IIFT 2022 Slot 1 Question Paper (18th Dec)

It is given that 10th July 2010 was a Thursday.

Let's calculate what day it is on 10th July 2003; from that; we can calculate 6th July 2003.

One year = 365 days = 52 weeks + 1day 

Hence, For any given date, the day will change by one place (i.e. Monday to Tuesday,...Wednesday to Thursday,...) for the succeeding year.

One Leap year = 366 days 52 weeks + 2 days.

Hence, If the year is a leap year, then the day will change by two places (i.e. Monday to Wednesday,...Wednesday to Friday,...) for the succeeding year.

We get the following table,

As 10th July 2003 is Tuesday, 6th July 2003 will be Friday.

Option (C) is the answer.

Let $$'y'$$ be the principal amount invested at a 6.5% per annum rate.

Let $$'x'$$ be the principal amount invested at a 7.5% per annum rate.

It is given that, $$\frac{y}{4}=\frac{x}{5}$$............(1)

It is also given that without reinvestment, his income from interest after 3 years is Rs. 400

i.e., $$y\times\ 3\times\ 0.065+x\times\ 3\times\ 0.075=400$$

$$\frac{4x}{5}\times\ 3\times\ 0.065+x\times\ 3\times\ 0.075=400$$

$$0.381x=400$$

Hence, $$x=Rs.\ 1049.87$$, then $$y=\frac{4x}{5}=\frac{1049.87\times\ 4}{5}=Rs.\ 839.90$$

Total invested sum = $$y+x=$$ = $$1889.77\approx Rs.1890$$

Option (A) is correct.

Let $$'x'$$ be the true time when the clock indicates 7 PM on 17th January 2022.

Let $$'y'$$ be the number of hours elapsed between 7 AM on 12th January, 2022 and $$'x'$$

The clock loses 18 minutes per 24 hours. 

For one hour, it loses$$=\frac{18}{24\ \times \ 60}=\frac{1}{80}\ \ hrs$$

For $$'y'$$ hours it loses = $$\frac{y}{80}\ \ hrs$$

Hence, the number of hours elapsed between 7 AM on 12th January 2022 and 7 PM on 17th January 2022 = $$y-\frac{y}{80}$$

$$y-\frac{y}{80}=5\times\ 24+12$$

$$\frac{79y}{80}=132$$

$$y=133.671\ hrs$$

$$y=24\times\ 5.5+1.67\ hrs$$

Hence, the approximate true time when the clock indicates 7 PM on 17th January 2022 = 7 AM, 12th January 2022 + $$'y' hrs$$

= 7 AM, 12th January 2022 + 5days +12 hrs +1.67 hrs

= 7 PM on 17th January 2022 + 1.67 hrs

= 08:40:15 PM 17th January 2022.

Option (D) is the answer.

Let the volume of the tank be L.C.M of (12,18) = $$72 m^3$$

The rate of inflow of the pipes are, $$\frac{72}{12}=6\ \frac{m^3}{hr}\ \&\ \frac{72}{18}=4\ \frac{m^3}{hr}$$

Let the rate of leakage be $$x\ \frac{m^3}{hr}$$

The total time it takes for both pipes to fill the tank = $$\frac{72}{6+4}=\frac{72}{10}=7.2\ hr$$

But due to leakage, the pipes took 36 min more to fill the tank, $$\frac{72}{6+4-x}=7.2\ +\frac{36}{60}$$

$$\frac{72}{10-x}=7.8$$

on solving, we get $$x=\frac{10}{13}\ \frac{m^3}{hr}$$

The total time taken by the leak to empty the full tank = $$\frac{72}{\frac{10}{13}}=93.6\ hr$$

The total marks = $$100\times\ 6=600$$

The recruit got 55% of the total marks, i.e. = $$55\%\times\ 600=330$$.

The recruit gets marks in the ratio of 10:9:8:7:6:5 in each parameter.

Let the marks in each parameter be $$10x,\ 9x,\ ...,\ 5x$$.

we have, $$10x+\ 9x+8x+7x+6x+5x=330$$

$$45x=330$$

$$x=\frac{22}{3}$$

Hence, the marks in each parameter are $$73.33,\ 66,\ 58.67,\ 51.33,\ 44,\ 36.67$$

The criteria for clearing the performance score is 55% of the maximum marks for first 3 and for the next 3 parameters is 50% of the maximum marks.

Hence the recruit cleared four parameters. Option (c) is the answer.

Let the lengths of BD, CD and AD be $$x,\ y\ \&\ h$$, respectively.

=> $$ BD+CD = x+y=21$$.....(1) 

As AD is altitude, using Pythagoras' theorem, we get

$$13^2=x^2+h^2$$ .....(2)

$$20^2=y^2+h^2$$.....(3)

Eq (3) - (2) gives 

$$20^2-13^2=y^2-x^2$$ 

$$231=\left(y-x\right)\ \left(y+x\right)$$

 $$\left(y-x\right)=\frac{231}{x+y}=\frac{231}{21}=\ 11$$......(4)

Adding Eq(1) & (4), we get $$2y=32$$  => $$y=16 cm$$ 

we get $$x+16=21$$ => $$x=5 cm$$

From Eq(2) => $$13^2=5^2+h^2$$ 

$$h^2=13^2-5^2$$ => $$h=12 cm$$

The area of the smaller triangle ABD = $$\ \frac{\ BD\times\ AD}{2}$$

= $$\ \frac{\ 5\times\ 12}{2}$$ = $$30 cm^2$$

30% of area of smaller triangle = $$30\%\ \times\ 30=9\ cm^2$$

Option (C) is the answer

It is given that 900 bags were made, and only 785 were sold.

Cost of manufacturing 900 bags $$=1200+\frac{125}{105}\times\ 900+0.80\times\ 900$$ $$=\ Rs. \frac{20940}{7}$$.....(1)

Revenue of the factory = Amount from sales + Amount from advertisements......(2)

Let the amount from advertisements be $$ Rs. 'y'$$

Amount from sales = $$785\times\ 3.25= Rs. 2551.25$$

Factory's profit percentage = 12%.....(3)

From Eq (1), (2) & (3)

We get, $$2551.25\ +y=\frac{20940}{7}\ \times\ 1.12$$

On solving, we get $$y=Rs.\ 799.15$$

Hence the approximate revenue from advertisements is = $$Rs.\ 799$$

Option (C) is the answer.

Let '$$p$$' be the principal amount and '$$r$$' be the interest rate per annum.

Simple Interest for t years = $$ptr$$

Compound Interest for t years = $$p\left(1+r\right)^t-p$$

From the given information,

$$\frac{\left(p\left(1+r\right)^4-p\right)-4pr}{\left(p\left(1+r\right)^3-p\right)-3pr}=\frac{20}{8}$$

$$\frac{\left(\left(1+r\right)^4-1\right)-4r}{\left(\left(1+r\right)^3-1\right)-3r}=\frac{5}{2}$$

$$2\times\ \left(\left(1+r\right)^4-1-4r\right)=5\times\ \left(\left(1+r\right)^3-1-3r\right)$$

$$2\times\ \left(1+r\right)^4-2-8r=5\times\ \left(1+r\right)^3-5-15r$$

$$2\times\ \left(1+r\right)^4=5\times\ \left(1+r\right)^3-3-7r$$

$$\ 2r^4+8r^3+12r^2+8r+2=\ 5r^3+15r^2+15r+5-3-7r$$

$$\ 2r^4+8r^3+12r^2=\ 5r^3+15r^2$$

$$\ 2r^2+3r-3=0$$

Hence, $$r=\frac{-3\pm\ \sqrt{\ 9+24}}{2\times\ 2}$$

$$r=\frac{-3\pm\ \sqrt{\ 33}}{4}$$

As interest rate cannot be negative,

$$r=\frac{-3+\ \sqrt{\ 33}}{4}$$ 

$$r=0.6861$$

Hence, the approximate value of $$r=69\%$$

Option (A) is correct.

