Quantitative Aptitude PYQs

Shortcut Trick
Identify Cost Price (CP) = 800 and Selling Price (SP) = 740.
Loss = CP − SP = 800 − 740 = 60.
Loss% = (Loss ÷ CP) × 100.
Loss% = (60 ÷ 800) × 100 = 60 ÷ 8 = 7.5%.
∴ The correct answer is 7.5%.

Alternate Method
Given: CP = 800, SP = 740.
Formulas: Loss = CP − SP | Loss% = (Loss ÷ CP) × 100

⇒ Loss = 800 − 740 = 60
⇒ Loss% = (60 ÷ 800) × 100
⇒ Loss% = 6000 ÷ 800 = 7.5%
∴ The correct answer is 7.5%.

Additional Information
Cost Price and Selling Price
Cost Price (CP) is the price at which an item is purchased. Selling Price (SP) is the price at which it is sold.
Loss Condition
A loss occurs when the Cost Price is greater than the Selling Price (CP > SP).

Shortcut Trick

The age difference between two people remains constant throughout their lives.

15 years ago, Ratio of ages (Girish : Rohan) = 2 : 1

Difference in ratio = 2 − 1 = 1 unit → Given as 10 years.

Girish's age 15 years ago = 2 units = 2 × 10 = 20 years.

Present age of Girish = 20 + 15 = 35 years.

∴ The correct answer is 35 years.

Alternate Method

Given: Age difference = 10 years; 15 years ago, Girish = 2 × Rohan.

Let Rohan's age 15 years ago be x.

⇒ Girish's age 15 years ago = 2x.

⇒ Present age of Rohan = x + 15.

⇒ Present age of Girish = 2x + 15.

⇒ (2x + 15) − (x + 15) = 10

⇒ x = 10.

⇒ Girish's present age = 2(10) + 15 = 20 + 15 = 35 years.

∴ The correct answer is 35 years.

Additional Information

Constant Difference Property

The difference between the ages of two individuals is a constant value and never changes with time.

Ratio Method for Ages

If the ratio of ages changes from a:b to c:d over 'n' years, the value of 1 unit can be found by equating the difference in parts if the difference is same.

Algebraic Representation

Always define variables for the past or present age consistently to avoid calculation errors in age-based problems.

Shortcut Trick

Selling Price (SP) = Marked Price (MP) × (100 − Discount%) ÷ 100

SP = 6,370 × (100 − 60) ÷ 100

SP = 6,370 × 40 ÷ 100 = 63.7 × 40

SP = 2,548

∴ The correct answer is ₹2,548.

Alternate Method

Given: Marked Price (MP) = ₹6,370, Discount Percentage = 60%.

Formula: Selling Price (SP) = MP − Discount | Discount = (Discount% ÷ 100) × MP

⇒ Discount Amount = (60 ÷ 100) × 6,370

⇒ Discount Amount = 0.6 × 6,370 = 3,822

⇒ Selling Price = 6,370 − 3,822

⇒ Selling Price = 2,548

∴ The correct answer is ₹2,548.

Additional Information

Marked Price & Discount

The Marked Price is the price printed on the label. Discount is always calculated on the Marked Price unless stated otherwise.

Selling Price Calculation

If the discount is d%, the buyer pays (100 − d)% of the MP. SP = MP × Multiplier.

Shortcut Trick

Identify the smallest difference between the given numbers.

Differences: 110 − 44 = 66, 198 − 110 = 88.

Smallest difference is 66. The HCF must be a factor of 66 (1, 2, 3, 6, 11, 22, 33, 66).

Check the largest factor of 66 that divides 44, 110, and 198: 22 × 2 = 44, 22 × 5 = 110, 22 × 9 = 198.

∴ The correct answer is 22.

Alternate Method

Given: Numbers are 44, 110, and 198.

Method Used: Prime Factorization Method

Perform prime factorization for each number:

⇒ 44 = 2 × 2 × 11 = 2² × 11¹

⇒ 110 = 2 × 5 × 11 = 2¹ × 5¹ × 11¹

⇒ 198 = 2 × 3 × 3 × 11 = 2¹ × 3² × 11¹

The Highest Common Factor (HCF) is the product of the lowest powers of common prime factors.

⇒ Common prime factors are 2 and 11.

