Quantitative Aptitude PYQs

Shortcut Trick

First part time = 84 ÷ 24 = 3.5 hr

New speed = 24 × 1.25 = 30 km/hr

Second part time = 150 ÷ 30 = 5 hr

Total time = 3.5 + 5 = 8.5 hr

∴ The correct answer is 8.5 hr.

Alternate Method

Given: First distance (D₁) = 84 km, First speed (S₁) = 24 km/hr, Second distance (D₂) = 150 km.

Formula Used: Time = Distance ÷ Speed

⇒ Time for first part (T₁) = D₁ ÷ S₁

⇒ T₁ = 84 ÷ 24 = 3.5 hr

⇒ Increased speed (S₂) = S₁ + (25% of S₁)

⇒ S₂ = 24 + (25/100 × 24) = 24 + 6 = 30 km/hr

⇒ Time for second part (T₂) = D₂ ÷ S₂

⇒ T₂ = 150 ÷ 30 = 5 hr

⇒ Total Time = T₁ + T₂ = 3.5 + 5 = 8.5 hr

∴ The correct answer is 8.5 hr.

Additional Information

Average Speed Formula

Average Speed = Total Distance ÷ Total Time. For this journey, Average Speed = (84 + 150) ÷ 8.5 ≈ 27.53 km/hr.

Percentage Increase in Speed

When speed increases by x%, the new speed is S × (1 + x/100). Here, 24 × (1 + 0.25) = 24 × 1.25 = 30 km/hr.

Unit Conversion

To convert km/hr to m/s, multiply by 5/18. To convert m/s to km/hr, multiply by 18/5.

Let the total salary be S.

Amount spent on rent =20% of S.

Remaining salary after rent = 80% of S.

Amount spent on Food = 30% of the remaining salary.

Amount spent on Transport = 25% of the remaining salary.

Amount spent on Entertainment = Remaining percentage of the remaining salary.

Total percentage spent on Food and Transport = 30% + 25% = 55% 

Percentage spent on Entertainment = 100% - 55% = 45%.

Entertainment = 45%of the remaining salary.

Given that Entertainment = ₹16,560:

45% of 80% of S = 16,560

0.45 × 0.8 × S = 16,560

0.36 × S = 16,560

S = 16,560 / 0.36

S = 46,000

Thus, the total salary of the person is ₹46,000.

Shortcut Trick

Area of circle 616 cm² → Radius (r) = 14 cm, Diameter (d) = 28 cm.

Ratio Diameter : Breadth = 7 : 4 ⇒ 7 units = 28 cm ⇒ 1 unit = 4 cm.

Breadth (b) = 4 × 4 = 16 cm; Length (l) = 16 + 6 = 22 cm.

Area of rectangle = 22 × 16 = 352 cm².

∴ The correct answer is 352 cm².

Alternate Method

Given: Area of circle = 616 cm², Ratio d : b = 7 : 4, l = b + 6.

Formula Used: Area of Circle = πr²; Area of Rectangle = l × b.

⇒ πr² = 616

⇒ (22/7) × r² = 616

⇒ r² = (616 × 7) ÷ 22 = 28 × 7 = 196

⇒ Radius (r) = 14 cm ∴ Diameter (d) = 28 cm

⇒ Diameter : Breadth = 7 : 4 ⇒ 28 : b = 7 : 4

⇒ Breadth (b) = (28 × 4) ÷ 7 = 16 cm

⇒ Length (l) = 16 + 6 = 22 cm

⇒ Area of rectangle = 22 × 16 = 352 cm²

∴ The correct answer is 352 cm².

Additional Information

Circle Properties

Circumference = 2πr and Diameter = 2r. For r = 7, Area = 154; for r = 14, Area = 616.

Rectangle Perimeter

Perimeter of a rectangle is given by 2(l + b), where l is length and b is breadth.

Ratio Applications

If a : b = x : y, then a = kx and b = ky where k is a constant proportionality factor.

Calculation:

Total shows in February = 160 + 210 + 200 = 570

Total shows in March = 150 + 220 + 240 = 610

Difference = Total shows in March - Total shows in February

⇒ 610 - 570 = 40

∴ The difference between the total number of movie shows in February and March across all multiplexes is 40.

Calculation:

Number of movie shows in January at multiplex C = 140

Number of movie shows in January at D is 5/7^th of that at C.

We need to find the number of movie shows at D in January.

Shows at D = (5/7) × Shows at C

⇒ (5/7) × 140

⇒ 700 / 7

⇒ 100

∴ The number of movie shows in January at D is 100.

