Quiz 1

Finding H.C.F. by division method:
H.C.F. of 513, 1134 and 1215 = 3 × 3 × 3 = 27.

Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.

Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.

HCF of numerators = 1
LCM of denominators = 48
=> 1/48

Required number = (L.C.M. of 12, 15, 20, 54) + 8 = 540 + 8 = 548.

Let the required numbers be 15.x and 11x.
Then, their H.C.F. is x. So, x = 13.
The numbers are (15 x 13 and 11 x 13) i.e., 195 and 143.

The Largest number of four digits is 9999.
Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540.
On dividing 9999 by 540,we get 279 as remainder.
Required number = (9999-279) = 9720.

Explanation:

For any this type of question, remember
Product of two numbers = Product of their HCF and LCM

So Other number = (693*11)/99 = 77

Easy trick is

We all know

Product of two numbers = Product of their HCF and LCM

So, 693*11 = 99*X

now look closely at the last digits on LHS, which are 3 and 1. Thus, if you definitely multiply 693*11 the units digit of the result will be 3 only.

Out of the which, options should 99 which has units digit as 9 be multiplied by so you also get units digit as 3. The only option is 7

As 9*7 = 63 which has units digit as 3.

Now, this is how smart people will solve it as it doesn't need any pen or paper. You can solve this in mind within 5 seconds.

L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together30 + 1 = 16 times.

Finding H.C.F. by division method:
H.C.F. of 513, 1134 and 1215 = 3 × 3 × 3 = 27.

Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.

Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.

HCF of numerators = 1
LCM of denominators = 48
=> 1/48

Required number = (L.C.M. of 12, 15, 20, 54) + 8 = 540 + 8 = 548.

Let the required numbers be 15.x and 11x.
Then, their H.C.F. is x. So, x = 13.
The numbers are (15 x 13 and 11 x 13) i.e., 195 and 143.

The Largest number of four digits is 9999.
Required number must be divisible by L.C.M. of 12,15,18,27 i.e. 540.
On dividing 9999 by 540,we get 279 as remainder.
Required number = (9999-279) = 9720.

Explanation:

For any this type of question, remember
Product of two numbers = Product of their HCF and LCM

So Other number = (693*11)/99 = 77

Easy trick is

We all know

Product of two numbers = Product of their HCF and LCM

So, 693*11 = 99*X

now look closely at the last digits on LHS, which are 3 and 1. Thus, if you definitely multiply 693*11 the units digit of the result will be 3 only.

Out of the which, options should 99 which has units digit as 9 be multiplied by so you also get units digit as 3. The only option is 7

As 9*7 = 63 which has units digit as 3.

Now, this is how smart people will solve it as it doesn't need any pen or paper. You can solve this in mind within 5 seconds.

L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes, they will toll together30 + 1 = 16 times.