SSC CGL Tier 2 Quant Previous Year Paper (Held On: 3 Feb 2022)

Shortcut Trick

Savings = 33¹/₃% = 1/3 of Income → Ratio of Saving : Expenditure = 1 : 2.

Percentage increase in Income = [ (Saving Part × % Increase) + (Expenditure Part × % Increase) ] ÷ Total Parts.

% Increase in Income = [ (1 × 22%) + (2 × 10%) ] ÷ (1 + 2)

% Increase in Income = (22 + 20) ÷ 3 = 42 ÷ 3 = 14%.

∴ The correct answer is 14%.

Alternate Method

Given: Saving = 33¹/₃% of Income, Saving increase = 22%, Expenditure increase = 10%.

Formula: Income = Savings + Expenditure

⇒ Let total Income = ₹300

⇒ Savings = 33¹/₃% of 300 = 1/3 × 300 = ₹100

⇒ Expenditure = Income − Savings = 300 − 100 = ₹200

⇒ New Savings = 100 + (22% of 100) = 100 + 22 = ₹122

⇒ New Expenditure = 200 + (10% of 200) = 200 + 20 = ₹220

⇒ New Income = New Savings + New Expenditure = 122 + 220 = ₹342

⇒ Increase in Income = 342 − 300 = ₹42

⇒ Percentage Increase in Income = (42 ÷ 300) × 100 = 14%

∴ The correct answer is 14%.

Additional Information

Income Balance Equation

Total Income = Expenditure + Savings. Any change in the parts must balance the total change.

Weighted Average Formula

Overall % Change = (Value₁ × Change₁ + Value₂ × Change₂) ÷ (Value₁ + Value₂).

Fractional Conversion

33¹/₃% = 1/3, 66²/₃% = 2/3, 16²/₃% = 1/6, 12.5% = 1/8.

Shortcut Trick

Remaining work for P = 18 − 5 = 13 days of P's work.

(P + Q) completes this in 13/5 days ⇒ (P + Q) × 13/5 = P × 13.

Efficiency Ratio (P + Q) : P = 5 : 1 ⇒ Efficiency of Q = 4 (if P = 1).

Total work = 18 × 1 = 18 units; 66 2/3% of work = 2/3 × 18 = 12 units.

Time taken by Q alone = 12 ÷ 4 = 3 days.

∴ The correct answer is 3 days.

Alternate Method

Given: P can finish work in 18 days. Worked for 5 days. Rem. work done with Q in 13/5 days.

Formula: Efficiency = Total Work ÷ Time | Time = Work ÷ Efficiency

⇒ Work completed by P in 5 days = 1 × 5 = 5 units.

⇒ Remaining work = 18 − 5 = 13 units.

⇒ Efficiency of (P + Q) = Remaining Work ÷ Time = 13 ÷ (13/5) = 5 units/day.

⇒ Efficiency of Q = (Efficiency of P + Q) − (Efficiency of P) = 5 − 1 = 4 units/day.

⇒ 66 2/3% of Total Work = (200/3)% × 18 = 2/3 × 18 = 12 units.

⇒ Time taken by Q = 12 ÷ 4 = 3 days.

∴ The correct answer is 3 days.

Additional Information

Work and Efficiency Relation

Efficiency is inversely proportional to time taken. If P is thrice as efficient as Q, time taken ratio P:Q = 1:3.

Work and Wages

Wages are distributed in the ratio of work done or efficiency (if time is same). Wage = Efficiency × Time.

Fraction to Percentage Conversion

Common values: 33 1/3% = 1/3, 66 2/3% = 2/3, 16 2/3% = 1/6, and 12 1/2% = 1/8.

Combined Work Formula

If A takes 'a' days and B takes 'b' days, together they take (a × b) ÷ (a + b) days.

Shortcut Trick

Ratio of corresponding sides k = AB/CA = 20/15 = 4/3.

In similar triangles ΔDAB ∼ ΔDCA, the ratio of segments on the produced side is DB/DC = k².

DB/DC = (4/3)² = 16/9.

The difference in ratio parts is 16 − 9 = 7 parts, which corresponds to BC = 7 cm.

Since 7 units = 7 cm, then 1 unit = 1 cm.

Length of DC = 9 units = 9 cm.

∴ The correct answer is 9 cm.

Alternate Method

Given: AB = 20 cm, BC = 7 cm, CA = 15 cm. Side BC is produced to D such that ΔDAB ∼ ΔDCA.

Formula: For similar triangles ΔDAB ∼ ΔDCA, the ratio of corresponding sides is equal.

⇒ DA/DC = DB/DA = AB/CA

⇒ AB/CA = 20/15 = 4/3

⇒ DA/DC = 4/3 ⇒ DA = (4/3)DC

⇒ DB/DA = 4/3 ⇒ DB = (4/3)DA

⇒ DB = (4/3) × (4/3)DC = (16/9)DC

⇒ Since D is on BC produced, DB = BC + DC

⇒ (16/9)DC = 7 + DC

⇒ (16/9)DC − DC = 7

⇒ (7/9)DC = 7

⇒ DC = 9 cm

∴ The correct answer is 9 cm.

Additional Information

AA Similarity Criterion

Two triangles are similar if two corresponding angles are equal; in this case, ∠D is common and ∠DAB = ∠DCA.

Side Ratios in Similarity

When triangles are similar, the ratio of any two corresponding sides is constant (e.g., AB/AC = BD/AD = AD/CD).

Geometric Mean Property

In this configuration, the side AD is the geometric mean of the segments on the produced line: AD² = DB × DC.

Shortcut Trick

Volume of cuboid = √(216 × 96 × 144) = 1728 cm³

Volume of 1 small cube = 1728 ÷ 8 = 216 cm³

Side of small cube (a) = (216)^1/3 = 6 cm

Lateral Surface Area (LSA) of 3 cubes = 3 × 4a² = 12 × 36 = 432 cm²

∴ The correct answer is 432 cm².

Alternate Method

Given:

Area of face 1 (lb) = 216 cm²

Area of face 2 (bh) = 96 cm²

Area of face 3 (lh) = 144 cm²

Formula Used:

Volume of cuboid (V) = √(Area₁ × Area₂ × Area₃)

Lateral Surface Area of cube = 4 × side²

Calculations:

⇒ Volume of cuboid = √(216 × 96 × 144)

⇒ Volume = √(2,985,984) = 1728 cm³

⇒ Volume of one small cube = 1728 ÷ 8 = 216 cm³

⇒ Side³ = 216

⇒ Side (a) = 6 cm

⇒ Lateral Surface Area of 1 cube = 4 × 6² = 4 × 36 = 144 cm²

⇒ Lateral Surface Area of 3 cubes = 3 × 144 = 432 cm²

∴ The correct answer is 432 cm².

Additional Information

Volume and Face Areas of a Cuboid

If A₁, A₂, and A₃ are the areas of three adjacent faces of a cuboid, then its Volume V = √(A₁ × A₂ × A₃).

Surface Areas of a Cube

For a cube with side 'a', the Lateral Surface Area (area of 4 vertical walls) is 4a² and the Total Surface Area is 6a².

Volume Conservation

When a solid object is melted and recast into several smaller objects, the total volume of all smaller objects equals the volume of the original solid.

Shortcut Trick

Income A : B (2020) = 5 : 4

Change ratios: A (4 : 5) → A (2021) = 5 × (5 ÷ 4) = 6.25 units

Change ratios: B (2 : 3) → B (2021) = 4 × (3 ÷ 2) = 6 units

Income A : B (2021) = 6.25 : 6 = 25 : 24 (Total = 49 units)

B's 2021 income = (24 ÷ 49) × 7,05,600 = 24 × 14,400 = Rs. 3,45,600

∴ The correct answer is Rs. 3,45,600.

Alternate Method

Given:

Income ratio A : B (2020) = 5 : 4

Individual income ratio A (2020 : 2021) = 4 : 5

Individual income ratio B (2020 : 2021) = 2 : 3

Total income of A and B in 2021 = Rs. 7,05,600

Formula Used:

New Income = Old Income × (New Ratio Part ÷ Old Ratio Part)

Calculations:

⇒ To make the 2020 income of A (5) divisible by 4 and B (4) divisible by 2, multiply the A:B (2020) ratio by 4.

⇒ New A:B (2020) = (5 × 4) : (4 × 4) = 20 : 16

⇒ A's income in 2021 = (20 ÷ 4) × 5 = 25 units

⇒ B's income in 2021 = (16 ÷ 2) × 3 = 24 units

⇒ Total income in 2021 = 25 + 24 = 49 units

⇒ Given 49 units = 7,05,600

⇒ 1 unit = 7,05,600 ÷ 49 = 14,400

⇒ Income of B in 2021 = 24 × 14,400 = 3,45,600

∴ The correct answer is Rs. 3,45,600.

Additional Information

Ratio Normalization

When connecting ratios across different time frames, find a common multiple for the shared term (usually the base year value) to simplify calculations.

Compound Ratios

The final ratio of two variables after changes can be found by multiplying the initial ratio parts by their respective fractional change factors.

Division by Ratio

To find an individual part, use the formula: Share = (Individual Ratio ÷ Sum of Ratio Parts) × Total Value.

The correct answer is Gain 12.5%

Key Points

• Let the Cost Price of the medical item be represented as CP and the Selling Price be represented as SP.
• The problem provides a linear relationship: 17 × CP = 8 × (CP + SP).
• By distributing the 8 on the right side of the equation, we get: 17 CP = 8 CP + 8 SP.
• To isolate the terms, subtract 8 CP from both sides: 17 CP − 8 CP = 8 SP, which results in 9 CP = 8 SP.
• The ratio of the Selling Price to the Cost Price is SP / CP = 9 / 8.
• Since the Selling Price (9 units) is higher than the Cost Price (8 units), the transaction results in a Gain.
• The Gain amount is SP − CP = 9 − 8 = 1 unit.
• The Gain Percentage is calculated using the formula: (Gain / CP) × 100.
• Substituting the values: (1 / 8) × 100 = 12.5%.
• Thus, the medical transaction resulted in a Gain of 12.5%. Options 1, 2, and 4 are mathematically inconsistent with the given equation.

Additional Information

• Profit and Loss concepts are fundamental in Financial Arithmetic. Profit occurs when SP > CP, while Loss occurs when CP > SP.
• The formula for Loss Percentage is ((CP − SP) / CP) × 100.
• In many competitive exams, reciprocal values are useful for quick calculations: 1/8 = 12.5%, 1/7 ≈ 14.28%, 1/9 ≈ 11.11%, and 1/6 ≈ 16.67%.
• Always remember that Gain or Loss is calculated on the Cost Price unless the question specifically mentions it is calculated on the Selling Price.

