Quantitative Aptitude PYQs

Shortcut Trick

Profit ratio of A : B is derived from the statement 1/6^th less: If B = 6, then A = 6 − 1 = 5.

Ratio of profits P_A : P_B = 5 : 6. Since time is the same (1 year), Profit ratio = Investment ratio.

Difference in ratio parts = 6 − 5 = 1 unit. Given investment difference = ₹500.

∴ 1 unit = ₹500. Value of x (A's investment) = 5 units = 5 × 500 = ₹2500.

∴ The correct answer is ₹2500.

Alternate Method

Given: A's Investment = ₹x, B's Investment = ₹(x + 500), Time = 1 year.

Formulas: Profit = Investment × Time | Profit Ratio = (I₁ × T₁) : (I₂ × T₂)

⇒ Let B's profit be P_B. Then A's profit P_A = P_B − (1/6)P_B = (5/6)P_B.

⇒ P_A / P_B = 5 / 6.

⇒ Since time is equal, x / (x + 500) = 5 / 6.

⇒ 6x = 5(x + 500) ⇒ 6x = 5x + 2500.

⇒ x = 2500.

∴ The correct answer is ₹2500.

Additional Information

Partnership Ratios

When the time period for all partners is the same, the profit is distributed solely in the ratio of their investments.

Fractional Change

If a value is 1/n less than another, it is represented as (1 − 1/n) of that value, or (n − 1)/n.

Shortcut Trick

Initial Milk = 3/10 × 70 = 21 litres; Initial Water = 7/10 × 70 = 49 litres.

New Water = 49 + 56 = 105 litres.

Ratio 4:5 means 5 units = 105 litres → 1 unit = 21 litres.

New Milk = 4 units = 4 × 21 = 84 litres.

Value of x = 84 − 21 = 63 litres.

∴ The correct answer is 63 litres.

Alternate Method

Given: Initial Volume = 70 litres, Ratio M:W = 3:7, Water added = 56 litres, New ratio = 4:5.

Calculations:

⇒ Initial Milk = 70 × (3 / (3 + 7)) = 21 litres

⇒ Initial Water = 70 − 21 = 49 litres

⇒ Final Water = 49 + 56 = 105 litres

⇒ Let final Milk be M_f. Then M_f / 105 = 4 / 5

⇒ M_f = (4/5) × 105 = 4 × 21 = 84 litres

⇒ x = Final Milk − Initial Milk = 84 − 21 = 63 litres

∴ The correct answer is 63 litres.

Additional Information

Ratio Distribution

To divide a quantity Q in ratio a:b, the parts are Q × a/(a+b) and Q × b/(a+b).

Mixture Alligation

When adding a pure component (like 100% milk), the concentration changes based on the weighted average of the components.

Constant Quantity Concept

If only one component is added, the other component remains proportionally constant relative to its own previous absolute value.

Shortcut Trick

Let the common ratio for A and B be x. Ages are 6x and 5x.

C age → 2 × 5x − 5 = 10x − 5.

Difference (C − A) → (10x − 5) − 6x = 11 ⇒ 4x = 16 ⇒ x = 4.

Age of A → 6 × 4 = 24 years.

∴ The correct answer is 24 years.

Alternate Method

Given: Ratio of A:B = 6:5. Age of C = 2 × B − 5. C − A = 11.

Calculations:

⇒ Let age of A = 6x and age of B = 5x.

⇒ Age of C = 2(5x) − 5 = 10x − 5.

⇒ According to the problem: C − A = 11

⇒ (10x − 5) − 6x = 11

⇒ 4x − 5 = 11

⇒ 4x = 16

⇒ x = 4

⇒ Age of A = 6 × 4 = 24.

∴ The correct answer is 24 years.

Additional Information

Ratio to Actual Value

If ratio is a:b, actual values are ax and bx where x is the highest common factor (HCF).

Age Problems Logic

Always define the ages of all individuals in terms of a single variable to simplify linear equations.

Difference Property

The difference between the ages of two people remains constant throughout their lives.

