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Most Expected CAT Quant Questions PDF:

Download Top 100 Quant Questions for CAT PDF. Very important and most expected Quantitavie aptitude Questions and answers for CAT based on asked questions in previous exam papers.

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Question 1: In a right triangle ABC, $\angle\ ABC\ is\ 90^{\circ\ }$. The ratio of sides AB and AC is 2:5. What is the ratio of side BC and the circumradius of the triangle?

a) $\sqrt{21}:5$

b) $\sqrt{84}:5$

c) $\sqrt{21}:10$

d) $\sqrt{21}:20$

Question 2: In a car race, car A beats car B by 45 km. car B beats car C by 50 km. and car A beats car C by 90 km. The distance (in km) over which the race has been conducted is

a) 475

b) 450

c) 500

d) 550

Question 3: Two alcohol solutions, A and B, are mixed in the proportion 1:3 by volume. The volume of the mixture is then doubled by adding solution A such that the resulting mixture has 72% alcohol. If solution A has 60% alcohol, then the percentage of alcohol in solution B is

a) 90%

b) 94%

c) 92%

d) 89%

Question 4: For all real value of x and y, f(x+y) = f(x)+f(y). If f(3)+f(4)+f(5)+…………f(11) = 189, then find the value of f(15).

a) 15

b) 45

c) 90

d) 60

Question 5: The graph of y = $x^2 – 8x + 13$ is symmetric wrt to the line x = c, then the value of c is

Question 6: Ravi has three alloys A, B and C with copper percentages as 36%, 50% and 60% respectively in them. In which of the following ratios could he melt the three alloys and mix them so as to get a percentage of 45% in the mixture?

a) 3:2:1

b) 4:3:2

c) 5:3:2

d) 3:4:6

Question 7: A and B are two types of Alloy formed by mixing Zinc and Iron in the ratio 4:11 and 7 : 3 respectively. A third alloy C is formed by mixing A and B in a particular ratio such that the final alloy has 60% Zinc. What is the ratio in which A and B are mixed?

a) 1 : 2

b) 3 : 10

c) 4 : 5

d) 3: 7

Question 8: Students in a college have to choose at least two subjects from chemistry, mathematics and physics. The number of students choosing all three subjects is 18, choosing mathematics as one of their subjects is 23 and choosing physics as one of their subjects is 25. The smallest possible number of students who could choose chemistry as one of their subjects is

a) 22

b) 21

c) 20

d) 19

Question 9: In a company there are three departments A,B,C. The average age of employees in A,B and C are $48$ years, $50$ years, $52$ years respectively.The average age of employees in A and C together is $50.5$ years. The average age of employees in B and C together is $51\frac{1}{9}$ years. What is the average age of employees of A,B,C taken together?

a) $50\frac{1}{9}$years

b) $50\frac{1}{3}$years

c) $51\frac{1}{9}$years

d) $51\frac{1}{3}$years

Question 10: A batsman played n + 2 innings and got out on all occasions. His average score in these n + 2 innings was 29 runs and he scored 38 and 15 runs in the last two innings. The batsman scored less than 38 runs in each of the first n innings. In these n innings, his average score was 30 runs and lowest score was x runs. The smallest possible value of x is

a) 4

b) 3

c) 2

d) 1

Question 11: Haritha, Anita and Vasanta have red, blue and green coloured dice respectively. All three of them rolled their dice simultaneously. Then the probability that the sum of the numbers turned up on the blue and green dice is equal to the number on the red dice is

a) $\frac{5}{72}$

b) $\frac{5}{12}$

c) $\frac{5}{36}$

d) $\frac{5}{216}$

Question 12: There are 2 jars A and B. Jar A contains water and alcohol in the ratio 3:2. Jar B contains alcohol-water mixture in some ratio. When 1 unit of the mixture from jar A is mixed with 2 units of mixture from jar B, the resultant mixture contains twice as much water as alcohol. The ratio of alcohol to water in jar B is

a) 3:10

b) 3:13

c) 3:7

d) 3:2

Question 13: The average weight of a class of 10 students increased by 2 kg when Ravi left and Suri joined. After a few months, Suri left and Raghu joined. The weight of Ravi was 30 Kgs and Raghu weighed 10 Kg less than Suri.
What is the difference in the final average weight of the class and the initial average weight of the class?

Question 14: In a class of 90 students, 35 students drink orange juice, 46 students drink apple juice and 25 students drink lemon juice. A is the maximum number of students who drink exactly 2 of the 3 drinks and B is the maximum number of students who drink all the 3 drinks. What is the ratio of A : B?
It is known that all the students drink atleast one drink.

a) 2 : 1

b) 3 : 1

c) 1 : 1

d) 3 : 2

Question 15: Leaving home at the same time, Amal reaches the office at 10:15 am if he travels at 8 km/hr, and at 9:40 am if he travels at 15 km/hr. Leaving home at 9.10 am, at what speed, in km/hr, must he travel so as to reach office exactly at 10 am?

a) 13

b) 12

c) 14

d) 11

Question 16: A train starts from Q with speed of 30 km per hour towards P which is at a distance of 240 km from Q. An insect starts flying with speed of 45 km per hour at the same time from P towards Q and meets the train at some point. It turns back immediately and flies towards P. When it reaches P, it turns back and flies towards the train. It continues to do so until the train reaches P. What is the sum of all time intervals when its is moving in the same direction as the train during the journey. It is given that the speeds of both the train and the insect remain constant?

a) 3 hours

b) 4 hours

c) 2 hours

d) 6 hours

Question 17: Anand and Sanand walk on an escalator with a constant speed of 3 steps/second and 1 step/second. Anand takes 120 steps, and Sanand takes 80 steps to reach the bottom from the top of the escalator in a downward moving escalator. How many steps would be visible in the stationary escalator?

a) 160

b) 180

c) 140

d) 150

Question 18: If a and b are the roots of $x^2+9x+5=0$ where a>b, what is the value of $(\frac{5}{2a+9})^2+(\frac{6}{2b+9})^2$?

Question 19: The value of $\log_{a}({\frac{a}{b}})+\log_{b}({\frac{b}{a}})$, for $1<a\leq b$ cannot be equal to

a) 0

b) -1

c) 1

d) -0.5

Question 20: Rahul and Seema can complete a work in 12 days and 15 days respectively. They started the work and completed 16.66% after which Deepa joined them and the work was completed in 2 days less than the stipulated time. In how many days could Deepa alone do the complete work?

a) $23\dfrac{11}{27}$ days

b) $9\dfrac{23}{27}$ days

c) $11\dfrac{23}{27}$ days

d) $16\dfrac{20}{27}$ days

e) $18\dfrac{7}{27}$ days

Question 21: A contractor agreed to construct a 6 km road in 200 days. He employed 140 persons for the work. After 60 days, he realized that only 1.5 km road has been completed. How many additional people would he need to employ in order to finish the work exactly on time?

Question 22: If $\log_{4}{5}=(\log_{4}{y})(\log_{6}{\sqrt{5}})$, then y equals

Question 23: An alloy is prepared by mixing three metals A, B and C in the proportion 3 : 4 : 7 by volume. Weights of the same volume of the metals A. B and C are in the ratio 5 : 2 : 6. In 130 kg of the alloy, the weight, in kg. of the metal C is

a) 48

b) 84

c) 70

d) 96

Question 24: The sum of the perimeters of an equilateral triangle and a rectangle is 90cm. The area, T, of the triangle and the area, R, of the rectangle, both in sq cm, satisfying the relationship $R=T^{2}$. If the sides of the rectangle are in the ratio 1:3, then the length, in cm, of the longer side of the rectangle, is

a) 27

b) 21

c) 24

d) 18

Question 25: A solid sphere is cut into 8 identical pieces by three mutually perpendicular cuts. By what percentage is the sum of the total surface areas of the eight pieces more than the total surface area of the original sphere?

a) 110

b) 150

c) 70

d) 90

Question 26: Anil buys 12 toys and labels each with the same selling price. He sells 8 toys initially at 20% discount on the labeled price. Then he sells the remaining 4 toys at an additional 25% discount on the discounted price. Thus, he gets a total of Rs 2112, and makes a 10% profit. With no discounts, his percentage of profit would have been

a) 50

b) 60

c) 54

d) 55

Question 27: The number of real-valued solutions of the equation $2^{x}+2^{-x}=2-(x-2)^{2}$ is:

a) 1

b) 2

c) infinite

d) 0

Question 28: The points (2,1) and (-3,-4) are opposite vertices of a parallelogram.If the other two vertices lie on the line $x+9y+c=0$, then c is

a) 12

b) 13

c) 15

d) 14

Question 29: Let $f(x)=x^{2}+ax+b$ and $g(x)=f(x+1)-f(x-1)$. If $f(x)\geq0$ for all real x, and $g(20)=72$. then the smallest possible value of b is

a) 16

b) 4

c) 1

d) 0

Question 30: If $f(x+y)=f(x)f(y)$ and $f(5)=4$, then $f(10)-f(-10)$ is equal to

a) 14.0625

b) 0

c) 15.9375

d) 3

Question 31: Among 100 students, $x_1$ have birthdays in January, $X_2$ have birthdays in February, and so on. If $x_0=max(x_1,x_2,….,x_{12})$, then the smallest possible value of $x_0$ is

a) 8

b) 9

c) 10

d) 12

Question 32: John takes twice as much time as Jack to finish a job. Jack and Jim together take one-thirds of the time to finish the job than John takes working alone. Moreover, in order to finish the job, John takes three days more than that taken by three of them working together. In how many days will Jim finish the job working alone?

Question 33: In a farm, thirty men can do a work in fifteen days.All of them started working on the first day.On every alternate day starting from the second day, some men didn’t turn up. The number of number of men who did not turn up is equal to number of the day(i.e. number of men who didn’t turn up on second day was 2 and on fourth day was 4 and so on). In how many days can they complete total work?

a) 20

b) 18

c) 16

d) 19

Question 34: If $x=(4096)^{7+4\sqrt{3}}$, then which of the following equals to 64?

a) $\frac{x^{7}}{x^{2\sqrt{3}}}$

b) $\frac{x^{\frac{7}{2}}}{x^{\frac{4}{\sqrt{3}}}}$

c) $\frac{x^{\frac{7}{2}}}{x^{2{\sqrt{3}}}}$

d) $\frac{x^{7}}{x^{4\sqrt{3}}}$

Question 35: Ramesh and his son were born on the same date and the age of Ramesh was 4 times that of his son when he was as old as his son is now. What could be the sum of present ages of Ramesh and his son on their birthdays, if Ramesh is at-least 45 years old and his son is not more than 30 years old?

Question 36: Two circular tracks T1 and T2 of radii 100 m and 20 m, respectively touch at a point A. Starting from A at the same time, Ram and Rahim are walking on track T1 and track T2 at speeds 15 km/hr and 5 km/hr respectively. The number of full rounds that Ram will make before he meets Rahim again for the first time is

a) 5

b) 3

c) 2

d) 4

Question 37: How many pairs(a, b) of positive integers are there such that $a\leq b$ and $ab=4^{2017}$ ?

a) 2018

b) 2019

c) 2017

d) 2020

Question 38: If Y is a negative number such that $2^{Y^2({\log_{3}{5})}}=5^{\log_{2}{3}}$, then Y equals to:

a) $\log_{2}(\frac{1}{5})$

b) $\log_{2}(\frac{1}{3})$

c) $-\log_{2}(\frac{1}{5})$

d) $-\log_{2}(\frac{1}{3})$

Question 39: If the volume of a right circular cylinder that can be cut out from a sphere of radius 6 cm is maximum possible, then find the radius of the cylinder?

a) $4\sqrt{3}$ cm

b) $2\sqrt{6}$ cm

c) $6\sqrt{3}$ cm

d) $4\sqrt{6}$ cm

Question 40: An amount of 300 is divided into three numbers such that the first number is thrice of the second number. Find the maximum possible product of all the three numbers.

Question 41: For real values of x and y, a function f($x^2-y^2,\ x+y$) = $\ \dfrac{\ x}{y}$, then the value of f(x,y) is

a) $\ \dfrac{\ \ \ x-y^2}{\ \ x-y^2}$

b) $\ \dfrac{\ \ \ x^2+y}{\ \ x^2-y}$

c) $\ \dfrac{\ \ \ x^2-y^2}{\ \ x^2+y}$

d) $\ \dfrac{\ \ \ x+y^2}{\ \ y^2-x}$

Question 42: A solid right circular cone of height 27 cm is cut into two pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in volume of the two pieces is 225 cc, the volume, in cc, of the original cone is

a) 243

b) 232

c) 256

d) 264

Question 43: Let C be a circle of radius 5 meters having center at O. Let PQ be a chord of C that passes through points A and B where A is located 4 meters north of O and B is located 3 meters east of O. Then, the length of PQ, in meters, is nearest to

a) 8.8

b) 7.8

c) 6.6

d) 7.2

Question 44: The vertices of a triangle are (0,0), (4,0) and (3,9). The area of the circle passing through these three points is

a) $\frac{14\pi}{3}$

b) $\frac{123\pi}{7}$

c) $\frac{12\pi}{5}$

d) $\frac{205\pi}{9}$

Question 45: In a company XYZ, there are three departments, P, Q and R. The number of employees in department P is half of the number of employees in department R. The total number of employees of Q and R is 280. If 20 employees of the department R are transferred to the department Q, then the ratio of the number of employees in department P to that of Q will be the same as the ratio of the number of employees in department Q to that of department R. Find the number of employees in department R.

