In a battle, the commander-in-chief arranges his soldiers in a formation of three concentric circles. The radii of the circles are in an arithmetic progression: the smallest radius is 70m (meters) and the largest is 140m.
If each soldier is to be separated from the adjacent soldiers standing on the same circle by 1m, how many soldiers are required to complete the formation? (Consider π = 22/7.)
XAT 2026 Question Paper - QADI
For the following questions answer them individually
Solution
Three concentric circles C1, C2 and C3 with the radius OA, OB and OC are drawn.
OA, OB and OC are in AP (given).
OA = 70m, OB = (70 + d)m, OC = (70 + 2d)m , where, d is the common difference.
OC = 70+2d = 140 (given)
d = 35m
OB = 70 + 35 = 105m
The soldiers are standing on the circumference of these three circles at a distance of 1m. Hence, to find the total number of soldiers, we have the find the length of the circumference of all three circles.
Total circumference = $$2\pi\ \left(r1+r2+r3\right)$$
= 2 $$\times\ $$ $$\ \frac{\ 22}{7}\times\ \left(70+105+140\right)$$ = 1980
Hence, the total number of soldiers are 1980.
$$\therefore\ $$ The required answer is D.
In a business school, only three specializations are offered: marketing, finance and operations. A student can opt for either one, or two, or no specializations. In the current batch of 120 students, 60 are specializing in marketing, 50 are specializing in finance and 30 are specializing in operations. During the placement process, all students specializing either in marketing or in finance are shortlisted for consulting job interviews.
If 45 students are not shortlisted for consulting job interviews, what is the MINIMUM possible number of students specializing in both marketing and finance?
Solution
Share
In the above figure, let the number of students specializing in :
Only marketing = a, Only finance = b, Only operations = c
Both marketing and finance but not operations = d, Both marketing and operations but not finance = e, Both finance and operations but not marketing = f
All three subjects = g, None of the subjects = h
Now it is given that only students having an specialization in marketing or finance gets a consulting shortlist. So, we can say that students doing specialization only in operations and students not doing specialization in any of the 3 subjects will not get a shortlist.
Hence, c + h = 45 (given) $$\longrightarrow\ i$$
Number of students doing specialization in marketing = a + d + e + g = 60 (given) $$\longrightarrow\ ii$$
Number of students doing specialization in finance = b + d + g + f = 50 (given) $$\longrightarrow\ iii$$
Number of students doing specialization in operations = c + e + g + f = 30 (given) $$\longrightarrow\ iv$$
Number of students who received the consulting shortlist = a + b + d + e + f + g = 75 $$\longrightarrow\ v$$
By using equation iv and v, we get,
a + b + d + 30 - c = 75
a + b + d = 45 + c $$\longrightarrow\ vi$$
Add equation ii and iii,
a + b + d + (d + g) + e + f + g = 110 $$\longrightarrow\ vii$$
Now, using equation iv, vi and vii, we get,
45 + c + (d + g) + 30 - c = 110
d + g = 35
Hence, the number of students specializing in marketing and finance = 35
$$\therefore\ $$ The required answer is A.
An archer, from the point A, is aiming at a target, placed on the top of a 56 feet tall tree. The base of the tree is at the point C. From the archer’s position, the angle of elevation to the target is 45° from his eye level. The archer, facing the tree, moves backwards on the straight line joining the points A and C, to a new position at the point B. From the point B, the angle of elevation from his eye level to the target becomes 30°.
How far did the archer move from A to B (in feet) if his eye level is at a height of 6 feet from the ground?
Solution
Let the tree be PC and the archer be RA.
Given : Height of the tree = 56ft, height of the archer up to the eye level = 6ft, angle of elevation at point A = 45$$^{\circ\ }$$, angle of elevation at point B = 30$$^{\circ\ }$$
To find : Length of AB
In $$\triangle\ $$PQR, $$\tan\ 45^{\circ\ }=\ \frac{\ PQ}{QR}=\ \frac{\ 50}{QR}$$
QR = 50ft (as $$\tan\ 45^{\circ\ }=\ 1$$)
Now, in $$\triangle\ PQS$$,
$$\tan\ 30^{\circ\ }=\ \frac{\ PQ}{QS}=\ \frac{\ PQ}{QR+RS}=\ \frac{\ 50}{50+RS}$$
$$\ \frac{\ 1}{\sqrt{3\ }}=\frac{\ 50}{50+RS}$$
RS = $$50\left(\sqrt{3\ }-1\right)$$ ft
Since RS = AB, AB = $$50\left(\sqrt{3\ }-1\right)$$ ft
Hence, the length of AB is $$50\left(\sqrt{3\ }-1\right)$$ feet
$$\therefore\ $$ The required answer is C.
Two buses travel between Jamshedpur and Kolkata in the opposite directions, on the same road. On that road, the maximum allowed speeds are different (but constant) for the opposite directions. Usually, both buses travel at the respective maximum allowed speeds to their respective destinations: the bus from Jamshedpur to Kolkata takes 4 hours, while the bus from Kolkata to Jamshedpur takes 3 hours.
One day, the two buses start at the same time. However, one hour after starting, the bus from Jamshedpur to Kolkata reduces its speed to half of its maximum allowed speed due to congestion on the road.
If both buses do not stop anywhere in between, how many hours after starting do they meet?
Solution
Let the maximum allowed speed from Jamshedpur to Kolkata be s1 kmph and the maximum allowed speed from Kolkata to Jamshedpur be s2 kmph. Since the bus is moving at the maximum allowed speed then we can say that the speed of the bus moving from Jamshedpur to Kolkata (Bus 1) is s1 kmph and the speed of the bus moving from Kolkata to Jamshedpur (Bus 2) is s2 kmph.
Time taken by Bus 1 to travel from Jamshedpur to Kolkata = 4 hrs
Distance travelled by Bus 1 = 4 $$\times\ $$s1
Time taken by Bus 2 to travel from Kolkata to Jamshedpur = 3 hrs
Distance travelled by Bus 1 = 3 $$\times\ $$s2
Since both the bus is travelling on the same road, the distance travelled by both the buses will be same.
4 $$\times\ $$s1 = 3 $$\times\ $$s2
$$\ \frac{\ s1}{s2}=\ \frac{\ 3}{4}$$
The ratio of speed of Bus 1 to the speed of Bus 2 is 3:4.
Hence, let the speed of Bus 1 be 3x kmph and the speed of Bus 2 be 4x kmph.
Total distance travelled by the buses = $$4\times\ 3x\ =3\times\ 4x=12x$$
Now, both the buses started at the same time and travelled at the maximum allowed speed for 1 hr.
