Ques that can be challenged:
Question 1: Log(1/4) of (n² − 7n + 11) > 0 is _. Answer 0 not 2 as both values of n² − 7n + 11 are negative and log cant be negative!
https://iimking.com/modern-maths-cat-2025-slot-1-actual-questions/
Question 2: Let 3 ≤ x ≤ 6 and [x²] = ([x])² – Options 1 2 4 are possible. Correct answer should be (3, √10] ∪ [5, √26) ∪ [4, √17] ∪ {6}
https://iimking.com/equations-algebra-cat-2025-slot-1-actual-questions/
Breakup G Strategy | Percentages + Equations | EASY | CAT 2025 Slot 1
Kamala divided her investment of Rs 100000 between stocks, bonds, and gold. Her investment in bonds was 25% of her investment in gold. With annual returns of 10%, 6%, 8% on stocks, bonds, and gold, respectively, she gained a total amount of Rs 8200 in one year. The amount, in rupees, that she gained from the bonds, was _____
Answer
(Total investment 100000; B = 25% of G; returns: stocks 10%, bonds 6%, gold 8%; total gain 8200) — find gain from bonds.
Let bonds = B, gold = G, stocks = S. Given B = G/4 ⇒ G = 4B.
Total money: S + B + G = 100000 ⇒ S + B + 4B = 100000 ⇒ S = 100000 − 5B.
Total gain: 0.10·S + 0.06·B + 0.08·G = 8200.
Substitute S and G: 0.10(100000 − 5B) + 0.06B + 0.08(4B) = 8200.
Compute: 10000 − 0.5B + 0.06B + 0.32B = 8200 ⇒ 10000 − 0.12B = 8200.
Solve: 0.12B = 1800 ⇒ B = 1800/0.12 = 15000.
Gain from bonds = 6% of B = 0.06 × 15000 = 900.
Answer (gain from bonds) = Rs 900.
Breakup 1.2/0.8 | Percentages | MEDIUM | CAT 2025 Slot 1
A shopkeeper offers a discount of 22% on the marked price of each chair, and gives 13 chairs to the customer for the discounted price of 12 chairs to earn a profit of 26% on the transaction. If the cost price of each chair is Rs 100, then the marked price, in rupees, of each chair is
Answer
Percentages 1.2 / 0.8
Let marked price be M. Discount 22% ⇒ selling price per chair = 0.78M.
Customer pays for 12 discounted chairs but receives 13 chairs, so total revenue = 12 × 0.78M = 9.36M.
Total cost = 13 × 100 = 1300. For 26% profit, revenue = 1.26 × 1300 = 1638.
So 9.36M = 1638 ⇒ M = 1638 / 9.36.
Compute M = 163800 / 936 = 2275 / 13 = 175.
Answer: Rs 175 per chair.
Breakup G Strategy | Percentages + Maxima Minima | MEDIUM | CAT 2025 Slot 1
In a class, there were more than 10 boys and a certain number of girls. After 40% of the girls and 60% of the boys left the class, the remaining number of girls was 8 more than the remaining number of boys. Then, the minimum possible number of students initially in the class was _____.
Answer
Let B = number of boys, G = number of girls (both integers), and B > 10.
After departures: remaining girls = 60% of G = 3G/5, remaining boys = 40% of B = 2B/5.
Given 3G/5 = 2B/5 + 8. Multiply by 5: 3G = 2B + 40. (Equation ★)
For 3G to be integer, G must be integer. For the remaining counts 3G/5 and 2B/5 to be integers (no fractional people), G and B should both be multiples of 5.
Let B = 5m (m integer) with B > 10 ⇒ m ≥ 3. Try smallest multiples of 5 for B:
• B = 15 ⇒ 2B + 40 = 70 ⇒ G = 70/3 (not integer)
• B = 20 ⇒ 2B + 40 = 80 ⇒ G = 80/3 (not integer)
• B = 25 ⇒ 2B + 40 = 90 ⇒ G = 90/3 = 30 (integer, works)
So the smallest B (multiple of 5) that satisfies (★) is B = 25, with G = 30.
Total initial students = B + G = 25 + 30 = 55.
