Explore topic-wise InterviewSolutions in Current Affairs.

Correct option:

(A) (0,0,0)

How many tangents can have a circle  infinity

The Ratio between side and diagonal of a sauare is 1:2

x² – 2x + 4y – 11 = 0
⇒ x² – 2x – 11 = -4y 
⇒ x² – 2x + 1 - 12 = -4y 
⇒ (x - 1)² = 4(-1)(y - 3) 
From here we can identify the focus.
As in (x - h)² = 4p (y - k) the focus is given by (h, k + p)
The directrix is y = k - p
Here focus is (1, 4) and directrix is y = 4

Now it's easy to solve as the locus formed by that point will also be a parabola with same focus and distance will be half. 

Let A = (-3, 1) and B = (7, 5). Let P = (x, y).

Then AP = PB ⇔ (AP)² = (PB)²

^⇔ (x + 3)² + (y -1)² = (x -  7)² + (y - 5)²

^⇔  6x - 2y + 10 = 14x -  10y + 74

⇔ 20 x + 8y - 64 = 0

⇔ 5x + 2y - 16 = 0 

Hence, the equation of  5x + 2y - 16 = 0

Take the two perpendicular lines as coordinate axes. P(x, y) is a point on the locus ⇔|x| + |y| = 1. This implies that

x + y = 1

x - y = 1

-x + y = 1

 -x - y = 1

 These lines form a square.

Equation of the line is

x/4 + y/b = 1

It passes through (2, 3). This implies that

2/4  + -3/b = 1 ⇒ -3/b = 1/2 ⇒ b = -6

Hence, the equation of the line is

x/4 - y/6 = 1

3x - 2y = 12= 0

Answer: (A) → (q)

(B) Let  x/a + y/b = 1

It passes through (4, 1) and forms a triangle with positive axes having area 8. Therefore

4/a + 1/b = 1  ...(1)

and 1/2(ab) = 8

From Eqs. (1) and (2), a = 8 and b = 2. Hence, the equation of the line is

x/8 + y/2 = 1  or x + 4y - 8 = 0

Answer: (B) → (p)

(C) Equation of the line with equal intercepts on the axes is

x/a + y/a = 1 or  x + y = a

This passes through the point (2, 5) implies that a = 7. Hence, the line is

x + y - 7 = 0

Answer: (C) → (r)

(D) Let the line be y =  (tan 135°)x + c where c = 8. That is, y = -x + 8 or x + y = 0. 

Answer: (D) → (t)

Correct option  (i) (B),(ii) (D),(iii) (A)

Explanation :

 (i)  Let A = (5 ,6) and B = (3, 2). The slope of AB is

and the midpoint of AB = (-1, 4). Hence, the perpendicular bisector of the segment (bar)AB is y - 4 = 2(x + 1) or 2x - y + 6 = 0 . Solving this equation and the given line equations, we have x = -2 and y = 2. Thus, (-2, 2) is the point on the given line which is equidistant from both A(-5 6) and B(3, 2).

 (ii) Line perpendicular to the given line is of the form

y = 1/3x + c

This line passes through (1, 1). It implies that

1 = 1/3 + c ⇒ c = 2/3

Thus, the required line is

y = x/3 + 2/3 or x - 3y + 2 = 0

(iii) The line y + 5 = k(n - 3) is parallel to the given line ⇒ k = -3. That is, 

3x + y = 4

or  x/(4/3) + y/4 = 1

Hence, the area of the triangle is

1/2(4/3)(4) = 8/3

Correct option  (C)  11 : 1

Explanation :

L ≡ 3x - y + 6 = 0, A = (3,4),B = (-2,1)

L₁₁ = 3(3) - 4 + 6 = 11, L₂₂ = 3(-2) -1 + 6 = -1

 Hence, the ratio AP : PB is

-L₁₁: L₂₂ = (-11) : (-1) = 11 : 1

The duration of Kabaddi game is 20-5-20 and for junior boys and girls is 15-5-15.

