Quantitative Aptitude PYQs

Shortcut Trick

Ratio of Profit = Ratio of (Investment × Time).

Profit Ratio P : Q = (25,000 × 12) : (75,000 × 6) = (1 × 12) : (3 × 6) = 12 : 18 = 2 : 3.

Given Q’s share (3 units) = Rs. 6,000 ⇒ 1 unit = Rs. 2,000.

Total profit (2 + 3 = 5 units) = 5 × 2,000 = Rs. 10,000.

∴ The correct answer is Rs. 10,000.

Alternate Method

Given: P’s investment = Rs. 25,000, Time = 12 months; Q’s investment = Rs. 75,000, Time = 6 months; Q’s profit = Rs. 6,000.

Formula: Profit₁ ÷ Profit₂ = (Investment₁ × Time₁) ÷ (Investment₂ × Time₂)

⇒ Ratio of Profit P : Q = (25000 × 12) : (75000 × 6)

⇒ Ratio P : Q = 300,000 : 450,000 = 30 : 45 = 2 : 3

⇒ Let P’s profit be 2x and Q’s profit be 3x.

⇒ 3x = 6000 ⇒ x = 2000

⇒ Total Profit = 2x + 3x = 5x = 5 × 2000 = Rs. 10,000

∴ The correct answer is Rs. 10,000.

Additional Information

Partnership Basics

In a partnership, if the period of investment is different, it is called a compound partnership where profit is shared in the ratio of the products of capital and time.

Ratio Proportionality

If Investment is constant, Profit ∝ Time. If Time is constant, Profit ∝ Investment.

Shortcut Trick

Length of bus = (Speed × Time) − Bridge Length = (15 × 10) − 50 = 100 m.

Relative Speed (Opposite direction) = 15 + 25 = 40 m/s.

Total distance to cross similar bus = 100 + 100 = 200 m.

Time taken = 200 ÷ 40 = 5 sec.

∴ The correct answer is 5 sec.

Alternate Method

Given: Bridge length = 50 m, crossing time = 10 sec, Bus Speed₁ = 15 m/s, Bus Speed₂ = 25 m/s.

Formula Used: Distance = Speed × Time; Relative Speed (Opposite) = S₁ + S₂.

⇒ Case 1: (Length of Bus + Length of Bridge) = Speed × Time

⇒ L + 50 = 15 × 10

⇒ L + 50 = 150 ⇒ L = 100 m

⇒ Case 2: Crossing a similar bus (Length = 100 m) from opposite direction

⇒ Total Distance = L + L = 100 + 100 = 200 m

⇒ Relative Speed = 15 + 25 = 40 m/s

⇒ Time = 200 ÷ 40 = 5 sec

∴ The correct answer is 5 sec.

Additional Information

Relative Speed Concept

When two objects move in opposite directions, relative speed is the sum of their speeds (S₁ + S₂). In the same direction, it is the difference (|S₁ − S₂|).

Crossing Distance Rule

When a vehicle crosses a platform/bridge or another vehicle, the total distance covered is the sum of the lengths of both entities.

Unit Conversion

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

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

Step-1∶ Parts of an equation enclosed in 'Brackets' must be solved first and 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 least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated

(8 + 6 × 7) ÷ 10 × 3

⇒ (8 + 42) ÷ 10 × 3

⇒ 50 ÷ 10 × 3

⇒ 5 × 3 = 15

25% of 800 + ? = 40% of 1250

⇒ 200 + ? = 500

⇒ ? = 500 - 200 = 300

35% of (32 × 8 ÷ 16 ÷ 4)% of 10000

35% of (32 × 8 × 1/16 × 1/4)% of 10000

⇒ 35% of (32 × 1/8)% of 10000

⇒ 35% of 4% of 10000

⇒ 0.35 × 0.04 × 10000 = 140

Given

(220 ÷ 20 × 135 ÷ 15) /3 = ?

⇒ (11 × 9)/3 = ?

