Chapter-wise representative MCQs
Below are representative MCQs covering the high-yield topics. Solve them under time pressure to mimic exam conditions.
Quantitative Aptitude — Arithmetic: Percentage & Ratio
Q1. Rajesh and Vimal own 20 hectares and 30 hectares of agricultural land, respectively, which are entirely covered by wheat and mustard crops. The cultivation area of wheat and mustard in the land owned by Vimal are in the ratio of $5: 3$. If the total cultivation area of wheat and mustard are in the ratio $11: 9$, then the ratio of cultivation area of wheat and mustard in the land owned by Rajesh is
1) $4: 3$
2) $3: 7$
3) $1: 1$
4) $7: 9$
Solution: Let the wheat and mustard areas on Vimal's 30 hectares be $5x$ and $3x$ respectively.
So, total area for Vimal = $5x + 3x = 8x = 30 \Rightarrow x = \frac{30}{8} = 3.75$
Therefore, wheat on Vimal’s land = $5x = 18.75$ hectares
Mustard on Vimal’s land = $3x = 11.25$ hectares
Now, total land = $20$ (Rajesh) $+$ $30$ (Vimal) = $50$ hectares
Total land for wheat = $\frac {11}{20} \times 50$ = 27.5 hectares
So, Total land for mustard = $\frac {9}{20} \times 50$ = 22.5 hectares
Now, for Rajesh land for wheat = $27.5-18.75=8.75$ hectares
land for mustard for Rajesh = $22.5-11.25=11.25$ hectares
So, ratio of wheat ($w$) to mustard ($m$) on Rajesh’s land:
$\frac{w}{m} = \frac{8.75}{11.25} = \frac{7}{9}$
Thus, the ratio is $7:9$
Hence, the correct answer is option 4.
Q2. After two successive increments, Gopal's salary became $187.5 \%$ of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was
1) 25
2) 20
3) 27.5
4) 30
Solution: Let initial salary = $100$
Let the first increment be $x%$
Then salary becomes: $100 \times \left(1 + \frac{x}{100}\right) = 100 + x$
Second increment = $2x%$
Final salary = $(100 + x) \times \left(1 + \frac{2x}{100}\right)$
Given final salary = $187.5%$ of original salary = $187.5$
So: $
(100 + x)\left(1 + \frac{2x}{100}\right) = 187.5
$
$⇒
(100 + x)\left(\frac{100 + 2x}{100}\right) = 187.5
$
$⇒
(100 + x)(100 + 2x) = 18750
$
$⇒
10000 + 200x + 100x + 2x^2 = 18750$
$
\Rightarrow 2x^2 + 300x + 10000 = 18750$
$
\Rightarrow 2x^2 + 300x - 8750 = 0
$
$⇒
x^2 + 150x - 4375 = 0
$
$⇒
x^2 + 175x-25x - 4375 = 0
$
$⇒
(x + 175)(x-25) = 0
$
Take positive root: $
x = 25
$
$25\%$ increase in the first increment.
Hence, the correct answer is option 1.
Quantitative Aptitude — Number Systems / Remainders
Q3. The number of all positive integers up to 500 with non-repeating digits is
Solution: We want to count all positive integers $\leq 500$ with non-repeating digits.
We'll consider $1$-digit, $2$-digit, and $3$-digit numbers, ensuring they are $\leq 500$ and digits do not repeat.
Case 1: 1-digit numbers (1 to 9):
All digits are unique. So, the total possible such numbers = 9
Case 2: 2-digit numbers (10 to 99):
Digit at Tens place can be from 1 to 9, so possible ways to fill tens place = 9
Digit at Units place can be any digit except the digit at tens place ie. 9 choices
So, the total possible such numbers = $9 \times 9 = 81$
Case 3: 3-digit numbers (100 to 499):
We limit to numbers from $100$ to $499$ (since $\leq 500$). So hundreds digit = $1$ to $4$ (4 choices)
For each:
Tens digit: choose from $0$–$9$, excluding hundreds digit $\Rightarrow$ 9 choices
Unit digit: choose from remaining digits $\Rightarrow$ 8 choices
So, the total possible such numbers = $4 \times 9 \times 8 = 288$
Case 4: 500
Digits: $5, 0, 0$ → Repeated digits → Not allowed.