Shortcut Trick

Let the marks of C = 100 units.

A = 10% less than C → A = 90 units.

D = 20% more than C → D = 120 units.

D is more than A by = 120 − 90 = 30 units.

Required percentage = (30 ÷ 90) × 100 = 1/3 × 100 = 33 1/3%.

∴ The correct answer is 33 1/3%.

Alternate Method

Given: B = A + 20% of A; A = C − 10% of C; D = C + 20% of C.

Let marks obtained by C = 100

⇒ Marks of A = 100 − (10% of 100) = 90

⇒ Marks of D = 100 + (20% of 100) = 120

⇒ Difference between D and A = 120 − 90 = 30

⇒ Percentage more = (Difference ÷ A) × 100

⇒ Percentage more = (30 ÷ 90) × 100 = 100/3%

⇒ Percentage more = 33 1/3%

∴ The correct answer is 33 1/3%.

Additional Information

Percentage Change Formula

Percentage Increase = [(New Value − Original Value) ÷ Original Value] × 100. Always identify the base value after the word 'than'.

Common Base Method

When multiple people are compared, identify the common person (in this case, C) and assume their value as 100 or the LCM of denominators for easier calculation.

Fractional Equivalents

1/3 = 33.33% or 33 1/3%. Recognizing these standard fractions helps in solving competitive exams faster without actual division.

Shortcut Trick

Identify algebraic identity: a³ + b³ = (a + b)(a² − ab + b²).

Let a = 2.53 and b = 2.47. Then a + b = 5.

Denominator = (10a)² − (10a × 10b) + (10b)² = 100(a² − ab + b²).

Expression = [(a + b)(a² − ab + b²)] ÷ [100(a² − ab + b²)] = (a + b) ÷ 100.

Expression = 5 ÷ 100 = 0.05 = 5 × 10⁻².

∴ The correct answer is −2.

Alternate Method

Given Expression: [(2.53)³ + (2.47)³] ÷ [25.3 × 25.3 − 624.91 + 24.7 × 24.7]

Let a = 2.53 and b = 2.47

⇒ Numerator = a³ + b³

⇒ Numerator = (a + b)(a² − ab + b²)

⇒ Numerator = (2.53 + 2.47)(2.53² − 2.53 × 2.47 + 2.47²) = 5 × (2.53² − 2.53 × 2.47 + 2.47²)

Now, analyze the Denominator:

⇒ 25.3 = 10 × 2.53 and 24.7 = 10 × 2.47

⇒ Denominator = (10 × 2.53)² − 624.91 + (10 × 2.47)²

⇒ Denominator = 100 × (2.53²) − 100 × (2.53 × 2.47) + 100 × (2.47²) [Since 100 × 2.53 × 2.47 = 100 × 6.2491 = 624.91]

⇒ Denominator = 100 × (2.53² − 2.53 × 2.47 + 2.47²)

Combining Numerator and Denominator:

⇒ Value = [5 × (2.53² − 2.53 × 2.47 + 2.47²)] ÷ [100 × (2.53² − 2.53 × 2.47 + 2.47²)]

⇒ Value = 5 ÷ 100 = 0.05

⇒ 5 × 10^k = 0.05

⇒ 5 × 10^k = 5 × 10⁻²

⇒ k = −2

∴ The correct answer is −2.

Additional Information

Algebraic Cube Identities

The sum of cubes is a³ + b³ = (a + b)(a² − ab + b²) and the difference of cubes is a³ − b³ = (a − b)(a² + ab + b²).

Decimal Scaling in Expressions

When terms are scaled by 10, the squared terms are scaled by 10² = 100 and cubed terms are scaled by 10³ = 1000.

Laws of Indices

For any non-zero number a, 1/aⁿ = a⁻ⁿ. In this case, 1/100 = 1/10² = 10⁻².

Shortcut Trick

Total relative distance covered by car = Initial distance ahead + Final distance behind = 80 + 40 = 120 m.

Relative speed in m/s = Distance ÷ Time = 120 ÷ 15 = 8 m/s.

Convert relative speed to km/h = 8 × (18/5) = 28.8 km/h.

Speed of the bus = Car speed − Relative speed = 75 − 28.8 = 46.2 km/h.

∴ The correct answer is 46.2 km/h.

Alternate Method

Given:

Speed of car (V_c) = 75 km/h

Initial distance of bus ahead = 80 m

Final distance of bus behind = 40 m

Time taken (t) = 15 seconds

Formula Used:

Relative Speed (Same direction) = V₁ − V₂

Distance = Speed × Time

Calculations:

⇒ Let the speed of the bus be V_b km/h.

⇒ Total relative distance to be covered by the car = 80 + 40 = 120 m.

⇒ Relative speed = (V_c − V_b) × (5/18) m/s.

⇒ Relative speed = Distance ÷ Time = 120 ÷ 15 = 8 m/s.

⇒ (75 − V_b) × (5/18) = 8

⇒ 75 − V_b = 8 × (18/5)

⇒ 75 − V_b = 144 ÷ 5

⇒ 75 − V_b = 28.8

⇒ V_b = 75 − 28.8

⇒ V_b = 46.2 km/h.

∴ The correct answer is 46.2 km/h.

Additional Information

Relative Speed Concepts

When two objects move in the same direction, relative speed = V₁ − V₂. When they move in opposite directions, relative speed = V₁ + V₂.

Unit Conversion

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

General Formula

Distance = Speed × Time. In overtaking problems, the distance is the total gap closed between the two objects.

Given:

$17\frac{1}{2}$% of $3\frac{1}{4}$% of $33\frac{1}{3}$% of 7200

Formula used:

x% of y = $\frac{x}{100} \times y$

Calculations:

$7200 \times \frac{1}{3} \times \frac{13}{400} \times \frac{35}{200}$ = ?

⇒ ? = $2400 \times \frac{13}{400} \times \frac{35}{200}$

⇒ ? = $6 \times 13 \times \frac{35}{200}$

⇒ ? = $78 \times \frac{35}{200}$

⇒ ? = $\frac{2730}{200}$

⇒ ? = 13.65

∴ The correct answer is option (3).

Shortcut Trick

For A₁: (n₁ − 2) × 180° = 1080° ⇒ n₁ − 2 = 6 ⇒ n₁ = 8.

For A₂: Interior − Exterior = 132° and Interior + Exterior = 180°.

2 × Exterior angle = 180° − 132° = 48° ⇒ Exterior angle = 24°.

n₂ = 360° ÷ 24° = 15.

Total sides = n₁ + n₂ = 8 + 15 = 23.

∴ The correct answer is 23.

Alternate Method

Given: Sum of interior angles of A₁ = 1080°. For polygon A₂, Interior angle − Exterior angle = 132°.

Formula Used: Sum of interior angles = (n − 2) × 180°; Interior + Exterior = 180°; Number of sides (n) = 360° ÷ Exterior angle.

Calculations for A₁:

⇒ (n₁ − 2) × 180° = 1080°

⇒ n₁ − 2 = 1080° ÷ 180°

⇒ n₁ − 2 = 6

⇒ n₁ = 8

Calculations for A₂:

⇒ Let Interior angle be I and Exterior angle be E.

⇒ I + E = 180° (Linear pair at vertex)

⇒ I − E = 132° (Given)

⇒ Subtracting the second from the first: (I + E) − (I − E) = 180° − 132°

⇒ 2E = 48°

⇒ E = 24°

⇒ n₂ = 360° ÷ E = 360° ÷ 24° = 15

Sum of the number of sides:

⇒ Total sides = n₁ + n₂

⇒ Total sides = 8 + 15 = 23

∴ The correct answer is 23.

Additional Information

Interior Angle Sum

The sum of all interior angles of any n-sided convex polygon is always given by (n − 2) × 180°.

Exterior Angle Sum

For any convex polygon, the sum of all exterior angles (one at each vertex) is always 360°.

Regular Polygon Properties

In a regular polygon, each interior angle is [(n − 2) × 180°] ÷ n and each exterior angle is 360° ÷ n.

Calculation:

The production of cement by company C in 2018 = 350

The production of cement by company C in 2019 = 400

The percentage increased of cement = [($\frac{400-350}{{350}} × 100)$]

⇒ ($\frac{50}{{350}} × 100)$

⇒ $\frac{100}{{7}}$%

Now,

The production of of cement by company C in 2020 = (400 + 400 × $\frac{1}{{7}}$)

⇒ (400 + $\frac{400}{{7}}$)

⇒ ($\frac{2800 + 400}{{7}}$)

⇒ $\frac{3200}{{7}}$

⇒ 457.1

∴ The production of of cement by company C in 2020 is 457.1

The correct answer is −2

Key Points

• To solve this expression, we utilize trigonometric identities and the properties of complementary angles where two angles sum to 90°.
• First, let us simplify the Numerator: 3(cosec² 26° − tan² 64°) + (cot² 42° − sec² 48°).
• Using the complementary angle rule tan(90° − A) = cot A: tan 64° = tan(90° − 26°) = cot 26°. Thus, (cosec² 26° − tan² 64°) becomes (cosec² 26° − cot² 26°). Since cosec² A − cot² A = 1, this term equals 1.
• Using the rule sec(90° − A) = cosec A: sec 48° = sec(90° − 42°) = cosec 42°. Thus, (cot² 42° − sec² 48°) becomes (cot² 42° − cosec² 42°). Since cot² A − cosec² A = −1, this term equals −1.
• The Numerator value is: 3(1) + (−1) = 3 − 1 = 2.
• Next, simplify the Denominator: cot(22° − θ) − cosec²(62° + θ) − tan(θ + 68°) + tan²(28° − θ).
• Observe that (22° − θ) and (θ + 68°) are complementary because (22 − θ) + (θ + 68) = 90°. Therefore, tan(θ + 68°) = cot(90° − (θ + 68°)) = cot(22° − θ). These two terms cancel each other out: cot(22° − θ) − cot(22° − θ) = 0.
• Observe that (62° + θ) and (28° − θ) are complementary because (62 + θ) + (28 − θ) = 90°. Therefore, tan(28° − θ) = cot(90° − (28° − θ)) = cot(62° + θ).
• The remaining terms in the Denominator are −cosec²(62° + θ) + cot²(62° + θ). Using the identity cot² A − cosec² A = −1, the denominator equals −1.
• Finally, calculate the value of the Expression: Numerator ÷ Denominator = 2 ÷ (−1) = −2.

