Quantitative Aptitude PYQs

Common Calculation

Let the number of lab assistants in College A be L _A.

Then, lab assistants in College B = (5/4) * L _A.

Lab assistants in College C = (7/5) * L _A.

According to the question: L _A + (5/4)L _A + (7/5)L _A = 73.

=> L _A × (1 + 1.25 + 1.4) = 73.

=> L _A × 3.65 = 73.

=> L _A = 20.

So, L _B = 25, LC = 28.

Professors in College A: P _A = L _A + 20 = 40.

Professors in College B: P _ B = 1.375 × P _A = 55.

Professors in College C: PC = (7/4) * L _C = 49.

Final distribution:

College A: 20 lab assistants, 40 professors

College B: 25 lab assistants, 55 professors

College C: 28 lab assistants, 49 professors

 

 

Calculation

Lab Assistants (A) = 20

Professors (A) = 40

Ratio = 20 : 40 = 1 : 2

Common Calculation

Let the number of lab assistants in College A be L _A.

Then, lab assistants in College B = (5/4) * L _A.

Lab assistants in College C = (7/5) * L _A.

According to the question: L _A + (5/4)L _A + (7/5)L _A = 73.

=> L _A × (1 + 1.25 + 1.4) = 73.

=> L _A × 3.65 = 73.

=> L _A = 20.

So, L _B = 25, LC = 28.

Professors in College A: P _A = L _A + 20 = 40.

Professors in College B: P _ B = 1.375 × P _A = 55.

Professors in College C: PC = (7/4) * L _C = 49.

Final distribution:

College A: 20 lab assistants, 40 professors

College B: 25 lab assistants, 55 professors

College C: 28 lab assistants, 49 professors

 

 

Calculation

Professors (B) = 55

Lab Assistants (B) = 25

Difference = 55 – 25 = 30

Common Calculation

Let the number of lab assistants in College A be L _A.

Then, lab assistants in College B = (5/4) * L _A.

Lab assistants in College C = (7/5) * L _A.

According to the question: L _A + (5/4)L _A + (7/5)L _A = 73.

=> L _A × (1 + 1.25 + 1.4) = 73.

=> L _A × 3.65 = 73.

=> L _A = 20.

So, L _B = 25, LC = 28.

Professors in College A: P _A = L _A + 20 = 40.

Professors in College B: P _ B = 1.375 × P _A = 55.

Professors in College C: PC = (7/4) * L _C = 49.

Final distribution:

College A: 20 lab assistants, 40 professors

College B: 25 lab assistants, 55 professors

College C: 28 lab assistants, 49 professors

 

 

Calculation

Total Professors = 40 + 55 + 49 = 144

Professors in C = 49

% = (49 / 144) × 100 ≈ 34.03%

Common Calculation

Let the number of lab assistants in College A be L _A.

Then, lab assistants in College B = (5/4) * L _A.

Lab assistants in College C = (7/5) * L _A.

According to the question: L _A + (5/4)L _A + (7/5)L _A = 73.

=> L _A × (1 + 1.25 + 1.4) = 73.

=> L _A × 3.65 = 73.

=> L _A = 20.

So, L _B = 25, LC = 28.

Professors in College A: P _A = L _A + 20 = 40.

Professors in College B: P _ B = 1.375 × P _A = 55.

Professors in College C: PC = (7/4) * L _C = 49.

Final distribution:

College A: 20 lab assistants, 40 professors

College B: 25 lab assistants, 55 professors

College C: 28 lab assistants, 49 professors

 

 

Calculation

Total Lab Assistants = 20 + 25 + 28 + 31 = 104

→ Average = 104 ÷ 4 = 26

Total Professors = 40 + 55 + 49 + 60 = 204

→ Average = 204 ÷ 4 = 51

Difference = 51 – 26 = 25

Common Calculation

Let the number of lab assistants in College A be L _A.

Then, lab assistants in College B = (5/4) * L _A.

Lab assistants in College C = (7/5) * L _A.

According to the question: L _A + (5/4)L _A + (7/5)L _A = 73.

=> L _A × (1 + 1.25 + 1.4) = 73.

=> L _A × 3.65 = 73.

=> L _A = 20.

So, L _B = 25, LC = 28.

Professors in College A: P _A = L _A + 20 = 40.

Professors in College B: P _ B = 1.375 × P _A = 55.