Question 46: A certain amount is given at compound interest and the amount increased by 27% between the 2nd year and 4th year. After how many years does the amount double itself?

a) 8 years

b) 7 years

c) 6 years

d) Can’t be determined

Question 47: The mean of all 4-digit even natural numbers of the form ‘aabb’,where $a>0$, is

a) 4466

b) 5050

c) 4864

d) 5544

Question 48: A train travelled at one-thirds of its usual speed, and hence reached the destination 30 minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for 5 minutes but then stopped for 4 minutes for an emergency. The percentage by which the train must now increase its usual speed so as to reach the destination at the scheduled time, is nearest to

a) 50

b) 58

c) 67

d) 61

Question 49: In a company, the ratio of male and female employees is 2:3. Each employee earns either 30000 or 60000 per month. The ratio of the average male salary to the female salary is 15:16. The ratio of the number of males who make Rs 60000 to that of females who make Rs 30000 is 1:4. Find the average salary of all the employees.

a) 41000

b) 39000

c) 45000

d) 48000

Question 50: On a rectangular metal sheet of area 135 sq in, a circle is painted such that the circle touches two opposite sides. If the area of the sheet left unpainted is two-thirds of the painted area then the perimeter of the rectangle in inches is

a) $3\sqrt{\pi}(5+\frac{12}{\pi})$

b) $4\sqrt{\pi}(3+\frac{9}{\pi})$

c) $3\sqrt{\pi}(\frac{5}{2}+\frac{6}{\pi})$

d) $5\sqrt{\pi}(3+\frac{9}{\pi})$

Question 51: Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is ₹ 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is

a) 33

b) 27

c) 30

d) 36

Question 52: A circle is inscribed in a rhombus with diagonals 12 cm and 16 cm. The ratio of the area of circle to the area of rhombus is

a) $\frac{6\pi}{25}$

b) $\frac{5\pi}{18}$

c) $\frac{3\pi}{25}$

d) $\frac{2\pi}{15}$

Question 53: Veeru invested Rs 10000 at 5% simple annual interest, and exactly after two years, Joy invested Rs 8000 at 10% simple annual interest. How many years after Veeru’s investment, will their balances, i.e., principal plus accumulated interest, be equal?

Question 54: Raghav earns salary 10000x, where x is an integer. His salary is increased by x%, then the salary becomes 10000p, where p is a perfect square. Then the salary of Raghav after the increase is

a) Rs. 2250000

b) Rs. 1960000

c) Rs. 1440000

d) Rs. 2560000

Question 55: In an equilateral triangle ABC, a point P is marked on AB. The side AC is extended to Q such that APQ is a right triangle with P = 90$^{\circ\ }$. The area of ABC is equal to the area of APQ. If BP = 6-$3\sqrt{2}$, then find the value of AB?

a) $6\sqrt{2}$ cm

b) 3 cm

c) $3\sqrt{2}$ cm

d) 6 cm

Question 56: Three vessels P, Q and R contain a solution of water and alcohol. The ratio of water and alcohol in vessel P and vessel Q are 2:3 and 1:4. After combining Q and R in the ratio 1:2, the resulting mixture contains the water and alcohol in the ratio 4:11. Find the percentage of water in a  mixture which has P, Q and R combined in a ratio 3:5:2.

Question 57: Amol started from X at 5 PM and travelled towards Y. Bhuvan started from Y at 7 PM and travelled towards X. If Amol and Bhuvan met at 9 PM and Amol reached his destination 20 minutes after Bhuvan reached his destination, when did Amol reach his destination?

a) 11:30PM

b) 11PM

c) 12AM

d) 1AM

Question 58: Jay has 3 types of boxes Large, medium and small. He first puts 10 large boxes on a table. He leaves some of these boxes empty and in all the other boxes puts 7 empty medium sized boxes. He then leaves some of the medium sized boxes empty and places 7 empty small boxes in the other medium sized boxes. If 82 boxes on the table were empty, then what is the total number of boxes he used? (All large boxes are of the same size, all medium boxes are of the same size and all small boxes are of the same size)

a) 88

b) 90

c) 94

d) 98

Question 59: In a school, every student likes one of the three sports – Cricket, Football, Rugby. 65% of the students like Cricket, 86% like Football and 57% of the students like Rugby. What is the maximum percentage of students who like exactly two sports?

a) 92%

b) 80%

c) 54%

d) 46%

Question 60: 207 people who attend “Bold Gym” in Kondapur take four types juices Apple, Orange, Pomegranate and Mango. There are a few people who do not take any of the juices. It is known that for every person in the Gym who takes atleast ‘N’ types of juices there are 2 persons who take atleast ‘N-1’ juices for N = 2, 3 and 4. If the number of people who take all four types of juices is equal to the number of people who do not take any juice at all, what is the number of people who take exactly 2 types of juices?

a) 23

b) 46

c) 69

d) 92

Question 61: All the three sides of an equilateral triangle ABC are divided into 3 equal parts using 2 points on each side. Each side has length $6\sqrt{3}$ cm. The points on side AB are A$_1$ and A$_2$ with point A$_1$ being the nearest to A. The points on side BC are B$_1$ and B$_2$ with point B$_1$ being the nearest to B. The points on side AC are C$_1$ and C$_2$ with point C$_1$ being the nearest to C. A circle is drawn inside $A_1B_1C_1$ such that it touches all its sides. Find the radius of the circle.

a) 3

b) $3\sqrt{3}$

c) $2\sqrt{3}$

d) $\sqrt{3}$

Question 62: If f(x) = $x^2+7x+10$, then the number of real roots of f(f(f(x))) = 0 is

a) 0

b) 4

c) 2

d) 3

Question 63: A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially?

Question 64: How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

Question 65: Consider two equations $y_{1} = |2x+5|$ and $y_{2} = |6x+7|$, what is the length of the segment where $|y_{1}| > |y_{2}|$?

a) $3\sqrt{5}$ units

b) $\sqrt{3}$ units

c) $\sqrt{5}$ units

d) $5\sqrt{3}$ units

Question 66: A trapezium ABCD has sides AD and BC parallel to each other. The diagonal AC divides the trapezium into two parts such that ABC is an equilateral triangle and ACD is a right triangle right angled at C. If the area of ABC is 100 square cm, then the area of the trapezium (in square cm)is

a) 200

b) 100+100$\sqrt{3}$

c) 300

d) 100+200$\sqrt{3}$

Question 67: A loan amount of Rs. 7200 was repaid in two equal half yearly installments. If the rate of compound interest is 50% p.a compounded semi annually, what is the amount paid in each installment?

Question 68: The cost price of 6 apples is equal to the marked price of 5 mangoes. The cost price of 5 mangoes is equal to the marked price of 2 apples. If the shopkeeper gives 2 mangoes free on the selling of 6 apples and makes a profit of 87.5%, find the profit percentage on selling 1 mango at its marked price.

a) 50

b) 40

c) 30

d) 20

Question 69: Amar, Akbar, Anthony and Arya are four friends who come to visit Jurassic park. Each of the four is accompanied by his father and grandfather to the park. In how many ways can these 12 people stand in the queue at the entry gate, if no father wants to stand ahead of his son?

a) $^{12}C_{3}$*$^{9}C_{3}$*$^{3}C_{1}$

b) $^{12}C_{9}$*$^{3}C_{1}$*$^{3}C_{1}$

c) $^{12}C_{6}$*$^{6}C_{3}$*$^{3}C_{3}$

d) $^{12}C_{3}$*$^{9}C_{3}$*$^{6}C_{3}$

Question 70: For a cubic function f(x), f(1)=f(2)=f(3)=2 and f(4)=14, then find the value of f(5)

Question 71: The area of the region satisfying the inequalities $\mid x\mid-y\leq1,y\geq0$ and $y\leq1$ is

Question 72: In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by

a) 5

b) 4

c) 3

d) 6

Question 73: A prisoner escapes from the Shaw shank prison. He runs away at a speed of 20 km/hr. Two hours later the warden sees that a prisoner is missing. He immediately sends a team to catch him. The police team chases him in a car which runs at a speed of 40 km/hr. The police team has a dog with them which runs at a speed of 50 km/hr. As soon as the police start the chase, the dog runs towards the prisoner. Once he reaches the prisoner, he turns back and returns to the police car. The dog continues this to and fro journey till the prisoner is caught. What is the distance run by the dog in the direction of the police(from prisoner to police)?

a) 90 km

b) 10 km

c) 80 km

d) 20 km

Question 74: How many integral values of x satisfy the following inequality: $\ \dfrac{\ \left(x-2\right)\left(x-4\right)^2\left(x-6\right)^3………\left(x-20\right)^{10}}{\ \left(x+2\right)\left(x+4\right)^2\left(x+6\right)^3………\left(x+20\right)^{10}}<0$

a) 10

b) 11

c) 12

d) 13

Question 75: In the given figure A-C-O-D-B are collinear and C, O and D are the centers of the three circles. If the length of the tangents AE,BF is $4\sqrt{2}$ units and $2\sqrt{35}$ units respectively and $DO-CO= 3$ units, then which of the following statements is true regarding the radii of the three circles?

a) They form an HP.

b) They form a GP.

c) They form an AP.

d) None of the above

Question 76: In the final year class of mechanical engineering at IIT-H, there are 90 students. The students have a compulsory course of fluid dynamics in the final semester. Other than this, the students can opt for at most two humanities optionals among History, Economics, Sociology and Psychology. Each student has to take at least 1 optional. 9 students from the class opted for both History and Economics. 20 students opted for both Economics and Sociology. None of the students in the class opted for both History and Sociology. In the test on fluid dynamics, the average score of those who opted for history is 45 and of those opted for Sociology was 65. However, the combined average marks in fluid dynamics of the students who opted for these subjects is 50. What is the maximum possible number of students who opted for only Economics?

a) 10

b) 15

c) 5

d) 20

e) 25

Question 77: The sum of all two-digit numbers that give a remainder of A when they are divided by 7 is 654. What is the value of A?

Question 78: Let the m-th and n-th terms of a geometric progression be $\frac{3}{4}$ and 12. respectively, where $m < n$. If the common ratio of the progression is an integer r, then the smallest possible value of $r + n – m$ is

a) 6

b) 2

c) -4

d) -2

Question 79: If $f(5+x)=f(5-x)$ for every real x, and $f(x)=0$ has four distinct real roots, then the sum of these roots is

a) 0

b) 40

c) 10

d) 20

Question 80: Pawan randomly selected a 4 digit number and he found out that the sum of the digits of the number is 32. What is the probability that the number selected by Pawan is divisible by 11?

a) $\dfrac{9}{35}$

b) $\dfrac{1}{4}$

c) $\dfrac{1}{7}$

d) None of these

Question 81: The number of pairs of integers $(x,y)$ satisfying $x\geq y\geq-20$ and $2x+5y=99$

Question 82: A number ‘n’ when divided by 6 leaves a remainder of ‘k’ and when divided by 12 leaves a remainder of ‘3k’. How many values can ‘n’ take if it is known that ‘n’ and ‘k’ are natural numbers less than 100?

Question 83: How many disticnt positive integer-valued solutions exist to the equation $(X^{2}-7x+11)^{(X^{2}-13x+42)}=1$ ?

a) 8

b) 4

c) 2

d) 6

Question 84: How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?

Question 85: In how many ways can a pair of integers (x , a) be chosen such that $x^{2}-2\mid x\mid+\mid a-2\mid=0$ ?

a) 6

b) 5

c) 4

d) 7

Question 86: Find the number of consecutive zeros after the decimal in the number $\frac{1}{40^{34}}$, given that log 2 = 0.3010

Question 87: Ram has been given the value of the first term as 6 and the common difference as 2 of an arithmetic progression. He is asked to write the sum of the first 20 terms of the series. After writing first x terms, he takes the common difference as 12 by mistake. As a result, the sum of all the terms increases by 56%, find the value of x

Question 88: Anil bought a second hand bike for some price from Ankit. Anil spent 5000 on repairs and then sold the bike at 25 percent profit to Sushil. Sushil found that the tyres of the bike were very old so he spent 5000 in replacing the tyres. He then sold the bike at a price which was 20,000 more then the price at which Ankit sold the bike to Anil. If Sushil earned a profit of 16.67 percent then find the price at which Anil sold the bike.

