LCM and HCF MCQ Quiz - Objective Question with Answer for LCM and HCF - Download Free PDF

Given:

Fractions: 1/3, 7/6, 5/9, 4/27, 8/15

Formula used:

The LCM of fractions is calculated as:

LCM of fractions = LCM of numerators / HCF of denominators

Calculations:

Step 1: Identify the numerators and denominators:

Numerators: 1, 7, 5, 4, 8

Denominators: 3, 6, 9, 27, 15

Step 2: Calculate LCM of numerators:

LCM(1, 7, 5, 4, 8) = 280

Step 3: Calculate HCF of denominators:

HCF(3, 6, 9, 27, 15) = 3

Step 4: Calculate the LCM of fractions:

LCM = LCM of numerators / HCF of denominators

LCM = 280 / 3

The LCM of the given fractions is 280/3.

Given:

Intervals of tolling for the five bells: 6, 5, 7, 10, and 12 seconds.

Formula Used:

The number of times the bells toll together is determined by the least common multiple (LCM) of their intervals within the given time frame.

LCM = Least Common Multiple

Calculation:

LCM of 6, 5, 7, 10, and 12

Prime factorization:

6 = 2 × 3

5 = 5

7 = 7

10 = 2 × 5

12 = 2² × 3

LCM = 2² × 3 × 5 × 7

⇒ LCM = 4 × 3 × 5 × 7

⇒ LCM = 420 seconds

1 hour = 3600 seconds

Number of times they toll together in 1 hour:

Excluding the start:

3600 / 420 = 8.57

⇒ 8 times

The correct answer is option 2

Given:

GCD(a, b) = 4

LCM(a, b) = 400

Formula used:

Product of two numbers = GCD × LCM

a × b = GCD(a, b) × LCM(a, b)

Let a = 4m and b = 4n, where GCD(m, n) = 1.

Calculation:

a × b = GCD(a, b) × LCM(a, b)

⇒ 4m × 4n = 4 × 400

⇒ 16mn = 1600

⇒ mn = 100

Find pairs (m, n) such that GCD(m, n) = 1 and mn = 100:

Possible pairs are:

(m, n) = (1, 100), (100, 1), (4, 25), (25, 4).

Each pair gives distinct values of a and b:

(a, b) = (4 × 1, 4 × 100) = (4, 400)

(a, b) = (4 × 100, 4 × 1) = (400, 4)

(a, b) = (4 × 4, 4 × 25) = (16, 100)

(a, b) = (4 × 25, 4 × 4) = (100, 16)

∴ The number of pairs of (a, b) is 2.

The correct answer is option (3).

Given:

Two numbers x and y are co-primes.

Formula used:

LCM of two co-prime numbers = Product of the numbers.

Calculation:

LCM of x and y = x × y

⇒ LCM = xy

∴ The correct answer is option (4).

Given:

The L.C.M. of ¹/₄ and ²/₅

Formula used:

LCM of fractions = LCM of numerators / HCF of denominators

Calculations:

Numerators: 1 and 2

Denominators: 4 and 5

LCM of 1 and 2 = 2

HCF of 4 and 5 = 1

⇒ LCM = LCM of numerators / HCF of denominators

⇒ LCM = 2 / 1

⇒ LCM = 2

∴ The correct answer is option (4).

Given:

Length of timber₁ = 143 m

Length of timber2 = 78 m

Length of timber3 = 117 m

Calculation:

Greatest possible length of each plank = HCF of 143, 78 and 117

143 = 13 × 11

78 = 13 × 2 × 3

117 = 13 × 3 × 3 

HCF is 13

∴ Greatest possible length of each plank is 13 m.

GIVEN:

Four bells ring simultaneously at starting and an interval of 6 sec, 12 sec, 15 sec and 20 sec respectively.

CONCEPT:

LCM: It is a number which is a multiple of two or more numbers.

CALCULATION:

LCM of (6, 12, 15, 20) = 60

All 4 bells ring together again after every 60 seconds

Now,

In 2 Hours, they ring together = [(2 × 60 × 60)/60] times + 1 (at the starting) = 121 times

∴ In 2 hours they ring together for 121 times

Mistake Points

In these type of question we assume that we have started counting the time after first ringing. Due to this when we calculate the LCM it gives us the ringing at 2nd time not the first time. So, we needed to add 1.

