math word problem?

Question

a bank contains 100 coins, each is either a nickel, a dime, or a quarter. there are twice as many quarters as nickels. the value of the collection is $13.75. how many quarters are in the collection?

how do you solve this? thanks

Answer 1

[14]
Let the no of nickels be x
So the no of quarters is 2x
And the no of dimes=100-(x+2x)=100-3x
According to the problem,
5x+10(100-3x)+25*2x=1375
5x+1000-30x+50x=1375
25x=1375-1000=375
x=375/25=15
Therefore,the bank has 15 nickels,30 quarters and 100-45 or 55 dimes

Answer 2

N= # of nickels
D = # of dimes
Q = # of quarters

there are twice as many quarters as nickels

Q = 2N

100 coins

N + D + Q = 100
N + D + 2N = 100
D = 100 - 3N

the value of the collection is $13.75

5N + 10D + 25Q = 1375 {pennies}
5N + 10(100 - 3N) + 25(2N) = 1375
5N + 1000 - 30N + 50N = 1375
25N = 375
N = 15

Q = 2(15) = 30

D = 100 - 45 = 55

5(15) + 10(55) + 25(30) = 1375
75 + 550 + 750 = 1375
1375 = 1375

Answer 3

Hello,

Let n = number of nickels
let d = number of dimes
let q = number of quarters

Then
n + d + q = 100
q = 2n
5n + 10d + 25q = 1375

Replace q by 2n giving us
n + d + 2n = 100
5n + 10d + 25(2n) = 1375

so 3n + 1d = 100
55n + 10d = 1375
Multiply the first one by -10 giving us

-30n - 10d = -1000

Add the two
25n = 375 so n = 15

Then since q = 2n q = 30

Hope This Helps!!

Answer 4

a = number of nickel
b = number of dimes
c = number of quarter

there areo 100 coins
a + b + c = 100

are twice as many quarters as nickels, tells you:
c = 2a

a nickel is $0.05
a dime is $0.10
and quarter is $0.25

given the total amount of money is 13.75

.05a + .1b + .25c = 13.75

now solve:
-.10(a + b + c = 100)
.05a + .1b + .25c = 13.75

-.10a - .10b - .10c = -10
.05a + .1b + .25c = 13.75
-----------------------------------
-.05a + .15c = 3.75

recall that c = 2a
-.05a + .15(2a) = 3.75
-.05a + .3a = 3.75
.25a = 3.75
a = 15

c = 2a
c = 2(15)
c = 30

b = 100 - (30 + 15)
b = 55


15 nickels
55 dimes
30 quarters


hope it helps
Rec

Answer 5

what?