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LCM and HCF MCQ for all Competitive exams with detailed explanations

LCM and HCF

The highest common factor (HCF) of two or more numbers is the largest number that divides each number without leaving a remainder. The least common multiple (LCM) of two or more numbers is the smallest number that is divisible by each of the numbers.

You can use the Euclidean algorithm to find the HCF of two numbers. This algorithm involves repeatedly dividing the larger number by the smaller number and taking the remainder. The last non-zero remainder is the HCF.

To find the LCM of two numbers, you can use the formula LCM = (a * b) / HCF, where a and b are the two numbers.

The HCF and LCM are important concepts in number theory and have many applications, such as finding the greatest common divisor of two polynomials, simplifying fractions, and solving systems of linear equations.

HCF and LCM Formula

HCF (Highest Common Factor):

The HCF or GCD (Greatest Common Divisor) of two or more numbers is the largest positive integer that divides each given number without leaving a remainder.

The formula for HCF:

The HCF of two numbers, a and b, can be found using the Euclidean algorithm:

Divide the larger number (a) by the smaller number (b) to get the quotient (q) and the remainder (r).
If the remainder (r) is 0, then the smaller number (b) is the HCF.
If the remainder (r) is not 0, then repeat the process with the smaller number (b) and the remainder (r).

Example:

Find the HCF of 12 and 18.

Divide 18 by 12 to get the quotient 1 and the remainder 6.
Since the remainder is not 0, repeat the process with 12 and 6.
Divide 12 by 6 to get the quotient 2 and the remainder 0.
Since the remainder is 0, the HCF of 12 and 18 is 6.

LCM (Least Common Multiple):

The LCM of two or more numbers is the smallest positive integer that is divisible by each of the given numbers.

The formula for LCM:

The LCM of two numbers, a and b, can be found using the formula:

LCM(a, b) = (a * b) / HCF(a, b)

Example:

Find the LCM of 12 and 18.

First, find the HCF of 12 and 18 using the Euclidean algorithm. The HCF is 6.
Then, substitute the values of a, b, and HCF into the LCM formula:

LCM(12, 18) = (12 * 18) / 6 = 36

Therefore, the LCM of 12 and 18 is 36.

MCQ Questions on LCM & HCF

1. The sum of the H.C.F. and L.C.M of two numbers is 680 and the L.C.M. is 84 times the H.C.F. If one of the number is 56, the other is :

a) 84                     b) 8                       c) 12                     d) 96


Correct Answer: (d)
Let HCF be h and LCM be l.
Then, l = 84h and l + h = 680
→ 84h + h = 680
→ h = 680/85 = 8
∴ l = 680 – 8 = 672
∴ Other number = (672×8)/56 = 96

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2. The smallest five-digit number which is divisible by 12,18 and 21 is :

a) 10224              b) 10080              c) 30256              d) 50321


Correct Answer: (b)

2 12 18 21 — divide by 2
3 6 9 21 — divide by 3
2 3 7

LCM of 12, 18 and 21 = 2 × 3 × 2 × 3 × 7 = 252
Of the options, 10080 ÷ 252 = 40

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3. If A and B are the H.C.F. and L.C.M. respectively of two algebraic expressions x and y, and A + B = x + y, then the value of A3 + B3 is

a) x3 – y3              b) y3                      c) x3                      d) x3 + y3


Correct Answer: (d)
Let no. are x and y and HCF = A, LCM = B
Using Rule, we have
xy = AB
⇒ x + y = A + B (given) …(i)
(x–y)2 = (x+y)2 – 4xy
or, (x–y)2 = (A+B)2 – 4 AB
(x–y)2 = (A–B)2
(x–y) = A – B …(ii)
Using (i) and (ii), we get
x = A and y = B
∴ A3+B3=x3+y3

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4. A number between 1000 and 2000 which when divided by 30, 36 and 80 gives a remainder of 11 in each case is

a) 1451                b) 1712                c) 1641                 d) 1523

Correct Answer: (a)

We find LCM of 30, 36 and 80.
2 30, 36, 80 — divide by 2
2 15, 18, 40 — divide by 2
3 15, 9, 20 — divide by 3
5 5, 3, 20 — divide by 5
1, 3, 4

LCM = 2 × 2 × 3 × 3 × 4 × 5 = 720
∴ Required number = 2 × 720 + 11 = 1440 + 11 = 1451

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5. The sum of a pair of positive integers is 336 and their H.C.F. is 21. The number of such possible pairs is

a) 2                       b) 4                       c) 3                       d) 5

Correct Answer: (b)
Let the numbers be 21x and 21y
where x and y are prime to each other.
21x + 21y = 336
21 (x + y) = 336
x + y = 336/21 = 16
∴ Possible pairs = (1, 15), (5, 11), (7, 9), (3, 13)

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