**DOWNLOAD FREE PDF** **<<CLICK HERE>>**

## Aptitude for GATE

Congratulations - you have completed *Aptitude for GATE*.

You scored %%SCORE%% out of %%TOTAL%%.

Your performance has been rated as %%RATING%%

Question 1 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

$1,50,000$ | |

$2,00,000$ | |

$1,75,000$ | |

$1,51,000$ |

**Civics Test Questions answers**

Let no. of senior female be $y$

$x + y$ = $300$

Total money given $=\left(\frac{8x}{9}\times 750= \frac{2000x}{ 3}\right) + \left(\frac{2y}{3}\times 1000= \frac{2000y}{3}\right)$

$\quad = \frac{2000}{3} \times (x +y)$

$\quad = \frac{2000}{3} \times 300$

$\quad =200000.$

So, $B$ is correct.

Question 2 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

10.1 | |

23.1 | |

22.5 | |

21.2 |

**Crack any exam**

Area ∆ABC/Area ∆CDE =(AB/DE)^2 =(2/7)^2 ,Now we know that Area ∆ABC = 1/2 *h1*AB and Area ∆CDE = 1/2 *h2*DE where h1 and h2 are height of triangle ABC and CDE respectively

4/49 =(h1*2)/(h2*7) --> h1/h2 =2/7 Now h1+h2 =7.2 we get h1 = 1.6 and h2= 5.6 and sum of both triangle area =21.2

Question 3 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

Water pours into a rectangular tank of $20\:metres$ depth which was initially half-filled. The rate at which the height of the water rises is inversely proportional to the height of the water at that instant. If the height of the water after an hour is observed to be $12\:metres$, how much time, in hours, will be required to fill up the tank?

$\frac{75}{11}$ | |

$\frac{125}{11}$ | |

$\frac{25}{3}$ | |

$5$ |

**NTA NET study material**

Then according to the question, $\frac{dH}{dt} =\frac{K}{H}$ -----$\left ( 1 \right )$

Givent at $t=0$, $H=10$ and at $t=1$, $H=12$

So, integrating by limits mentioned above, we get $\int_{10}^{12}HdH=K\int_{0}^{1}dt$

$\Rightarrow$ $K=\frac{144-100}{2}=22$

To find the complete time to fill up the tank, we will use again equation $\left ( 1 \right )$

Now, $\int_{10}^{20}HdH=22\times \int_{0}^{t}dt$

$\therefore$ $t=\frac{\left ( 400-100 \right )}{44}$ $\Rightarrow$ $t=\frac{75}{11}$

Option A is correct

Question 4 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

3/12 | |

3/11 | |

1/2 | |

None of these |

**NTA NET study material**

There are $4$ mutually exclusive possibilities to consider for ways Player A can draw the red ball:

(i) The first ball drawn is red. Probability of this is $\dfrac3{10}$

(ii) The first two balls drawn are black and the third one is red. Probability of this is $\dfrac7{10}\cdot\dfrac69\cdot\dfrac38=\dfrac{7\cdot 6\cdot 3}{10\cdot 9\cdot 8}$.

(iii) The first four balls drawn are black and the fifth one is red. Probability of this is $\dfrac7{10}\cdot\dfrac69\cdot\dfrac58\cdot\dfrac47\cdot\dfrac36=\dfrac{7\cdot 6\cdot 5\cdot 4\cdot 3}{10\cdot 9\cdot 8\cdot 7\cdot 6}$.

(iv) The first six balls drawn are black and the seventh is red. Probability of this is $\dfrac7{10}\cdot\dfrac69\cdot\dfrac58\cdot\dfrac47\cdot\dfrac36\cdot\dfrac25\cdot\dfrac34=\dfrac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 3}{10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4}$.