⇒ HCF = 2¹ × 11¹

⇒ HCF = 22

∴ The correct answer is 22.

Additional Information

Highest Common Factor (HCF)

HCF (or GCD) is the largest positive integer that divides each of the integers without leaving a remainder.

HCF and LCM Relation

For two numbers a and b, the product of the numbers is equal to the product of their HCF and LCM: a × b = HCF(a, b) × LCM(a, b).

HCF of Fractions

HCF of fractions = HCF of Numerators ÷ LCM of Denominators.

Co-prime Numbers

Two numbers are said to be co-prime if their HCF is exactly 1.

Shortcut Trick

Directly apply the formula for Total Surface Area of a Cylinder: 2πr(r + h).

Here, Radius (r) = 24 ÷ 2 = 12 ft and Height (h) = 9 ft.

TSA = 2 × (22/7) × 12 × (12 + 9) = 2 × (22/7) × 12 × 21

TSA = 44 × 12 × 3 = 1584 ft².

∴ The correct answer is 1584 ft².

Alternate Method

Given:

Base Diameter of the Cylinder = 24 ft

Height (h) of the Cylinder = 9 ft

Formula Used:

Total Surface Area (TSA) = 2 × π × r × (r + h)

Where r is the Radius and h is the Height.

Calculations:

⇒ Radius (r) = Diameter ÷ 2

⇒ r = 24 ÷ 2 = 12 ft

⇒ TSA = 2 × (22/7) × 12 × (12 + 9)

⇒ TSA = 2 × (22/7) × 12 × 21

⇒ TSA = 2 × 22 × 12 × 3

⇒ TSA = 44 × 36

⇒ TSA = 1584 ft²

∴ The correct answer is 1584 ft².

Additional Information

Curved Surface Area (CSA)

The area of only the vertical surface of the Cylinder, calculated as CSA = 2 × π × r × h.

Volume of Cylinder

The total space occupied inside the Cylinder, given by the formula V = π × r² × h.

Total Surface Area of Open Cylinder

If the Cylinder is open at one end, the area is 2πrh + πr². If open at both ends, it is just 2πrh.

Shortcut Trick

Successive discount multiplier method: Net multiplier = (1 − d₁/100) × (1 − d₂/100) × (1 − d₃/100).

Net Multiplier = 0.89 × 0.88 × 0.98 = 0.7832 × 0.98 = 0.767536.

Single discount% = (1 − Net Multiplier) × 100.

Discount% = (1 − 0.767536) × 100 = 0.232464 × 100 = 23.2464%.

∴ The correct answer is 23.25%.

Alternate Method

Given → Three consecutive discounts of 11%, 12% and 2%.
Formulas: Successive discount = a + b − (a × b) ÷ 100.

⇒ Step 1: Calculate single discount for 11% and 12%.
⇒ Discount₁₂ = 11 + 12 − (11 × 12) ÷ 100 = 23 − 1.32 = 21.68%.
⇒ Step 2: Combine result with the 3^rd discount of 2%.
⇒ Total Discount = 21.68 + 2 − (21.68 × 2) ÷ 100.
⇒ Total Discount = 23.68 − (43.36 ÷ 100) = 23.68 − 0.4336 = 23.2464%.
⇒ Approximately 23.25%.

∴ The correct answer is 23.25%.

Additional Information

Single Equivalent Discount

It is the single discount rate that provides the same Selling Price as multiple consecutive discounts on the Marked Price.

Order of Discounts

The final Equivalent Discount remains the same regardless of the order in which the successive discounts are applied (Commutative Property).

Shortcut Trick

For any quadrilateral, the angle formed by the angle bisectors of two adjacent angles is equal to half the sum of the other two interior angles.

∠TES = (∠Q + ∠R) ÷ 2

∠TES = (96° + 43°) ÷ 2

∠TES = 139° ÷ 2 = 69.5°

∴ The correct answer is 69.5°.

Alternate Method

Given:

In quadrilateral QRST:

∠Q = 96°

∠R = 43°

SE and TE are the angle bisectors of ∠S and ∠T respectively, meeting at E.