Additional Information

• Fundamental Trigonometric Identities:
• sin² θ + cos² θ = 1
• 1 + tan² θ = sec² θ
• 1 + cot² θ = cosec² θ
• Complementary Angle Ratios:
• sin(90° − θ) = cos θ
• cos(90° − θ) = sin θ
• tan(90° − θ) = cot θ
• cot(90° − θ) = tan θ
• sec(90° − θ) = cosec θ
• cosec(90° − θ) = sec θ

Given:

$11.\overline{4} + 22.5\overline{67} - 33.5\overline{9}$

Formula used:

Recurring decimal to fraction conversion:

a.b$\overline{c}$ = $\frac{\text{abc - ab}}{90}$

Calculations:

$11.\overline{4} = 11 + \frac{4}{9} = 11 + \frac{440}{990}$

$22.5\overline{67} = 22 + \frac{567-5}{990} = 22 + \frac{562}{990}$

$33.5\overline{9} = 33 + \frac{59-5}{90} = 33 + \frac{54}{90} = 33 + \frac{594}{990}$

⇒ $(11 + 22 - 33) + (\frac{440 + 562 - 594}{990})$

⇒ $0 + \frac{408}{990}$

⇒ $\frac{412 - 4}{990}$

⇒ $0.4\overline{12}$

∴ The correct answer is option (3).

The correct answer is 40,000

Key Points

• To find the amount of salary, we must first calculate the Profit Sharing Ratio between P and Q based on their initial investments.
• The Capital invested by P is Rs. 1,50,000 and by Q is Rs. 4,50,000.
• The Investment Ratio (P : Q) = 1,50,000 : 4,50,000 = 15 : 45 = 1 : 3.
• Since both partners invested for the same duration (one year), the Total Profit of Rs. 2,00,000 is divided in the same ratio as their capital (1 : 3).
• The total number of parts in the ratio is 1 + 3 = 4 parts.
• The share of profit for P = (1 / 4) multiplied by 2,00,000 = Rs. 50,000.
• The problem states that P receives a total of Rs. 90,000, which consists of his share of the profit plus a Salary that is not included in the profit pool.
• Total Amount Received by P = Profit Share + Salary.
• 90,000 = 50,000 + Salary.
• Salary = 90,000 − 50,000 = Rs. 40,000.

Additional Information

• In a Partnership, if the time period of investment is the same for all partners, the profit is distributed directly in the ratio of their Capitals.
• If the time periods are different, the profit ratio is calculated as (Capital 1 multiplied by Time 1) : (Capital 2 multiplied by Time 2).
• A Working Partner is a partner who actively manages the business and may receive extra compensation, such as a Salary or a Commission, in addition to their share of the profit.
• A Sleeping Partner is one who only invests money but does not take part in the daily operations of the business.

Given:

91% of A = 39% of B

B = x% of A

Formula used:

y% of Z = $\frac{y}{100} \times Z$

Calculation:

91% × A = 39% × B

⇒ $\frac{91}{100} \times A = \frac{39}{100} \times B$

⇒ 91 × A = 39 × B

⇒ B = $\frac{91}{39} \times A$

⇒ B = $\frac{7}{3} \times A$

Given B = x% of A

⇒ B = $\frac{x}{100} \times A$

Comparing both equations for B:

⇒ $\frac{x}{100} = \frac{7}{3}$

⇒ x = $\frac{7 \times 100}{3}$

⇒ x = $\frac{700}{3}$

∴ The correct answer is option (2).

Shortcut Trick

Net deviation in average = (Percentage₁ × Change₁) + (Percentage₂ × Change₂)

Net deviation = (60% × +5) + (40% × −10)

Net deviation = 3 − 4 = −1

New Average = Old Average + Net deviation = 45 − 1 = 44

∴ The correct answer is 44.

Alternate Method

Given:

Average of n numbers = 45

Group 1 = 60% of n, increase = +5 per number

Group 2 = 40% of n, decrease = −10 per number

Calculations:

⇒ Let the total count of numbers (n) be 100.

⇒ Initial sum of n numbers = n × Average = 100 × 45 = 4500

⇒ Number of elements in 60% group = 0.60 × 100 = 60

⇒ Total increase in sum for this group = 60 × 5 = 300

⇒ Number of elements in remaining 40% group = 0.40 × 100 = 40

⇒ Total decrease in sum for this group = 40 × 10 = 400

⇒ New total sum = 4500 + 300 − 400 = 4400

⇒ New Average = New sum ÷ n = 4400 ÷ 100 = 44

∴ The correct answer is 44.

Additional Information

Weighted Average Formula

When a group is divided into parts with different averages or changes, the overall change is given by (n₁×c₁ + n₂×c₂) ÷ (n₁ + n₂).

Effect of Uniform Change

If every number in a set is increased or decreased by a constant value 'k', the average of the set also increases or decreases by exactly 'k'.

Deviation Method

The new average can always be calculated as Old Average + (Net Sum of Deviations ÷ Total Count).

Shortcut Trick

In a cyclic quadrilateral ABCD, where sides meet at E and F:

In ΔADE: ∠A = 180° − (∠ADC + ∠AED) = 180° − (76° + 55°) = 49°

In cyclic quad ABCD, ∠ABC = 180° − ∠ADC = 180° − 76° = 104°

In ΔFAB (formed by intersection F): ∠AFB = 180° − (∠DAB + ∠ABC) = 180° − (49° + 104°) = 27°

∴ The correct answer is 27°.

Alternate Method

Given: ∠ADC = 76° and ∠AED = 55° in cyclic quadrilateral ABCD.

⇒ In ΔEAD (where AB and DC produced meet at E):

⇒ ∠AED + ∠ADE + ∠DAE = 180°

⇒ 55° + 76° + ∠A = 180°

⇒ ∠A = 180° − 131° = 49°

⇒ In cyclic quadrilateral ABCD, opposite angles are supplementary:

⇒ ∠ABC = 180° − ∠ADC = 180° − 76° = 104°

⇒ Now, in ΔFAB (where AD and BC produced meet at F):

⇒ ∠AFB + ∠FAB + ∠ABF = 180°

⇒ Here, ∠FAB = ∠A = 49° and ∠ABF = ∠ABC = 104°

⇒ ∠AFB + 49° + 104° = 180°

⇒ ∠AFB = 180° − 153° = 27°

∴ The correct answer is 27°.

Additional Information

Cyclic Quadrilateral Properties

The sum of opposite internal angles of a cyclic quadrilateral is always 180°. Also, the exterior angle at any vertex is equal to the interior opposite angle.

Angle Sum Property

The sum of all interior angles in any triangle is exactly 180°. This property is used to find the third angle when two angles of a triangle are known.

Intersection of Produced Sides

When opposite sides of a cyclic quadrilateral are produced to meet at points E and F, they form two distinct triangles sharing one vertex of the quadrilateral.

Shortcut Trick

Identify triplet (5, 12, 13) since cos θ = 12/13 (Base/Hypotenuse), so Perpendicular = 5.

Simplify Expression: [sin θ (1 − tan θ)] / [tan θ (1 + cosec θ)] = [cos θ (1 − tan θ)] / (1 + cosec θ).

Substitute values: [(12/13) × (1 − 5/12)] / (1 + 13/5).

Calculation: [(12/13) × (7/12)] ÷ (18/5) = (7/13) × (5/18) = 35/234.

∴ The correct answer is 35/234.

Alternate Method

Given: cos θ = 12/13

Formula: cos θ = Base / Hypotenuse | sin θ = Perp / Hyp | tan θ = Perp / Base | cosec θ = Hyp / Perp

⇒ Base (B) = 12, Hypotenuse (H) = 13, Perpendicular (P) = 5

⇒ sin θ = 5/13

⇒ tan θ = 5/12

⇒ cosec θ = 13/5

⇒ Numerator = sin θ(1 − tan θ) = (5/13) × (1 − 5/12) = (5/13) × (7/12) = 35/156

⇒ Denominator = tan θ(1 + cosec θ) = (5/12) × (1 + 13/5) = (5/12) × (18/5) = 18/12 = 3/2

⇒ Result = (35/156) ÷ (3/2) = (35/156) × (2/3) = 35 / (78 × 3) = 35/234

∴ The correct answer is 35/234.

Additional Information

Trigonometric Ratios

Standard ratios: sin θ = P/H, cos θ = B/H, tan θ = P/B, cot θ = B/P, sec θ = H/B, cosec θ = H/P.

Reciprocal Identities

sin θ = 1/cosec θ, cos θ = 1/sec θ, tan θ = 1/cot θ = sin θ / cos θ.

Pythagorean Triplets

Common sets like (3, 4, 5), (5, 12, 13), (8, 15, 17), and (7, 24, 25) satisfy P² + B² = H².

Shortcut Trick

Given x² − 3x + 1 = 0, divide by x to get x + 1/x = 3.

Divide both numerator and denominator of the target expression by x.

Expression = (x³ + 1/x³) ÷ (x + 5 + 1/x).

Since x + 1/x = 3, then x³ + 1/x³ = 3³ − 3(3) = 18.

Expression = 18 ÷ (3 + 5) = 18/8 = 9/4.

∴ The correct answer is 9/4.

Alternate Method

Given:

x² − 3x + 1 = 0

Expression to find: (x⁴ + 1/x²) ÷ (x² + 5x + 1)

Method/Formula:

If x + 1/x = k, then x³ + 1/x³ = k³ − 3k

Calculations:

⇒ x² + 1 = 3x

⇒ (x² + 1) ÷ x = 3x ÷ x

⇒ x + 1/x = 3

⇒ Now, divide the numerator and denominator of the expression by x:

⇒ Numerator = (x⁴ + 1/x²) ÷ x = x³ + 1/x³

⇒ x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x)

⇒ x³ + 1/x³ = 3³ − 3 × 3 = 27 − 9 = 18

⇒ Denominator = (x² + 5x + 1) ÷ x = x + 5 + 1/x

⇒ Denominator = (x + 1/x) + 5 = 3 + 5 = 8

⇒ Final value = 18 ÷ 8 = 9/4

∴ The correct answer is 9/4.

Additional Information

Reciprocal Sum Identities

If x + 1/x = k, then x² + 1/x² = k² − 2 and x³ + 1/x³ = k³ − 3k.

Difference and Addition Relation

(x + 1/x)² − (x − 1/x)² = 4. This is used to find the difference of reciprocals when the sum is known.

Higher Powers

x⁵ + 1/x⁵ = (x² + 1/x²)(x³ + 1/x³) − (x + 1/x). This formula allows solving for higher odd powers systematically.

Shortcut Trick

Ratio of CP : MP = (100 − Discount%) : (100 + Profit%)

CP : MP = (100 − 12.5) : (100 + 15) = 87.5 : 115 = 7 : 9.2 = 35 : 46

Given 35 units = ₹385 ⇒ 1 unit = ₹11

MP = 46 × 11 = ₹506 | SP = 385 × 1.15 = ₹442.75

Sum = MP + SP = 506 + 442.75 = ₹948.75

∴ The correct answer is ₹948.75.

