UPSC CSE Prelims
Factors & Multiples Previous Year Questions (PYQs)
Practice solved questions for Factors & Multiples with detailed step-by-step solutions, key insights, and trend analysis for UPSC CSE PRELIMS.
Solved Previous Year Questions
Filter & practice questions topic-wise
If $x$, $y$ and $z$ are integers, each greater than $1$, then is $x$ a prime number? Statement I : $xy^2 = 116$ Statement II : $xz = 261$
Detailed Explanation:
Statement I: $xy^2 = 116 = 2^2 \times 29$. Since $y > 1$ is an integer, $y^2$ must be a perfect square divisor of $116$. The only such divisor is $4$, so $y^2 = 4$ and $y = 2$. Thus $x = 116/4 = 29$, which is prime. Statement I alone is sufficient.
Statement II: $xz = 261 = 3^2 \times 29$. Possible values for $x$ are $3, 9, 29, 87$. Since $x = 3$ or $29$ gives prime, while $x = 9$ or $87$ gives composite, we cannot determine definitively. Statement II alone is insufficient.
Therefore, the question can be answered using Statement I alone but not Statement II alone, making Option 1 the correct answer.
In an objective type question paper, $5$ marks are awarded for a correct answer and $2$ marks are deducted for a wrong answer. A student attempted all the questions and got a score of $69$. Had he been awarded $4$ marks for a correct answer and $1$ mark deducted for a wrong answer, he would have scored $84$. How many questions were there in the question paper?
Detailed Explanation:
Let correct answers = x, wrong answers = y. First condition: 5x - 2y = 69. Second condition: 4x - y = 84.
From second equation, y = 4x - 84. Substituting into first: 5x - 2(4x - 84) = 69 → 5x - 8x + 168 = 69 → -3x = -99 → x = 33. Thus y = 132 - 84 = 48.
Total questions = 33 + 48 = 81, making Option 2 the correct answer.
In a sequence of numbers, each number other than the first two is the sum of the two immediately preceding numbers from it. If the first two numbers in the sequence are 4 and 7, then the sixth number is
Detailed Explanation:
This is a Fibonacci-like sequence where each term equals the sum of the two preceding terms.
Starting with a₁ = 4 and a₂ = 7: a₃ = 4 + 7 = 11; a₄ = 7 + 11 = 18; a₅ = 11 + 18 = 29; a₆ = 18 + 29 = 47.
Therefore, the sixth number is 47, making Option 4 the correct answer.
🧐 Not Sure What to Study Next?
Get a personalised study plan based on your goals, time and revision needs.
For $\frac{1}{3} < x < y < 2$, which of the following statements is/are always correct? I. $x + \frac{1}{x} < y + \frac{1}{y}$ II. $\frac{\sqrt{1 + y^2}}{y} < \frac{\sqrt{1 + x^2}}{x}$ Select the answer using the code given below.
Detailed Explanation:
Statement I: Define $f(t) = t + \frac{1}{t}$. Its derivative $f'(t) = 1 - \frac{1}{t^2}$ is negative on $(0,1)$ and positive on $(1,\infty)$. Thus $f(t)$ decreases on $(\frac{1}{3}, 1)$ and increases on $(1, 2)$—not monotonic throughout.
Counterexample: Take $x = \frac{1}{2}, y = \frac{3}{4}$. Then $x + \frac{1}{x} = 2.5$ but $y + \frac{1}{y} \approx 2.08$. So $x + \frac{1}{x} > y + \frac{1}{y}$, contradicting Statement I.
Statement II: Rewrite $\frac{\sqrt{1+t^2}}{t} = \sqrt{\frac{1}{t^2} + 1}$. For $t > 0$, as $t$ increases, $\frac{1}{t^2}$ decreases, making $\sqrt{\frac{1}{t^2}+1}$ strictly decreasing. Since $x < y$, we have $\frac{\sqrt{1+y^2}}{y} < \frac{\sqrt{1+x^2}}{x}$, confirming Statement II is always true.
Therefore, only Statement II is correct, making Option 2 the correct answer.
Related Topics in CSAT (Civil Services Aptitude Test)
Frequently Asked Questions
Common questions about Factors & Multiples in UPSC CSE PRELIMS