**HCF and LCM Aptitude Question and Answer – Formulas & Tricks**

**HCF(Highest Common Factor) Methods**

Highest Common Factor (HCF) or Greatest Common Divisor(GCD) or Greatest Common Measure (GCM) of two or more numbers is the greatest number which divides each of them exactly.

**Division Method**: This method uses successive division to find the HCF. To find the HCF of two numbers N1 and N2, the smaller of the two, Say N1, divides N2. The remainder R1 then divides N1 followed by the remainder R2 dividing R1, and so on till a remainder of 0 is reached. The last divisor is HCF.**Prime Factorization Method:**Express each one of the given numbers as a product of prime numbers. The product of the least powers of common prime factor gives HCF.**Fractions Method:**HCF for Fractions = (HCF of Numerators / LCM of Denominators)

**HCF Tricks**

- Find HCF of 12 and 16, Find
**the difference b/w 12 and 16**.**The difference is 4**. Now, check whether the**numbers are divisible by the difference**. 12 and 16 are divisible by 4. Hence The**HCF is 4** - Find HCF of 18 and 22, Find the difference between 18 and 22. The difference is 4. Now, check whether the numbers are divisible by the difference. Both 18 and 22 are not divisible y 4. So take the factors of the difference. The factor of 4 are 2*2*1. Then, Check the number are divisible by factors. 18 and 22 are divisible by factor 2. Hence the HCF is 2.
- Find HCF of 10, 14 and 34, Find the difference between 10 and 14. The difference is 4. Now find the difference between 14 and 34. The difference is 20. Now find the difference between 10 and 34. The difference is 24. Out of 4, 20 and 24, 4 is the least difference. So, we take the number 4 and check whether the numbers are divisible. 10, 14 and 34 are not divisible by 4. So take the factors of the difference. The factors of 4 are 2*2*1. Now, check whether the numbers are divisible by the factors. 10, 14 and 34 are divisible by factor 2. Hence the HCF is 2.
- Find HCF of 15, 25 and 35. Find the difference between 15, 25, and 35. The least difference is 10. But 10 is not divisible by 15, 25, and 35. So, we take the factor of the difference. The factor of 10 are 5*2*1. Now, check whether the number are divisible by the factors. 15, 25 and 35 are divisible by 5.
- Find HCF of 20, 36 and 50. Find the difference between 20, 36 and 52. The least difference is 14. But 14 is not divisible by 20, 36 and 52. So, take the factor of the difference. The factors of 14 are 7*2*1. Now, check whether the numbers are divisible by the factors. 20, 36 and 50 are divisible by 2. Hence the HCF is 2.

**LCM(Least Common Method)** **Methods**

Least Common Methods (LCM) of two or more numbers is the smallest number that is a multiple of all the numbers.

**Prime Factor Method**: We can find LCM using Prime Factorization Method in the following steps- Express each number as a product of prime factors
- LCM = The product of highest powers of all prime factors.

**Division Method**: We can find LCM using Division Method in the following steps- write the given numbers in a horizontal line separated by commas.
- Divide the given numbers by the smallest prime number which can exactly divide at least two of the given numbers.
- Write the quotients and undivided numbers in a line below the first
- Repeat the process until we reach a stage where no prime factor is common to any two numbers in the row.
- LCM = Product of all the divisors and the numbers in the last line.

**Fractions Method**: LCM for Fractions = (LCM of Numerators / HCF of Denominators)

**LCM Ticks**

- Find LCM of 2, 4, 8, 16. Choose the largest number, The largest Number is 16 is divisible by all other remaining numbers. 16 is divisible by 2, 4, 8. Hence, the LCM is 16.
- Find LCM of 2, 3, 7 21. Choose the largest number. The largest number is 21. Check whether 21 is divisible by all other remaining numbers. 21 is divisible by 3 and 7 but not by 2. So multiply 21 and 2. The result is 42. Now, Check whether 42 is divisible by 2, 3, 7. Yes, 42 is divisible. Hence, the LCM is 42.
- Find the LCM of 3, 5, 15, 30. Choose the largest number. The largest number is 30. Check whether 30 is divisible by all other remaining numbers. 30 is divisible by 3, 5 and 15. Hence, the LCM is 30.
- Find the LCM of 8, 24, 48, 96. Choose the largest number. The largest number is 96. Check whether 96 is divisible by all other remaining numbers. 96 is divisible by 8, 24 and 48. Hence, the LCM is 96.
- Find the LCM of 12, 36, 60, 108. Choose the largest number. The largest number is 108. Check whether 108 is divisible by all other remaining numbers. 108 is divisible by 12, 36 and 60. Hence, the LCM is 108.