Question 89: If the real root of the cubic equation $8a^3-12a^2-6a-1=0$ is expressed as $\frac{\sqrt[3]{p}+\sqrt[3]{q}+1}{r}$ where $p,q,r$ are natural numbers, what is the value of $p+q+r$

Question 90: $F_n$ is defined as the set {$2n+6, 3n+5$} where n is a natural number less than or equal to 500. For how many values of $n$ will $F_n$ and $F_{n+1}$ have one element each which is divisible by 7?

Question 91: The students of a class, from roll number 1 to n (n $\ge$ 1),  are to be given chocolates such that nth roll numbered student gets n chocolates. 100 chocolates are to be distributed such that no chocolate should remain in the end. More than one round of distribution is possible only if all the students get chocolates according to their roll numbers. How many values can n assume?

Question 92: Find the number of ways in which the number $6561^2$ can be written as a product of 3 factors.

a) 153

b) 120

c) 60

d) 30

Question 93: A milkman has 100L of milk with him. He bought this 100L for 10000 Rs. He sells 20L of it at 80Rs per liter and replaces the sold milk by water. He then again sells 20L of the mixture at 85 Rs per liter and replaces it with water. He then sells 25L of the mixture at 90 Rs per liter and replaces it water. He then finally sells 25L at 95 Rs per liter. What is his overall profit/loss in these transactions, assume water comes free of cost? (if you feel he had a loss of X Rs the put -X as the answer and if you feel he had an overall profit of X Rs then put X as the answer)

Question 94: The number of integer values of ‘a’ for which both the roots of the equation $ax^2+(a-4)x+a+1=0$ are greater than 0 is

Question 95: 10 boxes are kept in a row. In how many ways can we select three boxes such that no two boxes are next to each other?

Question 96: PQRXYZ is a hexagon in which all its interior angles are equal. If the lengths of PQ= 1cm, QR = 4cm, RX =2cm, XY = 2cm .Then lengths of PZ and ZY are

a)  3, 3

b) 4, 2

c) 2, 4

d) 4, 4

Question 97: What is the number of distinct points of intersection of $x^2 + y^2$ = 169, $5x + 12 y = 169$ and $x = 0$ .

a) 2

b) 4

c) 3

d) 1

Question 98: $x$ and $y$ are integers such that $\dfrac{3}{x} + \dfrac{2}{y} = \dfrac{1}{12}$. What is the sum of all such integral values of $y$?

Question 99: A container in a dairy had 100 litres of milk. The owner of the dairy added 20 litres of water to it. The mixture was then separated in 2 containers, A and B, having volume in the ratio 3:2. The owner then added a total of 10 litres of water to A and B in the ratio 2:3. He withdrew 10 litres each from both the containers at the same time. If the ratio of milk in A and B respectively is expressed as m:n (where m and n are co-primes), find the value of m+n.

Question 100: In how many ways can 9 identical bowls be kept inside 3 identical boxes such that the boxes can hold any number of bowls?

Assuming AB = 2x and AC = 5x

Using pythagoras theorem, BC = $x\sqrt{5^2-2^2}$  =$x\sqrt{21}$

Now the circumradius of any right triangle = Half of hypotenuse = 2.5x

Hence, the required ratio = $\ \frac{\ x\sqrt{21}}{2.5x}$ = $\sqrt{\ 84}:5$

Now car A beats car B by 45km. Let the speed of car A be v(a) and speed of car B be v(b).

$\frac{v\left(a\right)}{v\left(b\right)}=\frac{m}{m-45}$ …..(1)where ‘”m” is the entire distance of the race track.

Moreover $\frac{v\left(b\right)}{v\left(c\right)}=\frac{m}{m-50}$…….(2)

and finally $\frac{v\left(a\right)}{v\left(c\right)}=\frac{m}{m-90}$……(3)

Multiplying (1) and (2) we get (3). $\frac{m}{m-90}=\frac{m}{m-45}\left(\frac{m}{m-50}\right)$

Solving we get m=450 which is the length of the entire race track

Initially let’s consider A and B as one component

The volume of the mixture is doubled by adding A(60% alcohol) i.e they are mixed in 1:1 ratio and the resultant mixture has 72% alcohol.

Let the percentage of alcohol in component 1 be ‘x’.

Using allegations , $\frac{\left(72-60\right)}{x-72}=\frac{1}{1}$ => x= 84

Percentage of alcohol in A = 60% => Let’s percentage of alcohol in B = x%

The resultant mixture has 84% alcohol. ratio = 1:3

Using allegations , $\frac{\left(x-84\right)}{84-60}=\frac{1}{3}$

=> x= 92%

We have, f(x+y) = f(x)+f(y)

Let us try to find out the value of f(2)

f(2) = f(1)+f(1) = 2 f(1)

Similarly f(3) = f(2+1) = f(2)+ f(1) = 3f(1)

…….
Hence, f(k) = kf(1)

Now, f(3)+f(4)+f(5)+…………f(11) = 189
=> 3f(1) + 4f(1) + 5f(1)……..11f(1) = 189

=> f(1)*(3+4+5+…………….11)  = 189
=> 63f(1)= 189
=> f(1)=3
Hence, f(15)=15f(1) = 15*3=45

y = $x^2 – 8x + 13$
=$x^2-2*x*4 + 16-3$
=$(x-4)^2-3$
The graph of y at x = 4+p and x = 4-p is the same.
Hence the graph of y = $x^2 + 8x + 13$ is symmetric about x = 4

Let the ratio of the alloys A, B and C melted be x : y : z.
$\frac{36x+50y+60z}{x+y+z}$ = 45
=> 36x + 50y + 60z = 45x + 45y + 45z
=> 9x = 5y + 15z

From the single equation, we can’t find the exact ratio, but we can identify whether a given ratio satisfies the equation.
In this case, 9x = 5y+15z is satisfied only by (5,3,2) and is not satisfied by the other three (3,2,1); (4,3,2) and (3,4,6)

Hence, option (c) is the correct answer.

Let the ratio in which A and B are mixed be m : n
The proportion of Z in A = 4/15
The proportion of Zinc in B = 7/10
The proportion of Z in C = 6/10 = 3/5
When, A and B are mixed in the ratio m:n, the final mixture has Zinc in the ratio 3 : 5
$\frac{4m/15 + 7n/10}{m + n} = 3/5$
4m/3 + 7n/2 = 3(m + n)
8m + 21n = 18(m + n)
=> 10m = 3n
Thus, m/n = 3/10

Now 23 students choose maths as one of their subject.

This means (MPC)+ (MC) + (PC)=23 where MPC denotes students who choose all the three subjects maths, physics and chemistry and so on.

So MC + PM =5 Similarly we have PC+ MP =7

We have to find the smallest number of students choosing chemistry

For that in the first equation let PM=5 and MC=0. In the second equation this PC=2

Hence minimum number of students choosing chemistry will be (18+2)=20 Since 18 students chose all the three subjects.

Let us take the number of employees in A,B and C be a,b,c respectively.
As the average age of A and C taken together is 50.5
$48a+52c$ = $50.5(a+c)$  => $1.5c = 2.5a$
We get a:c = 3:5  ….(1)
$50b+52c$ = $51\frac{1}{9}(b+c)$   =>   $\frac{8}{9}c = 1\frac{1}{9}b$  =>  $\frac{8}{9}c = \frac{10}{9}b$
b:c = 4:5  …(2)
Combining (1) and (2) we get a:b:c = 3:4:5

This means that number of 48, 50 and 52 year old are in the ratio 3:4:5.
The total age of all of them is given by $48\times a+50\times b+52\times c$

Lets assume average age A, B  & C as 3k, 4k and 5k respectively
Average age of the group is the ratio of total age to total number of people.
i.e Average age=$\frac{48a+50b+52c}{a+b+c} = \frac{48\times 3k+50\times 4k+52\times 5k}{3k+4k+5k} = \frac{604k}{12k} = 50\frac{1}{3}$years

Given, $\frac{\text{sum of scores in n matches+38+15}}{n+2}=29$

Given, $\frac{\text{sum of scores in n matches}}{n}=30$

=> 30n + 53 = 29(n+2) => n=5

Sum of the scores in 5 matches = 29*7 – 38-15 = 150

Since the batsmen scored less than 38, in each of the first 5 innings. The value of x will be minimum when remaining four values are highest

=> 37+37+37+37 + x = 150

=> x = 2

Let a, b be the numbers turned up on blue and green dice respectively.

The number turned up on red dice = a+b

a+b $\leq$ 6 , a$\geq$ 1, b $\geq$ 1

Let a be c+1, b = d+1

c+1+d+1 $\leq$ 6

c+d$\leq$ 4

c+d+t = 4   ($t\ge\ 0$)

Number of non negative solutions = $^6C_2$ = 15

Total sample space = 6*6*6 = 216

Required probability = $\frac{15}{216}$ = $\frac{5}{72}$

We know that the resultant mixture contains twice as much as water as alcohol. Therefore, the ratio of alcohol to total volume is 1:3.

In jar A, the ratio of alcohol to total volume is 2:5. This means that for 1 unit of the total mixture from A, we will get 2/5 as the quantity of alcohol.

In jar B, let the ratio be represented by x. This means that for 1 unit of mixture from B will have x units of alcohol.

We know that 1 unit of mixture from jar A and 2 units of mixture from jar B yields the resultant mixture.

Quantity of alcohol from A=2/5 units

Quantity of alcohol from B=2x units

Total volume of mixture=3 units

$\frac{2/5 + 2x}{3}$ = $\frac{1}{3}$

2/5 + 2x = 1

2x = 1-(2/5) = 3/5

x =3/10

Therefore, the ratio of volume of alcohol to the total volume is 3/10.

The ratio of alcohol to water is 3:7. Therefore, option C is the right answer.

Change in total weight of 10 students = difference in weight of the student who joined and the student who left

=> weigth of first student who left = 30 + (10×2) = 50

weight of the student who joined last = 50 – 10 = 40

Thus change in average weight = (40 – 30)/10 = 1

Alternate Solution:

The average weight increases by 2 when a student is replaced by another student. Since, there are 10 students, total change(increase as +2) in weight is 2 x10 = 20kg

Since, the weight of the new student is 30kg, the weight of the replaced student would be 30 – 20 = 10kg. Also after the 2nd replacement a new student having weight 10kg less than the 30kg student joins the class. Thus, the weight of the new student = 30 -10 = 20kg.

The difference in total weight between the first case and after two replacement is thus 20 – 10 = 10kg. Thus, difference is average weight = 10/10 = 1kg

Let s, d, and t be the number of students who drink exactly 1, exactly 2 and all the 3 drinks respectively.
According to the given information,
s+d+t = 90
s+2d+3t = 35+46+25 = 106
Thus, 2t+d = 16
Thus, the maximum values of d and t are 16 and 8 respectively.
Hence, A = 16 and B = 8.
Thus, A:B = 2:1.
Hence, option A is the correct answer.

The difference in the time take to traverse the same distance $’d’$ at two different speeds is 35 minutes. Equating this: $\frac{d}{8}-\frac{d}{15}\ =\ \frac{35}{60}$

On solving, we obtain $d = 10 kms$. Let $x kmph$ be the speed at which Amal needs to travel to reach the office in 50 minutes; then

$\frac{10}{x}=\frac{50}{60}\ or\ x\ =\ 12\ kmph$.Hence, Option B is the correct answer.

The insect and the train will repeat the same process after the insect reaches its initial point.

Consider the distance between the insect and the train when the insect starts flying from P be d.

Distance between point where both meet and P = $\frac{d}{(30+45)} \times 45 = \frac{3d}{5}$

Then both will move in the same direction.

Time taken by the insect to reach P again=$\large\frac{\frac{3d}{5}}{45} = \frac{d}{75}$ (which is the required time interval)

Distance traveled by the train when the insect reaches P = $\frac{30*d}{75}$ = $\frac{2d}{5}$

Distance between the train and the insect when the insect reaches P = 3d/5-2d/5 = d/5

The same process will be repeated again with distance $d_{1}=\frac{d}{5}, d_{2}=\frac{d}{25}, d_{3}=\frac{d}{125}$ and so on.

Then time intervals will be $d_{1}/75,d_{2}/75,d_{3}/75$ and so on.

Sum of the time intervals = (d/75)(1+1/5+1/25+1/125………)

Now the initial distance,d =240 km

Sum of the time intervals = $\frac{240}{75}(\frac{1}{1-\frac{1}{5}})$ = 4 hours

Alternate Solution:

Total time taken by the train to reach Q=240/30=8 hours

The insect will move towards the train meet at some place and return back. Since the speed of insect is constant, no matter what time the train takes, the time taken by insect towards the train and and away from the train will remain same each time because the insect travels same distance both towards the train and away from the train.