The largest remainder we can get, = smallest of {largest remainder of '4', the largest remainder of '11', the largest remainder of '9'}

= smallest of {3,10,8} = 3.

The number which gives the remainder '3', when divided by 4,9& 11, will be of the form = L.C.M of (4,9&11) x n + 3.

L.C.M of (4,9&11) = 396.

The number which gives the remainder '3', when divided by 4,9& 11, will be of the form = 396n+3.

The largest 5-digit number of the form, 396n+3 = 396 x 252 + 3

=99792+3=99795.

Option (D) is correct.

From the above figure,

Using Pythagoras theorem

$$In\ \triangle\ AQO\ ,\ 55^2=a^2+c^2$$.....(1)

similarly. 

$$In\ \triangle\ BQO\ ,\ 65^2=a^2+d^2$$.....(2)

$$In\ \triangle\ CSO\ ,\ 75^2=b^2+d^2$$.....(3)

$$In\ \triangle\ DSO\ ,\ m^2=b^2+c^2$$.....(4)

From observation, Eq(1)-Eq(2)  = Eq(4)-Eq(3) = $$c^2-d^2$$

Hence, $$55^2-65^2=m^2-75^2$$

$$3025-4225=m^2-5625$$

$$m^2=4425$$

$$m=\pm\ 66.52$$

As 'm' cannot be negative 

$$m\approx\ \ 66.5$$

Option (A) is correct.

$$64\times\ \frac{-11}{16}+4\times\ -\frac{11}{16}\times\ 4^2+32\times\ -\frac{11}{16}\times\ 4$$It is given that b=-14.

Then the eq $$5x^2 + bx + 8 = 0$$ becomes $$5x^2 -14x + 8 = 0$$.

$$x=\frac{-b\pm\ \sqrt{\ b^2-4ac}}{2a}=\frac{14\pm\ \sqrt{\ 196-160}}{10}$$

$$=\frac{14\pm\ 6}{10}$$

$$x=2\ or\ 0.8$$

Hence $$\beta=2$$.

If $$\beta=2$$, Then $$\alpha = \beta^2$$ = $$4$$.

As $$\alpha$$ is a root of $$ax^2 + 2x + 3 = 0$$

$$4^2a+8+3=0$$

$$a=-\frac{11}{16}$$.

By verifying options, Option D satisfies for the a & b values.

= $$64\times\ \frac{-11}{16}+4\times\ -\frac{11}{16}\times\ \left(-14\right)^2+32\times\ -\frac{11}{16}\times\ -14$$

$$=\ -275$$

Hence Option (D) is the answer.

Let $$\sqrt{\ 1-x}=a\ \&\ \sqrt{\ 1+x}=b$$

R.H.S = $$(2 + 2\sqrt{1 - x^2})^{\frac{3}{2}}$$

=$$(1+x+1-x + 2\sqrt{(1 - x)(1+x)})^{\frac{3}{2}}$$

=$$(a^2+b^2+ 2ab)^{\frac{3}{2}}$$

=$$\left(a+b\right)^{2\times\ \frac{3}{2}}$$ 

=$$\left(a+b\right)^3$$

L.H.S = $$(1 - x)^{\frac{3}{2}} + (1 + x)^{\frac{3}{2}} + 2(\sqrt{1 - x^2})$$

= $$a^3+b^3+2ab$$

=$$\left(a+b\right)^3-3ab(a+b)+2ab$$

Equating L.H.S & R.H.S, we get

$$\left(a+b\right)^3-3ab(a+b)+2ab = \left(a+b\right)^3$$

$$a+b=\frac{2}{3}$$

substituting the values of 'a' & 'b' we get

$$\sqrt{\ 1-x}+\sqrt{\ 1+x}=\frac{2}{3}$$

squaring on both sides, we get

$$1-x+1+x+2\sqrt{\ 1-x^2}=\frac{4}{9}$$

$$2\sqrt{\ 1-x^2}=\frac{4}{9}-2$$

$$\sqrt{\ 1-x^2}=\frac{-7}{9}$$

Squaring on both sides

$$1-x^2=\frac{49}{81}$$

$$x^2=\frac{32}{81}$$

$$x=\frac{4\sqrt{\ 2}}{9}$$

Hence, option (D) is the answer.

The given equation is $$(x - 12)^2 + (y - 10)^2 = 64$$

$$(x - 12)^2 + (y - 10)^2 = 8^2$$

It is an equation of a circle with a radius of 8 units & its centre is at (12,10).

It implies that the extreme coordinate of 'x' is = 12+8 = 20.

Similarly, the extreme coordinate of 'y' is = 10+8 = 18.

The square numbers which are less than 20 & 18 are 4, 9 & 16.

For a point (x,y) to lie inside the circle, $$(x - 12)^2 + (y - 10)^2 = 64$$, the value of $$(x - 12)^2 + (y - 10)^2$$ should be less than 64.

Case (i):- (4,4)

$$(4-12)^2+(4-10)^2=8^2+6^2=100 > 64$$. 

Hence, the point lies outside the circle.

As $$(4-12)^2=64$$, all the square points with '4' as x coordinate i.e (4,9), (4,16),  lie outside the circle.

Case (ii):- (9,9)

$$(9-12)^2+(9-10)^2=3^2+1^2=10 < 64$$.

Hence, the point lies inside the circle.

Case (iii):- (16,16)

$$(16-12)^2+(16-10)^2=4^2+6^2=52 < 64$$.

Hence, the point lies inside the circle.

Case (iv):- (9,4)

$$(9-12)^2+(4-10)^2=3^2+6^2=45 < 64$$.

Hence, the point lies inside the circle.

Case (v):- (9,16)

$$(9-12)^2+(16-10)^2=3^2+6^2=45 < 64$$.

Hence, the point lies inside the circle.

Case (vi):- (16, 4)

$$(16-12)^2+(4-10)^2=4^2+6^2=52 < 64$$.

Hence, the point lies inside the circle.

Case (vii):- (16, 16)

$$(16-12)^2+(16-10)^2=4^2+6^2=52 < 64$$.

Hence, the point lies inside the circle.

A total of 6 perfect square points lie inside the circle.

Option 'C' is the answer.

Let's consider an n-digit number.

$$_{-------------n\ digits}$$

The 900 distinct numbers have to be formed using digits 2, 3, 4, 5 and 7.

The first and last digit numbers are the same & the first and third are prime numbers

$$abc_{-----------}a\ total\ 'n'\ digits.$$

the values of a & c can be anything from 2,3,5, and 7. i.e 4 values.

The second digit should be greater than or equal to 4. Hence, b can take any value from 4, 5 & 7. i.e. 3 values.

There are no conditions from the 4th digit to the 'n-1' digit.

Hence those digits can take any value from 2, 3, 4, 5 and 7, i.e. 5 values.

For n = 4, the number of digits that can be formed $$=4\times\ 4\times\ 3=48 < 900$$

For n = 5, the number of digits that can be formed $$=4\times\ 4\times\ 3\times\ 5 = 240 < 900$$

For n = 6, the number of digits that can be formed $$=4\times\ 4\times\ 3\times\ 5\times\ 5=1200 > 900$$

Hence a minimum of 6-digit number is required to form 900 distinct numbers with given conditions.