Therefore the minimum possible number of students initially in the class (with realistic integer people and integer remaining counts) is 55.
Breakup G Strategy | Interest | MEDIUM | CAT 2025 Slot 1
At a certain simple rate of interest, a given sum amounts to Rs 13920 in 3 years, and to Rs 18960 in 6 years and 6 months. If the same given sum had been invested for 2 years at the same rate as before but with interest compounded every 6 months, then the total interest earned, in rupees, would have been nearest to
1. 3221
2. 3180
3. 3096
4. 3156
Answer
Let P be the principal and r the annual simple rate (in decimal).
Amount after 3 years = P + 3·(P·r) = P(1 + 3r) = 13,920.
Amount after 6.5 years = P(1 + 6.5r) = 18,960.
Difference = interest for 3.5 years = 18,960 − 13,920 = 5,040.
So interest for 1 year = 5,040 ÷ 3.5 = 1,440.
Thus P·r = 1,440. Use A3: P + 3·1,440 = 13,920 ⇒ P = 13,920 − 4,320 = 9,600.
So annual simple rate r = 1,440 ÷ 9,600 = 0.15 = 15% p.a.
For 2 years compounded half-yearly: number of half-year periods n = 4; half-year rate = 15%/2 = 7.5% = 0.075.
Amount = P(1 + 0.075)^4 = 9,600 × (1.075)^4 ≈ 9,600 × 1.335469 ≈ 12,820.50.
Compound interest = 12,820.50 − 9,600 ≈ 3,220.50 → nearest rupee = 3,221. (Option 1)
Breakup G Strategy | Ratio + Table | MEDIUM | CAT 2025 Slot 1
The ratio of the number of students in the morning shift and afternoon shift of a school was 13 : 9. After 21 students moved from the morning shift to the afternoon shift, this ratio became 19 : 14. Next, some new students joined the morning and afternoon shifts in the ratio 3 : 8 and then the ratio of the number of students in the morning shift and the afternoon shift became 5 : 4. The number of new students who joined is
- 88
- 121
- 99
- 110
Answer
Solve using Matrix table | Answer 3. 99
| Stage | Morning Shift | Afternoon Shift | Ratio | Explanation |
| Initial | 13k | 9k | 13 : 9 | Given initial ratio |
| After 21 move M → A | 13k − 21 | 9k + 21 | 19 : 14 | 13k – 21/9k + 21 = 19/14 k=63 |
| After solving k = 63 | 798 | 588 | 19 : 14 | Substitute k = 63 into expressions |
| New students join (3x, 8x) | 798 + 3x | 588 + 8x | 5 : 4 | 798 + 3x/588 + 8x = 5/4 x = 9 |
| Total new students | 3x | 8x | Total: 11x | 11x = 11 × 9 = 99 |
Breakup | Mixtures + Percentages | HARD | CAT 2025 Slot 1
A container holds 200 litres of a solution of acid and water, having 30% acid by volume. Atul replaces 20% of the resulting solution with water, and finally replaces 10% of this solution with acid, then 15% of the solution obtained with water. The percentage of acid by volume in the final solution obtained after these three replacements, is nearest to
1. 25
2. 27
3. 29
4. 23
Answer
Initial: 200 L solution with 30% acid ⇒ acid = 0.30×200 = 60 L.
Replace 20% with water: remove 40 L (20% of 200). Acid removed = 20% of 60 = 12 L → acid left = 60 − 12 = 48 L. Add 40 L water → volume = 200 L, acid = 48 L.
Replace 10% with acid: remove 20 L. Acid removed = (48/200)×20 = 4.8 L → acid left = 48 − 4.8 = 43.2 L. Add 20 L pure acid → acid = 43.2 + 20 = 63.2 L. Volume = 200 L.
Replace 15% with water: remove 30 L. Acid removed = (63.2/200)×30 = 9.48 L → acid left = 63.2 − 9.48 = 53.72 L. Add 30 L water → acid = 53.72 L, volume = 200 L.
Final percentage acid = (53.72/200)×100 = 26.86% ≈ 26.9% → nearest = 27%.
Answer: 27 (option 2).
