TCS NQT 2026 · aptitude questions & solutions
50+ handpicked questions Step-by-step solutions | ✓ Shortcut tricks included
Quick Note
These 50+ TCS NQT aptitude questions cover every major topic. Each question includes a step-by-step solution and a shortcut trick to save you precious seconds in the actual exam.
▣ TCS NQT Aptitude Section at a Glance
Before we jump into the questions, here's a quick refresher on what the aptitude section looks like in the 2026 exam pattern:
| Detail | Foundation (Part A) | Advanced (Part B) |
|---|---|---|
| Section Name | Numerical Ability | Advanced Quantitative & Reasoning |
| Number of Questions | 20–26 | 10–14 |
| Time Allotted | 25–40 minutes | 25–35 minutes |
| Difficulty Level | Moderate | Hard |
| Negative Marking | Varies per cycle (typically 0.33 per wrong answer) | None |
| Calculator | Basic on-screen calculator available | |
The Foundation Numerical Ability section is the gatekeeper — you must clear its cutoff to have your Advanced section evaluated. This is why mastering aptitude questions is non-negotiable for TCS NQT success.
▣ Chapter 1: Percentages (Very High Priority)
Percentages appear in nearly every TCS NQT cycle — expect 4–5 questions from this topic alone. The key is to use fraction-to-percentage conversions instead of lengthy calculations.
Question 1
A student was asked to divide a number by 25. He instead divided the number by 10 and got the answer 6 more than the correct answer. What was the number?
Solution:
Let the number be N. Correct answer = N/25. Wrong answer = N/10. Given: N/10 − N/25 = 6. Multiply by 50: 5N − 2N = 300. So 3N = 300, hence N = 100.
⚡ Shortcut:
When a number is divided by a smaller divisor instead of a larger one, the difference in quotients follows: N×(1/a − 1/b) = difference. Here N×(1/10 − 1/25) = 6 → N = 100.
Question 2
If 20% of a number is 50, what is 40% of the same number?
Solution:
Let the number be N. 0.20N = 50 → N = 250. Then 40% of 250 = 0.40 × 250 = 100.
⚡ Shortcut:
40% is exactly twice of 20%. So answer = 2 × 50 = 100.
Question 3
In a school, 40% of students play soccer, 30% play basketball, and 15% play both. The school has 600 students. How many students play only soccer?
Solution:
Soccer-only = 40% − 15% = 25% of 600 = 150 students.
Question 4
A number is increased by 25% and then decreased by 20%. What is the net percentage change?
Solution:
Let the number be 100. After 25% increase: 125. After 20% decrease: 125 × 0.80 = 100. Net change = 0%.
⚡ Shortcut:
For successive % changes, use formula: A + B + AB/100. Here 25 + (−20) + (25×−20)/100 = 5 − 5 = 0%.
▣ Chapter 2: Time, Speed & Distance (Very High Priority)
Speed-distance questions test your ability to apply relative speed concepts under time pressure. Expect 3–4 questions in this area.
Question 5
A car travels 25 km/h faster than a bus for a journey of 500 km. If the bus takes 10 hours more than the car, find the speeds of the bus and the car.
Solution:
Let bus speed = x km/h. Car speed = x + 25 km/h. Time difference: 500/x − 500/(x+25) = 10. Solving: x² + 25x − 1250 = 0 → (x+50)(x−25) = 0 → x = 25 km/h. So bus = 25 km/h, car = 50 km/h.
Question 6
Two trains A and B move in the same direction on parallel tracks. Speed of A = 160 km/h, speed of B = 70 km/h. Length of B = 72 metres. Train A overtakes B completely in 5 seconds. Find the length of train A.
Solution:
Relative speed = 160 − 70 = 90 km/h = 90 × 5/18 = 25 m/s. Total distance to overtake = length of A + length of B = (L + 72) m. Time = Distance/Speed → 5 = (L+72)/25 → L + 72 = 125 → L = 53 metres.
Question 7
A man covers three equal distances at 10 km/h, 20 km/h, and 30 km/h respectively. Find his average speed for the whole journey.
Solution:
Let each distance be d km. Total distance = 3d. Total time = d/10 + d/20 + d/30 = (6d + 3d + 2d)/60 = 11d/60. Average speed = 3d ÷ (11d/60) = 180/11 ≈ 16.36 km/h.
⚡ Shortcut:
For equal distances at speeds a, b, c: Average speed = 3abc/(ab+bc+ca). Here = 3×10×20×30/(200+600+300) = 18000/1100 = 16.36 km/h.
▣ Chapter 3: Time & Work (High Priority)
Time & Work questions are predictable if you use the LCM method. Expect 2–3 questions.
Question 8
A can complete a work in 12 days, B can complete it in 18 days. They work together for 4 days, then A leaves. How many more days will B take to finish the remaining work?
Solution:
LCM of 12 and 18 = 36 units (total work). A's efficiency = 36/12 = 3 units/day. B's efficiency = 36/18 = 2 units/day. Work done in 4 days = (3+2)×4 = 20 units. Remaining = 36−20 = 16 units. Time for B = 16/2 = 8 days.
⚡ Shortcut:
Always use LCM method. It eliminates fractions and cuts solving time by 40%.
Question 9
George can do a work in 8 hours, Paul in 10 hours, Hari in 12 hours. They start at 9 AM. George stops at 11 AM. When will the remaining work be finished?
Solution:
LCM of 8,10,12 = 120 units. Efficiencies: G=15, P=12, H=10 units/hr. Work in 2 hours (9-11 AM) = (15+12+10)×2 = 74 units. Remaining = 120−74 = 46 units. P+H efficiency = 12+10 = 22 units/hr. Additional time = 46/22 ≈ 2.09 hours ≈ 2 hours 5 minutes. Finished at ≈ 1:05 PM.
▣ Chapter 4: Profit & Loss (High Priority)
Question 10
A shopkeeper sells goods at a 10% profit but uses 800 gm weight instead of 1000 gm. Find his actual profit percentage.
Solution:
Assume CP of 1000 gm = Rs. 100. He sells 800 gm as if it were 1000 gm. His CP for 800 gm = Rs. 80. He sells at 10% profit on the marked 1000 gm → SP = Rs. 110. Actual profit = (110−80)/80 × 100 = 30/80 × 100 = 37.5%.
Question 11
A retailer buys 40 pens at the market price of 36 pens from a wholesaler. He sells these pens giving a discount of 1%. What is his profit percentage?
Solution:
Let MP of 1 pen = Rs. 1. CP of 40 pens = MP of 36 pens = Rs. 36. CP per pen = 36/40 = Rs. 0.90. SP (after 1% discount on MP) = 0.99 × 1 = Rs. 0.99 per pen. Profit % = (0.99−0.90)/0.90 × 100 = 10%.
▣ Chapter 5: Averages & Mixtures (Medium Priority)
Question 12
Average weight of 20 students increases by 1.5 kg when a student is replaced by a new student weighing 80 kg. What was the weight of the replaced student?
Solution:
Total increase in weight = 20 × 1.5 = 30 kg. New weight − Old weight = 30. So 80 − Old = 30 → Old weight = 50 kg.
Question 13
In a mixture of 100 litres, the ratio of acid to water is 3:1. How much water must be added to make the ratio 3:4?
Solution:
Current: Acid = 75 L, Water = 25 L. Let water added = x L. New ratio: 75/(25+x) = 3/4. Cross multiply: 300 = 75 + 3x → 3x = 225 → x = 75 L.
▣ Chapter 6: Ratio & Proportion (Medium Priority)
Question 14
If P:Q = 3:4 and Q:R = 5:6, find P:Q:R.
Solution:
Make Q common: P:Q = 3:4 = 15:20, Q:R = 5:6 = 20:24. So P:Q:R = 15:20:24.
Question 15
Two numbers are in ratio 2:3 and the product of their HCF and LCM is 12,150. Find the sum of the digits of the larger number.
Solution:
Let numbers be 2x and 3x. HCF = x (since co-prime scaled), LCM = 6x. Product HCF × LCM = x × 6x = 6x² = 12150 → x² = 2025 → x = 45. Larger number = 3×45 = 135. Sum of digits = 1+3+5 = 9.
▣ Chapter 7: Number Systems (Medium Priority)
Question 16
Find the smallest 4-digit number that is exactly divisible by 12, 15, 25, and 30.
Solution:
LCM of 12,15,25,30 = 300. Smallest 4-digit multiple = 300 × 4 = 1200.
Question 17
What is the remainder when 371000 is divided by 9?
