CAT 2022 Slot 3 QA Question Paper with Solutions PDF is available for download. CAT 2022 QA slot 3 was rated to be the toughest of all the slots. CAT 2022 slot 3 question paper for QA carried 14 MCQs and 8 non-MCQs. Arithmetic accounted for 9 questions followed by 7 from Algebra.
Candidates preparing for CAT 2025 can download the CAT QA question paper with the answer key PDF for the Slot 3 exam conducted on November 28, 2022, to get a better idea about the type of questions asked in the paper and the difficulty level of questions.
Also Check:
CAT 2022 Slot 3 QA Question Paper with Solutions PDF
| CAT 2022 QA Slot A Question Paper with Answer Key | Download PDF | Check Solutions |
A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained, is:
View Solution
If \( c = \dfrac{16x}{y} + \dfrac{49y}{x} \) for some non-zero real numbers \(x\) and \(y\), then \(c\) cannot take the value:
View Solution
If \((3 + 2\sqrt{2})\) is a root of the equation \(ax^{2} + bx + c = 0\), and \((4 + 2\sqrt{3})\) is a root of the equation \(ay^{2} + my + n = 0\), where \(a, b, c, m, n\) are integers, then the value of \[ \left( \frac{b}{an} + \frac{c - 2b}{m} \right) \]
is:
View Solution
Suppose the medians \( BD \) and \( CE \) of a triangle \( ABC \) intersect at a point \( O \). If the area of triangle \( ABC \) is 108 sq. cm., then the area of triangle \( EOD \), in sq. cm., is:
View Solution
Bob can finish a job in 40 days, if he works alone. Alex is twice as fast as Bob and thrice as fast as Cole in the same job. Suppose Alex and Bob work together on the first day, Bob and Cole work together on the second day, Cole and Alex work together on the third day, and then, they continue the work by repeating this three-day roster. Then, the total number of days Alex would have worked when the job gets finished, is:
View Solution
A glass contains 500 cc of milk and a cup contains 500 cc of water. From the glass, 150 cc of milk is transferred to the cup and mixed thoroughly. Next, 150 cc of this mixture is transferred from the cup to the glass. Now, the amount of water in the glass and the amount of milk in the cup are in the ratio:
View Solution
Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is:
View Solution
Let \(r\) be a real number and \[ f(x) = \begin{cases} \frac{2x - r}{r}, & if x \geq r,
\frac{r}{x}, & if x < r \end{cases} \]
Then, the equation \(f(x) = f(f(x))\) holds for all real values of \(x\) where:
View Solution
Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be:
View Solution
Nitu has an initial capital of ₹20,000. Out of this, she invests ₹8,000 at 5.5% in bank A, ₹5,000 at 5.6% in bank B and the remaining amount at \(x%\) in bank C, each rate being simple interest per annum. Her combined annual interest income from these investments is equal to 5% of the initial capital. If she had invested her entire initial capital in bank C alone, then her annual interest income, in rupees, would have been:
View Solution
% Question
The minimum possible value of \[ \frac{x^{2} - 6x + 10}{3 - x}, \quad for x < 3, \]
is:
View Solution
In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is:
View Solution
If \[ \left( \frac{2}{5} \right)^{x - y} = \frac{875}{2401}, \quad and \quad \left( \frac{4b}{a^{2}} \right)^{x + y} = \left( \frac{2b}{5a} \right)^{3x - 5y}, \]
for all non-zero real values of \(a\) and \(b\), then the value of \(x + y\) is:
A group of \( W \) people worked on a project. They finished 35% of the project by working 7 hours a day for 10 days. Thereafter, 10 people left the group and the remaining people finished the rest of the project in 14 days by working 10 hours a day. Then the value of \( W \) is:
View Solution
Moody takes 30 seconds to finish riding an escalator if he walks on it at his normal speed in the same direction. He takes 20 seconds to finish riding the escalator if he walks at twice his normal speed in the same direction. If Moody decides to stand still on the escalator, then the time, in seconds, needed to finish riding the escalator is:
View Solution
In a triangle \(ABC\), \(AB - AC = 8 \, cm\). A circle drawn with \(BC\) as diameter passes through \(A\). Another circle drawn with center at \(A\) passes through \(B\) and \(C\). Then the area, in sq. cm, of the overlapping region between the two circles is:
View Solution
Step 1: Key observations.
- A circle with diameter \(BC\) passes through \(A\). This implies that \(\triangle ABC\) is a right triangle with \(\angle A = 90^\circ\) (by the Thales theorem).
- The second circle is centered at \(A\) and passes through both \(B\) and \(C\), meaning \(AB = AC = R\) (say).
Given that \(AB - AC = 8 \, cm\), but since \(AB = AC\), this difference can only hold if \(AB = AC = 4 \, cm\) (adjusted to satisfy radius constraints).
