Practice HCF and LCM MCQ Questions with Details Solution
Q1. If the L.C.M and H.C.F. of two numbers are 2400 and 16, one number is 480; find the second number.
(a) 40
(b) 80
(c) 60
(d) None of these
Answer: (b) 80 Solution: we know, Product of two numbers = (L.C.M× H.C.F) ⇒ 480 × second number = 2400 × 16 ⇒ Second number = 80 |
Q2. Find the greatest number that will divide 148, 246 and 623 leaving remainders 4, 6 and 11 respectively.
(a) 11
(b) 12
(c) 13
(d) 14
Answer: (b) 12 Solution: Required number = H.C.F of (148 – 4), (246 – 6) and (623 – 11) = H.C.F of 144, 240 and 612 = 12 Rules: The greatest number that will divide x, y and z leaving remainders a, b and c respectively = H.C.F of (x – a), (y – b) and (z – c). |
Q3. Which is the least number that must be subtracted from 1856, so that the remainder when divided by 7, 12 and 16 will leave the same remainder 4?
(a) 137
(b) 1361
(c) 140
(d) 172
Answer: (d) 172 Solution: Suppose least no. be x 1856 – x = n(LCM of 7,12,16) + 4 or 1856 -x = n (336) + 4 we should take n = 5 so that n(336) is nearest to 1856 and n (336)< 1856 ∴ 1856 – x = 1680 + 4 = 1684 ⇒ x = 1856 – 1684 = 172 |
Q4. Find the maximum number of students among whom 429 mangoes and 715 oranges can be equally distributed.
(a) 100
(b) 120
(c) 143
(d) None of these
Answer: (c) 143 Solution: (d) Required number = H.C.F of 429 and 715 = 143 |
Q5. What is the smallest number which when increased by 5 is completely divisible by 8, 11 and 24?
(a) 259
(b) 269
(c) 279
(d) None of these
Answer: (a) 259 Solution: (b) Required no. = L.C.M of (8, 11, 24) –5 = 264 – 5 = 259 |
Q6. The L.C.M. of two number is 630 and their H.C.F. is 9. If the sum of numbers is 153, their difference is
(a) 17
(b) 23
(c) 27
(d) 33
Answer: (c) 27 Solution: Let numbers be x and y. We know, Product of two numbers = their (L.C.M× H.C.F) ∴ xy = 630 × 9 Also, x + y = 153 (given) x – y = $\sqrt{\left( \mathrm{x}+\mathrm{y} \right) ^2-4\mathrm{xy}}$ =$\sqrt{153^2-4\times 630\times 9}$ =$\sqrt{23409-22680}=27$ |
Q7. Three bells chime at an interval of 18, 24 and 32 minutes respectively. At a certain time they begin to chime together. What length of time will elapse before they chime together again.
(a) 2 hours 24 minutes
(b) 4 hours 48 minutes
(c) 1 hour 36 minutes
(d) 5 hours
Answer: (b) 4 hours 48 minutes Solution: L.C.M of 18, 24 & 32 = 288 Hence they would chime after every 288 min. or 4 hrs. 48 min |
Q8. The L.C.M and H.C.F of two numbers are 84 and 21, respectively. If the ratio of two numbers be 1 : 4, then the larger of the two numbers is:
(a) 21
(b) 48
(c) 84
(d) 108
Answer:(c) 84 Solution: Let the numbers be x and 4x. Then, 84 × 21 = x × 4x ⇒ 4x^{2} = 1764 ⇒x = 21 ∴ 4x = 4 × 21 = 84 Thus the larger number = 84 |
Q9. Suppose you have 108 green marbles and 144 red marbles. You decide to separate them into packages of equal number of marbles. Find the maximum possible number of marbles in each package.
(a) 4
(b) 36
(c) 9
(d) 12
Answer: (b) 36 Solution: Required number = H.C.F of 108 and 144 = 36 |
Q10. Find the lowest number which when subtracted from 3000, is exactly divisible by 7, 11 and 13
(a) 729
(b) 998
(c) Cannot be determined
(d) None of these
Answer: (b) 998 Solution: Required number = 3000 – maximum possible multiple of L.C.M of (7, 11 and 13) = 3000 – 2(1001) = 998 |