Quantitative Aptitude PYQs

Quantity 1-

1(1/4)X2 + 3(1/4)X + 2 = 0

5/4X2 + 13/4X + 2 = 0

5X2 + 13X + 8 = 0

5X² + 5X + 8X + 8 = 0

5X(X + 1) + 8(X + 1) = 0

(5X + 8) (X + 1) = 0

X = -8/5 , -1

Quantity 2-

 Y2 + 1(1/2)Y + 1/2 = 0

Y2 + 3/2Y + 1/2 = 0

2Y2 + 3Y + 1 = 0

2Y2 + 2Y + Y + 1 = 0

2Y(Y + 1) + 1(Y + 1) = 0

(2Y + 1)(Y + 1)

Y = -1/2, -1

So, Quantity 1  ≤  Quantity 2

Hence, correct answer is option 4.

16(a) + 216 = 232

232 + 25 = 257

257 + 64 = 321

321 + 9 = 330

330 + 8 = 338

So, a = 16

Now, 

120 + 125 = 245

245 + 16 = 261

261 + 27 = 288

288 + 4 = 292

292 + 1 = 293

Hence, correct answer is option 2.

16(a) + 216 = 232

232 + 25 = 257

257 + 64 = 321

321 + 9 = 330

330 + 8 = 338

So, a = 16

Now, 

120 + 125 = 245

245 + 16 = 261

261 + 27 = 288

288 + 4 = 292

292 + 1 = 293

Statement 1 - 

a + c = 16 + 261 = 277

277 is a prime number.

So, statement 1 is correct.

Statement 2 -

2b + c = 2 × 245 + 261 = 751.

So, statement 2 is false.

Statement 3 - 

b - 3a = 245 - 48 = 197

So, statement 3 is false.

Hence, the correct answer is option 1.

Series 1-

Series 2-

10         30          68           130           222              350            520   

      20          38           62             92             128              170

             18           24            30             36                  42

                     6                6               6               6 

Hence, correct answer is option 3.

Let, Total selling price for all fresh mangoes sold in each month = 2x

And, Total selling price for all rotten mangoes sold in each month = x

Now, Total selling price of all mangoes sold in all four months = (108 × 2x + 108 × x ) + (97.2 × 2x + 64.8 × x ) + (283.5 × 2x + 94.5 × x ) + (226.8 × 2x + 97.2 × x ) 

Total selling price of all mangoes sold in all four months = 1795.5x

Therefore, Total selling price of all mangoes sold in all four months = 14364 × 125/100 = 17955

So, 1795.5x = 17955

x = 10

Total selling price for all fresh mangoes sold in each month = 2x = 20

Total selling price for all rotten mangoes sold in each month = x = 10

So, total selling price for all mangoes sold in the months of August = (226.8 × 20 + 97.2 × 10 ) = 5508

Hence, correct answer is option 1.

Total number of rotten mangoes sold in the month of May and July = (1080 × 20/100 × 50/100 ) + (1080 × 35/100 × 25/100) = 108 + 94.5 = 202.5

Total number of fresh mangoes sold in the month of June and August = (1080 × 15/100 × 60/100 ) + (1080 × 30/100 × 70/100) = 97.2 + 226.8 = 324

So, required ratio = 202.5 : 324 = 405 : 648.

Hence, correct answer is option 3.

Total mangoes sold in June = 1080 × 15/100 = 162

Rotten mangoes sold in June = 162 × 40/100 = 64.8

CP of all the mangoes sold in June = ₹1200

MP of all the mangoes sold in June = 1200 × 150/100 = ₹1800

SP of all the mangoes sold in June  = 1800 × 90/100 × 80/100 = ₹1296

SP of rotten mangoes sold in June = 1296 - 972 = ₹324

SP of each rotten mango sold in June = 324 ÷ 64.8 = ₹5

Hence, correct answer is option 3.