Option (B) is correct.

Let the probability of an even outcome of dice be $$'x'$$.

Then the probability of the odd outcome of the dice is $$'2x'$$.

The sum of all probabilities of all events is 1.

It implies $$x+x+x+2x+2x+2x=1$$

$$x=\frac{1}{9}$$

Probability of drawing a green ball from pot 'A' = Probability of selecting pot 'A' x probability of drawing a green ball.

= $$\left(\frac{2}{9}+\frac{1}{9}+\frac{2}{9}\right)\times\ \frac{10}{15}$$  $$=\frac{5}{9}\times\ \frac{10}{15}=\frac{10}{27}$$

Similarly, the Probability of drawing a green ball from pot 'B' = $$\left(\frac{1}{9}+\frac{2}{9}+\frac{1}{9}\right)\times\ \frac{2}{9}$$

 $$=\frac{4}{9}\times\ \frac{2}{9}=\frac{8}{81}$$

Probability of drawing a green ball = Probability of drawing a green ball from pot 'A' +  Probability of drawing a green ball from pot 'B'.

 $$=\frac{10}{27}+\ \frac{8}{81}=\frac{38}{81}$$

Option (A) is the answer.

The total number of members of Sahitya Shakti is 500.

Total members after removing people who read Shakespeare = 500-200=300

Total members after removing people who read Tolstoy = 300-100=200

Society added back people who read Shakespeare but not Tolstoy, and the number increased to 350.

Hence society added back 150 people. But initially, society removed 200 members.

Therefore, the other 50 members who were not added back also read Tolstoy. Those 50 members read all three books.

Option (A) False. Only 150 original members read both Premchand and Tolstoy.

Option(B) True. 50 original members read together Premchand and Shakespeare and Tolstoy

Option(C) False. 350 original members read both Premchand and Shakespeare but not Tolstoy

Option (D) False. 50 original members read both Shakespeare and Tolstoy

It is given that, $$P(A \cup B) < \frac{3}{4}, P(A) > \frac{1}{8}, P\left(\frac{A}{B}\right) < \frac{1}{2}$$

$$P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P\left(B\right)}< \frac{1}{2}$$

$$2\ P(A \cap B) < P(B)$$

 $$P(A \cap B) < P(B)-P(A \cap B)$$.........(1)

$$ P(A \cup B) = P(A) + P(B) - P(A \cap B) < \frac{3}{4}$$

$$ P(A) + P(B) - P(A \cap B) < \frac{3}{4}$$

$$ P(A) + (P(A \cap B) <) < \frac{3}{4}$$      From Eq (1),

$$ P(A) + P(A \cap B) < \frac{3}{4}$$

Hence Options (A), (B) & (D) are false.

$$ P(A) + P(A \cap B) < \frac{3}{4}$$

$$ (> \frac{1}{8}) + P(A \cap B) < \frac{3}{4}$$       From  $$(P(A) > \frac{1}{8})$$

$$P(A \cap B) < \frac{3}{4}-\frac{1}{8}$$

$$P(A \cap B) < \frac{5}{8}$$

Option (C) is correct.

Given that,

$$\tan\left(\theta\ -\gamma\ \right)=\frac{1}{\sqrt{\ 2}}$$

$$\frac{\tan\theta-\tan\gamma}{1+\tan\theta\ \tan\gamma\ }\ =\frac{1}{\sqrt{\ 2}}$$

$$\frac{\tan\theta-\tan\gamma}{1+\tan^2\alpha\ }\ =\frac{1}{\sqrt{\ 2}}$$

$$\frac{\tan\theta-\tan\gamma}{\sec^2\alpha\ }\ =\frac{1}{\sqrt{\ 2}}$$

$$\ \tan\theta\ -\tan\gamma\ =\frac{1}{\sqrt{\ 2}\cos^2\alpha\ }$$

$$\ \frac{\tan^2\alpha\ }{\tan\gamma\ }-\tan\gamma\ =\frac{1}{\sqrt{\ 2}\cos^2\alpha\ }$$

$$\ \tan^2\alpha\ -\tan^2\gamma\ =\frac{\tan\gamma\ }{\sqrt{\ 2}\cos^2\alpha\ }$$

Squaring on both sides,

$$\tan^4\alpha\ +\tan^4\gamma-2\ \tan^2\alpha\ \tan^2\gamma\ =\frac{\tan^2\gamma\ }{2\cos^4\alpha\ }$$

$$\tan^4\alpha\ +\tan^4\gamma\ =2\ \tan^2\alpha\ \tan^2\gamma+\frac{\tan^2\gamma\ }{2\cos^4\alpha\ }$$

$$\tan^4\alpha\ +\tan^4\gamma\ =\tan^2\gamma\ \left(2\ \tan^2\alpha\ +\frac{1}{2\cos^4\alpha\ }\right)$$

$$\tan^4\alpha\ +\tan^4\gamma\ =\tan^2\gamma\ \left(\ \frac{4\sin^2\alpha\ \cos^2\alpha}{2\cos^4\alpha}\ +\frac{1}{2\cos^4\alpha\ }\right)$$

$$\tan^4\alpha\ +\tan^4\gamma\ =\frac{\tan^2\gamma}{2}\ \left(\ \frac{4\sin^2\alpha\ \cos^2\alpha+1}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \frac{6\sin^2\alpha\ \cos^2\alpha+1-2\sin^2\alpha\ \cos^2\alpha}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \frac{6\sin^2\alpha\ \cos^2\alpha+\left(\sin^2\alpha\ +\cos^2\alpha\right)^2-2\sin^2\alpha\ \cos^2\alpha}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \frac{6\sin^2\alpha\ \cos^2\alpha+\sin^4\alpha\ +\cos^4\alpha+2\sin^2\alpha\ \cos^2\alpha-2\sin^2\alpha\ \cos^2\alpha}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \frac{6\sin^2\alpha\ \cos^2\alpha+\sin^4\alpha\ +\cos^4\alpha}{\cos^4\alpha}\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ 6\tan^2\alpha+\tan^4\alpha\ +1\ \right)$$

$$=\frac{\tan^2\gamma}{2}\ \left(\ \left(\tan^2\alpha+3\right)^2\ -8\ \right)$$

Option (A) is the answer.

It is given that $$\angle\ OAM=60^{\circ\ },\ \angle\ OBM=45^{\circ},\ \angle\ OCM=30^{\circ\ },\ AB=y,\ BC=15-5\sqrt{\ 3}\ $$

Let the height of the tree be OM = $$'h'$$.

Then $$AM\ =\dfrac{OM}{\tan\ 60^{\circ\ }}=\dfrac{h}{\sqrt{\ 3}}$$

From $$\triangle\ OBM,\ \tan\ 45^{\circ\ }=\dfrac{h}{\dfrac{h}{\sqrt{\ 3}}+y}$$

$$\dfrac{h}{\sqrt{\ 3}}+y=h$$

$$y=h-\dfrac{h}{\sqrt{\ 3}}$$

From $$\triangle\ OCM,\ \tan30^{\circ\ }=\dfrac{h}{15-5\sqrt{\ 3}+y+\dfrac{h}{\sqrt{\ 3}}}$$

$$\dfrac{1}{\sqrt{\ 3}}=\dfrac{h}{15-5\sqrt{\ 3}+h-\dfrac{h}{\sqrt{\ 3}}+\dfrac{h}{\sqrt{\ 3}}}$$

$$\dfrac{1}{\sqrt{\ 3}}=\dfrac{h}{15-5\sqrt{\ 3}+h}$$

$$15-5\sqrt{\ 3}+h=h\sqrt{\ 3}$$

$$15-5\sqrt{\ 3}=h\left(\sqrt{\ 3}-1\right)$$

$$h=\dfrac{15-5\sqrt{\ 3}}{\left(\sqrt{\ 3}-1\right)}$$

$$h=\dfrac{5\sqrt{\ 3}\times\ \left(\sqrt{\ 3}-1\right)}{\left(\sqrt{\ 3}-1\right)}=5\sqrt{\ 3}$$

Option (C) is the answer.