So, exclude $500$
Total count$ = 9 + 81 + 288 = 378$
Hence, the correct answer is $378$.
Quantitative Aptitude — Algebra (Linear & Quadratic)
Q4. The sum of all distinct real values of $x$ that satisfy the equation $10^x+\frac{4}{10^x}=\frac{81}{2}$, is
1) $4 \log _{10} 2$
2) $3 \log _{10} 2$
3) $2 \log _{10} 2$
4) $\log _{10} 2$
Solution: Let $y = 10^x$
Then the equation becomes:
$y + \frac{4}{y} = \frac{81}{2}$
$⇒y^2 + 4 = \frac{81}{2}y$
$⇒2y^2 + 8 = 81y$
$⇒2y^2 - 81y + 8 = 0$
Since $y = 10^x$ must be positive, both roots are valid as long as they are positive real numbers.
Let the roots be $y_1$ and $y_2$. Then:
So, the roots of original equation, $x_1 = \log_{10} y_1$ or $x_2= \log_{10} y_2$
We know: $y_1 y_2 = \frac{8}{2} = 4$
So: $x_1 + x_2 = \log_{10}(y_1y_2)= \log_{10}(4) = \log_{10} 4$
The sum of all distinct real values of $x$ is $2 \log _{10} 2$.
Hence, the correct answer is option 3.
Q5. For some constant real numbers $p, k$ and $a$, consider the following system of linear equations in $x$ and $y$ :
$\begin{aligned}
& p x-4 y=2 \\
& 3 x+k y=a
\end{aligned}$
A necessary condition for the system to have no solution for ( $x, y$ ), is
1) $2 a+k \neq 0$
2) $k p+12 \neq 0$
3) $a p+6=0$
4) $a p-6=0$
Solution: The given equations are:
$px - 4y = 2$
$3x + ky = a$
For a system of two linear equations $a_1x+b_1y=c_1$ and $a_2x+b_2y=c_2$, the condition for no solution is given by
$\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
So, from the given set of equations
$\frac{p}{3}=\frac{-4}{k} \neq \frac{2}{a}$
$⇒\frac{-4}{k} \neq \frac{2}{a}$
$⇒-4a \neq 2k$
$⇒2k+4a \neq 0$
$⇒k+2a \neq 0$
Hence, the correct answer is option 1.
Quantitative Aptitude — Modern Maths
Q6. All the values of $x$ satisfying the inequality $\frac{1}{x+5} \leq \frac{1}{2 x-3}$ are
1) $-5<x<\frac{3}{2}$ or $\frac{3}{2}<x \leq 8$
2) $x<-5$ or $x>\frac{3}{2}$
3) $x<-5$ or $\frac{3}{2}<x \leq 8$
4) $-5<x<\frac{3}{2}$ or $x>\frac{3}{2}$
Solution: Given: $\frac{1}{x+5} \leq \frac{1}{2x - 3}$
$⇒2x-3 \leq x+5$
$⇒x \leq 8$
But $x \neq -5$ and $x \neq \frac 32$ as at these values fractions are not defined.
So, the set of solutions becomes
$x<-5$ or $\frac{3}{2}<x \leq 8$
Hence, the correct answer is option 3.