Solution:

Quantity I:

Difference between 20% discount and 10% is 45

Therefore, 100% as the marked price and the difference between 20% and 10 % is 45

So

90% - 80% = 45

10% = 45

100% = 450

Then the marked price = Rs. 450

Quantity I:

Sugar A is purchased for 35 × 10 = 350

Then there is the remaining = 450 – 350 = Rs. 100 

For x = 100/20 = 5 kg of sugar, B is purchased

Quantity II:

The, marked price is increased by 20% then the amount is 540

Price of the sugar (A + B) = Rs.  50

Then in 540/50 = 10.8 kg purchased

Thus, Quantity I < Quantity II

Given:

​9.99% of 29.906 + 299.84% of 54.908 – 59.86% of 149.59 = ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

​9.99% of 29.906 + 299.84% of 54.908 – 49.86% of 149.59 = ?

⇒ ​10% of 30 + 300% of 55 – 50% of 150 = ?

⇒ ​(10/100) × 30 + (300/100) × 55 – (50/100) × 150 = ?

⇒ 3 + 165 – 75 = ?

⇒ ? = 93

∴ The value of ? is 93.

Given:

5.78% of 799.94 + ?% of 9.67 = 10.94 × 2.99 + √100 × 2.98

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

5.78% of 799.94 + ?% of 9.67 = 10.94 × 2.99 + √100 × 2.98

⇒ 6% of 800 + ?% of 10 = 11 × 3 + √100 × 3

⇒ (6/100) × 800 + (?/100) × 10 = 33 + 10 × 3

⇒ 48 + (?/10) = 33 + 30

⇒ 48 + (?/10) = 63

⇒ (?/10) = 63 - 48 

⇒ (?/10) = 15 

⇒ ? = 15 × 10 = 150

∴ The value of ? is 150.

Given:

728.921/3 + √1155.78 + 6.042 ÷ 2.962 + 1.990 = √?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

Since we need to find out the approximate value, we can write these values to their nearest integers.

728.821/3 + √1155.98 + 6.142 ÷ 2.992 + 1.970 = √?

⇒ ∛729 + √1156 + 62 ÷ 32 + 2 = √?

⇒ 9 + 34 + 36/9 + 2 = √?

⇒ 9 + 34 + 4 + 2 = √?

⇒ √? = 49

⇒ ? = 49²

⇒ ? = 2401

∴ The value of ‘?’ Is 2401.

Solution:

Quantity I:

Number of boys in the class A = 5x

Number of girls in the class B = 4x

If 20% of boys and 25%of the girls were removed and 20 more boys were added 

To find the number of students in the class A

[5x - 5x × (20/100) + 20]/[4x - 4x × 25/100] = 2/1

(4x + 20)/(3x) = 2/1

(4x + 20) = 6x

2x = 20

x = 10

Total number of boys and girls in the class = (9)×  10 = 90

Quantity II:

Number of boys in the class B = x

Number of girls in the class B = (x - 20) 

To find the value of x

[(x × 50) + (20 × (x - 20))]/(x + x - 20) = 37.5

[50x + 20x - 400] =  [75x - 750]

[70x - 400] = [75x - 750]

-5x = - 350

x = 70

Total number of students in the class = 70 + 50 = 120

Thus, Quantity I < Quantity II

Shortcut Trick

Given Average(A, D) = y + 10 and D = 65 ⇒ A + 65 = 2y + 20 ⇒ A = 2y − 45.

From Average(A, B, C) = y: A + (A + 10) + (A + 5) = 3y ⇒ 3A + 15 = 3y ⇒ A + 5 = y.

Equating A: y − 5 = 2y − 45 ⇒ y = 40 ∴ A = 35, B = 45.

Average of B and A = (45 + 35) ÷ 2 = 40.

∴ The correct answer is 40.

Alternate Method

Given: Score of A = x, then B = x + 10 and C = x + 5.