Since these events are mutually exclusive, the probability that Player A gets the red ball is simply the sum of these probabilities. The least common denominator is clearly $10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4=\frac{10!}{3!}$. It seems your book chose to use $10!$ as the common denominator, instead (probably to reduce the length of the answer). Rewriting the probabilities with $10!$ as denominator, we therefore have $$\begin{align}\Bbb P(A) &= \frac{3\cdot9!}{10!}+\frac{7\cdot 6\cdot 3\cdot 7!}{10!}+\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 5!}{10!}+\frac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 3\cdot 3!}{10!}\\ &= \frac{3\cdot9!+7\cdot 6\cdot 3\cdot 7!+7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 5!+7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 3\cdot 3!}{10!}.\end{align}$$

Question 5 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

Consider a sequence of numbers $\large (\epsilon _{n}: n= 1, 2,...)$, such that $\epsilon _{1}=10$ and

$\large \epsilon _{n+1}=\dfrac{20\epsilon _{n}}{20+\epsilon _{n}}$

for $n\geq 1$. Which of the following statements is true?

Hint: Consider the sequence of reciprocals.

The sequence $\large (\epsilon _{n}: n= 1, 2,...)$ converges to zero. | |

$\large \epsilon _{n}\geq 1$ for all $n$ | |

The sequence $\large (\epsilon _{n}: n= 1, 2,...)$ is decreasing and converges to $1.$ | |

The sequence $large (epsilon _{n}: n= 1, 2,...)$ is decreasing and then increasing. Finally it converges to $1.$ |

**Aptitude test Questions answers**

$\large\epsilon_1$ is positive.

In the formula for $\large\epsilon_{n+1}$, we only add, multiply and divide positive numbers. Thus, all $\large \epsilon_n$ are positive.

Also, $\large\epsilon_{n+1} < \epsilon_n$

**Proof: **

$\large\begin{align}

\epsilon_{n+1} - \epsilon_n &= \frac{20 \cdot \epsilon_n}{20+\epsilon_n} - \color{red}{\epsilon_n}\\[1em]

&= \frac{20 \cdot \epsilon_n \color{red}{-20\cdot \epsilon_n - (\epsilon_n)^2}}{20+\epsilon_n}\\[1em]

&= \frac{{\Large\color{red}-}(\epsilon_n)^2}{20+\epsilon_n}\\[1em]

&< 0\\[1em]

\hline

\epsilon_{n+1} - \epsilon_n &< 0\\[1em]

\epsilon_{n+1} &< \epsilon_n

\end{align}$

Thus, the sequence is decreasing.

Since the sequence is decreasing and is bounded below by $0$, we know that the sequence converges (Monotone Convergence Theorem).

The only fixed point of the sequence can be found as follows:

$\large\epsilon_f = \dfrac{20 \cdot \epsilon_f}{20+\epsilon_f}$

$\large20 \cdot \epsilon_f + (\epsilon_f)^2 = 20 \cdot \epsilon_f$

$\large(\epsilon_f)^2= 0$

$\large \epsilon_f = 0$

Hence, the sequence converges to $0$.

**Option a is correct.**

Question 6 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

There are multiple routes to reach from node $1$ to node $2$, as shown in the network.

The cost of travel on an edge between two nodes is given in rupees. Nodes 'a', 'b', 'c', 'd', 'e', and 'f' are toll booths. The toll price at toll booths marked 'a' and 'e' is Rs. $200$, and is Rs. $100$ for the other toll booths. Which is the cheapest route from node $1$ to node $2$?

$1-a-c-2$ | |

$1-f-b-2$ | |

$1-b-2$ | |

$1-f-e-2$ |

**Crack any exam**

**1-a-c-2 :**

1-a = 200

tax at a =200

a-c = 100

tax at c = 100

c-2 : 100

total = 200+200+100+100+100 = 700

**1-f-e-2 :**

1-f = 100

tax at f =100

f-e = 100

tax at e = 200

e-2 : 200

total = 100+100+100+200+200 = 700

**1-f-b-2 :**

1-f = 100

tax at f =100

f-b = 0

tax at b = 100

b-2 : 200

total = 100+100+0+100+200 = 500

**1-b-2 :**

1-b = 300

tax at b = 100

b-2 : 200

total = 300+100+200 = 600

**cheapest route = 1-f-b-2**

Question 7 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

1/5 | |

1/3 | |

1/4 | |

2/7 |

**English grammar Questions answers**

*x*litres of this liquid be replaced with water.