Given that $$x = a(a - d) (a + d) (a + 2d) + d^4$$ 

$$x=a\left(a-d\right)\left(a+d\right)\left(a+2d\right)+d^4$$

$$=a\left(a+d\right)\left(a-d\right)\left(a+2d\right)+d^4$$

$$=\left(a^2+ad\right)\left(a^2+2ad-ad-2d^2\right)+d^4$$

Let $$t=a^2+ad$$

$$x=\left(t\right)\left(t-2d^2\right)+d^4$$

$$x=t^2-2td^2+d^4$$

$$x=\left(t-d^2\right)^2$$

Substituting $$t=a^2+ad$$

$$x=\left(a^2+ad-d^2\right)^2$$.........(1)

Given that,  $$y = a(a + d) (a + 2d) (a + 3d) + d^4$$

$$y=a\left(a+d\right)\ \left(a+2d\right)\ \left(a+3d\right)+d^4$$

$$=a\left(a+3d\right)\ \left(a+d\right)\ \left(a+2d\right)+d^4$$

$$=\left(a^2+3ad\right)\ \left(a^2+2ad+ad+2d^2\right)+d^4$$

$$=\left(a^2+3ad\right)\ \left(a^2+3ad+2d^2\right)+d^4$$

Let  $$t=a^2+3ad$$

$$=\left(t\right)\ \left(t+2d^2\right)+d^4$$

$$=t^2+2td^2+d^4$$

$$=\left(t+d^2\right)^2$$

Substituting  $$t=a^2+3ad$$

$$y=\left(a^2+3ad+d^2\right)^2$$.........(2)

From Eq (1) & (2)

We get, $$x+y=\left(a^2-d^2+ad\right)^2+\left(a^2+d^2+3ad\right)^2$$

Option (D) is correct.

Let the cost of blue liquid per litre be $$'x'$$

then the cost of green liquid per litre is $$'x+4'$$

The S.P. of the mixture is Rs 20 per litre, and the profit % is 20%.

Then C.P. of the mixture per litre = $$\frac{20}{1+0.2}=\frac{20}{1.2}=\frac{50}{3}$$......(1)

Also, the C.P. of the mixture per litre is given by = $$\frac{4\left(x+4\right)+5\left(x\right)}{5+4}=\frac{9x+16}{9}$$.....(2)

Equating (1) to (2)

we get $$\frac{50}{3}=\frac{9x+16}{9}$$

$$150=9x+16$$

$$9x=134$$

$$x=14.88$$

Hence, the cost of green liquid per litre is = 14.88+4= 18.88.

Option (B) is correct.

Let the length of both trains be $$'x'$$. 

The speeds of trains are 54Kmph & 40Kmph.

The faster train passes the length of the slower train in 36 seconds.

It implies, $$\frac{x+x}{54-40}=\frac{36}{60\times\ 60}$$

$$\frac{2x}{14}=\frac{36}{60\times\ 60}$$

$$x=0.07\ Km$$

$$x=70\ m$$.

Hence, Option (B) is correct.

It is given that Ram covers ED in one-fifth of the time taken for AE, with a 2km/hr constant speed.

It implies, $$\frac{1}{5}\times\ \frac{AE}{2}\ =\frac{ED}{2}$$

$$AE\ =5\times\ ED$$......(1)

Let the speed of Shyam be 's' Kmph. He covers the whole distance from C to B through point E at the same speed in 18 hours.

It implies, $$\frac{CE+BE}{S}=18 Hrs$$.......(2)

From the properties of circles, $$AE\times\ BE=CE\times\ DE$$

$$5DE\times\ BE=CE\times\ DE$$        [From EQ(1)]

$$CE=5BE$$......(3)

From EQ(2) & (3)

$$\frac{5BE+BE}{S}=18$$

$$\frac{6BE}{S}=18$$

$$\frac{BE}{S}=3\ Hrs.$$

Option (C) is the answer.

For Triangle ECD,

$$EC=DE=\sqrt{\ b^2+\left(\frac{b}{2}\right)^2}=\frac{\sqrt{\ 5}b}{2}$$

CD = b

$$Area\ of\ \triangle\ ECD=r\times\ s$$....(1)

$$s=\frac{\left(\frac{\sqrt{\ 5}b}{2}+\frac{\sqrt{\ 5}b}{2}+b\right)}{2}=\frac{b\left(\sqrt{\ 5}+1\right)}{2}$$...(2)

$$Area\ of\ \triangle\ ECD\ =\ \frac{B\times\ H}{2}=\frac{b\times\ b}{2}=\frac{b^2}{2}$$.......(3)
From EQ (1), (2) & (3)

we get, $$\frac{b^2}{2}=\frac{b\left(\sqrt{\ 5}+1\right)}{2}\times\ r$$

$$\ r=\frac{b}{\sqrt{\ 5}+1}=\frac{b\left(\sqrt{\ 5}-1\right)}{4}$$

Area of the incircle of ECD = $$\ \pi\ r^2$$ 

= $$\ \pi\ \times\ \left(\frac{b\left(\sqrt{\ 5}-1\right)}{4}\right)^2$$

$$=\frac{\pi\ b^2\left(5+1-2\sqrt{\ 5}\right)}{16}=\frac{\pi\ b^2\left(3-\sqrt{\ 5}\right)}{8}$$....(4)

For Triangle AED,

Inradius, $$r=\frac{AE+AD-DE}{2}=\frac{b+\frac{b}{2}-\frac{\sqrt{\ 5}b}{2}}{2}=\frac{b\left(3-\sqrt{\ 5}\right)}{4}$$

Area of the incircle of Triangle AED, = $$\frac{\pi\ b^2\ \left(3-\sqrt{\ 5}\right)^2}{4^2}$$

$$=\pi\ b^2\ \frac{\left(9+5-6\sqrt{\ 5}\right)}{16}=\frac{\pi\ b^2\ \left(14-6\sqrt{\ 5}\right)}{16}$$......(5)

As, Triangle AED $$\simeq\ $$ Triangle BEC

Area of the incircle of $$\triangle\ AED$$ = Area of the incircle of $$\triangle\ BEC$$.

Hence, the area of the shaded region $$=\frac{\pi\ b^2\left(14-6\sqrt{\ 5}\right)}{16}+\frac{\pi\ b^2\left(14-6\sqrt{\ 5}\right)}{16}+\frac{\pi b^2\ \left(3-\sqrt{\ 5}\right)}{8}$$

$$=\frac{\pi\ b^2}{16}\times\ \left(14-6\sqrt{\ 5}+14-6\sqrt{\ 5}+6-2\sqrt{\ 5}\right)$$

$$=\frac{\pi\ b^2\left(34-14\sqrt{\ 5}\right)}{16}$$

Volume of large sphere$$=\frac{4\pi\ }{3}\times\ \left(\ \frac{\ 10}{2}\right)^3=\frac{4\pi\times\ 5^3\ }{3}$$

2/5 of the volume is recast into several spheres of 2m diameter.

