HCF and LCM

Sponsor Area

Question
SSCCGLENQA12042724

Find the greatest number which will exactly divide 200 and 320.

  • 10

  • 20

  • 16

  • 40

Solution

D.

40

Required number  = HCF of 200 and 320 = 40


Sponsor Area

Question
SSCCGLENQA12042719

Find the least number which when divided separately by 15, 20, 36 and 48 leaves 3 as remainder in each case.

  • 183
  • 243

  • 483

  • 723

Solution

D.

723

Required number  = (L.C.M. of 15, 20, 36 and 48) + 3
                           = 720 + 3 = 723

Question
SSCCGLENQA12042940

H.C.F. of 2 over 3 comma space space 4 over 5 space and space space 6 over 7 is

  • 48 over 105
  • 2 over 105
  • 1 over 105
  • 24 over 105

Solution

B.

2 over 105

H.C.F. of 2 over 3 comma space 4 over 5 space and space 6 over 7
equals space fraction numerator straight H. straight C. straight F. space of space 2 comma space 4 space and space 6 over denominator straight L. straight C. straight M. space of space 3 comma space 5 space and space 7 end fraction
equals space 2 over 105

Question
SSCCGLENQA12042657

If the L.C.M. and H.C.F. of two expressions are (x2 + 6x + 8) (x + 1) and (x + 1) respectively and one of the expressions is x2 + 3x + 2, find the other.

  • x2 + 5x + 4

  • x2 - 5x + 4

  • x2 + 4x + 5

  • x2 - 4x + 5

Solution

A.

x2 + 5x + 4

x2 + 6x + 8 = x2 + 4x + 2x +8 = x(x + 4) + 2(x + 4) = (x + 4) (x + 2)

x2 + 3x + 2 = x2 + 2x + x +2=  x(x + 2) + 1(x + 2) = (x + 2) (x + 1)
Now, 
First expression x Second expression = LCM X HCF
⟹  (x2 + 3x + 2) x Second expression = (x2 + 6x + 8) (x + 1) x (x + 1)
⟹ (x + 2) (x + 1) x Second expression = (x + 4) (x + 2) (x + 1) (x + 1)
rightwards double arrow space space Second space expression space equals space fraction numerator left parenthesis straight x plus 2 right parenthesis thin space left parenthesis straight x plus 4 right parenthesis thin space left parenthesis straight x plus 1 right parenthesis thin space left parenthesis straight x space plus 1 right parenthesis over denominator left parenthesis straight x space plus space 2 right parenthesis thin space left parenthesis straight x plus 1 right parenthesis end fraction space equals space left parenthesis straight x space plus space 4 right parenthesis thin space left parenthesis straight x space plus space 1 right parenthesis
⟹ (x + 4) (x + 1) = x2 + 4x + x + 4 = x2 + 5x + 4
 


 

Question
SSCCGLENQA12042739

If the students of 9th class are arranged in a row of 6, 8, 12 or 16, no student is left behind. Then the possible number of students in the class is

  • 60

  • 72

  • 80

  • 96

Solution

D.

96

Required number of students  = L.C.M. of 6, 8, 12 and 16 is 96.