Quantity of water in new mixture = $(3-\frac{3x}{8}+x)$ litres

Quantity of syrup in new mixture = $(5-\frac{5x}{8})$ litres

$(3-\frac{3x}{8}+x)$ = $(5-\frac{5x}{8})$

5x + 24 = 40 - 5x

10x = 16

$\therefore$ x = $(\frac{8}{5})$

So, the part of mixture replaced = $(\frac{8}{5} * \frac{1}{8})$ = $\frac{1}{5}$

Question 8 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

There is no fixed relation between food and famine$;$ famines can occur with or without a substantial _______ in food output$.$

aberration | |

weakening | |

decline | |

deterioration |

**Latest Current affairs Questions answers**

"aberration" means sharp change -- not necessarily a reduction.

"weakening" and "deterioration" must be followed by "of" than "in".

"decline" perfectly fits here.

Question 9 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

A large community practices birth control in the following peculiar fashion. Each set of parents continues having children until a son is born; then they stop. What is the ratio of boys to girls in the community if, in the absence of birth control, 51% of the babies are born male?

$51:49$ | |

$1:1$ | |

$49:51$ | |

$51:98$ |

**NTA NET study material**

Let, $X$ be the expected no of child a parent has.

So, no of boys $= 1.$

No. of girls $= X - 1.$

The probability of having a baby boy $= 0.51.$

And the probability of having a baby girl $= 0.49.$

So,

$X = 1 \times (0.51) + 2 \times (0.49) \times (0.51) + 3 \times (0.49)^2 \times (0.51) + 4 \times (0.49)^3 \times (0.51) $

$0.49X= 1 \times (0.49) \times (0.51) + 2 \times (0.49)^2 \times (0.51) + 3 \times (0.49)^3 \times (0.51) $

$X - 0.49X = (0.51)[ 1 + (0.49) \times (0.51) + (0.49)^2 \times (0.51) + (0.49)^3 \times (0.51) + \ldots ]$

$0.51X = (0.51) [ 1 / 0.51 ]$

$\implies X = 100 / 51.$

So, No of girl children $= X - 1.$

$\quad \quad = [100 / 51 ] - 1 = 49 / 51.$

No. of boy children $= 1.$

Hence, Ratio, Boys : Girls $= 51 : 49.$** **

Question 10 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

$\frac{1 + \sqrt{2}}{3}$ | |

$ \frac{1 + \sqrt{2}}{2}$ | |

$ \frac{1 + \sqrt{3}}{2}$ | |

$ \frac{2 + \sqrt{2}}{2}$ |

**GK (GENERAL KNOWLEDGE) Questions answers**

Let the lengths of the sides of square be $b$ and the sides of octagon be $a$.

**Given:** $4b=8a$

$ \implies b=2a$

A regular octagon can be divided into $4$ triangles and $1$ square .

$\newcommand{deg}{\,^\circ}$ **Area of the square within the octagon can be calculated as follows:** $$\begin{align}

&\Bigl ( 2a \cos{(22.5\deg)} \Bigr ) \times \Bigl ( 2a \cos{(22.5\deg)} \Bigr )\\[1em]

&= 4a^2 \cos^2{(22.5 \deg)}\\[1em]

&= 4a^2 \sin^2{(67.5 \deg)}\\[1em]

&= 2a^2 \Bigl (1 - \cos {(135 \deg)} \Bigr ) \qquad \Bigl \{\sin^2 x = \frac{1- \cos{2x}}{2}\\[1em]

