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LCM and HCF Questions Answers

Q1 :

Which one of the following numbers will divide 129 and 281 leaving 3 and 5 remainders respectively?

A
  

23

B
  

8

C
  

7

D
  

6

View Answer
Correct Answer: D

6

Description:

129 − 3 = 126
and 281 − 5 = 276
Now, the H.C.F. of 126 and 276 are required number i.e., 6

Q2 :

Find the LCM of 12,35,47 and 521frac{1}{2}, frac{3}{5}, frac{4}{7} text{ and } frac{5}{21}

A
  

227frac{2}{27}

B
  21
C
  

160frac{1}{60}

D
  60
View Answer
Correct Answer: D 60
Description:

LCM of the Fractions

=LCM of 1,3,4,5HCF of 2,7,21=601=60= frac{text{LCM of } 1,3,4,5}{text{HCF of } 2,7,21} = frac{60}{1} = 60

Q3 :

The product of two numbers is 396×576396 times 576 and the L.C.M. is 6336. Their G.C.M. is

A
  

36

B
  

26

C
  

59

D
  

76

View Answer
Correct Answer: A

36

Description:

G.C.M =396×5766336=36text{G.C.M } =frac{396times576}{6336} = 36

Q4 :

The greatest number that will divide 2930 and 3250, and will leave as remainder 7 and 11 respectively is:

A
  

37

B
  

53

C
  

59

D
  

79

View Answer
Correct Answer: D

79

Description:

Subtract 7 from 2930 and 11 from 3250, we have 2923 and 3239. The greatest number is the H.C.F. of 2923 and 3239 which is 79.

Q5 :

What is the smallest number which when increased by 5 is divisible by 36, 48, 21 and 28?

A
  

341

B
  

314

C
  

431

D
  

443

View Answer
Correct Answer: A

341

Description:

Add 5 to the L.C.M. of 36, 48, 21 and 28.

Q6 :

What is the least number which when divided by 12, 21 and 35 will leave in each case the same remainder 6?

A
  

420

B
  

526

C
  

414

D
  

426

View Answer
Correct Answer: D

426

Description:

Add 6 to the L.C.M. of 12, 21, 35.

Q7 :

Find the least number which must be subtracted from 6156 to make it a perfect square.

A
  

52

B
  

82

C
  

72

D
  

65

View Answer
Correct Answer: C

72

Description:

=6156=(72)2+72, so 72 is subtracted=6156 = (72)^2 + 72, text { so } 72 text { is subtracted}

Q8 :

The LCM of the polynomials x3+1,x3−3x2+3x−2x^3 + 1, x^3 - 3x^2 + 3x - 2 and x2−x−2x^2 - x - 2 is

A
  

(x+1)(x2+x+1)(x + 1)(x^2 + x + 1)

B
  

(x+1)(x−2)(x2−x+1)(x + 1)(x - 2)(x^2 - x + 1)

C
  

(x−1)(x−2)(x - 1)(x - 2)

D
  

(x−1)(x−2)(x2−x+1)(x - 1)(x - 2)(x^2 - x + 1)

View Answer
Correct Answer: B

(x+1)(x−2)(x2−x+1)(x + 1)(x - 2)(x^2 - x + 1)

Description:

x3+1=(x+1)(x2−x+1)x^3 + 1 = (x + 1) (x^2 - x + 1)

x3−3x2+3x−2=(x−2)(x2−x+1)x^3 - 3x^2 + 3x -2 = (x - 2)(x^2 - x + 1)

x2−x−2=(x−2)(x+1)x^2 - x - 2 = (x - 2)(x + 1)

∴LCM =(x+1)(x−2)(x2−x+1)therefore text{LCM } = (x + 1)(x - 2)(x^2 - x + 1)

Q9 :

The LCM of the polynomials x2+3x+2x^2 +3x + 2 and x3+3x2+3x+1x^3 + 3x^2 + 3x + 1 is

A
  

(x+1)(x+2)(x + 1)(x + 2)

B
  

(x+1)2(x+2)(x + 1)^2(x + 2)

C
  

(x+1)2(x+2)2(x + 1)^2(x + 2)^2

D
  

(x+1)3(x+2)(x + 1)^3(x + 2)

View Answer
Correct Answer: D

(x+1)3(x+2)(x + 1)^3(x + 2)

Description:

x2+3x+2=(x+1)(x+2)x^2 + 3x + 2 = (x + 1)(x + 2)

x3+3x2+3x+1=(x+1)3x^3 + 3x^2 + 3x + 1 = (x + 1)^3

∴LCM =(x+1)3(x+2)therefore text{LCM } = (x + 1)^3(x + 2)

Q10 :

The LCM of two numbers is 12 times their HCF. The sum of the HCF and LCM is 403. If one of the number is 93, then the other number is:

A
  

124

B
  

128

C
  

134

D
  

138

View Answer
Correct Answer: A

124

Description:

We know that
LCM ×times HCF = The product of two numbers

∴Other no. =LCM × HCF 1st no.therefore text{Other no. } = frac{text{LCM } times text{ HCF }}{1text{st no.}}

=31×(12×31)93= frac{31 times (12 times 31)}{93}

=124= 124

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