LCM and HCF MCQ for SSC CGL, CHSL, CPO, MTS & GD Exams

Detailed Answer Explanations

1. Q1. Answer: (C) $64$, $80$
   Let numbers be $16x$ and $16y$ (co-prime). $16x$ × $16y$ = $16$ × $480$ → xy = $30.$ Possible pairs $(5)$,$6$ gives numbers $80$ and $96.$
2. Q2. Answer: (A) $3$
   LCM of $4$,$6$,$10$,$15$ = $60.$ Least 6-digit number = $100000.$ Remainder when divided by $60$ is $40.$ Least number divisible by $60$ is $100020.$ Add $2$ → $100022.$ Sum of digits = $5.$
3. Q3. Answer: (C) $48$
   LCM = Product $/$ HCF = $216$ $/$ $6$ = $36.$
4. Q4. Answer: (C) $2500$
   LCM of $10$,$16$,$24$ = $240$ = $2^{4}$×$3$×$5.$ To make perfect square, multiply by $3$×$5$ = $15$ → $240$×$15$ = $3600.$
5. Q5. Answer: (C) $184$
   LCM of $6$,$9$,$15$,$18$ = $90.$ Number = $90k$ + $4.$ For k=4, $364$ divisible by $7.$
6. Q6. Answer: (A) $1$
   Let numbers be $12x$ and $12y$ (co-prime). 144xy = $2160$ → xy = $15.$ Pairs: $(1)$,$15$ and $(3)$,$5$ → $2$ pairs. But answer given is $2$? Actually options: $(2)$ $2.$ Let's check: $15$ = $1$×$15$, $3$×$5$ → $2$ pairs.
7. Q7. Answer: (A) $48$
   Using Rule $1$, other number = $(16)$ × $160$ $/$ $32$ = $80.$
8. Q8. Answer: (A) $312$
   LCM of $3$,$5$,$6$,$8$,$10$,$12$ = $120.$ Number = $120k$ + $2.$ For k=8, $962$ divisible by $13.$
9. Q9. Answer: (B) $186$
   LCM of $12$,$10$,$8$ = $120.$ Required number = $120$ + $6$ = $126.$
10. Q10. Answer: (C) $18002$
    LCM of $16$,$18$,$20$,$25$ = $3600.$ Required number = $3600k$ + $4.$ For k=5, $18004$ divisible by $7.$
11. Q11. Answer: (A) $1$
    Let numbers be $13x$ and $13y$ (co-prime). 169xy = $2028$ → xy = $12.$ Pairs: $(1)$,$12$, $(3)$,$4$ → $2$ pairs (since $2$,$6$ not co-prime).
12. Q12. Answer: (A) $36$
    $84$ × other = $12$ × $336$ → other = $48.$
13. Q13. Answer: (B) $3$
    Let numbers be $15x$ and $15y$ (co-prime). Product = 225xy = $6300$ → xy = $28.$ Pairs: $(4)$,$7$ and $(2)$,$14$ but $(2)$,$14$ not co-prime. So only $1$ pair? Actually explanation says $(7)$,$4$ and $(14)$,$2$ but $14$,$2$ not co-prime, so only $1$ pair? However answer given is $(3)$ $2$? Let's check: $28$ = $1$×$28$, $4$×$7$, $2$×$14.$ Co-prime pairs: $(1)$,$28$ and $(4)$,$7$ → $2$ pairs. So correct.
14. Q14. Answer: (A) $6557$
    LCM of $15$,$18$,$21$,$24$ = $2520.$ Largest 4-digit number $9999.$ Remainder when divided by $2520$ is $2439.$ Difference between divisors and remainders is $4.$ Required number = $9999$ - $2439$ - $4$ = $7556.$
15. Q15. Answer: (A) $144$
    LCM of $4$,$6$,$8$,$9$ = $72.$ $72$ divided by $13$ leaves remainder $7.$ So $72$ is the answer.
16. Q16. Answer: (B) $1$
    Let numbers be $12x$ and $12y$ (co-prime). LCM = 12xy = $924$ → xy = $77.$ Pairs: $(1)$,$77$ and $(7)$,$11$ → $2$ pairs.
17. Q17. Answer: (C) $540$
    LCM of $15$,$12$,$20$,$54$ = $540.$ Number = $540$ + $4$ = $544.$
18. Q18. Answer: (A) $60$
    LCM of $6$,$8$,$10$ = $120.$
19. Q19. Answer: (B) $842$
    LCM of $4$,$5$,$6$,$7$,$8$ = $840.$ Required number = $840k$ + $2$, divisible by $13.$ For k=3, $840$×$3+2$ = $2522$, divisible by $13.$
20. Q20. Answer: (A) $5$
    Let HCF = H, LCM = $4H.$ H+4H=125 → H=25, LCM=100. Other = $(100)$×$25\frac{}{100}$ = $25.$
21. Q21. Answer: (B) $539$
    Here divisor-remainder = $1$ each. LCM(10,$9$,8)=360. Required number = $360$ - $1$ = $359.$
22. Q22. Answer: (C) $24$
    $24$ × other = $8$ × $48$ → other = $16.$
23. Q23. Answer: (C) $723$
    LCM of $5$,$6$,$8$ = $120.$ Required number = $120k$ + $3.$ Check k=2 → $243$, which is divisible by $9.$ So $243.$
24. Q24. Answer: (A) $78$
    Numbers: $13x$ and $13y$, LCM = 13xy = $455$ → xy = $35$ = $5$×$7.$ Numbers: $65$ and $91.$ $91$ lies between $75$ and $125.$
25. Q25. Answer: (A) $10$:$16$:$54$ AM
    LCM of $24$,$36$,$54$ = $216$ seconds = $3$ min $36$ sec. Time = $10$:$18$:$36$ AM.
26. Q26. Answer: (B) $36$
    LCM of $9$,$10$,$15$ = $90.$ Numbers leaving remainder $7$: $90k+7.$ Largest less than $1936$: $90$×21+7=1897. Subtraction = 1936-1897=39.
27. Q27. Answer: (C) $99960$
    LCM of $3$,$5$,$8$,$12$ = $120.$ Largest 5-digit number $99999.$ Remainder when divided by $120$ is $39.$ Greatest divisible number = $99960.$ Add $2$ → $99962.$
28. Q28. Answer: (A) $55$
    12-5=7, 16-9=7. LCM(12,16)=48. Required number = $48$ - $7$ = $41.$
29. Q29. Answer: (A) $1677$
    LCM of $5$,$6$,$7$,$8$ = $840.$ Number = $840k$ + $3.$ For k=2, $1683$ divisible by $9.$
30. Q30. Answer: (C) $60$
    $75$ × other = $15$ × $225$ → other = $45.$
31. Q31. Answer: (A) $204$
    Using Rule $1$, other number = $(1920)$ × $16$ $/$ $128$ = $240.$
32. Q32. Answer: (C) $24$
    HCF must divide LCM. $35$ does not divide $120$, so cannot be HCF.
33. Q33. Answer: (B) $528$
    First number = $2$×$44$ = $88.$ Then $88$ × other = $44$ × $264$ → other = $132.$
34. Q34. Answer: (A) $99279$
    LCM of $16$,$24$,$30$,$36$ = $720.$ Largest 5-digit number $99999.$ Remainder when divided by $720$ = $639.$ Required number = $99999$ - $639$ + $10$ = $99370.$
35. Q35. Answer: (C) $345$
    Numbers = $23$×$13$ = $299$ and $23$×$14$ = $322.$ Larger = $322.$
36. Q36. Answer: (B) $48$
    LCM of $4$,$6$,$8$,$12$,$16$ = $48.$ Required number = $48$ + $2$ = $50.$
37. Q37. Answer: (C) $56$
    LCM must be a multiple of HCF $(8).$ $60$ is not a multiple of $8$, so cannot be LCM.
38. Q38. Answer: (B) $1260$
    LCM of $12$,$18$,$21$,$30$ = $1260.$ Required number = $1260$ $/$ $2$ = $630.$
39. Q39. Answer: (B) $25$
    LCM × HCF = 1st Number × 2nd Number → $225$ × $5$ = $25$ × x → x = $45.$
40. Q40. Answer: (A) $50$
    Using Rule $1$, other = $(15)$ × $300$ $/$ $60$ = $75.$
41. Q41. Answer: (C) $45$ minutes
    LCM of $252$,$308$,$198$ = $2772$ seconds = $46$ minutes $12$ seconds.
42. Q42. Answer: (B) $1296$
    Using Rule $1$, Required number = (LCM × HCF) $/$ First number = $(864)$ × $144$ $/$ $288$ = $432.$
43. Q43. Answer: (A) $189$
    LCM of $5$,$10$,$12$,$15$ = $60.$ Number = $60k$ + $2.$ For k=3, $182$ divisible by $7.$
44. Q44. Answer: (B) $25$
    $10$ × other = $30$ × $5$ → other = $15.$
45. Q45. Answer: (C) $140$
    $864$ × other = $96$ × $1296$ → other = $144.$
46. Q46. Answer: (B) $214434$
    LCM of $21$,$36$,$66$ = $2772$ = $2^{2}$ × $3^{2}$ × $7$ × $11.$ To make it a perfect square, multiply by $7$×$11$ = $77.$ $2772$×$77$ = $213444.$
47. Q47. Answer: (B) $1690$
    Using Rule $1$, other number = $(1820)$ × $26$ $/$ $130$ = $364.$

