Explore topic-wise InterviewSolutions in Current Affairs.

Set 1 :

(a) The poet drove away the painful thought of the distressing reality that her mother was getting old and she might die anytime. 

(b) When she looked out of the car, she saw young trees on the roadside, which appeared to be moving. She also saw a group of children, merrily rushing out of their homes to play. 

(c) As the poet passed through security check at the airport and happened to look at her mother, she was again haunted by the same fear of losing her to death. This shows that the joyful scene earlier didn’t help drive away the painful thought from her mind. 

(d) The merry children are symbolic of the exuberance of youth. The energetic and lively children present a contrast to the poet’s mother who has grown old and pale.

Set 2 : 

(a) The poet put away the thought of the-distressing reality of her mother getting old and of her impending death. 

(b) The ‘sprinting trees’ signify time that has passed at a fast pace. 

(c) The merry children epitomise bubbly youth. They represent the exuberance and liveliness of young age. 

(d) The poet makes use of these images to emphasise the contrast between old age and youth.

Set 3 :

(a) The poet Kamala Das looked out at young trees. 

(b) Seeing her aged mother, she felt insecure about the fact that she might be separated from her mother. The poet was also feeling guilty for neglecting her. She wondered if she would see her mother alive next time. However, she soon put these thoughts away. 

(c) The young sprinting trees symbolise happiness, strength and vigour which are the characteristics of youth in contrast to the dullness of old age.

(d) As the poet looked outside the window of her moving car, the trees appeared to be moving fast in the opposite direction. So, they are described as sprinting.

1. A) fast moving appearance

2. B) Simile

3. A) Poet’s fear of losing her old mother

4. D) to show the speed of the car

1. A) of her mother’s declining health

2. D) Both (A) and (B)

3. D) all of the above

4. A) Losing her mother

Statistics are the eyes of Govt. administration.

(a) The poet exhorts each one of us to count to twelve and then be quiet, silent and motionless for a brief moment. 

(b) He hopes to achieve and realise the value of quiet introspection. In this silence, we shall feel that all are together and will experience a strange feeling of togetherness. 

(c) By this, he means that we should not make any physical movement, as physical activity will stop dr interrupt our introspection. 

(d) The poet wants us to simply be silent for a moment and utilise that time to understand ourselves as well as others. Besides, language differences often lead to conflict, which the poet, perhaps, wants to avoid.

(a) It will be an ecstatic moment of tranquillity without rush or running of engines. 

(b) It would be called an exotic moment because it will be an instance of universal peace and brotherhood. In that moment, all of us would initiate introspection through meditation and the whole world will be enveloped in quietness. 

(c) We would feel very strange at that moment, because at that time everyone will have a feeling of oneness with their fellow human beings. It will be a new feeling altogether. 

(d) The poem is ‘Keeping Quiet’, and the poet is Pablo Neruda.

i) b) because of a government order to teach only German

ii) b) regretful

iii) d) both a and b

iv) a) He feels that they were his friends

i) c) He feels sorry for M Hamel as it was his last French lesson.

Ii) b) too little, too late

iii) c) M Hamel often sent students to water his flowers and gave a holiday when he wanted to go fishing

iv) d) (i) forgotten (ii) transformed (iii) hold on (iv) identity

Correct option is D) rough

i) b) Franz

ii) a) Rule of grammar

iii) b) He had not prepared his lesson

iv) d) Pitied him

i. a) Scared and reddened

ii. b) commotion in the class

iii. b) B, C

iv. c) M. Hamel made him stand outside the classroom

Given:

x = 7, y = 4, z = 9

Calculation:

I. x + y + z = 20

LHS,

x + y + z = 7 + 4 + 9

⇒ x + y + z = 20

LHS = RHS

II. x – y + z = 13

LHS,

x – y + z = 7 – 4 + 9

⇒ x – y + z = 12

LHS ≠ RHS

III. x + y – z = 3

LHS,

x + y – z = 7 + 4 – 9

⇒ x + y – z = 2

LHS ≠ RHS

IV. -x + y – z = -12

LHS,

-x + y – z = -7 + 4 – 9

⇒ -x + y – z = -12

LHS = RHS

∴ I and IV are TRUE.

Definition:

Composite number = Composite numbers can be defined as the whole numbers that have more than two factors.

Odd number = An odd number is a number which is not divisible by 2.

Even number = A number which is divisible by 2 and generates a remainder of 0 is called an even number.

Prime number = A number that can be divided exactly only by itself and 1.

∴ The required answer is even number

Calculation:

Prime numbers between 50 and 80 are 53, 59, 61, 67, 71, 73, and 79

Sum of prime numbers between 50 and 80 = 53 + 59 + 61 + 67 + 71 + 73 + 79

∴ Sum of prime numbers between 50 and 80 is 463

Given-

When positive numbers x, y and z are divided by 31, the remainders are 17, 24 and 27

Concept Used-

If a number (x) is divided by (a) leaves remainder (b), then x = a × n + b       [where n is an integer]

Calculation- 

Let the numbers be,

 x = 31 × 1 + 17 = 48

y = 31 × 1 + 24 = 55

z = 31 × 1 +  27 = 58

⇒ (4x - 2y + 3z) = 256

Now,

256 = 31 × 8 + 8

∴ When (4x - 2y + 3z) is divided by 31, the remainder will be 8

Solution:
7 x 13 x 11 + 11 = (7 x 13 + 1) x 11
= (91 + 1) x 11 = 92 x 11
= (2 x 2 x 23) x 11
= 2 x 2 x 11 x 23
Since 7 x 13 x 11 + 11 can be expressed as a product of primes, it is a composite number.
(7 x 6 x 5 x 4 x 3 x 2 x 1) + 3
= (7 x 6 x 5 x 4 x 2 x 1 + 1) x 3
= (1680 + 1) x 3 = 1681 x 3
= (41 x 41) x 3 = 3 x 41 x 41
Since (7 x 6 x 5 x 4 x 3 x 2 x 1) + 3 can be expressed as a product of primes, it is a composite number.