Shortcut Trick

Discount = 20% ⇒ SP = 80% of MP ⇒ 2400 = 0.8 × MP ⇒ MP = 3000.

Profit = 20% of MP = 0.2 × 3000 = 600.

CP = SP − Profit = 2400 − 600 = 1800.

New SP = 2700.

New Profit = New SP − CP = 2700 − 1800 = 900.

∴ The correct answer is Rs 900.

Alternate Method

Given → SP₁ = 2400, Discount = 20%, Profit = 20% of MP, SP₂ = 2700. Formulas → MP = SP ÷ (1 − D%), CP = SP − Profit.

⇒ MP = 2400 ÷ (1 − 0.20) = 2400 ÷ 0.8 = 3000. ⇒ Profit_initial = 20% of 3000 = 600. ⇒ CP = SP₁ − Profit_initial = 2400 − 600 = 1800. ⇒ New Profit = SP₂ − CP = 2700 − 1800 = 900.

∴ The correct answer is Rs 900.

Additional Information

Relationship between MP and CP

Marked Price is the price listed on the label, and Discount is always calculated on the Marked Price. Profit is calculated on the Cost Price unless stated otherwise.

Mark-up Percentage

Mark-up % = [(MP − CP) ÷ CP] × 100. It represents how much higher the MP is relative to the CP.

Shortcut Trick

Ratio of Upstream Speed (U) to Downstream Speed (D) is 2 : 3.

Total time = (Distance ÷ U) + (Distance ÷ D) ⇒ 7.5 = (180 ÷ 2x) + (180 ÷ 3x).

7.5 = 90/x + 60/x ⇒ 7.5 = 150/x ⇒ x = 150 ÷ 7.5 = 20.

Downstream Speed = 3x = 3 × 20 = 60 km/h.

∴ The correct answer is 60 km/h.

Alternate Method

Given: Upstream Speed : Downstream Speed = 2 : 3. Total Distance = 180 km. Total Time = 7.5 hours.

Formula Used: Time = Distance ÷ Speed

⇒ Let Upstream Speed = 2x and Downstream Speed = 3x.

⇒ Total Time = (180 ÷ 2x) + (180 ÷ 3x) = 7.5

⇒ 90/x + 60/x = 7.5

⇒ 150/x = 7.5

⇒ x = 150 ÷ 7.5 = 20

⇒ Downstream Speed = 3 × 20 = 60 km/h.

∴ The correct answer is 60 km/h.

Additional Information

Speed in Still Water

It is the average of downstream and upstream speeds: B = (D + U) ÷ 2.

Speed of Current (Stream)

It is half the difference between downstream and upstream speeds: S = (D − U) ÷ 2.

Average Speed for Round Trip

Average Speed = (2 × U × D) ÷ (U + D), which is also equal to Total Distance ÷ Total Time.

Calculation:

Difference between Monday and Tuesday

⇒ 180 - 160 = 20

So, the number of sweets sold on Monday is 20 less than on Tuesday.

Difference between Wednesday and Friday

⇒ 200 - 120 = 80

So, the number of sweets sold on Wednesday is 80 more than on Friday.

∴ The correct answer is: 20, 80.

Calculation:

Number of sweets sold on Thursday = x + 45

From the provided data, the number of sweets sold on Thursday = 140

⇒ x + 45 = 140

⇒ x = 140 - 45

⇒ x = 95

∴ The value of x is 95.

Calculation:

Number of sweets sold on Friday = 120

Ratio of sweets sold on Friday to Saturday = 4 : 7

We need to find the number of sweets sold on Saturday.

Since the number of sweets sold on Friday is given as 120:

⇒ 4x = 120

⇒ x = 120 / 4

⇒ x = 30

Now, the number of sweets sold on Saturday:

⇒ 7x = 7 × 30

⇒ 210

∴ The number of sweets sold on Saturday is 210.

Calculation:

Total sweets sold on Wednesday and Thursday

⇒ 200 + 140 = 340

Total sweets sold on Monday and Tuesday

⇒ 160 + 180 = 340

Required difference

⇒ 340 - 340 = 0

∴ The difference between the number of sweets sold on Wednesday and Thursday together and Monday and Tuesday together is 0.