Breakup G Strategy Time Speed + APGP | MEDIUM | CAT 2025 Slot 1
Shruti travels a distance of 224 km in four parts for a total travel time of 3 hours. Her speeds in these four parts follow an arithmetic progression, and the corresponding time taken to cover these four parts follow another arithmetic progression. If she travels at a speed of 960 meters per minute for 30 minutes to cover the first part, then the distance, in meters, she travels in the fourth part is
- 86400
- 96000
- 112000
- 76800
Answer
Work in minutes and meters: total time = 3 hr = 180 min, total distance = 224 km = 224000 m.
Times are 30,40,50,60 (sum 180 using AP). If speeds are v, v+d, v+2d, v+3d then total distance = v·180 + d·(40 + 2·50 + 3·60) = v·180 + d·320. With v = 960 m/min, v·180 = 960×180 = 172800, so the extra distance to be made up by the d-terms = 224000 − 172800 = 51200.
So d = 51200/320 = 160 m/min ⇒ v4 = 960 + 3·160 = 1440 m/min ⇒ distance in 4th part = 1440 × 60 = 86,400 m.
Breakup G Strategy | Time & Work | HARD | CAT 2025 Slot 1
Arun, Varun and Tarun, if working alone, can complete a task in 24, 21, and 15 days, respectively. They charge Rs 2160, Rs 2400, and Rs 2160 per day, respectively, even if they are employed for a partial day. On any given day, any workers may or may not be employed to work. If the task needs to be completed in 10 days or less, then the minimum possible amount, in rupees, required to be paid for the entire task is
1. 38400
2. 38880
3. 34400
4. 47040
Answer
Work rates (job/day): Arun = 1/24, Varun = 1/21, Tarun = 1/15.
Daily charges: Arun = 2160, Varun = 2400, Tarun = 2160.
Cost per unit work = daily charge ÷ (work per day) = daily_charge × time:
Arun = 2160×24 = 51,840; Varun = 2400×21 = 50,400; Tarun = 2160×15 = 32,400.
So Tarun is cheapest per unit, then Varun, Arun is most expensive.
Use Tarun for all 10 calendar days to minimize cost: work done = 10/15 = 2/3.
Remaining work = 1 − 2/3 = 1/3. Fill with next cheapest (Varun).
Days of Varun needed: smallest integer d with d/21 ≥ 1/3 ⇒ d ≥ 7. So take d = 7.
Check total work: 10/15 + 7/21 = 2/3 + 1/3 = 1 (done within 10 days since Varun can work on 7 of those 10 days).
Total cost = Tarun: 10×2160 = 21,600; Varun: 7×2400 = 16,800.
Sum = 21,600 + 16,800 = 38,400 rupees.
Hence minimum possible payment = Rs 38,400 (Option 1).
Breakup G Strategy | Simple Equations | EASY | CAT 2025 Slot 1
Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds, is _______
Answer
Let the trader hold x shares of stock B.
Then stock C shares = 20 − x.
Prices:
A = 120
B = 90
C = 150
Portfolio value equation:
10(120) + x(90) + (20 − x)(150) = 3300
1200 + 90x + 3000 − 150x = 3300
1200 + 3000 + (90x − 150x) = 3300
4200 − 60x = 3300
60x = 900
x = 15
Answer: 15 shares of stock B.
Breakup G Strategy | Rhombus + Algebra | MEDIUM | CAT 2025 Slot 1
If the length of a side of a rhombus is 36 cm and the area of the rhombus is 396 sq. cm, then the absolute value of the difference between the lengths, in cm, of the diagonals of the rhombus is ____
Answer
Let the diagonals be d1 and d2.
Area of a rhombus = 1/2 × d1 × d2 = 396 → d1 × d2 = 792.
In a rhombus, diagonals bisect each other at right angles.
So by Pythagoras: (d1/2)^2 + (d2/2)^2 = 36^2.
Multiply by 4: d1^2 + d2^2 = 4 × 1296 = 5184.
Use identity: (d1 − d2)^2 = (d1^2 + d2^2) − 2(d1 × d2).
Substitute: (d1 − d2)^2 = 5184 − 2×792 = 5184 − 1584 = 3600.
