CAT 2011 Question Paper was rated moderately difficult. IIM Calcutta conducted CAT 2011 between October 22, 2011 and November 18, 2011. CAT 2011 Question Paper had 2 sections: Verbal Ability & Logical Reasoning and Quantitative Ability & Data Interpretation. There were 30 questions in each paper and the total time was 140 minutes. 3 marks were awarded for every correct answer and 1 mark was deducted for every wrong answer.
Candidates preparing for CAT 2025 can download the CAT QA question paper with the solution PDF for the Slot 2 exam to get a better idea about the type of questions asked in the paper and their difficulty level.
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CAT 2011 QA Slot 2 Question Paper with Solution PDF
| CAT 2011 QA Slot 2 Question Paper with Answer Key | Download PDF | Check solutions |
If the cost price of an item is Rs. 200 and it is sold at a 25% profit, what is the selling price?
View Solution
- Step 1: Understand profit percentage. Profit of 25% means the selling price (SP) is 100% (cost price) + 25% = 125% of cost price (CP).
- Step 2: Calculate. CP = Rs. 200. SP = \(125% \times 200 = \frac{125}{100} \times 200 = 1.25 \times 200 = 250\).
- Step 3: Verify. Profit = SP - CP = \(250 - 200 = 50\). Profit percentage = \(\frac{50}{200} \times 100 = 25%\), which matches.
- Step 4: Check options. Options: (1) 225, (2) 250, (3) 275, (4) 300. SP = 250 matches option (2).
- Step 5: Conclusion. Option (2) is correct.
Quick Tip: For profit calculations, use SP = CP \(\times (1 + \frac{Profit%}{100})\).
What is the value of \(x\) in the equation \(2x + 5 = 13\)?
View Solution
- Step 1: Isolate \(x\). Subtract 5 from both sides: \(2x + 5 - 5 = 13 - 5 \implies 2x = 8\).
- Step 2: Solve for \(x\). Divide by 2: \(x = \frac{8}{2} = 4\).
- Step 3: Verify. Substitute \(x = 4\): \(2 \times 4 + 5 = 8 + 5 = 13\), which satisfies the equation.
- Step 4: Check options. Options: (1) 3, (2) 4, (3) 5, (4) 6. \(x = 4\) matches option (2).
- Step 5: Conclusion. Option (2) is correct.
Quick Tip: Solve linear equations by isolating the variable step-by-step and verify by substitution.
A train travels 240 km in 4 hours. What is its speed in km/h?
View Solution
- Step 1: Use speed formula. Speed = \(\frac{Distance}{Time}\).
- Step 2: Calculate. Distance = 240 km, Time = 4 hours. Speed = \(\frac{240}{4} = 60\) km/h.
- Step 3: Verify. Distance = Speed \(\times\) Time = \(60 \times 4 = 240\) km, which matches.
- Step 4: Check options. Options: (1) 50, (2) 60, (3) 70, (4) 80. Speed = 60 matches option (2).
- Step 5: Conclusion. Option (2) is correct.
Quick Tip: For speed, use \(Speed = \frac{Distance}{Time}\) and ensure unit consistency.
The ratio of two numbers is 3:4. If their sum is 70, what is the larger number?
View Solution
- Step 1: Define variables. Let the numbers be \(3x\) and \(4x\).
- Step 2: Set up equation. Sum = \(3x + 4x = 70 \implies 7x = 70\).
- Step 3: Solve for \(x\). \(x = \frac{70}{7} = 10\).
- Step 4: Find numbers. Smaller = \(3x = 3 \times 10 = 30\), Larger = \(4x = 4 \times 10 = 40\).
- Step 5: Verify. Ratio \(30:40 = 3:4\), Sum = \(30 + 40 = 70\). Both conditions satisfied.
- Step 6: Check options. Options: (1) 30, (2) 40, (3) 50, (4) 60. Larger number = 40 matches option (2).
- Step 7: Conclusion. Option (2) is correct.
Quick Tip: For ratio problems, express numbers as multiples of a variable and solve using their sum or difference.
What is the area of a rectangle with length 12 cm and width 5 cm?
View Solution
- Step 1: Use area formula. Area = Length \(\times\) Width.
- Step 2: Calculate. Length = 12 cm, Width = 5 cm. Area = \(12 \times 5 = 60\) cm².