Professors in College C: PC = (7/4) * L _C = 49.

Final distribution:

College A: 20 lab assistants, 40 professors

College B: 25 lab assistants, 55 professors

College C: 28 lab assistants, 49 professors

 

 

Calculation

Professors in B = 55 → 20% = 11

→ transfer = 11

→ New Professors in A = 40 + 11 = 51

→ New Professors in B = 55 – 11 - 10 = 34

→ ​Ratio = 51:34 = 3:2

Shortcut Trick

Let initial investments be 8x, 5x, and 6x. After changes, ratio is 5 : 3 : 4.

Using A and B: (8x + 5300) ÷ (5x − 4200) = 5 ÷ 3.

24x + 15900 = 25x − 21000 ⇒ x = 36900.

Initial investment of A = 8x = 8 × 36900 = 295200.

Check with C: (6 × 36900 + 19000) = 240400. Ratio B:C = (5 × 36900 − 4200) : 240400 = 180300 : 240400 = 3 : 4 (Satisfied).

∴ The correct answer is 295200.

Alternate Method

Given → Initial ratio A:B:C = 8 : 5 : 6. Final ratio 5 : 3 : 4.
Formulas: New Investment = Initial Investment + Addition − Withdrawal.

⇒ Let the initial investments of A, B, and C be 8x, 5x, and 6x respectively.
⇒ New investment of A = 8x + 5300.
⇒ New investment of B = 5x − 4200.
⇒ According to the question: (8x + 5300) / (5x − 4200) = 5 / 3.
⇒ 3 × (8x + 5300) = 5 × (5x − 4200).
⇒ 24x + 15900 = 25x − 21000.
⇒ 25x − 24x = 15900 + 21000 ⇒ x = 36900.
⇒ Initial amount invested by A = 8x = 8 × 36900 = 295200.

∴ The correct answer is 295200.

Additional Information

Partnership Ratios

In a Partnership, the ratio of profits is generally the ratio of the product of (Investment × Time). If time is constant, Profit Ratio = Investment Ratio.

Variable Management

When solving problems with multiple changes, equate any two parts of the ratio (e.g., A/B or B/C) to find the common multiplier x.

Shortcut Trick

Let the cost of a Bike be 5 and a Car be 6.

Total cost for Person A = (9 × 5) + (x × 6) = 45 + 6x.

Total cost for Person B = (12 × 5) + (2x × 6) = 60 + 12x.

Since B paid 60% more than A, the ratio of B to A is 160 : 100 or 8 : 5.

Equating ratios: (60 + 12x) ÷ (45 + 6x) = 8 ÷ 5.

300 + 60x = 360 + 48x ⇒ 12x = 60 ⇒ x = 5.

∴ The correct answer is 5.

Alternate Method

Given: Ratio of Cost Price (CP) of Bike : Car = 5 : 6. Let CP_Bike = 5k and CP_Car = 6k.
Formulas: Total Cost = (Quantity_Bike × CP_Bike) + (Quantity_Car × CP_Car) | Amount_B = Amount_A × (1 + 60/100).

⇒ Amount paid by Person A = (9 × 5k) + (x × 6k) = k(45 + 6x).
⇒ Amount paid by Person B = (12 × 5k) + (2x × 6k) = k(60 + 12x).
⇒ According to the question: k(60 + 12x) = 1.6 × k(45 + 6x).
⇒ 60 + 12x = 72 + 9.6x.
⇒ 12x − 9.6x = 72 − 60.
⇒ 2.4x = 12 ⇒ x = 12 ÷ 2.4 = 5.

∴ The correct answer is 5.

Additional Information

Ratio and Proportion

When quantities are given in ratio (a:b), we can assume actual values as ax and bx to simplify calculation.

Percentage Increase

A value increased by 60% is equivalent to multiplying the original value by 1.6 or the fraction 8/5.

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

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

Given expression is

344.98 - [551.98 ÷ 23.08] + 27.98 = ? - 13.99

⇒ 345 - [552 ÷ 23] + 28 = ? - 14

⇒ 345 - 24 + 28 = ? - 14

⇒ 349 = ? - 14

⇒ 349 + 14 = ?

∴ ? = 363

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

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

Given expression is

(55.89 + 11.94 × 6.94) ÷ 27.92 + 65.99 = ?