Let the number of small spheres be '$$n$$'

$$n\times\ \frac{4\pi}{3}\times\ \left(\frac{2}{2}\right)^3\ =\frac{2}{5}\times\ \frac{4\pi\times\ 5^3\ }{3}$$

On solving, we get $$n=50$$.

The ratio of the total surface area of the smaller spheres and the surface area of the large sphere= $$\left(50\times\ 4\pi r^2\right)\ :\ \left(4\pi R^2\right)$$

= $$\left(50\times\ 4\pi\left(1\right)^2\right)\ :\ \left(4\pi\left(5\right)^2\right)$$

$$=2\ :\ 1$$

Option D is the most effective way to change negativity because it aligns with the passage's view of the subconscious mind. The passage describes the subconscious as a "natural computer" that readily accepts information without analysis. Bypassing the critical conscious mind with positive images and suggestions (uncritical acceptance) directly feeds this "computer" with the desired program for positivity, leading to a more positive outlook and behavior change.

Option A goes against the passage's emphasis on using positive and productive suggestions.

Option B: While analyzing suggestions can be helpful, the passage highlights the subconscious mind's uncritical nature.

Option C creates a contradiction. Critical acceptance implies analyzing, which goes against the idea of uncritical suggestions.

Option A aligns with the passage's ideas because it combines:

Self-hypnosis: This technique, according to the passage, allows for direct positive programming of the subconscious mind.

Daily practice: Repetition strengthens the new positive suggestions in the subconscious.

Targeted goal: Focusing on a specific goal during self-hypnosis helps channel the subconscious mind's power towards achieving that goal.

Option B: The passage argues that willpower alone is not enough. It can be weakened by focusing on past failures.

Option C: While feeding your conscious mind with positivity is a good start, the passage emphasizes directly influencing the subconscious, which bypasses the critical conscious mind.

Option D: The passage suggests that after the initial reprogramming with self-hypnosis, the subconscious mind itself helps maintain the positive changes. So, constant effort wouldn't be necessary.

Option C is incorrect based on the passage because the passage doesn't directly state that willpower leads to establishing one's total self-image as a failure. Instead, it suggests that relying solely on willpower can emphasize past failures, which in turn can hinder improvement. The passage highlights that the problem with the willpower approach is that it places too much emphasis on past failures, contributing to a mental mindset that is not conducive to improvement. However, it does not explicitly state that willpower directly leads to establishing one's total self-image as a failure.

The passage highlights that the subconscious mind is influenced by what we feed it from the conscious mind. It doesn't analyze this information critically and can be reprogrammed with positive self-image, making Option C the most accurate choice. The conscious mind, according to the passage, is the one resistant to change, not the subconscious.

Option A: The passage describes the subconscious as lacking critical thinking and not questioning information it receives.

Option B: While negativity can impact the subconscious, the passage doesn't say it happens instantly. It emphasizes the role of repeated failures in shaping self-image.

Option D:The passage argues against critical acceptance. It suggests bypassing the critical conscious mind with positive suggestions directly to the subconscious (through self-hypnosis).

Option B is the most accurate explanation of "currency manipulation" because it describes the core practice the passage criticizes. The passage argues that developing economies are intentionally keeping their currencies' value low (depreciated) through various financial tools. This manipulation gives their exports an unfair advantage by making them cheaper in the global market. Option B captures this essence of artificial undervaluation through financial instruments, which is the crux of the currency manipulation concern raised in the passage.

Option A doesn't directly address the manipulation of the domestic currency itself.

Option C: Pegging a currency to market forces would be the opposite of manipulation, which involves artificial control.

Option D: The passage focuses on manipulation by developing economies to gain an export advantage, not manipulation by developed countries.

The passage argues that currency manipulation by developing economies leads to a net drain on aggregate demand in the US and Eurozone. This means there's a decrease in overall spending in these economies. This can happen because:

Consumers in the US and Eurozone might find imported goods (from the manipulating countries) cheaper, leading them to buy less domestically produced goods.

Businesses in the US and Eurozone might struggle to compete with cheaper exports from manipulating countries, leading to lower production and potentially job losses. This reduces overall economic activity and spending.

Therefore, Option D captures the broader concept of lost economic opportunities due to the manipulation and its negative impact on domestic demand in the US and Eurozone.

Option C is the correct answer as the passage explicitly states this as the first measure the US would employ.

Option A is mentioned as a fourth potential measure, after countervailing currency intervention.

Option B is a third potential measure, following countervailing currency intervention.

Option D is presented as the final option, potentially pursued with other affected countries.

Option C aligns with the passage, which suggests that one of the proposed policy measures to address currency manipulation is to treat manipulated exchange rates as export subsidies for the purpose of levying countervailing import duties. Therefore, option c is not false according to the passage.

Option A: The passage suggests that the USA and the Euro area could enhance employment opportunities if other countries stop manipulating their currencies, which implies that they may still be able to increase employment even without currency manipulation by other countries.

Option B: The passage does not mention leveraging extraordinary banking policy instruments to reduce the current account deficit. It primarily focuses on policy measures related to currency manipulation.

Option D: The passage states that the United States has experienced a trade deficit as a result of currency manipulation, so it is not an exception to the trade deficit phenomenon.

According to the passage,Option D is not a notable cause of multi-polarity.

Option A is mentioned in the line, “The absence of a clear hegemonic economic power could add to the volatility.”

Option B is discussed in the line, “The international economy almost certainly will continue to be characterized by various regional and national economies moving at significantly different speeds…”

Option C is indicated in the line, “The contrasting speed across different regional economies are exacerbating global imbalances and straining governments and the international system.”

The passage mentions that “a growing number of diverse state and non-state actors, as well as subnational actors, such as cities, will play important governance roles” and “their discordant values will complicate decision-making.” This suggests that the decentralized decision structures and discordant values could lead to a chronic deficit in multilateral governance which aligns with Option B.

Option A,C and D are not explicitly mentioned in the passage.

The passage mentions that “Transitions to democracy are much more stable and long-lasting when youth bulges begin to decline and incomes are higher,” which indicates that maturing age structures (I) and rising incomes (II) trigger democratization.

Option III and Option IV are not mentioned as triggers for democratisation in the passage.

The passage mentions that “the characteristics of IT use; multiple and simultaneous action, near instantaneous responses, mass organization across geographic boundaries, and technological dependence; increase the potential for more frequent discontinuous change in the international system.”  This aligns with Option A.

Options B, C and D are not mentioned in the passage. 

The passage mentions that these countries are steadily developing their research and development capabilities to handle more and more of these phases. "As that continues, we really will see the beginning of what Satyam Cherukuri, of Sarnoff, an American research and development firm, has called "the globalization of innovation" and an end to the old model of a single American or European multinational handling all the elements of the product development cycle from its own resources. More and more American and European companies are outsourcing significant research and development tasks to India, Russia, and China."

This aligns mostly with Option B.

Option A is incorrect because the passage does not suggest that outsourcing is happening due to the incapability of these countries to handle other aspects of the product cycle.

Option C is incorrect because the passage does not suggest that American or European multinationals are unable to handle all the elements of the product development cycle. Instead, it suggests a shift in the model where tasks are being outsourced to leverage global talent.

Option D is is incorrect because the passage does not suggest that the development of capabilities in these countries would deter American or European multinationals from handling all elements of the product development cycle. 

The passage mentions that when the author asked his daughter Orly, “Where are cars made?”, she answered, “Japan.” The author’s reaction of “Ouch!” suggests that this was not the answer he had expected. Therefore Option C is the correct answer.

Option A is incorrect. The passage mentions that the U.S. executive greeted Mr. Rao with “Namaste,” a common Hindi greeting, indicating that the call was not ignored.

