## TNPSC Aptitude and Mental Ability Study Materials:

Aptitude and Mental Ability questions are more important for the TNPSC Group 2 Prelims Exam. You will get 25 marks from that Aptitude and Mental Ability portion. On this page, TNPSC Group 2, 2a, and Group 4 Aptitude and Mental Ability Books Study Materials questions with answers are uploaded. Go through TNPSC Aptitude Shortcuts & tricks in order to save your time in the prelims exam.

Students who are preparing for the Group exam concentrate more on the maths part. you will easily score more marks in the Mental Ability part. For students’ benefit, we upload aptitude and mental ability English and Tamil questions and answers in PDF for download.  A TNPSC aspirant can download and use it for the group prelims exam. kindly download TNPSC Maths PDF given below:

### Highest Common Factor (HCF):

• The Highest Common Factor (HCF) of the two numbers is the largest factor that is common to both of them. The Highest Common Factor of the numbers x and y can be written as HCF (x,y).
• The Highest Common Factor (HCF) is also called the Greatest Common Divisor (GCD) or the Greatest Common Factor (GCF).
– HCF (1, x) =1
– HCF (x, y) = x, if y is a multiple of x. For example, HCF (3, 6) = 3.
– If the HCF of two numbers is 1, then the numbers are said to be co-primes or relatively prime. Here, the two numbers can both be primes as (5, 7) or both can be composites as (14, 27) or one can be a prime, and the other a composite as (11, 12).

### Lowest Common Multiple (LCM):

• The Least Common Multiple of any two non-]ero whole numbers is the smallest or the lowest common multiple of both the numbers. The Least Common Multiple of the numbers x and y can be written as LCM (x,y).
• We can find the least common multiple of two or more numbers by the following methods.
1. Division Method 2. Prime Factorisation Method

Example: Find the LCM of 156 and 124.

Solution: By Division method
Step 1: Start with the smallest prime factor and go on dividing till all the numbers are divided as given below.

Step 2: LCM = product of all prime factors
= 2 x 2 x 3 x 13 x 31 = 4836
Thus, the LCM of 156 and 124 is 4836.

By Prime Factorisation method

Step 1: We write the prime factors of 156 and 124 as given below (use of divisibility test rules will also help).

156 = 2 x 78 = 2 x 2 x 39 = 2 x 2 x 3 x 13
124 = 2 x 62 = 2 x 2 x 31

Step 2: The product of common factors is 2 x 2 and also the product of the factors that are not common is 3 x 13 x 31.

Step 3: Now, LCM = product of common factors x product of factors that are not common
= (2 x 2) x (3 x 13 x 31) = 4 x 1209 = 4836
Thus, LCM of 156 and 124 is 4836.

TNPSC Aptitude and Mental Ability  – LCM and HCF topic question and answers solutions PDF, Notes and Study Materials were given below,

### 1. Fill in the blanks

(i) The HCF of 45 and 75 is ________.

Ans: 15

(ii) The HCF of two successive even numbers is _______.

Ans: 156

(iii) If the LCM of 3 and 9 is 9, then their HCF is _________.

Ans: 2

(iv) The LCM of 26, 39 and 52 is _________.

Ans: 3

(v) The least number that should be added to 57 so that the sum is exactly divisible by 2,3, 4, and 5 is _________.

Ans: 3

### 2. Say True or False

(i) The numbers 57 and 69 are co-primes. Ans: False

(ii) The HCF of 17 and 18 is 1. Ans: True

(iii) The LCM of two successive numbers is the product of the numbers. Ans: True

(iv) The LCM of two co-primes is the sum of the numbers. Ans: False

(v) The HCF of two numbers is always a factor of their LCM.Ans: True

### TNPSC HCF and LCM Objective Type Questions:

1. Which of the following pairs is co-prime”

a) 51, 63 b) 52, 91 c) 71, 81 d) 81, 99

2. The greatest 4 digit number which is exactly divisible by 8, 9, and 12 is

a) 9999 b) 9996 c) 9696 d) 9936

3. The HCF of the two numbers is 2 and their LCM is 154. If the difference between the numbers is 8, then the sum is

a) 26 b) 36 c) 46 d) 56

4. Which of the following cannot be the HCF of two numbers whose LCM is 120″

a) 60 b) 40 c) 80 d) 30

5. GCD of any two prime numbers is __________

a) −1  b) 0  c) 1  d) 2

6.) The remainder when ( x2-2x+7) is divided by ( x + 4) is

a) 28 b) 31 c) 30 d) 29

7. Euclid’s division lemma states that for positive integers a and b, there exist unique integers q and r such that a b = +q r, where r must satisfy.

(A) 1 <<r b (B) 0 <<r b (C) 0 ≤ r< b (D) 0 < ≤r b

8. Using Euclid’s division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are

(A) 0, 1, 8 (B) 1, 4, 8 (C) 0, 1, 3 (D) 1, 3, 5

9. If the HCF of 65 and 117 is expressible in the form of 65 117 m – , then the value of m is

(A) 4 (B) 2 (C) 1 (D) 3

10. The sum of the exponents of the prime factors in the prime factorization of 1729 is

(A) 1 (B) 2 (C) 3 (D) 4

11. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

(A) 2025 (B) 5220 (C) 5025 (D) 2520

12. If ( x-6) is the HCF of x²-2x-24 and x²-kx-6 then the value of k is

(A) 3 (B) 5 (C) 6 (D) 8

13. 3y – 3/y ÷7y−7/3y² is

(A) 9y/7     (B) 9y³/(21y-21)     (C) 21y²-42y+21/3y³   (D) 7(y²–2y+1)/y²

14. y2 +1/y2 is not equal to

(A) y4+1/y2   (B)(Y+1/Y)²     (C) (Y-1/y)²+2    (D) (Y-1/y)²-2

15. x/x²-25 – 8/x²+6x+5 gives

(A) x²-7x+40/(x-5)(x+5)               (B) x²+7x+40/(x-5)(x+5)(x+1)

(C)  x²-7x+40/(x²-25)(x+1)    (C) x²+10/(x²-25)(x+1)