Quantitative Aptitude PYQs

Shortcut Trick
Quantity I: Age of Y = 2 units = 34 years ⇒ 1 unit = 17 years. Age of Z = 3 units = 51 years. Age of X = 51 + 2 = 53 years.
Quantity II: Current age of Z is 51 years. If the ratio changes to 17 : 10, the problem implies finding the new age of Z. Assuming Y stays constant (34 years), Z would become 20 years.
Comparing both: 53 years > 20 years.
∴ The correct answer is Quantity I > Quantity II.
Alternate Method
Given: Ratio of age Y : Z = 2 : 3. Present age of Y = 34 years.

Calculations for Quantity I:
⇒ Age of Y = 2k = 34
⇒ k = 17
⇒ Age of Z = 3k = 3 × 17 = 51 years
⇒ Age of X = Z + 2 = 51 + 2 = 53 years
Calculations for Quantity II:
⇒ New ratio Y : Z = 17 : 10
⇒ Since Y is 34, then 17 units = 34
⇒ 1 unit = 2
⇒ Age of Z = 10 × 2 = 20 years
⇒ 53 > 20
∴ The correct answer is Quantity I > Quantity II.
Additional Information
Ratio and Proportion in Ages
When the age of one person is fixed, the ages of others in a ratio are calculated using a common multiplier (k). If A:B = x:y and A is known, then B = (Value of A ÷ x) × y.
Comparison of Quantities
In banking exams, Quantity Comparison involves solving two independent math problems and establishing the mathematical relationship (>, <, =, ≤, ≥) between their results.

Shortcut Trick

Quantity I: Total ratio = 7 + 7 + 6 = 20 units. 20 units = ₹10,000 → 1 unit = ₹500. Boy share (6 units) = 6 × 500 = ₹3,000.

Quantity II: Let Boy = x, then each Girl = x + 500. Total: (x + 500) + (x + 500) + x = 10,000 → 3x + 1000 = 10,000 → 3x = 9,000 → x = ₹3,000.

Comparing both: ₹3,000 = ₹3,000.

∴ The correct answer is Quantity 1 = Quantity 2.

Alternate Method

Given: Total Prize Money = ₹10,000. Distribution among Girl 1, Girl 2, and Boy.

Quantity I Calculation:

⇒ Ratio of Girl 1 : Girl 2 : Boy = 7 : 7 : 6

⇒ Sum of ratios = 7 + 7 + 6 = 20

⇒ Share of Boy = (6 ÷ 20) × 10,000

⇒ Share of Boy = 6 × 500 = ₹3,000

Quantity II Calculation:

⇒ Let Boy share be ₹x. Girls share = ₹(x + 500) each.

⇒ (x + 500) + (x + 500) + x = 10,000

⇒ 3x + 1,000 = 10,000

⇒ 3x = 9,000

⇒ x = ₹3,000

Conclusion:

⇒ Quantity I = ₹3,000 and Quantity II = ₹3,000

∴ The correct answer is Quantity 1 = Quantity 2.

Additional Information

Ratio and Proportion

To divide a quantity A in the ratio a:b:c, the shares are [a/(a+b+c)]×A, [b/(a+b+c)]×A, and [c/(a+b+c)]×A.

Linear Equations

When values are compared using more than or less than, setting the smallest value as 'x' usually simplifies the equation.

Comparison of Quantities

Always ensure both quantities are in the same units (₹ in this case) before comparing for equality or inequality.

Shortcut Trick
Total balls = 5 (O) + 3 (B) + 4 (R) = 12. Total ways to pick 3 balls = 12C3.
Quantity I: Picking 2 Brown and 1 Red → 3C2 × 4C1 = 3 × 4 = 12 ways.
Quantity II: Picking 3 Orange → 5C3 = 10 ways.
Since the denominator (12C3) is same, compare numerators: 12 > 10.
∴ The correct answer is Quantity I > Quantity II.
Alternate Method
Given: Orange = 5, Brown = 3, Red = 4. Total = 12. Samples = 3.
Formula Used: Probability P(E) = n(E) ÷ n(S); nCr = n! ÷ [r!(n−r)!]