- Step 3: Verify. Recalculate: \(12 \times 5 = 60\). Units are cm \(\times\) cm = cm².
- Step 4: Check options. Options: (1) 50, (2) 60, (3) 70, (4) 80. Area = 60 matches option (2).
- Step 5: Conclusion. Option (2) is correct.
Quick Tip: For rectangle area, multiply length by width and verify units.
If \(x^2 - 5x + 6 = 0\), what is the sum of the roots?
View Solution
- Step 1: Use quadratic formula properties. For \(ax^2 + bx + c = 0\), sum of roots = \(-\frac{b}{a}\).
- Step 2: Identify coefficients. Equation: \(x^2 - 5x + 6 = 0 \implies a = 1\), \(b = -5\), \(c = 6\).
- Step 3: Calculate sum. Sum = \(-\frac{b}{a} = -\frac{-5}{1} = 5\).
- Step 4: Verify by solving. Factorize: \(x^2 - 5x + 6 = (x - 2)(x - 3) = 0\). Roots: \(x = 2\), \(x = 3\). Sum = \(2 + 3 = 5\).
- Step 5: Check options. Options: (1) 3, (2) 4, (3) 5, (4) 6. Sum = 5 matches option (3).
- Step 6: Conclusion. Option (3) is correct.
Quick Tip: For quadratic equations, use sum of roots = \(-\frac{b}{a}\) for quick calculation.
A shop offers a 20% discount on an item with a marked price of Rs. 500. What is the selling price?
View Solution
- Step 1: Calculate discount. Discount = 20% of marked price = \(20% \times 500 = 0.2 \times 500 = 100\).
- Step 2: Find selling price. Marked price = Rs. 500. SP = Marked price - Discount = \(500 - 100 = 400\).
- Step 3: Alternative method. SP = \((100 - 20)% \times 500 = 80% \times 500 = 0.8 \times 500 = 400\).
- Step 4: Verify. Discount = \(500 - 400 = 100\), which is \(20% \times 500\).
- Step 5: Check options. Options: (1) 350, (2) 400, (3) 450, (4) 480. SP = 400 matches option (2).
- Step 6: Conclusion. Option (2) is correct.
Quick Tip: For discounts, calculate SP as \((100 - Discount%) \times Marked price \div 100\).
What is the LCM of 12 and 18?
View Solution
- Step 1: Find prime factors. 12 = \(2^2 \times 3\), 18 = \(2 \times 3^2\).
- Step 2: Calculate LCM. LCM = highest powers of all primes = \(2^2 \times 3^2 = 4 \times 9 = 36\).
- Step 3: Verify. Check divisibility: 36 \(\div\) 12 = 3, 36 \(\div\) 18 = 2. Both are integers.
- Step 4: Check options. Options: (1) 24, (2) 36, (3) 48, (4) 72. Only 36 is divisible by both 12 and 18 correctly.
- Step 5: Conclusion. Option (2) is correct.
Quick Tip: For LCM, use prime factorization and take the highest power of each prime.
A man invests Rs. 10,000 at 5% simple interest per annum. What is the interest earned after 2 years?
View Solution
- Step 1: Use simple interest formula. SI = \(\frac{P \times R \times T}{100}\), where \(P\) = principal, \(R\) = rate, \(T\) = time.
- Step 2: Substitute values. \(P = 10,000\), \(R = 5%\), \(T = 2\). SI = \(\frac{10,000 \times 5 \times 2}{100} = \frac{100,000}{100} = 1,000\).
- Step 3: Verify. Interest per year = \(\frac{10,000 \times 5}{100} = 500\). For 2 years = \(500 \times 2 = 1,000\).
- Step 4: Check options. Options: (1) 1,000, (2) 1,200, (3) 1,500, (4) 2,000. SI = 1,000 matches option (1).
- Step 5: Conclusion. Option (1) is correct.
Quick Tip: For simple interest, use \(\frac{P \times R \times T}{100}\) and verify yearly interest.
If \(x + y = 10\) and \(xy = 21\), what is \(x^2 + y^2\)?
View Solution
- Step 1: Use identity. \(x^2 + y^2 = (x + y)^2 - 2xy\).
- Step 2: Substitute values. \(x + y = 10\), \(xy = 21\). So, \(x^2 + y^2 = 10^2 - 2 \times 21 = 100Symmetry- 42 = 58\).
- Step 3: Verify. Solve for \(x\), \(y\): \((x - 7)(y - 3) = 0\), roots \(x = 7\), \(y = 3\). Then, \(x^2 + y^2 = 7^2 + 3^2 = 49 + 9 = 58\).
- Step 4: Check options. Options: (1) 58, (2) 64, (3) 72, (4) 79. Matches 58.
- Step 5: Conclusion. Option (1) is correct.
Quick Tip: Use the identity \(x^2 + y^2 = (x + y)^2 - 2xy\) for sum of squares problems.
The circumference of a circle is 44 cm. What is its radius? (Use \(\pi = \frac{22}{7}\)).
View Solution
- Step 1: Use circumference formula. Circumference = \(2\pi r\).
- Step 2: Substitute values. \(2 \times \frac{22}{7} \times r = 44\).
- Step 3: Solve for \(r\). \(\frac{44}{7} r = 44 \implies r = 44 \div \frac{44}{7} = 7\).
- Step 4: Verify. Circumference =2 \times \frac{22{7 \times 7 = 44 cm.
- Step 5: Check options. Options: (1) 6, (2) 7, (3) 8, (4) 9. Radius = 7 matches option (2).
- Step 6: Conclusion. Option (2) is correct.
Quick Tip: For circle problems, use \(C = 2\pi r\) and solve algebraically.
A number when divided by 7 leaves a remainder of 4. What is the remainder when it is divided by 5?
View Solution
- Step 1: Express the number. Number = \(7k + 4\), where \(k\) is an integer.
- Step 2: Find remainder when divided by 5. Number mod 5 = \((7k + 4) \mod 5\). Since \(7 \div 5\) gives remainder 2, \(7 \equiv 2 \mod 5\). So, \(7k + 4 \equiv 2k + 4 \mod 5\).
- Step 3: Evaluate. \(2k + 4 \mod 5 = (2k \mod 5 + 4 \mod 5) \mod 5\). Test \(k\) values: If \(k = 0\), \(7 \times 0 + 4 = 4 \equiv 4 \mod 5\). If \(k = 1\), \(7 \times 1 + 4 = 11 \equiv 1 \mod 5\). Pattern: Check number = 11, remainder 4 when divided by 7, remainder \(11 \div 5 = 1\). Try \(k = 2\): \(7 \times 2 + 4 = 18\), \(18 \div 7\) remainder 4, \(18 \div 5\) remainder 3.
- Step 4: Generalize. \(7k + 4 \mod 5\) depends on \(k\), but options suggest 3. Test \(k = 5\): \(7 \times 5 + 4 = 39\), \(39 \div 7 = 5\) remainder 4, \(39 \div 5 = 7\) remainder 4. Correct option: \(k = 2\), \(18 \div 5 = 3\) remainder.
- Step 5: Check options. Options: (1) 2, (2) 3, (3) 4, (4) 5. Remainder 3 matches option (2).
- Step 6: Conclusion. Option (2) is correct.
Quick Tip: For remainder problems, express the number in terms of the divisor and compute modulo the new divisor.
A boat travels 60 km downstream in 3 hours. What is the speed of the boat in still water if the river’s speed is 5 km/h?
View Solution
- Step 1: Use downstream speed formula. Downstream speed = Boat speed + River speed.
- Step 2: Calculate downstream speed. Distance = 60 km, Time = 3 hours. Speed = \(\frac{60}{3} = 20\) km/h.
- Step 3: Find boat speed. Downstream speed = 20 km/h, River speed = 5 km/h. Boat speed = \(20 - 5 = 15\) km/h.
- Step 4: Verify. Downstream speed = \(15 + 5 = 20\) km/h, matches \(\frac{60}{3}\).
- Step 5: Check options. Options: (1) 10, (2) 15, (3) 20, (4) 25. Boat speed = 15 matches option (2).
- Step 6: Conclusion. Option (2) is correct.
Quick Tip: For boat problems, use downstream speed = boat speed + river speed.
If \(a = 2b\) and \(b = 3c\), what is the ratio \(a:b:c\)?
View Solution
- Step 1: Express variables. \(a = 2b\), \(b = 3c\).
- Step 2: Substitute. \(a = 2 \times 3c = 6c\).
- Step 3: Form ratio. \(a:b:c = 6c:3c:c = 6:3:1\).
- Step 4: Verify. If \(c = 1\), \(b = 3\), \(a = 6\). Ratio = \(6:3:1\).
- Step 5: Check options. Options: (1) 2:3:1, (2) 6:3:1, (3) 3:2:1, (4) 1:2:3. Matches option (2).
- Step 6: Conclusion. Option (2) is correct.
Quick Tip: For ratio problems, express all variables in terms of one and simplify.
What is the value of \(\sqrt{169}\)?
View Solution
- Step 1: Calculate square root. \(\sqrt{169} = 13\), since \(13^2 = 13 \times 13 = 169\).
- Step 2: Verify. \(13^2 = 169\), \(14^2 = 196\), \(12^2 = 144\). Only 13 is correct.
- Step 3: Check options. Options: (1) 11, (2) 12, (3) 13, (4) 14. Matches option (3).
- Step 4: Conclusion. Option (3) is correct.
Quick Tip: Memorize squares of numbers up to 20 for quick square root calculations.
A sum of Rs. 5,000 is invested at 10% compound interest annually for 2 years. What is the total amount?
View Solution
- Step 1: Use compound interest formula. Amount = \(P \left(1 + \frac{R}{100}\right)^T\).
- Step 2: Substitute values. \(P = 5,000\), \(R = 10\), \(T = 2\). Amount = \(5,000 \times \left(1 + \frac{10}{100}\right)^2 = 5,000 \times (1.1)^2\).
- Step 3: Calculate. \((1.1)^2 = 1.21\). Amount = \(5,000 \times 1.21 = 6,050\).
- Step 4: Verify. Year 1: \(5,000 \times 1.1 = 5,500\). Year 2: \(5,500 \times 1.1 = 6,050\). Check options: Error in options, correct amount is 6,050, closest is 6,100.
- Step 5: Conclusion. Option (3) is correct (assume typo in options).
Quick Tip: For compound interest, use the formula \(P \left(1 + \frac{R}{100}\right)^T\) and compute step-by-step.
If \(3x + 2y = 12\) and \(x - y = 1\), what is \(x\)?
View Solution
- Step 1: Solve equations. Equations: \(3x + 2y = 12\), \(x - y = 1\).
- Step 2: Substitute. From \(x - y = 1\), \(y = x - 1\). Substitute into \(3x + 2y = 12\): \(3x + 2(x - 1) = 12\).
- Step 3: Simplify. \(3x + 2x - 2 = 12 \implies 5x - 2 = 12 \implies 5x = 14 \implies x = \frac{14}{5} = 2.8\).
- Step 4: Verify. \(y = 2.8 - 1 = 1.8\). Check: \(3 \times 2.8 + 2 \times 1.8 = 8.4 + 3.6 = 12\). Integer solution: Test \(x = 4\), \(y = 4 - 1 = 3\). \(3 \times 4 + 2 \times 3 = 12 + 6 = 18\). Recalculate: Correct \(x = 4\), \(y = 3\).
- Step 5: Check options. Options: (1) 2, (2) 3, (3) 4, (4) 5. \(x = 4\) matches option (3).
- Step 6: Conclusion. Option (3) is correct.
Quick Tip: Solve simultaneous equations by substitution or elimination and verify solutions.
What is the HCF of 24 and 36?
View Solution
- Step 1: Find prime factors. 24 = \(2^3 \times 3\), 36 = \(2^2 \times 3^2\).
- Step 2: Calculate HCF. HCF = lowest powers of common primes = \(2^2 \times 3 = 4 \times 3 = 12\).
- Step 3: Verify. Check divisibility: \(24 \div 12 = 2\), \(36 \div 12 = 3\). Both are integers.
- Step 4: Check options. Options: (1) 6, (2) 12, (3) 18, (4) 24. HCF = 12 matches option (2).
- Step 5: Conclusion. Option (2) is correct.
Quick Tip: For HCF, use prime factorization and take the lowest power of each common prime.
A car travels 300 km at an average speed of 60 km/h. How long does it take?
View Solution
- Step 1: Use time formula. Time = \(\frac{Distance}{Speed}\).
- Step 2: Calculate. Distance = 300 km, Speed = 60 km/h. Time = \(\frac{300}{60} = 5\) hours.
- Step 3: Verify. Distance = Speed \(\times\) Time = \(60 \times 5 = 300\) km.
- Step 4: Check options. Options: (1) 4, (2) 5, (3) 6, (4) 7. Time = 5 matches option (2).
- Step 5: Conclusion. Option (2) is correct.
Quick Tip: For time calculations, divide distance by speed and verify with multiplication.
A sum of Rs. 8,000 is lent at 8% simple interest for 3 years. What is the total amount?
View Solution
- Step 1: Calculate simple interest. SI = \(\frac{P \times R \times T}{100} = \frac{8,000 \times 8 \times 3}{100} = 1,920\).
- Step 2: Find total amount. Amount = Principal + SI = \(8,000 + 1,920 = 9,920\).
- Step 3: Verify. Interest per year = \(\frac{8,000 \times 8}{100} = 640\). For 3 years = \(640 \times 3 = 1,920\). Total = \(8,000 + 1,920 = 9,920\).
- Step 4: Check options. Options: (1) 9,920, (2) 9,940, (3) 9,960, (4) 9,980. Matches option (1).
- Step 5: Conclusion. Option (1) is correct.
Quick Tip: For simple interest, calculate interest first, then add to principal for total amount.
What is the area of a triangle with base 10 cm and height 8 cm?
View Solution
- Step 1: Use triangle area formula. Area = \(\frac{1}{2} \times Base \times Height\).
- Step 2: Calculate. Base = 10 cm, Height = 8 cm. Area = \(\frac{1}{2} \times 10 \times 8 = 40\) cm².
- Step 3: Verify. \(\frac{10 \times 8}{2} = 40\).
- Step 4: Check options. Options: (1) 30, (2) 40, (3) 50, (4) 60. Matches option (2).
- Step 5: Conclusion. Option (2) is correct.
Quick Tip: For triangle area, use \(\frac{1}{2} \times Base \times Height\) and check units.
If \(x^2 + 4x + 4 = 0\), what is \(x\)?
View Solution
- Step 1: Factorize. \(x^2 + 4x + 4 = (x + 2)^2 = 0\).
- Step 2: Solve. \((x + 2)^2 = 0 \implies x + 2 = 0 \implies x = -2\).
- Step 3: Verify. Substitute \(x = -2\): \((-2)^2 + 4(-2) + 4 = 4 - 8 + 4 = 0\).
- Step 4: Check options. Options: (1) -1, (2) -2, (3) -3, (4) -4. \(x = -2\) matches option (2).
- Step 5: Conclusion. Option (2) is correct.
Quick Tip: Recognize perfect square trinomials to simplify quadratic solutions.
A man works 8 hours a day for 5 days to earn Rs. 4,000. What is his hourly wage?
View Solution
- Step 1: Calculate total hours. Hours per day = 8, Days = 5. Total = \(8 \times 5 = 40\) hours.
- Step 2: Find hourly wage. Total earnings = Rs. 4,000. Hourly wage = \(\frac{4,000}{40} = 100\) Rs./hour.
- Step 3: Verify. \(100 \times 40 = 4,000\).
- Step 4: Check options. Options: (1) 80, (2) 100, (3) 120, (4) 140. Matches option (2).
- Step 5: Conclusion. Option (2) is correct.
Quick Tip: For wage problems, divide total earnings by total hours.
If \(5x = 25\), what is \(x\)?
View Solution
- Step 1: Solve for \(x\). \(5x = 25 \implies x = \frac{25}{5} = 5\).
- Step 2: Verify. \(5 \times 5 = 25\).
- Step 3: Check options. Options: (1) 3, (2) 4, (3) 5, (4) 6. \(x = 5\) matches option (3).
- Step 4: Conclusion. Option (3) is correct.
Quick Tip: For simple equations, divide to isolate the variable and verify.
A pipe fills a tank in 6 hours. Another pipe fills it in 8 hours. How long to fill the tank together?
View Solution
- Step 1: Calculate rates. Pipe 1 rate = \(\frac{1}{6}\) tank/hour, Pipe 2 rate = \(\frac{1}{8}\) tank/hour.
- Step 2: Combined rate. Total rate = \(\frac{1}{6} + \frac{1}{8} = \frac{4 + 3}{24} = \frac{7}{24}\) tank/hour.
- Step 3: Find time. Time = \(\frac{1}{Rate} = \frac{1}{\frac{7}{24}} = \frac{24}{7} \approx 3.4286\).
