Quantitative Aptitude for Bank Exams – H.C.F and L.C.M

Quantitative Aptitude for Bank Exams H.C.F and L.C.M

Quantitative Aptitude for Bank Exams – H.C.F and L.C.M

H.C.F and L.C.M for Quantitative Aptitude for Bank Exams – The concept of H.C.F and L.C.M is important from competitive exams point of view in which are given two or more numbers of which we have to find Highest Common Factor and Least Common Multiple which can be calculated either by Factorization method or division method.

  1. What are the Factors and Multiples:

If a number x divides another number y exactly, then we can say is the factor of y.

Example: Factors of 12 are 1, 2, 3, 4, 6, and 12.

On the other hand, multiple of a number is its table itself.

Example: Multiples of 4 are 4, 8, 12, 16, …..

 

  1. Highest Common Factor (H.C.F.) or Greatest Common Divisor (G.C.D.):

The H.C.F. of two or more than two numbers is the greatest number that divides each of the numbers exactly.

There are two methods of calculating the H.C.F. of numbers:

  • Factorization Method: In this method, all the given numbers are expressed as the product of their prime factors & then calculating the product of its least powers of common prime factors will give H.C.F.
  • Division Method: Divide the larger by the smaller one then; divide the divisor by the remainder. Repeat this process of dividing the preceding number by the remainder till zero is obtained as remainder. The last divisor is our H.C.F.

 

Calculating the H.C.F. of more than two numbers: Let us assume we have to find the H.C.F. of three numbers, so, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number. In the same way, we can get the H.C.F. of more than three numbers.

 

  1. Least Common Multiple (L.C.M.):

L.C.M is the least number which is exactly divisible by each one of the given numbers is called their L.C.M.

There are two methods of calculating the L.C.M. of numbers:

  • Factorization Method: Expressing each one of the given numbers into the product of its prime factors. Then, a product of the highest powers of all the factors gives L.C.M.
  • Division Method (short-cut): Arrange the given numbers in a row in any order & divide them by the smallest number which divides at least two of the given numbers exactly and carries forward the numbers which are not divisible. Repeat the same process till no number is further divisible except no 1. The product of the divisors and the undivided numbers is the L.C.M. of the given numbers.

 

  1. A product of H.C.F. and L.C.M = Product of two numbers.

 

  1. Co-primes: Co-primes are set of two numbers whose H.C.F. is 1.

 

 

  1. C.F. and L.C.M. of Fractions:

1. H.C.F. = H.C.F. of Numerators/L.C.M. of Denominators

2. L.C.M. = L.C.M. of Numerators/H.C.F. of Denominators

 

  1. C.F. and L.C.M. of Decimal numbers:

H.C.F and L.C.M of decimal numbers can be calculated by converting the decimal numbers into fractions & then following the same approach of finding the H.C.F and L.C.M of fractions as given above.

 

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Let us now discuss some questions based on this concept:

 

Example: Find the L.C.M of 24, 36 and 40.

2 | 24 – 36 – 40

——————–

2 | 12 – 18 – 20

——————–

2 | 6 –   9 – 10

——————-

3 | 3 –   9 – 5

——————-

| 1 –   3 – 5

 

L.C.M.  = 2 x 2 x 2 x 3 x 3 x 5 = 360.

 

Example: The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

Sol: As Product of two numbers = H.C.F x L.C.M

275 x a = 11 x 7700

So, a (11 X 7700)/275 =  308

So, the other number = 308

 

Example: The H.C.F of 9/10, 12/25, 18/35, 21/40 is:

 

Sol: Required H.C.F = (H.C.F of 9, 12, 18, 21 )/(L.C.M of 10, 25, 35, 40) = 3/1400

 

Example: Three numbers are in the ratio of 3: 4: 5 and their L.C.M. are 2400. Find their H.C.F:

Sol: Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = (3 x 4 x 5) x = 60x.

So, 60x = 2400 or x = 40.

The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

 

Example: Six bells start ringing together and ring at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. How many times will they ring together in 30 minutes?

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

So, the bells will ring together after every 120 seconds (2 minutes).

In 30 minutes, they will ring together 30/2 + 1 = 16 times

 

Example: The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively is:

Sol: Required number = H.C.F. of (1657 – 6) and (2037 – 5)

= H.C.F. of 1651 and 2032 = 127.

 

Example: The least number, which when divided by 2, 3, 5 and 7 leaves a remainder of 8 in each case:

Sol: Required number = (L.C.M. of 2, 3, 5, 7) + 8

= 210 + 8

= 218.

Quantitative Aptitude for practice papers on H.C.F and L.C.M for Banking & Insurance Exam.

 

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