⇒ (56 + 12 × 7) ÷ 28 + 66 = ?

⇒ (56 + 84) ÷ 28 + 66 = ?

⇒ 140 ÷ 28 + 66 = ?

⇒ 5 + 66 = ?

∴ ? = 71

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

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

Given expression is

689.09 – [23.982 – (122.98 + 156.97) ÷ 4.92] = ?2

⇒ 689 – [242 – (123 + 157) ÷ 5] = ?2

⇒ 689 – [242 – 280 ÷ 5] = ?2

⇒ 689 – [242 – 56] = ?2

⇒ 689 – [576 – 56] = ?2

⇒ 689 – 520 = ?2

∴ 13 = ?

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

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

Given expression is

(117.98 + 60.92 × 1.99) ÷ 47.99 × 4.93 = ? 

⇒ (118 + 61 × 2) ÷ 48 × 5 = ? 

⇒ (118 + 122) ÷ 48 × 5 = ? 

⇒ 240 ÷ 48 × 5 = ? 

⇒ 5 × 5 = ?

∴ ? = 25

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

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

Given expression is

√255.96 × 1.98⁵ ÷ 7.983 × √4095.96 = ? × 63.94

⇒ √256 × 2⁵ ÷ 83 × √4096 = ? × 64

⇒ 16 × 32 ÷ 512 × 64 = ? × 64 

⇒ 1 × 64 = ? × 64

⇒ 1 = ?

General Solution:

 

Calculations:

From table: Bag A total = 480 + 240 = 720; Bag B total = 450 + 300 = 750

⇒ A + B = 1470

Bag F total = 14.28% less than 1470 ≈ (1 − 1/7) × 1470 = (6/7) × 1470 = 1260

In F, Red : Blue = 5 : 2 (7 parts) ⇒ 1 part = 1260 ÷ 7 = 180

⇒ Blue(F) = 2 × 180 = 360

Red(E) from table = 640

Required difference = Red(E) − Blue(F) = 640 − 360 = 280

Hence, the correct answer is 280.

General Solution:

 

Calculations:

From the table values

Bag B → Red = 450, Blue = 300

Bag C → Red = 450, Blue = 270

Bag A → Red = 480, Blue = 240

Bag E → Red = 640, Blue = 480

Average number of red balls (B and C)

⇒ (450 + 450) ÷ 2

⇒ 900 ÷ 2

⇒ 450

Average number of blue balls (A and E)

⇒ (240 + 480) ÷ 2

⇒ 720 ÷ 2

⇒ 360

Required difference

⇒ 450 − 360 = 90

Hence, the correct answer is 90.

General Solution:

 

Calculations:

⇒ 30% of blue balls in E = 30% × 480 = 144

⇒ New blue in B = 300 + 144 = 444

⇒ Total in B = 450 (red) + 444 (blue) = 894

⇒ % red = (450 ÷ 894) × 100 ≈ 50.3356% ≈ 50.3%

Hence, the correct answer is 50.3%.

General Solution:

 

Calculations:

We already calculated for Bag C:

Red = 450

Blue = 270

Now apply changes:

Red increased by 20% ⇒ 450 × 1.20 = 540

Blue decreased by 10% ⇒ 270 × 0.90 = 243

So new ratio = 540 : 243

Simplify:

⇒ Divide both by 27

540 ÷ 27 = 20

243 ÷ 27 = 9

New ratio = 20 : 9

Hence, the correct answer is 20:9.

General Solution:

 

Calculations:

We are asked: Find the total number of blue balls in Bags A, C, and E together.

From the General Solution Table given:

Bag A → Blue = 240

Bag C → Blue = 270

Bag E → Blue = 480

Now, total = 240 + 270 + 480 = 990

Hence, the correct answer is 990.

Shortcut Trick

Work Equivalence: (15M + 10W) × 30 = (10M + 20W) × 20

⇒ 3(15M + 10W) = 2(10M + 20W) ⇒ 45M + 30W = 20M + 40W ⇒ 25M = 10W ⇒ 5M = 2W

Total Work = (15M + 10W) × 30 = (15M + 25M) × 30 = 40M × 30 = 1200M units

One-third Work = 1200M ÷ 3 = 400M units

Efficiency of (6M + 4W) = 6M + 10M = 16M

Time = 400M ÷ 16M = 25 days

∴ The correct answer is 25 days.