⇒ n(S) = 12C3 = (12 × 11 × 10) ÷ (3 × 2 × 1) = 220
⇒ Quantity I: n(E1) = 3C2 × 4C1 = 3 × 4 = 12
⇒ P(Quantity I) = 12 ÷ 220
⇒ Quantity II: n(E2) = 5C3 = (5 × 4) ÷ (2 × 1) = 10
⇒ P(Quantity II) = 10 ÷ 220
⇒ 12/220 > 10/220 ∴ Quantity I > Quantity II
∴ The correct answer is Quantity I > Quantity II.
Additional Information
Combination Formula
Used for selection where order does not matter: nCr = n! / [r!(n−r)!]. Note that nCr = nCn−r.
Probability of Compound Events
When multiple independent selections are made, use the product of individual combinations for specific color requirements divided by total possible combinations.

Shortcut Trick

Let Ranjeet's income be x.

Sonia's income = 4x.

Ram's savings = Income − Expenditure = 80,000 − 200% of x = 80,000 − 2x.

Given: 80,000 − 2x = 4x + 20,000 ⇒ 60,000 = 6x ⇒ x = 10,000.

Income Ratio = 4x : x : 80,000 = 40,000 : 10,000 : 80,000 = 4 : 1 : 8.

∴ The correct answer is 4 : 1 : 8.

Alternate Method

Given: Ram's income = ₹80,000. Let Ranjeet's income = R.

Formula: Income = Expenditure + Savings

⇒ Sonia's income = 4 × R

⇒ Ram's expenditure = 200% of R = 2R

⇒ Ram's saving = 80,000 − 2R

⇒ According to question: 80,000 − 2R = 4R + 20,000

⇒ 60,000 = 6R

⇒ R = 10,000

⇒ Sonia's income = 4 × 10,000 = 40,000

⇒ Ratio = 40,000 : 10,000 : 80,000

⇒ Ratio = 4 : 1 : 8

∴ The correct answer is 4 : 1 : 8.

Additional Information

Income and Expenditure Relation

Income − Savings = Expenditure. Percentage increases or decreases are always calculated on the base value provided.

Ratio Scaling

Ratios can be simplified by dividing all terms by their Greatest Common Divisor (GCD) to reach the simplest form.

Shortcut Trick

Use the direct relationship formula: CP ÷ MP = (100 − D%) ÷ (100 + P%).

CP ÷ MP = (100 − 70) ÷ (100 + 20) = 30 ÷ 120 = 1 ÷ 4.

Given Cost Price (CP) = Rs. 145, which represents 1 unit.

Therefore, Marked Price (MP) = 4 × 145 = Rs. 580. ∴ The correct answer is Rs. 580.

Alternate Method

Given: CP = Rs. 145, Profit% = 20%, Discount% = 70%.

Formulas: SP = CP × (100 + P%) ÷ 100 | MP = SP × 100 ÷ (100 − D%)

⇒ Step 1: Calculate Selling Price (SP) = 145 × (120 ÷ 100) = 145 × 1.2 = Rs. 174.

⇒ Step 2: Calculate Marked Price (MP) using SP and Discount%.

⇒ MP × (100 − 70) ÷ 100 = 174.

⇒ MP × 30 ÷ 100 = 174 ⇒ MP = (174 × 100) ÷ 30 = 17400 ÷ 30 = 580.

∴ The correct answer is Rs. 580.

Additional Information

Marked Price and Discount

Marked Price (MP) is the price printed on an article. Discount is always calculated on the MP to arrive at the Selling Price (SP). Formula: SP = MP − Discount.

Profit on Cost Price

Profit is always calculated on the Cost Price (CP) unless stated otherwise. Formula: SP = CP + Profit.

Shortcut Trick

Total Work = LCM(20, 24, 30) = 120 units.

Efficiencies: A = 6, B = 5, C = 4.

Work done in first 2 days (A+B+C) = (6+5+4) × 2 = 30 units.

Work done in next 2 days (B+C) = (5+4) × 2 = 18 units.

Remaining work = 120 − 30 − 18 = 72 units (to be done by B alone).

Time for B = 72 ÷ 5 = 14.4 days. Total days = 2 + 2 + 14.4 = 18.4 days.

∴ The correct answer is 18.4 days.

Alternate Method

Step-by-step Calculation:

⇒ Work done by A, B, and C in the first 2 days = (6 + 5 + 4) × 2 = 15 × 2 = 30 units.