Total time taken in the direction of the train=Total time taken against the direction of the train=8/2=4 hours

Let the speed of the escalator be ‘e’ steps/second
Anand takes 120 steps to reach the bottom, so the time taken = 120/3 = 40 seconds
Number of steps taken by the escalator in same time = 40e steps
So number of steps in the escalator = 120 + 40e
Sanand takes 80 steps to reach the bottom, so the time taken = 80/1 = 80 seconds
Number of steps taken by the escalator in same time = 80e steps
So number of steps in the escalator = 80 + 80e
Since the escalator in same in this case, we get,
120 + 40e = 80 + 80e
40 = 40e
e = 1 steps/second
So number of steps = 120 + 40*1 = 160 steps
Hence, option A is the right answer.

As a is a root of $x^2+9x+5=0$,
We have $a^2+9a+5=0$
=> $x^2+9x=-5$
Now $(2a+9)^2=4a^2+36a+81$
=> $(2a+9)^2=4(a^2+9a)+81$
=> $(2a+9)^2=4(-5)+81$
=> $(2a+9)^2=61$
Similarly, $(2b+9)^2=61$
Now the expression given in the question is
= $(\frac{5}{2a+9})^2+(\frac{6}{2b+9})^2$
= $\frac{25}{61}+\frac{36}{61}$
= $\frac{61}{61}$
= 1.

On expanding the expression we get $1-\log_ab+1-\log_ba$

$or\ 2-\left(\log_ab+\frac{1}{\log_ba}\right)$

Now applying the property of AM>=GM, we get that  $\frac{\left(\log_ab+\frac{1}{\log_ba}\right)}{2}\ge1\ or\ \left(\log_ab+\frac{1}{\log_ba}\right)\ge2$ Hence from here we can conclude that the expression will always be equal to 0 or less than 0. Hence any positive value is not possible. So 1 is not possible.

Let the total work be 60 units (lcm of 12 and 15)
Rahul can complete the work in 12 days.
=> Rahul’s 1 day’s work = 5 units.
Seema can complete the work in 15 days.
=> Seema’s 1 day’s work = 4 units.
Before Deepa joined Rahul and Seema, 16.66% work was completed.
Work left = 83.33% of 60 units = 50 units.
Stipulated time required to complete the remaining work by Rahul and Seema = $\dfrac{50}{4 + 5}$ days = $\dfrac{50}{9}$ days.
After Deepa joined, the time taken was 2 days less than the stipulated time.
After Deepa joined, time to complete the work = $(\dfrac{50}{9} – 2)$days = $\dfrac{32}{9}$days
Let Deepa’s 1 day’s work be x units.
Then, (4 + 5 + x) * $\dfrac{32}{9}$ = 50
On solving, we get x = $\dfrac{81}{16}$ units
So, Deepa’s 1 day’s work = $\dfrac{81}{16}$ units
Time required by Deepa to complete the work = $\dfrac{60}{\frac{81}{16}}$ days = $\dfrac{320}{27}$ days = $11\dfrac{23}{27}$ days
Hence, option C is the correct answer.

Let the desired efficiency of each worker ‘6x’ per day.

140*6x*200= 6 km …(i)

In 60 days 60/200*6=1.8 km of work is to be done but actually 1.5km is only done.

Actual efficiency ‘y’= 1.5/1.8 *6x =5x.

Now, left over work = 4.5km which is to be done in 140 days with ‘n’ workers whose efficiency is ‘y’.

=> n*5x*140=4.5 …(ii)

(i)/(ii) gives,

$\frac{\left(140\cdot6x\cdot200\right)}{\left(n\cdot5x\cdot140\right)}=\frac{6}{4.5}$

=> n=180.

.’. Extra 180-140 =40 workers are needed.

$\frac{\log\ 5}{2\log2}\ =\frac{\log\ y}{2\log2}\cdot\frac{\log\ 5}{2\log6}$

$\log\ 36\ =\ \log\ y;\ \therefore\ y\ =36$

Let the volume of Metals A,B,C we 3x, 4x, 7x

Ratio weights of given volume be 5y,2y,6y

.’. 12xy+8xy+42xy=130 => 65xy=130 => xy=2.

.’.`The weight, in kg. of the metal C is 42xy=84.

Let the sides of the rectangle be “a” and “3a” m. Hence the perimeter of the rectangle is 8a.

Let the side of the equilateral triangle be “m” cm. Hence the perimeter of the equilateral triangle is “3m” cm. Now we know that 8a+3m=90……(1)

Moreover area of the equilateral triangle is $\frac{\sqrt{\ 3}}{4}m^2$ and area of the rectangle is $3a^2$

According to the relation given $\left(\frac{\sqrt{\ 3}}{4}m^2\right)^{^2}=\ 3a^2$

$\frac{3}{16}m^4=\ 3a^2\ or\ a^2=\frac{m^4}{16}$

$a=\frac{m^2}{4}$

Substituting this in (1) we get $2m^2+3m-90\ =0$ solving this we get m=6 (ignoring the negative value since side can’t be negative)

Hence a=9 and the longer side of the rectangle will be 3a=27cm

3 mutually perpendicular cuts pass through the centre of the sphere.

Each cut produces 2 new surfaces each of which is a circle with the radius of the sphere.

3 cuts will produce 6 circles.

Additional surface area = 6$\pi\ r^2$

Percentage increase in area = $\frac{6\pi\ r^2}{4\pi\ r^2}\times\ 100\ =\ 150\%$

Let the CP of the each toy be “x”. CP of 12 toys will be “12x”. Now the shopkeeper made a 10% profit on CP. This means that

12x(1.1)= 2112 or x=160 . Hence the CP of each toy is ₹160.

Now let the SP of each toy be “m”. Now he sold 8 toys at 20% discount. This means that 8m(0.8) or 6.4m

He sold 4 toys at an additional 25% discount. 4m(0.8)(0.75)=2.4m  Now 6.4m+2.4m=8.8m=2112 or m=240

Hence CP= 160 and SP=240. Hence profit percentage is 50%.

The graphs of $2^{x}+2^{-x} and 2-(x-2)^{2}$ never intersect. So, number of solutions=0.

Alternate method:

We notice that the minimum value of the term in the LHS will be greater than or equal to 2 {at x=0; LHS = 2}. However, the term in the RHS is less than or equal to 2 {at x=2; RHS = 2}. The values of x at which both the sides become 2 are distinct; hence, there are zero real-valued solutions to the above equation.

The midpoints of two diagonals of a parallelogram are the same

Hence the midpoint of (2,1) and (-3,-4) lie on $x+9y+c=0$

midpoint of (2,1) and (-3,-4) = ($\frac{2-3}{2},\frac{1-4}{2}$) = (-1/2 , -3/2)

Keeping this cordinates in the above line equation, we get c = 14

$f\left(x\right)=\ x^2+ax+b$

$f\left(x+1\right)=x^2+2x+1+ax+a+b$

$f\left(x-1\right)=x^2-2x+1+ax-a+b$

$g(x)=f(x+1)-f(x-1)= 4x+2a$

Now $g(20) = 72$ from this we get $a = -4$ ; $f\left(x\right)=x^2-4x\ +b$

For this expression to be greater than zero it has to be a perfect square which is possible for $b\ge\ 4$

Hence the smallest value of ‘b’ is 4.

The given function is equivalent to f(x) = $a^x$

Given, f(5) = 4

=> $a^5=4=>\ a=4^{\frac{1}{5}}$

=> f(x) = $4^{\frac{x}{5}}$

f(10) – f(-10) = 16 – 1/16 = 15.9375

$x_0=max(x_1,x_2,….,x_{12})$

$x_0$ will be minimum if x1,x2…x12 are close to each other

100/12=8.33

.’. max$(x_1,x_2,….,x_{12})$ will be minimum if $(x_1,x_2,….,x_{12})$=(9,9,9,9,8,8,8,8,8,8,8,8,)

.’. Option B is correct.

Let Jack take “t” days to complete the work, then John will take “2t” days to complete the work. So work done by Jack in one day is (1/t) and John is (1/2t) .

Now let Jim take “m” days to complete the work. According to question, $\frac{1}{t}+\frac{1}{m}=\frac{3}{2t}\ or\ \frac{1}{m}=\frac{1}{2t\ }or\ m=2t$ Hence Jim takes “2t” time to complete the work.

Now let the three of them complete the work in “p” days. Hence John takes “p+3” days to complete the work.

$\frac{1}{2t}\left(m+3\right)=\left(\frac{4}{2t}\right)m$

$\frac{1}{2t}\left(m+3\right)=\left(\frac{4}{2t}\right)m$

or m=1. Hence JIm will take (1+3)=4 days to complete the work. Similarly John will also take 4 days to complete the work

Here, let us assume that each man will do 1 unit of work everyday.
Total work = 30*(15) = 450 units.
According to the question 28 men were there on second day,26 on fourth and so on
Work done on first 20 days = 30+28+30+26+30+24………… = (30+28)+(30+26)+(30+24)…………
=58+56……..+42 =$\frac{10}{2}*(2(58)+(10-1)(-2))$ = 490
But, 30units is done on 19th day and 10units on 20th day.
The total work of 450 units will be done in first 18 days.

$x=2^{12\left(7+4\sqrt{\ 3}\right)}$.

$x^{\frac{7}{2}}=2^{42\left(7+4\sqrt{\ 3}\right)}$

$x^{2\sqrt{\ 3}}=2^{24\sqrt{\ 3}\left(7+4\sqrt{\ 3}\right)}$

$\frac{x^{\frac{7}{2}}}{x^{2{\sqrt{3}}}}$ = $2^{\left(7+4\sqrt{\ 3}\right)\left(42-24\sqrt{\ 3}\right)}=2^{\left(7+4\sqrt{\ 3}\right)\left(7-4\sqrt{\ 3}\right)6}$ =$2^6$.

Let the present age of Ramesh’s son be x
When Ramesh was x, his son was x/4. Thus, the no. of years before from the present during which Ramesh’s son’s age was x = x – x/4 = 3x/4
Thus, the present age of Ramesh = x + 3x/4 = 7x/4
Thus, the ratio of ages of Ramesh and his son = 7x/4x = 7/4
Thus, since, Ramesh age is at-least 45, his ages could be 49, 56 etc(only integers are possible as we need the ratio on their birth days). When Ramesh is 49 his son is 4/7 x 49 = 28. If Ramesh’s age is greater than 49, his son’s age would be greater than 30 which is not possible .
Thus, the age of Ramesh is 49 and his son’s age is 28
Thus, the required sum = 49 + 28 = 77

To complete one round Ram takes 100m/15kmph and Rahim takes 20m/5kmph

They meet for the first time after L.C.M of (100m/15kmph , 20m/5kmph) = 100m/5kmph=20m/kmph.

Distance traveled by Ram =20m/kmph * 15kmph =300m.

So, he must have ran 300/100=3 rounds.

Note:

CAT gave both 2 and 3 as correct answers because of the word ‘before‘.

$ab\ =\ 4^{2017}=2^{4034}$

The total number of factors = 4035.

out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035/2] = 2017.

And since the given number is a perfect square we have one set of two equal factors.

.’. many pairs(a, b) of positive integers are there such that $a\leq b$ and $ab=4^{2017}$ = 2018.

$2^{Y^2({\log_{3}{5})}}=5^{Y^2(\log_3 2)}$

Given, $5^{Y^2\left(\log_32\right)}=5^{\left(\log_23\right)}$

=> $Y^2\left(\log_32\right)=\left(\log_23\right)=>Y^2=\left(\log_23\right)^2$

=>$Y=\left(-\log_23\right)^{\ }or\ \left(\log_23\right)$

since Y is a negative number, Y=$\left(-\log_23\right)=\left(\log_2\frac{1}{3}\right)$

Assume the radius of the cylinder is r and the height is h.

From the figure, $(2r)^2+h^2=\left(Diameter\right)^2$ => $4r^2+h^2=\left(12\right)^2$ = 144 ….(1)
Now, the volume of the cylinder = $\pi r^2h$
To maximize $r^2h\$, we will maximize $r^4h^2$.
From 1, $4r^2+h^2=\left(12\right)^2$ = 144
=> $2r^2+2r^2+h^2=\left(12\right)^2$ = 144
Since the sum of the three terms is constant, then the product will be maximum when all three terms are equal. (Using AM, GM)
Here the product is $2r^2\times\ 2r^2\times\ h^2=4r^4h^2$, if this is maximized, then $r^4h^2$ will be maximized too, which in turn will maximize $r^2h$
So, $2r^2=h^2$
Now, 6r$^2$=144 (Since 2r$^2$ is equal to h$^2$, substitute this value in equation 1)
r = $\sqrt{24}=2\sqrt{6}\ cm$

Alternate Explanation:
Assume the radius of the cylinder is r and the height is h.

