HCF and LCM Shortcut, tricks, and tips

Type 3: Tips , Tricks and Shortcuts when sum of two  numbers is given , LCM and HCF is given to find the sum of reciprocals.

Question:  Sum of two numbers is 60 and the H.C.F. and L.C.M. of these numbers are 5 and 100 respectively, then the sum of the reciprocals of the numbers is equals to:

Options

A. \frac{3}{25}

B. \frac{11}{220}

C. \frac{21}{120}

D. \frac{11}{320}

Solution :   Let the numbers be a and b.

Now , given a+b = 60

a × b = HCF × LCM =  5   ×   100

= 500

 \frac{1}{a} +\frac{1}{b} = \frac{a+b}{a × b}

\frac{1}{a} +\frac{1}{b} = \frac{60}{500}

\frac{3}{25}

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 Sugar and rice  whose measurements are 403   grams,    434 grams and  465 grams What biggest measure must be there to measure all the different quantities exactly?

Options :

A. 31 grams

B. 21 grams

C. 41 grams

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 :Tips , tricks and Shortcuts of 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 : Tips , tricks and Shortcuts of 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