Explore topic-wise InterviewSolutions in Current Affairs.

(B) Microcytic anaemia

Answer is (c) 3/4

Answer is (c) 13/24

Answer is (b) 4/3

Answer is (b) vector a is perpendicular to vector b

Answer is (b) 5/7

Answer is (b) vector(3i - j + 5k)

Answer is (a) 0

Step 1 : Framing the equation

Let one of the odd positive number be x. The other odd positive number will be (x + 2)

 The product of the numbers is x(x+2)=195

∴x² +2x−195=0 

Step 2: Solving the equation 

x² +2x−195=0

 x² +15x−13x−195=0∴(x+15)(x−13)=0 

x+15=0 or x−13=0∴x=−15 or x=13 

Step 3: Interpreting and finding the solution 

Since the numbers must be positive, x = -15 is not taken. 

∴x=13 and x+2=13+2=15 

∴ The two consecutive odd positive numbers are 13 and 15.

Let the numbers be x and y

A/Q,

x + y = 42 ........(i)

x - y = 16 ........,(ii)

add (i) and (ii) equations

2x = 58

x = 29

So, y = 42 - x

= 42 - 29

= 13

Therefore, x = 29

y = 13

Required sequence is 1²+1, 2²+1, 3²+1, 4²+1, ....

10^th term of sequence is a₁₀ = 10²+1 = 100+1 = 101.

Algebraic form of the sequence is a_n = n² + 1.

Answer is (c) - 5

\(\frac{2x+3}{(x+3)(x^2+4)}=\frac{A}{x+3}+\frac{Bx+C}{x^2+4}\)

⇒ 2x + 3 = A(x²+4) + (Bx+C) (x+3)

⇒ (A+B)x² + (3B+C)x + 4A+3C = 2x+3

By comparing corresponding components, we get

A+B = 0  ...(1)

3B+C = 2  ...(2)

4A+3C = 3  ...(3)

Put B = -A from equation (1) into equation (2), we get

-3A+C = 2

⇒ -9A+3C = 6  ...(4)

Subtract equation (4) from equation (3), we get

13A = -3

⇒ A = -3/13

∴ B = -A = 3/13

and C = 2 - 3B = 2 - 9/13 = 17/13

∴ \(\frac{2x+3}{(x+3)(x^2+4)}=\frac{-3/13}{x+3}+\cfrac{\frac{3}{13}x+\frac{17}{13}}{x^2+4}\)

Answer is (c) [(cosθ,sinθ),(-sinθ,cosθ)]

Direction ratios of given line are a = 2, b = 6 and c = -9

We know that the relation between Direction ratios and cosines is \(l=\frac{a}{\sqrt{a^2+b^2+c^2}},\) \(m=\frac{b}{\sqrt{a^2+b^2+c^2}},\) \(n=\frac{c}{\sqrt{a^2+b^2+c^2}}\)

∴ Direction cosines of given line are

\(l=\frac{2}{\sqrt{2^2+6^2+(-9)^2}}\) \(=\frac{2}{\sqrt{4+36+81}}\)\(=\frac{2}{\sqrt{121}}\) \(=\frac{2}{11},\)

\(m=\frac{6}{\sqrt{2^2+6^2+(-9)^2}}\) \(=\frac{6}{11}\)

and \(n=\frac{-9}{\sqrt{2^2+6^2+(-9)^2}}\) \(=\frac{9}{11}\)

Hence, direction cosines of given line are \(\frac{2}{11},\frac{6}{11}\,and\,\frac{-9}{11}.\)

Answer is (a) B'A'

Correct answer is: (d) 1/3

P=3/9=1/3

Correct answer is: (c) 4√3 cm

(altitude)²=(side)² -(side/2)²

=8² -4²= 64-16 =48

Altitude=4√3 cm

Correct answer is: (c) 17m

H²=P²+B²

H²=15²+8²

H=17m

Answer is (a) m = n

H² = P² + B²
H² = 15² + 8²
H = 17m
Thus, he 17m away from the starting point

So, option (c) 17 m is the correct answer.

