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

31.

The H.C.F. of $$\frac{a}{b}\text{,}\frac{c}{d}\text{,}\frac{e}{f}$$  is equal to = ?

• (A) $$\frac{\text{LCM of }a,c,e}{\text{HCF of }b,d.f}$$  
• (B) $$\frac{\text{HCF of }a,c,e}{\text{LCM of }b,d,f}$$  
• (C) $$\frac{\text{HCF of }a,c,e}{\text{HCF of }b,d.f}$$  
• (D) $$\frac{ace}{bdf}$$

Show Answer

 $$\eqalign{ & {\text{HCF of fractions}} \cr & {\text{ = }}\frac{{{\text{HCF of Numerators}}}}{{{\text{LCM of Denominators}}}} \cr & {\text{HCF of }}\frac{a}{b}{\text{,}}\frac{c}{d}{\text{,}}\frac{e}{f} \cr & = \frac{{{\text{HCF of }}a,c,e}}{{{\text{LCM of }}b,d,f}} \cr} $$

32.

Three numbers are in the ratio 2 : 3 : 4, If their LCM is 240 the smaller of the three numbers is = ?

• (A) 20
• (B) 60
• (C) 30
• (D) 80

Show Answer

 $$\eqalign{ & {\text{Let number are}} \cr & {\text{ = }}2x,3x,4x \cr & {\text{given,}} \cr & {\text{LCM of }}\left( {2 \times 3 \times 4} \right)x = 24x \cr & 24x = 240 \cr & x = 10 \cr & \therefore {\text{numbers are 2}} \times {\text{10 = 20}} \cr & {\text{3}} \times 10 = 30 \cr & {\text{4}} \times 10 = 40 \cr & \therefore {\text{Smaller is 20}} \cr} $$

33.

The G.C.D. (greatest common divisor) of 1.08, 0.36, 0.9 is = ?

• (A) 0.03
• (B) 0.9
• (C) 0.18
• (D) 0.108

Show Answer

 Given numbers are 1.08, 0.36 and 0.90 HCF of 108, 36 and 90 is 18 ∴ HCF of given numbers = 0.18

34.

The least perfect square, which is divisible by each of 21, 36 and 66 is:

• (A) 214344
• (B) 214434
• (C) 213444
• (D) 231444

Show Answer

 LCM of (21, 36, 66) = 21 × 12 × 11 = 7 × 3 × 4 × 3 × 11 = 7 × 3 × 2 × 2 × 3 × 11 For perfect square multiply by 7 × 11 So that pairs of number from perfect square ∴ 7×7 ⎵ × 3×3 ⎵ × 2×2 ⎵ × 11×11 ⎵ Required result is (which is perfect square) ⇒ 213444

35.

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

• (A) 40
• (B) 80
• (C) 120
• (D) 200

Show Answer

 Let the numbers be 3x, 4x and 5x Then, their L.C.M. = 60x So, 60x = 2400 or x = 40 ∴ The numbers are (3 x 40), (4 x 40) and (5 x 40) Hence, required H.C.F. = 40

36.

The LCM of two numbers is 44 times of their of HCF. The sum of the LCM and HCF is 1125. If one number is 25 then the other number is = ?

• (A) 1100
• (B) 975
• (C) 900
• (D) 800

Show Answer

 $$\eqalign{ & {\text{Let HCF = }}x \cr & {\text{LCM = 44}}x \cr & {\text{given HCF + LCM}} \cr & = 44x + x = 45x \cr & 45x = {\text{ }}112{\text{5}} \cr & x = \frac{{1125}}{{45}} = 25 \cr & \therefore {\text{HCF}} = {\text{25}} \cr & {\text{LCM = 25}} \times {\text{44}} \cr & {\text{also given that one number}} \cr & {\text{ = 25}} \cr & {\text{Let another number = }}y \cr & \therefore 25y = 25 \times 25 \times 44 \cr & y = \frac{{25 \times 25 \times 44}}{{25}} \cr & y = 1100 \cr} $$

37.

What is the least number which when divided by the number 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 22 leaves no remainder ?

• (A) 312
• (B) 242
• (C) 1562
• (D) 1586

Show Answer

 $$\eqalign{ & {\text{LCM of }}\left( {3,5,6,8,10,12} \right) \cr & = 3 \times 5 \times 2 \times 4 \cr & = 120 \cr & {\text{Required number is }} \cr & \frac{{120K + 2}}{{22}} = \frac{{10K + 2}}{{22}} \cr & {\text{at }}K{\text{ = 2,}}\frac{{10K + 2}}{{22}} \cr & \Rightarrow {\text{Remainder = 0}} \cr & {\text{The given condition satisfied}} \cr & {\text{ = }}120K + 2 \cr & = 240 + 2 \cr & = 242 \cr} $$

38.

If three number are 2a, 5a and 7a, what will be their LCM ?

• (A) 70a
• (B) 65a
• (C) 75a
• (D) 73a

Show Answer

 The given three numbers are 2a, 5a and 7a LCM of 2a, 5a and 7a = 2 × 5 × 7 × a = 70a

39.

The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is = ?

• (A) 1677
• (B) 1683
• (C) 2523
• (D) 3363

Show Answer

 LCM of 5, 6, 7, 8 = 840 ∴ Required number is of the from 840k + 3 Least value of k for which (840k + 3) is divisible by 9 is k = 2 ∴ Required number = (840 × 2 + 3) = 1683

40.

The LCM and the HCF of the numbers 28 and 42 are in the ratio ?

• (A) 6 : 1
• (B) 2 : 3
• (C) 3 : 2
• (D) 7 : 2

Show Answer

 Numbers x = 28, y = 42 HCF (28 , 42) Difference = 42 - 28 = 14 For HCF of any numbers take their difference. HCF will be either the factor of that difference or the difference itself. Now, LCM of 28, 42 ∴ 14 × 2 × 3 = 84 ⇒ LCM : HCF ⇒ 84 : 14 ⇒ 6 : 1