&= 2 \times a^2 \Bigl (1 + \sin{(45\deg)} \Bigr )

\end{align}$$

**Area of the triangle can be calculated as follows:** $$\begin{align}

&2 \times \frac{1}{2} \Bigl (a \cos{(22.5\deg)} \times a \sin{(22.5\deg)} \Bigr )\\[1em]

&= \frac{a^2 \sin{(45\deg)}}{2} \qquad \Bigl \{ \sin{2x} = 2 \sin x \cos x\\[1em]

&=\frac{a^2 \sin{(45\deg)}}{2}

\end{align}$$

Area of $4$ triangles $= 2 \times a^2 \sin{(45\deg)}$ **$\therefore$ Area of the octagon $= 2a^2 + 2a^2 \sqrt{2}$**

**Ratio of area of octagon to area of square:** $$\begin{align}

&= \frac{2a^2 + 2a^2 \sqrt{2}}{b^2}\\[1em]

&= \frac{2a^2 + 2a^2 \sqrt{2}}{4a^2}\\[1em]

&= \frac{1 + \sqrt{2}}{2}

\end{align}$$

Area of a regular polygon with $n$ sides is: $\dfrac{n}{4} a^2 \cot{\left ( \dfrac{\pi}{n} \right )}$, where $a$ is the length of the side.

More:

$$\displaystyle \begin{array}{c}

\text{Properties of a regular $n$-gon}\\[1em]

\hline

\begin{array}{l|l}

\text{circum-radius} & \frac 12 a \csc{\left ( \frac \pi n\right )}\\[1em]

\text{in-radius} & \frac 12 a \cot{\left ( \frac \pi n\right )}\\[1em]

\text{area}& \frac n4 a^2 \cot{\left ( \frac \pi n\right )}\\[1em]

\text{perimeter} & an\\[1em]

\text{center angle} & \frac{2\pi}{n} \mathrm{rad} = \frac{360^\circ}{n}\\[1em]

\text{interior angle} & \frac{(n-2)\pi}{n} = \frac{180(n-2)}{n}^\circ\\[1em]

\text{interior angle sum} & (n-2) \pi\; \mathrm{rad} = 180(n-2)^\circ\\[1em]

\text{exterior angle sum} & (n+2) \pi\; \mathrm{rad} = 180(n+2)^\circ

\end{array}\\[1em]

\hline

n: \text{ number of sides}\\

a: \text{ length of each side}

\end{array}$$

Question 11 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

How many triplets of real numbers $(x,y,z)$ are simultaneous solutions of the equations $x+y=2$ and $xy-z^2=1$?

$0$ | |

$1$ | |

$2$ | |

infinitely many |

**Latest Current affairs Questions answers**

$(x-y)^2 = 4 - 4(1+z^2)$

$(x-y)^2 = - 4z^2$

Now $ LHS >= 0$ and $RHS<=0$ . So both of them are equal only when $z = 0$ and $x = y$ .

Solving along with $ x+y = 2$ we get the triplet $(x,y,z) = (1,1,0)$

Question 12 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

$(3474)^{1793}*(225)^{317}*(451)^{491}$

0 | |

1 | |

2 | |

3 |

**KBC Questions answers**

4 ; 4*4=16 ; 4*4*4=64 ; 4*4*4*4=256 ; 4*4*4*4*4=1024

you can see Here if number of 4 in unit place is odd then unit digit will be 4 and number of 4 in unit place is even then unit digit will be 6.

Now

5 ; 5*5 =25 ; 5*5*5=125 ; 5*5*5*5=625

Here unit digit will be 5 always.

1 ; 1*1 =1 ; 1*1*1 =1 ; 1*1*1*1=1

Here unit digit will be 1 always.

(3474)^{1793}^{ } :- Here number of 4 in unit place is odd so unit digit will be 4.

