Explore topic-wise InterviewSolutions in Current Affairs.

Correct option is (B) \(\frac{n(n+1)(2n+1)}{6}\)

\(1^2+2^2+3^2\) + …… + \(n^2\)

= \(\frac{n(n+1)(2n+1)}{6}\)

 B) \(\frac{n(n+1)(2n+1)}{6}\)

Correct option is (A) \(\frac{n(n+1)}{2}\)

\(\sum n\) = \(\frac{n(n+1)}{2}\)

Correct option is   A) \(\frac{n(n+1)}{2}\)

(a) 17, 71

(b) 37, 73

(c) 79, 97

Correct option is  C) 9

Correct option is (B) 1331

(A) (1+1) - 5 = -3 is not divisible by 11.

(B) (1+3) - (3+1) = 0 is divisible by 11.

(C) (1+1) - (2+1) = 2 - 3 = -1 is not divisible by 11.

(D) (1+3) - (3+4) = 4 - 7 = -3 is not divisible by 11.

From divisibility test for 11, it is clear that 1331 is divisible by 11 while 151, 1211 and 1334 is not divisible by 11.

Correct option is  B) 1331

Since, the number is divisible by 12, it will also be divisible 1,2,3,4 and 6 are numbers other than 12 by which this number is also divisible.

(a) The sum of three odd numbers is even.False 

3 + 5 + 7 = 15. i,e. odd

(b) The sum of two odd numbers and one even number is even.True 

3 + 5 + 6 = 14 i,e = even

(c) The product of three odd numbers is odd.
True 

3 × 5 × 7 = 105 i.e, odd.

(d) If an even number is divided by 2, the quotient is always odd. False 

4 ÷ 2 = 2 i.e, even

(e) All prime numbers are odd.False; 

2 is even number.

(f) Prime numbers do not have any factors. False; 

1 and the numbers itself are factors of the number.

(g) Sum of two prime numbers is always even. False 

2 + 3 = 5 i,e odd

h) 2 is the only even prime number.True

(i) All even numbers are composite numbers. False, 

2 is a prime numbers.

(j) The product of two even numbers is always even.True; 

2 × 4 = 8, i.e even

Correct option is (C) 2n

Even numbers are those numbers which is divisible by 2.

\(\therefore\) General form of even number be 2n.

Correct option is  C) 2n

(a) 5 = 5 × 1= 5; 5 × 2 = 10;

 5 × 3 = 15; 5 × 4 = 20;

 5 × 5 = 25. 5, 10, 15, 2, 25

(b) 8 = 8 × 1 = 8; 8 × 2 = 16; 

8 × 3 = 24; 8 × 4 = 32; 

8 × 5 =40. 8,16,24,32,40

(c) 9 = 9 × 1 = 9; 9 × 2 = 18; 

9 × 3 = 27; 9 × 4 = 36; 

9 × 5 = 45. 9,18,27,36,45

Factors of 5 = 1, 5

Factors of 12 = 1, 2, 3, 4, 5, 6, 12

As the common factors of these number is 1, the given two numbers are co-prime and the number will also be divisible by their product , i,e 60 and the factors of 60.

i.e, 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

(a) 4,8, 12

Factors of 4 = 1 2 4

Factors of 8 = 1 2 4 8

Factors of 12 = 1 2 3 4 6 12

Common factors = 1 2 4

(b) 5, 15 and 25

Factors of 5 = 1 5

Factors of 15 = 1 3 5 15

Factors of 25 = 1 5 25

Common factors = 1 5

1 – b,

2 – d,

3 – a,

4 – f,

5 – e.

(a) The sum of two odd numbers is even

eg:- 1 + 5 = 6; 16 + 18 = 34

(b) The sum of two even numbers is even

eg:- 4 + 2 = 6; 10 + 18 = 28

(a) Factors of 18 = 1, 2, 3, 6, 9, 18

Factors of 35 = 1, 5, 7, 35

Common factors = 1

Therefore, the given two numbers are co-prime

(b) Factors of 15 = 1, 3, 5, 15

Factors of 37 = 1, 37

Common factors = 1

The given two numbers are co-prime

(c) Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30

Factors of 415 = 1, 5, 83, 415

Common factors = 1, 5

As these numbers have a common factors other than 1, the given two numbers arc not co-prime

(d) Factors of 17 = 1, 17

Factors of 68 = 1, 2, 4, 17, 34, 68

Common factors = 1, 17

As these numbers have a common factors other than 1, the given two numbers are not co-prime

(e) Factors of 216 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 17, 36, 54, 72, 108, 216

Factors of 215 = 1, 5, 43, 215

Common factors = 1

The given two numbers are prime.

(f) Factors of 81= 1, 3, 9, 27, 81

Factors of 16 = 1, 2, 4, 8, 16

Common factors = 1

Therefore the given two numbers are co-prime.