⇒ A leaves after 2 days. B and C work for another 2 days (2 days after A leaves).

⇒ Work done by B and C in those 2 days = (5 + 4) × 2 = 9 × 2 = 18 units.

⇒ Total work completed so far = 30 + 18 = 48 units.

⇒ Remaining Work = 120 − 48 = 72 units.

⇒ C leaves, so the remaining work is completed by B alone.

⇒ Time taken by B to finish remaining work = 72 ÷ 5 = 14.4 days.

⇒ Total days to complete the work = 2 (initial) + 2 (with C) + 14.4 (B alone) = 18.4 days.

∴ The correct answer is 18.4 days.

Additional Information

LCM Method for Work

The most efficient way to solve Time and Work problems is by assuming the total work as the LCM of individual days to find unit efficiencies.

Efficiency and Time

Efficiency is inversely proportional to the time taken. Total Work = Efficiency × Time.

Leaving Work Midway

If a person leaves 'x' days before completion, add their potential work (Efficiency × x) to the total work and divide by total efficiency.

Shortcut Trick

Profit ratio A : B = 1800 : (3300 − 1800) = 1800 : 1500 = 6 : 5.

Investment ratio A : B = 16000 : 20000 = 4 : 5.

Time ratio = Profit ratio ÷ Investment ratio.

Time ratio A : B = (6 ÷ 4) : (5 ÷ 5) = 1.5 : 1 = 3 : 2.

Since A stayed for 12 months, 3 units = 12 ⇒ 1 unit = 4.

Time of B (x) = 2 units = 2 × 4 = 8 months.

∴ The correct answer is 8 months.

Alternate Method

Given: A investment = Rs. 16000, B investment = Rs. 20000, A time = 12 months, B time = x months, Total Profit = Rs. 3300, A share = Rs. 1800.

Formula: Profit Ratio = (Capital_A × Time_A) : (Capital_B × Time_B)

⇒ Profit of B = Total Profit − Profit of A = 3300 − 1800 = Rs. 1500.

⇒ Ratio of profits (A:B) = 1800 / 1500 = 6 / 5.

⇒ (16000 × 12) / (20000 × x) = 6 / 5

⇒ (16 × 12) / (20 × x) = 6 / 5

⇒ 192 / 20x = 6 / 5

⇒ 120x = 192 × 5 ⇒ 120x = 960

⇒ x = 960 ÷ 120 = 8.

∴ The correct answer is 8 months.

Additional Information

Partnership Basics

When time periods are different, profit is distributed based on the Product of Capital and Time.

Ratio Conversion

If Capital Ratio is C₁:C₂ and Profit Ratio is P₁:P₂, then Time Ratio T₁:T₂ = (P₁/C₁) : (P₂/C₂).

Shortcut Trick

Ratio of A : B is 5 : 3. Let ages be 5x and 3x.

Equation: (5x + 5) / (3x − 5) = 3 / 1

⇒ 5x + 5 = 9x − 15 ⇒ 4x = 20 ⇒ x = 5.

A age = 5 × 5 = 25 years. C age = 25 + 10 = 35 years.

∴ The correct answer is 35 years.

Alternate Method

Given: Ratio of A:B = 5:3. Age of C = Age of A + 10.

Formula: Ratio = Value 1 / Value 2

⇒ Let A = 5x and B = 3x

⇒ (5x + 5) / (3x − 5) = 3 / 1

⇒ 1 × (5x + 5) = 3 × (3x − 5)

⇒ 5x + 5 = 9x − 15

⇒ 9x − 5x = 15 + 5

⇒ 4x = 20 ⇒ x = 5

⇒ Age of A = 5 × 5 = 25 years

⇒ Age of C = 25 + 10 = 35 years

∴ The correct answer is 35 years.

Additional Information

Ratio and Proportion in Ages

The age difference between two people remains constant over time, which can often be used to simplify ratio-based age problems.

Successive Age Changes

If the current ratio is a:b, then n years later it becomes (ax+n):(bx+n) and n years ago it was (ax−n):(bx−n).

Common Calculation

Total number of students [ Boys + Girls] in class together is 1200.

Difference between boys and girls in all class together is 200.

Total number of boys in all class together is [1200 + 200] /100 = 700

Number of students in class A is 1200 × 12 /100 = 144

Number of boys in class A is 700 × 18 / 100 = 126

So, Number of girls in class A is 144 – 126 = 18

Similarly, we can calculate others class values also.

