/tex]:
- The number of 5 rupee notes is [tex]\( x = 20 \)[/tex].
Next, we need to find the number of 2 rupee notes:
- The number of 2 rupee notes is [tex]\( 3x = 3 \times 20 = 60 \)[/tex].
Finally, we find the number of 1 rupee notes using the total number of notes equation:
[tex]\[ y = 160 - 4x \][/tex]
[tex]\[ y = 160 - 4 \times 20 \][/tex]
[tex]\[ y = 160 - 80 \][/tex]
[tex]\[ y = 80 \][/tex]
So, Abhimanyu has:
- 20 notes of 5 rupees,
- 60 notes of 2 rupees,
- 80 notes of 1 rupee.