Formula: Average = Sum of observations ÷ Number of observations

⇒ Average(A, B, C) = (x + x + 10 + x + 5) ÷ 3 = y

⇒ 3x + 15 = 3y ⇒ x + 5 = y --- (Eq 1)

⇒ Average(A, D) = (x + 65) ÷ 2 = y + 10

⇒ x + 65 = 2y + 20 ⇒ x − 2y = −45 --- (Eq 2)

⇒ Substitute y = x + 5 in Eq 2: x − 2(x + 5) = −45

⇒ x − 2x − 10 = −45 ⇒ −x = −35 ⇒ x = 35

⇒ A = 35 and B = 35 + 10 = 45

⇒ Average(B, A) = (45 + 35) ÷ 2 = 80 ÷ 2 = 40

∴ The correct answer is 40.

Additional Information

Weighted Average

When two groups with averages A₁ and A₂ and sizes n₁ and n₂ are combined, the new average is (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Net Deviation Method

The sum of deviations of all observations from their mean is always zero. This helps in finding missing values quickly.

Shortcut Trick

Let Total Work (Capacity) = 12 units (LCM of 6 and 12).

Efficiency: A = 2, B = 1, C = 0.8 × 1 = 0.8 (emptying → −0.8).

Net Work: (A+B+C) × 4 + (−C) × x + (A+B) × 4 = 12.

⇒ (2+1−0.8) × 4 − 0.8x + (2+1) × 4 = 12 ⇒ 8.8 − 0.8x + 12 = 12.

⇒ 0.8x = 8.8 ⇒ x = 11.

∴ The correct answer is 11 min.

Alternate Method

Given: Time for Pipe A = 6 min, Pipe B = 12 min. Pipe C efficiency = 80% of Pipe B.

Calculations:

⇒ Efficiency of Pipe C (emptying) = 80% of 1 = 0.8 units/min.

⇒ Total work done by (A + B + C) in 4 minutes = (2 + 1 − 0.8) × 4 = 2.2 × 4 = 8.8 units.

⇒ Total work done by (A + B) in the last 4 minutes = (2 + 1) × 4 = 12 units.

⇒ Since total capacity is 12 units, the equation for the whole process is:

⇒ 8.8 − (0.8 × x) + 12 = 12

⇒ 8.8 − 0.8x = 0

⇒ 0.8x = 8.8

⇒ x = 8.8 ÷ 0.8 = 11

∴ The correct answer is 11 min.

Additional Information

Pipes & Cisterns Efficiency

Efficiency is inversely proportional to time. If an emptying pipe is present, its efficiency is taken as negative in calculations.

Total Work Calculation

Total work is generally assumed to be the LCM of the individual times taken by the pipes to fill or empty the tank.

Combined Rate Rule

When multiple pipes work together, the net rate is the algebraic sum of their individual efficiencies (Filling = +, Emptying = −).

Shortcut Trick

Total distance for first train = x + 3x = 4x.

4x = Speed × Time ⇒ 4x = 20 × 48 ⇒ 4x = 960 ⇒ x = 240 m.

Second train speed = 108 × 5/18 = 30 m/s. Length = 1.5 × 240 = 360 m.

Distance for second train = 360 + y = 30 × 20 = 600 ⇒ y = 240 m.

Value of (x + y) = 240 + 240 = 480 m.

∴ The correct answer is 480 m.

Alternate Method

Given:

Length of first train = x, Length of platform = 3x.

Speed of first train = 20 m/s, Time = 48 s.

Length of second train = 1.5x, Speed = 108 kmph, Time = 20 s.

Calculations:

⇒ Total distance in Case 1 = x + 3x = 4x

⇒ 4x = 20 × 48 = 960

⇒ x = 960 ÷ 4 = 240 m

⇒ Speed of second train in m/s = 108 × (5/18) = 30 m/s

⇒ Length of second train = 1.5 × 240 = 360 m

⇒ Total distance in Case 2 = 360 + y = 30 × 20

⇒ 360 + y = 600

⇒ y = 600 − 360 = 240 m

⇒ (x + y) = 240 + 240 = 480 m

∴ The correct answer is 480 m.

