UPSC CSE Prelims
HCF and LCM Previous Year Questions (PYQs)
Practice solved questions for HCF and LCM with detailed step-by-step solutions, key insights, and trend analysis for UPSC CSE PRELIMS.
Solved Previous Year Questions
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$X$, $Y$ and $Z$ jump forward $4'$, $6'$ and $5'$, respectively. At 8 AM, they all land on mark $199'$. How many times will they all land on the same mark (need not be at the same moment) between mark $195'$ and $1000'$, if all of them cross mark $1000'$ by 9 AM?
Detailed Explanation:
At 8 AM, all three are at mark 199'. Since X jumps 4', Y jumps 6', and Z jumps 5', they will all land on the same mark again when the distance from 199' is a multiple of LCM(4, 6, 5) = 60'.
Common landing marks are 199 + 60k where k = 0, 1, 2, ... Within the range 195' < m < 1000', we get: 199, 259, 319, 379, 439, 499, 559, 619, 679, 739, 799, 859, 919, 979. The next mark (1039') exceeds 1000'.
Therefore, counting from k=0 to k=13 gives 14 common marks, making Option 4 the correct answer.
If $10^m \times 1000 \times n = 75^{25} \times 25^{32} \times 32^{75}$, where $n$ is not divisible by $10$, then the value of $m$ is
Detailed Explanation:
Prime factorize the RHS: $75^{25} = (3 \times 5^2)^{25} = 3^{25} \times 5^{50}$, $25^{32} = 5^{64}$, and $32^{75} = 2^{375}$. Combined: $2^{375} \times 3^{25} \times 5^{114}$.
Maximum power of 10: Since $10 = 2 \times 5$, the limiting factor is $\min(375, 114) = 114$. Thus, RHS = $10^{114} \times 2^{261} \times 3^{25}$, where $(2^{261} \times 3^{25})$ is not divisible by $10$ (no factor of $5$).
LHS comparison: $10^m \times 1000 \times n = 10^{m+3} \times n$. Since $n$ is not divisible by $10$, we equate: $m + 3 = 114 \Rightarrow m = 111$.
Therefore, m = 111, making Option 2 the correct answer.
If the product of the HCF and LCM of two distinct numbers is the cube of one of the numbers, then which of the following statements is/are correct?
I. The difference of the numbers is an even number. II. One of the numbers is a perfect square.
Select the answer using the code given below.
Detailed Explanation:
Let two distinct numbers be a and b. Using HCF × LCM = a × b, and given this product equals the cube of one number (say a³): a × b = a³, so b = a².
Statement I: Difference = a² − a = a(a − 1), the product of two consecutive integers, which is always even. ✓
Statement II: One number is a², which is a perfect square. ✓
Therefore, both statements I and II are correct, making Option 3 the correct answer.
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The ratio of male to female workers in two companies $A$ and $B$ is $13 : 10$ and $7 : 5$, respectively. If both the companies have the same number of female workers, then what is the ratio of the total number of workers in $A$ to those in $B$?
Detailed Explanation:
Company A has male : female = 13 : 10, and company B has male : female = 7 : 5. Since both companies have the same number of female workers, equalize the female ratio parts by multiplying B's ratio by 2 → B becomes 14 : 10.
Now both companies have 10 units of female workers. Total workers in A = 13 + 10 = 23 units; total workers in B = 14 + 10 = 24 units.
Therefore, the ratio of total workers in A to B is 23 : 24, making Option 2 the correct answer.
There are three types of rectangular tiles : 3' * 3', 3' * 7' and 3' * 11'. An area of rectangular shape of dimensions 3' * 100' is to be covered using these tiles without breaking them. If x and y are the maximum and minimum numbers of tiles of various sizes, respectively, that can be used to cover the area exactly, then x - y is
Detailed Explanation:
The area 3' × 100' must be covered by tiles 3' × 3', 3' × 7', and 3' × 11'. Since all tiles have width 3', they fit along the length, giving the equation 3a + 7b + 11c = 100.
Maximum tiles (x): Use smallest tiles. 31 tiles of 3' give 93', leaving 7' for 1 tile of 7'. Total = 32 tiles. Minimum tiles (y): Use largest tiles. 8 tiles of 11' give 88', leaving 12' for 4 tiles of 3'. Total = 12 tiles.
Therefore, x − y = 32 − 12 = 20, making Option 1 the correct answer.
$A$ 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.
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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