Distance travelled by Bus 1= JA = $$3x\times\ 1=3x$$
Distance travelled by Bus 2= KB = $$4x\times\ 1=4x$$
Distance left to be travelled= AB = $$12x-\left(3x+4x\right)=5x$$
Now, the speed of Bus 1 is reduced to half due to congestion.
New speed of Bus 1 = $$\ \frac{\ 3x}{2}$$
After the reduction of speed, let the two buses meet at point C and the time taken by both the buses to reach C is the same.
Let the distance travelled by Bus 1 from A to C be y km.
BC = 5x-y km
Hence, $$\ \frac{\ y}{\ \frac{\ 3x}{2}}=\ \frac{\ 5x-y}{4x}$$
y = $$\ \frac{\ 15x}{11}$$
Time required = $$\ \frac{\ \ \frac{\ 15x}{11}}{\ \frac{\ 3x}{2}}=\ \frac{\ 10}{11}$$ hrs
Total time required for both the buses to meet = $$\frac{\ 10}{11}+1=\ \frac{\ 21}{11}$$ hrs
Hence, both the buses will meet after $$\frac{\ 21}{11}$$ hrs from starting.
$$\therefore\ $$ The required answer is D.
Read the following scenario and answer the THREE questions that follow.
Brijbhushan, a microfinancier, lends money at the rate of Rs.10 per square meter to small farmers at a village. He charges an annual interest rate of 10%. All the farming plots in that village are rectangular, with areas varying between a minimum of 1000 square meters and a maximum of 10,000 square meters.
This year, Brijbhushan has lent money only to five farmers: Aditya, Binod, Chhuttan, Dabloo and Govind. The perimeter of Chhuttan’s plot is 250 meters, with the length and width being at a ratio of 4:1. Aditya’s plot has an area three times the area of Govind’s plot. The area of Aditya’s plot is also the average of the areas of Govind’s plot and Dabloo’s plot. The plots belonging to Aditya, Binod and Dabloo are of the same width, but of different lengths. Moreover, the length of Binod’s plot is the sum of the lengths of Aditya’s plot and Dabloo’s plot.
What is the interest owed by Chhuttan to Brijbhushan at the end of the year?
Solution
For Chhuttan,
Perimeter of his land = 2 (l + b) = 250m
l + b = 125m $$\longrightarrow\ i$$
$$\ \frac{\ l}{b}=\ \frac{\ 4}{1}$$ (Given) $$\longrightarrow\ ii$$
From i and ii, we get,
4b + b = 125
b = 25m
l = 100m
Area of Chhuttan's land = 100 $$\times\ $$25 = 2500 $$m^2$$
Loan given to Chhuttan by Brijbhushan = 2500 $$\times10$$ = Rs 25000
Interest rate of the loan = 10%
Interest owed by Chhuttan at the end of one year = 10% $$\times\ $$Rs 25000 = Rs 2500
$$\therefore\ $$ The required answer is E.
What is the MAXIMUM possible value of the total loan Brijbhushan has given to these five farmers?
Solution
Let the name of farmers be A,B,C,D,G for simplicity.
Let the area of G's plot be x $$m^2$$.
Area of A's plot = 3x $$m^2$$
Let area of D's plot be y $$m^2$$.
The area of A's plot is the average of the area of D's and G's plot.
3x = $$\ \frac{\ x+y}{2}$$
y = 5x $$m^2$$
The width of the plot of A,B and D is the same and the lengths are different. Let the length of the plot of A,B and D be l1,l2 and l3 respectively and the width be y.
We got the following table.
From A's plot, $$l1\times\ y=3x$$
$$l1=\ \frac{\ 3x}{y}$$
From D's plot, $$l3\times\ y=5x$$
$$l3=\ \frac{\ 5x}{y}$$
The length of B's plot is the sum of the length of A's plot and D's plot.
l2 = l1 + l3 = $$\frac{\ 8x}{y}$$
Hence, the area of B's plot = $$l2\times\ y=8x$$
The required final table is :
Total area of all plots = 2500 + 17x
Total loan given by Brijbhushan = $$Rs\ \left(2500+17x\right)\times\ 10$$
Since, we have to maximize the loan amount, we have to maximize the value of x.
The largest area of a land = $$10000\ m^2$$ (Given)
Among the farmers, B will have the land with the largest area.
8x = 10000 $$m^2$$
x = 1250 $$m^2$$
Hence, maximum loan given by Brijbhushan = $$Rs\ \left(\left(1250\times\ 17\right)+2500\right)\times\ 10$$ = Rs 237500
$$\therefore\ $$ The required answer is C.
If the width of the Aditya’s plot is 25 meters, what is the MINIMUM possible length of Binod’s plot?
Solution
Let the name of farmers be A,B,C,D,G for simplicity.
Let the area of G's plot be x $$m^2$$.
Area of A's plot = 3x $$m^2$$
Let area of D's plot be y $$m^2$$.
The area of A's plot is the average of the area of D's and G's plot.
3x = $$\ \frac{\ x+y}{2}$$
y = 5x $$m^2$$
The width of the plot of A,B and D is the same and the lengths are different. Let the length of the plot of A,B and D be l1,l2 and l3 respectively and the width be y.
We got the following table.
From A's plot, $$l1\times\ y=3x$$
$$l1=\ \frac{\ 3x}{y}$$
From D's plot, $$l3\times\ y=5x$$
$$l3=\ \frac{\ 5x}{y}$$
The length of B's plot is the sum of the length of A's plot and D's plot.
l2 = l1 + l3 = $$\frac{\ 8x}{y}$$
Hence, the area of B's plot = $$l2\times\ y=8x$$
The required final table is :
The width of A's plot = 25m
y = 25m
For B's plot, $$l2\times\ 25=8x$$
To minimize the value of l2, we have to minimize the value of x.
Minimum possible area of land among the farmers = 1000 $$m^2$$
Smallest area of land = G's land = x $$m^2$$
x = 1000
Hence, minimum value of l2 = $$\ \frac{\ 8\times\ 1000}{25}=320\ m$$
$$\therefore\ $$ The required answer is B.
For the following questions answer them individually
Consider a seven-digit number 735x6y4, divisible by 44, where the two digits x and y are unknown.
Consider the following two additional pieces of information:
I. x and y are even numbers
II. x and y are equal
To determine the values of x and y UNIQUELY, which of the above pieces of information is/are MINIMALLY SUFFICIENT?
Solution
A seven-digit number 735x6y4 is divisible by 44. It means that the number is also divisible by 4 and 11.
