Quantitative Aptitude PYQs

Cars in E = 160 − 100 = 60

Bikes in F = 60 + 25% of 60 = 60 + 15 = 75

Total vehicles in F = Average of vehicles in A and B = (180 + 210)/2 = 195

Cars in F = 195 − 75 = 120

Bikes in B = 120

Required percentage more/less = (120 – 120)/120 x 100 = 0%

Cars in A = 180 − 90 = 90

Cars in B = 210 − 120 = 90

Average number of cars in A and B = (90 + 90)/2 = 90

Bikes in C = 80

Bikes in E = 100

Average number of bikes in C and E = (80 + 100)/2 = 90

Required difference = 90 – 90 = 0

Total bikes in D = 150

Unsold bikes = 150 x (1/3) = 50

Unsold bikes are 37.5% less than cars sold.

Therefore, Cars sold by B = 50/0.625 = 80

Total cars in B = 210 – 120 = 90

Unsold cars in B = 90 – 80 = 10

Cars booked online in A = 60

Cars booked offline in A = 90 – 60 = 30

Bikes booked online in A = 20% of 90 = 18

Bikes booked offline in A = 90 – 18 = 72

Total vehicles booked online = 60 + 18 = 78

Total vehicles booked offline = 30 + 72 = 102

Required difference = 102 – 78 = 24

Defective cars in D = (1/9) x 90 = 10

Non-defective cars in D = 90 – 10 = 80

Non-defective bikes in D = (1/2) x 80 = 40

Defective bikes in D = 150 – 40 = 110

Total defective vehicles in D = 10 + 110 = 120

Shortcut Trick

Initial Ratio (Honey : Syrup) = 5 : 4 (Total = 9 units)

Final Ratio (Honey : Syrup) = 4 : 5 (Total = 9 units)

Since the total quantity remains the same (180 litres), the units match.

Quantity of Honey remaining = Initial Honey × (1 − Removed/Total)

4 = 5 × (1 − x/180) ⇒ 4/5 = 1 − x/180 ⇒ x/180 = 1/5 ⇒ x = 36 litres.

∴ The correct answer is 36 litres.

Alternate Method

Given: Total mixture = 180 litres. Ratio H:S = 5:4. Final Ratio H:S = 4:5.

Step 1: Initial quantities

⇒ Honey = (5/9) × 180 = 100 litres

⇒ Syrup = (4/9) × 180 = 80 litres

Step 2: After removing 'x' litres of mixture

⇒ Honey removed = (5/9)x

⇒ Syrup removed = (4/9)x

Step 3: Final Honey quantity

⇒ Final Honey = (4/9) × 180 = 80 litres

⇒ 100 − (5/9)x = 80

⇒ 20 = (5/9)x

⇒ x = (20 × 9) ÷ 5 = 36 litres

∴ The correct answer is 36 litres.

Additional Information

Mixture Replacement Rule

When a part of a mixture is replaced by one of the ingredients, the component not being added decreases in a geometric progression. Formula: Final = Initial × (1 − x/V)ⁿ.

Alligation in Replacement

Replacement can be solved by treating the removed part as 0% concentration of the component and the pure replacement as 100%.

Shortcut Trick

Let CP_A be 100 units. MP_A = 140 units. SP_A = 140 × 0.9 = 126 units.

Profit on A = 26 units. Since Profit_A = Profit_B, Profit on B = 26 units.

SP_A − SP_B = (CP_A + 26) − (CP_B + 26) = CP_A − CP_B = Rs. 600. Given CP_A + CP_B = Rs. 3,600 and CP_A − CP_B = Rs. 600. Solving both: 2 × CP_B = 3,000 ⇒ CP_B = Rs. 1,500.

∴ The correct answer is Rs 1500.

Alternate Method

Given → CP_A + CP_B = 3600 | MP_A = 1.4CP_A | SP_A = MP_A × 0.9 | SP_A − SP_B = 600 | Profit_A = Profit_B.
Formulas: Profit = SP − CP | SP = CP × (100 + P%) ÷ 100 | SP = MP × (100 − D%) ÷ 100

⇒ SP_A = CP_A × 1.4 × 0.9 = 1.26CP_A.
⇒ Profit_A = 1.26CP_A − CP_A = 0.26CP_A.
⇒ Since Profit_A = Profit_B, then SP_B = CP_B + 0.26CP_A.
⇒ SP_A − SP_B = 600 ⇒ 1.26CP_A − (CP_B + 0.26CP_A) = 600.
⇒ CP_A − CP_B = 600.
⇒ We have (CP_A + CP_B) = 3600 and (CP_A − CP_B) = 600.
⇒ Subtracting equations: 2 × CP_B = 3000 ⇒ CP_B = 1500.

