Q 31.

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

Options:

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.

Options:

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.

Options:

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.

Options:

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.

Options:

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.

Options:

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.

Options:

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.

Options:

2a

4a

8a

16a

LCM = 2x2x2xa = 8a

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