Question

A box contains $7.45 in nickels, dimes, and quarters. There are 43 coins in all, and the sum of the numbers of nickels and dimes is 3 less than the number of quarters. How many coins of each kind are there?

Answer 1

Answer:

Number of nickels = 6

Number of dimes = 14

Number of quarters = 23

Step-by-step explanation:

Total number of coins = 43

Let number of nickels = n

Let number of dimes = d

Let number of quarters = q

$n+d+q=43 ...... (1)$

Money in n nickels = $n \times 0.05$

Money in d dimes = $d \times 0.10$

Money in q quarters = $q \times 0.25$

Total money = $7.45

$0.05n +0.10d+0.25q = 7.45\\\Rightarrow 5n +10d+25q = 745 .... (2)$

The sum of the numbers of nickels and dimes is 3 less than the number of quarters.

i.e.

$n+d=q-3$

Let us put the value  in (1):

$q-3+q=43\\\Rightarrow 2q= 43+3\\\Rightarrow q =23$

Putting the value of q in (1) and (2):

n+d = 20 .... (3)

$5n+10d+575 =745\\\Rightarrow 5n+10d =170\\\Rightarrow n+2d =34 .... (4)$

(4) - (3):

d = 14

Now, from (3):

n = 6

Number of nickels = 6

Number of dimes = 14

Number of quarters = 23