LCM MCQ Quiz - Objective Question with Answer for LCM - 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:

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:

The LCM of four consecutive numbers is 60.

The sum of the first two numbers is equal to the fourth number.

Find the sum of the four numbers.

Calculations:

Let the four consecutive numbers be n, n + 1, n + 2, and n + 3.

LCM of n, n + 1, n + 2, n + 3 = 60.

Given: n + (n+1) = (n+3).

⇒ 2n + 1 = n + 3.

⇒ n = 2.

Substitute n into the numbers:

Numbers = 2, 3, 4, 5.

Sum of the numbers:

⇒ 2 + 3 + 4 + 5 = 14.

∴ The correct answer is option (2).

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.

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.

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:

Numbers given = 0.126, 0.36 and 0.96

Concept Used:

LCM of fractions = \({ LCM (Numerators) \over HCF (Denominators)}\)

Calculation:

0.126 = \({126 \over 1000}\)

0.36 = \({360 \over 1000} \)

0.96 = \({960 \over 1000}\)

LCM (\({126 \over 1000}\), \({360 \over 1000} \), \({960 \over 1000}\)) = \({ LCM (126, 360, 960) \over HCF (1000, 1000, 1000)}\) = \({20160 \over 1000}\)

LCM (0.126, 0.36, 0.96) = 20.16

∴ The LCM of 0.126, 0.36 and 0.96 is 20.16.

Given values: A number divisible by 12, 15, and 25

Concept:

The number is the least common multiple (LCM) of 12, 15, and 25. The remainder when a number is divided by 9 is the same as the remainder when the sum of its digits is divided by 9.

Calculation:

⇒ The LCM of 12, 15, and 25 is 300. The smallest 6-digit number divisible by 300 is 100200.

⇒ Sum of digits of 100200 = 1 + 0 + 0 + 2 + 0 + 0 = 3

⇒ Remainder when 3 is divided by 9 = 3

Therefore, the remainder when the smallest 6-digit number divisible by 12, 15, and 25 is divided by 9 is 3.
Alternate Method LCM of 12, 15 and 25 is = 300.

and

300 ×  334 = 100200

So, the smallest 6-digit number which is divisible by 12, 15 and 25 is = 100200

When 100200 is divided by 9 then the remainder is:

100200 = 9 × 11133 + 3

remainder is = 3.

Calculation:

16 = 2 × 2 × 2 × 2 

10 = 2 × 5

12 = 2 × 2 × 3

27 = 3 × 3 × 3

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 = 2160

Before leaving remainder 9, the number required is 2169.

2169/13 = 166.84

So, this is not a divisible by 13.

2160 × 2 = 4320 + 9 = 4329

⇒ 4329 ÷ 13 = 333

Correct number is 4329.

Sum of the digit = 4 + 3 + 2 + 9

Given:

7, 2, 10, 4, 3, 12, 8, 4, 6, 4

Formula used:

Mode - The mode is the value that appears most frequently in a data set.

Mean = sum of data/number of data

Median = When data set is even = {(n/2)th + (n/2 + 1)th}/2

Calculation:

7, 2, 10, 4, 3, 12, 8, 4, 6, 4

Firstly data arrange in ascending order

⇒ 2, 3, 4, 4, 4, 6, 7, 8, 10, 12

The mode is the value that appears most frequently in a data set.

⇒ Mode = 4

Mean = sum of data/number of data

⇒ (2 + 3 + 4 + 4 + 4 + 6 + 7 + 8 + 10 + 12)/10

⇒ 60/10

⇒ 6

Median = When data set is even = {(n/2)th + (n/2 + 1)th}/2

⇒ {(10/2)th + (10/2 + 1)th}/2

⇒ (5th + 6th)/2

here 5th term is 4 and 6th term is 6

⇒ (4 + 6) /2

⇒ 10/2

⇒ 5

LCM of Median, Mean and Mode

⇒ LCM of 5, 6, 4

⇒ 3 × 4 × 5

⇒ 60

∴ The LCM of Median, Mean and Mode is 60.

 Confusion PointsHere the value of 5th term and 6th term is divided by 2 during Median calculation.

Given:

We need to find the sum of numbers in the range of 400 - 600 such that they are divisible by each 6, 12 and 16. 

Concept:

LCM (Lowest Common Multiple)

Calculation:

LCM (6, 12, 16) = 48 

The required numbers will be in the form of 48k, where k is a natural number.

For k = 9, 48k = 48 × 9 = 432 

For k = 10, 48k = 48 × 10 = 480 

For k = 11, 48k = 48 × 11 = 528

For k = 12. 48k = 48 × 12 = 576

∴ The sum of these 4 numbers that is, 432, 480, 528, and 576 is 2016.

Given:

The toll intervals of 6 bells = 2, 4, 6, 8, 10 and 12 seconds respectively

Concept:

L.C.M. of the least common multiple of the two or more numbers

Calculation:

Let the total number of times the bell will commence 

The LCM of 2, 4, 6, 8, 10, and 12 = 120 seconds = \({120\over 60}\ =\ 2\ minutes \)

They will commence tolling = 1 + \({30\over 2}\) = 1 + 15 = 16

∴ The required result will be 16.