**Type 2: How to Solve LCM and HCF Questions Quickly.**

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(11,7,5) }**

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

= 600

\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 : **The HCF Of 403 , 434 and 465 is 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**

**Read Also** –** Formulas to solve HCF & LCM questions **

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