### 29 Mar When it comes to quantitative aptitude section of the Bank PO exam, arithmetic is considered by many exam aspirants to be the most scoring area. Within arithmetic, LCM & HCF is a topic which can help you score at least a couple of sure shot marks. But, this can happen only when you know the shortcuts for solving the tricky questions asked from this topic.

Keeping this in mind, the institute providing the Best Bank PO Coaching in South Delhi has covered below all the concepts and time saving methods related to LCM & HCF.

LCM (Lowest Common Multiple):

The least number which is completely divisible by the given numbers is called their lowest common multiple (LCM) e.g. LCM of (24 & 36) = 72

To calculate LCM, first of all we have to express the given numbers as products of prime numbers (2, 3, 5, 7, 11, 13, 17…). This is known as Prime Factorization Method.

After doing so, prime factors with the highest powers need to be chosen and multiplied to get the LCM.

24 = 2×2×2×3 = 23×31

36 = 22×32

So, LCM of 24 & 36 = 23×32 = 72 (we have taken the highest powers)

Application of LCM:

TYPE 1: The least number which is completely divisible by x, y & z.

Solution: Find the LCM of x, y & z.

Question – Find the least number which is completely divisible by 24, 36 & 72.

Answer – 24 = 2×2×2×3 = 23×31

36 = 22×32

72 = 23×32

So, LCM of 24, 36 & 72 = 23×32 = 72

TYPE 2: The smallest number which when divided by x, y & z leaves the same remainder k in each case.

Solution: Find the LCM of {x, y & z} + k.

Question Find the least number which when divided by 3, 4 and 5 leaves remainder 2 in each case.

Answer – The least number which when divided by 3, 4 and 5 leaves remainder 2 in each case.

= LCM (3, 4, 5) + 2

= 60+2

= 62

TYPE 3: The smallest number which when divided by x, y & z leaves remainders p, q & r respectively.

Solution: Find the value of x – p = y – q = z – r = k

Find the LCM of {x, y & z} – k

Question Find the smallest number which when divided by 3, 4 and 5 leaves remainders 1, 2 and 3 respectively.

Answer – Here k = 3 – 1 = 4 – 2 = 5 – 3 =2

So, the least number which when divided by 3, 4 and 5 leaves a remainder of 1, 2 and 3 respectively = LCM (3, 4 & 5) – k

= 60 – 2

= 58

TYPE 4: Some persons start running around a circular stadium and complete one round in x seconds, y seconds and z seconds & so on. In how much time will they meet again at the starting point?

OR

A few bells toll at intervals of x sec, y sec, z sec & so on. Find after what interval they will toll again together.

Solution: Take the LCM of (x, y & z…)

Question A, B and C start running around a circular track and complete one round in 27 seconds, 9 seconds and 36 seconds respectively. In how much time will they meet again at the starting point?

Answer – LCM of 27, 9 and 36 = 108

So, they will meet again at the starting point after 108 seconds. i.e. 1 min 48 sec.

Question 5 bells commence tolling together and toll at intervals 2, 4, 6, 8 and 10 seconds respectively. Find in 40 minutes, how many times do they toll together?

Answer – LCM of 2, 4, 6, 8 and 10 sec = 120 sec = 2 min

Hence, the bells toll together after 120 seconds = 2 min HCF (Highest Common Factor):

The greatest number which can completely divide the given numbers is called their highest common factor (HCF). It is also known as GCD (greatest common divisor).

Finding HCF

(I) Prime Factorization Method

• To find HCF, we will express the given numbers as products of prime numbers (prime factors).
• After that, we will choose only that factor which is common in the given numbers.
• When choosing the common factor, we must take the smallest power of the common factor.

E.g. HCF of (24 & 32) =?

24 = 2×2×2×3 = 23×31

32 = 2×2×2×2×2 = 25

So, HCF of 24 & 32 = 23 = 8 (taking the lowest power of common factor)

So, 8 is the HCF.

Application of HCF:

TYPE 1: The greatest number which can divide x, y & z completely.

Solution: Find the HCF of x, y & z.

Question Find the greatest number which can divide 286 & 616 completely.

TYPE 2: The greatest number which on dividing x, y & z leaves remainders p, q & r respectively.

Solution: Find the HCF of {x – p, y – q & z – r}.

Question Find the greatest number that will divide 148, 246 and 623 leaving remainders 4, 6 and 11 respectively.

Answer – HCF (148 – 4, 246 – 6, 623 – 11)

= HCF (144, 240, 612)

144 = 12×12

240 = 12×20

612 = 12×51

So, HCF (144, 240 & 612) = 12

TYPE 3: The greatest number which on dividing x, y & z leaves the same remainder in each case.

Solution: Find the HCF of {x – y, y – z & z – x}

Question Find the greatest number which leaves the same remainder on dividing 333, 555 & 777.

Answer – The greatest number which when divides 333, 555 & 777 leaving the same remainder = HCF (555 – 333, 777 – 555, 777 – 333)

= HCF (222, 222, 444)

= 222

The tips & tricks you have learnt out here must be followed by self study and online test series offered at the Best Bank Coaching Classes. This will ensure that your preparation is systematic and fully exam oriented.

Summary

Quantitative aptitude experts at Vidya Guru coaching center have written this quality article for candidates who aspire to appear in IBPS Bank PO/Clerk exams. For getting more inputs from them, you can write to vidyagurudelhi@gmail.com.