How many pennies from each time period are there in the bag?

Question

In 1982, the US Mint changed the composition of pennies from all copper to zinc with copper coating. Pennies made prior to 1982 weigh 3.1 grams. Pennies made since 1982 weight 2.5 grams. If you have a bag of 1287

pennies, and the bag weighs 3601.5 grams, how many pennies from each time period are there in the bag?

Answer 1

See this as a quantity-amount problem. Amount can be money or, in this case, weight. Label the variables with letters that will help you distinguish them: C for all-copper and Z for mostly-zinc. Set up equations for the quantity and weight:

C + Z = 1287

3.1 C + 2.5 Z = 3601.5

Use the elimination method to get rid of the Zs: multiply all three items of the first equation by -2.5 and place under the second equation, draw a line, and subtract.

3.1 C + 2.5 Z = 3601.5

-2.5 C - 2.5 Z = -3217.5

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

.6 C = 384 Divide both sides by .6.

C = 640 Subtract 640 from 1287.

Z = 647

Answer 2

Given by the problem statement, there are two conditions we must satisfy: Weight and Quantity. We can create equations with the information given:

Lets make A represent pennies after 1982 and make B represent pennies before 1982.

Now lets make some equations (remember that units are important!, don't forget them if they are part of the problem!)

The total number of coins:

A + B = 1287 [pennies] ------------(Equation 1)

The total weight of the bag:

2.5 [grams/penny] * A + 3.1 [grams/penny] * B = 3601.5 [grams] ------------(Equation 2)

We have 2 equations and 2 unknowns. This is a solvable problem! (I omitted the units in my work, but it's good practice to either write them in as you work to keep track of them or at least remember to include it in your final answer. Keeping track of units will help you on complex science-type problems)

Lets work for pennies made before 1982, or what we call 'B'. To do that we need to find out what A is in terms of B.

A = 1287 - B

Plug in A into the Equation 2:

2.5 (1287-B) + 3.1B = 3601.5

Now Solve for B:

3217.5 - 2.5B +3.1B = 3601.5

0.6B = 384

:: B = 640

Now that B is known, use Equation 1 to find A:

A + 640 = 1287

:: A = 647

So there are 647 Pennies made after 1982 and 640 Pennies made before 1982 in the bag. Let's check our work using equation 2.

2.5 [grams/penny]*647 [pennies] + 3.1 [grams/penny]*640 [pennies] = 3601.5 [grams]

(True statement. Our solution is correct)

Answer 3

Important Information:

Mass prior to 1982 = 3.1g

Mass since 1982 = 2.5g

Total pennies = 1287

Total weight = 3601.5g

Variables/Coefficients:

Pennies prior to 1982 = a

Pennies since 1982 = b

Weight prior to 1982 = 3.1a

Weight since 1982 = 2.5b

Summarizing Equations:

3601.5 = 3.1a + 2.5b

1287 = a + b

Process:

We need to eliminate one of the variables.

Take 1287 = a + b and multiply both sides of the equation by 3.1

3989.7 = 3.1a + 3.1b

Subtract 3601.5 = 3.1a + 2.5b to cancel out a

388.2 = 0.6b

Divide both sides by 0.6

647 = b

Remember 1287 = a + b so substitute 647 for b

1287 = a + 647

Subtract 647 from both sides

640 = a

Conclusion:

The bag contains 640 pennies minted prior to 1982 and 647 minted since then.