Cognizant LCM & HCF Quiz-1

Let the numbers be a and b.

Then, a + b = 55 and ab = 5 x 120 = 600.

The required sum = \frac{1}{a} +\frac {1}{b} =\frac{b+a}{ab}

= \frac{55}{600}

= \frac{11}{120}

 

Explanation:

L.C.M. of 21, 36, 66 = 2772

Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11

To make it a perfect square, it must be multiplied by 7 x 11.

So, required number = 2 x 2 x 3 x 3 x 7 x 7 x 11 x 11 = 213444

Explanation:-

All numbers except 124 is factors of 1024 , so answer is 124

Explanation:

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

So Other number = \frac{(693\times 11)}{99} = 77

Easy trick is

We all know

Product of two numbers = Product of their HCF and LCM

So, 693\times 11 = 99\times X

now look closely at the last digits on LHS, which are 3 and 1. Thus, if you definitely multiply 693 x 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 x 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.

Explanation:

We need to find out the HCF for given length.
15 meter 75 cm = 1575 cm.
11 meter 25 cm = 1125 cm
7 meter 65 cm = 765 cm.
1575 = 5 \times 5 \times 3 \times 3 \times 7
1125 = 5 \times 5 \times 5 \times 3 \times 3
765 = 5 \times 3 \times 3 \times 17

HCF of 1575, 1125 and 765 is 5\times 3\times 3 =45 .

OR

All lengths are multiple of 5 , so LCM must be multiple of 5 and only multiple of 5 in options is 45

In this type of question, its obvious we need to calculate the HCF, trick is
HCF of (1661 - 10) and (2045 -13)
= HCF (1651, 2032) = 127

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 seconds respectively.

Hence, using LCM of the given numbers which is 120 we conclude that bell will toll together after 120 seconds = 2 minutes.

In 60 minutes they will toll together in \frac{60}{2}and 1 (at the starting).

Hence, total number of bells together is 30 + 1 = 31

As given the questions these numbers are co primes, so there is only 1 as their
common factor.
It is also given that two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;

So first number is = \frac {551}{29} = 19
Third number = \frac{1073}{29} = 37

So sum of these numbers is = (19 + 29 + 37) = 85

L.C.M of 8,12,16 = 48
Now, 48 x 1 + 3 = 51 which is divisible by 17
Required number = 51

Note that 20 -12 = 8; 25 -17 = 8;35-27= 8;40- 32 = 8
Required number = L.C.M. of ( 20, 25, 35 and 40) - 8 = 1392