(225)^{317} :- Here unit digit will be 5.

(451)^{491} :- Here unit place result will be 1.

So unit digit in ( 3474)1793 *( 225 )317 * ( 451 )491}

4*5*1=20

Hence unit digit in ( 3474)1793 *( 225 )317 * ( 451 )491 will be **ZERO.**

Question 13 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

Which of the following statements is true?

There are three consecutive integers with sum $2015$ | |

There are four consecutive integers with sum $2015$ | |

There are five consecutive integers with sum $2015$ | |

There are three consecutive integers with product $2015$ |

**Latest Current affairs Questions answers**

**A) There are $3$ consecutive integers with sum $2015$**

$3$ integers are $(x-1),x,(x+1)$

$(x-1)+x+(x+1) = 2015$

$\implies 3x = 2015$

$\implies x = 2015/3$

$=671.66$

*$670.66, 671.66, 672.66$* * are not integers.*

**B) There are $4$ consecutive integers with sum $2015$**

$4$ integers are $(x+1),(x+2),(x+3),(x+4)$

$(x+1)+(x+2)+(x+3)+(x+4) = 2015$

$\implies 4x +10 = 2015$

$\implies 4x = 2015-10 = 2005$

$\implies x = \dfrac{2005}{4}$

$=501.25$

*$502.25, 503.25, 504.25, 505.25$ are not integers.*

**C) There are $5$ consecutive integers with sum $2015$**

$5$ integers are $(x-2),(x-1),x,(x+1),(x+2)$

$ (x-2)+(x-1)+x+(x+1)+(x+2) = 2015$

$\implies 5x = 2015$

$\implies x = 2015/5$

$=403$

*$401, 402, 403, 404, 405$ are integers.*

The integers are $(403-2),(403-1),403,(403+1),(403+2) \implies 401,402,403,404,405$

Therefore,sum = $401+402+403+404+405 = 2015$

**D) There are $3$ consecutive integers with product $2015$**

$3$ integers are $(x-1),x,(x+1)$

$(x-1) \times x \times (x+1) = 2015$

$\implies (x^2-x)(x+1) = 2015$

$\implies x^3-x^2+x^2-x = 2015$

$\implies x^3-x = 2015$

$\implies x^3-x-2015 = 0$

it clearly shows that we cannot get integer from this equation.

Question 14 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

statements:-

- some authors are teachers
- No teacher is a lady.

conclusion:-

I.some teachers are not ladies

II.some ladies are teachers

I follows | |

II follows | |

Both I and II follows | |

None of I and II follows |

**KBC Questions answers**

Question 15 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

Begging has to be banned because it adversely affects welfare of the state | |

Begging is an offence the has to dealt firmly | |

Beggar are created due to lack of welfare scheme of state | |

Beggar are criminals |

**Computer science (GATE/NET) Questions answers**

The last sentence of the text mentions

the underlying disease(cause) is the failure of state to protect citizens who fall through the social security net.

which means beggars are created due to lack of social welfare schemes by the state.

- Beggars are lazy people who beg because they are unwilling to work - may or may not be true but cannot be inferred from the passage.
- Beggars are created because of the lack of social welfare schemes - Can be inferred from the last sentence.
- Begging is an offense that has to be dealt with firmly - It is given that High Court dispelled the idea of treating begging as an offense.
- Begging has to be banned because it adversely affects the welfare of the state - may or may not be true but cannot be inferred from the passage.

Question 16 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

**statement 1**. Is paying ransom or agreeing to the conditions of kidnappers of political figures, a proper course of action?

argument i. Yes, The victims must be saved at all.

argument ii. No, It encourages the kidnappers to continue their sinister activities

Argument I is Strong | |

Argument II is Strong | |

I or II is Strong | |

I and II are strong |

**UPSC FREE STUDY**

Question 17 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

Choose the word most similar in meaning to the given word:

Awkward

Inept | |

Graceful | |

Suitable | |

Dreadful |

**Latest Current affairs Questions answers**

Dreadful is not suitable here as it means horrible and is more negative than "awkward".