Multiples of 3 = 3, 6, 0, 12, 15, 18, 21, 24, 27, 30

Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40

Common multiples = 12, 24, 36, 48, 60, 72, 84, 96

(a) 6 and 8:-
Multiple of 6 = 6, 12, 18, 24 30, 36, 42, 48 54, 60, 66, 72

Multiple of 8 = 8, 16, 24 32, 40, 48 56, 64, 72 80
3 common multiples = 24, 48, 72

(b) 12 and 18:-

Multiples of 12 = 12, 24, 36, 48, 60, 72

Multiples of 18= 18, 36, 54, 72

3 Common multiples = 36, 72, 108

Correct option is (C) 45

(A) Sum of digits in, 87 = 8+7 = 15 is not divisible by 9.

(B) Sum of digits in, 33 = 3+3 = 6 is not divisible by 9.

(C) Sum of digits in, 45 = 4+5 = 9 is divisible by 9.

(D) Sum of digits in, 50 = 5+0 = 5 is not divisible by 9.

\(\therefore\) Among all given numbers, only 45 is divisible by 9.

Thus, 45 is a multiple of 9 in among all given numbers.

Correct option is  C) 45

Correct option is (D) None

\(\frac{a^5+b^5}{a-b}\) \(=a^4+a^3b+a^2b^2+ab^3+b^4+\frac{2b^5}{a-b}\)

Thus, \((a^5+b^5)\) is not completely divisible by (a-b).

Correct option is  D) None

Correct option is (A) 24

\(\because\) 24 = 24 \(\times\) 1,

24 = 12 \(\times\) 2,

24 = 8 \(\times\) 3,

24 = 6 \(\times\) 4.

\(\therefore\) 1, 2, 3, 4, 6, 8, 12 and 24 are the factors of 24.

Correct option is  A) 24

Correct option is  B) 10

Correct option is (B) 40

Smallest number which is divisible by 2,5 and 10

= LCM (2, 5, 10) = 10

\(\therefore\) Multiple of 10 is a number which is divisible by 2,5 and 10.

\(\because\) 40 is a multiple of 10.

\(\therefore\) 40 is divisible by 2,5 and 10.

Correct option is  B) 40

\(\over35a64\) is divisible by 3

And, if a number is divisible by 3 then sum of digits must be a multiple of 3.

i.e., 3 + 5 + a + 6 + 4 = multiple of 3

a + 18 = 0, 3, 6, 9, 12, 15…..

Here ‘a’ is a digit, where, ‘a’ can have values between 0 and 9.

a + 18 = 18 which gives a = 0.

a + 18 = 21 which gives a = 3.

a + 18 = 24 which gives a = 6.

a + 18 = 27 which gives a = 9.

∴ a = 0, 3, 6, 9

The given numbers are 985, 859 and 598.

The quotient obtained when the sum of these three numbers is divided by:

111

As we know that when the sum of three digit numbers in cyclic order is done and then divided by 111, quotient is sum of digits of a number.

Quotient = Sum of digits = 22

We know that when the sum of three digit numbers in cyclic order is done and then divided by sum of digits, quotient is 111.

Quotient = 111

Here, 3 × 37 = 111

∴ Quotient = 3 × (Sum of the digits) 

= 3 × 22 

= 66

The difference between the numbers 985 and 958 when divided by 9, 

As we know that when ten’s and unit’s place is interchanged we get quotient as a difference of unit’s and ten’s place.

∴ Quotient is 8 – 5 

= 3

Here our given numbers are 985, 859 and 598. 

Hence, the quotient obtained when the sum of these three numbers is divided by:

(i) 111

We know that when sum of three digit numbers, in cyclic order, is done, and then divided by 111, quotient is sum of digits of a number. 

Quotient = Sum of digits = 22

(2) 22 (Sum of digits) 

We know that when sum of three digit numbers, in cyclic order, is done, and then divided by sum of digits , quotient is 111. 

Quotient = 111

(3) 37

Here, 3 × 37 = 111 

∴ Quotient = 3 × ( Sum of the digits) = 3 × 22 = 66

Correct option is (A) 28

\(2^2+2^3+2^4\) = 4+8+16

= 28

Correct option is  A) 28

Correct option is  A) 3

Correct option is  D) 9

Correct option is (D) 1024

\(32^2=(30+2)^2\)

\(=30^2+2\times30\times2+2^2\)

= 900+120+4 = 1024

Correct option is D) 1024

The correct answer is option (b) 10x + y

Explanation:

The numbers are expressed as the sum of the product of it digits with the respective place value. So the generalised form of xy is 10x+y

Correct option is (C) 166815

Numbers from 11 to 1000 which are divisible by 3 are 12, 15, 18, ..... , 999.