Calculation

Sum of boys = 126 + 140 + 154 + 175 + 105 = 700

Average across 5 classes = 700/5 = 140.

Remove Class B

New sum = 700−140 = 560.

New average across remaining 4 = 560/4 = 140.

So, the average does not change; the decrease k = 140 − 140 =  0

Common Calculation

Total number of students [ Boys + Girls] in class together is 1200.

Difference between boys and girls in all class together is 200.

Total number of boys in all class together is [1200 + 200] /100 = 700

Number of students in class A is 1200 × 12 /100 = 144

Number of boys in class A is 700 × 18 / 100 = 126

So, Number of girls in class A is 144 – 126 = 18

Similarly, we can calculate others class values also.

Calculation

Quiz Competition Participants from Class B

Boys in B = 140

→ 40% participate = 0.40 × 140 = 56 boys

Girls in B = 76

→ 50% participate = 0.50 × 76=38 girls

Total participants from B = 56+38=94

Answer: 94 students

Common Calculation

Total number of students [ Boys + Girls] in class together is 1200.

Difference between boys and girls in all class together is 200.

Total number of boys in all class together is [1200 + 200] /100 = 700

Number of students in class A is 1200 × 12 /100 = 144

Number of boys in class A is 700 × 18 / 100 = 126

So, Number of girls in class A is 144 – 126 = 18

Similarly, we can calculate others class values also.

Calculation

Probability that a randomly chosen student from Class D is a girl

Girls in D = 125

Total in D = 300

Probability P(girl)=125/300 = 5/12​

Answer: 5/12

Common Calculation

Total number of students [ Boys + Girls] in class together is 1200.

Difference between boys and girls in all class together is 200.

Total number of boys in all class together is [1200 + 200] /100 = 700

Number of students in class A is 1200 × 12 /100 = 144

Number of boys in class A is 700 × 18 / 100 = 126

So, Number of girls in class A is 144 – 126 = 18

Similarly, we can calculate others class values also.

Calculation

Transfer of students between Class A and Class E

Before transfer:

Class A → Boys = 126, Girls = 18

Class E → Boys = 105, Girls = 75

After transfer:

15 boys move A → E

Class A boys = 126 – 15 = 111

10 girls move E → A →

Class A girls = 18 + 12 = 30

New Ratios:

Class A ratio (boys : girls) = 111:30 = 37:10

Shortcut Trick

Net change in marks = (Sum of Correct marks) − (Sum of Incorrect marks)

Net change = (75 + 50) − (72 + 52) = 125 − 124 = +1 mark

Difference in average = Net change ÷ Total students

Difference in average = 1 ÷ 25 = 0.04

∴ The correct answer is 0.04.

Alternate Method

Given:

Number of students = 25

Old average = 60

Incorrect marks = 72, 52

Correct marks = 75, 50

Calculations:

⇒ Old Total Sum = 25 × 60 = 1500

⇒ New Total Sum = 1500 − (72 + 52) + (75 + 50)

⇒ New Total Sum = 1500 − 124 + 125 = 1501

⇒ New Average = 1501 ÷ 25 = 60.04

⇒ Difference = New Average − Old Average = 60.04 − 60 = 0.04

∴ The correct answer is 0.04.

Additional Information

Average Correction Formula

Correct Average = [ (Incorrect Average × n) + (Correct Value − Incorrect Value) ] ÷ n

Net Deviation Method

Change in Average = (Total Correct Marks − Total Incorrect Marks) ÷ Number of Observations

Weighted Average

When groups of sizes n₁ and n₂ have averages a₁ and a₂, Combined Average = (n₁a₁ + n₂a₂) ÷ (n₁ + n₂)

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

Since, we need to find out the approximate value,

So, we can write these values to their nearest integers.

Given expression is

39.98% of (14.87 × 40.04) - ? = 55.96 × 9.93 ÷ 7.96

⇒ 40% of (15 × 40) - ? = 56 × 10 ÷ 8

⇒ 40% of (600) - ? = 70

⇒ 240 - ? = 70

⇒ ? = 240 - 70 = 170

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

Since, we need to find out the approximate value,

So, we can write these values to their nearest integers.