Calculation:

Total shows in January and February at B = 270 + 210 = 480

Total shows in February and March at C = 200 + 240 = 440

Required Ratio = (Total shows in B for January & February) : (Total shows in C for February & March)

⇒ 480 : 440

⇒ 12 : 11

∴ The required ratio is 12 : 11.

Calculation:

Number of movie shows in March at multiplex C = 240

Number of movie shows in April at C is 25% more than in March.

Number of shows in April at C:

⇒ 240 × (1 + 25/100)

⇒ 240 × (1 + 0.25)

⇒ 240 × 1.25

⇒ 300

∴ The number of movie shows in April at C is 300.

Calculation:

Total shows at A:

⇒ 180 + 160 = 340

Total shows at B:

⇒ 270 + 210 = 480

Difference = Total shows at B - Total shows at A

⇒ 480 - 340 = 140

∴ The number of movie shows in January and February at A is 140 less than at B.

Shortcut Trick

Let the total principal be 100 units.

Weighted Interest Rate = (30% × 12%) + (70% × 8%) = 3.6% + 5.6% = 9.2%.

9.2% of Principal = 2438

Principal = 2438 ÷ 0.092 = 26500.

∴ The correct answer is 26,500.

Alternate Method

Given: Interest 1 (at 12%) + Interest 2 (at 8%) = 2438.

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

⇒ Let Principal be P.

⇒ SI₁ = (0.30P × 12 × 1) ÷ 100 = 0.036P

⇒ SI₂ = (0.70P × 8 × 1) ÷ 100 = 0.056P

⇒ Total Interest = 0.036P + 0.056P = 0.092P

⇒ 0.092P = 2438

⇒ P = 2438 ÷ 0.092 = 26500

∴ The correct answer is 26,500.

Additional Information

Weighted Average Rate of Interest

When a principal is divided, the effective rate R_eff = (w₁r₁ + w₂r₂ + ...), where w is the weightage of each part.

Simple Interest Properties

Simple interest is calculated only on the original principal. Interest remains constant every year if the rate and principal are unchanged.

Shortcut Trick

Total Work (LCM of 48, 72, 54) = 432 units.

Efficiencies: A = 9, B = 6, C = 8.

Work done by (A + B) in 16 days = (9 + 6) × 16 = 15 × 16 = 240 units.

Remaining Work = 432 − 240 = 192 units.

Time taken by C = 192 ÷ 8 = 24 days.

∴ The correct answer is 24 days.

Alternate Method

Given: Individual time of A = 48 days, B = 72 days, C = 54 days.

Formula Used: Efficiency = Total Work ÷ Time; Work Done = Efficiency × Time.

⇒ Total work = LCM(48, 72, 54) = 432 units.

⇒ Combined Efficiency of A and B = 9 + 6 = 15 units/day.

⇒ Work completed by A and B in 16 days = 15 × 16 = 240 units.

⇒ Remaining Work = 432 − 240 = 192 units.

⇒ Days taken by C to finish remaining work = 192 ÷ 8 = 24 days.

∴ The correct answer is 24 days.

Additional Information

Pipes & Cisterns

Conceptually identical to Time & Work, where inlet pipes have positive efficiency and outlet pipes have negative efficiency.

Work and Wages

Wages are distributed in the ratio of the actual work done by each person, or in the ratio of their efficiencies if they work for the same duration.

Alternate Day Work

Calculate the work done in one cycle (sum of efficiencies over the cycle period) and then divide the total work by the cycle total.

Calculation 

⇒ 15 + 2 = 17 

⇒ 17 + 3 = 20

⇒ 20 + 5 = 25 

⇒ 25 + 7 = 32 

⇒ 32 + 11 = 43 

∴ The missing number is 20

Solution:

2, 9, 30, ?, 282, 849

The series follows the following pattern:

⇒ 2 x 3 + 3 = 9

⇒ 9 x 3 + 3 = 30

⇒ 30 x 3 + 3 = 93

⇒ 93 x 3 + 3 = 282

⇒ 282 x 3 + 3 = 849

⇒ The missing term in the series will be 93.

Hence, 93 is the correct answer.

Given:

110, 120, 135, 155, 180, ?

Calculation:

Considering the above series, the logic can be explained as follows

∴ The required value of "?" is 210

Shortcut Trick

Ratio of weights: Chinu : Arshi : Bhavya = 3 : 2 : 1

Total ratio parts = 3 + 2 + 1 = 6 units

Total weight = Average × Count = 20 × 3 = 60 kg

1 unit = 60 ÷ 6 = 10 kg

Chinu (3 units) = 3 × 10 = 30 kg

∴ The correct answer is 30 kg.

Alternate Method

Given:

Average weight of 3 children = 20 kg

Let the weight of Bhavya be x kg.