So |d1 − d2| = √3600 = 60.
Final Answer: 60
Breakup G Strategy | Simple Equations | MEDIUM | CAT 2025 Slot 1
If a -6b + 6c = 4 and 6a + 3b – 3c = 50, where a, b, and c are real numbers, the value of 2a + 3b – 3c is
1. 14
2. 15
3. 16
4. 18
Answer
Step 1 From eq(2): 6a + 3b − 3c = 50
→ take one-third → 2a + b − c = 50/3
Step 2 Compare target with this:
Target = (2a + b − c) + 2(b − c)
Step 3 Find (b − c) from eq(1):
a − 6b + 6c = 4 → rewrite → a = 4 + 6(b − c)
Step 4 Substitute into step 1:
2(4 + 6x) + x = 50/3
8 + 12x + x = 50/3
13x = 26/3 → x = 2/3
So (b − c) = 2/3
Step 5 Plug in:
Target = 50/3 + 2*(2/3) = 18
Breakup G Strategy | PnC + Factors + Maxima Minima | HARD | CAT 2025 Slot 1
In a 3-digit number N, the digits are non-zero and distinct such that none of the digits is a square, and only one of the digits is a prime number. Then, the number of factors of the minimum possible value of N is ______
Answer
Square digits forbidden: 1, 4, 9. Allowed nonzero digits: 2,3,5,6,7,8.
Exactly one digit must be prime, so the other two must be composite. Allowed composites here are 6 and 8, so the three digits must be {prime, 6, 8}.
To minimize the 3-digit number, put the smallest available digit in the hundreds place. The smallest prime available is 2.
So hundreds = 2, and place the remaining digits in increasing order: tens = 6, units = 8. Minimum N = 268.
Factorize 268 = 2^2 × 67.
Number of factors = (2 + 1) × (1 + 1) = 3 × 2 = 6.
Breakup G Strategy | Simple Equations | EASY | CAT 2025 Slot 1
Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds, is _______
Answer
Let the trader hold x shares of stock B.
Then stock C shares = 20 − x.
Prices:
A = 120
B = 90
C = 150
Portfolio value equation:
10(120) + x(90) + (20 − x)(150) = 3300
1200 + 90x + 3000 − 150x = 3300
1200 + 3000 + (90x − 150x) = 3300
4200 − 60x = 3300
60x = 900
x = 15
Answer: 15 shares of stock B.
Breakup G Strategy | Simple Equations | MEDIUM | CAT 2025 Slot 1
If a -6b + 6c = 4 and 6a + 3b – 3c = 50, where a, b, and c are real numbers, the value of 2a + 3b – 3c is
1. 14
2. 15
3. 16
4. 18
Answer
Step 1 From eq(2): 6a + 3b − 3c = 50
→ take one-third → 2a + b − c = 50/3
Step 2 Compare target with this:
Target = (2a + b − c) + 2(b − c)
Step 3 Find (b − c) from eq(1):
a − 6b + 6c = 4 → rewrite → a = 4 + 6(b − c)
Step 4 Substitute into step 1:
2(4 + 6x) + x = 50/3
8 + 12x + x = 50/3
13x = 26/3 → x = 2/3
So (b − c) = 2/3
Step 5 Plug in:
Target = 50/3 + 2*(2/3) = 18
X maro G Strategy | Concept of Quadratic Equations | CAT 2025 Slot 1
The number of non-negative integer values of k for which the quadratic equation x² – 5x + k = 0 has only integer roots, is
Answer
Quadratic: x² – 5x + k = 0.
For integer roots, discriminant must be a square:
D = 25 – 4k = n².
Try n = 1, 3, 5:
n = 1 → k = 6
n = 3 → k = 4
n = 5 → k = 0
All are non-negative.
Total values = 3.
Answer: 3
X maro | Visual | Inequality | Sets | EASY | CAT 2025 Slot 1
Let 3 ≤ x ≤ 6 and [x²] = ([x])², where [x] is the greatest integer not exceeding x.