Divisibility rule of 4 = Last two digits of the number is divisible by 4.
Divisibility rule of 11 = Sum of odd digits starting from left - Sum of even digits starting from left
For the given number to be divisible by 4, y4 should be divisible by 4.
So, the possible values of y = 0,2,4,6,8
For the number to be divisible by 11, (7+5+6+4)-(3+x+y) = 19-(x+y) should be divisible by 11.
Case 1 : Both x and y are even numbers.
If y = 0, x = 8
If y = 2, x = 6
If y = 4, x = 4
If y = 6, x = 2
If y = 8, x = 0
In this case, we got 5 different combinations of values.
Case 2 : Both x and y are equal.
If y = 4, x = 4
In this case, we got a unique solution.
Hence, we only need condition 2 to uniquely determine the values of x and y.
$$\therefore\ $$ The required answer is C.
The swimming pool in an exclusive club is a circular strip. A water flow is artificially maintained in the strip, along a clockwise direction, at a constant speed of 3km/h (kilometers/hour). One morning, Ayub and Rana start swimming at the same time from two diametrically opposite points on the strip. Ayub swims in a counterclockwise direction, while Rana swims in a clockwise direction. They swim at constant, but different, speeds. When they meet for the first time, Ayub has covered 60m (meters). They meet again after Rana has covered 180m from the first meeting point.
If Rana swims at the speed of 3 km/h in still water, how long does Ayub take to complete one round of the circular strip, if he swims in the clockwise direction?
Solution
Let us denote Ayub and Rana as A and R respectively for simplicity.
Let the speed of A and B in still water be Sa and Sr respectively and the speed of river be r.
r = 3 kmph = 50 $$\ \frac{\ m}{\min}$$ (clockwise)
Sr = 3 kmph = 50 $$\ \frac{\ m}{\min}$$ (clockwise)
Since A is swimming in counter clockwise direction and R is swimming in clockwise direction, let C be the point where they meet for the first time.
Distance travelled by A = AC = 60m
Speed of A = (Sa-50) $$\ \frac{\ m}{\min}$$ (as he is travelling in counter clockwise direction which is upstream)
Time taken by A to reach C = $$\ \frac{\ 60}{Sa-50}\min$$
Speed of R = (Sr+50) $$\ \frac{\ m}{\min}$$ (as he is travelling in clockwise direction which is downstream)
= 100 $$\ \frac{\ m}{\min}$$
Since, the time taken by R to reach C is the same as the time taken by A to reach C (as they started swimming at the same time), distance travelled by R = RC = $$\ \frac{\ 100\times\ 60}{Sa-50}m$$
Total distance travelled by both A and R = AC + RC = $$\ \frac{\ 6000}{Sa-50}+60\ m$$
As A and R travelled in opposite directions from exactly diametrical ends, the distance A to R will be half of the circumference of the circle.
$$\pi\ r$$ = $$\ \frac{\ 6000}{Sa-50}+60\ m$$ $$\longrightarrow\ i$$
Now, let's assume that they meet again at D after R travelled for 180m,
Time taken for them to meet again = $$\ \frac{\ 180}{100}\min$$
Distance travelled by A = CD (in counter clockwise direction) = $$\ \frac{\ \left(Sa-50\right)\times\ 180}{100}m$$
Distance travelled by R = CD (in clockwise direction) = 180m
Total distance travelled = $$\ \frac{\ \left(Sa-50\right)\times\ 180}{100}m$$ + 180 m
This distance will be equal to the circumference of the circle.
Hence, $$2\pi\ r$$ = $$\ \frac{\ \left(Sa-50\right)\times\ 180}{100}m$$ + 180 m $$\longrightarrow\ ii$$
Take the value of $$\pi\ r$$ from equation i and substitute it in equation ii.
$$\ \frac{\ 12000}{Sa-50}+120=\ \frac{\ 180\left(Sa-50\right)}{100}+180$$
Let the value of Sa-50 be t.
$$\ \frac{\ 12000}{t}+120=\ \frac{\ 180\left(t\right)}{100}+180$$
$$\ 3t^2+100t-20000=0$$
t = $$\ \frac{\ 200}{3}$$, -100 (Not possible as the speed can't be negative)
Sa-50 = $$\ \frac{\ 200}{3}$$
Sa = $$\ \frac{\ 350}{3}\ \frac{\ m}{\min}$$
Use this value in equation ii to get, $$2\pi\ r$$ = 300m
Speed of A when he is swimming in clockwise direction = $$\ \frac{\ 350}{3}+50$$
= $$\ \frac{\ 500}{3}\ \frac{\ m}{\min}$$
Time taken by A to complete one round of the circular strip = $$\ \frac{300}{\ \frac{\ 500}{3}\ }\min$$
= $$\ \frac{\ 9}{5}\min$$ = 1 min 48 sec
Hence, the time taken by A to complete one round of the circular strip is 1 min 48 sec.
$$\therefore\ $$ The answer is D.
125 is multiplied by either 10 or 15. The resultant number is again multiplied by either 10 or 15. This process continues.
Which of the following CANNOT be a resultant number at any point in time?
Solution
125 = $$5^3$$
10 = $$2\times\ 5$$
15 = $$3\times\ 5$$
In each step, we are multiplying 125 by either 10 or 15. So, the power of 5 and the power of either 2 or 3 will only increase by one in every step. So at any point of time, the difference between the power of 5 and the power of 2 and 3 combined will remain constant (which is 3).
Option A : Power of 5 = 691
Power of 2 = 235
Power of 3 = 453
Difference of powers = 691 - (235+453) = 3
Option B : Power of 5 = 1080
Power of 2 = 253
Power of 3 = 824
Difference of powers = 1080 - (235+824) = 3
Option C : Power of 5 = 1604
Power of 2 = 689
Power of 3 = 912
Difference of powers = 1604 - (689+912) = 3
Option D : Power of 5 = 1034
Power of 2 = 476
Power of 3 = 455
Difference of powers = 1034 - (476+455) = 103
Option E : Power of 5 = 1145
Power of 2 = 689
Power of 3 = 453
Difference of powers = 1145 - (689+453) = 3
Only in option D the difference of power is different than 3.
$$\therefore\ $$ The answer is D.
Three categories of candidates appear for an admission test: diligent (10%), lazy (30%) and confused (60%). A diligent candidate is 10 times more likely to clear the admission test compared to a lazy candidate.
If 40% of the candidates clearing the admission test are confused, what is the MAXIMUM possible value of the probability of a confused candidate clearing the test?