Formula Used:

Sum of interior angles of a quadrilateral = 360°

Sum of angles of a triangle = 180°

Calculations:

In quadrilateral QRST:

⇒ ∠Q + ∠R + ∠S + ∠T = 360°

⇒ 96° + 43° + ∠S + ∠T = 360°

⇒ 139° + ∠S + ∠T = 360°

⇒ ∠S + ∠T = 360° − 139°

⇒ ∠S + ∠T = 221°

Since SE and TE are the angle bisectors of ∠S and ∠T respectively:

⇒ ∠EST = ∠S ÷ 2

⇒ ∠ETS = ∠T ÷ 2

In triangle ΔTES:

⇒ ∠TES + ∠EST + ∠ETS = 180°

⇒ ∠TES + (∠S ÷ 2) + (∠T ÷ 2) = 180°

⇒ ∠TES + (∠S + ∠T) ÷ 2 = 180°

⇒ ∠TES + 221° ÷ 2 = 180°

⇒ ∠TES + 110.5° = 180°

⇒ ∠TES = 180° − 110.5°

⇒ ∠TES = 69.5°

Why other options are incorrect:

⇒ Option (1) is 56°: Incorrect because it does not satisfy the angle sum property of the triangle formed by the bisectors.

⇒ Option (2) is 70.5°: Incorrect because it would require the sum of ∠Q and ∠R to be 141°, whereas the actual sum is 139°.

⇒ Option (4) is 78°: Incorrect because it would require the sum of ∠Q and ∠R to be 156°, which contradicts the given values.

∴ The correct answer is 69.5°.

Additional Information

Angle Bisector Theorem in Quadrilaterals

The angle formed by the bisectors of two adjacent angles of a quadrilateral is always equal to half the sum of the remaining two angles. For example, in quadrilateral ABCD, the angle between bisectors of ∠A and ∠B is equal to (∠C + ∠D) ÷ 2.

Sum of Interior Angles of a Polygon

The sum of the interior angles of any n-sided polygon is given by the formula (n − 2) × 180°. For a quadrilateral (n = 4), the sum is (4 − 2) × 180° = 360°.

Angle Bisectors of Opposite Angles

If the bisectors of two opposite angles of a quadrilateral are drawn, the angle between them is equal to half the difference of the other two opposite angles.

Shortcut Trick

Total Food Supply is constant: M₁ × D₁ = M₂ × D₂

Initially, 102 men have food for 30 weeks.

New number of men = 102 + 78 = 180 men.

102 × 30 = 180 × Weeks ⇒ Weeks = (102 × 30) ÷ 180 = 102 ÷ 6 = 17 weeks.

∴ The correct answer is 17 weeks.

Alternate Method

Given: Initial men (M₁) = 102, Initial time (T₁) = 30 weeks, Men joining = 78.

Formula: M₁T₁ = M₂T₂ (since the total quantity of food remains the same).

⇒ Total Food Units = 102 × 30 = 3060 man-weeks.

⇒ New total number of men (M₂) = 102 + 78 = 180 men.

⇒ New time (T₂) = Total Food Units ÷ M₂

⇒ T₂ = 3060 ÷ 180

⇒ T₂ = 306 ÷ 18

⇒ T₂ = 17 weeks.

∴ The correct answer is 17 weeks.

Additional Information

Chain Rule Concept

For work or food consumption, the relationship is M₁D₁H₁/W₁ = M₂D₂H₂/W₂, where M is men, D is days/weeks, H is hours, and W is work or total food.

Inverse Proportion

When the quantity of food is constant, the number of men is inversely proportional to the number of days (M ∝ 1/D). If men increase, days decrease.

Shortcut Trick

LCM of 6, 8, 10 = 120 units (Total Capacity).

Efficiencies: A = 20, B = 15, C = −12.

Part filled by (A+B) in 3 hours = (20 + 15) × 3 = 105 units.

Remaining part = 120 − 105 = 15 units.

Time for (B+C) = 15 ÷ (15 − 12) = 15 ÷ 3 = 5 hours.

Percentage increase = [(5 − 3) ÷ 3] × 100 = 200/3 %.

∴ The correct answer is 200/3%.

Alternate Method

Given: Pipe A fills in 6h, Pipe B fills in 8h, Pipe C empties in 10h.