If S represents all feasible values of x, then a possible subset of S is
1. [3, √10] ∪ [4, √17] ∪ {6}
2. [3, √10) ∪ [5, √26)
3. (4, √18] ∪ [5, √27) ∪ {6}
4. (3, √10] ∪ [5, √26) ∪ {6}
Answer
Using X maro
For n = 3 → x must be before 4 → [3, √10)
For n = 4 → x must be before 5 → [4, √17)
For n = 5 → x must be before 6 → [5, √26)
For n = 6: only x = 6 (since x ≤ 6)
Only Option 4 matches: (3, √10] ∪ [5, √26) ∪ {6}
Note: ideally Set should include 4 as well.
X maro Functions + Maxima Minima | EASY | CAT 2025 Slot 1
A value of c for which the minimum value of f(x) = x² – 4cx + 8c is greater than the maximum value of
g(x) = –x² + 3cx – 2c, is
A. 1/2
B. –1/7
C. 2
D. –2
Answer
Let B = number of boys, G = number of girls (both integers), and B > 10.
After departures: remaining girls = 60% of G = 3G/5, remaining boys = 40% of B = 2B/5.
Given 3G/5 = 2B/5 + 8. Multiply by 5: 3G = 2B + 40. (Equation ★)
For 3G to be integer, G must be integer. For the remaining counts 3G/5 and 2B/5 to be integers (no fractional people), G and B should both be multiples of 5.
Let B = 5m (m integer) with B > 10 ⇒ m ≥ 3. Try smallest multiples of 5 for B:
• B = 15 ⇒ 2B + 40 = 70 ⇒ G = 70/3 (not integer)
• B = 20 ⇒ 2B + 40 = 80 ⇒ G = 80/3 (not integer)
• B = 25 ⇒ 2B + 40 = 90 ⇒ G = 90/3 = 30 (integer, works)
So the smallest B (multiple of 5) that satisfies (★) is B = 25, with G = 30.
Total initial students = B + G = 25 + 30 = 55.
Therefore the minimum possible number of students initially in the class (with realistic integer people and integer remaining counts) is 55.
Breakup G Strategy | Simple Equations | EASY | CAT 2025 Slot 1
Stocks A, B and C are priced at rupees 120, 90 and 150 per share, respectively. A trader holds a portfolio consisting of 10 shares of stock A, and 20 shares of stocks B and C put together. If the total value of her portfolio is rupees 3300, then the number of shares of stock B that she holds, is _______
Answer
Let the trader hold x shares of stock B.
Then stock C shares = 20 − x.
Prices:
A = 120
B = 90
C = 150
Portfolio value equation:
10(120) + x(90) + (20 − x)(150) = 3300
1200 + 90x + 3000 − 150x = 3300
1200 + 3000 + (90x − 150x) = 3300
4200 − 60x = 3300
60x = 900
x = 15
Answer: 15 shares of stock B.
Breakup G Strategy | Simple Equations | MEDIUM | CAT 2025 Slot 1
If a -6b + 6c = 4 and 6a + 3b – 3c = 50, where a, b, and c are real numbers, the value of 2a + 3b – 3c is
1. 14
2. 15
3. 16
4. 18
Answer
Step 1 From eq(2): 6a + 3b − 3c = 50
→ take one-third → 2a + b − c = 50/3
Step 2 Compare target with this:
Target = (2a + b − c) + 2(b − c)
Step 3 Find (b − c) from eq(1):
a − 6b + 6c = 4 → rewrite → a = 4 + 6(b − c)
Step 4 Substitute into step 1:
2(4 + 6x) + x = 50/3
8 + 12x + x = 50/3
13x = 26/3 → x = 2/3
So (b − c) = 2/3
Step 5 Plug in:
Target = 50/3 + 2*(2/3) = 18
X maro G Strategy | Concept of Quadratic Equations | CAT 2025 Slot 1
The number of non-negative integer values of k for which the quadratic equation x² – 5x + k = 0 has only integer roots, is
Answer
Quadratic: x² – 5x + k = 0.
For integer roots, discriminant must be a square:
D = 25 – 4k = n².
Try n = 1, 3, 5:
n = 1 → k = 6
n = 3 → k = 4
n = 5 → k = 0
All are non-negative.
Total values = 3.
Answer: 3
X maro | Visual | Inequality | Sets | EASY | CAT 2025 Slot 1
Let 3 ≤ x ≤ 6 and [x²] = ([x])², where [x] is the greatest integer not exceeding x.
If S represents all feasible values of x, then a possible subset of S is
1. [3, √10] ∪ [4, √17] ∪ {6}
2. [3, √10) ∪ [5, √26)
3. (4, √18] ∪ [5, √27) ∪ {6}
4. (3, √10] ∪ [5, √26) ∪ {6}
Answer
Using X maro
For n = 3 → x must be before 4 → [3, √10)
For n = 4 → x must be before 5 → [4, √17)
For n = 5 → x must be before 6 → [5, √26)
For n = 6: only x = 6 (since x ≤ 6)
Only Option 4 matches: (3, √10] ∪ [5, √26) ∪ {6}
Note: ideally Set should include 4 as well.
X maro Functions + Maxima Minima | EASY | CAT 2025 Slot 1
A value of c for which the minimum value of f(x) = x² – 4cx + 8c is greater than the maximum value of
g(x) = –x² + 3cx – 2c, is
A. 1/2
B. –1/7
C. 2
D. –2
Answer
Let B = number of boys, G = number of girls (both integers), and B > 10.
After departures: remaining girls = 60% of G = 3G/5, remaining boys = 40% of B = 2B/5.
Given 3G/5 = 2B/5 + 8. Multiply by 5: 3G = 2B + 40. (Equation ★)
For 3G to be integer, G must be integer. For the remaining counts 3G/5 and 2B/5 to be integers (no fractional people), G and B should both be multiples of 5.
Let B = 5m (m integer) with B > 10 ⇒ m ≥ 3. Try smallest multiples of 5 for B:
• B = 15 ⇒ 2B + 40 = 70 ⇒ G = 70/3 (not integer)
• B = 20 ⇒ 2B + 40 = 80 ⇒ G = 80/3 (not integer)
• B = 25 ⇒ 2B + 40 = 90 ⇒ G = 90/3 = 30 (integer, works)
So the smallest B (multiple of 5) that satisfies (★) is B = 25, with G = 30.
Total initial students = B + G = 25 + 30 = 55.
Therefore the minimum possible number of students initially in the class (with realistic integer people and integer remaining counts) is 55.
Concepts of Indices | HARD | CAT 2025 Slot 1
For any natural number k, let aₖ = 3ᵏ. The smallest natural number m for which (a₁)¹ × (a₂)² × … × (a₂₀)²⁰ < (a₂₁ × a₂₂ × … × a₂₀₊ₘ)⁵ is:
1. 56
2. 59
3. 58
4. 57
Answer
Given: aₖ = 3ᵏ.
Then (aᵢ)ⁱ = (3ᵢ)ⁱ = 3⁽ⁱ²⁾.
Left side:
L = (a₁)¹ × (a₂)² × … × (a₂₀)²⁰ = 3^(1² + 2² + … + 20²)
Sum of squares 1² + 2² + … + 20² = 2870 → So L = 3²⁸⁷⁰.
Right side inside product:
P = a₂₁ × a₂₂ × … × a₂₀₊ₘ = 3²¹ × 3²² × … × 3^(20+m) = 3^(sum of integers from 21 to 20+m).
Raise to power 5: R = P⁵ = 3^(5 × sum of integers from 21 to 20+m)
We need → 3²⁸⁷⁰ < 3^(5 × sum)
So: 2870 < 5 × (sum of 21 to 20+m)
Let N = 20 + m.
Sum from 21 to N = (N(N+1))/2 − 210.
So: 2870 < 5 × [(N(N+1))/2 − 210]
Divide and simplify:
(N(N+1))/2 > 784 → N(N+1) > 1568
Check values:
39×40 = 1560 (too small)
40×41 = 1640 (works)
So the smallest N = 40.
Thus:
m = N − 20 = 40 − 20 = 20 → Final Answer: m = 20
PnC + Factors + Maxima Minima | HARD | CAT 2025 Slot 1
In a 3-digit number N, the digits are non-zero and distinct such that none of the digits is a square, and only one of the digits is a prime number. Then, the number of factors of the minimum possible value of N is ______
Answer
Square digits forbidden: 1, 4, 9. Allowed nonzero digits: 2,3,5,6,7,8.
