Please help on this math problem!

Question

Micheal's bank contains only nickels, dimes, and quarters. There are 63 coins in all, valued at $5.55. The number of nickels is 9 short of being 3x the sum of the number of dimes and quarters. How many dimes are in the bank?

Answer 1

Let's let N be the number of nickels, D the number of dimes and Q the number of quarters.

So we have three equations that result from the information given:

(a) N + D + Q = 63

(b) 0.05N + 0.10D + 0.25Q = 5.55. If we divide both sides by 0.05, we get: N + 2D + 5Q = 111

(c) N = 3 (D + Q) - 9 = 3D + 3Q - 9

Let's take take the information from (c) and plug it into (b):

(3D + 3Q - 9) + 2D + 5Q = 111

So 5D + 8Q = 111 + 9 = 120.

Let's subtract 8Q from both sides to get:

5D = 120 - 8Q

Divide both sides by 5 to get:

D = 24 - (8/5)Q

Now let's use this in equation (a):

3D + 3Q - 9 + D + Q = 63

So combining like terms, we have:

4D + 4Q = 72

And 4( 24 - (8/5) Q) + 4Q = 72

96 - (32/5)Q + 4Q = 72

(-12/5)Q = 72 - 96 = -24

Q = -24(-5/12) = 10

So we know that there are 10 quarters in his bank.

But recall that D = 24 - (8/5)Q

So if Q=10, then we have D = 24 - (8/5)(10) = 24 - 16 = 8

So there are 8 dimes in his bank.

Finally, (a) gives us that N + D + Q = 63, so N + 8 + 10 = 63, and N = 63 - 18 = 45

So there are 45 nickels in his bank.

We can go back and double-check that our answer meets the conditions given in the problem:

45 nickels + 8 dimes + 10 quarters = 63 coins

45($0.05) + 8($0.10) + 10($0.25) = $2.25 + $0.80 + $2.50 = $5.55

And 3(8 dimes + 10 quarters) - 9 = 3(18) - 9 = 54 - 9 = 45 nickels

Therefore, Michael's bank has 45 nickels, 8 dimes and 10 quarters.

Answer 2

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

Since there are 63 total coins, we can say n + d + q = 63.

Since the value is $5.55, we can say 0.05n + 0.10d + 0.25q = 5.55 (the total value is equal to 5 cents times the number of nickles plus 10 cents times the number of dimes plus 25 cents times the number of quarters)

Finally, 3x the sum of the number of dimes and quarters is 3(d + q) and they tell us that if t=we had nine more nickels the number of nickles would be the same as that. So n + 9 = 3(d + q).

Summarizing:

(1) n + d + q = 63

(2) 0.05n + 0.10d + 0.25q = 5.55

(3) n + 9 = 3(d + q)

I'm going to multiply both sides of equation 2 by 100 to get rid of the decimals to get 5n + 10d + 25q = 555. Now, to simplify, I can divide both sides by 5 to get:

(2) n + 2d + 5q = 111

I'm going to multiply equation 3 out and combine like-terms to get:

n + 9 = 3d + 3q or:

(3) n - 3d - 3q = -9

Now, re-summarizing, we have:

(1) n + d + q = 63

(2) n + 2d + 5q = 111

(3) n - 3d - 3q = -9

I like to use elimination to solve three equations in three unknowns. I can eliminate the "n" by multiplying equation 3 by -1 making it -n + 3d + 3q = 9 and then first adding it to equation 1 and then adding it to equation 2:

(1) n + d + q = 63

(3) -n + 3d + 3q = 9

---------------------------

(4) 4d + 4q = 72

(2) n + 2d + 5q = 111

(3) -n + 3d + 3q = 9

--------------------------

(5) 5d + 8q = 120

Now we have:

(4) 4d + 4q = 72

5) 5d + 8q = 120

I can eliminate the "q" by multiplying equation 4 by -2 and adding them together:

(4) -8d + -8q = -144

5) 5d + 8q = 120

-------------------------

-3d = -24

or d = 8

I can solve for q by plugging d = 8 into either equation 4 or equation 5, I'll pick equation 4

4(8) + 4q = 72

32 + 4q = 72

4q = 40

q = 10

Now, I can plug both d = 8 and q = 10 into either equation 1, 2, or 3. I'll pick equation 1:

n + 8 + 10 = 63

n + 18 = 63

n = 45