⇒ Efficiency of (A + B) = 20 + 15 = 35 units/h

⇒ Work done in 3 hours = 35 × 3 = 105 units

⇒ Remaining Work = 120 − 105 = 15 units

⇒ Efficiency of (B + C) = 15 − 12 = 3 units/h

⇒ Time taken by (B + C) = 15 ÷ 3 = 5 hours

⇒ Difference in time = 5 − 3 = 2 hours

⇒ Percentage increase = (2 ÷ 3) × 100% = 200/3 %

∴ The correct answer is 200/3%.

Additional Information

Pipes and Cisterns Efficiency

Inlet pipes have positive efficiency, while outlet (leakage) pipes have negative efficiency. Total efficiency is the algebraic sum of individual efficiencies.

Percentage Increase Formula

Percentage change = [(Final Value − Initial Value) ÷ Initial Value] × 100.

Work and Time Relationship

Total Work = Efficiency × Time. If work is constant, time is inversely proportional to efficiency.

Shortcut Trick

Calculate Total Days: Feb(17) + Mar(31) + Apr(25) = 73 days.

Time (T) in years = 73 ÷ 365 = 1/5 years.

Interest for 1 year = 7% of ₹2000 = ₹140.

Interest for 1/5 year = 140 ÷ 5 = ₹28.

∴ The correct answer is ₹28.

Alternate Method

Given:

Principal (P) = ₹2,000

Rate (R) = 7% per annum

Time Period: 11 February 2023 to 25 April 2023

Formula Used:

Simple Interest (SI) = (P × R × T) ÷ 100

Calculations:

⇒ Total days in February (non-leap year) from 11^th = 28 − 11 = 17 days

⇒ Days in March = 31 days

⇒ Days in April = 25 days

⇒ Total Days (T) = 17 + 31 + 25 = 73 days

⇒ T (in years) = 73 ÷ 365 = 1/5 years

⇒ SI = (2000 × 7 × (1/5)) ÷ 100

⇒ SI = (2000 × 7) ÷ 500

⇒ SI = 4 × 7 = 28

∴ The correct answer is ₹28.

Additional Information

Time Calculation in Simple Interest

While calculating days, the day of deposit is generally excluded and the day of withdrawal is included. For a normal year, use 365 days, and for a leap year, use 366 days if February 29 is included.

Standard Days Shortcut

73 days is exactly 1/5 (20%) of a standard year. Similarly, 146 days = 2/5 and 219 days = 3/5 of a year.

Compound Interest Formula

Amount (A) = P(1 + R/100)ⁿ, where n is the number of conversion periods. Compound Interest (CI) = A − P.

Shortcut Trick

When the distance traveled is the same for both speeds, use the Harmonic Mean formula.

Average Speed = (2 × x × y) ÷ (x + y)

Average Speed = (2 × 20 × 12) ÷ (20 + 12) = 480 ÷ 32 = 15 km/h.

∴ The correct answer is 15 km/h.

Alternate Method

Given: Speed to school (x) = 20 km/h, Speed coming back (y) = 12 km/h.

Formula Used: Average Speed = Total Distance ÷ Total Time

⇒ Let the distance between home and school be D.

⇒ Total distance = D + D = 2D

⇒ Time taken to go (t₁) = D ÷ 20

⇒ Time taken to return (t₂) = D ÷ 12

⇒ Total Time = (D/20) + (D/12) = (3D + 5D) ÷ 60 = 8D/60

⇒ Average Speed = 2D ÷ (8D/60) = (2D × 60) ÷ 8D = 120 ÷ 8

⇒ Average Speed = 15 km/h

∴ The correct answer is 15 km/h.

Additional Information

Average Speed for Different Distances

If the distances are different, use the basic formula: (Total Distance) ÷ (Total Time taken for all segments).

Three Equal Distances

If a journey is covered in three equal parts at speeds u, v, and w, Average Speed = (3uvw) ÷ (uv + vw + wu).

Relative Speed

If two bodies move in the same direction, relative speed = S₁ − S₂. If in opposite directions, relative speed = S₁ + S₂.

Shortcut Trick

Fourth Proportion Rule: Product of Means = Product of Extremes

Given values: 17, X, 7, 91 are in proportion.

⇒ 17 × 91 = X × 7

⇒ X = (17 × 91) ÷ 7 = 17 × 13 = 221

∴ The correct answer is 221.