Weight of Chinu = 3x kg

Weight of Arshi = 2x kg

Formula Used:

Total Sum = Average × Number of entities

Calculations:

⇒ Total Weight = 20 × 3 = 60 kg

⇒ x + 3x + 2x = 60

⇒ 6x = 60

⇒ x = 10 kg

⇒ Weight of Chinu = 3x = 3 × 10 = 30 kg

∴ The correct answer is 30 kg.

Additional Information

Average Property

Average is the sum of observations divided by the total number of observations. Formula: Average = Σx_i / n.

Ratio & Proportion

If A:B = a:b and B:C = b:c, then A:B:C = a:b:c. This is used to distribute a total value among multiple parties.

Weighted Average

Used when different groups have different averages. Formula: (n₁×Avg₁ + n₂×Avg₂) / (n₁ + n₂).

Shortcut Trick

Let Total People = 100 units. Females = 60 units.

Postgraduate Females = 60% of 60 units = 36 units.

36 units → 360 ⇒ 1 unit = 10.

Males = 40 units = 40 × 10 = 400.

∴ The correct answer is 400.

Alternate Method

Given:

Percentage of males = 40%

Percentage of females = 100% − 40% = 60%

Percentage of Postgraduate females = 100% − 40% = 60% of females.

Total Postgraduate females = 360.

Calculations:

⇒ Let total number of people be 'x'.

⇒ Number of females = 0.60x

⇒ Number of Postgraduate females = 60% of 0.60x = 360

⇒ 0.60 × 0.60x = 360

⇒ 0.36x = 360

⇒ x = 360 ÷ 0.36 = 1000

⇒ Number of males = 40% of 1000 = 0.40 × 1000 = 400

∴ The correct answer is 400.

Additional Information

Percentage Breakdown

When a subgroup is defined as a percentage of another subgroup, multiply the percentages: Final % = Sub-percentage × Main percentage.

Base Change Rule

Always identify the correct base for percentage calculations (e.g., 40% of females means the base is the total number of females, not the total group).

Shortcut Trick

John salary = ₹12000. Rahul salary = 80% of 12000 = ₹9600.

Spent on Food (20%) + Furniture (25%) + Rent (30%) = 75%.

Remaining for Books = 100% − 75% = 25%.

Total for Books + Furniture = 25% + 25% = 50%.

Required Amount = 50% of ₹9600 = ₹4800.

∴ The correct answer is ₹4800.

Alternate Method

Given: John salary = ₹12000, Rahul salary = 20% less than John, Expenses: Food = 20%, Furniture = 25%, Rent = 30%.

Calculations:

⇒ Rahul salary = 12000 − (20% of 12000)

⇒ Rahul salary = 12000 − 2400 = ₹9600

⇒ Percentage spent on Books = 100 − (20 + 25 + 30) = 25%

⇒ Total % for Books and Furniture = 25% + 25% = 50%

⇒ Required Amount = 50% of 9600 = 9600 ÷ 2 = ₹4800

∴ The correct answer is ₹4800.

Additional Information

Percentage Change

To find a value x% less than a base B, use the formula: B × (100 − x)/100.

Successive Percentage Expenditure

If expenses are given as a % of the 'total' salary, they are additive. If given as % of 'remaining', they are multiplicative.

Base Value Concept

Always identify the correct base (Rahul's salary vs John's salary) before applying percentage calculations.

Shortcut Trick

Total Work (LCM of 15, 18, 10) = 90 units.

Efficiency of (A+B+C) = 90 ÷ 10 = 9 units/day.

Efficiency of (B+C) = 90 ÷ 18 = 5 units/day ⇒ Efficiency of A = 9 − 5 = 4 units.

Efficiency of (A+B) = 90 ÷ 15 = 6 units/day ⇒ Efficiency of B = 6 − 4 = 2 units.

Wage ratio = Efficiency ratio ⇒ A : B = 4 : 2 = 2 :

1. If 2 units = ₹600, then 1 unit (B’s wage) = ₹300.

∴ The correct answer is ₹300.

Alternate Method

Given: Time taken by (A+B) = 15 days, (B+C) = 18 days, (A+B+C) = 10 days. A’s wage = ₹600.

Formula: Efficiency = Total Work ÷ Time; Wage ∝ Efficiency.

⇒ Efficiency of A = Efficiency(A+B+C) − Efficiency(B+C) = 9 − 5 = 4

⇒ Efficiency of B = Efficiency(A+B) − Efficiency(A) = 6 − 4 = 2

⇒ Wage is distributed in the ratio of efficiencies of A : B = 4 : 2 = 2 : 1

⇒ Since A’s wage (2 units) = ₹600

⇒ B’s wage (1 unit) = 600 ÷ 2 = ₹300

∴ The correct answer is ₹300.