Why Practice LCM and HCF MCQs for SSC?

LCM and HCF is one of the most scoring and predictable topics in the Mathematics section of SSC exams. Regular practice helps you simplify radicals quickly, avoid arithmetic traps, and solve comparison problems accurately. At Shortcut Maths in Jadavpur, Kolkata, we recommend solving topic-wise MCQs to improve speed and accuracy.

Key LCM and HCF Topics Covered in These 47 MCQs

• Basic Surd Operations: Addition, subtraction, multiplication and division of radicals.
• Comparison of LCM and HCF: Ordering surds by converting to a common index.
• Remainders & Divisors: Solving remainder-based application problems.
• Evaluation: Simplifying expressions using given values.

Recommended Study Plan for SSC Maths

• Step 1 - Learn Concepts: Understand the core lcm and hcf rules and methods.
• Step 2 - Practice MCQs: Solve this 47-question set and review the detailed explanations.
• Step 3 - Attempt Mock Tests: Test your speed on exam-like mocks. Visit our Online Mock Tests section for SSC, Bank and Railway practice tests.

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Frequently Asked Questions

Last updated: July 23, 2026. For maths coaching and doubt clearing, contact Shortcut Maths, Jadavpur, Kolkata 700032. Call +91-9804490328 or WhatsApp us.