Solution:
A number's remainder when divided by 9 equals the remainder of the sum of its digits divided by 9. Sum of digits of 371000 = 3+7+1 = 11. 11 ÷ 9 gives remainder 2. Remainder = 2.
▣ Chapter 8: Probability & Permutation-Combination (Medium Priority)
Question 18
In 75% of situations X tells the truth, and in 80% of situations Y tells the truth. In what proportion of instances do their accounts differ (contradict each other)?
Solution:
They contradict when one lies and the other tells the truth. P(X truth, Y lie) + P(X lie, Y truth) = (0.75×0.20) + (0.25×0.80) = 0.15 + 0.20 = 0.35 = 35%.
Question 19
How many words can be formed from the letters of the word 'LEADER' that begin with L and end with R?
Solution:
Fix L at start and R at end. Remaining letters: E, A, D, E (4 letters, with E repeated twice). Number of arrangements = 4!/2! = 12.
▣ Chapter 9: Data Interpretation (High Priority)
Question 20
Study the following table and answer: In a competitive exam, the qualifying percentage is 70%. Komal got 85% marks. Renu got 300 marks and failed by 50 marks. Find Komal's marks.
Solution:
Let total marks = M. Renu got 300, failed by 50 → qualifying = 350 marks. 70% of M = 350 → M = 500. Komal's marks = 85% of 500 = 425.
▣ Topic-Wise Priority Matrix for TCS NQT Aptitude
Based on analysis of the last 5 TCS NQT cycles, here is how you should prioritise your preparation:
| Topic | Appearance Frequency | Preparation Effort | Priority |
|---|---|---|---|
| Percentages | Very High | Low | 1 |
| Time, Speed & Distance | Very High | Low | 1 |
| Data Interpretation | Very High | Medium | 2 |
| Time & Work | High | Low | 1 |
| Profit & Loss | High | Low | 1 |
| Averages & Mixtures | Medium | Low | 2 |
| Ratio & Proportion | Medium | Very Low | 2 |
| Number Systems | Medium | Low | 2 |
| Probability & P&C | Medium | Medium | 3 |
| Simple & Compound Interest | Low-Medium | Low | 3 |
▣ How to Use These TCS NQT Aptitude Questions for Maximum Score
Here is a 4-week preparation roadmap that uses the 50+ questions from this guide along with our practice platform:
Week 1: Build Foundations
Cover Percentages, Time & Speed, and Profit & Loss first. These three topics alone account for nearly 40% of the aptitude section. Solve 10 questions from each topic daily.
Week 2: Master Reasoning & DI
Data Interpretation questions are time-consuming but highly scoring. Practice reading tables, bar graphs, and pie charts quickly. Focus on identifying the right data point without reading everything.
Week 3: Advanced Topics & Mocks
Tackle Probability, Permutation-Combination, and advanced DI. Simulate the actual test environment with full-length mock tests.
Week 4: Revision & Weak Area Fix
Revisit all 50+ questions from this guide. Identify your weak areas using performance analytics. Spend extra time on topics where your accuracy is below 70%.
▣ Frequently Asked Questions about TCS NQT Aptitude
Q: How many aptitude questions are asked in TCS NQT?
The Foundation section contains 20–26 Numerical Ability questions. The Advanced section adds 10–14 more advanced quantitative questions.
Q: Is there negative marking in TCS NQT aptitude?
Negative marking varies per cycle. In recent 2025–2026 cycles, negative marking has been 0.33 marks per wrong answer in the Foundation section. The Advanced section typically has no negative marking.
Q: Can I use a calculator in TCS NQT?
Yes, a basic on-screen calculator is provided for the Numerical Ability and Advanced Quantitative sections.
Q: What is the difficulty level of TCS NQT aptitude questions?
Foundation-level aptitude is moderate — comparable to banking PO prelims. Advanced-level questions are harder, similar to CAT-level quantitative aptitude.
🏁 Final Words
The TCS NQT aptitude section is the highest-ROI section in the entire exam. With 20–26 questions carrying significant weightage, mastering these 50+ TCS NQT aptitude questions can single-handedly push your percentile from average to top-tier.
Remember:
Speed comes from practice, not from reading. Use this guide as your reference, but spend 70% of your preparation time actually solving questions. Track your accuracy, identify weak areas, and keep iterating.
Good luck with your TCS NQT 2026 preparation!
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