Calculation:

Number of sweets sold on Monday = 160

Only 40% of the total sweets made were sold on Monday.

Total sweets made on Monday:

⇒ (160 × 100) / 40

⇒ 16000 / 40 = 400

Unsold sweets on Monday:

⇒ 400 - 160 = 240

Number of sweets sold on Tuesday = 180

Only 60% of the total sweets made were sold on Tuesday.

Total sweets made on Tuesday:

⇒ (180 × 100) / 60

⇒ 300

Unsold sweets on Tuesday:

⇒ 300 - 180 = 120

Total unsold sweets = Unsold sweets on Monday + Unsold sweets on Tuesday

⇒ 240 + 120 = 360

∴ The total number of unsold sweets on Monday and Tuesday together is 360.

Given:

162 + (5/8) of 344 + ? = 500

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

162 + (5/8) of 344 + ? = 500

⇒ 256 + (5/8) of 344 + ? = 500

⇒ 256 + 215 + ? = 500

⇒ ? = 500 - 471

⇒ ? = 29

∴ 29 should come in place of the question mark (?).

Given:

10% of 450 - (8 × 5) = ? × 5

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

10% of 450 - (8 × 5) = ? × 5

⇒ 45 - 40 = ? × 5

⇒ 5 = ? × 5

⇒ ? = 1

∴ The value of '?' is 1.

Given:

78 ÷ 6 × 5 + √625 = ? × 9

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

78 ÷ 6 × 5 + √625 = ? × 9

⇒ 13 × 5 + 25 = ? × 9

⇒ 65 + 25 = ? × 9

⇒ ? × 9 = 90

⇒ ? = 10

∴ 10 should come in place of the question mark (?).

Given:

122 × 5 - √400 + ? × 5 = 850

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

122 × 5 - √400 + ? × 5 = 850

⇒ 144 × 5 - 20 + ? × 5 = 850

⇒ 720 - 20 + ? × 5 = 850

⇒ ? × 5 = 150

⇒ ? = 30

∴ 30 should come in place of the question mark (?).

Given:

20 - 102 = 41 - ?2 

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

20 - 102 = 41 - ?2 

⇒ 20 - 100 = 41 - ?²

⇒ -80 = 41 - ?²

⇒ ?² = 121

⇒ ? = 11

∴ The value of ? is 11.

Given:

5656 ÷ 28 × 11 - 333 = ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

5656 ÷ 28 × 11 - 333 = ?

⇒ 202 × 11 - 333 = ?

⇒ 2222 - 333 = ?

⇒ ? = 1889

∴ The value of ? is 1889.

Given:

90 × 20% of 105 ÷ 3 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

90 × 20% of 105 ÷ 3 = ?

⇒ 90 × (1/5) × 105/3 = ?

⇒ 18 × 35 = ?

⇒ ? = 630

∴ The value of ‘?’ Is 630.

Hint

20% = 1/5

Given:

440 + 1825 ÷ 25 + 102 = 215 + ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

440 + 1825 ÷ 25 + 102 = 215 + ?

⇒ 440 + (1825/25) + 102 = 215 + ?

⇒ 440 + 73 + 102 - 215 = ?

⇒ ? = 400

∴ The value of ‘?’ Is 400.

Given:

(4/9) of (729/2) + 400 = ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

(4/9) of (729/2) + 400 = ?

⇒ 81 × 2 + 400 = ?

⇒ 162 + 400 = ?

⇒ ? = 562

∴ The value of ? is 562.

Given :

 378 + ?^1/2 = 25% of 2400 - 192

 Calculation :

  378 + ?^1/2  = 25% of 2400 - 192

 ⇒ 378 + ?^1/2 = (2400 × 1 / 4 ) - 192

 ⇒ 378 + ?^1/2 = 600 - 192

 ⇒ ?^1/2  = 408 - 378

 ⇒ ?^1/2  = 30

 ⇒ (?)^2×1/2 = 30²

 ⇒ ? = 900

 Hence,900 will come in the place of '?'.