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The difference between the squares of two consecutive odd natural numbers is 88. What is the sum of these two numbers?
42
44
46
48
44
Let the consecutive odd numbers be (2n-1) and (2n+1). Their squares are (2n+1)2 - (2n-1)2 = 88. Simplifying this yields 8n = 88, so n = 11. The numbers are 2(11)-1 = 21 and 2(11)+1 = 23, and their sum is 21 + 23 = 44.
Let the consecutive odd numbers be (2n-1) and (2n+1). Their squares are (2n+1)2 - (2n-1)2 = 88. Simplifying this yields 8n = 88, so n = 11. The numbers are 2(11)-1 = 21 and 2(11)+1 = 23, and their sum is 21 + 23 = 44.
In number theory, the difference between the squares of two consecutive odd numbers (2n+1)2 - (2n-1)2 is always 8n, which is a multiple of 8, a principle often applied in quantitative aptitude sections of competitive exams.
Consecutive odd integers are represented as 2n-1 and 2n+1.
The square of an odd number is always of the form 8k+1.
The sum of two consecutive odd numbers is always an even number.
The difference between squares of any two consecutive odd numbers is always equal to 4 times the middle even number between them.
For the numbers 21 and 23, the middle even number is 22; 4 multiplied by 22 equals 88.
Option B (44) is correct because 21 + 23 = 44.
Options A, C, and D are incorrect as they do not satisfy the algebraic requirement of the difference of squares equaling 88.
B is correct тАФ The sum of the consecutive odd numbers 21 and 23 is 44, satisfying the difference of squares condition.
Recognizing patterns like 'difference of squares of consecutive odd numbers equals 8n' saves significant time in RRB, SSC, and Banking exams compared to solving quadratic equations manually.
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