∴ The correct answer is Rs 1500.

Additional Information

Marked Price and Selling Price

Marked Price is the price listed by the seller. Selling price is obtained by subtracting the discount from the Marked Price. SP = MP × (100 − Discount%) ÷ 100.

Profit Calculation

Profit in absolute terms (Rs.) is SP − CP. If profits are equal for two items, the difference in their Selling Prices is exactly equal to the difference in their Cost Prices.

Shortcut Trick

Let Vipin (V), Father (F), and Mother (M) be their present ages.

Total age = 34 × 3 = 102 years. Ratio F : M = 110 : 100 = 11 : 10.

Using V + 2 = (F − 4)/2 ⇒ F = 2V + 8. Substitute: V + (2V + 8) + (2V + 8)/1.1 = 102.

Solving for V gives Vipin's present age = 18 years.

Vipin's age 7 years hence = 18 + 7 = 25 years.

∴ The correct answer is 25 years.

Alternate Method

Given: Average age = 34, F = M + 10% of M, V + 2 = (F − 4) ÷ 2.

⇒ Total sum = V + F + M = 34 × 3 = 102

⇒ F = 1.1M → M = F ÷ 1.1

⇒ 2(V + 2) = F − 4 ⇒ F = 2V + 4 + 4 = 2V + 8

⇒ V + (2V + 8) + (2V + 8) ÷ 1.1 = 102

⇒ 1.1V + 1.1(2V + 8) + 2V + 8 = 102 × 1.1

⇒ 1.1V + 2.2V + 8.8 + 2V + 8 = 112.2

⇒ 5.3V + 16.8 = 112.2

⇒ 5.3V = 95.4

⇒ V = 18 years

⇒ Age after 7 years = 18 + 7 = 25 years

∴ The correct answer is 25 years.

Additional Information

Average Age Concept

The sum of ages is calculated by multiplying the average age by the number of persons (Sum = Average × n).

Percentage and Ratio

10% more means a multiplier of 1.1 or a ratio of 11:10. Always identify the base value (Mother's age here) for percentage calculations.

Age Relation Shifting

When shifting time, add or subtract the same constant for everyone. If current age is x, age 'n' years hence is (x + n) and 'n' years ago was (x − n).

Shortcut Trick

Ratio of A:B:C = 7:5:4 ⇒ Total units = 16

Total pens = 960 ⇒ 1 unit = 960 ÷ 16 = 60

Pens with B = 5 × 60 = 300

Pens with D = 120% of 300 = 360

Ratio D:E = 9:8 ⇒ 9 units = 360 ⇒ 8 units (E) = 320

∴ The correct answer is 320 pens.

Alternate Method

Given: Total pens for A, B, C = 960; Ratio A:B:C = 7:5:4; D = B + 20% of B; D:E = 9:8.

Step 1: Calculate pens with B

⇒ Total Ratio Parts = 7 + 5 + 4 = 16

⇒ Number of pens with B = (5 ÷ 16) × 960

⇒ B = 5 × 60 = 300 pens

Step 2: Calculate pens with D

⇒ D = 300 + 20% of 300

⇒ D = 300 + 60 = 360 pens

Step 3: Calculate pens with E

⇒ Ratio D : E = 9 : 8

⇒ 9 units = 360

⇒ 1 unit = 360 ÷ 9 = 40

⇒ Number of pens with E = 8 × 40 = 320 pens

∴ The correct answer is 320 pens.

Additional Information

Ratio and Proportion

If a total quantity T is divided into ratio a:b:c, then the share of b is calculated as [b ÷ (a+b+c)] × T.

Percentage Increase

An increase of x% in a value V results in a new value of V × (100 + x) ÷ 100.

Direct Proportion

When two quantities are in a ratio x:y, if the value of x corresponds to K, then the value of y is (K ÷ x) × y.

Shortcut Trick

Assume Total Capacity of the cistern = LCM(8, 6, 12) = 24 units.

Efficiency of A + B = 24 ÷ 8 = 3 | Efficiency of B = 24 ÷ 12 = 2.

