# Aptitude - HCF and LCM

## MCQ

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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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