Solution
Let us assume that the total number of candidates who appeared for the admission test$$=100k$$
So, as per the data, the number of diligent candidates$$=10k$$
The number of lazy candidates$$=30k$$
and, the number of lazy candidates$$=60k$$
Now, let us assume that the probability of a lazy candidate to clear the test $$=p$$
$$\therefore$$ The probability of a lazy candidate to clear the test $$=10p$$
Let us also assume that the probability of a confused candidate to pass the exam $$=q$$
Now, we know that
Probability of a deligent candidate passing the test,$$10p=\dfrac{\text{No of deligent candidates who passed}}{\text{total deligent candidates who appeared}}$$
or, $$p=\dfrac{\text{No of deligent candidates who passed}}{10k}$$
or, number of the diligent candidates who passed the test $$=10k*10p=100kp$$
Similarly, the number of lazy candidates who passed the test $$=30k*p=30kp$$
and, the number of confused candidates who passed the test $$=60k*q=60kq$$
Now, we are told that 40% of the candidates who clear the admission test are confused.
$$\therefore \dfrac{60kq}{100kp+30kp+60kq}=\dfrac{40}{100}$$
On simplifying this equation, we get:
$$q=\dfrac{13p}{9}$$
Now, in order to maximise $$q$$, we need to maximise $$p$$.
Now we know that the probability of a diligent candidate passing the test, $$10p\leq1$$
or, $$p\leq\dfrac{1}{10}$$
So the maximum value of $$p=\dfrac{1}{10}$$
Now, calculating the maximum value of $$q$$.
Hence, the MAXIMUM possible value of the probability of a confused candidate clearing the test, $$q=\dfrac{13}{90}$$
Hence, option B is correct.
In a multiple-choice examination, there are 20 questions. Each correct answer is worth 4 marks, while 2 marks are to be deducted for every wrong answer. Further, 1 mark is to be deducted for every unattempted question. One student receives a total of 46 marks in the examination. However, before releasing the marks, the professor realizes that she has, by mistake, deducted 2 marks for every unattempted question and 1 mark for every wrong answer.
After correction, how many marks will the student get?
Solution
Let the number of correct and wrong questions be x and y respectively.
Number of unattempted questions = 20 - (x+y) $$\longrightarrow\ i$$
4 marks are awarded for every correct questions and 2 and 1 marks are deducted for every wrong and unattempted questions respectively.
Total original marks scored = 4x-2y-((20-(x+y)) = 5x-y-20 $$\longrightarrow\ ii$$
But the teacher deducted 1 and 2 marks for every wrong and unattempted questions respectively by mistake.
Total marks = 4x-y-2(20-x-y) = 6x+y-40
Total marks = 46 (given)
6x+y-40 = 46
6x+y = 86
The possible pairs of (x,y) are (14,2), (13,8), (12,14), ........ , (1,80), (0,86).
x+y = 16, 21, 26, ........ , 81,86
Only one case will be possible as (x+y) $$\le\ $$ 20 (as equation i $$\ge\ 0$$)
Hence, Number of correct questions = x = 14
Number of wrong questions = y = 2
Number of unattempted questions = 20 - (14+2) = 4
Hence, final marks = $$\left(5\times\ 14\right)-2-20$$ = 48 (from equation ii)
$$\therefore\ $$ The required answer is C.
Let $$f:R^{2}\rightarrow R$$ be a real-valued function defined as $$f\left(0, y\right)=y+1 \text{and} f\left(x+1, y\right)=f\left(x, f\left(x, y\right)\right)+ x$$. What is the value of $$f(2, 2)$$?
Solution
Given : f(0,y) = y+1 and f(x+1,y) = f(x,f(x,y)) + x
To find : f(2,2)
f(x+1,y) = f(x,f(x,y)) + x
Let x=0
f(1,y) = f(0,f(0,y))
f(1,y) = f(0,y+1) (as f(0,y) = y+1 (given))
f(1,y) = (y+1)+1 = y+2
Hence, f(1,y) = y+2 $$\longrightarrow\ i$$
Now,
f(x+1,y) = f(x,f(x,y)) + x
Let x=1
f(2,y) = f(1,f(1,y)) + 1
f(2,y) = f(1,y+2) + 1 (from equation i)
f(2,y) = (y+2) + 2 + 1 = y+5
Put y=2
f(2,2) = 2+5 =7
$$\therefore\ $$ The required answer is B.
Let $$a_{1}<a_{2}<.... <a_{n}$$ be the list of all prime numbers less than 25. Define $$X_{i}=\frac{b_{i}}{a_{i}}$$, where $$b_{i}$$ is the sum of all $$a_{k}$$ where k ranges from 1 to n, $$k \neq i$$. Let B be the set of all integer-valued $$X_{i}$$. What is the Smallest element of B?
Solution
List of all prime numbers less than 25 = 2,3,5,7,11,13,17,19,23 = 9 numbers
$$a_1$$ = 2, $$a_2$$ = 3, $$a_3$$ = 5, $$a_4$$ = 7, $$a_5$$ = 11, $$a_6$$ = 13, $$a_7$$ = 17, $$a_8$$ = 19, $$a_9$$ = 23
Sum of all the above prime numbers ($$a_1$$ + $$a_2$$ + ...... + $$a_9$$) = 100
$$X_i=\ \frac{\ b_i}{a_i}$$ ,where, $$b_i$$ = Sum of all prime numbers $$a_1\ to\ a_{n\ }$$ except $$a_i$$
Example : $$b_3$$ is the sum of all the prime numbers $$a_1$$ to $$a_9$$ except $$a_3$$.