Alternate Method

Given: Case 1: 15 men and 10 women in 30 days. Case 2: 10 men and 20 women in 20 days. Required: Time for 6 men and 4 women for 1/3 work.

Formula Used: Total Work = (Number of workers × Efficiency) × Time

⇒ (15M + 10W) × 30 = (10M + 20W) × 20

⇒ 450M + 300W = 200M + 400W

⇒ 250M = 100W ⇒ 1W = 2.5M

⇒ Total Work = (15M + 10 × 2.5M) × 30 = (15M + 25M) × 30 = 40M × 30 = 1200M

⇒ Target Work = 1200M ÷ 3 = 400M

⇒ Efficiency of Group 3 = 6M + 4W = 6M + 4(2.5M) = 6M + 10M = 16M

⇒ Required Time = 400M ÷ 16M = 25 days

∴ The correct answer is 25 days.

Additional Information

Work-Efficiency Relation

Total work is the product of the number of people, their individual efficiencies, and the time taken (W = n × e × t).

MDH Rule

For comparing two work scenarios: (M₁D₁H₁)/W₁ = (M₂D₂H₂)/W₂.

Man-Woman Efficiency

When multiple types of workers are involved, always equate the total work equations to find the ratio of their efficiencies.

Shortcut Trick

Ratio of Milk : Water initially = 3 : 2. After removing 25 L, ratio remains 3 : 2.

Final ratio = 1 :

1. To equate milk (3 units), final ratio becomes 3 : 3.

Water increase = 3 − 2 = 1 unit. Given 1 unit = 7 litres.

Quantity after removal = (3 + 2) × 7 = 35 litres.

Initial quantity = 35 + 25 = 60 litres.

∴ The correct answer is 60 litres.

Alternate Method

Given: Initial ratio 3 : 2. Removal = 25 L. Addition = 7 L Water. Final ratio 1 :

1. Let the quantities of Milk and Water after removing 25 litres be 3x and 2x respectively.

⇒ According to question: 3x / (2x + 7) = 1 / 1

⇒ 3x = 2x + 7

⇒ x = 7

⇒ Mixture after removal = 3x + 2x = 5x = 5 × 7 = 35 litres

⇒ Initial quantity = 35 + 25 = 60 litres

∴ The correct answer is 60 litres.

Additional Information

Mixture Property

When a portion of a mixture is removed, the ratio of the ingredients in the remaining mixture remains the same as the initial ratio.

Replacement Method

If Milk : Water is a : b and x litres of mixture is replaced by x litres of water, the final ratio changes, but the amount of milk removed is (a/(a+b)) × x.

Alligation Rule

Quantity of Cheaper / Quantity of Dearer = (CP of Dearer − Mean Price) / (Mean Price − CP of Cheaper).

Shortcut Trick

Lateral Surface Area (LSA) of cube = 4a² = 676 ⇒ a² = 169 ⇒ Side (a) = 13 cm.

Height of cylinder (h) = Side − 4 = 13 − 4 = 9 cm.

Volume of cylinder = πr²h = 3 × r² × 9 = 972 ⇒ 27r² = 972 ⇒ r² = 36 ⇒ Radius (r) = 6 cm.

Curved Surface Area (CSA) = 2πrh = 2 × 3 × 6 × 9 = 324 cm².

∴ The correct answer is 324 cm².

Alternate Method

Given: LSA of cube = 676 cm², Cylinder Volume = 972 cm³, π = 3.

Formula Used: LSA = 4a²; Volume = πr²h; CSA = 2πrh.

⇒ 4 × a² = 676

⇒ a² = 169 ⇒ a = √169 = 13 cm

⇒ Height of cylinder (h) = a − 4 = 13 − 4 = 9 cm

⇒ πr²h = 972

⇒ 3 × r² × 9 = 972

⇒ 27 × r² = 972 ⇒ r² = 36 ⇒ r = 6 cm

⇒ Curved Surface Area = 2 × 3 × 6 × 9

⇒ CSA = 324 cm²

∴ The correct answer is 324 cm².

Additional Information

Cube Mensuration Formulas

Total Surface Area (TSA) = 6a² and Volume = a³, where 'a' is the side length.

Cylinder Mensuration Formulas

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

Relationship Tip

CSA of a cylinder is also equal to (Perimeter of base × height) = 2πrh.