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

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).