Option B is incorrect. The passage does not provide information on whether Americans knew or did not know how to say hello in Hindi in the past.

Option D is incorrect. The author speculates about a future where his hypothetical granddaughter might associate software with India, but he does not express a wish for this to happen. He actually says, “No, not yet, honey,” indicating that he doesn’t believe this is currently the case.

The author uses the term “flat world” to describe a global environment where opportunities and resources are becoming more evenly distributed and accessible, regardless of geography. This is largely due to advancements in technology and communication, which allow for collaboration and competition on a global scale. So, it’s a metaphor for a world where geographical boundaries are becoming less relevant in business. Therefore, Option B is the correct answer.

Options A,C and D are incorrect

The passage discusses the shift in the global business landscape, where geographical boundaries are becoming less relevant. It talks about how countries like India, China, and Russia are developing their capabilities and how American and European companies are outsourcing significant research and development tasks to these countries. This represents a convergence of technological and other forces that allow businesses to connect and collaborate irrespective of geography, enabling them to reach farther, faster, and deeper than ever before which is mentioned in Option B.

Options A, C and D are either incorrect or distorted.

Flotsam (i) matches with People or things that have been rejected or discarded as worthless (e). The term “Flotsam” is used to describe debris in the water that was not deliberately thrown overboard, often resulting from a shipwreck or accident. This can include parts of a ship, its equipment, cargo, and anything else that floats. However, the term has been metaphorically extended to refer to people or things that have been rejected or discarded as worthless.

Coterie (ii) matches with People with common interest who do things together in small group and do not like to include others (a). A coterie is a small group of people with shared interests or tastes, excluding others.

Insouciant (iii) matches with Showing a casual lack of concern (d). Insouciant means showing a casual lack of concern; indifferent.

Effrontery (iv) matches with Insolent or impertinent behaviour (b). Effrontery is insolent or impertinent behaviour; it’s being bold and disrespectful.

Phlegmatic (v) matches with Not easily made angry or upset (c). Phlegmatic refers to having an unemotional and stolidly calm disposition.

The correct meaning for the idiom “Bring someone to book” is Option A

This idiom is used when someone is made to face punishment or is held accountable for their actions. It’s often used in the context of formal systems like law, school, or work, where there are clear rules and consequences for breaking them. The phrase suggests that the person’s actions are being recorded in a ‘book’ or record, and they are being called to answer for them. It does not literally involve a book. The ‘book’ in the phrase is metaphorical, referring to the recording or noting down of someone’s wrong actions.

Irons in the fire means to be involved with many activities or jobs at the same time. 

Option D is closest to this and, hence, would be the correct answer. 

Fish out of water refers to an individual feeling awkward or anxious because they are in a situation that is not familiar.  

Option B is closest to this and, hence, is the correct answer. 

Looking at the options available for the first blank, we can eliminate option C since 'opposing ally' is an oxymoron and would not fit. 

While filling the second blank, we should look for a positive term to complement 'clarity'. Thus, we can eliminate options B and D, as both are negative and wouldn't fit in the second blank.

This leaves us with only option A. By using the third and fourth words from option A, we see that they fit in the third and fourth blanks.

Therefore, option A is the correct answer.

Inveterate is an adjective referring to having a particular habit, activity, or trait that is long-established and unlikely to change.

Notorious means well-known for something bad or an unfavorable quality, deed, etc.; infamous.

Congenital is an adjective used to refer to a trait or abnormality present from birth.

Glib is an adjective used to refer to the words of a speaker, fluent but insincere and shallow.

Therefore, Option A is the correct answer. 

Salubrious is an adjective used for health-giving or healthy.

Chronic is an adjective used for something persisting for a long period of time.

Egregious is an adjective used for something outstandingly bad or shocking.

Opprobrious is an adjective used for expressing scorn, disgrace, or contempt.

Option C is the correct answer.

Senescent refers to the process of growing old, ageing, or becoming elderly.

The form '-escent' comes from the Latin participle-forming suffix '—ēscēns', which roughly means 'becoming.'

Of all the given options, Option D would best explain the meaning of 'escent' in the word 'senescent.'

Therefore, Option D would be the correct answer. 

A demagogue is a political leader who seeks support by appealing to the desires and prejudices of ordinary people rather than by using rational arguments.

The form '-agogue' comes from the Greek '-agōgos', meaning 'leading.'

Of all the given options, Option B would best explain the meaning of 'agogos' in the word 'demagogue'.

Therefore, Option B is the correct answer. 

"Venal" means capable of being bought or influenced by money or other valuable considerations. 

"Avaricious" means having extreme greed for wealth or monetary gains.
"Mercenary" refers to professional individuals who join armed conflicts for their own profit.
"Untrustworthy" is an adjective describing someone negatively. 

"Incorruptible," on the other hand, is an adjective used to describe someone or something not subject to corruption or bribery. 

Of all the given options, "incorruptible" serves best as an antonym for "venal."

Therefore, Option D is the correct answer. 

"Laconic" is an adjective used to describe someone or something using a few words. 

"Compendious," "aphoristic," and "apothegmatic" mean stating facts or information in a concise manner. "Pleonastic," on the other hand, means the repetition of the same sense in different words, like the phrase "a true fact." 

Of all the given options, "pleonastic" would best fit as an antonym for "laconic."

Therefore, Option C is the correct answer. 

The correct answer is Option A as the correct spelling is Iniquitous, which means grossly unfair and morally wrong.

The collective noun 'shoal' is used to describe a group of species that swim together. Options A,B and C are types of fish whereas gnats are a type of insects. So, Option D is the correct answer.

The collective noun 'herd' is used to describe a group of grazing animals like sheep and cattle. Gulls are a type of seabird, which is why it is the odd ones out.

Option A: Chamois are a type of goat-antelope that live in mountainous regions. Since they are grazing animals, they are correctly referred to as a herd.

Option C:  While walruses are marine mammals, they are often referred to as a herd when grouped together on land or ice. 

Option D: Wrens are small birds. The correct collective nouns for wrens include a "flock" or a "chime", not a "herd."

A thalassocracy or thalattocracy, sometimes also maritime empire, is a state with primarily maritime realms, an empire at sea, or a seaborne empire.

A virtuoso is an individual who possesses outstanding talent and technical ability in a particular art or field such as fine arts, music, singing, playing a musical instrument, or composition.

An imbroglio is an extremely confused, complicated, or embarrassing situation. So, for 3 down, xv is the answer.

A decrepit is someone or something that is worn out or ruined because of age or neglect. So, the answer for 8 down is iv.

An incendiary is designed to cause fires. So, the correct answer for 2 across is vii.

To inveigle is to persuade (someone) to do something by means of deception or flattery. So, the correct answer for 7 across is  viii.

A whodunit is a mystery story that keeps the criminal's identity a secret until the very end.

Volacious means able or fit to fly.

From the given information, we get the following table.

In round 5, there is only one winner and one loser. Hence Smita & Atharv must play against each other.

In rounds 2,3 &4, there is one draw each. Hence those players must play against each other in their respective rounds.

From the above conclusions, we get the following table.

In round 1, Atharv cannot play against Cheena as both of them are winners. Hence, he plays against Sumedha in round 1 & against Cheena in round 4.

In round 4, only Smita & Parul are left. So they play against each other.

Then In round 3, Sumedha & Smita and Parul & Cheena will play against each other, respectively.

As Sumedha played against Smita in round 3, she will play against Cheena in round 2.

Then Isha plays against Smita in round 2.

Smita played against everyone except Cheena. Hence, she plays against Cheena in round 1.

