Practice MCQ Questions and Answer on Problems on H.C.F and L.C.M

181.

Find the greatest number of 4 digits which when divided by 4, 5, 6, 7 and 8 leaves 1, 2, 3, 4 and 5 as remainders = ?

• (A) 9237
• (B) 9240
• (C) 9840
• (D) 9999

Show Answer

 Clearly, (4 - 1) = 3, (5 - 2) = 3, (6 - 3) = 3, (7 - 4) = 3 and (8 - 5) = 3 LCM of 4, 5, 6, 7, 8 = 840 Greatest number of 4 digits = 9999 On dividing 9999 by 840, the remainder is 759 So, required number = (9999 - 759) - 3 = 9237

182.

Find the LCM of 186.6 and 373.2.

• (A) 373.2
• (B) 398.2
• (C) 186.6
• (D) 276.6

Show Answer

 \[\begin{array}{*{20}{c}} {{\text{LCM of}}}&{186.6}&{{\text{and}}}&{373.2} \\ {}&\begin{gathered} \,\,\, \downarrow \hfill \\ 186.6 \hfill \\ \end{gathered} &{}&\begin{gathered} \,\,\, \downarrow \hfill \\ 186.6 \times 2 \hfill \\ \end{gathered} \end{array}\] LCM = 186.6 × 2 = 373.2 Answer

183.

A number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, and when divided by 8 leaves a remainder of 7, is = ?

• (A) 1539
• (B) 539
• (C) 359
• (D) 1359

Show Answer

 10 - 9 = 1 9 - 8 = 1 8 - 7 = 1 ∴ LCM of (10, 9, 8) = 5 × 2 × 9 × 4 = 360 ∴ Required result = 360 - 1 = 359

184.

A gardener has to plant trees in rows containing equal number of trees. If he plants in rows of 6, 8, 10 or 12 then five trees are left unplanted. But if he plants in rows of 13 trees each, then no tree is left. What is the number of trees that the gardener plants ?

• (A) 485
• (B) 725
• (C) 845
• (D) None of these

Show Answer

 LCM of 6, 8, 10, 12 = 120 ∴ Required number is of the from 120k + 5 Least value of k for which (120k + 5) is divisible by 13 is k = 7 ∴ Required number = (120 × 7 + 5) = 845

185.

Let x be the smallest number, which when added to 2000 make the resulting number divisible by 12, 16, 18 and 21. The sum of the digits of x is = ?

• (A) 6
• (B) 5
• (C) 7
• (D) 8

Show Answer

 LCM of 12, 16, 18, 21 = 1008 Next number = 1008 × 2 = 2016 Divisible by all ∴ 16 is added Sum of digits = 1 + 6 = 7

186.

Find the least multiple of 23, which when divided by 18, 21 and 24 leaves the remainder 7, 10 and 13 respectively.

• (A) 3013
• (B) 3024
• (C) 3002
• (D) 3036

Show Answer

 18 - 7 = 11 21 - 10 = 11 24 - 13 = 11 Take LCM of (18, 21, 24) ⇒ 9 × 2 × 7 × 4 = 504 ⇒ Required number = (504k - 11) which is divided by 23 ∴ For $$\frac{{504k - 11}}{{23}},$$   remainder should be zero Put minimum value of k so that it completely divides 23 ⇒ at k = 6, 504k - 11 = 3013 completely divisible by 23 ∴ Required number is = 3013

187.

The H.C.F of 2 × 3 × 5 × 7, 2 × 3 × 5 × 7 and 3 × 5 × 7 × 11 is = ?4233322

• (A) 105
• (B) 1155
• (C) 2310
• (D) 27720

Show Answer

 H.C.F. = Product of lowest powers of common factors            = 3 × 5 × 7            = 105