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Chapter 1: Problem 2

You go to a convenience store to buy candy and find the owner to be rather odd. He allows you to buy pieces in multiples of four, and to buy four, you need $$\$ 0.23$$. He only allows you to do this by using 3 pennies and 2 dimes. You have a bunch of pennies and dimes, and instead of counting them, you decide to weigh them. You have \(636.3 \mathrm{~g}\) of pennies, and each penny weighs \(3.03 \mathrm{~g}\). Each dime weighs \(2.29 \mathrm{~g}\). Each piece of candy weighs \(10.23 \mathrm{~g}\). a. How many pennies do you have? b. How many dimes do you need to buy as much candy as possible? c. How much should all these dimes weigh? d. How many pieces of candy could you buy? (number of dimes from part b) e. How much would this candy weigh? f. How many pieces of candy could you buy with twice as many dimes?

Short Answer

Expert verified
a. Number of pennies \(= \frac{636.3}{3.03} = 210\) b. Number of dimes needed \(= \frac{2}{3} \times 210 = 140\) c. Total weight of dimes \(= 2.29 \mathrm{~g} \times 140 = 320.6 \mathrm{~g}\) d. Number of candy pieces \(= \frac{140}{2} = 70\) e. Total weight of candy \(= 10.23 \mathrm{~g} \times 70 = 716.1 \mathrm{~g}\) f. Number of candy pieces with double dimes \(= 2 \times 70 = 140\)

Step by step solution

01

Find the number of pennies.

We are given that the total weight of pennies is \(636.3 \mathrm{~g}\) and each penny weighs \(3.03 \mathrm{~g}\). To find the number of pennies, we will divide the total weight by the weight of one penny: Number of pennies \(=\frac{636.3}{3.03}\) ##Step 2: Calculate the number of dimes needed##
02

Find the number of dimes needed.

To purchase a piece of candy, we need 2 dimes. In other words, for every 3 pennies, we require 2 dimes. We can now use the result of step 1 to find the number of dimes needed. Number of dimes needed \(= \frac{2}{3}\) 脳 Number of pennies ##Step 3: Calculate the total weight of dimes##
03

Find the total weight of the dimes.

We know that each dime weighs \(2.29 \mathrm{~g}\). Now we will multiply the weight of one dime by the number of dimes needed (from step 2) to find the total weight of dimes. Total weight of dimes \(= 2.29 \mathrm{~g}\) 脳 Number of dimes needed ##Step 4: Calculate how many pieces of candy we could buy##
04

Find the number of candy pieces.

Divide the number of dimes needed (from step 2) by 2, since we need 2 dimes to buy one piece of candy. Number of candy pieces \(= \frac{\text{Number of dimes needed}}{2}\) ##Step 5: Calculate the total weight of the candy##
05

Find the total weight of the candy.

Each candy piece weighs \(10.23 \mathrm{~g}\). Multiply the weight of one candy piece by the number of candy pieces (from step 4) to find the total weight of the candy. Total weight of candy \(= 10.23 \mathrm{~g}\) 脳 Number of candy pieces ##Step 6: Calculate how many pieces of candy we could buy with twice as many dimes##
06

Find the number of candy pieces with double dimes.

We know that for every 2 dimes, we can buy one piece of candy. If we double the number of dimes needed (from step 2), we can buy twice as many candy pieces as we could buy initially (step 4 result). Number of candy pieces with double dimes \(= 2\) 脳 Number of candy pieces