∴ Efficiency of A = 3 − 2 = 1.

Efficiency of B − C = 24 ÷ (−6) = −4 (since they empty it) ⇒ 2 − C = −4 ⇒ Efficiency of C = 6 (emptying).

Combined efficiency of A + B − C = 1 + 2 − 6 = −3 (emptying).

Time to empty 3/4^th tank = (3/4 × 24) ÷ 3 = 18 ÷ 3 = 6 hours.

∴ The correct answer is 6 hr.

Alternate Method

Given: Time (A+B) = 8 hr, Time (B+C emptying) = 6 hr, Time (B filling) = 12 hr.

Calculations:

⇒ Efficiency of Pipe A = Efficiency(A+B) − Efficiency(B) = 3 − 2 = 1 unit/hr.

⇒ Efficiency of Pipe C = Efficiency(B) − Efficiency(B+C emptying) = 2 − (−4) = 6 units/hr (emptying).

⇒ Combined efficiency of A, B and C = Efficiency(A) + Efficiency(B) − Efficiency(C) = 1 + 2 − 6 = −3 units/hr.

⇒ (3/4)th part of the tank = (3/4) × 24 = 18 units.

⇒ Time taken to empty 18 units = 18 ÷ 3 = 6 hours.

∴ The correct answer is 6 hr.

Additional Information

Pipes & Cisterns Sign Convention

Always take the efficiency of an inlet pipe as positive (+) and an outlet (emptying) pipe as negative (−).

Total Work and Efficiency

Total Capacity = LCM of individual times. Efficiency = Total Capacity ÷ Time taken by the pipe.

Time taken for Partial Work

Time = (Required Work Part × Total Capacity) ÷ Combined Efficiency.

Common solution:

Let the value of x be x.

The number of wired chargers sold by Shop A = 12x

Given, {Ratio of Wired: Wireless in A} = 12:11

Therefore, {Wireless chargers sold by A} = {11/12} × 12x = 11x

Also,

Wireless chargers sold by B = (10x + 8)

Wireless chargers sold by C = ((10x + 8) – 12= 10x – 4

According to the question,

11x + (10x + 8) + (10x – 4) = 190

31x + 4 = 190

31x = 186

So, x = 6

Now,

 

Calculation:

Required difference = 140 – 138 = 2

Common solution:

Let the value of x be x.

The number of wired chargers sold by Shop A = 12x

Given, {Ratio of Wired: Wireless in A} = 12:11

Therefore, {Wireless chargers sold by A} = {11/12} × 12x = 11x

Also,

Wireless chargers sold by B = (10x + 8)

Wireless chargers sold by C = ((10x + 8) – 12= 10x – 4

According to the question,

11x + (10x + 8) + (10x – 4) = 190

31x + 4 = 190

31x = 186

So, x = 6

Now,

 

Calculation:

Required ratio = 68:72 = 17:18

Common solution:

Let the value of x be x.

The number of wired chargers sold by Shop A = 12x

Given, {Ratio of Wired: Wireless in A} = 12:11

Therefore, {Wireless chargers sold by A} = {11/12} × 12x = 11x

Also,

Wireless chargers sold by B = (10x + 8)

Wireless chargers sold by C = ((10x + 8) – 12= 10x – 4

According to the question,

11x + (10x + 8) + (10x – 4) = 190

31x + 4 = 190

31x = 186

So, x = 6

Now,

 

Calculation:

Required sum = 140 + 72 = 212

Common solution:

Let the value of x be x.

The number of wired chargers sold by Shop A = 12x

Given, {Ratio of Wired: Wireless in A} = 12:11

Therefore, {Wireless chargers sold by A} = {11/12} × 12x = 11x

Also,

Wireless chargers sold by B = (10x + 8)

Wireless chargers sold by C = ((10x + 8) – 12= 10x – 4

According to the question,

11x + (10x + 8) + (10x – 4) = 190

31x + 4 = 190

31x = 186

So, x = 6

Now,

 

Calculation:

Given, ratio of wired chargers: wireless chargers in D = 2:1

Total parts = 2 + 1 = 3

Therefore, Wired chargers sold by D = 2/3 x 168 = 112

Required difference = 112 – 66 = 46

Common solution:

Let the value of x be x.

