How To Solve LCM and HCF Questions Quickly

Type 2: Find HCF

Question 1. Three numbers are in the ratio of 4: 3: 6 and their L.C.M. are 3600. Find their H.C.F:

Options

A. 20

B. 30

C. 25

D. 50

Solution:     Let the numbers be 4x, 3x and 6x

Then, their L.C.M. = (4x * 3x *6x) = 72x

So, 72x = 3600 or x = 50

Therefore, numbers are (4 x 50), (3 x 50) and (6 x 50) = 200, 150, 300

The factors of 150 are: 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150

The factors of 300 are: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60,  100, 150, 300

The factors of 200 are: 1, 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 200

Then the highest common factor is 50.

Hence, required H.C.F. = 50

Correct option: D

Question 2. Find the HCF of 34, 48, 56, and 74

Options

A. 2

B. 4

C. 34

D. 8

Solution:    The factors of 34 are: 1, 2, 17, 34

The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

The factors of 56 are: 1, 2, 4, 7, 8, 14, 28, 56

The factors of 74 are: 1, 2, 37, 74

Therefore, the highest common factor is 2.

Correct option:A

Question 3. Find the HCF of \frac{2}{11},\frac{4}{17},\frac{6}{5}.

Options

A. \frac{1}{935}

B. \frac{2}{935}

C. \frac{2}{93}

D. \frac{2}{35}

Solution:    We know that

HCF = \frac{HCF of numerator}{LCM of Denominator}

HCF = \frac{HCF (2,4,6)}{LCM(11,17,5) }

HCF = \frac{2}{935}

Correct option:B

Type 3: How to Solve when sum of two  numbers is given , LCM and HCF is given to find the sum of reciprocals.

Question:  Sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equals to:

Options

A. \frac{11}{120}

B. \frac{11}{220}

C. \frac{21}{120}

D. \frac{11}{320}

Solution :   Let the numbers be a and b.

Now , given a+b = 55

a × b = HCF × LCM =  5     ×     120
HCF × LCM = 600

Now, as we know that \frac{1}{a} +\frac{1}{b} = \frac{a+b}{a\times b}

\frac{1}{a} +\frac{1}{b} = \frac{55}{600}

\frac{11}{120}

Correct Option : A

Type 4: How to Solve HCF, LCM Problems related to finding the biggest container to measure quantities

Question : Suppose there are three different containers contain different quantities of a mixture of milk and water whose measurements are 403  litres,   434  litres and 465 litres What biggest measure must be there to measure all the different quantities exactly?

Options :

A. 31 litres

B. 21 litres

C. 41 litres

D. 30 litres

Solution : Prime factorization of 403,434 and 465 is

                  403=13×31

                  434=2×7×31

                  465=3×5×31

                  H.C.F of 403, 434 and 465=31

                  Correct Option : A

Type 5 : How to Solve HCF, LCM Problems related to Bell ring.

Question:  Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

Options :

A. 8

B. 16

C. 9

D. 10

Solution :   L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

Hence, the bells will toll together after every 120 seconds(2 minutes).

Therefore, in 30 minutes ,number of times bells toll together is \frac{30}{2} + 1 = 16

Correct Option B

Type 6 : How to Solve HCF, LCM Problems related to Circle Based Runner Problem.

Question: Two people P and Q start running towards a circular track of length 400 m in opposite directions with initial speeds of 10 m/s and 40 m/s respectively. Whenever they meet, P’s speed doubles and Q’s speed halves. After what time from the start will they meet for the third time?

Options

A. 30 seconds

B.  26 seconds

C. 10 seconds

D. 20 seconds

Solution :    Time taken to meet for the 1^st time= \frac{400}{40+10}=8 sec.

Now P’s speed = 20m/s and Q’s speed=20 m/s.

Time taken to meet for the 2^nd time= \frac{400}{20+20} = 10 sec.

Now P’s speed =40 m/sec and Q’s speed = 10 m/sec.

Time taken to meet for the 3^rd time= \frac{400}{10+40}=8 sec.

Therefore, Total time= (8+10+8) = 26 seconds.

Correct Option B