Graceful and suitable are positive words and hence not applicable.

Question 18 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

55650 | |

49550 | |

75550 | |

85550 |

**Bank exam Questions answers**

2012 --- let it be x so x+15%x = 93771 so x =81540

2011---let it be y so y+8%y = 81540 so y will be **75500**

Question 19 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

55,440 | |

1,66,320 | |

4.790E+08 | |

39,91,680 |

**English grammar Questions answers**

But 3 of them will be green, 2 of them pink, 2 of them yellow and 5 of them white.

Answer = 12! /(3!* 2 ! * 2! * 5!) = 166320

Question 20 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

23% | |

21% | |

26% | |

20% |

**Aptitude test Questions answers**

Selling price of 1 article = Rs. 11/10

So profit = 11/10 - 10/11 = 21/110.

profit% = $\frac{\left(\frac{21}{110}\right)}{\left(\frac{10}{11}\right)}*100 = 21\%$

In general, for these type of questions, find CP and SP of one item and find profit/loss accordingly.

Question 21 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

212 | |

612 | |

651 | |

623 |

**Biology Questions answers**

$\text{CP}= \text{SP} \times \frac{100}{120} = \frac{5x}{6}$

$\text{MP} = \text{SP} \times \frac{100}{85} = \frac{20x}{17}$

$\text{Profit}=\frac{x}{6}$

$\text{Discount}=\frac{3x}{17}$

$\frac{3x}{17} - \frac{x}{6}=6$

$\implies x=612.$

Question 22 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

In the given diagram, teachers are represented in the triangle, researchers in the circle and administrators in the rectangle. Out of the total number of the people, the percentage of administrators shall be in the range of _______

0 to 15 | |

16 to 30 | |

31 to 45 | |

46 to 60 |

**Civics Test Questions answers**

$\rightarrow$ Total no of person $=80+20+20+40=160$

$\rightarrow$ Required $\%=(50/160)*100=31.25\%$

$\therefore$ Option $C$. $31$ to $45$ is the correct answer.

Question 23 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

4! $\times$ 4! | |

7P3 $\times$ 4! | |

7P4 $\times$ 3! | |

none of these |

**Aptitude test Questions answers**

positions are 1,2,3,4,5,6,7 Vowels may occuppy only odd positions (4 places {1,3,5,7})

3 vowels can be arranged in 4 odd positions in ^{4}P_{3 }ways

4 consonants can be arranged in the remaining 4 positions in 4! ways

**Total number of ways in which the letters of the word MACHINE can be arranged so that the vowels may occupy only odd positions = ^{4}P_{3 * }4! = 4! * 4! = 24 * 24 = 576**

Question 24 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

**Given the sequence $A,B,B,C,C,C,D,D,D,D,\ldots$ etc$.,$ that is one $A,$ two $B's,$ three $C's,$ four $D's,$ five $E's$ and so on, the $240^{th}$ latter in the sequence will be $:$**

$V$ | |

$U$ | |

$T$ | |

$W$ |

**Question 24 Explanation: **

**GK (GENERAL KNOWLEDGE) Questions answers**Given Sequence is :

$A,B,B,C,C,C,D,D,D,D,\ldots$ etc.