**HCF and LCM HCF and LCM Aptitude Question and Answer** **Formula and Tricks**

- Product of two numbers = Product of thir HCF and LCM.
- Two numbers are said to be co-prime if their HCF is 1.
- HCF = (HCF of Numerators / LCM of Denominators)
- LCM = (LCM of Numerators / HCF of Denominators)

HCF and LCM Aptitude Question and Answer **With Detailed Explanation** – Quantitative Aptitude

## HCF and LCM

Question 1 |

The Greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

9000 | |

9400 | |

9600 | |

9800 |

Question 1 Explanation:

**Answer: Option C**

__Explanation:__

Greatest Number of 4 digits is 9999.

L.C.M of 15, 25, 40 and 75 is 600.

On dividing 9999 by 600, the remainder is 399

Therefore, Required Number (9999 - 399) = 9600

Question 2 |

The Least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

3 | |

13 | |

23 | |

33 |

Question 2 Explanation:

**Answer: Option C**

__Explanation:__

LCM of 5, 6, 4 and 3 = 60.

On dividing 2497 by 60, the remainder is 37.

Therefore Number to be added = (60 - 37) = 23.

Question 3 |

A, B and C Start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the Starting point?

26minutes and 18 seconds | |

42minutes and 36 seconds | |

45minutes | |

46minutes and 12 seconds |

Question 3 Explanation:

**Answer: Option D**

__Explanation:__

LCM of 252, 308 and 198 = 2772.

So, A, B and C will again meet at the starting point in 2772 Sec

**That is, 46min 12Sec**

Question 4 |

Find the highest common factor of 36 and 84.

4 | |

6 | |

12 | |

18 |

Question 4 Explanation:

**Answer: Option C**

__Explanation:__

36 = 2

^{2}* 3

^{2}

84 = 2

^{2}* 3 * 7

Therefore, HCF = 2

^{2}* 3

**HCF = 12.**

Question 5 |

Two Numbers are in the ratio of 15:11. If the HCF of numbers is 13, Find the numbers.

75, 55 | |

105, 77 | |

15, 11 | |

195, 143 |

Question 5 Explanation:

**Answer: Option D**

__Explanation:__

Let the two numbers are 15x and 11x

HCF = x

x= 13

Therefore, the numbers are (15 * 13 and 11 * 13)

**Ans = 195 and 143**

Question 6 |

The LCM of two numbers is 7700, and their HCF is 11, if one of these numbers is 275, what is the other number?

279 | |

283 | |

308 | |

318 |

Question 6 Explanation:

**Answer: Option C**

__Explanation:__

Apply Formula

HCF * LCM = First Number * Second Number

11 * 7700 = 275 * Second Number

Therefore, Second Number = (11 * 7700) / 275

=(84700 / 275)

**Ans = 308**

Question 7 |

What is the greatest three-digit number which when divided by 6, 9 and 12 leaves a remainder of 3 in each case?

975 | |

996 | |

939 | |

903 |

Question 7 Explanation:

**Answer: Option A**

__Explanation:__

Greatest three digit number = 999

LCM of 6, 9 and 12

LCM = 2 * 2 * 3 * 3 = 36

On dividing 999 by 36,

Remainder = 27

Therefore, Greatest Three digit number divisible by 6, 9 and 12

=> (999 - 27) = 972

The required number is (972 + 3) = 975

Question 8 |

The LCM of two prime number x and y is 161. If x>y, find the value of 15y-7x.

-32 | |

-56 | |

38 | |

74 |

Question 8 Explanation:

**Answer: Option B**

__Explanation:__

The HCF of two Prime Numbers = 1

LCM / HCF = 161/1 = 161

161 = 7 * 23

The Ratio of numbers = 7 : 23

Hence, the expression gives 15 * 7 - 7 * 23 = -56

Question 9 |

The traffic lights at three different road crossings change after every 40 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 5 : 20 : 00hrs, then find the time at which they will change simultaneously.

5 : 28 : 00hrs | |

5 : 30 : 00hrs | |

5 : 38 : 00hrs | |

5 : 40 : 00hrs |

Question 9 Explanation:

**Answer: Option B**

__Explanation:__

Traffic lights at three different road crossing change after every 40sec, 72 sec and 108sec respectively

Therefore, Find the LCM of 40, 72 and 108.

LCM of 40, 72 and 108 = 1080

The traffic lights will change again after 080sec = 18min

The next simultaneous change takes place at 5 : 38 : 00hrs

Question 10 |

Three number are in the ratio 3:4:5 and their LCM is 2400. Their HCF is:

40 | |

80 | |

120 | |

200 |

Question 10 Explanation:

**Answer: Option A**

__Explanation:__

Let the numbers be 3x, 4x and 5x

Then, their LCM = 60x.

60x = 2400

x = 40

The Numbers are (3 * 40), (4 * 40) and (5 * 40)

Hence, Required HCF = 40.

There are 10 questions to complete.