$$X_1$$ = $$\ \frac{\ b_1}{a_1}$$ = $$\ \frac{\ 100-a_1}{a_1}$$ = $$\ \frac{\ 100-2}{2}$$ = 49
$$X_2$$ = $$\ \frac{\ b_2}{a_2}$$ = $$\ \frac{\ 100-a_2}{a_2}$$ = $$\ \frac{\ 100-3}{3}$$ = $$\ \frac{\ 97}{3}$$
$$X_3$$ = $$\ \frac{\ b_3}{a_3}$$ = $$\ \frac{\ 100-a_3}{a_3}$$ = $$\ \frac{\ 100-5}{5}$$ = 19
$$X_4$$ = $$\ \frac{\ b_4}{a_4}$$ = $$\ \frac{\ 100-a_4}{a_4}$$ = $$\ \frac{\ 100-7}{7}$$ = $$\ \frac{\ 93}{7}$$
$$X_5$$ = $$\ \frac{\ b_5}{a_5}$$ = $$\ \frac{\ 100-a_5}{a_5}$$ = $$\ \frac{\ 100-11}{11}$$ = $$\ \frac{\ 89}{11}$$
$$X_6$$ = $$\ \frac{\ b_6}{a_6}$$ = $$\ \frac{\ 100-a_6}{a_6}$$ = $$\ \frac{\ 100-13}{13}$$ = $$\ \frac{\ 87}{13}$$
$$X_7$$ = $$\ \frac{\ b_7}{a_7}$$ = $$\ \frac{\ 100-a_7}{a_7}$$ = $$\ \frac{\ 100-17}{17}$$ = $$\ \frac{\ 83}{17}$$
$$X_8$$ = $$\ \frac{\ b_8}{a_8}$$ = $$\ \frac{\ 100-a_8}{a_8}$$ = $$\ \frac{\ 100-19}{19}$$ = $$\ \frac{\ 81}{19}$$
$$X_9$$ = $$\ \frac{\ b_9}{a_9}$$ = $$\ \frac{\ 100-a_9}{a_9}$$ = $$\ \frac{\ 100-23}{23}$$ = $$\ \frac{\ 77}{23}$$
B is the set of all integer-valued $$X_i$$ = {$$X_1$$, $$X_3$$} = {49, 19}
The smallest element of B = 19
$$\therefore\ $$ The required answer is B.
Consider two circles, each having radius of 5cm (centimeters), touching each other at a point P. A direct tangent QR is drawn touching one circle at a point Q and the other circle at a point R. Inside the region PQR inscribed by the two circles and the tangent, a square ABCD is inscribed with its base AB on the tangent and the other side touching the two circles at points D and C, respectively.
Find the area of the square ABCD.
Solution
The image will be drawn as:
We will assume the side of the square to be 2a. XY will be half of CD i.e. 2a/2 = a. This means that OX = radius - XY ==> 5 - a.
Similarly, CX = radius - AC ==> 5 - 2a and OC = radius.
Now, we can use pythagoras theorem in triangle OCX:
$$\left(5-a\right)^2+\left(5-2a\right)^2=5^2$$ ==> $$25\ +\ a^2-10a\ +\ 25\ +\ 4a^2\ -20a=25$$
==> $$5a^2-30a+25\ =\ 0$$ ==> $$a^2\ -\ 6a\ +\ 5\ =\ 0\ $$.
This gives a = 5 or 1.
a can not be 5 as that will make CX i.e. 5 - 2a to be of negative length.
This means that side of the square is 2*1 = 2 cm and Area of the square is 2*2 = 4 sq. cm.
Rajan is a fruit seller. On any day, he sells only one kind of fruit. On the first day, he buys 9 kg of blueberries. On the second day, he buys 22 kg of kiwis. On the third day, he buys 50 kg of peaches. The per kg purchase price of each fruit is an integer. Further, on each of these three days, he spends the same amount to purchase fruits. On the fourth day, he buys mangoes at Rs. 35/kg and spends Rs. 15 less than any of the previous three days.
If he then sells all the mangoes at Rs. 50/kg, what is his MINIMUM possible profit on the fourth day?
Solution
Let the price of blueberries, kiwis and peaches be Rs b, k and p respectively.
Money spent on blueberries = Rs 9b
Money spent on kiwis = Rs 22k
Money spent on peaches = Rs 50p
Since, money spent everyday is the same,
9b = 22k = 50p = M , where, M is the total money spent everyday.
b = $$\ \frac{\ M}{9}$$
k = $$\ \frac{\ M}{22}$$
p = $$\ \frac{\ M}{50}$$
Since, b, k and p are integers, M should be the LCM of (9,22,50).
M = LCM (9,22,50) = LCM (22,450) = Rs 4950
Total money spent on the 4th day = Rs 4950 - Rs 15 = Rs 4935
Weight of mangoes bought on the 4th day = $$\ \frac{\ Rs\ 4935}{Rs\ \frac{35}{kg}}$$ = 141 kg
Cost Price of mangoes = Rs 35/kg
Sell Price of mangoes = Rs 50/kg
Profit on 4th day = $$\ \frac{\ Rs\ \left(50-35\right)}{kg}$$ $$\times\ $$141 kg = Rs 2115
Hence, the minimum possible profit on the 4th day is Rs 2115.
$$\therefore\ $$ The required answer is B.
How many solutions $$\left(x, y, z\right)$$ of the equation $$x+y^{2}+z^{3}=50$$ exist, where x, y and z are positive integers?
Solution
$$x+y^2+z^3=50$$
Let z = 1, $$x+y^2=49$$
Pairs of (x,y) = (48,1),(45,2),(40,3),(33,4),(24,5),(13,6) = 6 solutions
Let z = 2, $$x+y^2=42$$
Pairs of (x,y) = (41,1),(38,2),(33,3),(26,4),(17,5),(6,6) = 6 solutions
Let z = 3,$$x+y^2=23$$
Pairs of (x,y) = (22,1),(19,2),(14,3),(7,4) = 4 solutions
Total possible solutions = 16
$$\therefore\ $$ The required answer is C.
A triangular plot is such that two of its sides, of lengths 90m (meter) and 60m, are perpendicular to each other. There is a housing complex in a rectangular region within the plot. The area of the rectangular region is 4/9th of the area of the triangular plot. Additionally, two sides of the rectangular region lie on the two perpendicular sides of the triangle, and one vertex is on the hypotenuse. The members of the housing complex want to construct a wall along the perimeter of the rectangular region.
If the cost of construction is Rs. 5000/m, what is the MINIMUM possible cost of building the wall?
During Durga Puja, for the purpose of lighting, one puja pandal in Kolkata used many identical structures made of wooden sticks. The design of the structures was as follows: each structure was constructed with the help of six wooden sticks by combining an isosceles triangular structure, and a square structure, with the bases of both structures being the same. Let us take one such structure. Call the triangle PAB, with PA = PB, and the square ABCD, with AB being the same wooden stick as a common base for the triangle and the square. To make the structure strong, the two equal sides of the triangular structure were tied with the opposite side of square’s base, i.e., CD, at points E and F, in such a way that CE = EF = FD. The structure was hung from P.
If AB = 0.5m (meter), the total length of wooden sticks required for twenty such structures is:
Solution
The figure of one of the structure is given above.
Since, ABCD is a square, AB = CD = 0.5 m
Now, CE=EF=DF= (Given) = CD/3 = $$\ \ \ \frac{\ 0.5}{3}=\ \frac{\ 1}{6}m$$
Let PE = x m and EB = y m respectively.