Officials in Kabaddi are:

1. One referee
2. Two umpires
3. Two lines-men
4. One scorer.

Any team which gets more points shall be winner. In case of draw 5 extra minutes will be given until the match is decided.

(3) at least 7

First row with exactly one zero; total number of cases = 6

First row 2 zeros we get more cases

Total we get more than 7.

Directions: Questions Number 72 to 76 are Assertion – Reason type questions. Each of these questions contains two statements.

Statement-1: (Assertion) and Statement-2: (Reason)

Each of these questions also has four alternative choices, only one of which is the correct answer.

You have to select the correct choice.

(2) x = –1 

The locus of perpendicular tangents is directrix. i.e, x = –a; x = –1

Given: Power = 2kW = 2000W 

Current = 5A 

Voltage = 220V 

∴ Power = Current x Voltage

⇒ Current = \(\frac{Power}{Voltage}\) = \(\frac{2000}{220}\) = 9 A

Thus, the current exceeds the current rating of 5A. Therefore, either the fuse will burn out and if there is no fuse, then overloading will cause melting of wires which may result in short-circuiting or a fire.

It is the book review for the famous book, THE INVISIBLE MAN

Ralph Ellison’s The Invisible Man:

An extremely powerful story of a young Southern Negro, from his late high school days through three years of college to his life in Harlem.

His early training prepared him for a life of humility before white men, but through injustices- large and small, he came to realize that he was an "invisible man". People saw in him only a reflection of their preconceived ideas of what he was, denied his individuality, and ultimately did not see him at all. This theme, which has implications far beyond the obvious racial parallel, is skillfully handled. The incidents of the story are wholly absorbing. The boy's dismissal from college because of an innocent mistake, his shocked reaction to the anonymity of the North and to Harlem, his nightmare experiences on a one-day job in a paint factory and in the hospital, his lightning success as the Harlem leader of a communistic organization known as the Brotherhood, his involvement in black versus white and black versus black clashes and his disillusion and understanding of his invisibility- all climax naturally in scenes of violence and riot, followed by a retreat which is both literal and figurative. Parts of this experience may have been told before, but never with such freshness, intensity and power.

Youth unemployment is the situation of young people who are looking for a job, but cannot find a job, with the age range being that defined by the United Nations as 15–24 years old. An unemployed person is defined as someone who does not have a job but is actively seeking work. In order to qualify as unemployed for official and statistical measurement, the individual must be without employment, willing and able to work, of the officially designated "working age" (often from the teens to the mid-60s) and actively searching for a position. Youth unemployment rates tend to be higher than the adult rates in every country in the world.

Causes -There are multiple and complex causes behind youth unemployment. Among them, the quality and relevance of education, inflexible labour market and regulations, which in turn create a situation of assistance and dependency, are the main causes discussed today.

Effect - Unemployed youth has been called "a lost generation": not only because of productivity lost but also because of the long-term direct and indirect impact unemployment has on young people and their families. Unemployment has been said to affect earnings for about 20 years. Because they aren't able to build up skills or experience during their first years in the workforce, unemployed youth see a decrease in lifetime earnings when compared to those who had steady work or those who were unemployed as an adult. A lower salary can persist for 20 years following the unemployed period before the individual begins earning competitively to their peers.[10] Widespread youth unemployment also leads to a socially excluded generation at great risk for poverty.

Correct Answer is : (c) Tragedy

One of the most dangerous threats to life on earth is the increase in pollution which is caused because of the increase in the concentration of carbon dioxide and carbon monoxide in the environment. Since the last century, the immense increase in pollution in nature has been the cause of many harmful effects. Increase in Global warming, reduction in the quality of air and water and an increase in life-threatening diseases are some of the main effects of pollution. Hence, it is the need of the hour to look into the issue and find alternatives to them.  

