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Chapter 8: Problem 1

Suppose that in a bushel of 100 apples there are 20 that have worms in them and 15 that have bruises. Only those apples with neither worms nor bruises can be sold. If there are 10 bruised apples that have worms in them, how many of the 100 apples can be sold?

Short Answer

Expert verified
75

Step by step solution

01

- Determine Total Defective Apples

Count the apples that are either bruised or have worms. Start by combining the numbers: 20 apples with worms and 15 with bruises.
02

- Account for Overlap

Since 10 apples are both bruised and have worms, subtract these from the total defective count to avoid double-counting: Number of defective apples = 20 (worms) + 15 (bruises) - 10 (both) = 25
03

- Find the Sellable Apples

Subtract the number of defective apples from the total number of apples to get the number of sellable apples: 100 (total) - 25 (defective) = 75

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Key Concepts

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

Defective Apples Calculation
In this problem, we need to calculate the number of defective apples in a bushel of 100 apples.
Defective apples can either have worms, bruises, or both. Here鈥檚 a step-by-step guide on how to do this calculation:
First, we identify two main groups of defective apples:
  • 20 apples with worms
  • 15 apples with bruises
To find the total number of defective apples, we need to account for the overlap 鈥� the apples that have both worms and bruises.
According to the problem, 10 apples fall into this category. This means they have been counted twice, once in the 20 apples with worms and once in the 15 apples with bruises.
Therefore, to avoid double counting, we subtract these 10 overlap apples from our total defectives, leading to:
umber of defective apples = 20 (worms) + 15 (bruises) - 10 (both) = 25.
Now that we have the count of all defective apples, we can proceed to the next step of finding the sellable apples.
Set Intersection
In set theory, the intersection of two sets represents the elements that are common to both sets.
In this problem about apples, we consider:
  • Set A: Apples with worms
  • Set B: Apples with bruises
The intersection of Set A and Set B, denoted as \(A \cap B\), includes apples that have both worms and bruises.
From the problem, we know \(A \cap B\) = 10 apples.
Understanding this intersection is crucial as it provides the overlap that must be subtracted when calculating the total number of defective apples.
This step ensures we don't count those 10 apples twice in our final tally.
Set Subtraction
Set subtraction involves removing elements of one set from another to find the remaining elements.
In our apple problem, after identifying the intersection, we use set subtraction to get the number of sellable apples.
We start with the total set of 100 apples and subtract the set of defective apples, which we calculated to be 25.
Subtraction here looks like this: \(100 - 25 = 75\).
Therefore, 75 apples can be sold, meaning they have neither bruises nor worms.

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Most popular questions from this chapter

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