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What is the smallest of six consecutive odd integers whose average (ar GMAT Problem-Solving

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Sayantani Barman

Experta en el extranjero | Updated On - Mar 22, 2023

Question: What is the smallest of six consecutive odd integers whose average (arithmetic mean) is x + 2?

A. x - 5
B. x - 3
C. x - 1
D. x
E. x + 1

Answer: B

Solution and Explanation:

Approach Solution 1:
Apply the information in the question to the GMAT question at hand. These problems apply to numerous disciplines of mathematics. This question has to do with algebra. It is challenging to select the best option because of the way the options are presented. Candidates must be able to comprehend the appropriate approach to eliciting the desired response. Out of the five possible answers, there is only one that is correct.
If we set the first (or smallest) odd number equal to n, we get the following:
As the six successive odd numbers, n, n + 2, n + 4, n + 6, n + 8, and n + 10 are used.
Since (first number + last number)/2 is the average number of words in a collection with equal spacing:
(n + n + 10)/2 = x + 2
2n + 10 = 2x + 4
2n = 2x - 6
n = x - 3
Correct option:
B

Approach Solution 2:
Apply the information in the question to the GMAT question at hand. These problems apply to numerous disciplines of mathematics. This question has to do with fundamental math. It is challenging to select the best option because of the way the options are presented. Candidates must be able to comprehend the appropriate approach to eliciting the desired response. Out of the five possible answers, there is only one that is correct.
Using the smallest odd number as k
Thus, k + 2 equals the following odd integer (since each consecutive odd integer is always 2 greater than the odd integer before it)
And k + 4 is the following odd number.
And k + 6 is the following odd number.
And k + 8 is the following odd number.
And k + 10 = the final odd integer (number six).
As x+2 + 2 is the average of the six integers, we may write: (k+(k+2)+(k+4)+(k+6)+(k+8)+(k+10)) / 6 = x+2
Simplify: (6k+30)/ 6=x+2
multiply 6 to both sides of the equation to obtain: 6k+30=6x+12
What is the lowest odd integer among the following six?
We must find the solution to this equation for k because it is the lowest odd integer.
Consider: 6k+30=6x+12
30 is deducted from both sides: 6k=6x−18
K=x-3 is obtained by multiplying both sides by 6.
Correct option:
B

Approach Solution 3:
Apply the information in the question to the GMAT question at hand. These problems apply to numerous disciplines of mathematics. This question has to do with fundamental math. It is challenging to select the best option because of the way the options are presented. Candidates must be able to comprehend the appropriate approach to eliciting the desired response. Out of the five possible answers, there is only one that is correct.
Let's call the smallest odd integer "a." Since there are six consecutive odd integers, we know that the next five odd integers are a+2, a+4, a+6, a+8, and a+10.
To find the average of these six integers, we can add them all up and divide by 6:
(a + (a+2) + (a+4) + (a+6) + (a+8) + (a+10))/6 = x + 2
Simplifying this equation:
(6a + 30)/6 = x + 2
a + 5 = x + 2
a = x - 3
So the smallest odd integer is x - 3, which is answer choice B.
Correct option:
B

“What is the smallest of six consecutive odd integers whose average (ar" - is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been borrowed from the book “GMAT Official Guide Quantitative Review”.

To understand GMAT Problem Solving questions, applicants must possess fundamental qualitative skills. Quant tests a candidate's aptitude in reasoning and mathematics. The GMAT Quantitative test's problem-solving phase consists of a question and a list of possible responses. By using mathematics to answer the question, the candidate must select the appropriate response. The problem-solving section of the GMAT Quant topic is made up of very complicated math problems that must be solved by using the right math facts.

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*The article might have information for the previous academic years, please refer the official website of the exam.

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