Alternate Method

Given:

Cost Price (CP) = ₹385

Profit% = 15%

Discount% = 12.5% = 1/8

Calculations:

⇒ Selling Price (SP) = CP × (100 + Profit%)/100

⇒ SP = 385 × (115/100) = 385 × 1.15

⇒ SP = ₹442.75

⇒ Marked Price (MP) × (100 − Discount%)/100 = SP

⇒ MP × (87.5/100) = 442.75

⇒ MP = (442.75 × 100) ÷ 87.5

⇒ MP = 44275 ÷ 87.5 = ₹506

⇒ Sum = MP + SP

⇒ Sum = 506 + 442.75 = ₹948.75

∴ The correct answer is ₹948.75.

Additional Information

CP and MP Relationship

The direct ratio between Cost Price and Marked Price is given by CP/MP = (100 − Discount%) / (100 + Profit%).

Marked Price (MP)

It is the price displayed on the label. MP = SP + Discount = CP + Markup.

Effective Discount

Successive discounts of a% and b% result in a single discount of [a + b − (a×b)/100]%.

Shortcut Trick

Total daily wage of A and B = 18,000 ÷ 15 = ₹1,200.

Efficiency Ratio (A : B) = 5 : 1.

Wages are distributed in the ratio of efficiency: Total parts = 5 + 1 = 6.

Value of 6 parts = 1,200 → 1 part = 200.

Daily wage of A = 5 parts = 5 × 200 = ₹1,000.

∴ The correct answer is ₹1,000.

Alternate Method

Given:

Total amount received = ₹18,000

Number of days worked = 15 days

Efficiency of A = 5 × Efficiency of B

Calculations:

⇒ Daily wage of (A + B) = Total Wage ÷ Number of days

⇒ Daily wage of (A + B) = 18,000 ÷ 15 = ₹1,200

⇒ Let B's daily efficiency be 1x, then A's daily efficiency = 5x

⇒ Total efficiency = 5x + 1x = 6x

⇒ Since daily wage is proportional to efficiency: 6x = 1,200

⇒ x = 1,200 ÷ 6 = 200

⇒ Daily wage of A = 5x = 5 × 200 = ₹1,000

∴ The correct answer is ₹1,000.

Additional Information

Wage Distribution Rule

Wages are always distributed in the ratio of the work done, which is the product of Efficiency × Time. If time is the same for all workers, wages are distributed in the ratio of their efficiencies.

Efficiency and Work Relation

Work = Efficiency × Time. Efficiency is the amount of work done in a single unit of time (per day, per hour, etc.).

Ratio and Proportion in Wages

If the ratio of efficiencies of two workers is a : b, their shares of the total wage will be [a/(a+b)] × Total Wage and [b/(a+b)] × Total Wage respectively.

Shortcut Trick

Value of x³ + y³ + z³ − 3xyz = (Sum/2) × [Sum of square of differences]

Sum = 32.5 + 34.6 + 30.9 = 98

Sum of squared differences = (−2.1)² + (3.7)² + (−1.6)² = 4.41 + 13.69 + 2.56 = 20.66

Expression value = 49 × 20.66 = 1012.34

k = 1012.34 ÷ 0.98 = 1033

∴ The correct answer is 1033.

Alternate Method

Given:

x = 32.5, y = 34.6, z = 30.9

Formula Used:

x³ + y³ + z³ − 3xyz = 1/2(x + y + z)[(x − y)² + (y − z)² + (z − x)²]

Calculations:

⇒ Sum (x + y + z) = 32.5 + 34.6 + 30.9 = 98

⇒ (x − y) = 32.5 − 34.6 = −2.1

⇒ (y − z) = 34.6 − 30.9 = 3.7

⇒ (z − x) = 30.9 − 32.5 = −1.6

⇒ x³ + y³ + z³ − 3xyz = 1/2 × 98 × [(−2.1)² + (3.7)² + (−1.6)²]

⇒ 0.98k = 49 × [4.41 + 13.69 + 2.56]

⇒ 0.98k = 49 × 20.66

⇒ 0.98k = 1012.34

⇒ k = 1012.34 ÷ 0.98

⇒ k = 1033

∴ The correct answer is 1033.

Additional Information

Standard Algebraic Identity

x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx). This is the base form used for most expansions.

Zero Sum Property

If (x + y + z) = 0, then x³ + y³ + z³ = 3xyz. This is a very common shortcut in competitive exams.

Square Sum Variation

x³ + y³ + z³ − 3xyz = (x + y + z)[(x + y + z)² − 3(xy + yz + zx)]. This version is useful when the sum of products is given.

Shortcut Trick

By Componendo and Dividendo: sec θ / tan θ = (7 + 1) / (7 − 1) = 8 / 6 = 4 / 3.

Since sec θ / tan θ = (1/cos θ) / (sin θ/cos θ) = 1/sin θ = cosec θ.

∴ cosec θ = 4/3 and cot² θ = cosec² θ − 1 = (4/3)² − 1 = 7/9.

Required value = (4/3 + 7/9) / (4/3 − 7/9) = (19/9) / (5/9) = 19/5.

∴ The correct answer is 19/5.

Alternate Method

Given: (sec θ − tan θ) / (sec θ + tan θ) = 1/7

Formula Used: Componendo and Dividendo: If a/b = c/d, then (b+a)/(b-a) = (d+c)/(d-c). Also, cosec² θ − cot² θ = 1.

⇒ (sec θ − tan θ) / (sec θ + tan θ) = 1/7

⇒ [(sec θ + tan θ) + (sec θ − tan θ)] / [(sec θ + tan θ) − (sec θ − tan θ)] = (7 + 1) / (7 − 1)

⇒ (2 sec θ) / (2 tan θ) = 8 / 6

⇒ (1/cos θ) × (cos θ/sin θ) = 4 / 3

⇒ 1/sin θ = cosec θ = 4 / 3

⇒ cot² θ = cosec² θ − 1 = (4/3)² − 1 = 16/9 − 1 = 7/9

⇒ Value = (cosec θ + cot² θ) / (cosec θ − cot² θ)

⇒ Value = (4/3 + 7/9) / (4/3 − 7/9)

⇒ Value = [(12 + 7) / 9] / [(12 − 7) / 9] = 19 / 5

∴ The correct answer is 19/5.

Additional Information

Componendo and Dividendo Rule

If the ratio a/b = c/d, then (a + b) / (a − b) = (c + d) / (c − d). This is very useful in simplifying trigonometric fractions.

Trigonometric Identities

The three fundamental identities are: sin² θ + cos² θ = 1, sec² θ − tan² θ = 1, and cosec² θ − cot² θ = 1.

Quadrant Rule for Trigonometry

In the first quadrant (0 ° to 90 °), all trigonometric ratios (sin, cos, tan, cosec, sec, cot) are positive.

Shortcut Trick

Interest for 3 years (7 − 4) = ₹(6,750 − 6,000) = ₹750

Interest for 1 year = 750 ÷ 3 = ₹250

Interest for 4 years = 250 × 4 = ₹1,000

Sum = Amount (at 4 years) − SI (for 4 years) = 6,000 − 1,000 = ₹5,000

∴ The correct answer is ₹5,000.

Alternate Method

Given:

Amount after 4 years (A₁) = ₹6,000

Amount after 7 years (A₂) = ₹6,750

Formula Used:

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

Amount (A) = Principal (P) + Simple Interest (SI)

Calculations:

⇒ Interest for (7 − 4) years = A₂ − A₁

⇒ Simple Interest for 3 years = ₹6,750 − ₹6,000 = ₹750

⇒ Simple Interest for 1 year = 750 ÷ 3 = ₹250

⇒ Simple Interest for 4 years = 250 × 4 = ₹1,000

⇒ Principal (P) = Amount after 4 years − SI for 4 years

⇒ Principal (P) = 6,000 − 1,000 = ₹5,000

∴ The correct answer is ₹5,000.

Additional Information

Simple Interest Property

In Simple Interest, the interest remains constant every year because it is always calculated on the original Principal (P).

General Amount Formula

The amount after T years can be written as A = P × (1 + RT ÷ 100), where R is the rate per annum.

Finding the Rate of Interest

Once the sum is known, the rate R can be calculated using R = (SI × 100) ÷ (P × T). Here, R = (1000 × 100) ÷ (5000 × 4) = 5% p.a.

Shortcut Trick

Find the common difference (k): (36 − 16) = (72 − 52) = (80 − 60) = (88 − 68) = 20.

The least number is LCM(36, 72, 80, 88) − k.

LCM(36, 72, 80, 88) = 7920.

Required Number = 7920 − 20 = 7900.

Sum of digits = 7 + 9 + 0 + 0 = 16.

∴ The correct answer is 16.

Alternate Method

Given divisors and their respective remainders:

To find the Least Common Multiple (LCM) of 36, 72, 80, and 88:

⇒ Prime factorisation:

⇒ 36 = 2² × 3²

⇒ 72 = 2³ × 3²

⇒ 80 = 2⁴ × 5

⇒ 88 = 2³ × 11

⇒ LCM = 2⁴ × 3² × 5 × 11

⇒ LCM = 16 × 9 × 5 × 11 = 7920

⇒ The least number (N) = LCM − Common Difference

⇒ N = 7920 − 20 = 7900

⇒ Sum of digits of N = 7 + 9 + 0 + 0 = 16

∴ The correct answer is 16.

Additional Information

LCM with Constant Difference

When a number leaves remainders r₁, r₂, r₃ when divided by a, b, c such that (a−r₁) = (b−r₂) = (c−r₃) = k, the number is LCM(a, b, c) − k.

LCM with Same Remainder

If the number leaves the same remainder 'r' in each case when divided by a, b, c, then the least such number is LCM(a, b, c) + r.

Divisibility of Sum of Digits

If the sum of digits of a number is divisible by 3 or 9, then the number itself is divisible by 3 or 9 respectively.

Shortcut Trick Use the value substitution method by picking a specific angle, let θ = 30°. Substitute values: tan 30° = 1/√3, cot 30° = √3, sec 30° = 2/√3, sin 30° = 1/2. Expression = (1/√3 + √3) × (2/√3 + 1/√3) × (1 − 1/2) = (4/√3) × (3/√3) × (1/2). Expression = (4/√3) × √3 × 1/2 = 2. Check options for θ = 30°: cosec 30° = 1/sin 30° = 1/(1/2) = 2. ∴ The correct answer is cosec θ. Alternate Method Given: The expression (tan θ + cot θ) (sec θ + tan θ) (1 − sin θ) for 0° < θ < 90°. Formula Used: tan θ = sin θ/cos θ; cot θ = cos θ/sin θ; sec θ = 1/cos θ; sin2θ + cos2θ = 1.