⇒ ? = 33

∴ (220 ÷ 20 × 135 ÷ 15) /3 = 33

Concept used:

Calculation:

40% of 180 + 70% of ? = 121

⇒ (40/100 × 180) + (70/100 × ?) = 121

⇒ 72 + (70/100 × ?) = 121

⇒ (70/100 × ?) = 49

 ∴ ? = 49 × 100/70 = 70.

142 + 712 - 192 = ? + 123

⇒ 196 + 5041 - 361 = ? + 123 

⇒ ? = 4753

Smart Approach

As last digit in all options are different we need not to solve question fully

Taking only last digits, we get

6 + 1 - 1 = ? + 3

⇒ ? = 3

Hence, option 3 is correct.

80% of [80% of 245 – 56] = ? ÷ 5

⇒ 80% of [80% of 245 – 56] = ? ÷ 5

⇒ 80% of [(80/100) × 245 – 56] = ? ÷ 5

⇒ 80% of [196 – 56] = ? ÷ 5

⇒ 80% of 140 = ? ÷ 5

⇒ 112 × 5 = ?

∴ ? = 560

√? × 45 = 250 × 15 + 138 × 25

⇒ √? × 45 = 3750 + 3450

⇒ √? × 45 = 7200

⇒ √? = 160

⇒ ? = 25600

Given:

2? = 32 ÷ 128 × 1024 ÷ 8

Concept:

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

Calculation:

2? = 32 ÷ 128 × 1024 ÷ 8

⇒ 2? = (32/128) × (1024/8)

⇒ 2? = (1/4) × (128) = 32

⇒ 25 = 32

(?) = 5

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

25% of 1080 - ?% of 400 = 90

⇒ 270 - 4 × ? = 90

⇒ 270 - 90 = 4 × ?

⇒ 180 = 4 × ?

⇒ ? = 180/4

∴ ? = 45

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 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 least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated 

25% of 160 + 50% of 320 = ?

⇒ 40 + 160 = ?

∴ ? = 200

52 × 48 + 113 × 87 + 35 × 25 = ? - √196

⇒ (50² - 2²) + (100² - 13²) + (30² - 5²) = ? - 14

⇒ 2500 - 4 + 10000 - 169 + 900 - 25 + 14 = ?

⇒ ? = 13216 

Short trick

As all options has different unit digit. Hence, we can find the answer only by calculating unit digit.

52 × 48 + 113 × 87 + 35 × 25 = ? - √196

⇒ 6 + 1 + 5 = ? - 4

⇒ 12 + 4 = ?

⇒ ? = 16 = 6 (unit digit)

Hence, answer will be 13216.

√676 + 82 ×  5.5 + 73 = ? + 121

⇒ 26 + 64 × (55/10) + 343 - 121 = ?

⇒ 26 + 352 + 343 - 121 = ?

⇒ 721 - 121 = ?

⇒ ? = 600

Concept used:

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

Calculations:

2.5% of 40 + 4.5% of 80 = ?% of 10

⇒ (2.5/100) × 40 + (4.5/100) × 80 = (?/100) × 10

⇒ 1 + 3.6 = ?/10

⇒ 4.6 = ?/10

⇒ ? = 46

∴ The value of ? is 46.

450% of ? ÷ 90 + 350% of 200 ÷ 70 = 20

⇒ (450/100 × ?/90) + (350/100 × 200/70) = 20

⇒ ?/20 + 10 = 20

⇒ ?/20 = 10

⇒ ? = 200

Shortcut Trick

Total interest in 4 years = ₹2880 → Interest in 1 year = 2880 ÷ 4 = ₹720.

Overall interest rate = (720 ÷ 12000) × 100 = 6%.

Using Alligation: (Rate 1: 7%) vs (Rate 2: 4%) with Mean (6%).

Ratio of sums = (6 − 4) : (7 − 6) = 2 :

1. Sum at 4% = (1 ÷ 3) × 12000 = ₹4000.

∴ The correct answer is ₹4000.

