Aptitude | HCF and LCM

MCQ

Q 21.

Find the value of X if HCF of X, 3240 and 3600 is 36 and their LCM is 24x35x52x72 

  • 22x35x72

  • 25x52x72

  • 23x35x72

  • 22x53x72

Q 22.

Find GCD of 0.36, 0.9 and 1.08

  • 0.18

  • 0.90

  • 0.108

  • 0.27

Q 23.

Find the LCM of 22, 54, 108, 135 and 198

  • 330

  • 1980

  • 5940

  • 11880

Q 24.

Find the LCM of 24, 36 and 40

  • 120

  • 240

  • 360

  • 480

Q 25.

Find the HCF of

4x27x3125, 5x11x16x49x81 and 7x8x9x25

  • 1260

  • 540

  • 180

  • 360

Q 26.

Find the HCF of 3x5x7x11, 2x2x2x2x3x3x5x5x5x7 and 2x2x2x3x3x3x5x5x7x7

  • 105

  • 27720

  • 2310

  • 1155

Q 27.

Express 1095/1168 in simplest form

  • 17/26

  • 25/26

  • 15/16

  • 13/16

Q 28.

Express 252 as a product of primes

  • 2x3x3x3x7

  • 2x2x2x3x7

  • 3x3x3x3x7

  • 2x2x3x3x7

Q 29.

Jai and Veeru are running in opposite direction on a circular path at Eden Garden, Calcutta. Jai takes 5 minutes while Veeru takes 6 minutes to complete a lap. If they decided to run 6 laps then find the number of time they will cross each other.

  • 1

  • 2

  • 3

  • 4

Total time taken by Jai to complete a lap = 5 minutes

Total time taken by Veeru to complete a lap = 6 minutes

Total time taken by Jai to complete 6 laps = 5x6 = 30 minutes.

Total time taken by Veeru to complete 6 laps = 6x6 = 36 minutes.

HCF of 5 minutes and 6 minutes is 30 minutes.

So, they will cross only once.

Q 30.

Find the HCF of (x4 - y4) and (x6 - y6).

  • x4 - y4

  • x3 - y3

  • x2 - y2

  • x - y

f(x) = (x4 - y4)

= (x2)2 - (y2)2

= (x2 - y2)(x2 + y2)

= (x - y)(x + y)(x2 + y2)

 

g(x) = (x6 - y6)

= (x3)2 - (y3)2

= (x3 - y3)(x3 + y3)

= (x - y)(x2 - y2 + xy) (x + y)(x2 + y2 - xy)

= (x - y)(x + y) (x2 - y2 + xy)(x2 + y2 - xy)

 

Common factor in f(x) and g(x) = (x - y)(x + y) = (x2 - y2)

Therefore, HCF = (x2 - y2)

Q 31.

Find the least number which when divided by 12, 16 and 18 leaves 5 as remainder in each case.

  • 148

  • 149

  • 150

  • 151

LCM of 12, 16 and 18 = 144

The required least number = 144+5 = 149

Q 32.

Find the greatest 3 digit number which when divided by 6, 9 and 12 leaves 3 as remainder in each case.

  • 965

  • 975

  • 985

  • 995

The greatest 3 digit number is 999.

LCM of 6, 9 and 12 is 36.

When we divide 999 by 36 we get 27 as remainder.

So, the required 3 digit number is (999 - 27 + 3) i.e., 975.

Q 33.

Rocky has two numbers in the ratio 3:4. Find the LCM if their HCF is 4.

  • 12

  • 24

  • 36

  • 48

Let the two numbers be 3q and 4q where q is the HCF.

Given, HCF = 4.

So, the two numbers are 12 and 16.

Therefore, the required LCM = 48

Q 34.

If the HCF of two numbers is 8 then which of the following can never be their LCM.

  • 60

  • 48

  • 56

  • 24

LCM of two numbers is always a multiple of their HCF.

Given, HCF = 8

Now, from the list of options we see that 60 is not a multiple of 8.

Hence, 60 can never be the LCM of the two numbers.

Q 35.

The product of two coprime numbers is 117. Find their LCM.

  • 117

  • 111

  • 171

  • None of these

HCF of two coprime numbers = 1.

We know, that Product of two numbers = HCF of the numbers x LCM of the numbers.

Therefore, 117 = 1 x LCM

or, LCM = 117

Or, we can say that LCM of two coprime numbers is equal to the product of the numbers.

Q 36.

Find the HCF of 1/2 and 3/4.

  • 1/3

  • 1/4

  • 4/3

  • 2/3

Required HCF = HCF of numerators / LCM of denominators

= HCF(1,3) / LCM(2,4)

= 1/4

Q 37.

Find the LCM of 1/3, 2/9, 5/6 and 4/27.

  • 10/3

  • 20/3

  • 3/10

  • 3/20

Required LCM = LCM of numerators / HCF of denominators

= LCM(1,2,5,4) / HCF(3,9,6,27)

= 20/3

Q 38.

Jai, Veeru and Dhanno are friends and they have 3 numbers 2a, 4a and 8a respectively. Find the LCM of their numbers.

  • 2a

  • 4a

  • 8a

  • 16a

LCM = 2x2x2xa = 8a

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