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass and Weight
Mass and weight are often terms used interchangeably, but they describe different concepts. Mass is the measure of the amount of matter in an object, typically measured in grams or kilograms. Weight, on the other hand, is the force exerted by gravity on an object, which depends on both the mass and the gravitational field strength. On Earth, this is approximately 9.8 m/s虏.
In this context, mass plays a crucial role as we calculate how many pennies, dimes, or pieces of candy we have based on their mass. For example, knowing that each penny has a mass of 3.03 grams allows us to determine the total number of pennies by dividing the total mass by this single unit mass.
Unit Conversion
Unit conversion is an essential skill in problem-solving that allows students and scientists to translate information from one set of units to another. It requires understanding the equivalence between units, such as converting from pounds to grams or from inches to centimeters.
In our exercise, we need to relate weights to count, such as knowing how to convert the mass of many pennies to the number of individual pennies, which requires an understanding that mass is derived from multiplying number by the mass of one unit.
  • Convert total weight to item count by dividing total mass by mass per unit item.
  • Understanding these conversions will simplify complex calculations and prevent common mistakes.
Arithmetic Operations
Arithmetic operations include addition, subtraction, multiplication, and division鈥攕kills fundamental to mathematics. Using these operations, we can solve various parts of a problem, like computing the number of items or their total mass.
For instance, dividing total mass by the mass of one penny calculates how many pennies are present. This requires both division and the understanding of multiplication鈥檚 role in determining total mass: Mass = Number of Items 脳 Mass per Item.
Multiplication aids in determining the total weight of dimes where we multiply the number of dimes by the weight of one dime. These fundamental operations form the basis of problem-solving in algebra and everyday quantitative reasoning tasks.
Problem Solving Steps
Problem-solving involves a structured approach to finding solutions. Breaking down a problem into manageable steps helps clarify the process and identify necessary calculations or concepts required to achieve a solution.
  • Identify what is given and what needs to be found. For example, the weight of coins was given, and we needed to determine their count.
  • Break the problem into smaller, logical steps, as shown in the solution, where determining the number of pennies was the first step.
  • Use logical reasoning: For every 2 dimes, you get a piece of candy, which means division is used to determine how many candies can be purchased.
  • Finally, verify each step鈥檚 result to ensure the solution's correctness and make relationships and patterns clear.
This logical approach ensures not only a solution to the problem but also an understanding of the underlying concepts and processes.

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Most popular questions from this chapter

A sample containing \(33.42 \mathrm{~g}\) of metal pellets is poured into a graduated cylinder initially containing \(12.7 \mathrm{~mL}\) of water, causing the water level in the cylinder to rise to \(21.6 \mathrm{~mL}\). Calculate the density of the metal.

Perform the following mathematical operations, and express the result to the correct number of significant figures. a. \(\frac{2.526}{3.1}+\frac{0.470}{0.623}+\frac{80.705}{0.4326}\) b. \((6.404 \times 2.91) /(18.7-17.1)\) c. \(6.071 \times 10^{-5}-8.2 \times 10^{-6}-0.521 \times 10^{-4}\) d. \(\left(3.8 \times 10^{-12}+4.0 \times 10^{-13}\right) /\left(4 \times 10^{12}+6.3 \times 10^{13}\right)\) e. \(\frac{9.5+4.1+2.8+3.175}{4}\) (Assume that this operation is taking the average of four numbers. Thus 4 in the denominator is exact.) f. \(\frac{8.925-8.905}{8.925} \times 100\) (This type of calculation is done many times in calculating a percentage error. Assume that this example is such a calculation; thus 100 can be considered to be an exact number.)

The U.S. trade deficit at the beginning of 2005 was $$\$ 475,000,000$$. If the wealthiest \(1.00 \%\) of the U.S. population \((297,000,000)\) contributed an equal amount of money to bring the trade deficit to $$\$ 0$$, how many dollars would each person contribute? If one of these people were to pay his or her share in nickels only, how many nickels are needed? Another person living abroad at the time decides to pay in pounds sterling ( \(\mathrm{f}\) ). How many pounds sterling does this person contribute (assume a conversion rate of \(1 \mathrm{f}=\$ 1.869)\) ?

Precious metals and gems are measured in troy weights in the English system: $$ \begin{aligned} 24 \text { grains } &=1 \text { pennyweight (exact) } \\ 20 \text { pennyweight } &=1 \text { troy ounce (exact) } \\ 12 \text { troy ounces } &=1 \text { troy pound (exact) } \\ 1 \text { grain } &=0.0648 \mathrm{~g} \\ 1 \text { carat } &=0.200 \mathrm{~g} \end{aligned} $$ a. The most common English unit of mass is the pound avoirdupois. What is 1 troy pound in kilograms and in pounds? b. What is the mass of a troy ounce of gold in grams and in carats? c. The density of gold is \(19.3 \mathrm{~g} / \mathrm{cm}^{3}\). What is the volume of a troy pound of gold?

Many times errors are expressed in terms of percentage. The percent error is the absolute value of the difference of the true value and the experimental value, divided by the true value, and multiplied by \(100 .\) Percent error \(=\frac{\mid \text { true value }-\text { experimental value } \mid}{\text { true value }} \times 100\) Calculate the percent error for the following measurements. a. The density of an aluminum block determined in an experiment was \(2.64 \mathrm{~g} / \mathrm{cm}^{3}\). (True value \(2.70 \mathrm{~g} / \mathrm{cm}^{3}\).) b. The experimental determination of iron in iron ore was \(16.48 \% .\) (True value \(16.12 \% .)\) c. A balance measured the mass of a \(1.000-\mathrm{g}\) standard as \(0.9981 \mathrm{~g}\)
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