Exactly one digit must be prime, so the other two must be composite. Allowed composites here are 6 and 8, so the three digits must be {prime, 6, 8}.
To minimize the 3-digit number, put the smallest available digit in the hundreds place. The smallest prime available is 2.
So hundreds = 2, and place the remaining digits in increasing order: tens = 6, units = 8. Minimum N = 268.
Factorize 268 = 2^2 × 67.
Number of factors = (2 + 1) × (1 + 1) = 3 × 2 = 6.
Answer: 6.
Vedic Patterns G Strategy | Inequality | EASY | CAT 2025 Slot 1
The number of distinct pairs of integers (x, y) satisfying the inequalities x > y ≥ 3 and x + y < 14 is ____.
Answer
We fix y and find all possible x values that satisfy both conditions:
x > y
x + y < 14
| y | Allowed x values | Count |
| 3 | 4, 5, 6, 7, 8, 9, 10 | 7 |
| 4 | 5, 6, 7, 8, 9 | 5 |
| 5 | 6, 7, 8 | 3 |
| 6 | 7 | 1 |
| Total | 16 |
Vedic Pattern + Concepts of Logs DPAC G Strategy | EASY | CAT 2025 Slot 1
The number of distinct integers n for which log(1/4) of (n² − 7n + 11) > 0 is _____.
Answer
log(1/4) of (n² − 7n + 11) > 0 is _____.
Using DPAC Strategy: Cycle
n² − 7n + 11 < (1/4)0 → n² − 7n + 11 < 1
Now we need values of n which satisfy 0 < n² − 7n + 11 < 1
Compute E(n) = n² − 7n + 11 for nearby integers:
n = 1 → E = 5 (not in (0,1))
n = 2 → E = 1 (equals 1, excluded)
n = 3 → E = −1 (negative, excluded)
n = 4 → E = −1 (negative, excluded)
n = 5 → E = 1 (equals 1, excluded)
For n ≤ 0 or n ≥ 6, E ≥ 5 (too large).
Therefore no integer n satisfies 0 < E < 1. Final answer: 0.
Note: Official OG says answer is 2. Which is challenged by Cetking
PnC G Strategy | Concepts of PnC | EASY | CAT 2025 Slot 1
A cafeteria offers 5 types of sandwiches. Moreover, for each type of sandwich, a customer can choose one of 4 breads and opt for either small or large sized sandwich. Optionally, the customer may also add up to 2 out of 6 available sauces. The number of different ways in which an order can be placed for a sandwich, is
- 880
- 600
- 800
- 840
Answer
Given 5 sandwiches
4 breads
2 sizes
Sauces: choose 0, 1, or 2 from 6
Number of sauce choices:
0 sauces → 1 way
1 sauce → 6 ways
2 sauces → 15 ways
Total = 22 ways.
Total combinations =
5 × 4 × 2 × 22 = 880.
Answer: 880
Visual Lens | Odd even numbers | MEDIUM | CAT 2025 Slot 1
In the set of consecutive odd numbers {1, 3, 5, …, 57}, there is a number k such that the sum of all the elements less than k is equal to the sum of all the elements greater than k. Then, k equals
- 37
- 43
- 39
- 41
Answer
Let the diagonals be d1 and d2.
Area of a rhombus = 1/2 × d1 × d2 = 396 → d1 × d2 = 792.
In a rhombus, diagonals bisect each other at right angles.
So by Pythagoras: (d1/2)^2 + (d2/2)^2 = 36^2.
Multiply by 4: d1^2 + d2^2 = 4 × 1296 = 5184.
Use identity: (d1 − d2)^2 = (d1^2 + d2^2) − 2(d1 × d2).
Substitute: (d1 − d2)^2 = 5184 − 2×792 = 5184 − 1584 = 3600.
So |d1 − d2| = √3600 = 60.