Then Parul must have played against Isha in round 1.

Sumedha played against everyone except Parul. Hence she plays against Parul in round 5.

Cheena played against Isha in round 5.

Hence, The final arrangement is

1A) Cheena won in rounds 1 & 4. In those rounds, she played against Smita & Atharv.

Option (B) is correct.

From the given information, we get the following table.

In round 5, there is only one winner and one loser. Hence Smita & Atharv must play against each other.

In rounds 2,3 &4, there is one draw each. Hence those players must play against each other in their respective rounds.

From the above conclusions, we get the following table.

In round 1, Atharv cannot play against Cheena as both of them are winners. Hence, he plays against Sumedha in round 1 & against Cheena in round 4.

In round 4, only Smita & Parul are left. So they play against each other.

Then In round 3, Sumedha & Smita and Parul & Cheena will play against each other, respectively.

As Sumedha played against Smita in round 3, she will play against Cheena in round 2.

Then Isha plays against Smita in round 2.

Smita played against everyone except Cheena. Hence, she plays against Cheena in round 1.

Then Parul must have played against Isha in round 1.

Sumedha played against everyone except Parul. Hence she plays against Parul in round 5.

Cheena played against Isha in round 5.

Hence, The final arrangement is

The game between Smita & Parul played in round 4, i.e 9:45 AM.

Option (D) is correct.

From the given information, we get the following table.

In round 5, there is only one winner and one loser. Hence Smita & Atharv must play against each other.

In rounds 2,3 &4, there is one draw each. Hence those players must play against each other in their respective rounds.

From the above conclusions, we get the following table.

In round 1, Atharv cannot play against Cheena as both of them are winners. Hence, he plays against Sumedha in round 1 & against Cheena in round 4.

In round 4, only Smita & Parul are left. So they play against each other.

Then In round 3, Sumedha & Smita and Parul & Cheena will play against each other, respectively.

As Sumedha played against Smita in round 3, she will play against Cheena in round 2.

Then Isha plays against Smita in round 2.

Smita played against everyone except Cheena. Hence, she plays against Cheena in round 1.

Then Parul must have played against Isha in round 1.

Sumedha played against everyone except Parul. Hence she plays against Parul in round 5.

Cheena played against Isha in round 5.

Hence, The final arrangement is

Option (1) True. Parul won against Isha at 09:00 AM in round 1.

Option (2) True. Smita lost against Sumedha at 09:30 AM in round 3.

Option (3) False. Cheena won against Atharv at 09:45 AM in round 4.

Option (4) True. Atharv lost against Smita at 10:00 AM in round 5.

From the given information, we get the following table.

In round 5, there is only one winner and one loser. Hence Smita & Atharv must play against each other.

In rounds 2,3 &4, there is one draw each. Hence those players must play against each other in their respective rounds.

From the above conclusions, we get the following table.

In round 1, Atharv cannot play against Cheena as both of them are winners. Hence, he plays against Sumedha in round 1 & against Cheena in round 4.

In round 4, only Smita & Parul are left. So they play against each other.

Then In round 3, Sumedha & Smita and Parul & Cheena will play against each other, respectively.

As Sumedha played against Smita in round 3, she will play against Cheena in round 2.

Then Isha plays against Smita in round 2.

Smita played against everyone except Cheena. Hence, she plays against Cheena in round 1.

Then Parul must have played against Isha in round 1.

Sumedha played against everyone except Parul. Hence she plays against Parul in round 5.

Cheena played against Isha in round 5.

Hence, The final arrangement is

Smita played against Sumedha in round 3. Before round 3, Smita lost in rounds 1 & 2, i.e. 2 games.

Option (A) is the answer.

Out of A, B, C, D & E, three are developed countries, and two are developing countries.

a) Country A is a developed country.

b) It is given that no two developing countries will sit next to each other and country D will always be seated next to a developing country.

Hence country D is a developed country.

c) It is given in the question that 'C' is also a developed country. Hence 'B' and 'E' are developing countries.

Country A will always take a seat as far as possible from country B. Two arrangements are possible.

No two developing countries will sit next to each other. Hence, 'E' can take only one position.

Countries 'C' and 'D' can take either of the positions.

Option (A) is True.

Option (B) is True.

Option (C) is True.

Option (D) is False.

Hence Option (D) is the answer.

Out of A, B, C, D & E, three are developed countries, and two are developing countries.

a) Country A is a developed country.

b) The country A which is a developed country. will always take a seat a far as possible from country B

These are the two possible positions.

c) It is given that no two developing countries will sit next to each other and country D will always be seated next to a developing country.

Hence country D is a developed country.

It is given in the question that 'D' has a developing country to its right.

So, two  'B', 'C' and 'E' are developing countries.

Now, what can be the possible positions of D.

There are 4 possible positions for D.

Of these positions only at 3, 4 D has B to it right. So, The answer is 2.

Given A and B are as far as possible. And also E is sitting right to A. With this information there are 2 arrangements.

Now we have to arrange C, D in those two positions. 

From the given information,

1) Tom will sit on the left of Richard.

2) John and Jame will sit together.

3) Michael is not in extreme cabins.

4) John, Michael, Richard and Tom are readers.

5) Jame, Mary and Franka are non-readers.

6) Mary is in the third cabin, and as she doesn't like reading, on both sides of her cabin, non-readers must be seated.

It is given that Michael is not in extreme cabins. Hence he must take the fifth or sixth cabin.

Case (1):- Michael is in the sixth cabin.

In this configuration, Tom and Richard cannot sit together. Hence this case is not possible.

 Case (2):- Michael is in the fifth cabin.

Tom and Richard can take the sixth and seventh cabins only.

Since Jane and John have to sit together. Jane can only take the second cabin. Hence Franka will take the third cabin.

Therefore the farthest person from Mary is Richard.

Option (D) is correct.

From the given information,

1) Tom will sit on the left of Richard.

2) John and Jame will sit together.

3) Michael is not in extreme cabins.

4) John, Michael, Richard and Tom are readers.

5) Jame, Mary and Franka are non-readers.

6) Mary is in the third cabin, and as she doesn't like reading, on both sides of her cabin, non-readers must be seated.

It is given that Michael is not in extreme cabins. Hence he must take the fifth or sixth cabin.

Case (1):- Michael is in the sixth cabin.

In this configuration, Tom and Richard cannot sit together. Hence this case is not possible.

Case (2):- Michael is in the fifth cabin.

Tom and Richard can take the sixth and seventh cabins only.

Since Jane and John have to sit together. Jane can only take the second cabin. Hence Franka will take the third cabin.

Therefore Michael occupies the fifth cabin.

Option (C) is correct.

Let's name the 8 locations as L1, L2, L3, L4, L5, L6, L7, L8.

From each of the 8 locations, we will have a bus to all the other seven locations in the morning. So, there will be 8*7=56 buses scheduled in the morning.

Also, we will have the same number of buses scheduled in the evening. So, another 56 buses have to be scheduled.

The total number of buses is 56+56=112.

Let's name the 8 locations as L1, L2, L3, L4, L5, L6, L7, L8.

Given there are two core locations, from each of the core location there has to be a bus to 7 other location.

Also, there has to be a bus from the other locations to the core location. So, for the first core location there are 7*2=14.

Similarly, for the second core location there has to be 14 buses. But this includes bus from first core location to second core location and bus form second core location to first core location. These two are already counted in the first. So, there will be 12 additional buses for the second location.

This is all for the morning. The same number of buses will be there for the evening. So, the total number of buses will be 2*(14+12)=2*26=52

From the given information, we get following table

From the given information, we get the following table,

In the third round, the negative mark for the incorrect answer is high.