Alternate Method

Given: Total Principal (P) = ₹12000, Total Time (T) = 4 years, Total SI = ₹2880.

Formula Used: SI = (P × R × T) ÷ 100

⇒ Let sum lent at 4% be x. Then sum lent at 7% is (12000 − x).

⇒ [(12000 − x) × 7 × 4] ÷ 100 + [x × 4 × 4] ÷ 100 = 2880

⇒ (12000 − x) × 28 + 16x = 288000

⇒ 336000 − 28x + 16x = 288000

⇒ 12x = 48000

⇒ x = 4000

∴ The correct answer is ₹4000.

Additional Information

Simple Interest Concept

Simple Interest is calculated only on the principal amount. The interest remains constant every year if the rate and principal do not change.

Rule of Alligation in SI

When a sum is divided into two parts with different interest rates, the ratio of the two parts can be found using the difference between individual rates and the overall average rate.

Effective Rate of Interest

Effective Rate = (Total SI × 100) ÷ (Principal × Time). This helps in simplifying multi-rate problems into a single comparison.

Shortcut Trick

Let the total sum of money be the LCM of 21 and 28.

LCM(21, 28) = 84 units.

Efficiency of X = 84 ÷ 21 = 4 units/day.

Efficiency of Y = 84 ÷ 28 = 3 units/day.

Combined efficiency = 4 + 3 = 7 units/day.

Number of days = 84 ÷ 7 = 12 days.

∴ The correct answer is 12 days.

Alternate Method

Given:

Wages for X can be paid for = 21 days

Wages for Y can be paid for = 28 days

Formula Used:

If a sum pays A for 'a' days and B for 'b' days, then it pays both for (a × b) ÷ (a + b) days.

Calculations:

⇒ Let the total sum of money be S.

⇒ Daily wage of X = S ÷ 21

⇒ Daily wage of Y = S ÷ 28

⇒ Combined daily wage = (S ÷ 21) + (S ÷ 28)

⇒ Combined daily wage = (4S + 3S) ÷ 84 = 7S ÷ 84 = S ÷ 12

⇒ Number of days the sum lasts for both = S ÷ (S ÷ 12)

⇒ Number of days = 12 days

∴ The correct answer is 12 days.

Additional Information

Work and Wages Principle

Wages are generally distributed in the ratio of work done or in the ratio of efficiencies if the time period is the same.

Time and Work Relation

The logic used for sum of money and wages is identical to Time and Work problems where sum is Total Work and wages/day is Efficiency.

Reciprocal Method

For multiple people, the number of days D is found using: 1/D = 1/d₁ + 1/d₂ + ... + 1/d_n.

Shortcut Trick

Let the Marked Price (MP) be 100%.

Discount for Rohit = 10% ⇒ SP₁ = 90%.

Discount for Abhishek = 30% ⇒ SP₂ = 70%.

Difference in Selling Price (SP) = 90% − 70% = 20%.

According to the question, 20% = Rs. 240. Marked Price (100%) = (240 ÷ 20) × 100 = Rs. 1200.

∴ The correct answer is Rs. 1200.

Alternate Method

Given: Discount₁ = 10%, Discount₂ = 30%, Difference in SP = Rs. 240. Formulas: SP = MP × (100 − Discount%) ÷ 100.

⇒ Let Marked Price be x.

⇒ SP for Rohit = x × (90/100) = 0.9x.

⇒ SP for Abhishek = x × (70/100) = 0.7x.

⇒ Difference = 0.9x − 0.7x = 240.

⇒ 0.2x = 240.

⇒ x = 240 ÷ 0.2 = 1200.

∴ The correct answer is Rs. 1200.

Additional Information

Marked Price and Selling Price

The Marked Price (MP) is the price printed on an item. The price at which it is sold after a discount is the Selling Price (SP). SP = MP − Discount.

Discount Percentage

Discount is always calculated on the Marked Price. Formula: Discount% = (Discount ÷ MP) × 100.