There are 65 coins in nickels dimes and quarters. The number of nickels is two more than three times the number of dimes. How many of each is there if the value of the coins is 6.85?

Answer 1

There are 38 nickels, 12 dimes and 15 quarters. The 38 + 12 + 15 = 65 coins, and (as to value) $1.90 + $1.20 + $3.75 = $6.85 Curious as to how to work this? It will take three separate equations, because that's how many unknowns there are. Let's look. We have 3 unknowns, and they are the number of nickels (n), the number of dimes (d) and the number of quarters (q). The number of coins sums to 65, and that "translates" into an equation in "math speak" here. n + d + q = 65 We know the values of the coins. A nickel (n) is 5 cents, a dime (d) is 10 cents, and the quarter (q) is 25 cents. That is their "penny" or "cent" value, and if we multiply the number of nickels (n) by 5, we know how many cents we have in nickels. If we multiply the number dimes (d) we have by 10, we know how many cents we have in dimes. And if we multiply the number of quarters (q) by 25, we know how many cents we have in quarters. The total number of cents we have in all the coins was given as 685 cents, which is $6.85 coverted into cents so we can write an equation. Remember we need to keep everything in the same units, and in this case it's cents. Let's write that equation. 5n + 10d + 25q = 685 We also know that the number of nickels (n) is three times the number of dimes (d) plus two more. If you guessed that we could write an equation for this, you'd be correct, so let's do that. n = 3d + 2 That last equation is going to allow us to make substitutions into the other two equations. What we've done with that last equation is expressed one variable in terms of another. We'll just substitute that back into the other two equations. n + d + q = 65 (3d +2) + d + q = 65 [See how we substituted for the n here? Be sure to get that.] 3d + 2 + d + q = 65 4d + 2 + q = 65 See how we did this? Now we'll do it again to the second equation. 5n + 10d + 25q = 685 5(3d + 2) + 10d + 25q = 685 [See how we substituted here? Be sure to get that.] (15d + 10) + 10d + 25q = 685 15d + 10 + 10d + 25d = 685 25d + 10 + 25q = 685 We now have two equations with two unknowns (variables) in each one. We can solve by the method of simultaneous equations. To do that, we write one equation "over" the other one and subtract one from the other. We must apply one "math trick" to make this work, though. We may have to "manipulate" one or both to get the same multipliers on one of the variables so that when we subtract one equation from the other, one of the variables "disappears" or "cancels out" when we do. Let's do that. 4d + 2 + q = 65 25d + 10 + 25q = 685 We have one equation over the other, and we can subtract, but we need to "massage" one or both to get one of the multiples of the variables equal in each of the equations. About the most direct and "simple" route is to take the first equation and multiply through by 25 to make the q into 25q there. Let's do that. 25 (4d + 2 + q = 65) [Let's distribute that 25 among all the factors in the equation.] 100d + 50 + 25q = 1625 [Let's put the other equation back below this and subtract.] 25d + 10 + 25q = 685 75d + 40 + 0q = 940 75d + 40 = 940 75d = 900 d = 12 [Wow! We finally got something! The number of dimes. Let's plug this back in somewhere and get more!] 4d + 2 + q = 65 4 (12) + 2 + q = 65 48 + 2 + q = 65 50 + q = 65 q = 65 -50 q = 15 [This is the number of quarters, as you can see. We could have used another equation, but this one was handy. Let's get that last variable using the original equation with the total coin count.] n + d + q = 65 n + 12 + 15 = 65 n + 27 = 65 n = 65 - 27 = 38 [That's 38 nickels.] Take these numbers and plug them back into the equation with the dollar amount (the "cent" amount as we wrote it so our units would be consitent) and check the work. 5n + 10d + 25q = 685 5(38) + 10(12) + 25(15) = 190 + 120 + 375 = 685 [We check out! Whew!]