Hence to get the minimum possible score,  the two incorrect answers must be from round 3.

For all other cases, the minimum score will be greater than 120.

examples,

1) 

2) 

Therefore the minimum possible score is 120.

Option (A) is correct.

From the given information, we get the following table,

Only two answers are correct in the whole quiz. We can select only two correct answers in the following ways.

In any configuration total score is not equal to 40.

Therefore the probability is '0'.

Option (D) is correct.

It is given that the total number of athletes who participated from the given five countries is 32400. 

From figure 1, we can easily find the total number of athletes for each of the five countries.

Let the number of total female athletes from these five countries be 100x. Then we will get this table:

In the question, it's given that the total number of female athletes who participated in 2021 Olympics from the given five countries were 15000.

Hence, 100x = 15000 => x = 150

Hence, we get this following table:

From the table, we can see that 3 countries (Canada, China, and Italy) had the number of male athletes less than the number of female athletes.

The correct option is B

It is given that the total number of athletes who participated from the given five countries is 32400.

From figure 1, we can easily find the total number of athletes for each of the five countries.

Let the number of total female athletes from these five countries be 100x. Then we will get this table:

In the question, it's given that the average number of male athletes from all the given five countries 4000, which implies the total number of male athletes = (4000*5) = 20000

Hence, the number of total female athletes from these five countries = (32400 - 20000) = 12400

=> 100x = 12400 => x = 124

the average number of female athletes from Italy and Canada is (25x+14x)/2 = 19.5x =19.5 * 124 = 2418

The correct option is D

It is given that the total number of athletes who participated from the given five countries is 32400.

From figure 1, we can easily find the total number of athletes for each of the five countries.

Let the number of total female athletes from these five countries be 100x. Then we will get this table:

In the question, it's given that the proportion of male athletes to female athletes from Canada was 31:50.

Let the number of male athletes from Canada be 31y, which implies the number of female athletes from Canada is 50y. The total number of athletes from Canada is 81y, which is equal to 4860

=> 81y = 4860 => y = 60. Therefore, the number of female athletes from Canada is (50*60) = 3000, which is 25% of the total female athletes (Fig 2).

Hence, the total number of female athletes = 3000*(100/25) = 12000. Hence, 100x = 12000 => x = 120

The following table is given below:

Hence, the ratio for male to female athletes for China in the same year is 360:288, which can be simplified as 5: 4

The correct option is C

It is given that the total number of athletes who participated from the given five countries is 32400.

From figure 1, we can easily find the total number of athletes for each of the five countries.

Let the number of total female athletes from these five countries be 100x. Then we will get this table:

It is given that the total number of female athletes who participated from the given five countries were 18000.

=> 100x = 18000 => x = 180

The following table is given below:

From the table, we can see that the ratio of the number of female athlete from Canada to the total number of athletes from Brazil is 45: 81, which can be simplified into 5:9

The correct option is A

Figure 3 depicts the pie chart indicating the percentage distribution of per unit steel price over the period of eight years from 2005 to 2012.

The pie graph shows the percentages of unit steel prices over the eight years from 2005 to 2012.

So, in 2006 it was 15.91%; in 2008, it was 13.64%; in 2009, it was  9.09%; and in 2011 it was 13.64%.

=> The prices across the 4 years will be 15.91x, 13.64x, 9.09x, and 13.64x

=> The cost per unit of steel was the highest in 2006.

First, we need to find the prices of units of steel from 2005 to 2012. 88 is the sum of prices, and 11 is the average across 8 years.

In 2005, it is equal to 11.36/100 * 88 = 10.

Similarly, calculating for other years, we get:

The prices of steel for 2005 - 12 are 10, 14, 12, 12, 8, 10, 14, 8, respectively in each year, and the revenues are 1000, 1600, 1200, 1400, 800, 800, 1200, 1100.

=> The number of pieces produced each year is 1000/10, 1600/14, 1200/12, 1400/12, 800/8, 800/10, 1200/14, 1100/8

=> The average of all the above 8 quantities = 104.27, which is approx 104.

First, we need to find the prices of units of steel from 2005 to 2012. 88 is the sum of prices, and 11 is the average across 8 years.

In 2005, it is equal to 11.36/100 * 88 = 10.

Similarly, calculating for other years, we get:

The steel prices for 2005 - 12 are

2005 = 10

2006 = 14

2007 = 12

2008 = 12

2009 = 8

2010 = 10

2011 =14

2012 = 8

The lowest and highest per unit price of steel is 8 and 14

The required ratio is 8 : 14 = 4: 7.

First, we need to find the prices of units of steel from 2005 to 2012. 88 is the sum of prices, and 11 is the average across 8 years.

In 2005, it is equal to 11.36/100 * 88 = 10.

Similarly, calculating for other years, we get:

The steel prices for 2005 - 12 are 10, 14, 12, 12, 8, 10, 14, 8, respectively, and the revenues are 1000, 1600, 1200, 1400, 800, 800, 1200, 1100.

In 2013, the revenue increased by 25% => New revenue = 1100 * 1.25 = 1375.

The price increased by 10% => 8 * 1.1 = 8.8

=> The required quantity = Revenue / per unit price = 1375 / 8.8 = 156.25.

Total number of patients admitted in July 2021 = 1000

The number of covid positive patients = 64% of total patients

= 64% of 1000 = 640 patients.

Out of 640, 194 are female covid positive patients.

The number of male covid positive patients = 640 - 194 = 446 patients.

The ratio of the number of admitted Covid-19-positive male patients and the number of admitted Covid-19 positive female patients = $$\frac{446}{194}=\frac{223}{97}$$

Option (C) is the answer.

The number of admitted patients in March 2021 = 900

By August 2021, the number of admitted patients are increased by 40% from March 2021 

= $$(100+40)\%\times\ 900$$

= $$(1.4)\ \times\ 900$$ = $$1260$$

Out of 1260 admitted patients, 25% are covid positive.

Hence the number of covid positive patients are =$$(25)\%\times\ 1260$$ 

=$$\frac{25}{100}\times\ 1260$$

$$\frac{1}{4}\times\ 1260=315$$ Patients.

Option (D) is correct.

Let the number of admitted patients in April 2021 be $$5x$$

and  the number of patients found Covid-19 positive was be $$4x$$.

From the table, in April 2021 the ratio of the number of covid positive male patients to covid positive female patients is 

given as 5:3.

The number of covid positive female patients in April 2021 = $$\frac{3}{3+5}\times\ $$ total covid positive patients

= $$\frac{3}{3+5}\times\ 4x$$ = $$\frac{3}{8}\times\ 4x\ =\ \frac{3x}{2}$$.

Hence, the percentage of the number of covid positive female patients in the number of admitted patients =

= $$\ \frac{3x}{2\ \times\ 5x}\times\ 100$$ = $$\ 30$$

Option (A) is correct.

It is given that the ratio of number of male covid positive patients to female covid patients = 9:5

Let the number of male covid positive patients be '$$9x$$' and female covid positive patients be $$5x$$

The difference between number of male and female covid positive patients = 72

$$9x-5x=72$$

$$4x=72$$

$$x=18$$

Hence the total number of covid positive patients = $$9x+5x=14x=14\times\ 18=252$$

From the table, the total number of covid positive patients is 42% of the total admitted patients.

Let the total number of admitted patients be $$y$$

Then,  the total number of covid positive patients = 42% of $$y$$

$$0.42y=252$$

$$y=600\ $$

Therefore the total number of admitted patients is 600.

Option (D) is correct.