Previously asked Deloitte questions
Problems candidates reported from recent Deloitte tests, rewritten with worked solutions (memory-based).
1. Two towns are 72 km apart. Arun cycles from the first town towards the second at 17 km/h, and Varun starts at the same time at 19 km/h, reaches the second town, turns back and meets Arun on the way. How far has Varun travelled when they meet?
- 75 km
- 76 km
- 86 km
- 70 km
Answer
B. 76 km
How to solve: Distance = speed × time. Keep the units matching (km with hours, m with seconds).
- Let Arun cover z km.
- In the same time Varun covers 72 + (72 − z).
- So z/17 = (144 − z)/19, giving z = 68.
- Varun travels 144 − 68 = 76 km (4 hours each).
Shortcut: For a fixed distance, speed and time are inversely proportional: double the speed, half the time.
2. Without stops a bus averages 50 km/h; with four 15-minute stops its average falls to 40 km/h. What is the distance between the two cities?
- 360 km
- 200 km
- 160 km
- 450 km
Answer
B. 200 km
How to solve: Distance = speed × time. Keep the units matching (km with hours, m with seconds).
- The stops add 1 hour
- D/40 = D/50 + 1.
- That gives D/200 = 1
- D = 200 km.
Shortcut: For a fixed distance, speed and time are inversely proportional: double the speed, half the time.
3. Taps A and B together fill a cistern in 2.4 hours. Tap B alone takes 4 hours, and tap A delivers 100 litres per hour. What is the capacity of the cistern?
- 500 litres
- 1000 litres
- 600 litres
- 1200 litres
Answer
C. 600 litres
How to solve: Treat pipes like workers: a pipe that fills in a hours does 1/a of the tank each hour. Add fillers, subtract emptiers.
- A alone fills 1/2.4 − 1/4 = 1/6 of the cistern per hour
- It takes 6 hours: 6 × 100 = 600 litres.
Shortcut: Take the tank as LCM of the times (units), so each pipe fills a whole number of units per hour.
4. What is the third proportional to (a − b) and (a² − b²)?
- a³ + ab² − a²b − b³
- a³ + a²b − ab² − b³
- a³ − ab² − a²b + b³
- a³ − a²b + ab² − b³
Answer
B. a³ + a²b − ab² − b³
How to solve: Write the quantities as multiples of one unknown (ak, bk), use the given condition to find k, then answer the question.
- The third proportional is (a² − b²)² ÷ (a − b) = (a − b)(a + b)² = a³ + a²b − ab² − b³.
Shortcut: Ratios scale: multiply or divide all terms by the same number. Make the common term equal before combining ratios.
5. A finishes a job in 12 days working 8 hours a day. B finishes the same job in 8 days working 10 hours a day. Working together 8 hours a day, how many days do they need?
- 15/8 days
- 39/12 days
- 60/11 days
- 8 days
Answer
C. 60/11 days
How to solve: Rate = 1 ÷ days. Add rates for people working together; time = work ÷ total rate.
- A needs 12 days × 8 hours = 96 hours for the whole job
- A does 1/96 of it per hour.
- B needs 8 days × 10 hours = 80 hours
- B does 1/80 per hour.
- Together per hour: 1/96 + 1/80 = 5/480 + 6/480 = 11/480 of the job.
- So they need 480/11 hours.
- At 8 hours a day that is (480/11) ÷ 8 = 60/11 days.
Shortcut: Take total work as the LCM of the given days so every rate is a whole number of units per day.
6. Class B has 50% more students than class A. The number of girls in A equals the number of boys in B, and the percentage of girls is the same in both classes. What percentage of all the students are boys?
- 25%
- 40%
- 60%
- 33.33%
Answer
B. 40%
How to solve: Turn every percent into a fraction of the right base, write what you know as an equation and solve it step by step.
- Let class A have T students, g of them girls (so T − g boys).
- Class B has 50% more students: 1.5T.
- B's boys = A's girls = g
- B has 1.5T − g girls.
- Equal girl percentages: (1.5T − g)/1.5T = g/T.
- Multiply both sides by 1.5T: 1.5T − g = 1.5g
- 1.5T = 2.5g and g = 0.6T.
- Boys in A = T − 0.6T = 0.4T
- Boys in B = g = 0.6T.
- Total boys = 0.4T + 0.6T = T out of T + 1.5T = 2.5T students: T ÷ 2.5T × 100 = 40%.
Shortcut: Percent → fraction: 10% = 1/10, 12.5% = 1/8, 20% = 1/5, 25% = 1/4, 33⅓% = 1/3, 50% = 1/2. Fractions make the arithmetic quick.
7. A loan of ₹22,000 is repaid in two equal yearly instalments at 20% per annum compound interest. What is each instalment?
- ₹16,400
- ₹11,000
- ₹14,400
- ₹12,800
Answer
C. ₹14,400
How to solve: Amount = P × (1 + R/100)^n, and compound interest = amount − P. Each year earns interest on the previous year's interest too.
- Each instalment x is paid at the end of year 1 and year 2.
- Its value today is x ÷ 1.2 (one year early) and x ÷ 1.2² = x ÷ 1.44 (two years early).
- These add up to the loan: x/1.2 + x/1.44 = 22,000.
- Write x/1.2 as 1.2x/1.44, so (1.2x + x)/1.44 = 2.2x/1.44 = 22,000.
- So x = 22,000 × 1.44 ÷ 2.2 = ₹14,400.
Shortcut: For 2 years, CI = 2R + R²/100 percent of P (e.g. 10% → 21%). For 3 years at 10%, the amount is 1.331 × P.
8. 60 litres of oil at ₹32 a litre are mixed with oil that costs ₹9 a litre less. The mixture sells at ₹28 a litre with no profit and no loss. How many litres of the cheaper oil were used?
- 36 litres
- 48 litres
- 50 litres
- 24 litres
Answer
B. 48 litres
How to solve: Track the quantity that does not change, or use the weighted average: mean = Σ(quantity × value) ÷ Σ quantity.
- The cheaper oil costs 32 − 9 = ₹23 a litre.
- Let B litres of it be used.
- Total cost = 60 × 32 + 23B = 1920 + 23B.
- With no profit or loss this equals the selling value 28 × (60 + B) = 1680 + 28B.
- So 1920 + 23B = 1680 + 28B, which gives 1920 − 1680 = 28B − 23B: 240 = 5B and B = 48 litres.
Shortcut: Alligation gives mixing ratios instantly: ratio = (high − mean) : (mean − low).
9. A square is drawn inside a circle of circumference 100 cm with its corners on the circle. What is the side of the square?
- 50√2 cm
- 100/π cm
- 50√2/π cm
- 100√2/π cm
Answer
C. 50√2/π cm
How to solve: Pick the right formula for the shape and keep the units consistent.
- Diameter = 100/π, which is the square's diagonal.
- Side = diagonal/√2 = 100/(π√2) = 50√2/π cm.
Shortcut: Area scales with the square and volume with the cube: double the side → 4× area, 8× volume.