The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is || ADRE 1.0 SLRC 2022 PAPER-III SOLVED QUESTIONS

The sum of the four consecutive alternate numbers is 64. The smallest number amongst them is:

(A) 13
(B) 11
(C) 16
(D) 9

Solution:

To find the smallest number among four consecutive alternate numbers whose sum is 64, let's denote the smallest number as x.

Consecutive alternate numbers increase by 2, so the numbers can be written as:

• First number: x
• Second number: x + 2
• Third number: x + 4
• Fourth number: x + 6

According to the problem, their sum is 64:

x + (x + 2) + (x + 4) + (x + 6) = 64

Combine the terms:

x + x + 2 + x + 4 + x + 6 = 64
4x + 12 = 64

Subtract 12 from both sides:

4x = 52

Divide by 4:

x = 13

So, the smallest number is x = 13.

Thus, the answer is: 13

Importance of this Question:

• This question was previously asked in the ADRE SLRC-2022 Paper III for the posts of Grade-III.
• Understanding and solving such questions enhances problem-solving skills.
• Practicing these types of questions is essential for effective preparation for competitive exams.
• It helps in familiarizing with the type of questions that can be asked in the exams.