Alternate Method

Given: First term (a) = 17, Second term (b) = X, Third term (c) = 7, Fourth term (d) = 91.

Formula Used: a ÷ b = c ÷ d

⇒ 17 ÷ X = 7 ÷ 91

⇒ 17 ÷ X = 1 ÷ 13

⇒ X = 17 × 13

⇒ X = 221

∴ The correct answer is 221.

Additional Information

Fourth Proportion

If four quantities a, b, c, d are such that a : b = c : d, then d is called the fourth proportion to a, b, and c.

Third Proportion

If a, b, c are in continued proportion such that a : b = b : c, then c is the third proportion, given by c = b² ÷ a.

Mean Proportion

The mean proportion between two numbers a and b is √(a × b).

Shortcut Trick

Ratio of Cost Price (CP) to Selling Price (SP) = 40 : 50 = 4 : 5.

Profit units = 5 − 4 = 1 unit.

Profit % = (Profit units ÷ CP units) × 100 = (1 ÷ 4) × 100 = 25%.

∴ The correct answer is 25%.

Alternate Method

Given: Cost Price (CP) = ₹40, Selling Price (SP) = ₹50.

Formulas: Profit = SP − CP | Profit % = (Profit ÷ CP) × 100

⇒ Profit = 50 − 40 = ₹10.

⇒ Profit % = (10 ÷ 40) × 100.

⇒ Profit % = (1 ÷ 4) × 100 = 25%.

∴ The correct answer is 25%.

Additional Information

Profit and Loss Basics

Profit occurs when Selling Price (SP) is greater than Cost Price (CP). Formula: Profit = SP − CP.

Loss Calculation

Loss occurs when Cost Price (CP) is greater than Selling Price (SP). Formula: Loss % = (Loss ÷ CP) × 100.

Shortcut Trick

Assume all 150 questions were correct: Total marks = 150 × 4 = 600 marks.

Actual marks scored = 250 marks. Difference = 600 − 250 = 350 marks.

Penalty per wrong answer = 4 (loss of mark) + 1 (negative mark) = 5 marks.

Number of wrong questions = 350 ÷ 5 = 70.

Number of correct questions = 150 − 70 = 80.

∴ The correct answer is 80.

Alternate Method

Given: Total questions = 150, Correct marks = 4, Wrong marks = −1, Total score = 250.

Formula Used: Total Marks = (Correct × 4) − (Wrong × 1)

⇒ Let Correct questions be x and Wrong questions be (150 − x).

⇒ 4x − 1(150 − x) = 250

⇒ 4x − 150 + x = 250

⇒ 5x = 400

⇒ x = 80

∴ The correct answer is 80.

Additional Information

Alligation Method

Average marks per question = 250 ÷ 150 = 5/3. Using alligation between +4 and −1 gives the ratio of questions.

Negative Marking Impact

Every wrong answer reduces the potential score by (Positive Marks + Absolute Negative Marks).

Score Calculation

Total Score = N_c × M_c − N_w × M_w where N is count and M is marks per question.

Shortcut Trick

Let Total Work (LCM of 6, 9, 12) = 36 units.

Efficiencies: A = 6, B = 4, C = −3. Combined (A+B+C) = 7 units/hr.

Work done in 1.5 hrs: Total = 7 × 1.5 = 10.5 units; Pipe C = |−3 × 1.5| = 4.5 units.

Ratio of C's work to total work: (4.5 ÷ 10.5) × 100 = (3/7) × 100 = 300/7%.

Comparing with x/7%, we get x = 300. Factors of 300 = 2² × 3¹ × 5² → (2+1)(1+1)(2+1) = 18.

∴ The correct answer is 18.

Alternate Method

Calculations:

⇒ Net efficiency of (A + B + C) = 6 + 4 − 3 = 7 units/hr

⇒ Work done by all three in 1.5 hours = 7 × 1.5 = 10.5 units

⇒ Work done (magnitude) by Pipe C in 1.5 hours = 3 × 1.5 = 4.5 units

⇒ Percentage of C's work relative to total = (4.5 ÷ 10.5) × 100

⇒ Percentage = (45 ÷ 105) × 100 = (3 ÷ 7) × 100 = 300/7 %

⇒ Given percentage = x/7 %, therefore x = 300

⇒ Prime factorization of 300 = 2² × 3¹ × 5²

⇒ Number of factors = (2 + 1) × (1 + 1) × (2 + 1) = 3 × 2 × 3 = 18

∴ The correct answer is 18.