⇒ Simplifying (tan θ + cot θ) = (sin θ / cos θ + cos θ / sin θ) = (sin2θ + cos2θ) / (sin θ cos θ) ⇒ (tan θ + cot θ) = 1 / (sin θ cos θ) ⇒ Simplifying (sec θ + tan θ)(1 − sin θ) = (1/cos θ + sin θ/cos θ)(1 − sin θ) ⇒ ((1 + sin θ) / cos θ) × (1 − sin θ) = (1 − sin2θ) / cos θ ⇒ cos2θ / cos θ = cos θ ⇒ Multiplying both parts: [1 / (sin θ cos θ)] × cos θ = 1 / sin θ ⇒ 1 / sin θ = cosec θ ∴ The correct answer is cosec θ. Additional Information Reciprocal Identities Trigonometric ratios are related by reciprocals: cosec θ = 1/sin θ, sec θ = 1/cos θ, and cot θ = 1/tan θ. Pythagorean Identities Fundamental identities include sin2θ + cos2θ = 1, 1 + tan2θ = sec2θ, and 1 + cot2θ = cosec2θ. Quotient Identities The tangent and cotangent functions can be expressed as ratios: tan θ = sin θ/cos θ and cot θ = cos θ/sin θ.

Shortcut Trick

Simple Interest 1 (SI₁) = 11,046 − 8,400 = ₹2,646

Since Rate (R) and Time (T) are constant, Simple Interest is directly proportional to Principal (P).

Ratio of Principals (P₁ : P₂) = 8,400 : 10,800 = 7 : 9

Required Interest (SI₂) = (SI₁ ÷ 7) × 9 = (2,646 ÷ 7) × 9

SI₂ = 378 × 9 = ₹3,402

∴ The correct answer is ₹3,402.

Alternate Method

Given:

Case 1: P₁ = ₹8,400, A₁ = ₹11,046, R = 8.75%

Case 2: P₂ = ₹10,800, R = 8.75%, T is same as Case 1

Formula Used:

SI = (P × R × T) ÷ 100 | Amount (A) = P + SI

Calculations:

⇒ SI₁ = A₁ − P₁ = 11,046 − 8,400 = 2,646

⇒ Since R and T are same, SI₁ ÷ P₁ = SI₂ ÷ P₂

⇒ 2,646 ÷ 8,400 = SI₂ ÷ 10,800

⇒ SI₂ = (2,646 × 10,800) ÷ 8,400

⇒ SI₂ = (2,646 × 108) ÷ 84

⇒ SI₂ = (2,646 × 9) ÷ 7

⇒ SI₂ = 378 × 9

⇒ SI₂ = 3,402

∴ The correct answer is ₹3,402.

Additional Information

Direct Proportionality in Simple Interest

When rate (R) and time (T) are constant, the simple interest (SI) varies directly with the principal (P). If P increases by a factor, SI increases by the same factor.

Effective Rate of Interest

The effective simple interest over a period T is simply (R × T)%. For example, 8.75% for 4 years is an effective 35% of the principal.

Relationship between SI and Amount

Amount = P(1 + RT/100). The ratio of Amount to Principal is (100 + RT) : 100.

Shortcut Trick

Volume of cylinder is proportional to π, so check for divisibility by 11 in the options.

Radius (r) = Circumference ÷ (2 × π) = 264 ÷ (2 × 22/7) = 42 cm.

Volume = πr²h = (22/7) × 42 × 42 × 50 = 277200 cm³.

Capacity = 277200 ÷ 1000 = 277.2 litres.

∴ The correct answer is 277.2 litres.

Alternate Method

Given:

Circumference of the base = 264 cm

Height (h) of the vessel = 50 cm

π = 22/7

Formula Used:

Circumference = 2 × π × r

Volume of Cylinder = π × r² × h

1 litre = 1000 cm³

Calculations:

⇒ 2 × (22/7) × r = 264

⇒ (44/7) × r = 264

⇒ r = (264 × 7) ÷ 44

⇒ r = 6 × 7 = 42 cm

⇒ Volume = (22/7) × 42 × 42 × 50

⇒ Volume = 22 × 6 × 42 × 50

⇒ Volume = 132 × 2100 = 277200 cm³

⇒ Capacity in litres = 277200 ÷ 1000

⇒ Capacity = 277.2 litres

∴ The correct answer is 277.2 litres.

Additional Information

Volume of Cylinder

The volume or capacity of a cylinder is given by the formula V = πr²h, where r is the radius and h is the height.

Curved Surface Area (CSA)

The area of the curved surface of a cylinder is CSA = 2πrh.

Total Surface Area (TSA)

The total surface area including the top and bottom circles is TSA = 2πr(r + h).

Unit Conversions

Common units include 1 cm³ = 1 mL, 1000 cm³ = 1 litre, and 1 m³ = 1000 litres.

Given:

$9 \div [\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}\div(\frac{3}{4}-\frac{1}{3});of;\frac{2}{9}]$

Formula used:

BODMAS rule (Bracket, Of, Division, Multiplication, Addition, Subtraction)

Calculation:

⇒ $9 \div [\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} \div (\frac{9-4}{12});of;\frac{2}{9}]$

⇒ $9 \div [\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} \div \frac{5}{12};of;\frac{2}{9}]$

⇒ $9 \div [\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} \div (\frac{5}{12} \times \frac{2}{9})]$

⇒ $9 \div [\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} \div \frac{5}{54}]$

⇒ $9 \div [\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{6} \times \frac{54}{5}]$

⇒ $9 \div [\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{9}{5}]$

⇒ $9 \div [\frac{30 + 20 + 15 + 108}{60}]$

⇒ $9 \div \frac{173}{60}$

⇒ $9 \times \frac{60}{173}$

⇒ $\frac{540}{173}$

∴ The correct answer is option (1).

Shortcut Trick

Inner radius (r) = 3 m, Depth (h) = 6 m, Embankment Width = 2 m

Outer radius (R) = r + width = 3 + 2 = 5 m

Height of embankment (H) = (r² × h) / (R² − r²)

H = (3² × 6) / (5² − 3²)

H = 54 / (25 − 9) = 54 / 16 = 27 / 8 = 3 3/8 m

∴ The correct answer is 3 3/8 m.

Alternate Method

Given:

Inner radius of well (r) = 3 m

Depth of well (h) = 6 m

Width of embankment = 2 m

Formula Used:

Volume of Cylinder = π × r² × h

Volume of Embankment (Hollow Cylinder) = π × (R² − r²) × H

Calculations:

⇒ Volume of soil taken out = π × r² × h

⇒ Volume of soil = π × 3² × 6 = 54π m³

⇒ Outer radius of embankment (R) = Inner radius + width = 3 + 2 = 5 m

⇒ Volume of embankment = π × (R² − r²) × H

⇒ Volume of embankment = π × (5² − 3²) × H = π × (25 − 9) × H = 16πH

⇒ Volume of soil = Volume of embankment

⇒ 54π = 16πH

⇒ H = 54 / 16 = 27 / 8

⇒ H = 3 3/8 m

∴ The correct answer is 3 3/8 m.

Additional Information

Volume of Cylinder

The volume of a solid cylinder is given by V = πr²h, where r is the radius and h is the height.

Volume of Hollow Cylinder (Embankment)

The volume of a hollow cylinder is the difference between the volumes of the outer and inner cylinders: V = π(R² − r²)h.

Conservation of Volume

When an object is reshaped (like soil from a hole used to build a platform), the total volume remains constant.

Shortcut Trick

Substitute θ = 45° where tan 45° = 1, cot 45° = 1, and sec 45° = √2.

Numerator = 1⁶ − (√2)⁶ + 3(√2)²(1)² = 1 − 8 + 3(2) = 1 − 8 + 6 = −1.

Denominator = 1² + 1² + 2 = 1 + 1 + 2 = 4.

Simplified value = −1/4. Checking Option 4: −cos²45° sin²45° = −(1/2) × (1/2) = −1/4.

∴ The correct answer is −cos² θ sin² θ.

Alternate Method

Given: expression = (tan⁶ θ − sec⁶ θ + 3sec² θ tan² θ) ÷ (tan² θ + cot² θ + 2)

Formula Used: (a − b)³ = a³ − b³ − 3ab(a − b); sec² θ − tan² θ = 1; sin² θ + cos² θ = 1

⇒ Let x = tan² θ and y = sec² θ. We know x − y = −1.

⇒ Numerator = x³ − y³ + 3xy = x³ − y³ − 3xy(−1)

⇒ Substituting (−1) with (x − y): Numerator = x³ − y³ − 3xy(x − y) = (x − y)³

⇒ Numerator = (−1)³ = −1

⇒ Denominator = tan² θ + cot² θ + 2 = (tan θ + cot θ)²

⇒ Denominator = (sin θ ÷ cos θ + cos θ ÷ sin θ)² = [(sin² θ + cos² θ) ÷ (sin θ cos θ)]²

⇒ Since sin² θ + cos² θ = 1, Denominator = [1 ÷ (sin θ cos θ)]² = 1 ÷ (sin² θ cos² θ)

⇒ Final Expression = −1 ÷ [1 ÷ (sin² θ cos² θ)] = −sin² θ cos² θ

∴ The correct answer is −cos² θ sin² θ.

Additional Information

Pythagorean Identities

Fundamental relations used: sin² θ + cos² θ = 1, sec² θ − tan² θ = 1, and cosec² θ − cot² θ = 1.

Reciprocal and Quotient Relations

tan θ + cot θ = sec θ cosec θ = 1 ÷ (sin θ cos θ), which is a common simplification for sum of squares.

Algebraic Cube Identity

The form a³ ± b³ ± 3ab(a ± b) = (a ± b)³ is frequently used to simplify high-power trigonometric expressions.

Shortcut Trick

Apply the rule of Alligation on the profit/loss percentages.

Profit part = +20%, Loss part = −5%, Overall Profit = +15%.

Ratio (Profit : Loss) = (15 − (−5)) : (20 − 15) = 20 : 5 = 4 : 1.

Required ratio (Loss : Profit) = 1 : 4.

∴ The correct answer is 1 ∶ 4.

Alternate Method

Given:

Total quantity of rice = 640 kg

Profit on Part 1 = 20%

Loss on Part 2 = 5% (represented as −5% profit)

Overall transaction profit = 15%

Calculations:

⇒ Let quantity sold at 20% profit be Q₁ and quantity sold at 5% loss be Q₂.