The number of wired chargers sold by Shop A = 12x

Given, {Ratio of Wired: Wireless in A} = 12:11

Therefore, {Wireless chargers sold by A} = {11/12} × 12x = 11x

Also,

Wireless chargers sold by B = (10x + 8)

Wireless chargers sold by C = ((10x + 8) – 12= 10x – 4

According to the question,

11x + (10x + 8) + (10x – 4) = 190

31x + 4 = 190

31x = 186

So, x = 6

Now,

 

Calculation:

Cost price of each wired charger = Rs. 20

Profit earned = 10%

Profit on each wired charger = 20 x 10% = Rs 2

Hence, total profit = 144 x 2 = Rs 288

Shortcut Trick
Ratio of Upstream speed (U) to Downstream speed (D) = 9 : 16
Total time = (DistanceU ÷ U) + (DistanceD ÷ D) = 10 hours
⇒ (90 ÷ 9x) + (160 ÷ 16x) = 10 ⇒ 10/x + 10/x = 10 ⇒ 20/x = 10 ⇒ x = 2
Upstream speed = 18 kmph, Downstream speed = 32 kmph
Stream speed (s) = (32 − 18) ÷ 2 = 7 kmph; Boat speed (v) = (32 + 18) ÷ 2 = 25 kmph
New Downstream speed (New s = 14) = 25 + 14 = 39 kmph ⇒ Time = 390 ÷ 39 = 10 hr
∴ The correct answer is 10 hr.
Alternate Method

Given: Upstream (U) = 9x, Downstream (D) = 16x
Formula: Time = Distance ÷ Speed
⇒ 90/(9x) + 160/(16x) = 10
⇒ 10/x + 10/x = 10
⇒ 20/x = 10 ⇒ x = 2
⇒ Upstream speed (U) = 18 kmph, Downstream speed (D) = 32 kmph
⇒ Boat speed (v) = (32 + 18)/2 = 25 kmph
⇒ Stream speed (s) = (32 − 18)/2 = 7 kmph
⇒ New Stream speed = 7 × 2 = 14 kmph
⇒ New Downstream speed = 25 + 14 = 39 kmph
⇒ New Time taken = 390 / 39 = 10 hr
∴ The correct answer is 10 hr.
Additional Information
Speed of Boat and Stream
Boat speed in still water (v) = (Downstream + Upstream) ÷ 2. Stream speed (s) = (Downstream − Upstream) ÷ 2.
Average Speed of Boat
If a boat covers same distance upstream and downstream, Average Speed = (U × D) ÷ v = (v2 − s2) ÷ v.
Relative Speed in Water
Downstream speed = v + s; Upstream speed = v − s. Always v > s for the boat to move upstream.

Shortcut Trick

Apply Alligation on average weights: Boys (64 kg) and Girls (44 kg) with Mean (52 kg).

Ratio of Boys : Girls = (52 − 44) : (64 − 52)

Ratio = 8 : 12 = 2 : 3.

∴ The correct answer is 2:3.

Alternate Method

Given:

Total students = 90

Average weight (Group) = 52 kg

Average weight (Boys) = 64 kg

Average weight (Girls) = 44 kg

Calculations:

⇒ Let number of boys be B and girls be G.

⇒ Total weight = 52 × 90 = 4680 kg.

⇒ 64B + 44G = 52(B + G)

⇒ 64B + 44G = 52B + 52G

⇒ 64B − 52B = 52G − 44G

⇒ 12B = 8G

⇒ B/G = 8/12 = 2/3

∴ The correct answer is 2:3.

Additional Information

Alligation Rule

It is used to find the ratio in which two ingredients at given prices/weights are mixed to produce a mixture of a mean price/weight. Ratio = (Dearer − Mean) : (Mean − Cheaper).

Weighted Average Formula

Average = (n₁w₁ + n₂w₂) ÷ (n₁ + n₂), where n is the quantity and w is the weight/value.

Total Sum Calculation

Sum of all observations = Average × Number of observations.

Shortcut Trick

Area of square = 8100 m² → Side = 90 m → Perimeter = 360 m.

Perimeter of rectangle = 360 + 180 = 540 m.

Rectangle ratio L:B = 2:1 → Sum of ratio parts for perimeter = 2 × (2 + 1) = 6 units.

6 units = 540 m → 1 unit = 90 m.

Difference (L − B) = 2 units − 1 unit = 1 unit = 90 m.

