# GATE aptitude questions

## Aptitude for GATE

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 Question 1 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER]
In appreciation of the social improvements completed in a town, a wealthy philanthropist decided to gift $Rs\; 750$ to each male senior citizen in the town and $Rs\; 1000$ to each female senior citizen. Altogether, there were 300 senior citizens eligible for this gift. However, only $\dfrac{8}{9}^{th}$ of the eligible men and $\dfrac{2}{3}^{rd}$ of the eligible women claimed the gift.How much money (in Rupees) did the philanthropist give away in total?
 A $1,50,000$ B $2,00,000$ C $1,75,000$ D $1,51,000$
Question 1 Explanation:
Let no. of senior male be $x$
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]
Triangles ABC and CDE have a common vertex C with side AB of triangle ABC being parallel to side DE of triangle CDE. If length of side AB = 2 cm and length of side DE = 7 cm and perpendicular distance between sides AB and DE is 7.2 cm, then find the sum of areas of triangle ABC and triangle CDE.
 A 10.1 B 23.1 C 22.5 D 21.2
Crack any exam
Question 2 Explanation:
Proof ∆ABC ≂∆CDE (alternate angle and vertical angle are same ,so using AA property both are similar) therefore,

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?

 A $\frac{75}{11}$ B $\frac{125}{11}$ C $\frac{25}{3}$ D $5$
NTA NET study material
Question 3 Explanation:
Let the rate of change of height of the tank be denoted by $\frac{dH}{dt}$
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]
An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is no replacement of the balls drawn.)
 A 3/12 B 3/11 C 1/2 D None of these
NTA NET study material
Question 4 Explanation:

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.

 A The sequence $\large (\epsilon _{n}: n= 1, 2,...)$ converges to zero. B $\large \epsilon _{n}\geq 1$ for all $n$ C The sequence $\large (\epsilon _{n}: n= 1, 2,...)$ is decreasing and converges to $1.$ D The sequence $large (epsilon _{n}: n= 1, 2,...)$ is decreasing and then increasing. Finally it converges to $1.$
Question 5 Explanation:

$\large\epsilon_1$ 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 $2$?

 A $1-a-c-2$ B $1-f-b-2$ C $1-b-2$ D $1-f-e-2$
Crack any exam
Question 6 Explanation:

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]
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
 A 1/5 B 1/3 C 1/4 D 2/7
Question 7 Explanation:
Suppose the vessel initially contains 8 litres of liquid. Let 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]
Choose the most appropriate word from the options given below to complete the following sentence$:$

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

 A aberration B weakening C decline D deterioration
Question 8 Explanation:
famine is acute shortage of food and as per the sentence we require a word meaning "reduction in food output."
"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?

 A $51:49$ B $1:1$ C $49:51$ D $51:98$
NTA NET study material
Question 9 Explanation:

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]
square and regular octagon has same perimeter..find ratio of area octagon to square.....
 A $\frac{1 + \sqrt{2}}{3}$ B $\frac{1 + \sqrt{2}}{2}$ C $\frac{1 + \sqrt{3}}{2}$ D $\frac{2 + \sqrt{2}}{2}$
Question 10 Explanation:

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 $xy-z^2=1$?

 A $0$ B $1$ C $2$ D infinitely many
Question 11 Explanation:
$(x-y)^2 = (x+y)^2 - 4 xy$
$(x-y)^2 = 4 - 4(1+z^2)$

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

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]
what is the unit digit in

$(3474)^{1793}*(225)^{317}*(451)^{491}$
 A 0 B 1 C 2 D 3
Question 12 Explanation:

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?

 A There are three consecutive integers with sum $2015$ B There are four consecutive integers with sum $2015$ C There are five consecutive integers with sum $2015$ D There are three consecutive integers with product $2015$
Question 13 Explanation:

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:-

1. some authors are teachers
2. No teacher is a lady.

conclusion:-

 A I follows B II follows C Both I and II follows D None of I and II follows
Question 14 Explanation:
Pick one teacher from the set of teachers and check if that teacher is a lady or man. If there exists atleast one teacher who is not a lady then we can say I follows. Since its given that no teacher is a lady, so whoever we pick, we will see that the teacher is not a lady. So I follows.
 Question 15 [CLICK ON ANY CHOICE TO KNOW MCQ multiple objective type questions RIGHT ANSWER]
A recent high court judgment has sought to dispel idea of begging as a disease- which leads to its stigmatization and criminalization and regard it as symptom. The underlying disease is the failure of state to protect citizen who fall through social security net. Which of the following can be inferred above?
 A Begging has to be banned because it adversely affects welfare of the state B Begging is an offence the has to dealt firmly C Beggar are created due to lack of welfare scheme of state D Beggar are criminals
Question 15 Explanation:

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.

1. 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.
2. Beggars are created because of the lack of social welfare schemes - Can be inferred from the last sentence.
3. 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.
4. 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

 A Argument I is Strong B Argument II is Strong C I or II is Strong D I and II are strong
UPSC FREE STUDY
Question 16 Explanation:
Both the arguments are strong enough. The conditions have to be agreed to, in order to save the life of the victims, though actually they ought not to be agreed to, as they encourage the sinister activities of the kidnappers.
 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

 A Inept B Graceful C Suitable D Dreadful
Question 17 Explanation:
Awkward means unsuitable and is a synonym for inept.

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]
The population of a state in the year was 93771. If the rate of increase was 8% and 15% respectively from the previous 2 years before 2013. What was the population in the year 2011?
 A 55650 B 49550 C 75550 D 85550
Question 18 Explanation:
2013----- 93771

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]
In how many number of ways we can paint 12 offices so that 3 are green 2 pink 2 yellow and remaining ones are white?
 A 55,440 B 1,66,320 C 4.790E+08 D 39,91,680
Question 19 Explanation:
Total number of ways 12 offices can be painted = 12!

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]
A man purchases 11 articles @ 10 rs and sells 10 articles @11 rs. Find The Overall Loss or Gain% ?
 A 23% B 21% C 26% D 20%
Question 20 Explanation:
Now cost price of 1 article = Rs. 10/11

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]
An article was sold at a discount of 15% and there was a profit of 20%. If the gained was 6 rupees less than the discount offered, what was the selling price of the article ?
 A 212 B 612 C 651 D 623
Question 21 Explanation:
Let $\text{SP} =x$

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

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

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

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

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

$\implies x=612.$
 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 _______

 A 0 to 15 B 16 to 30 C 31 to 45 D 46 to 60
Question 22 Explanation:
$\rightarrow$ No of administrator $=10+20+20=50$

$\rightarrow$ 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]
Find the number of ways in which the letters of the word MACHINE can be arranged so that the vowels may occupy only odd positions.
 A 4! $\times$ 4! B 7P3 $\times$ 4! C 7P4 $\times$ 3! D none of these
Question 23 Explanation:
In word  MACHINE there are 3 vowels  {A,E,I} and 4 consonent {M,C,H,N}
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 4Pways
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 = 4P3 * 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 $:$

 A $V$ B $U$ C $T$ D $W$
Question 24 Explanation:

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

 A $1$ B $\frac{2}{5}$ C $\frac{3}{5}$ D None of these
NTA NET study material
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
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