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The sum of the squares of two consecutive odd natural numbers is 202. The sum of the numbers is:
18
20
22
24
20
Let the two consecutive odd natural numbers be x and x+2. Solving the equation x┬▓ + (x+2)2 = 202 results in x┬▓ + x┬▓ + 4x + 4 = 202, which simplifies to 2x┬▓ + 4x - 198 = 0 or x┬▓ + 2x - 99 = 0. Factoring gives (x+11)(x-9) = 0, identifying the positive odd numbers as 9 and 11, whose sum equals 20.
Let the two consecutive odd natural numbers be x and x+2. Solving the equation x┬▓ + (x+2)2 = 202 results in x┬▓ + x┬▓ + 4x + 4 = 202, which simplifies to 2x┬▓ + 4x - 198 = 0 or x┬▓ + 2x - 99 = 0. Factoring gives (x+11)(x-9) = 0, identifying the positive odd numbers as 9 and 11, whose sum equals 20.
In the context of Indian competitive exams, the properties of natural numbers (N) are studied under Number Theory, a core component of the Quantitative Aptitude syllabus for RRB, SSC, and UPSC CSAT.
Consecutive odd natural numbers are always separated by a difference of 2.
The sum of squares of two consecutive odd numbers 9 and 11 is 81 + 121 = 202.
Natural numbers are defined as the set of positive integers {1, 2, 3,...} and do not include zero or negative integers.
Algebraic identities like (a+b)2 = a┬▓ + 2ab + b┬▓ are essential for solving quadratic equations in competitive aptitude tests.
When dealing with consecutive odd or even numbers, representing them as 2n-1 and 2n+1 can also simplify the calculation.
Option A (18), C (22), and D (24) are incorrect because no two consecutive odd numbers satisfy the sum-of-squares condition of 202 for these respective sums.
B is correct тАФ The two consecutive odd natural numbers are 9 and 11, which sum to 20.
Mastering quadratic factorization is crucial not only for Algebra but also for solving time-speed-distance and profit-loss problems involving variable dependencies in RRB examinations.
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