One of the important element which keeps the life going on this earth is oxygen. The life-supporting oxygen is the product which we receive from the trees. Hence, trees are the basic manufacturer of oxygen which helps to grow life on Earth. Since last century there has been immense cutting down of the trees for different uses. The man has been cutting trees so as to make the life easy. In this run, the man himself has become the cause of his downfall. Trees purify the air and help in keeping the atmosphere cool. Cutting down of the trees has encouraged global warming which is a global concern. A rise in temperature has been noticed which is not a good symbol for life on Earth. Hence, tree plantation should be encouraged and the cutting of the trees should be minimized.  

Given AP is 7, 11, 15, …., 139. 

First term of AP is a = 7. 

And common difference of AP is d = a₂ – a₁ = 11 – 7 = 4. 

Let n^th term of AP is 139. 

∴ a + (n – 1) d = 139. (∵ n^th term of AP is given by a_n = a + (n – 1) d) 

⇒ 7 + (n – 1)4 = 139 (∵ a = 7 & d = 4) 

⇒ 4(n – 1) = 139 – 7 = 132 

⇒ n – 1 = 132 4 = 33 

⇒ n = 33 + 1 = 34. 

Hence, total term in given AP is 34.

The odd integers between 2 and 100 which is divisible by 3 are 3, 9, 15, ..., 99. 

First and last odd integer between 2 and 100 which is divisible by 3 are 3 and 99 respectively. 

First term of given sequence is a = 3. 

And common difference of AP is d = a₂ – a₁ = 9 –3 = 6. 

Let 99 is nth term of sequence 3, 9, 15, .………., 99. 

∴ a + (n – 1) d = 99. (∵ a_n = a + (n – 1) d) 

⇒ 3 + (n – 1)6 = 99 (∵ a = 3 & d = 6) 

⇒ 6(n – 1) = 99 – 3 = 96 

⇒ n – 1 = \(\frac{96}{6}\) = 16 

⇒ n = 16 + 1 = 17. 

Now, sum of sequence 3, 9, 15, ………, 99 = \(\frac{n}{2}\)[a + a_n]

= \(\frac{17}{2}\) [3 + 99] (∵ n = 17, a = 3 & a_n = 99) 

= \(\frac{17}{2}\) × 102 = 17 × 51 = 867. 

Hence, the sum of all odd integers between 2 and 100 which are divisible by 3 is 867.

S(x, y) = (0, x + y)

T(x, y) = (2x , 3y)

(S + T) (x, y) = S(x, y) + T(x, y)

\(\)= (0, x + y) + (2x, 3y)

= (2x, x + 4y)

U = (-1 , 5)

V = (6, -3)

U - 2V = (-1, 5) - 2(6, -3)

= (-1 -12, 5 + 6)

= (-13, 11)

T(U - 2V) = T(-13, 11) = (-26, 33).

(i) Total number of engineers = 40

Number of engineers living less than 7 km from their place of work = 9

Hence, required probability that an engineer lives less than 7 km from her place of work, P = 9/40

(ii) Number of engineers living more than or equal to 7 km from their place of work = 40 − 9 = 31

Hence, required probability that an engineer lives more than or equal to 7 km from her place of work, P = 31/40

(iii) Number of engineers living within 1/2 km from her place of work = 0

Hence, required probability that an engineer lives within 1/2 km from her place of work, P = 0

 i. Total no. of possible pairs = 50 × 40 = 2000 

No. of pairs in which both are girls = 20 × 25 = 500

Probability of both being girls = \(\frac{500}{2000}=\frac{1}{4}\)

ii. No. of pairs in which both are boys = 30 × 15 = 450

Probability of both being boys = \(\frac{450}{2000}=\frac{9}{40}\)

iii. No. of pairs in which one is boy and one is girl = 2000 – (500 + 450) = 2000 – 950 = 1050

