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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 \(= \frac{2}{3}\) × 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 \(= 2.29 \mathrm{~g}\) × 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 \(= 10.23 \mathrm{~g}\) × 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 \(= 2\) × Number of candy pieces

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Most popular questions from this chapter

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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