⇒ By Alligation rule: Q₁ / Q₂ = (Mean Value − Lower Value) / (Higher Value − Mean Value)

⇒ Q₁ / Q₂ = (15 − (−5)) / (20 − 15)

⇒ Q₁ / Q₂ = 20 / 5

⇒ Q₁ / Q₂ = 4 / 1

⇒ The ratio of Profit Quantity to Loss Quantity = 4 : 1

⇒ The ratio of Loss Quantity to Profit Quantity = 1 : 4

∴ The correct answer is 1 ∶ 4.

Additional Information

Rule of Alligation

It is used to find the ratio in which two ingredients at given prices/percentages are mixed to produce a mixture of a known average price/percentage. Ratio = (Cheaper / Difference) : (Dearer / Difference).

Profit and Loss signs

While using alligation for profit and loss, always treat profit as a positive value and loss as a negative value (e.g., 5% loss = −5% profit).

Total Quantity calculation

If the total quantity is known, individual quantities can be found by: Quantity = (Individual Ratio / Total Ratio) × Total Quantity.

Shortcut Trick

Take the lowest power 5⁷¹ as a common factor: 5⁷¹ (1 + 5¹ + 5² + 5³ + 5⁴)

Simplify the sum in the bracket: 1 + 5 + 25 + 125 + 625 = 781

Now, check divisibility of 781 by the options: 781 ÷ 71 = 11.

Since 781 is divisible by 71, the entire expression is divisible by 71.

∴ The correct answer is 71.

Alternate Method

Given Expression: 5⁷¹ + 5⁷² + 5⁷³ + 5⁷⁴ + 5⁷⁵

To find the divisor, we factorize the given sum of powers:

⇒ 5⁷¹ (5⁰ + 5¹ + 5² + 5³ + 5⁴)

⇒ 5⁷¹ (1 + 5 + 5² + 5³ + 5⁴)

⇒ 5⁷¹ (1 + 5 + 25 + 125 + 625)

⇒ 5⁷¹ × 781

Now, check the divisibility of 781 with the given options:

⇒ 781 ÷ 71 = 11

⇒ 781 ÷ 69 (Not divisible)

⇒ 781 ÷ 89 (Not divisible)

⇒ 781 ÷ 73 (Not divisible)

Since 781 = 71 × 11, the expression is a multiple of 71.

∴ The correct answer is 71.

Additional Information

Divisibility of Exponents

For a sum of consecutive powers aⁿ + aⁿ⁺¹ + ... + a^n+k, always factor out the smallest power aⁿ to find the numeric factor.

Geometric Progression Sum

The sum 1 + r + r² + ... + rⁿ can be calculated using the formula S_n = (rⁿ⁺¹ − 1) ÷ (r − 1).

Factor Theorem

If a number 'x' divides 'y', then 'x' will also divide any integer multiple of 'y' (i.e., y × k).

Shortcut Trick

Find y directly: y = 5/2 − 7/4 = 3/4.

Find (x + z) from total sum: (x + z) = 23/12 − 3/4 = 14/12 = 7/6.

Ratio z : x = 5 : 2 ⇒ Total 7 units = 7/6 ⇒ 1 unit = 1/6.

So, x = 2/6 = 4/12 and z = 5/6 = 10/12. Also, y = 3/4 = 9/12.

Ratio (z − x) : (y − x) = (10/12 − 4/12) : (9/12 − 4/12) = 6/12 : 5/12 = 6 : 5.

∴ The correct answer is 6 : 5.

Alternate Method

Given:

x < y < z

z ÷ x = 5/2

5/2 − y = 7/4

x + y + z = 1 11/12 = 23/12

Calculations:

⇒ y = 5/2 − 7/4 = 10/4 − 7/4 = 3/4

⇒ x + z = 23/12 − 3/4 = 23/12 − 9/12 = 14/12 = 7/6

⇒ From z/x = 5/2, let z = 5k and x = 2k

⇒ 2k + 5k = 7k = 7/6 ⇒ k = 1/6

⇒ x = 2 × 1/6 = 1/3 and z = 5 × 1/6 = 5/6

⇒ (z − x) = 5/6 − 1/3 = 5/6 − 2/6 = 3/6 = 1/2

⇒ (y − x) = 3/4 − 1/3 = (9 − 4)/12 = 5/12

⇒ Ratio = (1/2) ÷ (5/12) = 1/2 × 12/5 = 6/5

∴ The correct answer is 6 : 5.

Additional Information

Mixed Fractions

A mixed fraction a b/c is converted to an improper fraction as (a × c + b) / c.

Ratio of Fractions

To find the ratio of two fractions (a/b) : (c/d), multiply the first fraction by the reciprocal of the second: (a/b) × (d/c).

Comparing Fractions

To compare fractions, convert them to a common denominator using the LCM of the denominators.

Shortcut Trick

In ΔABC and ΔDBE, ∠B is common and ∠ACB = ∠DEB, so ΔABC ∼ ΔDBE by AA similarity.

Let BD = x, then BC = 3x (since BD:CD = 1:2).

Ratio of sides: AB/BD = BC/BE ⇒ 12/x = 3x/5.

3x² = 60 ⇒ x² = 20 ⇒ x = 2√5.

BC = 3x = 3 × 2√5 = 6√5 cm.

∴ The correct answer is 6√5 cm.

Alternate Method

Given: AB = 12 cm, BE = 5 cm, BD:CD = 1:2.

Formula Used: If two triangles are similar, the ratios of their corresponding sides are equal.

⇒ Let BD = x. Given BD : CD = 1 : 2, so CD = 2x.

⇒ Total length BC = BD + CD = x + 2x = 3x.

⇒ In ΔABC and ΔDBE:

⇒ ∠ABC = ∠DBE (Common angle)

⇒ ∠ACB = ∠DEB (Given)

⇒ ∴ ΔABC ∼ ΔDBE (By AA Similarity Criterion)

⇒ AB/DB = BC/BE = AC/DE

⇒ Using AB/DB = BC/BE:

⇒ 12/x = 3x/5

⇒ 12 × 5 = 3x × x

⇒ 60 = 3x²

⇒ x² = 20

⇒ x = √20 = 2√5

⇒ BC = 3x = 3 × 2√5 = 6√5 cm.

∴ The correct answer is 6√5 cm.

Additional Information

AA Similarity Criterion

If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar and their corresponding sides are in the same ratio.

Ratio of Sides in Similar Triangles

If ΔABC ∼ ΔPQR, then AB/PQ = BC/QR = AC/PR. This property is used to find unknown lengths in geometric figures.

Section Formula (Internal)

If a point D divides a line segment BC in the ratio m:n, then BD = [m/(m+n)] × BC and CD = [n/(m+n)] × BC.

Shortcut Trick

For radius r = 7 m, Circumference = 44 m.

Total distance for 8 rounds = 8 × 44 = 352 m.

Total time in seconds = (3 × 60) + 40 = 220 s.

Speed = (352 ÷ 220) × (18 ÷ 5) = 1.6 × 3.6 = 144/25 km/h.

∴ The correct answer is 144/25 km/h.

Alternate Method

Given:

Radius of circular field (r) = 7 m

Number of rounds (n) = 8

Time (t) = 3 minutes 40 seconds

Formula:

Circumference of circle = 2 × π × r

Speed = Total Distance ÷ Total Time

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

Calculations:

⇒ Circumference (C) = 2 × (22 ÷ 7) × 7 = 44 m

⇒ Total Distance (D) = 8 × 44 = 352 m

⇒ Total Time (t) in seconds = (3 × 60) + 40 = 220 s

⇒ Speed (v) in m/s = 352 ÷ 220 = 1.6 m/s

⇒ Speed in km/h = 1.6 × (18 ÷ 5)

⇒ Speed = (16 ÷ 10) × (18 ÷ 5) = (8 ÷ 5) × (18 ÷ 5)

⇒ Speed = 144 ÷ 25 km/h

∴ The correct answer is 144/25 km/h.

Additional Information

Unit Conversion

To convert speed from m/s to km/h, multiply the value by 18/5. Conversely, multiply by 5/18 to convert km/h to m/s.

Distance in Circular Motion

The total distance covered in n rounds around a circular path is n × (2πr), where r is the radius of the circle.

Average Speed

Average speed is defined as the total distance traveled divided by the total time taken (Total Distance ÷ Total Time).

Shortcut Trick

Convert fraction ratio to integer: LCM(2, 3, 4) = 12.

New Ratio → (1/2 × 12) : (2/3 × 12) : (3/4 × 12) = 6 : 8 : 9.

Difference between Greatest (9) and Smallest (6) = 3 units.

3 units = 33 ⇒ 1 unit = 11.

Since 6, 8, and 9 have no common factor other than 1, the value of 1 unit is the HCF.

∴ The correct answer is 11.

Alternate Method

Given:

Ratio of three numbers = 1/2 : 2/3 : 3/4

Difference (Greatest − Smallest) = 33

Calculations:

⇒ To simplify the ratio, multiply each term by the LCM of 2, 3, and 4 (LCM = 12).

⇒ Ratio = (1/2 × 12) : (2/3 × 12) : (3/4 × 12) = 6 : 8 : 9

⇒ Let the three numbers be 6k, 8k, and 9k.

⇒ Smallest number = 6k, Greatest number = 9k

⇒ Difference = 9k − 6k = 3k

⇒ Given: 3k = 33

⇒ k = 33 ÷ 3 = 11

⇒ The numbers are 6 × 11 = 66, 8 × 11 = 88, and 9 × 11 = 99.

⇒ HCF(66, 88, 99) = HCF(6×11, 8×11, 9×11)

⇒ Since HCF(6, 8, 9) = 1, the HCF of the numbers is 11.

∴ The correct answer is 11.

Additional Information

Ratio Simplification

When a ratio is given in fractions (a/b : c/d), multiply all terms by the LCM of denominators (b and d) to convert them into integers.

HCF of Numbers in Ratio

If three numbers are in the ratio x : y : z and their HCF is H, then the numbers are Hx, Hy, and Hz. Here, x, y, and z must be co-prime in their simplest form.

LCM and HCF Relationship

For any numbers in ratio a : b : c with HCF H, their LCM = H × (LCM of a, b, c).

Calculation:

Total production in A in 2016 = 280

Total production in C in 2016 = 350

Total of them = 280 + 350

⇒ 630

Total production in B in 2017 and 2019 = 200 + 360

⇒ 560

Ratio = 630 : 560

⇒ 9 : 8

∴ Required ratio is 9 : 8

Shortcut Trick

Ratio of Boys : Girls = 33(1/3)% : 66(2/3)% = 1 : 2

Ratio of Average Scores (B : G) = 100 : 166(2/3) = 3 : 5

Weighted average = [(1 × 3) + (2 × 5)] ÷ (1 + 2) = 13/3 units

13/3 units = 78 ⇒ 1 unit = (78 × 3) ÷ 13 = 18

Average score of girls = 5 units = 5 × 18 = 90

∴ The correct answer is 90.

