I have 40 coins. All dimes, nickels and quarters worth $4.05. Ihave 7 more nickels then dimes. How many quarters do i have?

Neveah Salazar

Neveah Salazar

Open question

2022-09-02

I have 40 coins. All dimes, nickels and quarters worth $4.05. Ihave 7 more nickels then dimes. How many quarters do i have?

Answer & Explanation

ce1ret3i

ce1ret3i

Beginner2022-09-03Added 12 answers

1 nickel = 5 cent
1 dime = 10 cent
1 quarter = 25 cent
$ 4.05 = 4.05*100 = 405 cent
total no. of coins = 40;
let X = no. of dimes,
given there are 7 more nickels than dimes, so X+7 = no. of nickels
so no. of quarters = 40 - {(X+7) + X}
= 40 - 2X - 7
= 33 - 2X
given, All dimes, nickels and quarters worth $4.05 = 405cent.
so, 5*(X + 7) + 10*(X) + 25*(33 - 2X) =405
=> 5X + 35 + 10X + 825 - 50X = 405
=> 35 + 825 - 405 = 50X - 5X - 10X
=> 455 = 35X
=> X = 465/35
= 13 = no. op dimes
so, X + 7 = 13 + 7 = 20 = no. of nickels
33 - 2X = 33 - 2*13 = 7 = no. of quarters
you can check your answer- 5 *20 + 10*13 +25*7 = 100 + 130+ 175= 405
Answer:- 20 nickels,
13dimes,
7 quarters.
Felix Fitzgerald

Felix Fitzgerald

Beginner2022-09-04Added 2 answers

Let x = number of nickels(1 nickel=5cents)
y = number of dimes(1 dime= 10cents)
z = number of quarters (1quarter= 25 cents)
(1) x + y + z = 40
(2) x = y + 7
(3) 5x + 10y + 25z =405 ($4.05100=405 cents)
From (1) & (2): y + 7 + y + z = 40
2y+ z + 7 = 40
2y + z = 40 - 7
2y+ z = 33
z= 33 - 2y
Substitute in(3): 5x + 10y + 25z = 405
5(y + 7) + 10y + 25(33 - 2y) = 405
5y+ 35 + 10y + (25)(33) - 25(2y) = 405
5y + 35 + 10y + 825 - 50y = 405
-35y + 860 = 405
35y = 860 - 405
35y = 455
y = 455 35
y= 13
x = y +7
x = 13 +7
x = 20
z= 33 - 2y
z= 33 - 2(13)
z= 33 - 26
z =7
We will have 7 quarters .

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