H.C.F. and L.C.M. of Numbers: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 4)

Let the required number be h, and the common remainder be r. Since all three numbers leave the same remainder r, their pairwise differences are exactly divisible by h. 283 – 235 = 48, and 319 – 283 = 36. So h = H.C.F.(48, 36) = 12. Check: 48/12 = 4 and 36/12 = 3, both whole numbers.

Let the required number be h. Subtract each remainder from its number first: 724 – 12 = 712 and 770 – 18 = 752. h must divide both differences exactly, so h = H.C.F.(712, 752). That H.C.F. is 8. Check: 712/8 = 89 and 752/8 = 94, both whole numbers.

Let the required number be h. Subtract each remainder from its number first: 703 – 8 = 695 and 752 – 17 = 735. h must divide both differences exactly, so h = H.C.F.(695, 735). That H.C.F. is 5. Check: 695/5 = 139 and 735/5 = 147, both whole numbers.

Let the required number be h. Subtract each remainder from its number first: 872 – 7 = 865 and 920 – 15 = 905. h must divide both differences exactly, so h = H.C.F.(865, 905). That H.C.F. is 5. Check: 865/5 = 173 and 905/5 = 181, both whole numbers.

Let the required number be h. Subtract each remainder from its number first: 473 – 18 = 455 and 499 – 4 = 495. h must divide both differences exactly, so h = H.C.F.(455, 495). That H.C.F. is 5. Check: 455/5 = 91 and 495/5 = 99, both whole numbers.

Let the required number be h. Subtract each remainder from its number first: 287 – 17 = 270 and 326 – 16 = 310. h must divide both differences exactly, so h = H.C.F.(270, 310). That H.C.F. is 10. Check: 270/10 = 27 and 310/10 = 31, both whole numbers.

Let the required number be h. Subtract each remainder from its number first: 818 – 8 = 810 and 855 – 5 = 850. h must divide both differences exactly, so h = H.C.F.(810, 850). That H.C.F. is 10. Check: 810/10 = 81 and 850/10 = 85, both whole numbers.

Let the required number be h. Subtract each remainder from its number first: 807 – 17 = 790 and 839 – 9 = 830. h must divide both differences exactly, so h = H.C.F.(790, 830). That H.C.F. is 10. Check: 790/10 = 79 and 830/10 = 83, both whole numbers.

Let the required number be h. Subtract each remainder from its number first: 443 – 11 = 432 and 487 – 15 = 472. h must divide both differences exactly, so h = H.C.F.(432, 472). That H.C.F. is 8. Check: 432/8 = 54 and 472/8 = 59, both whole numbers.

Check the H.C.F. of each pair listed. For (13, 49): H.C.F.(13, 49) = 1. Since that H.C.F. is 1, (13, 49) is co-prime. The other pairs share a common factor greater than 1: (11,55) has H.C.F. 11; (56,24) has H.C.F. 8; (38,54) has H.C.F. 2. So (13, 49) is the only co-prime pair.

Let the fraction be 285/437. Find their H.C.F.: H.C.F.(285, 437) = 19. Divide both terms by 19: 285/19 = 15 and 437/19 = 23. So 285/437 reduces to 15/23. Check: H.C.F.(15, 23) = 1, confirming the fraction is fully reduced.

Let the fraction be 434/837. Find their H.C.F.: H.C.F.(434, 837) = 31. Divide both terms by 31: 434/31 = 14 and 837/31 = 27. So 434/837 reduces to 14/27. Check: H.C.F.(14, 27) = 1, confirming the fraction is fully reduced.

Let the fraction be 242/660. Find their H.C.F.: H.C.F.(242, 660) = 22. Divide both terms by 22: 242/22 = 11 and 660/22 = 30. So 242/660 reduces to 11/30. Check: H.C.F.(11, 30) = 1, confirming the fraction is fully reduced.

Let the fraction be 234/975. Find their H.C.F.: H.C.F.(234, 975) = 39. Divide both terms by 39: 234/39 = 6 and 975/39 = 25. So 234/975 reduces to 6/25. Check: H.C.F.(6, 25) = 1, confirming the fraction is fully reduced.

