Practice Questions on LCM - Advanced

The LCM is the smallest number that all given integers can divide without leftovers. It's super useful for stuff like adding fractions or finding out when events sync up.

The LCM is the smallest positive integer that is a multiple of all the given integers.

For Example, Take 4 and 5. Their LCM is 20 because it's the smallest number divisible by both without a remainder.

1. Prime Factorization:

Break down each number into its prime factors (the basic building blocks of any number). Then, pick the highest power of each prime number found in either number and multiply them together.

2. Using Greatest Common Divisor (GCD):

Alternatively, divide the product of the two numbers by their GCD (the biggest number that divides both evenly). This gives you the LCM. This can be expressed mathematically as:

\text{LCM}(a, b) = \frac{|a \cdot b|}{\text{GCD}(a, b)}

Solved Examples on Least Common Multiple ( LCM ) - Advanced

Question 1: Three bells ring at intervals of 15 seconds, 20 seconds, and 30 seconds, respectively. If they all ring together at 10:00:00 hours, at what time will they ring together again?

Solution:

We need to find the least common multiple (LCM) of the times after which the bells ring: 15 seconds, 20 seconds, and 30 seconds.
LCM(15, 20, 30) = 60 seconds.
Calculate the next time the bells will ring together
The bells all ring together at 10:00:00 hours. Since the LCM of their timings is 60 seconds, the next time they will all ring together is 60 seconds after 10:00:00.
Adding 60 seconds (or 1 minute) to 10 : 00 : 00 gives:
10 : 00 : 00 + 1 : 00 = 10 :01 : 00.
The bells will ring together again at 10:01:00 hours.

Question 2: Four machines beep at intervals of 20 minutes, 30 minutes, 40 minutes, and 50 minutes, respectively. If they all beep together at 9:00 AM, at what time will they beep together again?

Solution:

We need to find the least common multiple (LCM) of the intervals 20, 30, 40, and 50.
LCM(20, 30, 40, 50) = 600 minutes.
Calculate the next time they will beep together:
600 minutes is equivalent to 10 hours (because 600 ÷ 60 = 10).
Adding 10 hours to 9:00 AM gives:
9:00 AM + 10 hours = 7:00 PM.
The machines will beep together again at 7:00 PM.

Question 3: The LCM of the two numbers is 2520, and their HCF is 35. If one of the numbers is 140, find the other number.

Solution:

We know the formula relating the LCM and HCF of two numbers a and b: LCM(a, b) × HCF(a, b) = a × b
We are given:
• LCM(a, b) = 2520
• HCF(a, b) = 35
• One of the numbers, a = 140
Let the other number be b.
Substituting the given values into the LCM-HCF formula:
2520 × 35 = 140 × b
88200 = 140 × b
b = 88200/140
b = 630
The other number is 630.

Question 4: The LCM of two numbers is 1260, and their HCF is 18. If one of the numbers is 72, find the other number.

Solution:

We know the formula relating the LCM and HCF of two numbers a and b: LCM(a, b) × HCF(a, b) = a × b
We are given:
LCM(a, b) = 1260,
HCF(a,b) = 18
a = 72
Let the other number be b.
Substituting the given values into the LCM-HCF formula:
1260 × 18 = 72 × b
22680 = 72 × b
b = 22680/72 =315
The other number is 315.

Question 5: If the ratio of two numbers is 5:6 and their H.C.F. is 2, find their L.C.M.

Solution:

Numbers are 5x and 6x.
x = 2 (from H.C.F.).
Numbers: 5x = 5 × 2 = 10, 6x = 6 × 2 = 12.
L.C.M of 10 and 12:
10 = 2 × 5,
12 = 2²× 3,
L.C.M = 2²× 3 × 5 = 60.

Question 6: The L.C.M. of two numbers is 360, and their H.C.F. is 12. If the sum of the two numbers is 132, find the difference between the two numbers.

Solution:

Given:
L.C.M. = 360,
H.C.F. = 12,
Sum of the two numbers = 132.
Let the two numbers be x and 132 − x.
Use the relationship: L.C.M. × H.C.F. = x × y
Substitute the given values:
360 × 12 = x · (132 - x).
4320 = x(132 -x)
x² - 132x + 4320 = 0
Fatorize the equation:
(x -72) (x - 60) = 0
x =72 or x =60
The two numbers are 72 and 60.
Difference of two numbers is |72 - 60| = 12.

Question 7: Find the LCM of the polynomials (5x²− 10x), (15x⁴ − 20x²), and (10x⁵− 10x⁴− 20x³).

Solution:

Given Polynomials: (5x²− 10x), (15x⁴ − 20x²), and (10x⁵− 10x⁴− 20x³).
Factors of First Polynomial: (5x²− 10x)
= 5x(x - 2)
Factors of Second Polynomial: (15x⁴ − 20x²)
= 5x²(3x² - 4)
Factors Third Polynomial: (10x⁵− 10x⁴− 20x³)
= 10x³(x² - x - 2)
= 10x³(x²+ x - 2x - 2)
=10x³(x (x + 1) -2(x + 1))
=10x³(x + 1)(x - 2)
LCM = 10x³(x + 1)(3x² - 4)(x - 2)

Question 8: Find the LCM of the polynomials (x²− 4), (x² - 5x- 6), and (x²+ x- 6).

Solution:

Given Polynomials:(x²− 4), (x² - 5x- 6), and (x²+ x- 6).
Factors First Polynomial: (x²− 4)
= (x -2)(x + 2)
Factors of Second Polynomials: (x² - 5x- 6)
=(x + 1)(x - 6)
Factors of Third Polynomials: (x²+ x- 6)
=(x - 2)(x + 3)
LCM = (x - 2)(x + 2)(x + 1)(x - 6 )(x + 3)

Also Check:

LCM - Practice Questions (Basic)
HCF and LCM – Aptitude Questions
LCM Quiz

Unsolved Practice Questions on LCM (Advanced)

Question 1: Three lights blink at intervals of 12 seconds, 16 seconds, and 24 seconds, respectively. If they all blink together at 6:00 PM, at what time will they blink together again?

Question 2: Five bells toll at intervals of 5 minutes, 10 minutes, 15 minutes, 20 minutes, and 25 minutes, respectively. If they all toll together at 8:00 AM, at what time will they toll together again?

Question 3: The LCM of two numbers is 840, and their HCF is 14. If one of the numbers is 42, find the other number.

Question 4: The LCM of two numbers is 990, and their HCF is 11. If one of the numbers is 66, find the other number.

Question 5: If the ratio of two numbers is 7:9 and their HCF is 4, find their LCM.

Question 6: The LCM of two numbers is 420, and their HCF is 14. If the sum of the two numbers is 154, find the difference between the two numbers.

Question 7: Find LCM of (2x²- 4x),(3x⁴- 12x²), (2x⁵- 2x⁴- 4x³).

Question 8: Find LCM of (x²- xy - 2y²), (2x²-xy-y²).

Answer key:
The three lights will blink together again at 6:00:00 PM + 48 seconds = 6:00:48 PM.
The five bells will toll together again at 8:00:00 AM + 300 minutes = 01:00:00 PM.
The other number is 280.
The other number is 165.
The LCM of the two numbers is 252.
The difference between the two numbers is 14.
LCM = 6x³(x - 2)(x + 2)(x + 1)
LCM = ( x + y)( x - 2 y)( 2 x + y)( x - y)

Practice Questions on LCM - Advanced

Important Formulas and Concepts Related to LCM

Solved Examples on Least Common Multiple ( LCM ) - Advanced

Also Check:

Unsolved Practice Questions on LCM (Advanced)

Answer key:

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