MHB Math Problem: Find # of Elements in 2nd/3rd Subsets

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The math problem involves a set of 200 elements partitioned into three subsets, with the second and third subsets containing the same number of elements. The relationship between the subsets is defined by the equations A + 2B = 200 and 4B = 3A, leading to the conclusion that B equals 60. Since B represents the number of elements in either the second or third subset, the total for both is 120. The confusion arises from interpreting the question, which asks for the total in the union of the second and third subsets, not the count of individual subsets.
elimeli
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I am practicing for my math exam next week and I came across this problem:

A set has 200 elements in it. It is partitioned into three subsets so that the second and third subsets have the same number of elements. If four times the number of elements in the second subset is three times as many as in the first, how many elements are in either the second or third subset?

The answer is 120, but I don't understand how to get to that answer.

I equaled 4 nx(2) = 3 nx(1), and so I got that 4nx(2)/3 = nx(1). I then plugged that into 200 = nx(2) + nx(2) + 4nx(2)/3 and got that nx(2)=60. The answer is 120, so I would have to multiply my answer by 2 to get it and I do not understand why, since the answer is asking for either the second or third subset and not the addition of the two.

Can anybody explain?
 
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I would let $A$ be the cardinality of the first subset, and $B$ be the cardinality of the second and third subsets, so that we have:

$$A+2B=200$$

$$4B=3A\implies A=\frac{4}{3}B$$

Substituting for $A$ into the first equation we get:

$$\frac{4}{3}B+2B=200$$

Multiply through by $$\frac{3}{2}$$:

$$2B+3B=300$$

$$5B=300$$

$$B=60$$

Thus, the cardinality of the union of the second and third subsets is:

$$2B=120$$
 
MarkFL said:
I would let $A$ be the cardinality of the first subset, and $B$ be the cardinality of the second and third subsets, so that we have:

$$A+2B=200$$

$$4B=3A\implies A=\frac{4}{3}B$$

Substituting for $A$ into the first equation we get:

$$\frac{4}{3}B+2B=200$$

Multiply through by $$\frac{3}{2}$$:

$$2B+3B=300$$

$$5B=300$$

$$B=60$$

Thus, the cardinality of the union of the second and third subsets is:

$$2B=120$$

But why is the answer the union of the second and third subsets if the question is asking for either one of them? Shouldn't it be 60? Or am I misinterpreting the question?
 
The question asks how many elements are in either the second or third subsets, so they are asking for the total number of elements in the union of the two subsets. If an element is in either the second or third subset, then it is to be counted. :)
 
MarkFL said:
The question asks how many elements are in either the second or third subsets, so they are asking for the total number of elements in the union of the two subsets. If an element is in either the second or third subset, then it is to be counted. :)

I see. Thank you so much for clarifying!
 
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