Let the fraction be 114/126. Find their H.C.F.: H.C.F.(114, 126) = 6. Divide both terms by 6: 114/6 = 19 and 126/6 = 21. So 114/126 reduces to 19/21. Check: H.C.F.(19, 21) = 1, confirming the fraction is fully reduced.

Let the fraction be 36/42. Find their H.C.F.: H.C.F.(36, 42) = 6. Divide both terms by 6: 36/6 = 6 and 42/6 = 7. So 36/42 reduces to 6/7. Check: H.C.F.(6, 7) = 1, confirming the fraction is fully reduced.

Let the fraction be 42/156. Find their H.C.F.: H.C.F.(42, 156) = 6. Divide both terms by 6: 42/6 = 7 and 156/6 = 26. So 42/156 reduces to 7/26. Check: H.C.F.(7, 26) = 1, confirming the fraction is fully reduced.

Let the fraction be 360/660. Find their H.C.F.: H.C.F.(360, 660) = 60. Divide both terms by 60: 360/60 = 6 and 660/60 = 11. So 360/660 reduces to 6/11. Check: H.C.F.(6, 11) = 1, confirming the fraction is fully reduced.

Let the fractions be 17/21 and 14/5. For fractions, L.C.M. = L.C.M. of numerators over H.C.F. of denominators. L.C.M. of numerators: L.C.M.(17, 14) = 238. H.C.F. of denominators: H.C.F.(21, 5) = 1. So the L.C.M. is 238/1. Reducing by their common factor 1 gives 238/1. Check: 238 x 1 = 238 and 1 x 1 = 1, confirming the reduction.

Let the fractions be 27/28 and 11/22. For fractions, H.C.F. = H.C.F. of numerators over L.C.M. of denominators. H.C.F. of numerators: H.C.F.(27, 11) = 1. L.C.M. of denominators: L.C.M.(28, 22) = 308. So the H.C.F. is 1/308. Reducing by their common factor 1 gives 1/308. Check: 1 x 1 = 1 and 308 x 1 = 308, confirming the reduction.

Let the fractions be 9/11 and 19/21. For fractions, H.C.F. = H.C.F. of numerators over L.C.M. of denominators. H.C.F. of numerators: H.C.F.(9, 19) = 1. L.C.M. of denominators: L.C.M.(11, 21) = 231. So the H.C.F. is 1/231. Reducing by their common factor 1 gives 1/231. Check: 1 x 1 = 1 and 231 x 1 = 231, confirming the reduction.

Let the fractions be 17/19 and 11/8. For fractions, H.C.F. = H.C.F. of numerators over L.C.M. of denominators. H.C.F. of numerators: H.C.F.(17, 11) = 1. L.C.M. of denominators: L.C.M.(19, 8) = 152. So the H.C.F. is 1/152. Reducing by their common factor 1 gives 1/152. Check: 1 x 1 = 1 and 152 x 1 = 152, confirming the reduction.

Let the fractions be 6/11 and 20/28. For fractions, H.C.F. = H.C.F. of numerators over L.C.M. of denominators. H.C.F. of numerators: H.C.F.(6, 20) = 2. L.C.M. of denominators: L.C.M.(11, 28) = 308. So the H.C.F. is 2/308. Reducing by their common factor 2 gives 1/154. Check: 1 x 2 = 2 and 154 x 2 = 308, confirming the reduction.

Let the fractions be 3/22 and 8/24. For fractions, L.C.M. = L.C.M. of numerators over H.C.F. of denominators. L.C.M. of numerators: L.C.M.(3, 8) = 24. H.C.F. of denominators: H.C.F.(22, 24) = 2. So the L.C.M. is 24/2. Reducing by their common factor 2 gives 12/1. Check: 12 x 2 = 24 and 1 x 2 = 2, confirming the reduction.

Let the fractions be 5/19 and 3/13. For fractions, H.C.F. = H.C.F. of numerators over L.C.M. of denominators. H.C.F. of numerators: H.C.F.(5, 3) = 1. L.C.M. of denominators: L.C.M.(19, 13) = 247. So the H.C.F. is 1/247. Reducing by their common factor 1 gives 1/247. Check: 1 x 1 = 1 and 247 x 1 = 247, confirming the reduction.