I have spent over 2 hours this past week on this Question - help!?

Question

They say this is basic Algebra!
There are $8.80 in nickles, dimes and quarters. If he has 4 nickels fewer than three times the number of dimes and one quarter more than three-fourths the number of dimes, how many coins of each type does he have??????

Answer 1

N = the number of nickels
D = the number of dimes
Q = the number of quarters

First we have to figure out our equations.

There is 5 cents in a nickel, 10 cents in a dime and 25 cents in a quarter, so
8.80 = .05N+ .1D + .25Q

4 nickels fewer than three times the number of dimes gives the equation
N = 3D - 4

one quarter more than three-fourths the number of dimes gives the equation
Q = 3/4 D + 1

We have the following three equations:
8.80 = .05N+ .1D + .25Q
N = 3D - 4
Q = 3/4 D + 1

In the three equations, D is the only variable that they have in common. We will have to find D first to figure it all out.

So we will substitute into our first equation leaving D in the equation everywhere.
8.80 = .05N+ .1D + .25Q
8.80 = .05(3D - 4) + .1D + .25(3/4D + 1)
8.80 = .15D - .2 + .1D + .1875D + .25
8.80 +.2 - .25 = .15D + .1D + .1875D
8.75 = .4375 D
20 = D

We will use D and substitute back into the other equations we have to find how many of the other coins.

N = 3D - 4
We know D = 20, so
N = 3*20 - 4
N = 56

Q = 3/4 D + 1
Again, we know D = 20, so
Q = 3/4 (20) + 1
Q = 15 + 1
Q = 16

We know now that there are
20 dimes
56 nickels
16 quarters

Answer 2

16 quarters
20 dimes
56 nickels

why?
we need to know the number of coins that satisfy the conditions.
We are given two clues. The first is that the number of quarters is 1 more than 3/4 the number of dimes. It makes the following equation...

Q = 3/4D + 1

The second clue is the the number of nickels is 4 fewer than 3 times the number of dimes, or...

N = 3D-4

There can only be one combination of coins that satisfies the equations while still totalling $8.80.
I just took an educated guess at what I thought D would have to be, then plugged in the equations, worked them out, and added the dollar amounts applicable to each number of each type of coin.

20 dimes = $2.00-- use that to solve the Q equation...

Q = 3/4*20 + 1
or Q = 15 + 1
Q = 16 expressed in $$--($4.00)

Solve the N equation the same way...

N = 3*20 - 4
or N = 60 - 4
N = 56 expressed in $$ --($2.80)

$2.00 + $4.00 + 2.8 = $8.80

Answer 3

Based on the information provided I came up with 2 basic equations. 3D-4=N, 1+3/4D=Q (D=# of Dimes), (N=# of Nickels), (Q=# of Quarters). From this you know that: $.1D+$.05N+$.25Q=$8.80. Then substitute above equations into total equation then $.1D+$.05(3D-4)+$.25(1+3/4D)=$8.80. From this you find that D=20 plug this in equations for N and Q,3*20+4=N=56, 1+3/4*20=Q=16.

Answer 4

2 hours? wow!

start by very carefully translating the words to a mathematical expression.

let n=number of nickels, d=number of dimes, q=number of quarters
then
4 nickels fewer than three times the number of dimes
is
n = 3d - 4
and
one quarter more than three-fourths the number of dimes
q = 1 + 3d/4

so far we have 2 equations in 3 unknowns. we are given the
value of the money, so a 3rd equation is:
n/20 + d/10 + q/4 = 8.8 (multiply through by 20 to get rid of fractions)

should be easy enough sailing from here.

n = 3d - 4
q = 1 + 3d/4
n + 2d + 5q = 176

3d - 4 +2d + 5(1 + 3d/4) = 176
5d + 15d/4 =175
35d/4 = 175
d = 4*175/35 = 20

n = 3(20) - 4 = 56
q = 1 + 3(20)/4 = 16

and i didn't even need a calculator.

Answer 5

Let n denote the number of nickels, d the number of dimes, and q the number of quarters.

From the given information:

.05n + .1d + .25q = 8.80
n = 3d - 4
q = 3d/4 + 1

Substituting in for n and q in the larger equation:

.05(3d - 4) + .1d + .25(3d/4 + 1) = 8.80

.15d - .2 + .1d + .1875d + .25 = 8.80

.4375d + .05 = 8.80

.4375d = 8.75

d = 20

Substituting d into the expressions for n and q:

n = 3d - 4 = 3(20) - 4 = 56

q = 3(20)/4 + 1 = 16

There are 20 dimes, 56 nickels, and 16 quarters.

Answer 6

alright i did it twice and i got an answer once and nothing the second time. It's either:
1. not possible (0 with a / through it)
2. 20 dimes, 54 nickels, and 16 quarters.

Answer 7

Yes. If you need this money, then you should definitely say something. Sometimes an employer might cut (or remove) hours if they're trying to get an employee to quit, while other times they might have just had their total payroll cut. You might even have lost hours to a fellow employee that made it vocal that they need more hours. Regardless, call up, BE POLITE, and ask if there is anything you can do that would help get you back on the schedule (time permitting).

Answer 8

5n+10d+25Q=880
3d-4=n
3/4d+1=Q
175d+20=3520

he has 20 dimes 16 quarters and 56 nickles i checked my answer if you have any other questions email me

Answer 9

n=3d-4
q=.75d+1
5n+10d+25q=880
5(3d-4)+10d+25(.75d+1)=880
3d-4+2d+5(.75d+1)=176
3d+2d+3.75d=176+4-5
8.75d=175
d=175/8.75=20 dimes $2.00
n=3d-4=60-4=56 nickles $2.80
q=.75d+1=16 quarters $4.00
total=$8.80

Answer 10

Find the liketerms, find how much money. Then solve.