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zmey [24]
3 years ago
9

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?
Mathematics
1 answer:
kari74 [83]3 years ago
5 0

Answer:

<em>Number of nickels = 6</em>

<em>Number of dimes = 14</em>

<em>Number of quarters = 23</em>

<em></em>

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

<em>Number of nickels = 6</em>

<em>Number of dimes = 14</em>

<em>Number of quarters = 23</em>

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