Plz teach how to solve Coin Problems ?

Question

Sylvia has $11.25 in nickels, dimes, and quarters. She has 3 times as many nickels as dimes and 5 more quarters than dimes. How many of each kind of coin does she have?

The Answers are 20 Dimes, 60 nickels , 25 Quarters

the problem comes with the answers.

But can some1 explain to me very well how to step by step get the answer
i tried for 20 mins on 1 problem still don't get it!
THX

Answer 1

Let N, D and Q be the number of nickels, dimes and quarters she has.

So we know that:
5N + 10D + 25Q = 1125*

(*I like to work in whole numbers, so do this all in cents, rather than dollars. A nickel is 5 cents, so 5N will be the value of the nickels. All together it adds to $11.25 or 1125 cents.)

We also know that she has 3 times as many nickels as dimes:
N = 3D

And 5 more quarters than dimes:
Q = D + 5

Now you can substitute N = 3D into the first equation:
5(3D) + 10D + 25Q = 1125
15D + 10D = 25Q = 1125
25D + 25Q = 1125

Do the same with Q = D + 5:
25D + 25(D + 5) = 1125
25D + 25D + 125 = 1125
50D = 1125 - 125
50D = 1000
D = 20

Now just solve for the other values:
N = 3D
N = 3(20)
N = 60

Q = D + 5
Q = (20) + 5
Q = 25

So all together:
20 dimes, 60 nickels and 25 quarters.

Answer 2

You should define your "unknowns" and then write down as many equations as you can, using the unknowns.

Let's say she has N nickels, D dimes and Q quarters

Clearly, .05N + 0.1D +.25Q = 11.25 from the first sentence of the problem.

Also, N=3D and Q=D+5 from the other two clues. By the way, we now have three equations and three unknowns, which is a useful indication that we can solve this!

We can use the second and third equations in the first like this:

.05(3D) + 0.1D + 0.25(D+5) = 11.25
So 0.50D + 1.25 = 11.25
So D = (11.25-1.25)/0.5 = 20

After that, N=3D =3(20) = 60
and Q=D+5 = 20 + 5 = 25

and you're done!

Answer 3

Say d=number dimes, n=number nickels, and q=numbers quarters. You are also given, 3d=n and d+5=q

Using the value of each coin, with the total equal to $11.25,

.1d+.05n+.25q=11.25

If you plug n=3d and q=d+5 into this formula

.1d+.05(3d)+.25(d+5) = 11.25

Solving this equation for d,

.1d+.15d+.25d+1.25 = 11.25, or
.5d+1.25=11.25, or
.5d=10, or
d=20

Using n=3d, n=3*20, n=60

Using q=d+5, q=20+5, q=25.

Answer 4

Well, dimes is your variable, describe it by x.
So if you have 3 times as many nickels as dimes then it is (3x)
If you have 5 more quarters than dimes it is (x+5)
Now construct an equation for the solving of the dollar amount.
11.25 = 0.05(3x) + 0.10(x) + 0.25(x+5)
11.25 = 0.15x + 0.10x + 0.25x + 1.25
10.00 = 0.50x
x = 20
so the amount of quarters is x+5 which is 25
and the amount of nickels is 3x which is 60

Answer 5

Let x=number of dimes,values of all dimes is 10x
number of nickles=3x, values of all nickles is 15x
number of quarters=5+x, values of all quarters is 25(5+x)
The value of all coins is 1125cents

You can write this equation according to their values
10x+15x+25(5+x)=1125
25x+125+25x=1125
50x=1000
x=20
number of dimes =20

Answer 6

It has only been 20 years since I have had this in school, but here goes.

X=# of dimes
3X=# of nickels
x+5=# of quarters

However, we need the value of these coins so

.1X = value of dimes
3X(.05)=value of nickels =.15X
.25(x+5)=quarters = .25X+1.25

thus

.1x+.15X+.25X+1.25=11.25
.5X+1.25=11.25
.5X=10
X=20

Thus the number of dimes equals 20
nickels equals 3(20)=60
quarters equals 20+5=25

Answer 7

p = how many pennies

n = how many nickles

d = how many dimes

q = how many quarters


.01p + .05n + .10d + .25q = how much money

typically we would not have all the coins at one time. and most times the quantity of one coin is given in terns of another coin.

for example, meg has only dimes and quarters. she has twice as many dimes as quarters. how many dimes and quarters does she have if she has 90 cents?

q = how many quarters

2q = now many dimes: becasue she has twice as many dimes as quarters

.25q + .10(2q) = .90

be carefull that you are using the same units either dollars or cents but do not mix them.

Answer 8

11.25 = .05n + .1d + .25q
3d = n
q = d + 5
substitution
11.25 = .05(3d) + .1d + .25(d+5)
11.25 = .15d + .1d + .25d + 1.25
10 = .5d
d = 20
3(20) = n
n=60
q = 20 + 5
q= 25