In $$\triangle\ $$PFE and $$\triangle\ $$PAB,
$$\angle\ P=\angle\ P$$ (common angle)
$$\angle\ PFE=\angle\ PAB$$ (since FE is parallel to AB)
Hence, $$\triangle\ $$PFE and $$\triangle\ $$PAB are similar.
$$\ \frac{\ PE}{PB}=\ \frac{\ FE}{AB}$$
$$\ \frac{\ PE}{PE+EB}=\ \frac{\ FE}{AB}$$
$$\ \ \ \frac{\ x}{x+y}=\ \frac{\ \ \frac{\ 1}{6}}{0.5}$$
y = 2x i.e. EB = 2 $$\times\ $$PE
Now, in $$\triangle\ $$ECB,
$$EB^2=EC^2+CB^2$$
$$\left(2x\right)^2=\ \left(\frac{1\ }{6}\right)^2+\ \left(\frac{1\ }{2}\right)^2$$
$$x=\ \frac{\ \sqrt{10\ }}{12}m$$
PB = PE + EB = x + 2x = 3x = $$\ \frac{\ \sqrt{10}}{4}\ $$m
Length of the structure = PA + PB + AB + BC + CD + DA
= $$\ \frac{\ \sqrt{10}}{4}\ $$ + $$\ \frac{\ \sqrt{10}}{4}\ $$ + $$\ \frac{\ 1}{2}$$ + $$\ \frac{\ 1}{2}$$ + $$\ \frac{\ 1}{2}$$ + $$\ \frac{\ 1}{2}$$
= $$\left(2+\ \frac{\ \sqrt{10\ }}{2}\right)$$m
Length of 20 such structures =$$\left(2+\ \frac{\ \sqrt{10\ }}{2}\right)\times\ 20$$
= $$10\left(4+\sqrt{10\ }\right)m$$
$$\therefore\ $$ The required answer is C.
A park has two gates, Gate 1 and Gate 2. These two gates are connected via two alternate paths. If one takes the first path from Gate 1, they need to walk 80m (meters) towards east, then 80m towards south, and finally 20m towards west to arrive at Gate 2. The second path is a semi-circle connecting the two gates, where the diameter of the semi-circle is the straight-line distance between the two gates.
A person walking at a constant speed of 5 kilometers/hour enters the park through Gate 1, walks along the first path to reach Gate 2 and then takes the second path to come back to Gate 1.
Which of the following is the CLOSEST to the time the person takes, from entering the park to coming back to Gate 1, if she never stops in between?
Solution
The figure of the path is given below. Gate 1 and Gate 2 is situated at A and B respectively.
We have drawn ED which is perpendicular to AB in order to get a right angled triangle AED where AE = 60m and ED = 80m.
From pythagoras theorem, $$AD^2=AE^2+ED^2$$
$$AD^2=60^2+80^2$$
AD = 100m
Length of the semi circular path AD = $$\pi\ r=\pi\ \left(\frac{AD\ }{2}\right)\ $$ = $$50\pi\ \ $$ = 157 m
Length of path ABCD = 80+80+20 = 180 m
A person entered through gate 1 and travelled along path 1 and path 2 and exited from gate 1.
Total distance covered = 180+157 = 337m
Speed of the person = 5 kmph = $$\ \frac{\ 5\times\ 1000}{60}=\ \frac{\ 250}{3}\ \frac{\ m}{\min}$$
Total time taken = $$\ \frac{\ 337}{\ \frac{\ 250}{3}}\min$$ = 4 min (approx)
So the total time taken for the person to travel across the park is 4 min.
$$\therefore\ $$ The required answer is A.
Ronny uses a 5-digit key for a combination lock, where 5 digits need to be entered in a fixed sequence. While he remembers that the 5 digits are 9, 8, 7, 5 and 4, he has forgotten the sequence he uses. He also remembers that the sum of the first three digits is a multiple of 3, and so is the sum of the last three digits. Further, the sum of the last four digits is a multiple of 4.
Which of the following is DEFINITELY FALSE?
Solution
The 5 digits which are used in the combination lock are 9,8,7,5 and 4.
The sum of first 3 digits and the last 3 digits are divisible by 3. Hence, we can say that the third digit is common in both the combinations.
The possible groups of 3 digits whose sum is divisible by 3 are (9,8,7),(9,8,4),(9,7,5),(9,5,4).
Out of these 4 possible groups, we have to select two groups which can be used as the first and last 3 digits.
The selection should be done in such a way that both the groups should only have one digit in common.
Example : If we select (9,8,7) and (9,8,4) and let the middle digit be 9, then we will have 8 in both the first 3 digits and the last 3 digits which is only possible for the 3rd digit which is occupied by 9 in this case. Hence, this case is not possible.
So the only possible combinations for the first and last 3 digits are :
Case 1: (9,8,7) and (9,5,4) where the 3rd digit will be 9.
Case 2: (9,8,4) and (9,7,5) where the 3rd digit will be 9.
Now, it is given that the sum of the last 4 digits is divisible by 4.
The possible cases of the above scenario are :
Case 3: (9,8,7,4) where the first digit will be 5.
Case 4: (8,7,5,4)where the first digit will be 9 which is not possible as 9 is fixed as the 3rd digit from Case 1 and 2.
Since, the first digit is 5 and the third digit is 9, the second digit will either be 4 or 7 and the last two digits will be either (8,7) or (8,4)
So, the possible combinations of the lock are 54987, 54978, 57984 and 57948.
Hence, the option which is definitely false is E as no possible combinations has 8 as its 2nd digit.
$$\therefore\ $$ The required answer is E.
There are three rectangular tanks in a building. The length, width and height of the first tank are m meters each, and the length, width and height of the second tank are n meters each. However, the length, width and height of the third tank are m meters, n meters and 1 meter, respectively.Initially, the first tank is full of water, while the second and the third are empty. When the second and the third tanks are completely filled with water transferred from the first tank, 85,000 liters of water is still left in the first tank.
If both m and n are positive integers, what is the value of m? (1 meter$$^{3}$$ =1000 liters)
Solution
The length, width and height of Tank 1 is m meters each.
Volume of Tank 1 = $$m^{3\ }$$ cubic meters
The length, width and height of Tank 2 is n meters each.
Volume of Tank 2 = $$n^{3\ }$$ cubic meters
The length, width and height of Tank 3 is m, n and 1 m respectively.
Volume of Tank 3 = mn cubic meters
Since, Tank 1 is filled first and after Tank 1 is full, water is flowing into Tank 2 and Tank 3,
Volume of water left in Tank 1 after Tank 2 and Tank 3 are filled = $$\left(m^{3\ }-n^3-mn\right)$$ cubic meters
Volume of water left in Tank 1 = 85000 L (given) = 85$$m^3$$
Hence, $$\left(m^{3\ }-n^3-mn\right)$$ = 85 , where, m and n are integers.
$$n^3+mn=m^3-85$$
From option B, we get $$n^3+7n=258$$ $$\longrightarrow\ $$ n=6
From option C, we get $$n^3+5n=40$$ $$\longrightarrow\ $$ No integer solution
From option D, we get $$n^3+6n=131$$ $$\longrightarrow\ $$ No integer solution
From option E, we get $$n^3+10n=915$$ $$\longrightarrow\ $$ No integer solution
Hence, we got m = 7 and n = 6.
$$\therefore\ $$ The required answer is B.
Read the following scenario and answer the THREE questions that follow.
Light Chemicals is an industrial paint supplier with presence in three locations: Mumbai, Hyderabad and Bengaluru. The sunburst chart below shows the distribution of the number of employees of different departments of Light Chemicals. There are four departments: Finance, IT, HR and Sales. The employees are deployed in four ranks: junior, mid, senior and executive. The chart shows four levels: location, department, rank and gender (M: male, F: female). At every level, the number of employees at any location/department/rank/gender are proportional to the corresponding area of the region represented in the chart.
Due to some issues with the software, the data on junior female employees have gone missing. Notice that there are junior female employees in Mumbai HR, Sales and IT departments, Hyderabad HR department, and Bengaluru IT and Finance departments: the corresponding missing numbers are marked u, v, w, x, y and z in the diagram, respectively.
It is also known that:
a) Light Chemicals has a total of 210 junior employees.
b) Light Chemicals has a total of 146 employees in the IT department.
c) Light Chemicals has a total of 77 employees in the Hyderabad office.
d) In the Mumbai office, the number of junior female employees is 55.
Based on the given information and the sunburst chart, the MOST LIKELY values of u, v and w, respectively, are:
Solution
It is given to us that Light Chemicals has a total of 210 junior employees.
This means that: 25 + w + 30 + v + u + 15 + 18 + z + 16 + y + x + 10 = 210 ==> u + v + w + x + y + z = 96.
Then, Light Chemicals has a total of 146 employees in the IT department.
This means that: 22 + 15 + 25 + w + 14 + 8 + 16 + y + 12 + 10 = 146 ==> w+y = 24.
Then, Light Chemicals has a total of 77 employees in the Hyderabad office.
This means that: 13 + 11 + 9 + 7 + x + 10 + 9 + 6 = 77 ==> x = 12.
From these 3 equations, we get u + v + z = 60.
Then, In the Mumbai office, the number of junior female employees is 55.
This means that: w + v + u = 55.
Subtracting this from the 1st equation, we get x + y + z = 41 and subtracting value of x from this, we get y+z = 29.
We are given that: At every level, the number of employees at any location/department/rank/gender are proportional to the corresponding area of the region represented in the chart.
it implies that v > u > w.
This means that the possible option is either B or D. But in option D, v is 4 time u which is not proportionately correct in the chart.
Thus, the possible values are 20, 25, 10 i.e. Option B.
Based on the given information and the sunburst chart, which of the following has the HIGHEST number of employees?
Solution
It is given to us that Light Chemicals has a total of 210 junior employees.
This means that: 25 + w + 30 + v + u + 15 + 18 + z + 16 + y + x + 10 = 210 ==> u + v + w + x + y + z = 96.
Then, Light Chemicals has a total of 146 employees in the IT department.
This means that: 22 + 15 + 25 + w + 14 + 8 + 16 + y + 12 + 10 = 146 ==> w+y = 24.
Then, Light Chemicals has a total of 77 employees in the Hyderabad office.
This means that: 13 + 11 + 9 + 7 + x + 10 + 9 + 6 = 77 ==> x = 12.
From these 3 equations, we get u + v + z = 60.
Then, In the Mumbai office, the number of junior female employees is 55.
This means that: w + v + u = 55.
Subtracting this from the 1st equation, we get x + y + z = 41 and subtracting value of x from this, we get y+z = 29.
Now, the values of:
A. IT department, all offices together is 146(b).
B. Bengaluru office : 22+20+18+z+16+y+12+10+8+6 = 102 + y+z = 102+29 = 131.
C. Sales Department of Mumbai Office: 30+v+12+10 = 52 + v. As w+v+u = 55, v is definitely smaller than 55 and thus 52+v is smaller than 107.
E. Mid-level, all offices together: 22+15+18+12+20+22+13+11 = 133.
The maximum value is of Option A.
Based on the given information and the sunburst chart, which of the following statements CANNOT BE TRUE?
Solution
It is given to us that Light Chemicals has a total of 210 junior employees.
This means that: 25 + w + 30 + v + u + 15 + 18 + z + 16 + y + x + 10 = 210 ==> u + v + w + x + y + z = 96.
Then, Light Chemicals has a total of 146 employees in the IT department.
This means that: 22 + 15 + 25 + w + 14 + 8 + 16 + y + 12 + 10 = 146 ==> w+y = 24.
Then, Light Chemicals has a total of 77 employees in the Hyderabad office.
This means that: 13 + 11 + 9 + 7 + x + 10 + 9 + 6 = 77 ==> x = 12.
From these 3 equations, we get u + v + z = 60.
Then, In the Mumbai office, the number of junior female employees is 55.
This means that: w + v + u = 55.
Subtracting this from the 1st equation, we get x + y + z = 41 and subtracting value of x from this, we get y+z = 29.
Number of female junior employees in Finance department of Bengaluru = z and number of female junior employees in IT department in Bengaluru = y.
$$y\ne\ z$$ because y+z = 29 and thus they have to be distinct otherwise they won't be having a natural number value.
thus, the answer is Option D.
Read the following scenario and answer the THREE questions that follow.
An investment company, WinLose, recruits employees to trade in the share market. For newcomers, they have a one-year probation period. During this period, the employees are given Rs. 1 lakh per month to invest the way they see fit. They are evaluated at the end of every month, using the following criteria:
1. If the total loss in any span of three consecutive months exceeds Rs. 20,000, their services are terminated at the end of that 3-month period,
2. If the total loss in any span of six consecutive months exceeds Rs. 10,000, their services are terminated at the end of that 6-month period.
Further, at the end of the 12-month probation period, if there are losses on their overall investment, their services are terminated.
Ratan, Shri, Tamal and Upanshu started working for WinLose in January. Ratan was terminated after 4 months, Shri was terminated after 7 months, Tamal was terminated after 10 months, while Upanshu was not terminated even after 12 months. The table below, partially, lists their monthly profits (In Rs. ‘000) over the 12-month period, where x, y and z are masked information.