Given expression is

49.96% of 320.05 + 74.96% of 119.97 + √120.99 = ?

⇒ 50% of 320 + 75% of 120 + √121 = ?

⇒ (50/100) × 320 + (75/100) × 120 + 11 = ?

⇒ 160 + 90 + 11 = ?

⇒ ? = 261

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

Since, we need to find out the approximate value,

So, we can write these values to their nearest integers.

Given expression is

[53.98 – {34.97 – (55.98 – 62.97 ÷ 8.99 × 6.96)}] = ?

⇒ [54 – {35 – (56 – 63 ÷ 9 × 7)}] = ?

⇒ [54 – {35 – (56 – 7 × 7)}] = ?

⇒ [54 – {35 – (56 – 49)}] = ?

⇒ [54 – {35 – 7}] = ?

⇒ [54 – 28] = ?

⇒ 26 = ?

Concept used:

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

Calculations:

[{(5.89)2 + (4.94)2 }× 9.95] ÷ 9.89 = (?)2 – 19.88

⇒ [(62 + 52) × 10)] ÷ 10 = (?)2 – 20

⇒ [(36 + 25) × 10)] ÷ 10 = (?)2 – 20

⇒ [61 × 10)] ÷ 10 = (?)2 – 20

⇒ 610 ÷ 10 = (?)2 – 20

⇒ 61 + 20 = (?)2

⇒ 81 = (?)2

⇒ 9 = ?

∴ The approx. value of ? is 9

A candidate must be careful to consider the rules of approximation while solving calculating the approximate values of a given equation. If not done so, the value of the final result (i.e. ?) shall be far away from the required value.

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

Since, we need to find out the approximate value,

So, we can write these values to their nearest integers.

Given expression is

14.98% of (23.98 × 24.89) - ? = 15.96 × 29.9 ÷ 7.99

⇒ 15% of (24 × 25) - ? = 16 × 30 ÷ 8

⇒ 15% of 600 - ? = 60

⇒ 15% of 600 - ? = 60

⇒ 90 - ? = 60

∴ ? = 30

Shortcut Trick

Using the relation between Total Surface Area (TSA) and Volume (V): V = (TSA × r) ÷ 3

TSA = 4 × π × r² = 5544 ⇒ πr² = 1386. For πr² = 1386, r = 21 m (Standard circle value).

Volume = 8t = (5544 × 21) ÷ 3 ⇒ 8t = 5544 × 7

t = (5544 × 7) ÷ 8 = 693 × 7 = 4851.

∴ The correct answer is 4851.

Alternate Method

Given: Total Surface Area = 5544 m², Volume = 8t

Formula Used: Surface Area = 4πr²; Volume = (4/3)πr³

⇒ 4 × (22/7) × r² = 5544

⇒ r² = (5544 × 7) ÷ (4 × 22)

⇒ r² = 63 × 7 = 441 ⇒ r = 21 m

⇒ Volume = (4/3) × (22/7) × 21 × 21 × 21

⇒ 8t = 4 × 22 × 21 × 21 ÷ 1 = 88 × 441 = 38808

⇒ t = 38808 ÷ 8 = 4851

∴ The correct answer is 4851.

Additional Information

Properties of Sphere

A sphere has only one surface, so its Curved Surface Area is equal to its Total Surface Area, given by 4πr².

Hemisphere Formulas

For a hemisphere: CSA = 2πr², TSA = 3πr², and Volume = (2/3)πr³.

Radius and Surface Area Relation

If the radius is doubled, the Surface Area increases by 4 times (2²) and the Volume increases by 8 times (2³).

Common Calculation

Total number of students [ Boys + Girls] in class together is 1200.

Difference between boys and girls in all class together is 200.

Total number of boys in all class together is [1200 + 200] /100 = 700

Number of students in class A is 1200 × 12 /100 = 144

Number of boys in class A is 700 × 18 / 100 = 126

So, Number of girls in class A is 144 – 126 = 18

Similarly, we can calculate others class values also.

Calculation

Group 1 = Classes A + B

Boys = 126 + 140 = 266

Girls = 18 + 76 = 94

Group 2 = Classes D + E

Boys = 175 + 105 = 280

Girls = 125 + 75 = 200

Required ratio = [ 266 + 94] : [ 280 +200] = 360:480 = 3:4