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

Given:

p(x) = 2x³ – 3x² – 2x + 3 and q(x) = 3x² + 8x + 5

Concept:

H.C.F. of two or more equations is the greatest factor that divides each of them exactly.

Calculation:

The factors of p(x) = 2x³ – 3x² – 2x + 3

⇒ x² × (2x – 3) – 1 × (2x – 3)

⇒ (x² – 1) × (2x – 3)

⇒ (x – 1) × (x + 1) × (2x – 3)

And, the factors of q(x) = 3x² + 8x + 5

⇒ 3x² + 5x + 3x + 5

⇒ x × (3x + 5) + 1 × (3x + 5)

⇒ (3x + 5) × (x + 1)

Given:

Number 1: 6

Number 2: 29

Formula Used:

Least Common Multiple (LCM) of a and b = (a × b) / Highest Common Factor (HCF) of a and b

Highest Common Factor (HCF) of 6 and 29 = 1 (since 6 and 29 are co-prime numbers)

Calculation:

LCM(6, 29) = (6 × 29) / HCF(6, 29)

LCM(6, 29) = (6 × 29) / 1

LCM(6, 29) = 174

HCF(6, 29) = 1

X = LCM(6, 29) = 174

Y = HCF(6, 29) = 1

X + 4Y = 174 + 4 × 1

X + 4Y = 174 + 4

X + 4Y = 178

The value of (X + 4Y) is 178.

Given:

Numbers: , ,

Formula Used:

HCF of fractions = HCF of numerators / LCM of denominators

Calculation:

Convert mixed fractions to improper fractions:

 = (4 × 5 + 1)/5 = 21/5

 = (5 × 7 + 2)/7 = 37/7

 = (7 × 9 + 4)/9 = 67/9

Extract numerators and denominators:

Numerators: 21, 37, 67

Denominators: 5, 7, 9

Find HCF of numerators:

HCF(21, 37, 67) = 1 (since all are co-prime)

Find LCM of denominators:

LCM(5, 7, 9):

Prime factorization:

5 = 5¹

7 = 7¹

9 = 3²

LCM = 5¹ × 7¹ × 3² = 315

HCF = HCF of numerators / LCM of denominators

HCF = 1 / 315

The HCF of , , and is 1/315 .

Given:

The first lightship flashes every 24 seconds.

The second lightship flashes every 60 seconds.

The third lightship flashes every 132 seconds.

All three flash together at 8:00 a.m.

Formula Used:

To find the next time all three lightships flash together, calculate the Least Common Multiple (LCM) of the three time periods.

Next Time = Starting Time + (LCM of time periods in seconds).

Calculation:

Find LCM of 24, 60, and 132:

Prime factorization:

24 = 2³ × 3

60 = 2² × 3 × 5

132 = 2² × 3 × 11

LCM = Highest powers of all prime factors = 2³ × 3 × 5 × 11

⇒ LCM = 8 × 3 × 5 × 11

⇒ LCM = 1320 seconds

Convert 1320 seconds to minutes:

1320 ÷ 60 = 22 minutes

Next flash time = 8:00 a.m. + 22 minutes

⇒ 8:22 a.m.

The three lightships will next flash together at 8:22 a.m.

Given:

LCM = 56 × HCF

HCF + LCM = 1710

Formula used:

HCF × LCM = Product of two numbers

Calculation:

Let, HCF = x and LCM = 56x

So, x + 56x = 1710

⇒ 57x = 1710

⇒ x = 1710/57 = 30

Then, HCF = 30 and LCM = 30 × 56 = 1680

One number = 240

Let, another number = a

Therefore, 30 × 1680 = 240 × a

⇒ a = (30 × 1680)/240 = 210

∴ The other number is 210

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.

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

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

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.