From the figure, $(2r)^2+h^2=\left(Diameter\right)^2$   => $4r^2+h^2=\left(12\right)^2$ = 144 ….(1)
Now, the volume of the cylinder = $\pi r^2h$ = $\pi\dfrac{(144-h^2)}{4}h$ = $\pi\dfrac{(144h-h^3)}{4}$             (From 1)
To find the maximum value, differentiating with respect to h, we get 144-$3h^2$ = 0
=> $h^2=48$
Hence, from 1, $4r^2=144-48$
=> r = $\sqrt{\frac{96}{4}}=2\sqrt{6}$ cm

Assuming the first number is 3x and the second number is x.

So the third number will be 300-4x.

Then the product of three numbers will be, x$\times\ 3x\times\ \left(300-4x\right)$ = $\ \frac{\ 3}{4}\times\ 2x\times\ 2x\times\ \left(300-4x\right)$

Since the numbers 2x, 2x and 300-4x is constant and equal to 300, the product will be maximized when all the numbers are equal. (Using AM-GM)

Hence, 2x = 300-4x   =>x=50

So the value of x$\times\ 3x\times\ \left(300-4x\right)$ = 50*150*100 = 750000

Assuming, $x^2-y^2=a$ and $x+y=b$ ………(1)
Hence, x-y = $\ \frac{\ a}{b}$……..(2)
Adding 1 and 2, we get, 2x = b+$\ \frac{\ a}{b}$  =>x = $\ \ \frac{\ 1}{2}\left(\frac{\ a}{b}+b\right)$
Similarly, y = $\frac{1}{2}\left(b-\frac{a}{b}\right)$
Given, f($x^2-y^2,\ x+y$) = $\ \dfrac{\ x}{y}$ = $\ \dfrac{\ \ \frac{\ a}{b}+b}{\ b-\frac{\ a}{b}}$
=> f(a,b)= $\ \dfrac{\ \ \frac{\ a}{b}+b}{\ b- \frac{\ a}{b}}$ = $\ \dfrac{\ \ \ a+b^2}{\ \ b^2-a}$
=> f(x,y)= $\ \dfrac{\ \ \ x+y^2}{\ \ y^2-x}$

Let the base radius be 3r.

Height of upper cone is 9 so, by symmetry radius of upper cone will be r.

Volume of frustum=$\frac{\pi}{3}\left(9r^2\cdot27-r^2.9\right)$

Volume of upper cone = $\frac{\pi}{3}.r^2.9$

Difference= $\frac{\pi}{3}\cdot9\cdot r^2\cdot25=225$ => $\frac{\pi}{3}\cdot r^2=1$

Volume of larger cone = $\frac{\pi}{3}\cdot9r^2\cdot27=243$

We can form the following figure based on the given information:

Since OA = 4 m and OB=3 m; AB = 5 m. OR bisects the chord into PC and QC.

Since AB = 5 m, we have $a+b = 5 …(i)$  Also, $4^2\ -k^2=a^2…\left(ii\right)$ and $3^2\ -k^2=b^2…\left(iii\right)$

Subtracting (iii) from (ii), we get: $a^2\ -b^2=7…\left(iv\right)$

Substituting (i) in (iv), we get $a – b = 1.4 …(v)$; $\left[\left(a+b\right)\left(a\ -b\right)=7;\ \therefore\ \left(a-b\right)=\frac{7}{5}\right]$

Solving (i) and (v), we obtain the value of $a=3.2$ and $b=1.8$

Hence, $k^2\ =\ 5.76$

Moving on to the larger triangle $\triangle\ POC$, we have $5^2-k^2=\left(x+a\right)^2$;

Substituting the previous values, we get: $(25-5.76)=\left(x+3.2\right)^2$

$\sqrt{19.24}=\left(x+3.2\right)$ or $x = 1.19 m$

Similarly, solving for y using $\triangle\ QOC$, we get $y=2.59 m$

Therefore, $PQ = 5+2.59+1.19 = 8.78 \approx\ 8.8 m$

Hence, Option A is the correct answer.

Equation of circle $x^2+y^2+2gx+2fy+c=0$

It passes through (0,0), (4,0) and (3,9). Substitute each point in the above equation:

=> On substituting the value (0,0) in the above equation, we obtain: $c=0$

=> On substituting the value (4,0) in the above equation, we obtain:  $16+0+8g+0 = 0$ ; $g=-2$

=> On substituting the value (3,9) in the above equation, we obtain: $9+81-12+18f = 0$ ; $f= -13/3$

Radius of the circle = $\sqrt{\ g^2+f^2-c}$ => $r^2=\frac{205}{9}$

Therefore, Area = $\pi\ r^2=\frac{205\pi\ }{9}$

Assuming the number of employees in department P = p
Then the number of employees in department R = 2p
Now, the total number of employees in Q and R is 280.
Since the transfer is from R to Q, the total number of employees will still be 280.
Hence, the number of employees of R after transfer = 2p-20
=> The number of employees of Q after transfer = 280-(2p-20) = 300-2p
Now, the ratio of the number of employees in department P to that of Q will be the same as the ratio of the number of employees in department Q to that of department R.
=> $\frac{p}{300-2p}=\frac{300-2p}{2p-20}$
=> $p\left(p-10\right)=\left(150-p\right)\left(300-2p\right)$
=> $p^2-10p=45000-600p+2p^2$
=> $p^2-590p+45000=0$
=> (p-90)(p-500)=0
=> p=90  (Rejecting the 500, because the total number of employees in R will become 2p = 1000 but there is a condition that the total number of employees in Q and R is 280)
The number of employees in department P = 90
Hence the number of employees in department R = 180

The ratio of the amounts at the end of the 2nd and 4th years is $(1+R)^2 = 1.27$
After 6 years, the ratio is $(1+R)^6 = 2.05$
It can be seen that $(1+R)^5 < 2$
So, the amount doubles every 6 years

The four digit even numbers will be of form:

1100,1122,1144…1188,2200,2222,2244….9900,9922,9944,9966,9988

Their sum ‘S’ will be (1100+1100+22+1100+44+1100+66+1100+88)+(2200+2200+22+2200+44+…)….+(9900+9900+22+9900+44+9900+66+9900+88)

=> S=1100*5+(22+44+66+88)+2200*5+(22+44+66+88)….+9900*5+(22+44+66+88)

=> S=5*1100(1+2+3+…9)+9(22+44+66+88)

=>S=5*1100*9*10/2 + 9*11*20

Total number of numbers are 9*5=45

.’. Mean will be S/45 = 5*1100+44=5544.

Option D

Let the total distance be ‘D’ km and the speed of the train be ‘s’ kmph. The time taken to cover D at speed d is ‘t’ hours. Based on the information: on equating the distance, we get $s\ \times\ t\ =\ \frac{s}{3}\times\ \left(t+\frac{1}{2}\right)$
On solving we acquire the value of $\ t\ =\frac{1}{4}$ or 15 mins. We understand that during the return journey, the first 5 minutes are spent traveling at speed ‘s’ {distance traveled in terms of s = $\ \frac{s}{12}$}. Remaining distance in terms of ‘s’ = $\ \frac{s}{4}-\frac{s}{12}\ =\frac{s}{6}$

The rest 4 minutes of stoppage added to this initial 5 minutes amounts to a total of 9 minutes. The train has to complete the rest of the journey in $15 – 9 = 6 mins$ or {1/10 hours}. Thus, let ‘x’ kmph be the new value of speed. Based on the above, we get $\frac{s}{\frac{6}{x}}\ =\frac{1}{10}\ or\ x\ =\frac{10s}{6}$

Since the increase in speed needs to be calculated: $\frac{\left(\frac{10s}{6}\ -s\right)}{s}\times\ 100\ =\frac{200}{3}\approx\ 67\%$ increase.

Hence, Option C is the correct answer.

Assuming the number of males is 2x and the number of females is 3x.

The number of males who make Rs 60000 can be assumed y and the number of females who make Rs 30000 can be assumed 4y.

The ratio of the average salary of males to the average salary of females = $\ \frac{\left(\ 60000y+30000\left(2x-y\right)\right)\times\ 3x}{2x\times\ \left(60000\left(3x-4y\right)\right)}$ = $\ \frac{\ 15}{16}$

=> $\ \frac{\ \left(2y+2x-y\right)\times\ 3}{\left(2\left(3x-4y\right)+4y\right)\times\ 2}=\ \frac{\ 15}{16}$

=> 8x+4y = 15x-10y

=> 7x = 14y  =>x = 2y ……….(1)

Number of employees earning Rs 60000 = y+3x-4y = 3x-3y = 6y-3y = 3y ……..(1)

Number of employees earning Rs 30000 =2x-y+4y = 2x+3y = 4y+3y  = 7y ………(2)

Average salary = $\ \frac{\ 60000\times\ 3y+30000\times\ 7y}{3y+7y}$ = 39000

Let ABCD be the rectangle with length 2l and breadth 2b respectively.

Area of the circle i.e. area of painted region = $\pi\ b^2$.

Given, 4lb-$\pi\ b^2$=(2/3)$\pi\ b^2$.

=> 4lb=(5/3)$\pi\ b^2$.

=>l=$\frac{5\pi}{12}b$.

Given, 4lb=135 => 4*$\frac{5\pi}{12}b^2$=135 => b= $\frac{9}{\sqrt{\ \pi\ }}$

=> l=$\frac{15}{4}\sqrt{\ \pi\ }$
Perimeter of rectangle =4(l+b)=4($\frac{15}{4}\sqrt{\ \pi\ }$+$\frac{9}{\sqrt{\ \pi\ }}$ )=$3\sqrt{\pi}(5+\frac{12}{\pi})$.

Hence option A is correct.

Let the number of pencils bought by Aron be “p” and the cost of each pencil be “a”.

Let the number of sharpeners bought Aron be “s” and the cost of each sharpener be “b”.

Now amount spent by Aron will be (pa)+(sb)

Aditya bought (2p) pencils and (s-10) sharpeners. Amount spent will be (2pa)+(s-10)b

Amount spent in both the cases is same

pa + sb = 2pa + (s-10)b or pa=10b

Now its given in the question that cost of sharpener is 2 more than pencil i.e. b=a+2

pa= 10a+20 or a=20/(p-10)

Now the number of pencils has to be minimum, for that we have to find smallest “p” such that both “p” and “a” are integers. The smallest such value is p=11 . Total number of pencils bought will be p+2p=11+22=33

Let the length of radius be ‘r’.

From the above diagram,

$x^2+r^2=6^2\$….(i)

$\left(10-x\right)^2+r^2=8^2\$—-(ii)

Subtracting (i) from (ii), we get:

x=3.6 => $r^2=36-\left(3.6\right)^2$ ==> $r^2=36-\left(3.6\right)^2\ =23.04$.

Area of circle = $\pi\ r^2=23.04\pi\$

Area of rhombus= 1/2*d1*d2=1/2*12*16=96.

.’. Ratio of areas = 23.04$\pi\$/96=$\frac{6\pi}{25}$

Let their individual Amounts be equal after ‘t’ years. Let their initial investments amount to $A_V$ and $A_J$ ;

$A_V\ =10,000\left(1+\frac{5t}{100}\right)$ and $A_J\ =8,000\left(1+\frac{10\left(t-2\right)}{100}\right)$

Equating both: $10,000\left(1+\frac{5t}{100}\right)\ =8,000\left(1+\frac{10\left(t-2\right)}{100}\right)$

On simplifying both sides, we get: $15t\ =\ 180\ ;\ t\ =\ 12$

Given,

Initial salary of Raghav = 10000x
Increase in salary = x% of 10000x
final salary of Raghav 10000p = 10000 ($x+ \frac{x^2}{100}$ )
=>p = $x+ \frac{x^2}{100}$
100p = $100 x+ x^2$
100p + 2500 =  $100 x+ x^2 + 2500$
100p + 2500 = $(x+50)^2$
$x+50$ = 10$(\sqrt{p+25}$)
$x$ = 10[$(\sqrt{p+25}$) – 5]
p = $n^2$
$x$ = 10[$(\sqrt{n^2+5^2}$) – 5]
For x to be integer, $(\sqrt{n^2+5^2}$ has to be integer.