Additional Information

Work and Wages Relationship

Wages are always distributed in the ratio of work done. If all persons work for the same duration, the wage ratio is identical to their efficiency ratio.

LCM Method for Total Work

Total work is assumed as the Least Common Multiple (LCM) of the individual days to ensure efficiencies are whole numbers, simplifying calculations.

Efficiency of Individual in a Group

Efficiency of one person can be found by subtracting the combined efficiency of others from the total combined efficiency: Eff(A) = Eff(A+B+C) − Eff(B+C).

Shortcut Trick

Initial average = 56.2 kg.

Calculate deviations of new weights from 56.2: (62.3 − 56.2) + (59.7 − 56.2) + (59.5 − 56.2) = 6.1 + 3.5 + 3.3 = 12.9 kg.

Net change in average = Total Deviation ÷ Total children = 12.9 ÷ 63 ≈ 0.20 kg.

New average = 56.2 + 0.20 = 56.40 kg.

∴ The correct answer is 56.40 kg.

Alternate Method

Given: Initial average weight of 60 children = 56.2 kg. Weights of 3 new children = 62.3 kg, 59.7 kg, and 59.5 kg.

Formula Used: Average = (Sum of all observations) ÷ (Total number of observations)

⇒ Total weight of 60 children = 60 × 56.2 = 3372 kg.

⇒ Total weight of 63 children = 3372 + 62.3 + 59.7 + 59.5

⇒ Total weight of 63 children = 3372 + 181.5 = 3553.5 kg.

⇒ New Average = 3553.5 ÷ 63

⇒ New Average = 56.408... ≈ 56.40 kg.

∴ The correct answer is 56.40 kg.

Additional Information

Deviation Method for Average

New Average = Old Average + (Sum of deviations ÷ New total count). This avoids large multiplications.

Weighted Average

When two groups are combined, Combined Average = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂), where n is the count and x is the average.

Shortcut Trick

Total Interest = ₹640

Interest from First Sum = 600 × 12% × 6 = ₹432

Interest from Second Sum = 640 − 432 = ₹208

Second Sum = (208 × 100) ÷ (10 × 6) = 2080 ÷ 6 = ₹346.66

∴ The correct answer is ₹346.66.

Alternate Method

Given:

Principal 1 (P₁) = ₹600, Rate 1 (R₁) = 12%, Time (T) = 6 years

Rate 2 (R₂) = 10%, Total Simple Interest (SI_total) = ₹640

Formula Used:

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

⇒ SI₁ = (600 × 12 × 6) ÷ 100

⇒ SI₁ = 6 × 12 × 6 = ₹432

⇒ SI₂ = SI_total − SI₁

⇒ SI₂ = 640 − 432 = ₹208

⇒ SI₂ = (P₂ × 10 × 6) ÷ 100

⇒ 208 = (P₂ × 60) ÷ 100

⇒ P₂ = (208 × 100) ÷ 60

⇒ P₂ = 2080 ÷ 6 = ₹346.66

∴ The correct answer is ₹346.66.

Additional Information

Simple Interest Property

Simple interest remains constant every year because it is always calculated on the original Principal amount.

Amount Formula

Total Amount (A) = Principal (P) + Simple Interest (SI). In this question, the interest was additive.

Shortcut Trick

Ratio of Ankit : Son now is 3 : 1 (Difference = 2 units).

Ratio 4 years ago was 13 : 3 (Difference = 10 units).

Equating differences: Multiply present ratio by 5 → 15 : 5.

Change in units: 15 − 13 = 2 units → 4 years. So, 1 unit = 2 years.

Ankit's present age = 15 × 2 = 30 years.

∴ The correct answer is 30 years.

Alternate Method

Given: Present age ratio is 3 :

1. Ratio 4 years ago was 13 : 3.

Let the present age of the son be x and Ankit be 3x.

⇒ (3x − 4) / (x − 4) = 13 / 3

⇒ 3 × (3x − 4) = 13 × (x − 4)

⇒ 9x − 12 = 13x − 52

⇒ 13x − 9x = 52 − 12

⇒ 4x = 40

⇒ x = 10

⇒ Ankit's present age = 3x = 3 × 10 = 30 years.

∴ The correct answer is 30 years.

Additional Information

Concept of Age Difference

The difference between the ages of two people always remains constant regardless of the number of years passed.

Ratio Balancing Method

To solve age ratio problems quickly, multiply the ratios such that the difference between the parts (e.g., 15−5=10 and 13−3=10) becomes equal.