Q7. For any non-zero real number $x$, let $f(x)+2 f\left(\frac{1}{x}\right)=3 x$. Then, the sum of all possible values of $x$ for which $f(x)=3$, is
1) -2
2) 2
3) -3
4) 3
Solution: We are given:
$
f(x) + 2f\left(\frac{1}{x}\right) = 3x \quad-------- \text{(1)}
$
Also, we are told: $ f(x) = 3 $
Substituting into (1):
$
3 + 2f\left(\frac{1}{x}\right) = 3x \Rightarrow f\left(\frac{1}{x}\right) = \frac{3x - 3}{2} \quad --------\text{(2)}
$
Now use (1) again but switch $ x $ with $ \frac{1}{x} $:
$
f\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x}
$
Substitute $ f(x) = 3 $:
$
f\left(\frac{1}{x}\right) + 6 = \frac{3}{x} \Rightarrow f\left(\frac{1}{x}\right) = \frac{3}{x} - 6 \quad------ \text{(3)}
$
Equating (2) and (3):
$
\frac{3x - 3}{2} = \frac{3}{x} - 6
$
Multiply both sides by 2:
$
3x - 3 = \frac{6}{x} - 12
$
Bring all terms to one side:
$
3x + 9 - \frac{6}{x} = 0
$
Multiply through by $ x $:
$
3x^2 + 9x - 6 = 0
$
Divide by 3:
$
x^2 + 3x - 2 = 0
$
Solve using quadratic formula:
$
x = \frac{-3 \pm \sqrt{3^2 + 4 \cdot 1 \cdot 2}}{2} = \frac{-3 \pm \sqrt{17}}{2}
$
Sum of all possible values of $ x $ is:
$
\frac{-3 + \sqrt{17}}{2} + \frac{-3 - \sqrt{17}}{2} = -3
$
Hence, the correct answer is option 3.
Quantitative Aptitude — Geometry and Mensuration
Q8. A circular plot of land is divided into two regions by a chord of length $10 \sqrt{3}$ meters such that the chord subtends an angle of $120^{\circ}$ at the center. Then, the area, in square meters, of the smaller region is
1) $20\left(\frac{4 \pi}{3}+\sqrt{3}\right)$
2) $20\left(\frac{4 \pi}{3}-\sqrt{3}\right)$
3) $25\left(\frac{4 \pi}{3}-\sqrt{3}\right)$
4) $25\left(\frac{4 \pi}{3}+\sqrt{3}\right)$
Solution: The chord subtends an angle of $120^\circ$ at the center.
Let the radius of the circle be $r$.
In a triangle formed by two radii and the chord, apply the Cosine Rule:
Using the cosine rule in $\triangle AOB$ (where $AB = 10\sqrt{3}$ and angle $\angle AOB = 120^\circ$):
$
AB^2 = AO^2 + BO^2 - 2 \cdot AO \cdot BO \cdot \cos(120^\circ)
$
$
⇒(10\sqrt{3})^2 = 2r^2 - 2r^2 \cdot \cos(120^\circ)
$
$
⇒300 = 2r^2 - 2r^2 \cdot \left(-\frac{1}{2}\right)
$
$
⇒300 = 2r^2 + r^2 = 3r^2 \Rightarrow r^2 = 100 \Rightarrow r = 10
$
Now, area of the smaller region = area of sector $AOB$ – area of triangle $AOB$
Angle at center = $120^\circ = \frac{2\pi}{3}$ radians
Radius = $10$
Area of sector: $
\frac{1}{2} \cdot r^2 \cdot \theta = \frac{1}{2} \cdot 100 \cdot \frac{2\pi}{3} = \frac{100\pi}{3}
$
Area of triangle (using formula for triangle with angle between two sides): $
\frac{1}{2} \cdot r^2 \cdot \sin(120^\circ) = \frac{1}{2} \cdot 100 \cdot \frac{\sqrt{3}}{2} = 25\sqrt{3}
$
Area of smaller region (segment) $ =
\frac{100\pi}{3} - 25\sqrt{3}
=25\left(\frac{4 \pi}{3}-\sqrt{3}\right)$ sq. meters.
Hence, the correct answer is option 3.