We have only one Pythagorean triplet which have 5 as one of the numbers ie (5, 12, 13)
=> n = 12
$x$ = 10[$(\sqrt{12^2+5^2}$) – 5]
$x$ = 10[13 – 5]
= 80
Final salary of Raghav = (80+64)*10000
= Rs 1440000

We have,

Denoting the length AB by $a$ and PB by $b$. (It is given that $b$ = 6-$3\sqrt{2}$ cm)
Now, the area of ABC = $\frac{\sqrt{3}}{4}a^2\$ ….(1)
In the triangle APQ, AP = $a-b$
$\tan\ 60^{\circ\ }=\frac{PQ}{AP}$
=> PQ = ($a-b$)$\sqrt{3}$
Now, area of APQ = $\frac{1}{2}\times\ \sqrt{3}\left(a-b\right)\left(a-b\right)$ ……(2)
Equating (1) and (2),
$\frac{1}{2}\times\ \sqrt{3}\left(a-b\right)\left(a-b\right)=\ \frac{\sqrt{3}}{4}a^2$
=> $a^2= 2(a-b)^2$
=> $a=\sqrt{2}(a-b)$
=> $b$ = $\frac{a}{\sqrt{2}}\left(\sqrt{2}-1\right)$
=> $a$ = $\frac{b\sqrt{2}}{\sqrt{2}-1}=b\sqrt{2}(\sqrt{2}+1)$ = $\sqrt{2}\left(6-3\sqrt{2}\right)\left(\sqrt{2}+1\right)$ = $6(\sqrt{2}-1)(\sqrt{2}+1)$ = 6 cm

Assume the amount of water per litre mixture of R = x
After Q and R in the ratio 1:2, the ratio of water and alcohol  = 4:11
Assuming the volume of Q = 5 litres, then volume of R will be 10 litres,
Hence, water in 5 litres of Q = $\frac{5}{1+4}\times\ 1$ = 1 litre
Water in 10 litres of R = 10x
Hence total water in the mixture = $\left(10+5\right)\times\ \frac{4}{4+11}$  = 10x+1
=> x = 0.3
Now, if P, Q and R are combined in 3:5:2, then the volume can be assumed as 30, 50 and 20.
So amount of water in P = $30\times\ \frac{2}{5}=\ 12$
Similarly,  amount of water in Q = $50\times\ \frac{1}{5}$ = 10
The amount of water in R = $20\times\ 0.3\ \ =6$
Hence, the total amount of water in (30+50+20) = 100 litres mixture = 12+10+6 = 28 litres
Hence, the percentage  = 28

Since Amol started at 5PM and meet Bhuvan at 9PM who starts at 7PM,
Total distance XY = 4A+2B where A and B are speeds of Amol and Bhuvan respetively.
Time taken by Amol = (4A+2B)/A = 4+2B/A
Time taken by Bhuvan = (4A+2B)/B = 2+4 A/B
4 + 2B/A = 2 + 4A/B + 2 + 1/3
2x – 4/x = 1/3 where x = B/A
$6x^2-12 = x$
$6x^2-x-12 = 0$
$6x^2-9x+8x-12$ = 0
$3(2x-3)+4(2x-3) = 0$
(3x+4)(2x-3) = 0
x = 3/2 = B/A
Time taken by Amol = (4A+2B)/A = 4+2B/A = 4+(2*3/2) = 4+3 = 7 hours
Amol will reach at 12AM.

Let the number of large boxes, in which Jay places medium boxes, be m. Hence, m<10.

Let the number of medium boxes, in which Jay places small boxes, be n. Hence, n<m

After adding 7*m medium boxes, the number of empty boxes = (10-m) large boxes + 7m medium boxes = 10+6m. Once a large box gets 7 medium boxes, it is no longer empty.

After adding the 7*n small boxes, the number of empty boxes = (10-m) large boxes + (7m-n) medium boxes + 7n small boxes = 10+6m+6n = 10+6(m+n)

We know that 10+6(m+n) = 82

6(m+n) = 72 => m+n = 12

Now the total number of boxes used = 10 + 7m + 7n = 10+7(m+n) = 10+ 7*12 = 94

Let the percentage of students who like all three sports be c, the percentage of students who like exactly two sports be b and the percentage of students who like exactly one sport be a.
So, a + b + c = 100
a + 2b + 3c = 65 + 86 + 57 = 208
=> b + 2c = 108
To get the maximum percentage of students who like exactly two sports, we have to maximize b, so the c has to minimum
b + 2c = 108
b + c = 100
So, we get C = 8
Thus the minimum value of c which satisfies these equations is 8
Thus the maximum value of b = 92
Hence the correct answer is 92

Let the number of persons who do not take any juice and the number of persons who take all 4 types of juices be ‘x’.
As there are ‘x’ people who take atleast 4 types of juices there should be ‘2x’ people who take atleast 3 types of juices.
This means the number of people who take exactly 3 types of juices = 2x – x = x.
As there are ‘2x’ people who take atleast 3 types of juices there should be ‘4x’ people who take atleast 2 types of juices.
This means the number of people who take exactly 2 types of juices = 4x – x – x = 2x.
As there are ‘4x’ people who take atleast 2 types of juices there should be ‘8x’ people who take atleast 1 type of juices.
This means the number of people who take exactly 1 type of juices = 8x – 4x = 4x.
Thus the total number of people in the Gym = x + 4x + 2x + x + x = 9x.
9x = 207 => x = 23.
The number of people who take exactly 2 types of juices = 2x = 46.

We have,

Given the length of the equilateral triangle as $6\sqrt{3}$.
In triangle AB$_1$A$_1$, AA$_1$ = $2\sqrt{3}$ (one-third of AC) and AB$_1$ = $4\sqrt{3}$ (two-third of AB)
In triangle AB$_1$A$_1$, cos$\angle\ A_1AB_1$ = $\frac{\ \left(AA_1\right)^2+\left(AB_1\right)^2-\left(A_1B_1\right)^2}{2\times\ AB_1\times\ AA_1}$
=> $\ \frac{1}{2}\ =\ \ \frac{\ \left(2\sqrt{3}\right)^2+\left(4\sqrt{3}\right)^2-\left(A_1B_1\right)}{2\times\ 2\sqrt{3}\times\ 4\sqrt{3}}$
$=>\ 2\sqrt{3}\times\ 4\sqrt{3}\ =\ \left(2\sqrt{3}\right)^2+\left(4\sqrt{3}\right)^2-\left(A_1B_1\right)^2$
=> $\left(A_1B_1\right)^2\ =\ 12+48-24\ =36$
$=>A_1B_1\ =6$
Inradius = $\frac{Area}{Semi\ \ perimeter}$
Area of AB$_1$A$_1$ = $\frac{\sqrt{3}}{4}\times\ 6^2\ =\ 9\sqrt{3}$
Semi-perimeter = $\frac{\left(6\times\ 3\right)}{2}$ = 9
Hence inradius = $\sqrt{3}$ cm

Alternate Solution:
Given the length of the equilateral triangle as $6\sqrt{3}$.
In triangle  AB$_1$A$_1$, AA$_1$ = $2\sqrt{3}$ (one-third of AC) and AB$_1$ = $4\sqrt{3}$ (two-third of AB)
Now, ${\cos\ \angle\ AA_1B_1\ =\ \frac{AA_1}{AB_1}}$ (This equation will only true for a right angle triangle).
=> $\cos\ 60^{\circ\ }$ = $\frac{2\sqrt{3}}{4\sqrt{3}}$
Since LHS = RHS = $\frac{1}{2}$
Hence,  AB$_1$A$_1$ is a right triangle.
=>  B$_1$A$_1$ =AB$_1$ sin 60 = $4\sqrt{3}\times\ \frac{\sqrt{3}}{2}$ = 6
Inradius = $\frac{Area}{Semi\ \ perimeter}$
Area of  AB$_1$A$_1$ = $\frac{\sqrt{3}}{4}\times\ 6^2\ =\ 9\sqrt{3}$
Semi-perimeter = $\frac{\left(6\times\ 3\right)}{2}$ = 9
Hence inradius = $\sqrt{3}$ cm

f(x) = $x^2+7x+10$
-2, -5 are the roots of f(x)
Now for f(f(x)) = 0,
$x^2+7x+10$ = -2 or $x^2+7x+10$ = -5
$x^2+7x+12$ = 0 or $x^2+7x+15$ = 0
x = -3, -4 and the second equation has imaginary roots
Now for f(f(x)) = 0,
$x^2+7x+10$ = -3 or $x^2+7x+10$ = -4
$x^2+7x+13=0$ or $x^2+7x+14$ = 0
Now both the equations has imaginary roots.

Let the initial number of chocolates be 64x.

First child gets 32x+1 and 32x-1 are left.

2nd child gets 16x+1/2 and 16x-3/2 are left

3rd child gets 8x+1/4 and 8x-7/4 are left

4th child gets 4x+1/8 and 4x-15/8 are left

5th child gets 2x+1/16 and 2x-31/16 are left.

Given, 2x-31/16=0=> 2x=31/16 => x=31/32.

.’. Initially the Gentleman has 64x i.e. 64*31/32 =62 chocolates.

Let the number be ‘abc’. Then, $2<a\times\ b\times\ c<7$. The product can be 3,4,5,6.

We can obtain each of these as products with the combination 1,1, x where x = 3,4,5,6. Each number can be arranged in 3 ways, and we have 4 such numbers: hence, a total of 12 numbers fulfilling the criteria.

We can factories 4 as 2*2 and the combination 2,2,1 can be used to form 3 more distinct numbers.

We can factorize 6 as 2*3 and the combination 1,2,3 can be used to form 6 additional distinct numbers.

Thus a total of 12 + 3 + 6 = 21 such numbers can be formed.

First equation is $y_{1}$ = 2x+5 or – (2x+5)
Second equation is $y_{2}$ = 6x+7 or – (6x+7)
Finding the intersection points:
2x+5 = -(6x+7) so x = -1.5 and y = 2 and
2x+5 = (6x+7) so x = -0.5 and y = 4
Plotting the equations on graph.

$|y_{1}| > |y_{2}|$ will be the region highlighted in the dark blue colour.
The coordinates of the boundary points is (-1.5, 2) and (-0.5, 4)
So by distance formula,
Length = $\sqrt{(-1.5+0.5)^2 + (2-4)^2}$ = $\sqrt{5}$ units
Hence C is the right answer.

We have,

Since BC and AD are parallel, $\angle\ ABC+\angle BAC\ +\angle\ DAC\ =\ 180^{\circ\ }$  (Sameside interior angles of a transversal)

=> $60^{\circ\ }+60^{\circ\ }\ +\angle\ DAC\ =\ 180^{\circ\ }$
=> $\angle\ DAC\ =\ 60^{\circ\ }$
=> $\angle\ ADC\ =\ 90^{\circ\ }-60^{\circ\ }\$ = $30^{\circ\ }$
Assume AB = AC = BC = a.
In the right triangle,  tan $\tan\angle\ ADC\ =\ \frac{AC}{DC}$   => $\tan30^{\circ\ }=\ \frac{a}{DC}$
=> $DC\ =\ a\sqrt{3}$
Area of ABC = $\frac{\sqrt{3}}{4}a^2$
Area of ACD = $\frac{1}{2}\times\ a\times\ a\sqrt{3}$
Since the area of ADC is 2 times the area of ABC, the area of ADC will become 200 sq cm.
Hence the area of trapezium = 100+200 = 300 square cm

Let the amount paid in each instalment be Rs x
The rate of interest is 50% paid semiannually,
The interest is 25%
The amount at the end of the first six months = 7200*1.25 = Rs 9000
(9000-x)1.25 = x
x = 5000

Assuming the cost price of 6 apples and marked price of 5 mangoes be 30x.
=> Cost price of 1 apple = 5x and marked price of 1 mango = 6x
Assuming the cost price of 5 mangoes and the marked price of 2 apples is 10y.
=> Cost price of 1 mango = 2y and marked price of 1 apple = 5y
The selling price of 6 apples and 2 free mangoes= 6*5y + 2*0= 30y
The cost price of 6 apples and 2 mangoes = 5*6x = 30x+2*2y = 30x+4y
Now, cost price$\left(1+\frac{\Pr ofit}{100}\right)$ = Selling price
=> (30x+4y)(1+$\left(1+\frac{87.5}{100}\right)$) = 30y
=> (30x+4y)(1+$\left(1+\frac{7}{8}\right)$) = 30y
=> 30x+4y = 16y
=> $\frac{x}{y}=\frac{2}{5}$
Hence, cost price of 1 mango = 2y, marked price of 1 mango = 6x = 6*$\frac{2y}{5}$ = 12y/5
=> Profit % = $\frac{\left(\frac{12y}{5}-2y\right)}{2y}\times\ 100\ =\ 20\%$

We can see that there are 12 positions in the queue. .

Let us first give 3 positions to Amar’s family, this can be done in $^{12}C_{3}$ ways.

No father wants to stand ahead of her son(given). Amar, his father and grandfather can be arranged in only one possible way(i.e. Amar will stand ahead of his father, and Amar’s father will stand ahead of his grandfather. However, there can be people standing between them).

Now, let us give 3 positions to Akbar’s family and this can be done in $^{9}C_{3}$ ways.

For Anthony, it can be done in $^{6}C_{3}$ ways. Only 3 positions are left for Arya to occupy.

