Acollection of dimes and quarters is worth $19.85. there are 128 coins in all. how many of each type of coin are in the collection?

Number of dimes were 81 and number of quarters were 47

Solution:

Let "d" be the number of dimes

Let "q" be the number of quarters

We know that,

value of 1 dime = $ 0.10

value of 1 quarter = $ 0.25

Given that There are 128 coins in all

number of dimes + number of quarters = 128

d + q = 128 ------ eqn 1

Also given that collection of dimes and quarters is worth $19.85

number of dimes x value of 1 dime + number of quarters x value of 1 quarter = 19.85

0.1d + 0.25q = 19.85  -------- eqn 2

Let us solve eqn 1 and eqn 2

From eqn 1,

d = 128 - q -------- eqn 3

Substitute eqn 3 in eqn 2

0.1(128 - q) + 0.25q = 19.85

12.8 - 0.1q + 0.25q = 19.85

12.8 + 0.15q = 19.85

0.15q = 7.05

Therefore from eqn 3,

d = 128 - q

d = 128 - 47

Thus number of dimes were 81 and number of quarters were 47

none of the above.

for one of the equations to be used in the process, there has to be a term in the equation since squared is .

since there is only a term in the first equation, we cannot derive a from completing the square.

solving with completing the square

the anwser is d) 75°

step-by-step explanation: