Quantitative Aptitude PYQs

Shortcut Trick

Ratio of Speed of Boat (u) : Speed of Stream (v) = 7 : 5.

Downstream speed ratio = 7 + 5 = 12 units.

Actual downstream speed = 156 km / 1 hr = 156 km/hr.

12 units = 156 → 1 unit = 13 km/hr.

Difference (u − v) = 7 − 5 = 2 units = 2 × 13 = 26 km/hr.

∴ The correct answer is 26 km/hr.

Alternate Method

Given:

Ratio of Speed of Boat (u) to Speed of Stream (v) = 7 : 5

Let u = 7x and v = 5x

Formula Used:

Downstream Speed = Speed of Boat + Speed of Stream

Speed = Distance ÷ Time

Calculations:

⇒ Downstream Speed = 156 km ÷ 1 hr = 156 km/hr

⇒ 7x + 5x = 156

⇒ 12x = 156

⇒ x = 156 ÷ 12 = 13

⇒ Difference = 7x − 5x = 2x

⇒ Difference = 2 × 13 = 26 km/hr

∴ The correct answer is 26 km/hr.

Additional Information

Downstream and Upstream Speed

Downstream Speed (D) = u + v and Upstream Speed (U) = u − v, where u is boat speed in still water and v is stream speed.

Calculating Component Speeds

Speed of Boat in still water (u) = (D + U) ÷ 2 and Speed of Stream (v) = (D − U) ÷ 2.

Average Speed Formula

If a boat travels a certain distance downstream and returns the same distance upstream, Average Speed = (D × U) ÷ u.

Shortcut Trick

Remaining quantity of milk = Initial quantity × (1 − Replaced quantity ÷ Total capacity)ⁿ

Here, Initial quantity = 60 liters, Replaced quantity = 6 liters, and number of operations n = 3.

Ratio of remaining milk = (1 − 6 ÷ 60)³ = (9 ÷ 10)³ = 0.729

Remaining milk = 60 × 0.729 = 43.74 liters.

∴ The correct answer is 43.74 litre.

Alternate Method

Given:

Initial quantity of pure milk = 60 liters

Quantity of milk replaced in each cycle = 6 liters

Total number of operations (n) = 1 (initial) + 2 (further) = 3 times

Formula Used:

Remaining Milk = PV × (1 − x ÷ V)ⁿ

Where, PV = Initial volume of milk, x = Volume replaced in each step, V = Total volume of the container, n = Number of times the process is repeated.

Calculations:

⇒ Remaining Milk = 60 × (1 − 6 ÷ 60)³

⇒ Remaining Milk = 60 × (1 − 1 ÷ 10)³

⇒ Remaining Milk = 60 × (9 ÷ 10)³

⇒ Remaining Milk = 60 × (729 ÷ 1000)

⇒ Remaining Milk = 6 × 7.29

⇒ Remaining Milk = 43.74 liters

∴ The correct answer is 43.74 litre.

Additional Information

Repeated Dilution Formula

When a quantity x is withdrawn from a vessel containing V volume of liquid, and replaced by water, after n operations, the quantity of pure liquid remaining is V × (1 − x/V)ⁿ.

Ratio Method for Mixture Replacement

The multiplying factor in each step is (Total Volume − Replaced Volume) ÷ Total Volume. For multiple cycles, raise this fraction to the power of the number of operations.

Shortcut Trick

Given Curved Surface Area of cylinder = 2πrh = 88 cm².

Let radius be r and height be r + 8.6. Then 2 × (22/7) × r × (r + 8.6) = 88 ⇒ r(r + 8.6) = 14.

By inspection, if r = 1.4, then 1.4 × (1.4 + 8.6) = 1.4 × 10 = 14 (Satisfied).

Radius of sphere = (3/2) × 1.4 = 2.1 cm. Volume = (4/3) × (22/7) × (2.1)³ = 38.808 ≈ 39 cm³.

∴ The correct answer is 39 cubic.cm.

Alternate Method

Given: CSA of cylinder = 88 cm², Height (h) = r + 8.6, Ratio r : R_s = 2 : 3.

Formula Used: CSA = 2πrh, Volume of Sphere = (4/3)πR³.

⇒ 2 × (22/7) × r × (r + 8.6) = 88

⇒ (44/7) × (r² + 8.6r) = 88

⇒ r² + 8.6r = 14 ⇒ r² + 8.6r − 14 = 0

⇒ Solving for r: r = 1.4 cm (taking positive root)

⇒ Radius of sphere (R_s) = (3/2) × 1.4 = 2.1 cm

⇒ Volume = (4/3) × (22/7) × 2.1 × 2.1 × 2.1

⇒ Volume = 4 × 22 × 0.1 × 4.41 ÷ 1 = 38.808 cm³

∴ The correct answer is 39 cubic.cm.

Additional Information

Surface Area and Volume of Sphere

Total Surface Area of sphere = 4πR² and Volume = (4/3)πR³.

Properties of Cylinder

CSA = 2πrh, Total Surface Area (TSA) = 2πr(h + r), and Volume = πr²h.

Mensuration Unit Conversion

Always ensure units are consistent; 1 liter = 1000 cm³ and 1 m³ = 1,000,000 cm³.

Shortcut Trick
Quantity A: Efficiency ratio P : Q = 2 : 1 ⇒ Time ratio = 1 : 2. Difference 1 unit = 40 days. P = 40, Q = 80. Together = (40 × 80) ÷ (40 + 80) = 80/3 = 26.66 days.
Quantity B: Efficiency ratio Avinash : Rahul = 100 : 75 = 4 : 3. Avinash takes 60 days. Total work = 4 × 60 = 240 units. Together = 240 ÷ (4 + 3) = 240/7 = 34.28 days.
Comparing A (26.66) and B (34.28) ⇒ A < B.
∴ The correct answer is Quantity A < Quantity B.
Alternate Method
Quantity A:
Given: Efficiency of P = 2 × Q. Let time taken by Q be x days, then P takes (x − 40) days.
⇒ Efficiency ∝ 1/Time ⇒ 2/1 = x / (x − 40)
⇒ 2x − 80 = x ⇒ x = 80 days (Q), P = 40 days.
Person Days Total Work Efficiency P 40 80 2 Q 80 1
⇒ Time taken together = 80 ÷ (2 + 1) = 80/3 = 26.66 days.
Quantity B:
Given: Avinash = 60 days. Rahul efficiency = 75% of Avinash.
⇒ If Avinash efficiency = 4, Rahul efficiency = 3.
⇒ Total Work = 60 × 4 = 240 units.
⇒ Time taken together = 240 ÷ (4 + 3) = 240/7 = 34.28 days.
Comparison: 26.66 days < 34.28 days.
∴ The correct answer is Quantity A < Quantity B.
Additional Information
Efficiency and Time Relationship
Efficiency is inversely proportional to the time taken to complete a task. If the ratio of efficiency is a:b, the ratio of time taken is b:a.
Combined Work Formula
For two people with individual times X and Y, the time taken together is (X × Y) ÷ (X + Y).
Work and Wages
Wages are distributed in the direct ratio of the work done by each person, which is equivalent to the ratio of their efficiencies when they work for the same duration.

Shortcut Trick
Total balls = 10 + 10 + 5 = 25.
Probability (Quantity A) = (Red / Total) × (Purple / Total) = (10 / 25) × (10 / 25).
Calculation: (2 / 5) × (2 / 5) = 4 / 25 = 0.16.
Quantity A (0.16) > Quantity B (0.06).
∴ The correct answer is Quantity A > Quantity B.
Alternate Method
Given:
Number of Red balls = 10, Purple balls = 10, Black balls = 5.
Total balls = 10 + 10 + 5 = 25.

Quantity A:
The event is with replacement (ball is kept back).
⇒ P(Red) = 10 / 25 = 2 / 5.
⇒ P(Purple) = 10 / 25 = 2 / 5.
⇒ Total Probability = (2 / 5) × (2 / 5) = 4 / 25.
⇒ 4 / 25 = 0.16.
Quantity B:
⇒ Value = 0.06.
Comparison:
⇒ 0.16 > 0.06.
∴ The correct answer is Quantity A > Quantity B.
Additional Information
Independent Events (With Replacement)
When an item is replaced, the probability of the second event remains unchanged by the first event: P(A ∩ B) = P(A) × P(B).
Dependent Events (Without Replacement)
If the ball was not replaced, the total sample space for the second draw would decrease to 24: P(A ∩ B) = P(A) × P(B|A).
Probability Range
The probability of any event always lies between 0 and 1 inclusive (0 ≤ P(E) ≤ 1).

Shortcut TrickQuantity A: Distance = (S1 × S2) / (S2 − S1) × ΔTDistance = (20 × 25) / (25 − 20) × (90 / 60) = (500 / 5) × 1.5 = 100 × 1.5 = 150 km.Quantity B: Distance = Speed × Time = 80 × 2 = 160 km.Comparing: 150 km < 160 km ⇒ Quantity A < Quantity B.∴ The correct answer is Quantity A < Quantity B.Alternate MethodQuantity A:Given Speed 1 (S1) = 20 km/hr, Speed 2 (S2) = 25 km/hr, Time saved = 90 minutes = 1.5 hours.⇒ Let distance be D.⇒ D/20 − D/25 = 1.5⇒ (5D − 4D) / 100 = 1.5⇒ D / 100 = 1.5⇒ D = 150 km.Quantity B:Given Speed = 80 km/hr, Time = 2 hours.⇒ Distance = Speed × Time⇒ Distance = 80 × 2 = 160 km.⇒ 150 km (Quantity A) < 160 km (Quantity B).∴ The correct answer is Quantity A < Quantity B.Additional InformationSpeed-Time Inverse RelationshipFor a constant distance, speed is inversely proportional to time (S1/S2 = T2/T1). This is useful for solving ratio-based distance problems.Distance Formula for Speed ChangeWhen the difference in time (ΔT) is given for two different speeds covering the same distance: Distance = [Product of Speeds / Difference of Speeds] × ΔT.Average SpeedIf a person covers equal distances at speeds x and y, the average speed is (2xy) / (x + y).

Shortcut Trick
Quantity I: Profit % = [Markup % − Discount % − (Markup × Discount)/100] = [50 − 20 − (50 × 20)/100] = 20%.
Profit = 20% of ₹2000 = ₹400.
Quantity II: SP = MP − Discount = 2800 − (10% of 2800) = 2800 − 280 = ₹2520.
Profit = SP − CP = 2520 − 2100 = ₹420.
Comparing both: ₹400 < ₹420.
∴ The correct answer is Quantity I < Quantity II.
Alternate Method

Quantity I:
⇒ Cost Price (CP) = ₹2000
⇒ Marked Price (MP) = 2000 + 50% of 2000 = 2000 + 1000 = ₹3000
⇒ Selling Price (SP) = 3000 − 20% of 3000 = 3000 − 600 = ₹2400
⇒ Profit = 2400 − 2000 = ₹400
Quantity II:
⇒ Marked Price (MP) = ₹2800
⇒ Discount = 10% of 2800 = ₹280
⇒ Selling Price (SP) = 2800 − 280 = ₹2520
⇒ Cost Price (CP) = ₹2100
⇒ Profit = 2520 − 2100 = ₹420
Comparison: Quantity I (₹400) < Quantity II (₹420)
∴ The correct answer is Quantity I < Quantity II.
Additional Information
Marked Price and Discount
MP is the price printed on the label. Discount is always calculated on MP: SP = MP × (100 − Discount%)/100.
Profit and Markup Relation
Profit % = Markup % − Discount % − (Markup % × Discount %)/100. This is useful for successive percentage calculations.
Cost Price and Profit
Profit = SP − CP. Profit % is always calculated on the CP unless otherwise stated: Profit % = (Profit/CP) × 100.

Shortcut Trick

Let Vani’s pocket money be 100%.

Saurabh receives 16%.

Saurabh’s remaining % after spending (15% + 45% = 60%) is 40% of his share.

40% of 16% = 3200 ⇒ 6.4% = 3200.

100% = (3200 ÷ 6.4) × 100 = 50000.

∴ The correct answer is Rs 50000.

Alternate Method

Given:

Vani’s contribution to Saurabh = 16%

Saurabh’s expenses = 15% (Books) + 45% (Watch) = 60%

Saurabh’s savings = 100% − 60% = 40%

Remaining amount = Rs 3200

Calculations:

Let Vani’s initial pocket money be Rs X.

⇒ Amount received by Saurabh = 0.16 × X

⇒ Remaining amount with Saurabh = 40% of (0.16 × X)

⇒ 3200 = (40 ÷ 100) × 0.16 × X

⇒ 3200 = 0.40 × 0.16 × X

⇒ 3200 = 0.064 × X

⇒ X = 3200 ÷ 0.064

⇒ X = 50000

∴ The correct answer is Rs 50000.

Additional Information

Percentage Change

To find the original value when a percentage is given: Original = (Result ÷ Final Percentage) × 100.

Successive Percentage

When multiple percentages are applied to a base, use the multiplying factor: Final = Initial × f₁ × f₂.

Base Change

Always ensure the percentage is calculated on the correct base (e.g., Vani's total vs Saurabh's share).

Shortcut Trick

Let the numbers be 3x, 5x and 5y, 3y.

Given: 3y − 5x = 8 and 8x + 8y = 192 ⇒ x + y = 24.

Solving equations: x = 8, y = 16.

Sum of 1^st and 3^rd = 3x + 5y = 3(8) + 5(16) = 24 + 80 = 104.

∴ The correct answer is 104.

Alternate Method

Given: Sum = 192; Ratio A:B = 3:5; Ratio C:D = 5:3; D − B = 8.

Calculations:

⇒ Let 1^st = 3x, 2^nd = 5x, 3^rd = 5y, 4^th = 3y.

⇒ 3x + 5x + 5y + 3y = 192 ⇒ 8x + 8y = 192 ⇒ x + y = 24 ---(Eq. 1)

⇒ 3y − 5x = 8 ---(Eq. 2)

⇒ Multiply (Eq. 1) by 5: 5x + 5y = 120

⇒ Add this to (Eq. 2): (5x + 5y) + (3y − 5x) = 120 + 8

⇒ 8y = 128 ⇒ y = 16

⇒ x = 24 − 16 = 8

⇒ Sum of 1^st and 3^rd = 3x + 5y = 3(8) + 5(16) = 24 + 80 = 104.

∴ The correct answer is 104.

Additional Information

Ratio and Proportion

When multiple ratios are given for different pairs of numbers, use common variables (like x and y) to represent the actual values before solving linear equations.

Linear Equations

System of two variables can be solved using the Elimination Method or Substitution Method to find the values of unknowns.

Shortcut Trick

Ratio of Profit = (Capital_A × Time_A) : (Capital_B × Time_B)

Ratio = (4000 × 10) : (5000 × 12) = 40000 : 60000 = 2 : 3.

B's share is 3 units = Rs. 450 ⇒ 1 unit = Rs. 150.

Distributed profit (75%) = 2 + 3 = 5 units = 5 × 150 = Rs. 750. Total profit (100%) = 750 × (100 ÷ 75) = Rs. 1000.

∴ The correct answer is Rs. 1000.

Alternate Method

Given: A's Investment = Rs. 4000, B's Investment = Rs. 5000, A's Time = 10 years, B's Time = 12 years, Charity = 25%, B's share = Rs. 450. Formulas: Profit Ratio = (C₁ × T₁) : (C₂ × T₂) | Total Profit = Distributed Profit ÷ (1 − Charity fraction).