\(\therefore\) Sum = 12+15+18+...+999

= (3+6+9+12+15+...+999) - (3+6+9)

= 3 (1+2+3+...+333) - 18

\(=3\times\frac{333\times334}2-18\)

\(=3\times333\times167-18\)

= 166833 - 18

= 166815

Correct option is  C) 166815

(i) (5 × 10) + (6 × 1); 

50 + 6 = 56 

(ii) (7 × 100) + (5 × 10)+ (8 × 1); 

700 + 50 + 8 = 758 

(iii) (6 × 1000) + (5 × 10) + (8 × 1); 

6000 + 50 + 8 = 6058 

(iv) (7 × 1000) + (6 × 1); 

7000 + 6 = 7006 

(v) (1 × 1000) + (1 × 10); 

1000 + 10 = 1010

39 = 30 + 9 = (3 × 10) + (9 × 1) 

52 = 50 +2 = (5 × 10) + (2 × 1) 

106 = 100 + 6 = (1 × 100) + (0 × 10)+(6 × 1) 

359 = 300 + 50 + 9 = (3 × 100) + (5 × 10) + (9 × 1) 

628 = 600 + 20 + 8 = (6 × 100) + (2 × 10) + (8 × 1) 

3458 = 3000 + 400 + 50 + 8 = (3 × 1000) + (4 × 100) + (5 × 10) + (8 × 1) 

9502 = 9000 + 500 + 2 = (9 × 1000) + (5 × 100) + (0 × 10) + (2 × 1) 

7000 = 7000 + 0 + 0 + 0 = (7 × 1000) + (1 × 100) + (0 × 10) + (0 × 1)

(1) Preparation of DNA sequences corresponding to A&B chain of human insulin.

(2) Introduce them in plasmid of E.coli. 

(3) Products of A&B Chains are separated. 

(4) Combine A&B chains by creating disulphide bonds

= \(\frac{4}{5} \times 15000\)

= 4 x 3000

= 12000

= \(\frac{4}{5} \times 15000\)

= 4 x 3000

= 12000

These conditions are satisfied exactly when we do as follows: First paint the central triangle in any one of the three colours. Next paint the remaining 3 triangles, with any one of the remaining two colours.

By the fundamental principle of counting, this can be done in 3 × 2 × 2 × 2 = 24 ways.

The term modernisation was coined by Daniel Lerner

Height of a conical vessel = 3.5 cm and Capacity of conical vessel is 3.3 litres or 3300 cm³ 

Now, 

We know, 

Volume of cone = \(\frac{1}{3}\)πr²h 

3300 = \(\frac{1}{3}\) x \(\frac{22}{7}\) x r² x 3.5 

or r² = 900 

or r = 30 

So, radius of cone is 30 cm 

Hence, diameter of its base = 2 Radius = 2 x 30 cm = 60 cm

Heterotrophic microbes naturally present in sewage are used, and vigorous growth of aerobic microbes as flocs uses organic matter in effluent and reduce BOD of waste water; other kinds of bacteria grow in it anaerobically and digest the bacteria and fungi called flocs (masses of bacteria associated with fungal filaments). As they digest flocs, a mixture of CH₄, H₂S, and CO₂ or biogas is evolved, which can be used as a fuel.

Answer is (a) [0,1]

Δ = |(1,0,0),(0,1,0),(0,0,1)| = 1.|(1,0),(0,1)| + 0|(0,0),(0,1)| + 0|(0,1),(0,0)|

= 1(1 - 0) + 0 + 0

= 1

Given: A(7, p, 2) and B(q, -2, 5), line segment joining these two points will be, AB = (q-7)i + (-2-p)j + 3k 

And the line segment joining C(2, -3, 5) and D(-6, -15, 11) will be, CD = -8i – 12j + 6k 

Then, the angle between these two line segments will be 0 degree. So, the cross product will be 0.

AB x CD = 0

((q-7)i + (-2-p)j + 3k) x ( -8i – 12j + 6k) = 0 

Thus, solving this we get,

p = 4 and q = 3

Mendeleev’s periodic law : 

“The physical and chemical properties of elements are the periodic function of their atomic masses.

Dmitri Mendeleev, John Newlands and Johann 

Doberiener were the scientists who made the classification of elements based on their atomic mass in the nineteenth century.

a) intelligent and playful

Correct option is: (C) both lanthanides and actinides

f-block elements (both Lanthanides and actinides) are also called rare earth elements.

(C) both lanthanides and actinides

c) humans and whales

Correct option is: (B) p-block

Aluminium belongs to Group - 13 elements. i.e. P-block elements

Correct option is (B) p-block

i. La and Hg both belongs to period 6. 

Therefore, 

n = 6.

ii. Elements of group 3 to group 12 belong to the d-block of the modem periodic table.

iii. The general outer electronic configuration of the d-block elements is ns^0-2 (n -1 )^1-10.

iv. Therefore, the outer electronic configuration of all ten elements from La to Hg is as given in the table below.

 [Note : There are 14 elements between La and Hf which are called lanthanides. 

Therefore, 

After La, electrons are filled in 4f subshell of lanthanide elements. 

Once all the 14 elements of lanthanide series are filled, next electron enters 5d subshell of Hf. 

Hence,

The outer electronic configurations of Hf to Hg often include completely filled 4f subshell. For example, the electronic configuration of Hf '5d²6s²' can also be written as '4f¹⁴5d²6s²'.]