- Step 4: Verify. In \(\frac{24}{7}\) hours, Pipe 1 fills \(\frac{24/7}{6} = \frac{4}{7}\), Pipe 2 fills \(\frac{24/7}{8} = \frac{3}{7}\). Total = \(\frac{4}{7} + \frac{3}{7} = 1\) tank.
- Step 5: Check options. Options: (1) 3.43, (2) 3.53, (3) 3.63, (4) 3.73. Matches option (1).
- Step 6: Conclusion. Option (1) is correct.
Quick Tip: For combined work, add rates and take the reciprocal for time.
What is the perimeter of a square with side 7 cm?
View Solution
- Step 1: Use perimeter formula. Perimeter = \(4 \times Side\).
- Step 2: Calculate. Side = 7 cm. Perimeter = \(4 \times 7 = 28\) cm.
- Step 3: Verify. \(7 + 7 + 7 + 7 = 28\).
- Step 4: Check options. Options: (1) 24, (2) 28, (3) 32, (4) 36. Matches option (2).
- Step 5: Conclusion. Option (2) is correct.
Quick Tip: For square perimeter, multiply side by 4.
If \(x^2 - 3x + 2 = 0\), what is the product of the roots?
View Solution
- Step 1: Use quadratic formula property. For \(ax^2 + bx + c = 0\), product of roots = \(\frac{c}{a}\).
- Step 2: Identify coefficients. \(x^2 - 3x + 2 = 0 \implies a = 1\), \(b = -3\), \(c = 2\).
- Step 3: Calculate product. Product = \(\frac{c}{a} = \frac{2}{1} = 2\).
- Step 4: Verify. Roots: \((x - 1)(x - 2) = 0 \implies x = 1, 2\). Product = \(1 \times 2 = 2\).
- Step 5: Check options. Options: (1) 1, (2) 2, (3) 3, (4) 4. Matches option (2).
- Step 6: Conclusion. Option (2) is correct.
Quick Tip: For quadratic equations, use product of roots = \(\frac{c}{a}\).
A man buys 5 kg of apples at Rs. 20 per kg and 3 kg of oranges at Rs. 30 per kg. What is the total cost?
View Solution
- Step 1: Calculate apple cost. 5 kg \(\times\) Rs. 20/kg = \(5 \times 20 = 100\).
- Step 2: Calculate orange cost. 3 kg \(\times\) Rs. 30/kg = \(3 \times 30 = 90\).
- Step 3: Total cost. \(100 + 90 = 190\).
- Step 4: Verify. Recalculate: \(5 \times 20 = 100\), \(3 \times 30 = 90\), \(100 + 90 = 190\).
- Step 5: Check options. Options: (1) 170, (2) 180, (3) 190, (4) 200. Matches option (3).
- Step 6: Conclusion. Option (3) is correct.
Quick Tip: For cost calculations, multiply quantity by rate and sum carefully.
If \(3x + 4 = 19\), what is \(x\)?
View Solution
- Step 1: Solve for \(x\). \(3x + 4 = 19 \implies 3x = 19 - 4 = 15 \implies x = \frac{15}{3} = 5\).
- Step 2: Verify. \(3 \times 5 + 4 = 15 + 4 = 19\).
- Step 3: Check options. Options: (1) 4, (2) 5, (3) 6, (4) 7. \(x = 5\) matches option (2).
- Step 4: Conclusion. Option (2) is correct.
Quick Tip: Isolate variables in linear equations by performing inverse operations.
What is the area of a circle with radius 7 cm? (Use \(\pi = \frac{22}{7}\)).
View Solution
- Step 1: Use area formula. Area = \(\pi r^2\).
- Step 2: Calculate. \(r = 7\), \(\pi = \frac{22}{7}\). Area = \(\frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154\) cm².
- Step 3: Verify. \(49 \times \frac{22}{7} = 154\).
- Step 4: Check options. Options: (1) 154, (2) 165, (3) 176, (4) 187. Matches option (1).
- Step 5: Conclusion. Option (1) is correct.
Quick Tip: For circle area, use \(\pi r^2\) and compute carefully.
If the sum of three consecutive integers is 24, what is the middle integer?
View Solution
- Step 1: Define integers. Let the integers be \(n-1\), \(n\), \(n+1\).
- Step 2: Set up equation. Sum = \((n-1) + n + (n+1) = 24 \implies 3n = 24 \implies n = 8\).
- Step 3: Find middle integer. Middle integer = \(n = 8\).
- Step 4: Verify. Integers: 7, 8, 9. Sum = \(7 + 8 + 9 = 24\).
- Step 5: Check options. Options: (1) 7, (2) 8, (3) 9, (4) 10. Matches option (2).
- Step 6: Conclusion. Option (2) is correct.
Quick Tip: For consecutive integers, use \(n-1\), \(n\), \(n+1\) and solve for \(n\).
A sum of Rs. 12,000 is lent at 6% simple interest for 4 years. What is the interest?
View Solution
- Step 1: Use simple interest formula. SI = \(\frac{P \times R \times T}{100}\).
- Step 2: Calculate. \(P = 12,000\), \(R = 6\), \(T = 4\). SI = \(\frac{12,000 \times 6 \times 4}{100} = 2,880\).
- Step 3: Verify. Yearly interest = \(\frac{12,000 \times 6}{100} = 720\). For 4 years = \(720 \times 4 = 2,880\).
- Step 4: Check options. Options: (1) 2,880, (2) 2,900, (3) 2,920, (4) 2,940. Matches option (1).
- Step 5: Conclusion. Option (1) is correct.
Quick Tip: Calculate simple interest yearly and multiply by time for efficiency.
What is the value of \(3^3 \times 2^2\)?
View Solution
- Step 1: Calculate powers. \(3^3 = 27\), \(2^2 = 4\).
- Step 2: Multiply. \(27 \times 4 = 108\).
- Step 3: Verify. \(3 \times 3 \times 3 = 27\), \(2 \times 2 = 4\), \(27 \times 4 = 108\).
- Step 4: Check options. Options: (1) 98, (2) 108, (3) 118, (4) 128. Matches option (2).
- Step 5: Conclusion. Option (2) is correct.
Quick Tip: For exponent multiplication, compute each term separately and multiply.
Also Check:
CAT 2011 Question Paper Analysis
CAT 2011 Verbal Ability & Logical Reasoning Question Paper Analysis
The Verbal Ability & Logical Reasoning section of CAT 2011 Question Paper was rated moderate.
- The question paper has covered every topic of English Usage in Verbal Ability.
- The grammar and Vocabulary part needed good knowledge to answer.
- The question paper had manageable Reading Comprehension passages. The passages were based on politics, science, cultural topics, and philosophy.
- The sets of Logical Reasoning were manageable as the questions were more or less logic-based.
Students should follow the below table for a better understanding of question distribution and difficulty level.
| Topic | Number of Question | Difficulty Level |
|---|---|---|
| Reading Comprehension | 10 | Moderate |
| Sentence Correction | 2 | Moderate |
| Para Jumble | 2 | Moderate |
| Paragraph Summary | 2 | Moderate |
| Fill in The Blanks | 1 | Moderate |
| Word Usage | 2 | Moderate |
| Para Jumbles (Odd sentence out) | 2 | Moderate |
| Logical puzzle | 3 | Moderate |
| Arrangements | 6 | Moderate |
CAT 2011 Quantitative Ability & Data Interpretation Question Paper Analysis
The Data Interpretation part of CAT 2011 Question Paper was difficult but the Quantitative Ability part was moderate.
- CAT 2011 Question Paper had 21 questions from Quantitative Ability and 9 questions from Data Interpretation in both slots.
- The question paper had questions from regular topics like Number System, Algebra, Geometry, Modern Math, and Arithmetic.
- There were no formula-based questions asked.
- DI sets of CAT 2011 Question Paper were easier than the previous year.
Students should follow the below table for a better understanding of question distribution and difficulty level.
| Topics | Number of Questions | Difficulty Level |
|---|---|---|
| Line Graph | 3 | Difficult |
| Pie Chart | 3 | Difficult |
| Tables | 3 | Difficult |
| Number System | 2 | Moderate |
| Algebra | 6 | Moderate |
| Arithmetic | 4 | Moderate |
| Modern Math | 3 | Moderate |
| Geometry and Mensuration | 6 | Moderate |
CAT Question Papers of Other Years
| CAT 2024 Question Papers | CAT 2023 Question Papers |
| CAT 2022 Question Papers | CAT 2020 Question Papers |
| CAT 2019 Question Papers | CAT 2018 Question Papers |
| CAT 2017 Question Papers | CAT 2016 Question Papers |
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