x³ + y³ = 3axy---(1) at (3a/2, 3a/2)

\(\therefore\) 3x² + 3y² \(\frac{dy}{dx}\) = 3a(x\(\frac{dy}{dx}\) + y)----(2)

(by differentiating given equation w.r.to x)

⇒ (3y² - 3ax)\(\frac{dy}{dx}\) = 3ay - 3x²

⇒ \(\frac{dy}{dx}\) = \(\frac{3ay-3x^2}{3y^2-3ax}\) 

⇒ \((\frac{dy}{dx})\)_{(x = 3a/2, y = 3a/2)} = \(\cfrac{\frac{9a^2}2-3\times\frac{9a^2}4}{3\times\frac{9a^2}4-\frac{9a^2}2}\) 

 = \(\frac{18a^2-27a^2}{27a^2-18a^2}=\frac{-9a^2}{9a^2}=-1\)

Differentiating equation(2) w.r.to x, we get

6x + 6y\((\frac{dy}{dx})^2\) + 3y²\(\frac{d^2y}{dx^2}\) = 3a(x\(\frac{d^2y}{dx^2}\) + \(\frac{dy}{dx}+\frac{dy}{dx}\))

⇒ (3y² - 3ax)\(\frac{d^2y}{dx^2}\) = -6y\((\frac{dy}{dx})^2\) - 6x - 6a\(\frac{dy}{dx}\)

⇒ \(\frac{d^2y}{dx^2}\)(at x = 3a/2, y = 3a/2) = \(\cfrac{-6\times\frac{3a}2\times(-1)^2-6\times\frac{3a}2-6a\times-1}{3\times\frac{9a^2}4-3a\times\frac{3a}2}\) 

 = \(\cfrac{-9a-9a+6a}{\frac{27a^2-18a^2}4}\) = \(\frac{-12a\times4}{9a^2}=\frac{-16}{3a}\)

\(\therefore\) Radius of curvature of given curve at (3a/2, 3a/2)

 = \(\left|\cfrac{(1+(\frac{dy}{dx})^2)^{3/2}}{\frac{d^2y}{dx^2}}\right|\) = \(\left|\cfrac{(1+(-1)^2)^{3/2}}{\frac{-16}{3a}}\right|\)

 = \(|\frac{-2\sqrt2\times3a}{16}|\) = \(|\frac{-3\sqrt2a}8|\) = \(\frac{3\sqrt2a}{8}\)(\(\because\) Radius is never negative)

Given 

The marked price (MP) for watch = 450 Rs.

Sell price = 390 Rs.

Discount = ?

Formula used

Discount = MP - S.P.

Discount = (Discount/MP) × 100 

Calculation 

Discount = 450 - 390 = 60 Rs.

The rate of discount = (discount/MP) × 100 

⇒ (60/450) × 100 = 40/3 = 13(1/3)%

∴ The rate of discount is 13(1/3)%  and discount is 60 Rs.

Correct option is (c) (1/3,1/6,-1)

Explantion :

We have

3x + 1 = 6y - 2 = 1 - z

(x + (1/3))/ (1/3) = (y - (1/3))/ (1/6) = (z - 1)/ -1

Therefore, the DRs are

(1/3,1/6,-1)

Concept:

Left coset:

Let G be a group and H be a subgroup of G. Let a be an element of G. Then the subset {ah: h ∈ H} is called a left coset of H in G and is denoted by aH.

Right coset:

Let G be a group and H be a subgroup of G. Let a be an element of G. Then the subset {ha: h ∈ H} is called a right coset of H in G and is denoted by Ha.

Normal Subgroup:

Let G be a group, H be a subgroup of G. Then H is said to be a normal subgroup of G, if x h x-1 ∈ H, ∀ x ∈ G and ∀ h ∈ H.

Note:

• For a normal subgroup left coset is equal to right coset.
• If G is an abelian group then every subgroup of G is a normal subgroup. 

Calculation:

Given: G = (Z, +) and H = (4Z, +) is a subgroup of G.

G = (Z, +) is an abelian group

As we know that, if G is an abelian group then every subgroup of G is a normal subgroup.

∴ H is a normal subgroup

So, all the left and right cosets of H are the same.

Now, let's find out the cosets of H in G.

The distinct cosets of H in G are:

0 + H = {4n : n ∈ Z} = H

1 + H = {4n + 1: n ∈ Z}

2 + H = {4n + 2: n ∈ Z}

3 + H = {4n + 3: n ∈ Z}

So, H, 1 + H, 2 + H, 3 + H are the cosets of H in G.

Hence, there are 4 cosets of H in G

Explanation:

\(\left[ A \right] = \left[ {\begin{array}{*{20}{c}} 6&{ - 2}&2\\ { - 2}&3&{ - 1}\\ 2&{ - 1}&3 \end{array}} \right]\)

As we know:

Product of Eigen values = Determinant of matrix

Determinant of matrix = 6 × (9 – 1) – (- 2) (- 6 + 2) + 2 (2 - 6)

Determinant of matrix = 48 – 8 – 8

∴ Determinant of matrix = 32

Now,

Let the third Eigen value be x

Product of Eigen value = Determinant of [A]

16 × x = 32

∴ x = 2

∴ Third Eigen value = 2

 The number days from 13^th May, 2005 to 18^rd August 2005 ( both inclusive)

The number of odd days = (14 x 7 days )+1 = 99

So there is 1 odd day in the given period.