Alternate Method

Given: Total students = 45, Boys % = 33(1/3)%, Overall average = 78.

Formula: Average = Total Score ÷ Number of Students

⇒ Number of boys = 45 × (1/3) = 15

⇒ Number of girls = 45 − 15 = 30

⇒ Ratio of B : G = 15 : 30 = 1 : 2

⇒ Let boys' average = 3x, then girls' average = 3x + (2/3 × 3x) = 5x

⇒ Total score = (1 × 3x) + (2 × 5x) = 13x

⇒ Combined Average = 13x ÷ 3 = 78

⇒ 13x = 78 × 3 = 234

⇒ x = 234 ÷ 13 = 18

⇒ Average score of girls = 5x = 5 × 18 = 90

∴ The correct answer is 90.

Additional Information

Weighted Average Formula

Weighted Avg = (n₁ × a₁ + n₂ × a₂) ÷ (n₁ + n₂), where n is quantity and a is average.

Percentage to Fraction Conversion

33(1/3)% = 1/3 and 66(2/3)% = 2/3. Using these fractions simplifies complex calculations significantly.

Alligation Method for Averages

The ratio of the quantities is inversely proportional to the deviations of their averages from the mean average.

Shortcut Trick

Let the side of the square be 1 unit. Then AB = 1, AC = √2, and AO = OB = 1/√2.

In ΔAOB, AF is the bisector of ∠OAB ⇒ OF/FB = AO/AB = (1/√2)/1 = 1/√2.

⇒ OF = [1/(1 + √2)] × OB = 1/[√2(1 + √2)].

In ΔABC, AG is the bisector of ∠CAB ⇒ CG/GB = AC/AB = √2/1 = √2.

⇒ CG = [√2/(1 + √2)] × BC = √2/(1 + √2) (since BC = 1).

Ratio OF/CG = [1/{√2(1 + √2)}] / [√2/(1 + √2)] = 1/(√2 × √2) = 1/2.

∴ The correct answer is 1 : 2.

Alternate Method

Given: ABCD is a square with side 'a'. Diagonals AC and BD intersect at O. AG is the angle bisector of ∠CAB.

Formula Used: Angle Bisector Theorem states that the bisector of an angle of a triangle divides the opposite side into segments that are proportional to the other two sides of the triangle.

⇒ Let side AB = a. Then diagonal AC = a√2. Half-diagonal AO = OB = (a√2)/2 = a/√2.

⇒ In ΔAOB, AF is the bisector of ∠OAB (45°). By Angle Bisector Theorem:

⇒ OF / FB = AO / AB = (a/√2) / a = 1 / √2

⇒ OF = [1 / (1 + √2)] × OB = [1 / (1 + √2)] × (a/√2) = a / (2 + √2)

⇒ In ΔABC, AG is the bisector of ∠CAB (45°). By Angle Bisector Theorem:

⇒ CG / GB = AC / AB = (a√2) / a = √2 / 1

⇒ CG = [√2 / (1 + √2)] × BC = (√2 × a) / (1 + √2)

⇒ Ratio OF : CG = [a / (2 + √2)] / [(a√2) / (1 + √2)]

⇒ OF / CG = [a / {√2(1 + √2)}] × [(1 + √2) / (a√2)] = 1 / (√2 × √2) = 1 / 2

∴ The correct answer is 1 : 2.

Additional Information

Angle Bisector Theorem

In any triangle, the angle bisector divides the opposite side into segments proportional to the adjacent sides: Side₁/Side₂ = Segment₁/Segment₂.

Square Diagonals

The diagonals of a square are equal, bisect each other at 90°, and bisect the corner angles into 45° each.

Trigonometric Ratio tan(22.5°)

The value of tan(22.5°) is √2 − 1, which can be used to solve problems involving half-angles of 45°.

Shortcut Trick

Total interest factor for first part (R₁ × T₁) = (20/3)% × (21/5) = 28%.

Total interest factor for second part (R₂ × T₂) = 4% × (11/4) = 11%.

Given: Interest on first part = 2 × Interest on second part.

⇒ P₁ × 28% = 2 × (P₂ × 11%) ⇒ 28 P₁ = 22 P₂.

⇒ P₂/P₁ = 28/22 = 14/11.

∴ The correct answer is 14 : 11.

Alternate Method

Given:

Total Sum = ₹5,000

For First Part (P₁): Rate (R₁) = 20/3% p.a., Time (T₁) = 21/5 years.

For Second Part (P₂): Rate (R₂) = 4% p.a., Time (T₂) = 11/4 years.

Condition: SI₁ = 2 × SI₂

Formula Used:

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

Calculations:

⇒ SI₁ = (P₁ × 20/3 × 21/5) ÷ 100

⇒ SI₁ = (P₁ × 4 × 7) ÷ 100 = 28 P₁ / 100

⇒ SI₂ = (P₂ × 4 × 11/4) ÷ 100

⇒ SI₂ = (P₂ × 11) ÷ 100 = 11 P₂ / 100

⇒ Since SI₁ = 2 × SI₂

⇒ 28 P₁ / 100 = 2 × (11 P₂ / 100)

⇒ 28 P₁ = 22 P₂

⇒ P₂ / P₁ = 28 / 22

⇒ P₂ / P₁ = 14 / 11

∴ The correct answer is 14 : 11.

Additional Information

Simple Interest Concept

Simple interest is calculated only on the principal amount. Formula: SI = (P × R × T) ÷ 100, where P is Principal, R is Rate, and T is Time.

Ratio of Principals

If interests on two parts are equal, P₁ : P₂ = (1 ÷ R₁T₁) : (1 ÷ R₂T₂). If SI₁ = n × SI₂, then P₁R₁T₁ = n × P₂R₂T₂.

Mixed Fraction Conversion

To convert a mixed fraction A b/c into an improper fraction, use (A × c + b) ÷ c. For example, 6 2/3 = (6×3 + 2)/3 = 20/3.

Shortcut Trick Relative speed in same direction = Total Distance ÷ Time = 140 ÷ 7 = 20 km/h Relative speed in opposite direction = Total Distance ÷ Time = 140 ÷ 1 = 140 km/h Speed of faster car (y) = (Sum of relative speeds + Difference of relative speeds) ÷ 2 Sy = (140 + 20) ÷ 2 = 160 ÷ 2 = 80 km/h ∴ The correct answer is 80 km/h.

Alternate Method Given: Distance (D) = 140 km Time taken in same direction (t1) = 7 hours Time taken in opposite direction (t2) = 1 hour Condition: Speed of y (Sy) > Speed of x (Sx)

Formula Used: Relative speed (Opposite direction) = S1 + S2 Relative speed (Same direction) = S1 − S2 (where S1 > S2) Speed = Distance ÷ Time

Calculations: ⇒ Case 1 (Opposite direction): Sy + Sx = 140 ÷ 1 ⇒ Sy + Sx = 140 — (Equation 1) ⇒ Case 2 (Same direction): Sy − Sx = 140 ÷ 7 ⇒ Sy − Sx = 20 — (Equation 2) Adding Equation 1 and Equation 2: ⇒ (Sy + Sx) + (Sy − Sx) = 140 + 20 ⇒ 2Sy = 160 ⇒ Sy = 160 ÷ 2 = 80 km/h Substituting Sy in Equation 1: ⇒ 80 + Sx = 140 ⇒ Sx = 140 − 80 = 60 km/h Checking condition: Sy (80) > Sx (60). Condition satisfied. ∴ The correct answer is 80 km/h.

Additional Information Relative Speed Concept When two objects move in the same direction, their relative speed is the difference (u − v). When they move in opposite directions, it is the sum (u + v). Speed Conversion To convert km/h to m/s, multiply by 5/18. To convert m/s to km/h, multiply by 18/5. Average Speed for Constant Distance If an object covers the same distance at speed u and then at speed v, the average speed is (2 × u × v) ÷ (u + v).

Calculation:

Number of students passed in A = 1200 × 22/100

⇒ 264

Number of boys who passed from A = 264 × 5/11

⇒ 120

Number of students passed in D = 1200 × 20/100

⇒ 240

Number of boys who passed from D = 240 × 2/5

⇒ 96

Ratio = 120 : 96

⇒ 5 : 4

∴ Required ratio is 5 : 4

Alternate MethodThe number of boys who passed from A = (22/100) × (5/11) × 1200

The number of boys who passed from D = (20/100) × (2/5) × 1200

The required ratio = [(22/100) × (5/11) × 1200]/[(20/100) × (2/5) × 1200]

⇒ (2 × 5 × 12) ∶ (4 × 2 × 12)

⇒ 5 ∶ 4

∴ Required ratio is 5 : 4

Given:

$\frac{3}{70}+ \frac{1}{42}+\frac{1}{66}+\frac{3}{286}+\frac{1}{130}+\frac{1}{170}$ = ?

Formula used:

Addition of fractions by taking L.C.M.

Calculation:

$\frac{3}{70}+ \frac{1}{42}+\frac{1}{66}+\frac{3}{286}+\frac{1}{130}+\frac{1}{170}$

⇒ $(\frac{3}{70} + \frac{1}{42}) + (\frac{1}{66} + \frac{3}{286}) + (\frac{1}{130} + \frac{1}{170})$

⇒ $\frac{9 + 5}{210} + \frac{13 + 9}{858} + \frac{17 + 13}{2210}$

⇒ $\frac{14}{210} + \frac{22}{858} + \frac{30}{2210}$

⇒ $\frac{1}{15} + \frac{1}{39} + \frac{3}{221}$

⇒ $(\frac{1}{15} + \frac{1}{39}) + \frac{3}{221}$

⇒ $\frac{13 + 5}{195} + \frac{3}{221}$

⇒ $\frac{18}{195} + \frac{3}{221}$

⇒ $\frac{6}{65} + \frac{3}{221}$

⇒ $\frac{6 \times 17 + 3 \times 5}{1105}$

⇒ $\frac{102 + 15}{1105}$

⇒ $\frac{117}{1105} = \frac{9}{85}$

∴ The correct answer is option (3).

Shortcut Trick

Savings % (Old) = 100% − [50% + (20% of 50%) + (15% of 10%)] = 100% − 61.5% = 38.5%

Savings 1 = 38.5% of 1,60,000 = ₹61,600

New Salary = 1.30 × 1,60,000 = ₹2,08,000

Savings % (New) = 100% − [50% + (30% of 50%) + (20% of 15%)] = 100% − 68% = 32%

Savings 2 = 32% of 2,08,000 = ₹66,560

Total Savings = 61,600 + 66,560 = ₹1,28,160

∴ The correct answer is ₹1,28,160.