Probability of one being boy and one girl = \(\frac{1050}{2000}=\frac{21}{40}\)

iv. No. of pairs in which atleast one is boy = 1050 + 450 = 1500

Probability in which atleast one is boy = \(\frac{1050}{2000}=\frac{3}{4}\)

i. Total no. of possible pairs = 50 × 40 = 2000 

No. of pairs in which both are girls = 20 × 25 = 500

Probability of both being girls = \(\frac{500}{2000}=\frac{1}{4}\)

ii. No. of pairs in which both are boys = 30 × 15 = 450

Probability of both being boys = \(\frac{450}{2000}=\frac{9}{40}\)

iii. No. of pairs in which one is boy and one is girl = 2000 – (500 + 450) = 2000 – 950 = 1050

Probability of one being boy and one girl = \(\frac{1050}{2000}=\frac{21}{40}\)

iv. No. of pairs in which atleast one is boy = 1050 + 450 = 1500

Probability in which atleast one is boy = \(\frac{1050}{2000}=\frac{3}{4}\)

Total number of pairs 6 x 6 = 36

a. Pairs of spots with total number 6 are (1,5) (2,4) (3,3) (4,2) (5,1)

∴ Probability = \(\frac{5}{36}\)

b. Pairs with 9, (3,6) (4,5) (5,4) (6,3)

∴ Probability = \(\frac{4}{36}=\frac{1}{9}\)

c. Probability to be 1 = \(\frac{0}{36}=0\)

d. Pairs with total number of spots as prime number (1,1) (1,2) (1,4) (1,6) (2,1) (2,3) (2,5) (3,2) (3,4) (4,1) (4,3) (5,2) (5,6) (6,1) (6,5)

∴ Probability = \(\frac{15}{36}\)

We have, A₉=7*a₂ 
a₁₂=5a₃+2 

So, 
a₉=(a+d)7 
a₁₂=5(a+2d)+2 

So, 
a+8d=7a+7d, 

d=6a 
a+11d=5a+10d+2, 
d=4a+2, 
2a=2, 
a=1 
d=6a=6 

So, first term is 1 & common difference is 6.

a. Total number of cards = 52

Number of black cards = 26

Probability of the picked card being black = \(\frac{26}{52}=\frac{1}{2}\)

b. No. of spade cards = 13

Probability of a picked card being spade = \(\frac{12}{52}=\frac{1}{4}\)

c. No. of cards spade or diamond =13 + 3 = 26

Probability of a picked card being spade or diamond = \(\frac{26}{52}=\frac{1}{2}\)

Total pairs = 50 × 40 = 2000

No. of pairs in which both are boys = 20 × 25 = 500

No. of pairs in which both are girls = 30 × 15 = 450

No. of pairs in which one is a boy and the other a girl = 30 × 25 + 20 × 15 = 750 + 300 = 1050

1/(2a+b+2x) - 1 / (2x) = 1 / 2a + 1 / b

(2x - 2a - b - 2x) / (2x)(2a+b+2x) = (2a + b) / 2ab

-(2a + b) / (2x)(2a+b+2x) = (2a + b) / 2ab

2a + b gets cancelled.

-2ab = (2x)(2a+b+2x)

-2ab = 4ax + 2bx + 4x²

4x2 + 2bx + 4ax + 2ab = 0

2x(2x + b) + 2a(2x + b) = 0

(2x +2a)(2x + b) = 0

2x + 2a = 0

x = - a

2x + b = 0

x = - b / 2

So, x = -a , -b/2

Solution:

If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5  
It can be observed that 5 is not in the prime factorisation of (2²x7)ⁿ or 28ⁿ.  
Hence, for any value of n, 28ⁿ will not be divisible by 5.  
Therefore, 28ⁿ cannot end with the digit 0 for any natural number n.

