harry has $2.25 in nickels dimes and quarters. if he had twice as many nickels half as many dimes and the sa

Question

harry has $2.25 in nickels dimes and quarters. if he had twice as many nickels half as many dimes and the same number of quarters he'd have 2.50.... he has 27 coins altogether, how many of each

Answer 1

10 nickels- 50 cents
5 dimes- 50 cents
5 quarters - 1.25

I am not sure of the wording of the question but i think this is right.

Answer 2

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

5n + 10d + 25q = 225
10n + 5d + 25q = 250
n + d + q = 27
Subtract first equation from the second to get rid of q.
5n - 5d = 25
n - d = 5
n = 5+d
Substitute for n in all equations
5(5+d) + 10d + 25q = 225
10(5+d) + 5d + 25q = 250
(5+d) +d + q = 27

25+ 15d + 25q = 225; 15d + 25q = 200; 3d + 5q = 40
50 + 15d + 25q = 250; 3d + 5q = 40
5 + 2d + q = 27; 2d + q = 22; q = 22 - 2d
Substitute for q in the 1st or 2d equation (which are now equivalent)
3d + 5(22-2d) = 40
3d + 110 - 10d = 40
70 = 7d
10 = d
Since n-d=5; n = 15
Since n+d+q = 27; q = 2
Check 5*15 + 10*10 + 2*25 = 75 + 100 + 50 = 225
Check 5*30 + 5*10 + 2*25 = 150 + 50 + 50 = 250
Checks.
15 nickels, 10 dimes, 2 quarters

Answer 3

This question is flawed because twice as many nickels and half as many dimes combine for the same value..

ie.. if i had 10 nickels and 10 dimes... I would have 1.50

twice as many nickels is 20 and half as many dimes is 5

20 nickels and 5 dimes = 1.50

If he has twice as many nickels and half as many dimes. and the same # of quarters, then he has to have the same value of 2.25, value wouldn't (COULDN't ) change.!

Answer 4

lets convert everything into cents.. fractions are a bit difficult to handle.

Let the number of quarters be x, number of dimes be y and quarters be z.

From the first statement:
25x + 10y + 5z = 225

i.e. 5x + 2y + z = 45 -- eq(1)

From the second statement
25x + 5y + 10z = 250

i.e. 5x + y + 2z = 50 -- eq(2)

From the third statement
x + y + z = 27 -- eq(3)

subtract eq(1) from eq(2)
5z - 5y = 25
There z = y + 5

Substitute the value of z in eq(1)

Now,
5x + 2y + (y+5) = 45
i.e. 5x + 3y = 40 -- eq(4)

Substitute value of z in eq(3)
x + y + (y+5) = 27
i.e. x + 2y = 22

i.e. 5x + 10y = 110 -- eq(5)

Subtract eq4 from eq5

Now,
7y = 70

Therefore y = 10
Therefore z = 15
Therefore x = 2

Number of nickels = 15
Number of dimes = 10
number of quarters = 2

Answer 5

unsolvable. you can not have all of these be true and have 27 coins.