Kevin has total of 7590 as currency notes in the denominations of 750, 320 and 10. The ratio of the number of 50 notes and 720 notes is 3:5. If he has a total of 25 notes, how many notes of each denomination are there with him?​

The number of 50 notes = 6

The number of 20 notes = 10

The number of 10 notes = 9

Step-by-step explanation:

The given parameters are;

The total amount Kevin has as currency = 590

The denominations of the notes Kevin has = 50, 20, and 10

The ration of the 50 notes and the 20 notes = 3:5

The total number of notes he has = 25

Therefore, we have;

Let 'x', 'y', and 'z' represent the numbers of 50, 20, and 10 notes Kevin has respectively, we are given the following;

x/y = 3/5...(1)

x + y + z = 25...(2)

50·x + 20·y + 10·z = 590...(3)

Therefore, from equation (1) we get;

x = 3·y/5

By substituting the value of x = 3·y/5 in equations (2) and (3), we get;

For equation (2), we get

x + y + z = 3·y/5 + y + z = 8·y/5 + z = 25

z = 25 - 8·y/5

For equation (3), we get;

50·x + 20·y + 10·z = 50·(3·y/5) + 20·y + 10·z = 50·y + 10·z = 590

∴ z = (590 - 50·y)/10 = 59 - 5·y

Equating the two values of 'z' gives;

25 - 8·y/5 = 59 - 5·y

5·y - 8·y/5 = 59 - 25 = 34

17·y/5 = 34

y = 34/17 × 5 = 10

y = 10

x = 3·y/5 = 3 × 10/5 = 6

x = 6

z = 59 - 5·y = 59 - 5 × 10 = 9

z = 9

Therefore, Kevin has 6, 50 notes, 10, 20 notes and 9, 10 notes

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

step-by-step explanation: