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Q1
Miscellaneous MULTIPLE CHOICE
There are 25 rooms in a hotel. Each room can accommodate at most three people. For each room, the single occupancy charge is Rs. 2000 per day, the double occupancy charge is Rs. 3000 per day, and the triple occupancy charge is Rs. 3500 per day.
If there are 55 people staying in the hotel today, what is the maximum possible revenue from room occupancy charges today?
Ramesh bought a mobile from a local store. He paid 1/6 of the price via UPI and 1/3 of the price via cash. He agreed to pay the balance amount a year later. While paying back the balance amount, Ramesh paid 10% interest on the balance amount.
If the interest paid was Rs. 6000, what was the original price of the mobile?
Adu and Amu have bought two pieces of land on the Moon from an e-store. Both the pieces of land have the same perimeters, but Adu’s piece of land is in the shape of a square, while Amu’s piece of land is in the shape of a circle.
The ratio of the areas of Adu’s piece of land to Amu’s piece of land is:
The market value of beams, made of a rare metal, has a unique property: the market value of any such beam is proportional to the square of its length. Due to an accident, one such beam got broken into two pieces having lengths in the ratio 4:9.
Considering each broken piece as a separate beam, how much gain or loss, with respect to the market value of the original beam before the accident, is incurred?
Asked: the percentage gain or loss when a beam whose value is proportional to (length)2 breaks in the ratio 4 : 9.
Take the pieces as \(4k\) and \(9k\), so the original beam was \(13k\). Value \(= c \times \text{length}^2\).
Original value: \(169ck^2\)
Pieces: \(16ck^2 + 81ck^2 = 97ck^2\)
Loss: \(169 - 97 = 72\) parts
Loss % \(= \frac{72}{169} \times 100 \approx 42.60\%\)
Answer: 42.60% loss
Q5
Geometry - Coordinate Geometry MULTIPLE CHOICE
ABCD is a rectangle, where the coordinates of C and D are (−2,0) and (2,0), respectively. If the area of the rectangle is 24, which of the following is a possible equation representing the line AB?
Asked: a possible equation of line AB for rectangle ABCD with C(−2, 0), D(2, 0) and area 24.
In a rectangle named ABCD the vertices go in order, so CD is a side and AB is the opposite side, parallel to CD.
CD lies on the x-axis and has length \(2 - (-2) = 4\).
The other side: \(24 \div 4 = 6\).
AB is parallel to CD (horizontal) and 6 units away from it: \(y = 6\) or \(y = -6\).
Among the options, \(y = 6\) fits. The lines \(4x + 6y = 24\) and \(x + y = 12\) are not horizontal, and \(x = 6\) is vertical, so none of these can be parallel to CD.
Answer: y = 6 (option 5)
Q6
Modern Math - Probability MULTIPLE CHOICE
Consider the quadratic function \(f(x) = ax^2 + bx + a\) having two irrational roots, with \(a\) and \(b\) being two positive integers, such that \(a, b \leq 9\). If all such permissible pairs \((a, b)\) are equally likely, what is the probability that \(a + b\) is greater than 9?
Asked: among pairs \((a, b)\), \(1 \le a, b \le 9\), for which \(ax^2 + bx + a\) has two irrational roots, the probability that \(a + b > 9\).
Condition: the discriminant \(D = b^2 - 4a^2\) must be positive and not a perfect square.
Step 1, D > 0: \(b > 2a\).
a = 1: b = 3 to 9 (7 pairs)
a = 2: b = 5 to 9 (5 pairs)
a = 3: b = 7, 8, 9 (3 pairs)
a = 4: b = 9 (1 pair)
That gives 16 pairs.
Step 2, remove perfect squares: check each \(b^2 - 4a^2\). Only (2, 5) gives \(25 - 16 = 9 = 3^2\), which means rational roots. No other pair gives a square (for example, a = 1 gives 5, 12, 21, 32, 45, 60, 77). So there are 15 permissible pairs.
Step 3, a + b > 9:
a = 1: (1, 9)
a = 2: (2, 8), (2, 9)
a = 3: (3, 7), (3, 8), (3, 9)
a = 4: (4, 9)
That is 7 favourable pairs. Probability = 7/15.
Answer: 7/15
Q7
Data Sufficiency MULTIPLE CHOICE
A farmer has a quadrilateral parcel of land with a perimeter of 700 feet. Two opposite angles of that parcel of land are right angles, while the remaining two are not. The farmer wants to do organic farming on that parcel of land. The cost of organic farming is Rs. 400 per square foot.
Consider the following two additional pieces of information:
Asked: which information fixes the area (and so the cost) of a quadrilateral with perimeter 700 ft and two opposite right angles.
Call it ABCD with \(\angle B = \angle D = 90^\circ\). The diagonal AC joins the two non-right corners. It is the common hypotenuse of right triangles ABC and ADC: \(AB^2 + BC^2 = AD^2 + DC^2 = AC^2\).
Area = \(\frac{1}{2}(AB \cdot BC + AD \cdot DC)\).
Statement I alone (one side = 110): perimeter and one side do not fix the shape, because the diagonal can vary. Not sufficient.
Statement II alone (AC = 265): let \(s_1 = AB + BC\) and \(s_2 = AD + DC\), so \(s_1 + s_2 = 700\). Then \(AB \cdot BC = \frac{s_1^2 - 265^2}{2}\) and \(AD \cdot DC = \frac{s_2^2 - 265^2}{2}\). The total depends on how 700 splits between \(s_1\) and \(s_2\). Not sufficient.
\(AD \cdot DC = \frac{s_2^2 - 265^2}{2}\) is fixed
So the area, and the cost at Rs. 400 per sq ft, is determined. By symmetry it does not matter which side is 110.
Answer: I and II together only
Q8
Geometry - Coordinate Geometry MULTIPLE CHOICE
A straight line \(L_1\) has the equation \(y = k(x - 1)\), where \(k\) is some real number. The straight line \(L_1\) intersects another straight line \(L_2\) at the point (5, 8). If \(L_2\) has a slope of 1, which of the following is definitely FALSE?
Asked: which statement about the two lines is definitely false.
Find the lines. \(L_1: y = k(x-1)\) passes through (5, 8), so \(8 = 4k\) and \(k = 2\). That gives \(L_1: y = 2x - 2\). \(L_2\) has slope 1 through (5, 8), so \(L_2: y = x + 3\).
x-intercepts: \(L_1\) at 1, \(L_2\) at −3. Distance = 4. Statement 2 is true.
y-intercepts: \(L_1\) at −2, \(L_2\) at 3. Distance = 5, not 6. Statement 3 is false.
\(L_1\) passes through (1, 0). Statement 4 is true.
Distance from the origin to \(L_2\) (\(x - y + 3 = 0\)): \(\frac{3}{\sqrt{2}}\). Statement 1 is true.
Distance from the origin to \(L_1\) (\(2x - y - 2 = 0\)): \(\frac{2}{\sqrt{5}}\). Statement 5 is true.
Answer: The distance between the y-intercepts of the two lines is 6
Q9
Algebra - Logarithms MULTIPLE CHOICE
For how many distinct real values of \(x\) does the equation below hold true? (Consider \(a > 0\)) \[ \frac{x^2 \log_a(16)}{\log_a(32)} - \frac{\log_a(64)}{\log_a(32)} - x = 0 \]
Asked: the number of real solutions of the given equation.
Key idea: a ratio of logs to the same base does not depend on the base: \(\frac{\log_a 16}{\log_a 32} = \frac{4\log_a 2}{5\log_a 2} = \frac{4}{5}\) and \(\frac{\log_a 64}{\log_a 32} = \frac{6}{5}\) (for any valid base \(a > 0, a \ne 1\)).
The equation becomes \(\frac{4}{5}x^2 - \frac{6}{5} - x = 0\).
Multiply by 5: \(4x^2 - 5x - 6 = 0\).
Factorise: \((4x + 3)(x - 2) = 0\), so \(x = 2\) or \(x = -\frac{3}{4}\).
There are two distinct real values, whatever the base.
Answer: 2
Q10
Miscellaneous MULTIPLE CHOICE
In a computer game, each move requires pressing a button. When the button is pressed for the first time, as a move, the computer randomly chooses a cell from a 4x4 grid of sixteen cells and puts an "X" mark on that cell. When the button is pressed subsequently, the computer randomly chooses a cell from the remaining unmarked cells and puts an "X" mark on that cell. This goes on till the end of the game. The game ends when either all the cells in any one row, or all the cells in any one column, are marked with "X". What is the maximum possible number of times a player has to press the button to finish the game?
Asked: the largest number of presses that may be needed before some row or column of the 4×4 grid is fully marked.
Key idea: find the most cells that can be marked without completing any row or column. The next press must then end the game.
Each of the 4 rows needs at least one unmarked cell, so at least 4 cells stay empty.
4 empty cells are enough if they sit in different rows and different columns, for example along a diagonal. Then every row and every column has exactly one gap.
So up to 16 − 4 = 12 cells can be marked with the game still running.
The 13th mark must fill one of these gaps, which completes that row (and that column).
Answer: 13
Q11
Geometry - Circles MULTIPLE CHOICE
There are five dustbins along a circular path at different places. Ramesh takes multiple rounds of the path every morning, always at the same speed. He noticed that it took him a different number of steps to walk between any two consecutive dustbins. Ramesh also noticed that starting from any of the dustbins, it took a minimum 360 steps to reach every second dustbin, and a maximum 1260 steps to reach every third dustbin. If Ramesh's step size is 0.77 meter, and the width of the path is negligible, which the following can be the radius of the circular path?
Let the five gaps between consecutive dustbins be \(g_1, \dots, g_5\) steps (all different), with circumference \(C = g_1 + \dots + g_5\) steps.
Reading the data as bounds: every walk to the second dustbin ahead (two consecutive gaps) takes at least 360 steps, and every walk to the third dustbin ahead (three consecutive gaps) takes at most 1260 steps.
Adding all five two-gap sums counts each gap twice: \(2C \ge 5 \times 360\), so \(C \ge 900\).
Adding all five three-gap sums counts each gap three times: \(3C \le 5 \times 1260\), so \(C \le 2100\). Equality would need all gaps equal to 420, which is not allowed because the gaps differ, so \(C < 2100\).
Convert to metres: \(C < 2100 \times 0.77 = 1617\) m, so \(r < \frac{1617}{2\pi} \approx 257.4\) m.
260 m would need about 2122 steps, which is impossible.
250 m needs \(2\pi \times 250 \approx 1570.8\) m, about 2040 steps. This works: for example, gaps of 400, 412, 404, 416, 408 have every adjacent pair at least 808 (so every three-gap sum is at most 2040 − 780 = 1260) and every two-gap sum above 360.
Answer: 250 meters (see the note below on the wording)
Q12
Arithmetic - Average MULTIPLE CHOICE
Eight employees of an organization have been rated on a scale of 1 to 50 for their performance. All ratings are integers. The overall average rating of the eight employees is 30. While the five employees with the highest ratings average 38, the five employees with the lowest ratings average 25. Which of the following, about the ratings obtained by the eight employees, is DEFINITELY FALSE?
\(r_4\) and \(r_5\) are counted in both sums, so \(r_4 + r_5 = 190 + 125 - 240 = 75\).
So \(r_1 + r_2 + r_3 = 50\) and \(r_6 + r_7 + r_8 = 115\).
Since \(r_4 \le r_5\) and they add to 75: \(r_5 \ge 38\).
Statement 5 (highest = 40): then \(r_6 + r_7 = 75\). But \(r_6, r_7 \ge r_5 \ge 38\) means \(r_6 + r_7 \ge 76\). This is impossible, so statement 5 is definitely false.
The others can hold:
Statement 1: 37, 38, 38, 38, 39 at the top works (\(r_8 = 39\)).
Statement 2: e.g. 1, 12, 37, 37, 38, ...
Statement 3: e.g. 6, 7, 37, 37, 38, ...
Statement 4: the median is always \(\frac{r_4 + r_5}{2} = 37.5\).
Answer: The highest rating obtained is 40.
Q13
Algebra - Number Theory MULTIPLE CHOICE
Arun selected an integer x between 2 and 40, both inclusive. He noticed that the greatest common divisor of the selected integer x and any other integer between 2 and 40, both inclusive, is 1. How many different choices for such an x are possible?
Asked: how many x in 2 to 40 have GCD 1 with every other integer in 2 to 40.
If x is composite, it shares a prime factor p with p itself (p is in the range and differs from x). So x must be prime.
If x is a prime with \(2x \le 40\), i.e. \(x \le 20\), then \(\gcd(x, 2x) = x > 1\). So x must exceed 20.
A prime above 20 has no multiple other than itself up to 40, so it is coprime to every other number in the range.
Primes from 21 to 40: 23, 29, 31, 37, so there are 4 choices.
Answer: 4
Q14
Geometry - Triangles MULTIPLE CHOICE
An industrial robot manufacturing company is tasked to design humanoid robots to be used in warehouses where the robots need to pick items from a stack of shelves. The height of the topmost shelf from the ground is 7 feet. To operate, the robot has to move on a track, running parallel to the stack of shelves. The track is fixed 1 foot away from the base of the stack of shelves. Further, the robot cannot bend its arms by more then 60o from the horizontal plane. If the robot's arms are attached to its shoulder, what should be the minimum height of the robot from the ground to the shoulder for its arms to reach the topmost shelf?
Asked: the minimum shoulder height that lets the arm reach a shelf 7 ft high, 1 ft away horizontally, when the arm can rise at most 60° above the horizontal.
The shoulder is 1 ft (horizontally) from the shelf face.
An arm at angle \(\theta\) gains height \(1 \times \tan\theta\) over that 1 ft.
The steepest allowed angle is 60°, so the greatest gain is \(\tan 60^\circ = \sqrt{3}\) ft.
Shoulder height \(h\) must satisfy \(h + \sqrt{3} \ge 7\), so \(h \ge 7 - \sqrt{3}\) ft (about 5.27 ft).
Answer: \(7 - \sqrt{3}\) feet
Q15
Geometry - Mensuration MULTIPLE CHOICE
A solid trophy, consisting of two parts, has been designed in the following manner: the bottom part is a frustum of a cone with the bottom radius 30 cm, the top radius 20 cm, and height 40 cm, while the top part is a hemisphere with radius 20 cm. Moreover, the flat surface of the hemisphere is the same as the top surface of the frustum. If the entire trophy is to be gold-plated at the cost of Rs. 40 per square cm, what would the cost for gold-plating be closest to?
Asked: the gold-plating cost of the exposed surface (frustum + hemisphere) at Rs. 40 per sq cm.
Surfaces exposed: the frustum's curved surface, its bottom base, and the hemisphere's curved surface. The frustum's top and the hemisphere's flat face are joined, so they are hidden.
Cost \(\approx 11818 \times 40 \approx\) Rs. 4,72,700, which is closest to Rs. 4,73,000.
Answer: Rs. 4,73,000
Q16
Algebra - Surds & Indices MULTIPLE CHOICE
If \(a, b\), and \(c\) are all positive integers, with \(4 a>b\), then which of the following conditions is BOTH NECESSARY AND SUFFICIENT for the expression \(\sqrt[3]{(3)^a(21)^{(3 a-b)}(49)^{(2 b+c)}}\) to be a positive integer?
Asked: the exact condition for \(\sqrt[3]{3^a\, 21^{3a-b}\, 49^{2b+c}}\) to be a positive integer.
Write everything in primes: \(21 = 3 \times 7\) and \(49 = 7^2\).
Power of 3: \(a + (3a - b) = 4a - b\)
Power of 7: \((3a - b) + 2(2b + c) = 3a + 3b + 2c\)
Both powers are positive (\(4a > b\) is given). The cube root is an integer exactly when both powers are multiples of 3.
\(4a - b = 3a + (a - b)\) is a multiple of 3 exactly when \(a - b\) is.
\(3a + 3b + 2c\) is a multiple of 3 exactly when \(2c\) is, i.e. when \(c\) is.
So the condition is "\(a - b\) and \(c\) are both divisible by 3". Option 3 is sufficient but not necessary (a = 4, b = 1 works). Options 1 and 2 are not sufficient (a = 2, b = 1, c = 1 satisfies option 1 but fails).
Answer: a − b and c are divisible by 3
Read the following scenario and answer the THREE questions that follow.
The diagram below represents a road network connecting five towns, namely Meereen, Lannisport, Winterfell, Oldtown, and Gulltown. The maximum speed limits along any stretch of road are as shown in the diagram. The straight road that connects Meereen to Gulltown passes through Oldtown. Another straight road, running west to east, connecting Meereen to Winterfell, passes through Lannisport. Further, two straight roads, one from Lannisport to Oldtown and another from Winterfell to Gulltown, are perpendicular to the road joining Meereen to Winterfell, and run from south to north.
Consider a car always travelling at the maximum permissible speed, and always taking the shortest route. It takes 1 hour to reach Oldtown from Meereen, 2 hours to reach Gulltown from Oldtown, and 45 minutes to reach Winterfell from Gulltown. (For this problem, always consider the shortest route in terms of distance.)
Q17
Arithmetic - Time, Speed & Distance MULTIPLE CHOICE
Tyrion Lannister drove from Meereen to Oldtown, then from Oldtown to Lannisport, and finally from Lannisport to Winterfell, always taking the shortest paths. He always drove at a speed 10 km/hr below the maximum speed limits for the stretches he took.
Distances from the data: M→O takes 1 h at 100 km/h, so MO = 100 km. O→G takes 2 h at 25 km/h, so OG = 50 km. G→W takes 45 min at 120 km/h, so GW = 90 km. So MG = 150 km, and triangle MWG is right-angled at W: \(MW = \sqrt{150^2 - 90^2} = 120\) km. LO is parallel to WG, so triangle MLO is similar to MWG with ratio 100/150: LO = 60 km, ML = 80 km, LW = 40 km.
Tyrion drives 10 km/h below each limit:
M→O: 100 km at 90 km/h = 1.111 h (66.7 min)
O→L: 60 km at 20 − 10 = 10 km/h = 6 h (360 min)
L→W: 40 km at 110 km/h = 0.364 h (21.8 min)
Total \(\approx 448.5\) min, about 7 h 28 min.
Answer: 7 hrs 28 mins (option 4)
Q18
Arithmetic - Time, Speed & Distance MULTIPLE CHOICE
Missandei starts from Gulltown towards Oldtown by the shortest path, driving at the maximum permissible speed. From Oldtown, she drives at a speed of 10 km/hr towards Lannisport. When Missandei starts from Gulltown, Varys starts at the same time from Lannisport to Oldtown along the shortest path, always driving at the maximum permissible speed.
If they don’t stop anywhere, at what point will they meet?
Distances from the data: M→O takes 1 h at 100 km/h, so MO = 100 km. O→G takes 2 h at 25 km/h, so OG = 50 km. G→W takes 45 min at 120 km/h, so GW = 90 km. So MG = 150 km, and triangle MWG is right-angled at W: \(MW = \sqrt{150^2 - 90^2} = 120\) km. LO is parallel to WG, so triangle MLO is similar to MWG with ratio 100/150: LO = 60 km, ML = 80 km, LW = 40 km.
Missandei: G→O is 50 km at 25 km/h, so she reaches Oldtown at t = 2 h. She then heads south towards L at 10 km/h.
Varys: leaves L at t = 0, heading north on LO at 20 km/h. At t = 2 h he is 40 km north of L, i.e. 20 km short of O.
From t = 2 they approach each other at 10 + 20 = 30 km/h across a 20 km gap, so they meet after \(\frac{2}{3}\) h.
In that time Varys covers \(20 \times \frac{2}{3} = 13.33\) km more, so he is 53.33 km north of L, which is 6.67 km south of Oldtown. (Check: Missandei covers \(10 \times \frac{2}{3} = 6.67\) km south of O.)
Answer: Approximately 6.67 km south of Oldtown (option 2)
Q19
Arithmetic - Time, Speed & Distance MULTIPLE CHOICE
The capital city, King’s Landing, located 40 km to the south of Gulltown on the road connecting Gulltown to Winterfell, did not have a straight road, connecting to Meereen. Now, a new expressway is being built to connect these two towns by a straight road.
What should be the maximum speed limit allowed on this expressway so that it cuts down the travel time, from Meeren to King’s Landing, from the fastest possible route through the road network shown in the diagram, by 20 minutes?
Distances from the data: M→O takes 1 h at 100 km/h, so MO = 100 km. O→G takes 2 h at 25 km/h, so OG = 50 km. G→W takes 45 min at 120 km/h, so GW = 90 km. So MG = 150 km, and triangle MWG is right-angled at W: \(MW = \sqrt{150^2 - 90^2} = 120\) km. LO is parallel to WG, so triangle MLO is similar to MWG with ratio 100/150: LO = 60 km, ML = 80 km, LW = 40 km.
King's Landing (K) is 40 km south of G on GW, i.e. WK = 90 − 40 = 50 km.
Fastest existing route M→K:
M→W→K: 120 km + 50 km, all at 120 km/h = 170/120 h = 85 min.
Via Oldtown and Gulltown: 1 h + 2 h + 20 min, far slower.
Expressway: straight MK = \(\sqrt{120^2 + 50^2} = 130\) km. It must take 85 − 20 = 65 min.
Speed = \(130 \div \frac{65}{60} = 120\) km/h.
Answer: 120 km/hr (option 4)
Read the following scenario and answer the THREE questions that follow.
The plots below depict and compare the average monthly incomes (in Rs. ’000) of males and females in ten cities of India in the years 2005 and 2015. The ten cities, marked A-J in the records, are of different population sizes. For a fair comparison, to adjust for inflation, incomes for both the periods are scaled to 2025 prices. Each red dot represents the average monthly income of females in a particular city in a particular year, while each blue dot represents the average monthly income of males in a particular city in a particular year. The gender gap for a city, for a particular year, is defined as the absolute value of the average monthly income of males, minus the average monthly income of females, in that year.
Q20
DI - Tables & Graphs MULTIPLE CHOICE
In which city did the gender gap, in terms of 2025 prices, change the least, from 2005 to 2015, in terms of percentage?
Key point: the cities have different populations, so a "combined" average for the 10 cities is a population-weighted average. Its exact value is unknown, but it must lie between the smallest and largest city values.
Option 2: men's 2005 city averages range from 30 to 40. A weighted average cannot fall below 30, so "less than Rs. 30,000" is impossible. Definitely false.
Option 1: gaps range from 2.5 to 17.5 in 2005 and from 5 to 10 in 2015 (simple means 11.25 and 7.75). A lower combined gap in 2015 is possible.
Option 3: men earned more in 2005 in A, B, C, D, H and J, which is 6 of 10 cities. True.
Option 4: women's 2015 values range from 20 to 27.5. A weighted average below 22 is possible if city D dominates, so this is not definitely false.
Option 5: in each year every city's male value exceeds its female value, and the median of the male values (32.5 in both years) exceeds that of the female values (22.5 in both years). True.
Answer: option 2 (men's combined average below Rs. 30,000 in 2005)
Q22
DI - Tables & Graphs MULTIPLE CHOICE
Rs.100 in 2025 is worth Rs. 60 in 2015 prices, and Rs. 25 in 2005 prices. Based on the given plots, which of the following statements, about the unscaled incomes, i.e., the incomes before scaling to 2025 prices, CANNOT be correct? (All statements refer to people represented in the given plots.)
Unscaling: 2015 value = scaled × 0.60 and 2005 value = scaled × 0.25. The same factors apply to the gaps.
Option 1: H women 2015: 25 × 0.6 = 15, about Rs. 15,000. Can be correct.
Option 2: G women: 30 × 0.25 = 7.5 → 27.5 × 0.6 = 16.5, a rise of 120%. Can be correct.
Option 3: E men: 32.5 × 0.25 = 8.125 → 32.5 × 0.6 = 19.5, a rise of 140%. Can be correct.
Option 4: G's gap goes from 2.5 × 0.25 = 0.625 to 5 × 0.6 = 3, so it rose. (A also rises: 3.125 → 4.5.) So the gaps did not shrink in all cities. Cannot be correct.
Option 5: unscaled values fall only if a scaled value drops by more than 58% (since 0.25/0.6 ≈ 0.42). No value drops that much, so all rise. Can be correct.
Answer: The unscaled gender gaps reduced in all 10 cities from 2005 to 2015
Read the following scenario and answer the THREE questions that follow.
GadRev is a firm that reviews different latest gadgets through a team of four reviewers (R1, R2, R3, and R4). Recently the reviewers reviewed four different tech gadgets (A, B, C, and D) on a scale of 1 to 5 (all integer values) where 1 denotes poor and 5 denotes excellent. These review ratings were then tabulated. However, due to a technical glitch, some of these ratings got deleted. The average rating given by each reviewer, and the average rating given to each gadget were earlier communicated to the team management in a separate email and hence can be useful to retrieve the deleted ratings. The available ratings along with the average ratings are represented in the following table:
Q23
DI - Tables & Graphs MULTIPLE CHOICE
What rating provided by Reviewer R1 to Gadget A can help determining the remaining ratings uniquely?
Set up: row sums (4 × average) are 16, 16, 16, 17 for R1 to R4, and column sums are 16, 16, 16, 17 for A to D.
R1: A + C = 16 − 4 − 3 = 9
R2: B + D = 9
R3: A + B + C = 11
R4: A + D = 17 − 9 = 8
Col A: R1A + R3A + R4A = 13
Col B: R2B + R3B = 8
Col C: R1C + R3C = 7
Col D: R2D + R4D = 9
Let R1A = x and R3A = p. Then:
R1C = 9 − x and R3C = 7 − (9 − x) = x − 2
R4A = R3B = 13 − x − p
R2B = R4D = x + p − 5
R2D = 14 − x − p
Bounds (each rating from 1 to 5):
R1C ≤ 5 gives x ≥ 4. With x ≤ 5, x is 4 or 5 (R3C = x − 2 is then fine).
13 − x − p ∈ [1, 5] gives x + p ∈ [8, 12].
x + p − 5 ∈ [1, 5] gives x + p ∈ [6, 10].
14 − x − p ∈ [1, 5] gives x + p ∈ [9, 13].
Together: x + p = 9 or 10.
#
R1A (x)
R3A (p)
R1C
R3C
R3B = R4A
R2B = R4D
R2D
1
4
5
5
2
4
4
5
2
5
4
4
3
4
4
5
3
5
5
4
3
3
5
4
Each row of this table satisfies all eight sums.
This question: R1A can only be 4 or 5. R1A = 4 leaves exactly one completion (row 1). R1A = 5 leaves two (rows 2 and 3). So the rating that fixes everything uniquely is 4.
Answer: 4 (option 5)
Q24
DI - Tables & Graphs MULTIPLE CHOICE
In how many different ways could Reviewer R2 have rated Gadget B so that the ratings lead to the same averages for the gadgets and the reviewers as shown in the table?
Set up: row sums (4 × average) are 16, 16, 16, 17 for R1 to R4, and column sums are 16, 16, 16, 17 for A to D.
R1: A + C = 16 − 4 − 3 = 9
R2: B + D = 9
R3: A + B + C = 11
R4: A + D = 17 − 9 = 8
Col A: R1A + R3A + R4A = 13
Col B: R2B + R3B = 8
Col C: R1C + R3C = 7
Col D: R2D + R4D = 9
Let R1A = x and R3A = p. Then:
R1C = 9 − x and R3C = 7 − (9 − x) = x − 2
R4A = R3B = 13 − x − p
R2B = R4D = x + p − 5
R2D = 14 − x − p
Bounds (each rating from 1 to 5):
R1C ≤ 5 gives x ≥ 4. With x ≤ 5, x is 4 or 5 (R3C = x − 2 is then fine).
13 − x − p ∈ [1, 5] gives x + p ∈ [8, 12].
x + p − 5 ∈ [1, 5] gives x + p ∈ [6, 10].
14 − x − p ∈ [1, 5] gives x + p ∈ [9, 13].
Together: x + p = 9 or 10.
#
R1A (x)
R3A (p)
R1C
R3C
R3B = R4A
R2B = R4D
R2D
1
4
5
5
2
4
4
5
2
5
4
4
3
4
4
5
3
5
5
4
3
3
5
4
Each row of this table satisfies all eight sums.
This question: exactly three valid completions exist (the table above).
Answer: 3 (option 4)
Read the following scenario and answer the THREE questions that follow.
In an 8-week course, a professor administered a test at the end of each week. Each of the eight tests was scored out of 4 marks, and a student could only receive a non-negative integer score.
Two students, Ravi and Sumana, took the eight tests.
In the first test, Ravi and Sumana scored the same marks. From the second to eighth tests, Ravi scored the exact same non-zero marks. Sumana scored the same marks as Ravi from the fifth test onwards. Ravi’s total marks in the first three tests was the same as Sumana’s total marks in the first two tests. Also, Sumana’s marks in the first test, total marks of the first two tests, and total marks of the eight tests are in a geometric progression.
Set up: let both score s in test 1, and let Ravi score r (1 to 4) in each of tests 2 to 8. Sumana also scores r in tests 5 to 8, and \(S_2, S_3, S_4\) in tests 2 to 4.
Ravi's first three tests = Sumana's first two: \(s + 2r = s + S_2\), so \(S_2 = 2r\). Since \(S_2 \le 4\), r is 1 or 2.
Sumana's total: \(T = s + 2r + S_3 + S_4 + 4r\).
GP: \((s + 2r)^2 = s \cdot T\). Here \(s \ne 0\), otherwise the GP would force \(2r = 0\).
Simplifying:
r = 1: \(s(2 + S_3 + S_4) = 4\)
r = 2: \(s(4 + S_3 + S_4) = 16\)
Case
r
s
S2
S3+S4
T
GP
a
1
1
2
2
9
1, 3, 9
b
1
2
2
0
8
2, 4, 8
c
2
2
4
4
18
2, 6, 18
d
2
4
4
0
16
4, 8, 16
(Other values fail: s = 4 with r = 1 needs a negative sum, and s = 1 with r = 2 needs \(S_3 + S_4 = 12\).)
Check each option:
Option 1: \(S_2 = 2r\) is 2 or 4, never 3. False.
Option 2: Sumana's 8th test = r ≤ 2, never 4. False.
Option 3: Ravi's 3rd test = r ≤ 2. False.
Option 4: s = 2 occurs in cases b and c. Can be true.
Option 5: r is non-zero. False.
Answer: Sumana scored 2 marks in the first test (option 4)
Q27
LR - Puzzles MULTIPLE CHOICE
If Ravi scored 4 marks in the first test, how many marks did Sumana score in the third test?
Set up: let both score s in test 1, and let Ravi score r (1 to 4) in each of tests 2 to 8. Sumana also scores r in tests 5 to 8, and \(S_2, S_3, S_4\) in tests 2 to 4.
Ravi's first three tests = Sumana's first two: \(s + 2r = s + S_2\), so \(S_2 = 2r\). Since \(S_2 \le 4\), r is 1 or 2.
Sumana's total: \(T = s + 2r + S_3 + S_4 + 4r\).
GP: \((s + 2r)^2 = s \cdot T\). Here \(s \ne 0\), otherwise the GP would force \(2r = 0\).
Simplifying:
r = 1: \(s(2 + S_3 + S_4) = 4\)
r = 2: \(s(4 + S_3 + S_4) = 16\)
Case
r
s
S2
S3+S4
T
GP
a
1
1
2
2
9
1, 3, 9
b
1
2
2
0
8
2, 4, 8
c
2
2
4
4
18
2, 6, 18
d
2
4
4
0
16
4, 8, 16
(Other values fail: s = 4 with r = 1 needs a negative sum, and s = 1 with r = 2 needs \(S_3 + S_4 = 12\).)
This question: Ravi's first test is 4, so s = 4. Only case d has s = 4: r = 2 and \(S_3 + S_4 = 0\). So Sumana scored 0 in the third test.
Check: Sumana's scores are 4, 4, 0, 0, 2, 2, 2, 2. Test 1 = 4, first two = 8, total = 16: a GP with ratio 2. Ravi's first three = 4 + 2 + 2 = 8 = Sumana's first two.
Answer: 0
Q28
LR - Puzzles MULTIPLE CHOICE
If Ravi scored 1 mark in the second test, what is the maximum possible value of Sumana’s total marks in all the eight tests together?
Set up: let both score s in test 1, and let Ravi score r (1 to 4) in each of tests 2 to 8. Sumana also scores r in tests 5 to 8, and \(S_2, S_3, S_4\) in tests 2 to 4.
Ravi's first three tests = Sumana's first two: \(s + 2r = s + S_2\), so \(S_2 = 2r\). Since \(S_2 \le 4\), r is 1 or 2.
Sumana's total: \(T = s + 2r + S_3 + S_4 + 4r\).
GP: \((s + 2r)^2 = s \cdot T\). Here \(s \ne 0\), otherwise the GP would force \(2r = 0\).
Simplifying:
r = 1: \(s(2 + S_3 + S_4) = 4\)
r = 2: \(s(4 + S_3 + S_4) = 16\)
Case
r
s
S2
S3+S4
T
GP
a
1
1
2
2
9
1, 3, 9
b
1
2
2
0
8
2, 4, 8
c
2
2
4
4
18
2, 6, 18
d
2
4
4
0
16
4, 8, 16
(Other values fail: s = 4 with r = 1 needs a negative sum, and s = 1 with r = 2 needs \(S_3 + S_4 = 12\).)
This question: r = 1 leaves cases a (T = 9) and b (T = 8). The maximum possible total is 9.
Example: Sumana scores 1, 2, 1, 1, 1, 1, 1, 1. Test 1 = 1, first two = 3, total = 9, a GP with ratio 3.
