Set X consists of eight consecutive integers. Set Y consists of all the integers that result from adding 4 to each of the integers in set X and all the integers that result from subtracting 4 from each of the integers in set X. How many more integers are there in set Y than in set X ?a)0b)4c)8d)12e)16Correct answer is option 'C'. Can you explain this answer?

Question

Answer

To solve this problem, let's start by considering set X, which consists of eight consecutive integers. Let's assume the first integer in set X is "n."

So, set X can be written as follows:
X = {n, n+1, n+2, n+3, n+4, n+5, n+6, n+7}

Now, we need to find set Y, which consists of all the integers obtained by adding 4 to each integer in set X and subtracting 4 from each integer in set X.

When we add 4 to each integer in set X, we get the following set:
X_add4 = {n+4, n+5, n+6, n+7, n+8, n+9, n+10, n+11}

Similarly, when we subtract 4 from each integer in set X, we get the following set:
X_subtract4 = {n-4, n-3, n-2, n-1, n, n+1, n+2, n+3}

Now, set Y consists of all the integers in X_add4 and X_subtract4. To find the total number of integers in set Y, we can combine these two sets and remove any duplicates.

Combining X_add4 and X_subtract4, we get:
Y = {n-4, n-3, n-2, n-1, n, n+1, n+2, n+3, n+4, n+5, n+6, n+7, n+8, n+9, n+10, n+11}

To count the number of integers in set Y, we subtract the smallest integer in set Y from the largest integer and add 1. In this case, the smallest integer is (n-4), and the largest integer is (n+11).

Therefore, the number of integers in set Y is (n+11) - (n-4) + 1 = n + 11 - n + 4 + 1 = 16.

Hence, there are 16 more integers in set Y than in set X.

Therefore, the correct answer is C.

Answer

Problem Analysis:
We are given a set X consisting of eight consecutive integers. Let's assume the first integer in set X is 'a'. Then the set X can be represented as {a, a+1, a+2, a+3, a+4, a+5, a+6, a+7}.
We are also given a set Y which consists of all the integers that result from adding 4 to each of the integers in set X and subtracting 4 from each of the integers in set X. So, the set Y can be represented as {a+4, a+5, a+6, a+7, a+8, a+9, a+10, a+11, a-4, a-3, a-2, a-1}.

Counting the Integers:
To determine how many more integers there are in set Y than in set X, we need to count the number of integers in each set.

Counting the Integers in Set X:
We can see that set X consists of eight consecutive integers. So, the number of integers in set X is 8.

Counting the Integers in Set Y:
Set Y consists of all the integers that result from adding 4 to each of the integers in set X and subtracting 4 from each of the integers in set X.
When we add 4 to each of the integers in set X, we get four additional integers: a+4, a+5, a+6, a+7.
When we subtract 4 from each of the integers in set X, we also get four additional integers: a-4, a-3, a-2, a-1.
So, the total number of integers in set Y is 8 + 4 + 4 = 16.

Calculating the Difference:
To find the difference between the number of integers in set Y and set X, we subtract the number of integers in set X from the number of integers in set Y:
16 - 8 = 8.

Conclusion:
There are 8 more integers in set Y than in set X. Therefore, the correct answer is option C) 8.

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