⇒ Profit sharing ratio of A and B = (4000 × 10) : (5000 × 12) = 40 : 60 = 2 : 3.
⇒ Let the profit shared between them be P_s.
⇒ B's share = [3 ÷ (2 + 3)] × P_s = 450.
⇒ (3 ÷ 5) × P_s = 450 ⇒ P_s = 450 × (5 ÷ 3) = Rs. 750. ⇒ Since 25% went to charity, P_s represents 75% of the total profit.
⇒ 0.75 × Total Profit = 750.
⇒ Total Profit = 750 ÷ 0.75 = Rs. 1000.

∴ The correct answer is Rs. 1000.

Additional Information

Partnership Profit Distribution

In a partnership, profit is always distributed based on the product of investment and time duration. Profit Ratio = (I₁T₁) : (I₂T₂).

Active vs. Sleeping Partner

Sometimes a portion of profit is set aside for management or charity before the remaining is divided according to the investment ratio.

General Solution:

Calculations:

E-books sold = 25% of total books = 0.25 × 4800 = 1200.

Non-e-books sold = Total books - E-books = 4800 - 1200 = 3600.

Let the average price of a non-e-book be P.

The price of an e-book is a 20% discount on P, so its price is P × (1 - 0.20) = 0.80P.

Total revenue = (Number of non-e-books× Price of non-e-book) + (Number of e-books × Price of e-book).

273,600 = (3600 × P) + (1200 × 0.80P).

273,600 = 3600P + 960P.

73,600 = 4560P.

Solve for the price of a non-e-book (P):

P = 273,600 / 4560.

P = 60.

The average price of a single non-e-book was Rs. 60.

Hence, the correct answer is Rs. 60.

General Solution:

Calculations:

Given:

2013 Books sold = 3500

2013 Magazines sold = 2000

Book: CP = ₹50, SP = ₹70 ⇒ Profit per Book = ₹70 − ₹50 = ₹20

Magazine: CP = ₹30, SP = ₹50 ⇒ Profit per Magazine = ₹50 − ₹30 = ₹20

Profit from Books = 3500 × ₹20 = ₹70,000

Profit from Magazines = 2000 × ₹20 = ₹40,000

Total Profit = ₹70,000 + ₹40,000 = ₹1,10,000

Hence, the correct answer is Rs. 1,10,000.

General Solution:

Calculations:

Magazines sold in 2014 = 1200

Magazines sold in 2016 = 1400

Total Magazines (2014 + 2016) = 1200 + 1400 = 2600

Books sold in 2015 = 4800

Required percentage = (2600 ÷ 4800) × 100

= (2600 × 100) ÷ 4800

= 260000 ÷ 4800

= 54.17%

Hence, the correct answer is 54.17%

General Solution:

Calculations:

Books 2012 = 6000

Books 2013 = 3500

Books 2014 = 1800

Total Books = 6000 + 3500 + 1800 = 11300

Magazines 2015 = 3000

Magazines 2016 = 1400

Total Magazines = 3000 + 1400 = 4400

Difference = 11300 − 4400 = 6900

Thus, total number of Books (2012–2014) is 6900 more than the total number of Magazines (2015–2016).

General Solution:

Calculations:

Total Books in 2016 = 2200

Let Paperback books = x

Hardcover books = 66.66% of Paperback books = (2/3) × x

Equation: Paperback + Hardcover = 2200

x + (2/3)x = 2200

(5/3)x = 2200

x = (2200 × 3) ÷ 5 = 1320

Hence, the correct answer is 1320.

General Solution:

Calculations:

Find total items sold in 2015:

Books in 2015 = 4800

Magazines in 2015 = 3000

Total in 2015 = 4800 + 3000 = 7800

Find total in 2017:

Given: 2017 total = 20% more than 2015

20% of 7800 = 0.2 × 7800 = 1560

So total in 2017 = 7800 + 1560 = 9360

Ratio Books : Magazines = 4 : 1 → total ratio parts = 5

One part = 9360 ÷ 5 = 1872

Books = 4 × 1872 = 7488

Hence, the correct answer is 7488 Books.

Shortcut Trick

Calculate Effective Principal for 1 year: (500 × 3) + (700 × 2) + (300 × 4) = 1500 + 1400 + 1200 = 4100.

Rate of Interest = (Total Interest ÷ Total Effective Principal) × 100.

Rate = (615 ÷ 4100) × 100 = 615 ÷ 41 = 15%.

∴ The correct answer is 15%.

Alternate Method

Given: Total Amount = ₹1500, P₁ = ₹500, T₁ = 3 years, P₂ = ₹700, T₂ = 2 years, Total SI = ₹615.

Formula Used: Simple Interest (SI) = (P × R × T) ÷ 100

⇒ Remaining Principal (P₃) = 1500 − (500 + 700) = ₹300.

⇒ Let the rate of interest be R% per annum.

⇒ [(500 × R × 3) ÷ 100] + [(700 × R × 2) ÷ 100] + [(300 × R × 4) ÷ 100] = 615

⇒ 15R + 14R + 12R = 615

⇒ 41R = 615

⇒ R = 615 ÷ 41 = 15.

∴ The correct answer is 15%.

Additional Information

Simple Interest Property

Simple Interest is calculated only on the Principal amount, meaning the interest earned remains constant every year if the rate and principal are unchanged.

Effective Principal Concept

When multiple investments are made at the same rate, the total interest is equivalent to a single investment of Σ(P × T) for 1 year at that rate.

Amount Formula

Total Amount (A) = Principal (P) + Simple Interest (SI). For a single rate: A = P(1 + RT/100).

Shortcut Trick

Change in delay = 40 minutes − (4/9 × 60) minutes = 40 − 26.67 = 13.33 minutes = 40/3 minutes.

Speed ratio = 7 : 9 ⇒ Time ratio = 9 : 7. Difference in time units = 2 units.

For the 70 km stretch: 2 units → 40/3 minutes ⇒ 7 units (normal time) → 140/3 minutes.

Speed = Distance ÷ Normal Time = 70 ÷ (140 / (3 × 60)) = 70 × 180 ÷ 140 = 90 km/hr.

∴ The correct answer is 90 km/hr.

Alternate Method

Given: Accident 1 at 90 km, Delay = 40 min. Accident 2 at 160 km, Delay = 4/9 hr = 26.67 min.

Formula: Speed = Distance ÷ Time | Time Difference = Distance × (1/v_reduced − 1/v_normal)

⇒ Let normal speed be s and reduced speed be 7s/9.

⇒ Difference in delay = 40 − 80/3 = 40/3 minutes.

⇒ This difference occurs over the 70 km stretch where speed was 7s/9 in Case 1 and s in Case 2.

⇒ 70/(7s/9) − 70/s = 40/(3 × 60) hours

⇒ 90/s − 70/s = 2/9

⇒ 20/s = 2/9

⇒ s = (20 × 9) ÷ 2 = 90 km/hr.

∴ The correct answer is 90 km/hr.

Additional Information

Ratio of Speed and Time

When distance is constant, Speed is inversely proportional to Time. If speed becomes a/b of original, time taken becomes b/a of original.

Accident Problems Shortcut

The difference in time delay is solely due to the distance between the two potential accident points. Analyze only that specific segment to find the speed.

GIVEN :

25, 26, 30, 39, 65, 80

CALCULATION :

The series follows the following pattern:

25 + 12 = 26

26 + 22 = 30

30 + 32 = 39

39 + 42 = 55

55 + 52 = 80

∴ The wrong term in the series is 65.

Calculation:

The series follows the following pattern:

24 × 1 + 1 = 25

25 × 2 + 1 = 51

51 × 3 + 1 = 154

154 × 4 + 1 = 617

617 × 5 + 1 = 3086

∴ The wrong term in the series is 157.