DILR — (Data Interpretation)
Q9. Comprehension:
Out of 10 countries -- Country 1 through Country 10 -- Country 9 has the highest gross domestic product (GDP), and Country 10 has the highest GDP per capita. GDP per capita is the GDP of a country divided by its population. The table below provides the following data about Country 1 through Country 8 for the year 2024.
- Column 1 gives the country's identity.
- Column 2 gives the country's GDP as a fraction of the GDP of Country 9.
- Column 3 gives the country's GDP per capita as a fraction of the GDP per capita of Country 10.
- Column 4 gives the country's annual GDP growth rate.
- Column 5 gives the country's annual population growth rate.
Country | GDP | GDP per Capita | GDP growth rate | Population growth rate |
Country 1 | 0.15 | 0.41 | 0.2% | -0.12% |
Country 2 | 0.14 | 0.25 | 0.9% | -0.41% |
Country 3 | 0.13 | 0.02 | 6.5% | 0.70% |
Country 4 | 0.12 | 0.38 | 0.5% | 0.49% |
Country 5 | 0.10 | 0.36 | 0.7% | 0.31% |
Country 6 | 0.08 | 0.08 | 3.2% | 0.61% |
Country 7 | 0.08 | 0.30 | 0.7% | -0.11% |
Country 8 | 0.07 | 0.41 | 1.2% | 0.71% |
Assume that the GDP growth rates and population growth rates of the countries will remain constant for the next three years.
Question: Which one among the countries 1 through 8 has the smallest population in $2024 ?$
1) Country 8
2) Country 3
3) Country 7
4) Country 5
Solution: When we need to compare two countries in the list, either by GDP, GDP per capita or population, we may simply use the values found in the table, as they all depend on a common reference point.
In this question, we need to compare the populations of the countries.
Population = $\frac{GDP}{\text{GDP per capita}}$
So, let's analyse the options:
Option A: Country 8
GDP = 0.07 and GDP Per Capita = 0.41
So, Population = $\frac{0.07}{0.41} \ or \ \frac{7}{41}$
Option B: Country 3
GDP = 0.13 and GDP Per Capita = 0.02
So, Population = $\frac{0.13}{0.02} \ or \ \frac{13}{2}$
Option C: Country 7
GDP = 0.08 and GDP Per Capita = 0.3
Population=$\frac{0.08}{0.3} \ or \ \frac{8}{30}$
Option D: Country 5
GDP = 0.1 and GDP Per Capita = 0.36
Population=$\frac{0.1}{0.36} \ or \ \frac{10}{36}$
Comparing the four values, we are looking for the smallest value among the four,
$\frac{7}{41}, \frac{13}{2}, \frac{8}{30}, \frac{10}{36}$
From the above four $\frac{7}{41}$ is the smallest.
Hence, the first option is correct.
DILR — (Logical Reasoning)
Q10. Comprehension:
Eight gymnastics players numbered 1 through 8 underwent a training camp where they were coached by three coaches - Xena, Yuki, and Zara. Each coach trained at least two players. Yuki trained only even numbered players, while Zara trained only odd numbered players. After the camp, the coaches evaluated the players and gave integer ratings to the respective players trained by them on a scale of 1 to 7 , with 1 being the lowest rating and 7 the highest.
The following additional information is known.
1. Xena trained more players than Yuki.
2. Player-1 and Player-4 were trained by the same coach, while the coaches who trained Player-2, Player-3 and Player-5 were all different.
3. Player-5 and Player-7 were trained by the same coach and got the same rating. All other players got a unique rating.
4. The average of the ratings of all the players was 4.
5. Player-2 got the highest rating.
6. The average of the ratings of the players trained by Yuki was twice that of the players trained by Xena and two more than that of the players trained by Zara.
7. Player-4's rating was double of Player-8's and less than Player-5's.
Question: What was the rating of Player-6?