Additional Information

Total Number of Factors

If a number N = p^a × q^b × r^c where p, q, r are prime factors, then the total number of factors is (a+1)(b+1)(c+1).

Efficiency Concept in Pipes

Inlet pipe efficiency is positive (+), and outlet pipe (leak) efficiency is negative (−). Net work is the algebraic sum of efficiencies.

LCM Method

Total capacity of the tank is assumed as the Least Common Multiple (LCM) of individual times to simplify unit calculations.

Shortcut Trick

To gain the maximum advantage, Dolly should choose the option with the highest equivalent discount percentage.

Successive discounts formula: a + b − (a × b ÷ 100).

Option 1: 7 + 7 − 0.49 = 13.51% | Option 3: 21 + 21 − 4.41 = 37.59%.

Option 4: Single discount = 47%.

Option 2: 43 + 38 − (43 × 38 ÷ 100) = 81 − 16.34 = 64.66%.

Comparing 13.51%, 64.66%, 37.59%, and 47%, the value 64.66% is the highest.

∴ The correct answer is Two successive discounts of 43% and 38%.

Alternate Method

Given → Marked Price (MP) = ₹5,300.
Formulas: Successive Discount = a + b − (ab ÷ 100) | Net Discount % = (Total Discount ÷ MP) × 100.

⇒ Calculate effective discount for Option 2: 43 + 38 − (43 × 38)/100.
⇒ 81 − (1634/100) = 81 − 16.34 = 64.66%.
⇒ Calculate effective discount for Option 3: 21 + 21 − (21 × 21)/100 = 42 − 4.41 = 37.59%.
⇒ Comparing all options, Option 2 offers the maximum discount.

∴ The correct answer is Two successive discounts of 43% and 38%.

Additional Information

Successive Discounts

When a series of discounts are applied one after another, the net discount is always less than the sum of individual percentages due to the reducing base.

Maximum Advantage

For a buyer, the best option is the one that results in the lowest Selling Price (SP), which occurs at the highest Effective Discount Rate.

Shortcut Trick

Relative Speed (Opposite Direction) = S₁ + S₂ = 90 + 60 = 150 km/h.

Convert speed to m/s: 150 × (5/18) = 125/3 m/s.

Total Distance = Length_A + Length_B = 250 + 150 = 400 m.

Time = 400 ÷ (125/3) = (400 × 3) ÷ 125 = 1200 ÷ 125 = 9.6 seconds.

∴ The correct answer is 10 sec.

Alternate Method

Given: Length of Train A = 250 m, Length of Train B = 150 m, Speed of A = 90 km/h, Speed of B = 60 km/h.

Formula Used: Time = (Total Distance) ÷ (Relative Speed)

⇒ Total Distance = 250 + 150 = 400 m

⇒ Relative Speed = 90 + 60 = 150 km/h

⇒ Relative Speed in m/s = 150 × (5/18) = 41.67 m/s

⇒ Time taken = 400 ÷ 41.67 ≈ 9.6 seconds

⇒ Rounding off to nearest seconds ≈ 10 seconds

∴ The correct answer is 10 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).

Unit Conversion

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

Crossing Objects

When a train crosses another train or a bridge, total distance is the sum of lengths of both objects.

Shortcut Trick

Income ratio = 5 : 9, Expenditure ratio = 1 : 3.

Difference in income parts = 9 − 5 = 4; Difference in expenditure parts = 3 − 1 = 2.

To equalize savings, multiply expenditure ratio by 2 → 2 : 6.

Savings (parts) = 5 − 2 = 3 parts. Given 3 parts = ₹48,000 → 1 part = ₹16,000.

Manjit income = 5 parts = 5 × ₹16,000 = ₹80,000.

∴ The correct answer is ₹80,000.

Alternate Method

Given:

Income of Manjit = 5x, Income of Keshav = 9x.

Monthly Savings of each = ₹48,000.