Therefore, there are $^{12}C_{3}$*$^{9}C_{3}$*$^{6}C_{3}$ ways in which they can stand.

Since f(1)=f(2)=f(3)=2, then the function can be assumed as f(x) = k(x-1)(x-2)(x-3)+2

Now, f(4) = k(3)(2)(1)+2 = 14    => 6k = 12  => k = 2

f(5) = 2(5-1)(5-2)(5-3)+2 = 2*4*3*2+2 = 50

The area of the region contained by the lines $\mid x\mid-y\leq1,y\geq0$ and $y\leq1$ is the white region.

Total area = Area of rectangle + 2 * Area of triangle = $2+\left(\frac{1}{2}\times\ 2\times\ 1\right)\ =3$

Hence, 3 is the correct answer.

Let x(1) be the least number and x(10) be the largest number. Now from the condition given in the question , we can say that

x(2)+x(3)+x(4)+……..x(10)= 47*9=423……………….(1)

Similarly x(1)+x(2)+x(3)+x(4)…………….+x(9)= 42*9=378……………(2)

Subtracting both the equations we get x(10)-x(1)=45

Now, the sum of the 10 observations from equation (1) is 423+x(1)

Now the minimum value of x(10) will be 47 and the minimum value of x(1) will be 2 . Hence minimum average 425/10=42.5

Maximum value of x(1) is 42. Hence maximum average will be 465/10=46.5

Hence difference in average will be 46.5-42.5=4 which is the correct answer

The prisoner gets a 2 hour head start. Hence, he would have travelled 40 km by then. The relative speed of the police and the prisoner is 20 km/hr. Hence, the police will catch the prisoner in 2 hrs. The dog keeps running for this entire duration. Hence, the distance run by the dog = 50*2 = 100 km.
Now when the dog is running away from the police, his relative speed is 10 km/hr. When he is running towards the police, the relative speed is 90 km/hr
Hence, the distance covered by him towards the police and away from the police will also be in the same ratio. i.e 1:9
Hence, the distance run by dog towards the police = 100/10 = 10 km

Alternatively: Consider this case:

P and Q are travelling with speeds s1 and s2.

P travels Distance d in time $\frac{D}{s1}.$

In the same time Q travels$\frac{Ds2}{s1}$ The remaining distance $D-\frac{Ds2}{s1}$ is covered by both P and Q in time $\frac{D-\frac{Ds2}{s1}}{s1+s2}$. Hence, the distance P covers towards Q would be $\frac{D-\frac{Ds2}{s1}}{s1+s2}*s1$

The ratio of distance covered would be $\frac{D}{\frac{D-\frac{Ds2}{s1}}{s1+s2}*s1}$.

This becomes (s1+s2)/(s1-s2).

We can see that the ratio does not depend on Distance at all. Hence, whatever be the distance when they travel at 50kmph and 40 kmph respectively, they will travel distances in ratio 9:1.

We have, $\ \dfrac{\ \left(x-2\right)\left(x-4\right)^2\left(x-6\right)^3………\left(x-20\right)^{10}}{\ \left(x+2\right)\left(x+4\right)^2\left(x+6\right)^3………\left(x+20\right)^{10}}<0$
Here x $\ne\$ -2,-4,-6,-8,-10,-12,-14,-16,-18,-20 as the denominator cannot be zero. …..(1)
Also, the value of inequality cannot be 0. Hence $x\ \ne\ 2,4,6,8,10,12,14,16,18,20$ …..(2)
Since, even powers will always be positive, hence these would not have any effect on the sign of inequality. So they can be ignored.
=> $\ \dfrac{\ \left(x-2\right)\left(x-6\right)^3\left(x-10\right)^5……..\left(x-18\right)^9}{\ \ \left(x+2\right)\left(x+6\right)^3\left(x+10\right)^5……..\left(x+18\right)^9}$<0
Also, the sign of $\left(x-6\right)^3\$ will be the same as x-6 because $\left(x-6\right)^3\ =\ \left(x-6\right)^2\left(x-6\right)$ and the even power will not have any effect on the sign and hence can be ignored. Similarly, other values can be replaced too.
=> $\ \dfrac{\ \left(x-2\right)\left(x-6\right)\left(x-10\right)…….\left(x-18\right)}{\left(x+2\right)\left(x+6\right)\left(x+10\right)……..\left(x+18\right)}$<0
=> x $\in\left\{\left(-18,-14\right)\ U\ \left(-10,-6\right)\ U\ \left(-2,2\right)\ U\ \left(6,10\right)\ U\ \left(14,18\right)\right\}$ – {-8,-16,8,16}  (From condtion 1 and 2)
Each set contains 3 integers. Hence total number of integers = 3*5-4 = 15-4=11
11 integers satisfying the inequality are -17,-15,-9,-7,-1,0,1,7,9,15,17

In the given figure, OE and OF are the radii of circle with centre O.
Also, $\angle AEO$ & $\angle BFO$ are right angles.
Let, $CO=x$ units.
Then $DO=x+3$ units

$\because$ $OE=OF$
On applying Pythagoras theorem to $\triangle AEO$ and $\triangle BFO$ we have
$AO^2-AE^2=BO^2-BF^2$
=> $(2x)^2-(4\sqrt{2})^2$ = $(2x+6)^2-(2\sqrt{35})^2$
=> $x=3$ units

$\therefore CO=3$ units & $DO=6$ units
$OE=\sqrt{AO^2-AE^2}=\sqrt{6^2-(4\sqrt{2})^2}=2$ units.

The radii 2, 3 and 6 units form an HP.

Hence, option A is correct.

We have been given that the strength of the class is 90. From 1, we know that 9 students opted for History and Economics. 20 students opted for economics and sociology. From this information we can, infer that there were at least 9 students who have opted for history. Now, the average marks in history is 60 and the average marks in Sociology is 65 and the combined average is 50. Thus, if ‘x’ students opted for history and y for sociology then
45x + 65y = 50(x + y)
=> 5x = 15y
=> x/y = 3
Thus, the number of students opting for history is three times those who opted for sociology. We have been given that 30 students opted for both sociology and economics. Hence, there are at least 20 students who have opted for sociology. Thus, the number of students opting for history at least be 60 (since we calculated that the ratio is 1:3). Hence, in order to maximize the number of people who opted for only economics, we must minimize the number of people who have opted for other subjects. None of the students have opted for both history and sociology (given). Hence, there are at least 60 + 20 – 80 students who have not opted for ‘only economics’. Since the strength of the class is 90, so at most 10 students can opt for only economics.

The sum of all two-digit numbers that give a remainder of A when they are divided by 7 is 654. This must be an AP of difference 7.

There are 90, two-digit numbers from 10 to 99. 90/7=12.85.

Thus there must be 12 or 13 terms.

Sum of n terms $\frac{n}{2}\left(2a+\left(n-1\right)d\right)$=654 Since, 13 is not a factor of 654

Thus n must be 12.

The first term = 14+A, Last term=  91+A

Sum of n terms=$\frac{n}{2}\left(2\left(14+A\right)+\left(n-1\right)7\right)$

$6\left(2\left(14+A\right)+77\right)$=654

A=2

Let the first term of the GP be “a” . Now from the question we can show that

$ar^{m-1}=\frac{3}{4}$    $ar^{n-1}=12$

Dividing both the equations we get $r^{m-1-n+1}=\frac{1}{16}\ or\ r^{m-n}=16^{-1\ }or\ r^{n-m}=16$

So for the minimum possible value we take Now give minimum possible value to “r” i.e -4 and n-m=2

Hence minimum possible value of r+n-m=-4+2=-2

Let ‘r’ be the root of the function. It follows that f(r) = 0. We can represent this as $f\left(r\right)=f\left\{5-\left(5-r\right)\right\}$

Based on the relation: $f\left(5-x\right)=f\left(5+x\right)$; $f\left(r\right)=f\left\{5-\left(5-r\right)\right\}=f\left\{5+\left(5-r\right)\right\}$

$\therefore\ f\left(r\right)=f\left(10-r\right)$

Thus, every root ‘r’ is associated with another root ‘(10-r)’ [these form a pair]. For even distinct roots, in this case four, let us assume the roots to be as follows: $r_1,\ \left(10-r_1\right),\ r_2,\ \left(10-r_2\right)$

The sum of these roots = $r_1\ +\left(10-r_1\right)+\ r_2+\ \left(10-r_2\right)\ =\ 20$

Hence, Option D is the correct answer.

Let us denote the 4 digit number by abcd.
We are given that a+b+c+d = 32.
a,b,c,d in any order can be (8,8,8,8), (7,8,8,9), (7,7,9,9), (8,6,9,9), or (9,5,9,9).
From these combinations, we will get 35 numbers whose sum is 32.
For the number to be divisible by 11, the sum of the unit’s and hundred’s digit should be equal to the sum of the ten’s and the thousand’s digit.(difference of 11 or 22 is not possible)
Thus, the sum of the unit’s and hundred’s digit = sum of the ten’s place and the thousand’s place digit = 16
79, 88 are the 2 number which will yield the sum of 16.
The possible cases are:- 7997, 9779, 9977, 7799 8789, 8987, 9878, 7898 and 8888.
Hence,the required probability = $\dfrac{9}{35}$
Hence, option A is the correct answer.

We have 2x + 5y = 99 or $x=\frac{\left(99-5y\right)}{2}$

Now $x\ge\ y\ \ge\ -20$ ; So $\frac{\left(99-5y\right)}{2}\ge\ y\ ;\ 99\ge7y\ or\ y\le\ \approx\ 14$

So $-20\le y\le14$. Now for this range of “y”, we have to find all the integral values of “x”. As the coefficient of “x” is 2,

then (99 – 5y) must be even, which will happen when “y” is odd. However, there are only 17 odd values of “y” be -20 and 14.

Hence the number of possible values is 17.

It has been given that the number leaves a remainder of ‘k’ when divided by 6 and 3k when divided by 12.
Let the number be 12x + 3k.
Now, 12x will always be divisible by 6. Therefore, out of 3k, 2k should be divisible by 6 for the number to leave a remainder of ‘k’ when divided by 6.
=> 2k should be equal to 6.
=> k = 3.
Now, we know that the number is of the form 12x+9.
x can vary from 0 to 7.
Therefore, x (and ‘n’) can take a total of 8 values.

$(X^{2}-7x+11)^{(X^{2}-13x+42)}=1$

if $(X^{2}-13x+42)$=0 or $(X^{2}-7x+11)$=1 or $(X^{2}-7x+11)$=-1 and $(X^{2}-13x+42)$ is even number

For X=6,7 the value $(X^{2}-13x+42)$=0

$(X^{2}-7x+11)$=1 for X=5,2.

$(X^{2}-7x+11)$=-1 for X=3,4 and for X=3 or 4, $(X^{2}-13x+42)$ is even number.

.’. {2,3,4,5,6,7} is the solution set of X.

.’. X can take six values.

Here there are two cases possible

Case 1: When 7 is at the left extreme

In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9)

So total ways 3(8)(7)= 168

Case 2: When 7 is not at the extremes

Here there are 3 cases possible. And the remaining two places can be filled in 7(7) ways.(Remember 0 can’t come on the extreme left)

Hence in total 3(7)(7)=147 ways

Total ways 168+147=315 ways

$x^{2}-2\mid x\mid+\mid a-2\mid=0$

where x>= 0 and x>=2

$x^2-2x+a-2\ =0$ Using quadratic equation we have $x=\ 1+\sqrt{\ 3-a}\ and\ x=1-\sqrt{\ 3-a}$ Only two integer values are possible

a=2 and a=3. So corresponding “x” values are x=1 and a=3, x=2 and a=2, x=0 and a=2

where x>=0 and x<2

Applying the above process we get x=1 and a=1

where x<0 and x>=2 we get a=3 and x=-1 , a=2 and x=-2

where x<0 and x<2 we get a=1 and x=-1

Hence there are total 7 values possible

Number of consecutive zeros after the decimal = [log n]+1
[log 40^{-34}] = [-34*(1+2log2)] = [-54.47] = -55
Number of consecutive zeros = -55+1 = -54
Hence there are 54 zeros after the decimal.

The sum of the first 20 terms = $\frac{20}{2}\left[2\times\ 6+\left(20-1\right)\times\ 2\right]220​[2× 6+(20−1)× 2]$  = 500
After the sum increases by 56%, the new sum increase by 500*56/100 = 280
The common difference increases by = 12-2 = 10
Here the remaining terms = 20-x
Until xth term both correct and wrong series will be the same.