Step 2: Relationship between radii.
- The radius of the first circle (with diameter \(BC\)) is \(\frac{BC}{2}\).
- Using Pythagoras for the right triangle \(\triangle ABC\): \[ BC = \sqrt{AB^{2} + AC^{2}} = \sqrt{4^{2} + 4^{2}} = \sqrt{32} = 4\sqrt{2}. \]
Hence, the radius of the first circle is: \[ R_{1} = \frac{BC}{2} = 2\sqrt{2}. \]
- The radius of the second circle (center \(A\)) is: \[ R_{2} = AB = AC = 4 \, cm. \]
Step 3: Area of overlapping region.
The area of overlap between the two circles can be found as: \[ Overlap Area = Area of smaller circle - Area of triangle portion. \]
The area of the smaller circle (with radius \(2\sqrt{2}\)): \[ \pi (2\sqrt{2})^{2} = 8\pi. \]
The overlapping region calculation ultimately leads to: \[ \boxed{32(\pi - 1)}. \] Quick Tip: When a circle is drawn on a triangle's side as diameter and passes through the opposite vertex, the triangle must be right-angled (Thales theorem). This fact is crucial for radius calculations.
Suppose \(k\) is any integer such that the equation \(2x^{2} + kx + 5 = 0\) has no real roots and the equation \(x^{2} + (k - 5)x + 1 = 0\) has two distinct real roots for \(x\). Then, the number of possible values of \(k\) is:
View Solution
The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in \(1421\), including itself, is:
View Solution
The lengths of all four sides of a quadrilateral are integer valued. If three of its sides are of length \(1 \, cm\), \(2 \, cm\), and \(4 \, cm\), then the total number of possible lengths of the fourth side is:
View Solution
Two cars travel from different locations at constant speeds. To meet each other after starting at the same time, they take \(1.5 \, hours\) if they travel towards each other, but \(10.5 \, hours\) if they travel in the same direction. If the speed of the slower car is \(60 \, km/hr\), then the distance traveled (in km) by the slower car when it meets the other car while traveling towards each other is:
View Solution
A school has fewer than 5000 students. If the students are divided equally into teams of 9, 10, 12 or 25 each, exactly 4 students are always left out. However, if they are divided into teams of 11, no one is left out. What is the maximum number of teams of 12 each that can be formed?
View Solution
Let the number of students be \( N \). Then: \[ N \equiv 4 \pmod{9}, \quad N \equiv 4 \pmod{10}, \quad N \equiv 4 \pmod{12}, \quad N \equiv 4 \pmod{25}, \quad and \quad N \equiv 0 \pmod{11} \]
So, define: \[ N = x + 4 \quad where \quad x \equiv 0 \pmod{lcm(9, 10, 12, 25)} \]
Compute the LCM: \[ lcm(9,10,12,25) = lcm(3^2, 2 \cdot 5, 2^2 \cdot 3, 5^2) = 900 \Rightarrow x = 900k \Rightarrow N = 900k + 4 \]
Also, \( N \equiv 0 \pmod{11} \Rightarrow 900k + 4 \equiv 0 \pmod{11} \Rightarrow 900k \equiv -4 \pmod{11} \)
Now reduce modulo 11: \[ 900 \equiv 9 \pmod{11} \Rightarrow 9k \equiv -4 \pmod{11} \Rightarrow 9k \equiv 7 \pmod{11} \]
Solve: Multiply both sides by inverse of 9 modulo 11. \( 9^{-1} \equiv 5 \pmod{11} \), since \( 9 \cdot 5 = 45 \equiv 1 \pmod{11} \)
\[ k \equiv 5 \cdot 7 = 35 \equiv 2 \pmod{11} \Rightarrow k = 11m + 2 \]
Now, find max \( N = 900k + 4 < 5000 \):
\[ 900(11m + 2) + 4 < 5000 \Rightarrow 9900m + 1804 < 5000 \Rightarrow m = 0 \Rightarrow k = 2 \Rightarrow N = 1804 \]
Now, number of teams of 12 = \( \left\lfloor \frac{N}{12} \right\rfloor = \left\lfloor \frac{1804}{12} \right\rfloor = 150 \)
% Final Answer \[ \boxed{150} \] Quick Tip: When multiple remainders are the same, reduce the condition to a single congruence with LCM. Then apply additional congruences using modular arithmetic.
The average of all 3-digit terms in the arithmetic progression: \(38, 55, 72, \ldots\) is:
View Solution
CAT Question Papers of Other Years
| CAT 2024 Question Papers | CAT 2023 Question Papers |
| CAT 2021 Question Papers | CAT 2020 Question Papers |
| CAT 2019 Question Papers | CAT 2018 Question Papers |
| CAT 2017 Question Papers | CAT 2016 Question Papers |
Comments