Total number of cars = 400

Number of defective toys = 12

Number of non-defective toys = 400 – 12 = 388

a. Probability of getting a defective toy = \(\frac{12}{400}=\frac{3}{100}\)

b. Probability of getting a non-defective toy = \(\frac{388}{400}=\frac{97}{100}\)

Total no. of balls = 10 

No. of black balls = 6 

No. of white balls = 4

Probability of it to be black = \(\frac{6}{10}=\frac{3}{5}\)

Probablitiy of to be white = \(\frac{4}{10}=\frac{2}{5}\)

Answer is 12 odd numbers

First multiple of 7 = 105 

Last multiple of 7 = 294 

Number of multiples = \(\frac{294-105}{7}+1=28\)

Leap year have 366 days. 

That is 52 weeks and 2 days. 

There two days are 

Sunday – Monday, Monday – Tuesday, Tuesday – Wednesday, Wednesday – Thursday, Thursday – Friday, Friday – Saturday, Saturday – Sunday. 

∴ Probability for to occur 53 Sunday is 2/7.

A non-leap year has 365 days

A year has 52 weeks. Hence there will be 52 Sundays.

52 weeks = 52 x 7 = 364 days .

365– 364 = 1day extra.

In a non-leap year there will be 52 Sundays and 1day will be left.

This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday,friday,Saturday, Sunday.

Of these total 7 outcomes, the favourable outcomes are 1.

Hence the probability of getting 53 Sundays = 1 / 7.

Total pairs = 20 x 15 = 300 

Number of pairs getting both black= 8 x 9 = 72

Probability of getting both black = \(\frac{72}{300}=\frac{6}{25}\)

Probability of getting one black and one white = \(\frac{8\times6+12\times9}{300}=\frac{156}{300}=\frac{39}{75}\)

Answer is 368, 386, 683, 638, 836, 863

a. Total number of balls in the first bag = 10 + 12 =22, Number of yellow balls = 12

Probability of getting a yellow ball = \(\frac{12}{22}=\frac{6}{11}\)

b. Total number of balls in the second bag = 15 + 7 = 22, Number of a yellow ball = 7

Probability of getting a yellow ball = \(\frac{7}{22}\)

c. Total number of balls = 22 + 22 = 44 Number of yellow balls = 12 + 7 = 19

Probability of getting a yellow ball = \(\frac{19}{44}\)

Total two-digit numbers = 90 

Product of the digits of a number drawn is a prime number 12, 13, 15, 17, 21, 31, 51, 71. 

(1 is not a prime number)

Total number whose product of the digits drawn is a prime number = 8 

Probability of product of the digits drawn is a prime number = \(\frac{8}{90}=\frac{4}{45}\)

Total three-digit numbers = 900 

Product of the digits of a number drawn is a prime number 112, 113, 115, 117, 121, 131, 151, 171, 211, 311, 511, 711.

Total number whose product of the digits drawn is a prime number = 12

Probability of product of the digits drawn is a prime number = \(\frac{12}{900}=\frac{4}{300} =\frac{1}{75}\)

Total no. of two-digit numbers = 90

Perfect squares from 10 to 99 are 16, 25, 36, 49, 64, 81.

There are 6 favourable numbers Probability = \(\frac{6}{90}=\frac{1}{15}\)

Probability of sum being odd = \(\frac{4}{12}=\frac{1}{3}\)

Probability of sum being even = \(\frac{8}{12}=\frac{2}{3}\)

Probablity of getting prime numbers from 1 to 50 = \(\frac{15}{50}=\frac{3}{10}\)

Probability of getting a multiple of 5 from 1 to 50 = \(\frac{10}{50}=\frac{1}{5}\)

∴ Better guess would be prime numbers as probability of that is more.

x=tany 

int (tanysecy -sec y)dy

sec y- ln(sec y+ tan y)

A quadratic equation as an equation of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic is y = ax² + bx + c, where a, b, and c are numbers and a cannot be 0. Examples of quadratic equations include all of these: y = x² + 3x + 1.