Additional Information

Crossing Stationary Objects

When a train crosses a platform or another stationary train, the total distance covered is the sum of the length of the train and the length of the object.

Unit Conversion

To convert speed from kmph to m/s, multiply by 5/18. To convert m/s to kmph, multiply by 18/5.

Relative Speed

If two objects move in the opposite direction, relative speed is the sum (S₁ + S₂); in the same direction, it is the difference |S₁ − S₂|.

Shortcut Trick

Relation between Marked Price (MP) and Cost Price (CP): MP÷CP = (100 + Profit%)÷(100 − Discount%).

MP÷CP = (100 + 12)÷(100 − 30) = 112÷70 = 8÷5.

Markup % (X) = (8 − 5)÷5 × 100 = 3÷5 × 100 = 60%.

New CP = X + 450 = 60 + 450 = Rs. 510. New SP = 510 × (100 − 20)÷100 = 510 × 0.8 = Rs. 408. ∴ The correct answer is Rs. 408.

Alternate Method

Given → CP₁ = Rs. 550, Profit = 12%, Discount = 30%, Loss = 20%.
Formulas: SP = CP × (100 + P%)÷100 | MP = SP × 100÷(100 − D%) | Markup % = (MP − CP)÷CP × 100.

⇒ First Selling Price (SP₁) = 550 × 112÷100 = Rs. 616. ⇒ Marked Price (MP) = 616 × 100÷(100 − 30) = 616 × 100÷70 = Rs. 880. ⇒ Markup Value = MP − CP₁ = 880 − 550 = Rs. 330. ⇒ X% = (330÷550) × 100 = 60%, so X = 60.
⇒ New Cost Price (CP₂) = X + 450 = 60 + 450 = Rs. 510. ⇒ New Selling Price (SP₂) = 510 × (100 − 20)÷100 = 510 × 80÷100 = Rs. 408. ∴ The correct answer is Rs. 408.

Additional Information

Marked Price & Markup

Marked price is the price displayed on the tag. Markup = MP − CP. Markup % = (Markup ÷ CP) × 100.

Relationship Formula

The ratio of MP to CP can be directly found using CP × (100 + P%) = MP × (100 − D%).

Solution:

Selling price of the article = 4500 × (130/100) = Rs.5850

Marked price of the article = (5850/90) × 100 = 6500

Profit amount of the article = (5850 - 4500) = Rs.1350

New profit amount of the article = (1350 + 50) = 1400

New selling price = (4500 + 1400) = 5900

New marked price = 5900/90 ×  100 = 6555.556 = ~ 6556 

So required change = (6556 - 6500) = Rs. 56

Solution:

Cost price of article C = 4000

Selling price of article C = 4000 × (120/100) = 4800

Marked price of article E = 4800

Selling price of article E = 4800 × (100 - 25)/100 = 3600

To find the cost price of the article E = (3600/120) × 100 = 3000

Solution:

Cost price of article B and D are in the ratio 25x: 24x

Profit percentage of article B and D is 40% and 25% respectively

Selling price of article B  = 25x × (140/100) = 35x

Selling price of article D = 24x × (125/100) = 30x

(35x - 30x) = 1000

5x = 1000

x = 200

Selling price of article B = 35 × 200 = 7000

Selling price of article D = 30 × 200 = 6000

Marked price of the article B = (7000/80) × 100 = 8750

Marked price of the article D = (6000/75) × 100 = 8000

Required ratio = 8750: 8000 = 35: 32

Solution:

Cost price of the article C = 4000

Selling price of the article  4000 × (120/100) = 4800

New selling price of the article C = (4800 + 200) = 5000 

Now, the profit percentage = (5000 - 4000)/4000 x 100 = 25%

If the discount percentage and the profit percentage of the article C is same, then the discount percentage of the article = 25%

Marked price of the article = (5000)/75 × 100 = Rs.6666.66