Answer is (d) 0

Correct answer is: (c) 1

(tanθcosecθ)² -(sinθsecθ)²

=tan²θcosec²θ-sin²θsec²θ

=(sin²θ/cos²θ)x1/ sin²θ - sin²θx1/cos²θ

=(1- sin²Ɵ)/ cos²Ɵ= cos²Ɵ/ cos²Ɵ =1

Answer is (d) 5π/6

Answer is (d) 8x

Answer is (c) 50

Concept:

Let z = x + iy be a complex number.

• Modulus of z = \(\left| {\rm{z}} \right| = {\rm{}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
• arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{y}{x}} \right)\)
• Conjugate of z = x – iy

Calculation:

Let z = (i - i²)³

= (i - (-1))³ = (1 + i) 3

Using (a + b) 3 = a3 + b3 + 3a2b + 3ab2

⇒ z = 13 + i3 + 3 × 12 × i + 3 × 1 × i2

= 1 – i + 3i – 3

= -2 + 2i

So, conjugate of (1 + i) 3 is -2 - 2i

Concept:

If z = a + ib, then |z| = \(\rm \sqrt{a^2 + b^2}\)

Solution:

The given equation can be solved as:

iz3 + z2 - z + i = 0

⇒ iz3 + i²z + z2 + i = 0

⇒ iz(z² + i) + (z2 + i) = 0

⇒ (z² + i)(zi + 1) = 0

⇒ z² + i = 0 OR zi + 1 = 0

⇒ z2 = -i OR z = i

|z| = 1 in both the cases.

Concept:

Triangle inequality:

| z1 + z₂  | \(\leq\)| z₁ | + | Z₂| 

| cos α | \(\leq\) 1

Calculations:

Given, | z2 + 2z cos α | \(\leq\)| z2 | + | 2z cos α |

We know that, | cos α | \(\leq\) 1

⇒ | z2 + 2z cos α | \(\leq\)| z2 | + | 2z |

Put the value of |z| as  √3 - 1

⇒ | z2 + 2z cos α | < (√3 - 1)² + 2 (√3 - 1)

⇒ | z2 + 2z cos α | < 2 

Hence, If |z| < √3 - 1, then |z2 + 2z cos α| is less than 2

Given:

Numbers = 21, 84 and 49

Concept used: 

If a, b and c are three numbers then their HCF will be the multiplication of the common factors taken from their prime factorization. 

Calculations:

Prime factorization of 21, 84 and 49 can be written as,

21 = 3 × 7

84 = 3 × 4 × 7

49 = 7 × 7

HCF = 7

∴ The HCF of 21, 84 and 49 is 7.

Given 

Total Order = 19000 

Firm produces the product every day = 1000 

Formula Used 

Percentage = (actual/Total) × 100

Calculation 

⇒ Quantity of product to be completed = 19000 

⇒ Actual production per day = 1000 - 5% of 1000 = 1000 - 50 = 950 

⇒ No. of days to completed the order = 19000/950 = 20 days 

∴ the no. of days to complete the order by firm in 20 days 

Given:

Six digits number is 25a64b

Concept used:

Divisibility rule of 11, says, in a number when we add even place digits and subtract it from the sum of odd place digits, results will be either 0 or multiple of 11

Calculation:

In digits 25a64b, the sum of odd place digits = 2 + a + 4

⇒ 6 + a

the sum of even place digits = 5 + 6 + b

⇒ 11 + b

According to the divisibility rule of 11,

The sum of odd place digits – the sum of even place digits = 0

⇒ 6 + a – (11 + b) = 0

⇒ a – b = 11 – 6

⇒ a – b = 5

∴ The value of (a – b) is 5.

Given:

The number = 16129

Concept Used:

Prime factorization method is used.

Calculation :

Prime factorization of 16129,

16129 = 127 × 127

√16129 = √(127 × 127)

⇒ 127

∴ The square root of 16129 is 127.