∴ The correct answer is 90 m.

Alternate Method

Given: Area of square = 8100 m², Ratio L:B = 2:1, Perimeter Difference = 180 m.

Formula: Perimeter of square = 4 × side; Perimeter of rectangle = 2(L + B).

⇒ Side of square (a) = √8100 = 90 m

⇒ Perimeter of square = 4 × 90 = 360 m

⇒ Perimeter of rectangle = 360 + 180 = 540 m

⇒ Let length = 2x and breadth = x

⇒ 2(2x + x) = 540

⇒ 6x = 540 ⇒ x = 90

⇒ Difference = 2x − x = x = 90 m

∴ The correct answer is 90 m.

Additional Information

Mensuration Formulas

Area of Rectangle = L × B and Diagonal = √(L² + B²).

Square Properties

Diagonal of a square = a√2, where 'a' is the side length.

Ratio and Perimeter

If L:B = m:n, then Perimeter = 2k(m+n) where k is the common multiplier.

Shortcut Trick

Profit Ratio = (Capital × Time) ratio.

A's capital: 60,000 for x months + 48,000 (80% of 60k) for (12 − x) months.

B's capital: 50,000 for x months + 30,000 (60% of 50k) for (12 − x) months.

Ratio ⇒ [60x + 48(12 − x)] ÷ [50x + 30(12 − x)] = 3 ÷ 2.

Solving: (12x + 576) ÷ (20x + 360) = 3 ÷ 2 ⇒ 24x + 1152 = 60x + 1080.

36x = 72 ⇒ x = 2.

∴ The correct answer is 2.

Alternate Method

Given → Initial Investments of A = Rs. 60,000 and B = Rs. 50,000. Time period = 1 year (12 months).
Formulas: Profit Ratio = (C₁ × T₁) : (C₂ × T₂) | New Investment = Old Investment × (100 − Withdrawal%) ÷ 100

⇒ A's total investment unit = 60,000 × x + 60,000 × 0.8 × (12 − x) = 60,000x + 48,000(12 − x)
⇒ B's total investment unit = 50,000 × x + 50,000 × 0.6 × (12 − x) = 50,000x + 30,000(12 − x)
⇒ Ratio = [60x + 576 − 48x] ÷ [50x + 360 − 30x] = 3 ÷ 2
⇒ (12x + 576) ÷ (20x + 360) = 3 ÷ 2
⇒ 24x + 1152 = 60x + 1080 ⇒ 36x = 72 ⇒ x = 2.

∴ The correct answer is 2.

Additional Information

Partnership Basics

Profit is shared in the ratio of the product of Capital and Time. If capital changes during the period, calculate effective capital for each interval.

Withdrawal Calculations

When a partner withdraws y% of capital, the remaining capital used for the rest of the period is Initial × (100 − y) ÷ 100.

Shortcut Trick

Let Total Notebooks bought = 9x and Sold = 5x.

Total CP = 9x × 6 = 54x and Total SP = 5x × 9 = 45x.

Loss = CP − SP = 54x − 45x = 9x.

Given Loss = CP of 12 notebooks = 12 × 6 = ₹72.

⇒ 9x = 72 ⇒ x = 8.

Notebooks Sold = 5x = 5 × 8 = 40.

∴ The correct answer is 40 notebooks.

Alternate Method

Given: Ratio of bought to sold = 9 : 5. CP = ₹6. SP = ₹9.

⇒ Let Total Notebooks bought = 9x.

⇒ Total Cost Price (CP) = 9x × 6 = ₹54x.

⇒ Total Notebooks sold = 5x.

⇒ Total Selling Price (SP) = 5x × 9 = ₹45x.

⇒ Loss = Total CP − Total SP = 54x − 45x = 9x.

⇒ According to question, Loss = 12 × CP per unit = 12 × 6 = ₹72.

⇒ 9x = 72 ⇒ x = 8.

⇒ Number of notebooks sold = 5x = 5 × 8 = 40.

∴ The correct answer is 40 notebooks.

Additional Information

Loss and Profit Calculation

Profit = SP − CP (when SP > CP). Loss = CP − SP (when CP > SP).

Defective Items Logic

In business math, unsold or defective items contribute to the Total CP but yield zero SP, effectively increasing the loss per unit.

Ratio and Proportion

If the ratio is a:b, the actual values are ax and bx where x is the common multiplier found using the given condition.