Final Answer: 60
Breakup G Strategy | Rhombus + Algebra | MEDIUM | CAT 2025 Slot 1
If the length of a side of a rhombus is 36 cm and the area of the rhombus is 396 sq. cm, then the absolute value of the difference between the lengths, in cm, of the diagonals of the rhombus is ____
Answer
Let the diagonals be d1 and d2.
Area of a rhombus = 1/2 × d1 × d2 = 396 → d1 × d2 = 792.
In a rhombus, diagonals bisect each other at right angles.
So by Pythagoras: (d1/2)^2 + (d2/2)^2 = 36^2.
Multiply by 4: d1^2 + d2^2 = 4 × 1296 = 5184.
Use identity: (d1 − d2)^2 = (d1^2 + d2^2) − 2(d1 × d2).
Substitute: (d1 − d2)^2 = 5184 − 2×792 = 5184 − 1584 = 3600.
So |d1 − d2| = √3600 = 60.
Final Answer: 60
Visual Lens | Coordinate Geometry | MEDIUM
The (x, y) coordinates of vertices P, Q and R of a parallelogram PQRS are (-3, -2), (1, -5) and (9, 1), respectively. If the diagonal SQ intersects the x-axis at (a, 0), then the value of a is
- 10/3
- 27/7
- 13/4
- 29/9
Answer
For a parallelogram P-Q-R-S in that order, S = P + R − Q.
P = (−3, −2), Q = (1, −5), R = (9, 1) ⇒ S = (−3+9−1, −2+1−(−5)) = (5, 4).
Line SQ passes through Q(1, −5) and S(5, 4). Slope = (4 − (−5)) / (5 − 1) = 9/4.
Equation: y + 5 = (9/4)(x − 1). Put y = 0 to find x-intercept: 5 = (9/4)(x − 1).
x − 1 = 20/9 ⇒ x = 1 + 20/9 = 29/9.
Answer: 29/9 (option 4).
Concepts of Rhombus + Algebra
If the length of a side of a rhombus is 36 cm and the area of the rhombus is 396 sq. cm, then the absolute value of the difference between the lengths, in cm, of the diagonals of the rhombus is ____
Answer
Let the diagonals be d1 and d2.
Area of a rhombus = 1/2 × d1 × d2 = 396 → d1 × d2 = 792.
In a rhombus, diagonals bisect each other at right angles.
So by Pythagoras: (d1/2)^2 + (d2/2)^2 = 36^2.
Multiply by 4: d1^2 + d2^2 = 4 × 1296 = 5184.
Use identity: (d1 − d2)^2 = (d1^2 + d2^2) − 2(d1 × d2).
Substitute: (d1 − d2)^2 = 5184 − 2×792 = 5184 − 1584 = 3600.
So |d1 − d2| = √3600 = 60.
Final Answer: 60
Concepts of Circles Trigonometry | HARD | CAT 2025 Slot 1
In a circle with center O and radius 6√2 cm, PQ and SR are two parallel chords separated by one of the diameters. If ∠PQC = 45°, and the ratio of the perpendicular distance of PQ and SR from C is 3:2, then the area, in sq cm, of the quadrilateral PQRS is 1. 20(3 + √14) 2. 4(3√3 + 14) 3. 20(√2 + √7) 4. 4(3√2 + 7)
Answer
Radius R = 6√2. Assume ∠POQ = 90° so PQ subtends 90° at center.
For PQ: d1 = OM = R cos(45°) = 6√2 · 1/√2 = 6.
PQ = 2R sin(45°) = 2·6√2·1/√2 = 12.
Given d1:d2 = 3:2, so d2 = 6·(2/3) = 4.
For SR: SN = sqrt(R^2 − d2^2) = sqrt((6√2)^2 − 4^2) = sqrt(72 − 16) = sqrt(56) = 2√14.
Hence SR = 2·SN = 4√14.
The chords PQ and SR are parallel, so PQRS is a trapezium with parallel sides PQ and SR.
Height h = d1 + d2 = 6 + 4 = 10.
Area = 1/2 · (PQ + SR) · h = 1/2 · (12 + 4√14) · 10.
Area = 5 · (12 + 4√14) = 60 + 20√14.
Therefore area = 20(3 + √14), which is Option A.
