UPSC CSE Prelims
Number System Previous Year Questions (PYQs)
Showing solved Previous Year Questions for Chapter: Number System
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$A$ is a 2-digit number with different digits. $B$ is also a 2-digit number and is obtained by reversing the digits of $A$. If $A - B$ is a multiple of $27$, where $A > B$, how many such different $A$'s are possible?
Detailed Explanation:
Let A = 10a + b and B = 10b + a, where a and b are digits with a ≠ b and b ≠ 0 (so B remains 2-digit).
A − B = 9(a − b). For this to be a multiple of 27, we need 9(a − b) divisible by 27, so (a − b) must be a multiple of 3. Since A > B, we have a > b, giving possible differences: 3, 6, or 9.
For a − b = 3: pairs (4,1), (5,2), (6,3), (7,4), (8,5), (9,6) = 6 numbers. For a − b = 6: pairs (7,1), (8,2), (9,3) = 3 numbers. For a − b = 9: pair (9,0) is invalid since B would not be 2-digit. Total = 6 + 3 = 9.
Therefore, 9 different values of A are possible, making Option 2 the correct answer.
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.
The class average $x$ in a test increases by $4$ when the score of a student is rectified, whose corrected score is $100$ instead of $0$. Later, the score of another student was found to have been recorded as $81$ in place of $56$. If there are no other corrections and the final corrected average is $y$, then $y - x$ is
Detailed Explanation:
Let the number of students be n. When one student's score is corrected from 0 to 100, the total increases by 100 and the average increases by 4. Thus: 100/n = 4, giving n = 25 students.
The second correction changes a score from 81 to 56, decreasing the total by 25. This decreases the average by 25/25 = 1.
After the first correction, average = x + 4. After the second correction, y = x + 4 − 1 = x + 3. Hence y − x = 3, making Option 2 the correct answer.
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There are four types of weights, namely $1$ kg, $2$ kg, $5$ kg and $10$ kg. What is the maximum number of different ways one can measure $20$ kg, if at least eight but not more than eleven weights of $1$ kg are to be used while measuring?
Detailed Explanation:
We need a + 2b + 5c + 10d = 20 where a ∈ {8, 9, 10, 11} (1 kg weights) and b, c, d ≥ 0 (2 kg, 5 kg, 10 kg weights).
Case a = 8: 2b + 5c + 10d = 12 → (b,c,d) = (6,0,0), (1,2,0), (1,0,1) → 3 ways
Case a = 9: 2b + 5c + 10d = 11 (odd, so c odd) → (b,c,d) = (3,1,0) → 1 way
Case a = 10: 2b + 5c + 10d = 10 → (b,c,d) = (5,0,0), (0,2,0), (0,0,1) → 3 ways
Case a = 11: 2b + 5c + 10d = 9 (odd, so c odd) → (b,c,d) = (2,1,0) → 1 way
Therefore, total = 3 + 1 + 3 + 1 = 8, making Option 2 the correct answer.
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