Formula Used:

Income − Savings = Expenditure

Calculations:

⇒ (5x − 48,000) / (9x − 48,000) = 1 / 3

⇒ 3 × (5x − 48,000) = 1 × (9x − 48,000)

⇒ 15x − 1,44,000 = 9x − 48,000

⇒ 15x − 9x = 1,44,000 − 48,000

⇒ 6x = 96,000

⇒ x = 16,000

⇒ Monthly income of Manjit = 5x = 5 × 16,000 = ₹80,000.

∴ The correct answer is ₹80,000.

Additional Information

Income, Expenditure and Savings

Basic relation: Income = Expenditure + Savings. If savings are equal for both, the difference between income units and expenditure units must be made equal.

Ratio Balancing Method

To balance ratios (a : b) and (c : d), multiply the terms by the difference of the other ratio: (a−b) and (c−d) to make the step-differences equal.

Cross Multiplication Method

For equations like (ax−k)/(bx−k) = m/n, the direct shortcut is x = k(n−m) / (an−bm).

Shortcut Trick

Use the standard algebraic identity for the difference of squares: a² − b² = (a − b)(a + b).

Given: a² − b² = 21 and a − b = 3.

Directly divide the product by the known factor: (a + b) = (a² − b²) ÷ (a − b).

Calculation: 21 ÷ 3 = 7.

∴ The correct answer is 7.

Alternate Method

Given:

a² − b² = 21

a − b = 3

Formula Used:

The identity for a² − b² is expressed as the product of the sum and difference of the terms: a² − b² = (a − b)(a + b).

Calculations:

⇒ a² − b² = (a − b)(a + b)

⇒ 21 = 3 × (a + b)

⇒ a + b = 21 ÷ 3

⇒ a + b = 7

∴ The correct answer is 7.

Additional Information

Difference of Squares

This identity states that the difference of two squares is equal to the product of the sum and the difference of the two terms: a² − b² = (a + b)(a − b).

Square of a Binomial (Sum)

The expansion of the square of a sum is: (a + b)² = a² + 2ab + b².

Square of a Binomial (Difference)

The expansion of the square of a difference is: (a − b)² = a² − 2ab + b².

Relationship between Identities

The identities can be linked: (a + b)² − (a − b)² = 4ab.

Shortcut Trick

Calculate semi-perimeter (s) = (90 + 11 + 97) ÷ 2 = 99 cm.

Area of triangle = √[99 × (99−90) × (99−11) × (99−97)] = √[99 × 9 × 88 × 2].

Area = √[9 × 11 × 9 × 8 × 11 × 2] = √[9² × 11² × 16] = 9 × 11 × 4 = 396 cm².

Altitude (h) on side 11 cm = (2 × Area) ÷ base = (2 × 396) ÷ 11 = 72 cm.

∴ The correct answer is 72 cm.

Alternate Method

Given:

Sides of the triangle are a = 90 cm, b = 11 cm, and c = 97 cm.

Formula Used:

Semi-perimeter (s) = (a + b + c) ÷ 2

Area (A) = √[s(s − a)(s − b)(s − c)]

Area (A) = (1 ÷ 2) × base × altitude (h)

Calculations:

⇒ s = (90 + 11 + 97) ÷ 2 = 198 ÷ 2 = 99 cm

⇒ Area = √[99 × (99 − 90) × (99 − 11) × (99 − 97)]

⇒ Area = √[99 × 9 × 88 × 2]

⇒ Area = √[11 × 9 × 9 × 11 × 8 × 2]

⇒ Area = √[11² × 9² × 16] = 11 × 9 × 4 = 396 cm²

⇒ Altitude (h) corresponds to base 11 cm:

⇒ 396 = (1 ÷ 2) × 11 × h

⇒ h = (396 × 2) ÷ 11

⇒ h = 36 × 2 = 72 cm

∴ The correct answer is 72 cm.

Additional Information

Heron's Formula

It is used to find the area of a triangle when all three sides are known: Area = √[s(s−a)(s−b)(s−c)], where s is the semi-perimeter.

Altitude of a Triangle

The altitude is the perpendicular distance from a vertex to the opposite side (base). Height = (2 × Area) ÷ Base.

Semi-Perimeter

The semi-perimeter of a triangle is half the sum of the lengths of its sides: s = (a + b + c) ÷ 2.