CAT 2025 Slot 1 Question Paper with Solutions
Candidates are advised to refer to this page for the complete memory-based questions and detailed solutions from CAT 2025 Slot 1. This section has been updated with memory-based questions.
CAT 2025 Slot 1 Questions with Solutions: Quantitative Ability (Memory-Based Questions)
Q 1. The number of distinct pairs of integers satisfying
$x > y \ge 3$ and $x + y < 14$ is __.
Solution:
We need integer pairs $(x, y)$ with $y \geq 3, x>y$, and $x+y<14$, i.e. $x+y \leq 13$.
For a fixed $y$ we need integers $x$ satisfying $y+1 \leq x \leq 13-y $
This requires $y+1 \leq 13-y$, so $2 y \leq 12$ and hence $y \leq 6$. Thus $y=3,4,5,6$.
Number of $x$ values for each $y$ is $(13-y)-(y+1)+1=13-2 y$ :
$y=3: 13-2 \cdot 3=7$ pairs $(4,3),(5,3), \ldots,(10,3)$.
$y=4: 13-2 \cdot 4=5$ pairs $(5,4), \ldots,(9,4)$.
$y=5$ : $13-2 \cdot 5=3$ pairs $(6,5),(7,5),(8,5)$.
$y=6: 13-2 \cdot 6=1$ pair $(7,6)$.
Total $7+5+3+1=16$.
Q 2. Arun, Tarun and Varun work for 24, 21 and 15 days respectively and get paid 2160, 2400 and 2160 rupees respectively.
They get paid the same even if they work for a partial day. If the work has to be completed within 10 days or less, what is the minimum amount that has to be paid to complete the entire task?
Solution:
Arun and Varun are the cheapest workers.
Daily pay:
Arun $=$ ₹ $90 /$ day, rate $=1 / 24$ job $/$ day
Varun = ₹144/day, rate = $1 / 15$ job/day
Tarun costs more, so exclude him.
Together in 10 days:
$\frac{10}{24}+\frac{10}{15}>1$
So the work can be completed within 10 days by Arun and Varun.
Since both cost ₹2160 per full job, any combination using only them will cost ₹2160.
Q 3. The number of integer values of $k$ for which the quadratic equation
$x^2 - 5x + k = 0$ has real roots is __.
Solution:
For real roots, the discriminant must be non-negative:
$\begin{gathered}b^2-4 a c \geq 0 \\ (-5)^2-4(1)(k) \geq 0 \\ 25-4 k \geq 0\end{gathered}$
$
-4 k \geq-25 \Rightarrow k \leq \frac{25}{4}
$
$k \leq 6.25 \Rightarrow k \leq 6$ (for integer values)
∴ All integer values of $k$ are: $\{\ldots,-3,-2,-1,0,1,2,3,4,5,6\}$
All integers $k \leq 6$
Number of integer values = infinitely many
Q 4. The number of non-negative values of $n$ for which
$\log_{1/4}(n^2 - 7n + 14) > 0$ is __.
Solution: We know:
- Base $1/4 < 1$, so
$\log_{1/4}(x) > 0 \Rightarrow 0 < x < 1$.
Here,
$x = n^2 - 7n + 14$.
So we need:
$0 < n^2 - 7n + 14 < 1$
Step 1: Solve
$n^2 - 7n + 14 > 0$
This quadratic is always positive because its discriminant is:
$D = (-7)^2 - 4(1)(14) = 49 - 56 = -7 < 0$
So this part is true for all real $n$.
Step 2: Solve
$n^2 - 7n + 14 < 1$
$n^2 - 7n + 13 < 0$
Discriminant:
$D = (-7)^2 - 4(1)(13) = 49 - 52 = -3 < 0$
Since the quadratic is always positive, it can never be < 0.
So no real $n$ satisfies the inequality.
Answer: 0
Q 5. In a cafeteria, there are 5 breads. One can choose 1 bread from the available breads, either small or large sized, and can choose up to 2 sauces from 6 available sauces. What is the number of different ways one can place an order?
Solution:
5 types of bread
For each bread, choose size: small or large → 2 choices
Sauce choice: choose 0, 1, or 2 sauces from 6 available
Number of ways to choose sauces:
$^6C_0 + ^6C_1 + ^6C_2 = 1 + 6 + 15 = 22$
Total ways:
$5 \times 2 \times 22 = 220$
Answer: 220
Q 6. If the length of each side of a rhombus is $36$ cm, and the area of the rhombus is $396$ cm$^2$, then what is the absolute value of the difference between the lengths of its diagonals?
Solution:
Area of rhombus:
$A = \frac{d_1 d_2}{2}$
So:
$\frac{d_1 d_2}{2} = 396$
$d_1 d_2 = 792$
Also, using rhombus property:
$36^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2$
$1296 = \frac{d_1^2 + d_2^2}{4}$
$d_1^2 + d_2^2 = 5184$
Now compute:
$(d_1 - d_2)^2 = d_1^2 + d_2^2 - 2d_1 d_2$
$(d_1 - d_2)^2 = 5184 - 1584 = 3600$
$|d_1 - d_2| = \sqrt{3600} = 60$
Answer: 60 cm
Q 7. A shopkeeper allows a discount of $22%$ on the market price of each chair, gives 13 chairs to a customer for the discounted price of 12 chairs, and wants to earn a profit of $26%$. The CP of each chair is $100$. Calculate the market price of each chair.
Solution:
Given:
ITM Business School MBA Admissions 2026
400+ Company Visits | Highest CTC Offered: 21 LPA | Average CTC Offered: 8.65 LPA | 100% Placement Assurance
UPES MBA Admissions 2026
Last Date to Apply: 26th March | Ranked #36 amongst institutions in Management by NIRF | 100% Placement
So required SP per chair:
$100 \times 1.26 = 126$
Discount offered = $22%$
Let market price be $M$.
After discount:
$M(1 - 0.22) = 0.78M$
But the customer gets 13 chairs for price of 12 ⇒ shopkeeper’s revenue per chair:
$\frac{0.78M \times 12}{13}$
This must equal desired SP = 126.
So:
$\frac{12 \times 0.78M}{13} = 126$
$0.72M = 126$
$M = \frac{126}{0.72} = 175$
Answer: Market Price = 175
CAT 2025 Slot 1 Questions with Solutions: Data Interpretation & Logical Reasoning
Will be updated soon.
CAT 2025 Slot 1 Questions with Solutions: Verbal Ability & Reading Comprehension
Will be updated soon.
Also Read:
CAT 2025 Slot 2 Question Paper with Solutions