Note:
1. A negative profit value indicates a loss.
2. The value in any cell is an integer.
Illustration: As Upanshu is continuing after March, that means his total profit during January-March (2z+2z+0) ≥ ─ Rs.20,000. Similarly, as he is continuing after June, his total profit during January-June ≥ ─ Rs.10,000, as well as his total profit during April-June ≥ ─ Rs.20,000.
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What BEST can be said about the value of x?
Solution
As Ratan was terminated after 4 months, he did not register a loss of more than 20000 at the end of 3 months but registered a loss of more than 20000 in 2nd, 3rd and 4th month. This implies: x - 30 >= -20 and x - 35 < -20 ==> 15 > x >= 10. As x is an integer and x/2 must also be an integer x could be 10, 12 or 14.
For Shri, as he was terminated after 7 months, his loss for the first 6 months did not exceed 10000. It implies that x + 1.5y - 65 >= -10 ==> x+1.5y >= 55.
Also, for the first 6 months, his loss did not exceed 20000 for any group of 3 consecutive months. It implies that: x+y-25 >= -20 ==> x+y >= 5
y - 45 >= -20 ==> y >= 25 and 0.5y - 45 >= -20 ==> y >= 50.
Now, he could be terminated for 2 reasons i.e. 1. his loss for May, June, July exceeded 20000 OR 2. his loss for last 6 months exceeded 10000.
Case 1: 0.5y - 50 < -20 ==> y < 60.
Case 2: 1.5y - 95 < -10 ==> y < 56.667.
As y is also even, possible values for y are 50, 52, 54, 56 and 58.
As Tamal was terminated after 10 months, in ay consecutive period of 3 months for the first 9 months, his loss did not exceed 20000 and for any consecutive period of 6 months, his loss did not exceed 10000.
This implies: 0.5x + y + x >= -20 ==> 1.5x + y >= -20.
There are 2 cases for Tamal also.
Case 1: 5 + 10 - z < -20 ==> z > 35
Case 2: x - 20 + 20 -z < -10 ==> z -10 > x.
Similarly, for Upanshu, the equations will be that the total profits should exceed 0 : 5z - x + y - 81 >= 0
z -48 >= -20 ==> z >= 28 and y - x - 58 >= -10 ==> y -x >= 48. This is only possible when x = 10 and y = 58.
So, value for x is 10 i.e. Option E.
What BEST can be said about the value of y?
Solution
As Ratan was terminated after 4 months, he did not register a loss of more than 20000 at the end of 3 months but registered a loss of more than 20000 in 2nd, 3rd and 4th month. This implies: x - 30 >= -20 and x - 35 < -20 ==> 15 > x >= 10. As x is an integer and x/2 must also be an integer x could be 10, 12 or 14.
For Shri, as he was terminated after 7 months, his loss for the first 6 months did not exceed 10000. It implies that x + 1.5y - 65 >= -10 ==> x+1.5y >= 55.
Also, for the first 6 months, his loss did not exceed 20000 for any group of 3 consecutive months. It implies that: x+y-25 >= -20 ==> x+y >= 5
y - 45 >= -20 ==> y >= 25 and 0.5y - 45 >= -20 ==> y >= 50.
Now, he could be terminated for 2 reasons i.e. 1. his loss for May, June, July exceeded 20000 OR 2. his loss for last 6 months exceeded 10000.
Case 1: 0.5y - 50 < -20 ==> y < 60.
Case 2: 1.5y - 95 < -10 ==> y < 56.667.
As y is also even, possible values for y are 50, 52, 54, 56 and 58.
As Tamal was terminated after 10 months, in ay consecutive period of 3 months for the first 9 months, his loss did not exceed 20000 and for any consecutive period of 6 months, his loss did not exceed 10000.
This implies: 0.5x + y + x >= -20 ==> 1.5x + y >= -20.
There are 2 cases for Tamal also.
Case 1: 5 + 10 - z < -20 ==> z > 35
Case 2: x - 20 + 20 -z < -10 ==> z -10 > x.
Similarly, for Upanshu, the equations will be that the total profits should exceed 0 : 5z - x + y - 81 >= 0
z -48 >= -20 ==> z >= 28 and y - x - 58 >= -10 ==> y -x >= 48. This is only possible when x = 10 and y = 58.
So, value of y is 58 i.e. Option A.
What BEST can be said about the profit generated by Upanshu at the end of his probation period?
Solution
As Ratan was terminated after 4 months, he did not register a loss of more than 20000 at the end of 3 months but registered a loss of more than 20000 in 2nd, 3rd and 4th month. This implies: x - 30 >= -20 and x - 35 < -20 ==> 15 > x >= 10. As x is an integer and x/2 must also be an integer x could be 10, 12 or 14.
For Shri, as he was terminated after 7 months, his loss for the first 6 months did not exceed 10000. It implies that x + 1.5y - 65 >= -10 ==> x+1.5y >= 55.
Also, for the first 6 months, his loss did not exceed 20000 for any group of 3 consecutive months. It implies that: x+y-25 >= -20 ==> x+y >= 5
y - 45 >= -20 ==> y >= 25 and 0.5y - 45 >= -20 ==> y >= 50.
Now, he could be terminated for 2 reasons i.e. 1. his loss for May, June, July exceeded 20000 OR 2. his loss for last 6 months exceeded 10000.
Case 1: 0.5y - 50 < -20 ==> y < 60.
Case 2: 1.5y - 95 < -10 ==> y < 56.667.
As y is also even, possible values for y are 50, 52, 54, 56 and 58.
As Tamal was terminated after 10 months, in ay consecutive period of 3 months for the first 9 months, his loss did not exceed 20000 and for any consecutive period of 6 months, his loss did not exceed 10000.
This implies: 0.5x + y + x >= -20 ==> 1.5x + y >= -20.
There are 2 cases for Tamal also.
Case 1: 5 + 10 - z < -20 ==> z > 35
Case 2: x - 20 + 20 -z < -10 ==> z -10 > x.
Similarly, for Upanshu, the equations will be that the total profits should exceed 0 : 5z - x + y - 81 >= 0
z -48 >= -20 ==> z >= 28 and y - x - 58 >= -10 ==> y -x >= 48. This is only possible when x = 10 and y = 58.
Profits generated by Upanshu at the end of his probation period are 5z + y-x - 81. Now, z >= 28 and y-x = 48 which gives 5z +y-x >= 188 and 5z +y-x -81 >= 107.
Thus, the correct answer is Option A i.e. atleast 107000.
