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

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:

Alarm 1 interval = 90 seconds

Alarm 2 interval = 44 seconds

First beep together at 6:00 PM

Formula Used:

Time for next beep together = LCM (Least Common Multiple) of the intervals

Calculation:

LCM of 90 and 44:

Prime factorization of 90 = 2 × 3² × 5

Prime factorization of 44 = 2² × 11

LCM = Product of the highest powers of all prime factors

LCM = 2² × 3² × 5 × 11 = 1980 seconds

1980 ÷ 60 = 33 minutes and 0 seconds

Next beep together = 6:00 PM + 33 minutes

⇒ Next beep together = 6:33 PM

∴ The correct answer is option 4.

Given:

Time taken by the first person to complete one cycle = 120 seconds

Time taken by the second person to complete one cycle = 150 seconds

Time taken by the third person to complete one cycle = 80 seconds

Formula used:

To find when the three people will meet at the starting point, we need to calculate the LCM (Least Common Multiple) of the times they take to complete one cycle.

Calculation:

Prime factorizations:

120 = 2³ × 3 × 5

150 = 2 × 3 × 5²

80 = 2⁴ × 5

LCM = Highest power of each prime factor:

LCM = 2⁴ × 3 × 5² = 16 × 3 × 25 = 1200 seconds

∴ The three people will meet at the starting point after 1200 seconds (or 20 minutes).

Given:

Lengths to measure = 48 m, 72 m, 96 m, and 120 m

Formula Used:

To find the highest length that can measure these lengths exactly, we calculate their Highest Common Factor (HCF).

Calculation:

Find the HCF of 48, 72, 96, and 120:

Prime factorization:

48 = 2⁴ × 3

72 = 2³ × 3²

96 = 2⁵ × 3

120 = 2³ × 3 × 5

Common prime factors: 2 and 3

Lowest powers of common factors:

2³

3

HCF = 2³ × 3

⇒ HCF = 8 × 3

⇒ HCF = 24

The highest length that can measure 48 m, 72 m, 96 m, and 120 m exactly is 24 m.

Given:

Least Common Multiple (LCM) = 364

Greatest Common Factor (GCF) = 26

One number = 26

Formula Used:

LCM × GCF = Product of the two numbers

Calculation:

Let the other number be x.

Product of the two numbers = 26 × x

LCM × GCF = 364 × 26

⇒ 26 × x = 364 × 26

⇒ x = (364 × 26) / 26

⇒ x = 364

The other number is 364.

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

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:

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.

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:

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.

Concept used:

LCM of Fraction = LCM of Numerator/HCF of Denominator

Calculation:

​​\(\frac{2}{4}, \frac{5}{6}, \frac{10}{8}\) = \(\frac{1}{2}, \frac{5}{6}, \frac{5}{4}\)

⇒ LCM of (1, 5, 5) = 5

⇒ HCF of (2, 6, 4) = 2

⇒ \(\dfrac{LCM\; of\;(1,5,5)}{HCF\;of\;(2,4,6)}\) = 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.

Given:

Ratio of numbers = 7 ∶ 11

HCF = 28

Calculation:

Let the numbers be 7x and 11x

HCF of 7x and 11x is x

HCF = x = 28

The numbers will be 7 × 28 and 11 × 28

⇒ The numbers will be 196 and 308

Sum of numbers = 196 + 308

⇒ Sum of numbers = 504

∴ Sum of numbers is 504

Shortcut Trick Note that the sum of two numbers is asked.

Let the numbers be 7x and 11x.

Add the numbers:

⇒ 7x + 11x

⇒ 18x

Now see, the final number must be the multiple of 18, so in options only 504 is multiple of 18.

∴ The sum of two number is 504.