Question 515404: Joe has a collection of nickels and quarters that is worth $7.40. If the number of nickels were doubled and the number of quarters were increased by 8, the value of the coins would be $10.80. How many quarters does he have?
I need step by step through the equation on how to solve it. Could you please help me with this as I am having troubles trying to solve it?
Found 2 solutions by drcole, mucklef:
Answer by drcole(72)   (Show Source):
You can put this solution on YOUR website! Let N be the original number of nickels, and let Q be the original number of quarters. N nickels are worth 5N cents, and Q quarters are worth 25Q cents, so N nickels and Q quarters are worth 5N + 25Q cents. We know that the original number of nickels and quarters are worth $7.40, or 740 cents together, so we can make the algebraic equation:
5N + 25Q = 740
That's our first equation. Now we double the number of nickels to 2N, and we increase the number of quarters to Q + 8. 2N nickels are worth 5(2N) = 10N cents, and Q + 8 quarters are worth 25(Q + 8) cents. Together, this new set of nickels and quarters are worth $10.80, or 1080 cents, giving us a second algebraic equation:
10N + 25(Q + 8) = 1080
10N + 25Q + 200 = 1080 (distributing 25 on the left side)
10N + 25Q = 880 (subtracting 200 from both sides)
Now we have two linear equation in two unknowns. We can solve this by a number of methods, but we'll try substitution. First, notice that we can simplify the first equation by dividing both sides by 5:
N + 5Q = 148
Now we can solve for N:
N = 148 - 5Q (subtract 5Q from both sides)
Now we'll substitute 148 - 5Q for N in the second equation and solve for Q:
10(148 - 5Q) + 25Q = 880 (substituting)
1480 - 50Q + 25Q = 880 (distributing 10 on the left side)
1480 - 25Q = 880 (combining like terms)
-25Q = -600 (subtracting 1480 from both sides)
Q = 24 (dividing both sides by -25)
So Q = 24. We can now substitute back to find N:
N = 148 - 5(24) = 148 - 120 = 28
So our solution is that Joe originally had 28 nickels and 24 quarters.
Let's see if this makes sense. 28 nickels would be worth 140 cents, or $1.40, and 24 quarters would be worth 600 cents, or $6.00, making $1.40 + $6.00 = $7.40, which is what we want. Now we double the number of nickels to 56 and increase the number of quarters by 8 to get 32. 56 nickels are worth 280 cents, or $2.80, and 32 quarters are worth 800 cents, or $8.00, making $2.80 + $8.00 = $10.80, which is also what we want. So we have the correct answer.
Answer by mucklef(9) (Show Source):
You can put this solution on YOUR website! For this problem, we need to break it down into pieces and work on it. We're given the information that the money increase from $7.40 to $10.80. By using subtraction:
$10.80 - $7.40 = $3.40
We find it changes by $3.40. I says the quarter are increased by 8. Since each quarter is worth $.25, we know that $2.00 of that increase is from quarters. That leaves us with $1.40. It also says that nickels are doubled. Since each nickle is worth $.05, we use division:
$1.40 / $.05 = 28 nickels
So there are $1.40 worth, or 28, nickels in the $7.40. Now subtraction:
$7.40 - $1.40 = $6.00
So, there is $6.00 worth of quarters:
$6.00 / $.25 = 24 quarters
Which is 24 quarters, which is your answer.
Hope this helped, all the best.
-mucklef
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