**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**

Login/Signup to comment