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

Detailed Answer Explanations

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

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.

Related Study Materials

• SSC CGL Practice Set PDF — Topic-wise and full-length mock tests
• SSC CHSL Practice Set PDF — Previous year pattern questions
• SSC CGL Syllabus 2026 — Complete topic-wise syllabus
• 200 Maths Tricks for Competitive Exams — Shortcut formulas for quantitative aptitude
• Average MCQ for SSC — Practice average problems with explanations

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.