Assuming xth term = a
Correct series: (x+1)th term, (x+2)th term………20th term = a+2, a+4, a+6………
Wrong series: (x+1)th term, (x+2)th term………20th term = a+12,a+24,a+36…….
Clearly, the difference is 10, 20, 30………..(20-x)*10

Hence, the increase = 10[1+2+3+…..20-x] = 10$\frac{\left(20-x\right)\left(21-x\right)}{2}$ = 280
=> (x-21)(x-20) = 56
=> $x^2-41x+364\ =\ 0$
=> x= 13, x=28
Rejecting the value greater than 20, we get x=13

Let Ankit sold the bike to Anil at Rupees x.
Then Cost price for Anil will be x + 5000
=> Hence, Anil sold it for 1.25(x + 5000) = 1.25x + 6250
=> Now Sushil further spent 5000. Hence the cost price for Sushil = 1.25x + 11250
We have been given that
x + 20000 – 1.25x – 11250 = $\frac{1}{6}*(1.25x + 11250)$
=> $-\frac{1}{4} x + 8750$ = $\frac{5x}{24}+ 1875$
=> $\frac{11x}{24} = 6875$
=> x = 15000.
Anil spent Rs. 5000 on repairs. And then, he sold it at a profit of 25%. => Anil sold it at 1.25*20000 = 25,000.

We know that $8a^3-12a^2-6a-1=0$
Therefore, $-8a^3 + 12a^2 + 6a + 1 = 0$
Or, $8a^3 + 12a^2+ 6a + 1 = 16a^3$
Therefore, $(2a+1)^3 = 16a^3$
Or, $(2+1/a)^3 = 16$
Therefore, $1/a = \sqrt[3]{16}-2$
Hence, $a = \frac{1}{\sqrt[3]{16}-2} = \frac{\sqrt[3]{256}+\sqrt[3]{128}+4}{16-8}$
Therefore, $a = \frac{\sqrt[3]{4}+\sqrt[3]{2}+1}{2}$
Hence, $p=4,q=2,r=2$ and the required sum equals 8

$F_n$ = {$2n+6, 3n+5$}

$F_1$= {8,8}

$F_2$= {10,11}

$F_3$= {12,14}

$F_4$= {14,17}

$F_5$= {16,20}

$F_6$= {18,23}

$F_7$= {20,26}

.

Values of n for which $2n+6$ is divisible of 7 is 4, 11, 18, …….

Similarly, the values of n for which $3n+5$ is divisible by 7 is 3, 10, 17, ………..

.

The values of n for which both $F_n$ and $F_{n+1}$ will have exactly one element which is divisible by 7 is 3, 10, 17……….. because in that case, ${n+1}$ will be 4, 11, 18,……

Let n be the number of terms
3 + (n-1)7 $\leq$ 500
(n-1)7 $\leq$ 497
(n-1) $\leq$ 71
n $\leq$ 72

So roll number 1 takes 1 chocolate, 2 takes 2 and so on. So total chocolates distributed are $\frac{n(n+1)}{2}$ which should be a factor of 100, so that no chocolates remain after the distribution.
If n=1, $\frac{n(n+1)}{2}=1$ which is a factor of 100. So, n=1 is possible.
n=2, $\frac{n(n+1)}{2}=3$ which is not possible.
The next and the last possible case is n=4, where $\frac{n(n+1)}{2}=10$. So n can assume only 2 values.

$6561^2$ can be written as $3^{16}$.

Let a, b and c be the three factors => abc = $3^{16}$

If x, y and z are the powers of 3 in a, b and c respectively, then x + y + z = 16 => $^{18}C_2$ solutions => 153 solutions.

But, we have to find the unordered solutions => remove cases (k,k,k) and (k,k,m), divide the resulting number by 6 and then add the removed cases again.

Case (k, k, k): As 16 is not divisible by 3, there will be no (k,k,k) cases.

Case (k, k, m): Here 2k + m = 16. k can go from 0 to 8 and correspondingly m will go from 16 to 0. Hence, there will be a total of 9 cases.

So, there are 0 (k,k,k) cases and 9 (k,k,m) cases => 9*3 = 27 solutions have to be removed => 153 – 27 = 126 cases are of the form (k,m,n).

126/6 = 21 => 21 + 9 = 30 unordered solutions.

Note: In the question we have not been asked to find the number of ordered solutions. Every solution of the form (k,m,n) can be arranged in 3! = 6 ways. But only one of them should be counted when counting the number of unordered solutions. So they have been divided by 6. So 126/6 = 21. Also every solution of the form (k,k,m) can be arranged in 3 ways. So they have been divided by 3 to get 27/3 = 9. Thus, the total number of solutions = 21+9 = 30.

The milkman buys milk at 100 Rs per liter.
He sells 20L of pure milk at 80 Rs per liter.
Thus, he faces of loss of 400 Rs I.e. -400.
After adding water he has 80L of milk and 20L of water out of which he sells 20L.
In this 20L there will be 16L milk and 4L water
Thus, CP of 16L of milk = 1600
SP of 20L of mixture = 20*85 = 1700
Thus, he makes a profit of 100 Rs here.
Overall loss/profit till now = -400+100 = -300 Rs
He now has 64L of milk and 36L of water.
He sells 25L (at 90 Rs per liter) of this mixture I.e. 16L of milk and 9L of water.
Thus, CP of 16L of milk = 1600 Rs
SP of 25L of mixture = 25*90= 2250 Rs
Thus, he makes a profit of 2250-1600 = 650 Rs here.
Overall loss/profit till now = -300+650 = 350 Rs
He now has 48L of milk and 52L of water.
He sells 25L (at 95 Rs per liter) of this mixture I.e. 12L of milk and 13L of water.
Thus, CP of 12L of milk = 1200 Rs
SP of 25L of mixture = 25*95= 2375 Rs
Thus, he makes a profit of 2375 -1200 = 1175 Rs here.
Thus, Overall Profit/Loss in these transactions = 1175+350 = 1525 Rs.

It is given that, both the root of the quadratic equation, $ax^2+(a-4)x+a+1=0$ are greater than 0. Therefore, the sum of roots should be greater than 0.

$\Rightarrow$ $-\dfrac{B}{A}$ > 0
$\Rightarrow$ $-\dfrac{(a-4)}{a}$ > 0
$\Rightarrow$ $\dfrac{a-4}{a}$ < 0
$\Rightarrow$ $0 < a < 4$   … (1)

It is given that the quadratic equation has both the roots greater than 0. Therefore D $\geq$ 0
$\Rightarrow$ $(a-4)^2-4*(a)*(a+1) \geq 0$

$\Rightarrow$ $a^2-8a+16-4a^2-4a \geq 0$

$\Rightarrow$ $-3a^2-12a+16 \geq 0$

$\Rightarrow$ $3a^2+12a-16 \leq 0$

$\Rightarrow$ $\dfrac{-6-2\sqrt{21}}{3} \leq a \leq \dfrac{-6+2\sqrt{21}}{3}$

$\Rightarrow$ $-5.05 \leq a \leq 1.05$  … (2)

From equation (1) and (2), we can see that ‘a’ can be equal to 1. Let’s figure out the value of both the roots explicitly at a = 1.
At a = 1, we can see that the equation reduces to $x^2-3x+2=0$. Therefore, x = 1 or 2. We can say that at a = 1, both the roots of the given equation are greater than 0.

Hence, we can say that there is only one integer value (a=1) integer value of ‘a’ for which both the root of the quadratic equation $ax^2+(a-4)x+a+1=0$ are greater than 0.

If we select three random boxes, then there will be 7 remaining boxes.

a+b+c+d=7

We can select three boxes such that no two boxes are next to each other.

b,c cannot be 0

b=1+b’

c=1+c’

a+1+b’+1+c’+d= 7

a+b’+c’+d=5

Number of non negative integral solution.

= $^{\left(5+4-1\right)}C_{\left(4-1\right)}$

= $^8C_3$

=56 cases

All the internal angles of the hexagon are equal.
Let’s calculate the measure of each internal angle.
Sum of the internal angles of a polygon of n sides = (n-2)*180
Each internal angle =$\frac{(n-2)*180}{n}$
=$120^0$
So the Exterior angle of angle P is $60^0$. APQ forms an equilateral triangle of side 1 cm.
Similarly, RBX forms an equilateral triangle of side 2 cm.
Length of AB =7 cm
ABC is an equilateral triangle of side 7 cm.
Length of CY = length of CB- length of YB
=7-4=3 cm
Length of PZ = Length of AC – (length of AP+ length of CZ)
7-(1+3)
=3 cm
Hence the length of ZY is 3 cm and PZ is 3 cm

$5x + 12 y = 169$ and $x = 0$ are straight lines and $x^2 + y^2$ = 169 is a circle with centre (0, 0)
The perpendicular distance from the centre (0, 0) to the line $5x + 12 y = 169$  is $|\frac{-169}{\sqrt{5^2+12^2}}|$ = 13 i.e the straight line is tangent to the circle
Substitute $x = \frac{169-12y}{5}$ in  $x^2 + y^2$ = 169
$[\frac{169-12y}{5}]^2 + y^2$ = 169
$169^2 + 144y^2 -2*12y*169 + 25y^2 = 169*25$
$169^2 + 169y^2 -2*12y*169 = 169*25$
Take 169  common on both sides and cancel them
$169 + y^2 -2*12y = 25$
$y^2 -24y + 144$ = 0
$(y-12)^2$ = 0
y = 12
On substituting the value of y in $5x + 12 y = 169$, we get x = 5
One intersecting point is (5, 12)
The lines x = 0 and $5x + 12 y = 169$ meet at (0, 169/12) –(Second intersecting point)
The line x = 0 and $x^2 + y^2$ = 169 meet at (0, 13) and (0, -13)

Hence they meet at 4 distinct points

$\dfrac{3}{x} + \dfrac{2}{y} = \dfrac{1}{12}$

$\Rightarrow \dfrac{2x+3y}{xy} = \dfrac{1}{12}$

$\Rightarrow 24x+36y = xy$

$\Rightarrow 24x-xy+36y = 0$

This can be expressed as

$\Rightarrow 24x-xy+36y+(36 \times 24)-(36 \times 24) = 0$

$\Rightarrow x(24-y)-36(24-y)+(36 \times 24)=0$

$\Rightarrow (x-36)(24-y)= -864$

For the integral values of $y$, $(24-y)$ must be an integer as well.

This means that $24-y$ will be one of the factors of 864.

We know that $864 = 2^5 \times 3^3$

Hence, Number of factors of 864 = (5+1) x (3+1) = 24

However, the factor can be both positive as well as negative. There are thus 24 positive and 24 negative factors.

Consider the case for example:

2 is a factor of 864. For 24-y to be equal to 2, y = 22

-2 is also a factor of 864. For 24-y to be equal to -2, y = 26

In general, say a and -a are factors of 864.

So 24-y1=a and 24-y2=-a

So, y1 = 24-a and y2 = 24+a

The sum of these two factors is y1+y2 = 24-a+24+a = 48

There are 24 such pairs and each of these add up to 48.

Thus, the sum of values of $y$ = $24 \times 48 = 1152$

Initially, the amount of Milk and Water in the container is 100, 20l respectively. They are in the ratio 5:1.

Now, this 120l is transferred to containers A, B in the ratio 3:2 => Amount of mixture in A $\frac{3}{5}\cdot120$ =72 and B = $\frac{2}{5}\cdot120$=48.

The ratio of Milk and water would still be in the ratio 5:1 => Amount of Milk and Water in A = 60l, 12 l and in B = 40l, 8l.

Now, the 10l of water added in A and B in ratio 2:3 => Total amount of water in A= 12+4=16 l and in B = 8+6=14l

The amount of (Milk, Water) in A, B = (60,16) & ( 40,14)

If, 10 l of the mixture is taken from A then the quantity of (Milk, Water) removed will be and then added to B is ($\left(\frac{60}{76}\cdot10,\frac{16}{76}\cdot10\right)\ =\ \left(\frac{150}{19},\ \frac{40}{19}\right)$

Similarly, the amount of (Milk, Water) emoved from B and then added to A is ($\left(\frac{200}{27},\frac{70}{27}\right)$

.’. The amount of Milk left in A = 60-150/19 = 990/19

& The amount of Milk left in B = 40-200/27 = 880/27

The ratio of milk in two containers = 243:152= m:n

Required sum = m +n = 395.

Three cases are possible.
Case A – All the three boxes contain an equal number of bowls, i.e. (3,3,3).
Thus, in one way we can distribute an equal number of bowls in 3 baskets.
Case B – Two boxes contain the same number of bowls.
Possibilities are (0,0,9), (1,1,7), (2,2,5), (4,4,1).
Thus there are 4 possibilities for this case.
Case C – Three boxes contain a distinct number of bowls.
Possibilities are
(0,1,8), (0,2,7), (0,3,6), (0,4,5);
(1,2,6), (1,3,5);
(2,3,4).
Thus there are 7 possibilities for this case.
In all, there are 1 + 4 + 7 = 12 ways in which 9 identical bowls be kept inside 3 identical boxes such that the boxes can hold any number of bowls.