Given:

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Calculation:

Four bells ringing timing is 12 sec, 15 sec, 20 sec,30 sec 

Now we have to take LCM of time interval

⇒ LCM of (12, 15, 20, 30) = 60

Total seconds in 8 hours = 8 × 3600 = 28800

Number of times bell rings = 28800/60

⇒ Number of times bell rings = 480

If four bells ring together in starting

⇒ 480 + 1 

∴ The bell ringing 481 times in 8 hours.

Mistake PointsThe bells start tolling together, the first toll also needs to be counted, that is the number of times of tolling since the first time.

We know that,

product of two numbers = L.C.M × H.C.F of those numbers

Let the second number be x.

24 × x = 168 × 6

x = 6 × 7

x = 42

Given:

24 mango trees, 56 apple trees and 72 orange trees have to be planted in rows such that each row contains the same number of trees of one variety only.

Calculations:

There are 24 mangoes trees, 56 apple trees & 72 Orange trees.

To get the minimum number of rows, we need maximum trees in each row.

In each row, we need the same number of trees

So we need to calculate HCF

HCF of 24, 56 & 72

⇒ 24 = 2³ × 3

⇒ 56 = 2³ × 7

⇒ 72 = 2³ × 3²

HCF = 2³ = 8

Number of minimum rows = (24 + 56 + 72)/8 = 152/8

⇒ 19

∴ The correct choice will be option 3.

Given:

HCF = 24

LCM = 168

Ratio of numbers = 1 ∶ 7.

Formula:

Product of numbers = LCM × HCF

Calculation:

Let numbers be x and 7x.

x × 7x = 24 × 168

⇒ x² = 24 × 24

⇒ x = 24

∴ Larger number = 7x = 24 × 7 = 168.

Given:

The number between 550 and 700 such that when they are divided by 12, 16, and 24, leave the remainder 5 in each case

Concept Used:

LCM is the method to find the Least Common Multiples

Calculations:

⇒ LCM of 12, 16, and 24 = 48

Multiple of 48 bigger than 550 which leaves remainder 5 are

⇒ 1st Number = 48 x 12 + 5 = 581

⇒ 2nd Number = 48 x 13 + 5 = 629

⇒ 3rd Number = 48 x 14 + 5 = 677

⇒ Sum of these numbers are = 581 + 629 + 677 = 1887

⇒ Hence, The sum of the numbers are 1887.

Shortcut Trick Option elimination method:  Subtract the remainder of 5 in every no means in the option 15 we have to subtract because the sum of the three numbers is given.

In this case only 3, no is a possible case

So we have to subtract 15 and then check the divisibility of 16 and 3.

Given:

H.C.F of numbers = 13

LCM of numbers = 585

Calculation:

 Let the number be 13a and 13b where a and b are co-prime.

L.C.M of 13a and 13b = 13ab

According to question, 13ab = 585

⇒ ab = 45

⇒ ab = 5 × 9

⇒ a = 5 and b = 9 or a = 9 and b = 5

⇒ First number = 13a

⇒ First number = 13 × 5

⇒ First number = 65

⇒ Second number = 13b

⇒ Second number =13 × 9

⇒ Second number = 117

Required difference = 117 - 65 = 52

∴ Required difference = 52

Formula used:

n(A∪B) = n(A) + n(B) - n(A∩B)

Calculation:

On dividing 100 by 3 we get a quotient of 33

The number of multiple of 3, n(A) = 33

On dividing 100 by 4 we get a quotient of 25

The number of multiple of 4, n(B) = 25

LCM of 3 and 4 is 12

On dividing 100 by 12 we get a quotient of 8

The number of multiple of 12, n(A∩B) = 8

The number which is multiple of 3 or 4 = n(A∪B)

Now, n(A∪B) = n(A) + n(B) - n(A∩B)

⇒ 33 + 25 - 8

⇒ 50

∴ The total number multiple of 3 or 4 is 50

Concept used:

LCM of Fraction = LCM of Numerator/HCF of Denominator

Calculation:

​​ = 

⇒ LCM of (1, 5, 5) = 5

⇒ HCF of (2, 6, 4) = 2

⇒  = 5/2

∴ The correct answer is 5/2.

Mistake Points Please note that LCM means the lowest common multiple. LCM is the lowest number which is completely divisible by all given numbers(2/4, 5/6, 10/8). 

In these types of questions, make sure that you reduce the fractions to their lowest forms before you use their formulae, otherwise, you may get the wrong answer.

If we don't reduce the fractions to their lowest forms then LCM is 5 but the LCM of these 3 numbers is 5/2.