Here, each alphabet is repeated as many times as its position in the alphabetical order. $$\begin{array}{|c|c|l|} \hline \textbf{Alphabet} & \textbf{Position in} & \textbf{Position in the Given Sequence} \\& \textbf{alphabetical order}&\\\hline A & 1 & 1^{st} \\\hline B & 2 & 2^{nd} \text{ and }\ \ 3^{rd} \\\hline C & 3 & 4^{th}, 5^{th} \text{ and } \ \ 6^{th} \\\hline D & 4 & 7^{th}, 8^{th}, 9^{th} \text{ and } \ \ 10^{th} \\\hline \ldots & \ldots& \ldots \\\hline \text{Some Alphabet} & n & \left ( \frac{n(n+1)}{2} - n+1 \right ),

\left ( \frac{n(n+1)}{2} - n+2 \right ),\\&&\ldots,240^{th},\ldots, \left(\frac{n.(n+1)}{2}\right)\\

\hline

\end{array}$$ Here, $\frac{n(n+1)}{2}$ is the last position in the given sequence for an alphabet whose position is $n$ in the alphabetical order. So, If we have to find an alphabet whose position is $i^{th}$ in the given sequence and whose position is $n$ in the alphabetical order then we have to find smallest integer value of $n$ such that $\frac{n(n+1)}{2} \geqslant i$

For example, if $i=5,$ it means we have to find alphabet whose position is $5^{th}$ in the given sequence, then we have to find minimum value of $n$ such that $\frac{n(n+1)}{2} \geqslant 5$, So, minimum value of $n = 3$ means it will be alphabet $C.$

Similarly, if $i=8$ means we have to find alphabet whose position is $8^{th}$ in the given sequence, then we have to find minimum value of $n$ such that $\frac{n(n+1)}{2} \geqslant 8$, So, minimum value of $n = 4$ means it will be alphabet $D.$

Now, in this question, we have to find **smallest** integer value of $n$ such that

$\quad\frac{n(n+1)}{2}$ $\geq$ $240$ $(\because$ Sum of first $n$ natural numbers = $\frac{n(n+1)}{2})$

$\quad \Rightarrow n*(n+1)\geq 480$

Options given are $V, U, T, W.$ So, value of $n$ can be $22,21,20$ or $23.$

$22*23 = 506$ $\geq$ $480$ $\Rightarrow$ $n= 22.$

The $22^{nd}$ alphabet is $V.$

$\therefore$ Option $(A).$ $V$ is the correct answer.

Question 25 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER] |

**If $\sin(\sin^{-1} \frac{2}{5} + \cos ^{-1} x) =1$, then $x$ equals**

$1$ | |

$\frac{2}{5}$ | |

$\frac{3}{5}$ | |

None of these |

**Question 25 Explanation: $sin \left (sin^{-1}\ \frac{2}{5} + cos^{-1}\ x \right) =1$ **

$\implies \left (sin^{-1}\ \frac{2}{5} + cos^{-1}\ x \right) =sin^{-1}\ 1$

$\implies sin^{-1}\ \frac{2}{5} + cos^{-1}\ x =\frac{\pi}{2}$

$\implies cos^{-1}\ x =\frac{\pi}{2}- sin^{-1}\ \frac{2}{5}$

$\implies cos^{-1}\ x =cos^{-1}\ \frac{2}{5}$ $( \because sin^{-1}\ \frac{2}{5} + cos^{-1}\ \frac{2}{5}=\frac{\pi}{2})$

$\implies x= \frac{2}{5}$

$\therefore$ Option $B.$ is correct

**NTA NET study material**$\implies \left (sin^{-1}\ \frac{2}{5} + cos^{-1}\ x \right) =sin^{-1}\ 1$

$\implies sin^{-1}\ \frac{2}{5} + cos^{-1}\ x =\frac{\pi}{2}$

$\implies cos^{-1}\ x =\frac{\pi}{2}- sin^{-1}\ \frac{2}{5}$

$\implies cos^{-1}\ x =cos^{-1}\ \frac{2}{5}$ $( \because sin^{-1}\ \frac{2}{5} + cos^{-1}\ \frac{2}{5}=\frac{\pi}{2})$

$\implies x= \frac{2}{5}$

$\therefore$ Option $B.$ is correct

**Once you are finished, click the button below. Any items you have not completed will be marked incorrect.**

** There are 25 questions to complete.**