Solution: From conditions 3 and 4, only two players got the same ratings. Thus, let $K$ be the repetitive number, then:
$\frac{(1+2+3+4+5+6+7+K)}{8} = 4$
It means $K$ = 4
Thus, players 5 and 7 got the same rating. As Xena trained more players than Yuki. So, there are two possibilities, which are as follows:
Trainer | Possibility 1 | Possibility 2 | Average |
Xena | 3 | 4 | X |
Yuki | 2 | 2 | 2X |
Zara | 3 | 2 | 2X - 2 |
Possibility 1, X = $\frac{38}{13}$, (Not possible)
Possibility 2, X = 3, (Possible)
Trainer | Number of players trained | Average Scores | Total Scores |
Xena | 4 | 3 | 12 |
Yuki | 2 | 6 | 12 |
Zara | 2 | 4 | 8 |
Step 2:
From condition 5, player 2 got the rating of 7. Also, player 4 got the rating of 2, and player 8 got the rating of 1.
So, we get:
Player | Rating | Trained by |
1 |
|
|
2 | 7 |
|
3 |
|
|
4 | 2 |
|
5 | 4 |
|
6 |
|
|
7 | 4 |
|
8 | 1 |
|
Yuki has 2 players, and their total score is 12. The only available combination that works is 7 + 5. Because player-2 got a score of 7, the other score can only be that of player-6, that is 5. This is also consistent with the fact that the only even-numbered player left is player-6, and Yuki trained only even-numbered players.
Therefore, Yuki trains player-2 and player-6.
The overall score of Zara is 8. The correct combinations used in 8 are 2+6 and 4+4.
The point of 2 is player 4, but Zara trained only those players who have an odd number; therefore, it is not possible.
The only valid move is 4 + 4, which can be given to player-5 and player-7, both belong to odd-numbered.
Therefore, Zara has to be above player-5 and player-7.
Thus, the table of the players and their trainers is as follows:
Player | Rating | Trained by |
1 | 3/6 | Xena |
2 | 7 | Yuki |
3 | 6/3 | Xena |
4 | 2 | Xena |
5 | 4 | Zara |
6 | 5 | Yuki |
7 | 4 | Zara |
8 | 1 | Xena |
Thus, the rating of player 6 is 5.
Hence, 5 is the correct answer.
VARC — Reading Comprehension
Q11. The passage given below is followed by four alternate summaries. Choose the option that best captures the essence of the passage.
When the tradwife puts on that georgic, pinstriped dress, she is not just admiring the visual cues of a fantastical past. She takes these dreams of storybook bliss literally, tracing them backward in time until she reaches a logical conclusion that satisfies her. And by doing so, she ends up delivering an unhappy reminder of just how much our lives consist of artifice and playacting. The tradwife outrages people because of her deliberately regressive ideals. And yet her behaviour is, on some level, indistinguishable from the nontradwife's. The tradwife's trollish genius is to beat us at our own dress-up game. By insisting that the idyllic cottage daydream should be real, right down to the primitive gender roles, she leaves others feeling hollow, cheated. The hullabaloo and headaches she causes may be the price we pay for taking too many things at face value: our just deserts, served Instagram-perfect by a manicured hand on a gorgeous ceramic dish, with fat, mouthwatering maraschino cherries on top.
1) The tradwife's vintage dress and adherence to traditional roles reveal the artificial nature of modern life and its superficial values.
2) By promoting an idealized past, the tradwife exposes the artifice of contemporary values and mocks societal norms.
3) The tradwife, with her vintage dress and traditional roles, highlights the superficiality of modern life and challenges current societal norms.
4) The tradwife's commitment to outdated gender roles and retro fashion critiques the superficiality of today's societal ideals.
Solution: Option 3 is the most accurate because it captures the two main ideas of the passage:
The tradwife uses old-fashioned clothing and traditional gender roles to create a staged version of the past. In doing so, she ends up exposing how shallow and artificial modern life and social behaviour have become, making people uncomfortable by reflecting their own pretence back at them.
The other options miss either the element of social critique or the idea of exposing superficiality. Option 3 includes both.