Alternate Method

Given:

Initial Salary = ₹1,60,000

Expenses: P = 50% of Income, E = 20% of P, T = 15% of E

New Salary = Initial × 130%

New Expenses: P = 50% of Income, E = 30% of P, T = 20% of E

Step 1: Calculate Initial Savings

⇒ P₁ = 50% of 1,60,000 = ₹80,000

⇒ E₁ = 20% of 80,000 = ₹16,000

⇒ T₁ = 15% of 16,000 = ₹2,400

⇒ Savings₁ = 1,60,000 − (80,000 + 16,000 + 2,400) = ₹61,600

Step 2: Calculate New Savings after 30% Raise

⇒ New Salary = 1,60,000 × 1.3 = ₹2,08,000

⇒ P₂ = 50% of 2,08,000 = ₹1,04,000

⇒ E₂ = 30% of 1,04,000 = ₹31,200

⇒ T₂ = 20% of 31,200 = ₹6,240

⇒ Savings₂ = 2,08,000 − (1,04,000 + 31,200 + 6,240) = ₹66,560

Step 3: Calculate the Sum

⇒ Total Savings = 61,600 + 66,560 = ₹1,28,160

∴ The correct answer is ₹1,28,160.

Additional Information

Percentage of Percentage

When a value is a percentage of another percentage, multiply the decimals: x% of y% of Total = (x/100) × (y/100) × Total.

Successive Increment

A salary raise of r% is calculated as: New Salary = Original Salary × (1 + r/100).

Savings Calculation

Savings = Total Income − Sum of all individual expenses (P + E + T + ...).

Shortcut Trick

Ratio of radii (r₁ : r₂) = 7 : 10.5 = 2 : 3

Surface area of sphere is proportional to r²

Ratio of surface areas (A₁ : A₂) = 2² : 3² = 4 : 9

Increase in units = 9 − 4 = 5

Percentage increase = (Increase ÷ Original value) × 100 = (5 ÷ 4) × 100 = 125%

∴ The correct answer is 125%.

Alternate Method

Given:

Initial radius (r₁) = 7 cm

Final radius (r₂) = 10.5 cm

Formula Used:

Surface area of a sphere = 4 × π × r²

Percentage increase = [(New Area − Old Area) ÷ Old Area] × 100

Calculations:

⇒ Initial Surface Area (S₁) = 4 × π × (7)² = 196π cm²

⇒ Final Surface Area (S₂) = 4 × π × (10.5)² = 4 × π × 110.25 = 441π cm²

⇒ Increase in Surface Area = 441π − 196π = 245π cm²

⇒ Percentage Increase = (245π ÷ 196π) × 100

⇒ Percentage Increase = (245 ÷ 196) × 100

⇒ Percentage Increase = (5 ÷ 4) × 100

⇒ Percentage Increase = 1.25 × 100 = 125%

∴ The correct answer is 125%.

Additional Information

Volume of a Sphere

The volume (V) is given by (4/3) × π × r³. If radius increases in ratio a:b, volume increases in ratio a³:b³.

Hemisphere Surface Areas

Curved Surface Area (CSA) = 2 × π × r² and Total Surface Area (TSA) = 3 × π × r².

Scaling Factor Rule

For any 3D object, if linear dimensions are scaled by factor 'k', area scales by k² and volume scales by k³.

Shortcut Trick

Ratio CSA/TSA = 1/3 ⇒ 2πrh / [2πr(h+r)] = 1/3 ⇒ h/(h+r) = 1/3 ⇒ r = 2h.

Volume = πr²h = π(2h)²h = 4πh³ = 4312.

4 × (22/7) × h³ = 4312 ⇒ h³ = 343 ⇒ h = 7 cm.

Radius r = 2 × 7 = 14 cm.

CSA = 2 × (22/7) × 14 × 7 = 616 cm².

∴ The correct answer is 616 cm².

Alternate Method

Given: Volume = 4312 cm³, Curved Surface Area (CSA) = 1/3 Total Surface Area (TSA).

Formula Used: Volume = πr²h; CSA = 2πrh; TSA = 2πr(h+r).

⇒ CSA = 1/3 TSA

⇒ 2πrh = 1/3 [2πr(h + r)]

⇒ 3h = h + r ⇒ r = 2h

⇒ Volume = πr²h = 4312

⇒ (22/7) × (2h)² × h = 4312

⇒ (22/7) × 4h³ = 4312

⇒ h³ = (4312 × 7) / 88

⇒ h³ = 343 ⇒ h = 7 cm

⇒ r = 2 × 7 = 14 cm

⇒ CSA = 2 × (22/7) × 14 × 7 = 616

∴ The correct answer is 616 cm².

Additional Information

Volume of Cylinder

It is the capacity of the cylinder, calculated as Base Area × Height, V = πr²h.

Surface Areas of Cylinder

Curved Surface Area (CSA) = 2πrh. Total Surface Area (TSA) includes the two circular bases, TSA = 2πrh + 2πr².

Special Case: CSA = 1/2 TSA

If CSA is half of TSA, then 2πrh = 2πr², which means the height of the cylinder is equal to its radius (h = r).

Shortcut Trick

From 8x + y = 15, we get y = 15 − 8x.

Substitute in 7x + 11y = 3: 7x + 11(15 − 8x) = 3 ⇒ 7x + 165 − 88x = 3 ⇒ −81x = −162 ⇒ x = 2.

Then y = 15 − 8(2) = −1. Point P is (2, −1).

Check options: 3(2) + 5(−1) = 6 − 5 = 1. This satisfies 3x + 5y = 1.

∴ The correct answer is 3x + 5y = 1.

Alternate Method

Given:

Equation 1: 7x + 11y = 3

Equation 2: 8x + y = 15

Calculation:

⇒ From 8x + y = 15, we get y = 15 − 8x

⇒ Substitute y in 7x + 11y = 3:

⇒ 7x + 11(15 − 8x) = 3

⇒ 7x + 165 − 88x = 3

⇒ −81x = 3 − 165

⇒ −81x = −162

⇒ x = 162 ÷ 81 = 2

⇒ Substitute x = 2 in y = 15 − 8x:

⇒ y = 15 − 8(2) = 15 − 16 = −1

⇒ Intersection point P = (2, −1)

⇒ Now, test P(2, −1) in the options:

⇒ (1) 2(2) + (−1) = 3 ≠ 2

⇒ (2) 2(2) − (−1) = 5 ≠ 1

⇒ (3) 3(2) + 5(−1) = 6 − 5 = 1 (Satisfied)

⇒ (4) 3(2) + 2(−1) = 4 ≠ 3

∴ The correct answer is 3x + 5y = 1.

Additional Information

Intersection of Lines

The coordinates of the point of intersection of two lines are the values of x and y that satisfy both linear equations simultaneously.

Point on a Line Rule

A point (x₁, y₁) lies on the graph of an equation if and only if substituting x = x₁ and y = y₁ into the equation results in a true statement (LHS = RHS).

System of Linear Equations

For a₁x + b₁y = c₁ and a₂x + b₂y = c₂, a unique solution exists if a₁/a₂ ≠ b₁/b₂.

Shortcut Trick

CP : MP ratio = (100 − Discount%) : (100 + Profit%) = 75 : 120 = 5 : 8.

Let initial CP = 500 and MP = 800.

New CP = 500 − 15% of 500 = 425.

New SP (to keep 20% profit) = 425 × 1.2 = 510.

New Discount = 800 − 510 = 290.

New Discount % = (290 ÷ 800) × 100 = 36.25%.

∴ The correct answer is 36.25%.

Alternate Method

Given: Initial Profit = 20%, Initial Discount = 25%, CP decrease = 15%.

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

⇒ Let the initial cost price (CP₁) be ₹100.

⇒ Initial selling price (SP₁) = 100 + 20% of 100 = ₹120.

⇒ Initial marked price (MP) = (SP₁ × 100) ÷ (100 − 25) = (120 × 100) ÷ 75 = ₹160.

⇒ New cost price (CP₂) = 100 − 15% of 100 = ₹85.

⇒ Required new profit percentage = 20%.

⇒ New required selling price (SP₂) = 85 + 20% of 85 = 85 + 17 = ₹102.

⇒ New discount amount = MP − SP₂ = 160 − 102 = ₹58.

⇒ New discount percentage = (58 ÷ 160) × 100 = 36.25%.

∴ The correct answer is 36.25%.

Additional Information

CP and MP Relationship

The ratio of Cost Price to Marked Price is always CP/MP = (100 − D%) / (100 + P%), where D is discount and P is profit.

Successive Discounts

If two discounts d₁% and d₂% are given, the single equivalent discount is (d₁ + d₂ − (d₁ × d₂)/100)%.

Markup Percentage

Markup % = [(MP − CP) / CP] × 100. It represents how much the price is raised above the cost price.

Calculation:

Total production in B in 2015, 2016, and 2018 = 150 + 180 + 300

⇒ 630

Average production = 630/3 = 210

Now,

Total production in C in 2015, and 2017 = 200 + 250

⇒ 450

Average production = 450/2 = 225

Difference = 225 - 210 = 15

Required % = (15/225) × 100

⇒ $6\frac{2}{3}$%

∴ Required answer is $6\frac{2}{3}$%​​

Shortcut Trick

Total Surface Area (TSA) of solid hemisphere = 3πr² = 942 cm²

⇒ πr² = 942 ÷ 3 = 314

⇒ 3.14 × r² = 314 ⇒ r² = 100 ⇒ r = 10 cm

Volume = (2/3)πr³ = (2/3) × 3.14 × 1000

Volume = 6280 ÷ 3 ≈ 2093.33 cm³

∴ The correct answer is 2093 cm³.

Alternate Method

Given:

Total Surface Area (TSA) = 942 cm²

π = 3.14

Formula Used:

TSA of solid hemisphere = 3πr²

Volume of hemisphere = (2/3)πr³

Calculations:

⇒ 3 × 3.14 × r² = 942

⇒ 9.42 × r² = 942

⇒ r² = 942 ÷ 9.42

⇒ r² = 100

⇒ r = √100 = 10 cm

⇒ Volume = (2/3) × 3.14 × (10)³

⇒ Volume = (2 × 3.14 × 1000) ÷ 3

⇒ Volume = 6280 ÷ 3

⇒ Volume = 2093.333... cm³

Rounding to the closest integer:

⇒ Volume ≈ 2093 cm³

∴ The correct answer is 2093 cm³.

Additional Information

Curved Surface Area of Hemisphere

The surface area of the curved part of a hemisphere is 2πr².

Full Sphere Properties

For a full sphere of radius r: Total Surface Area = 4πr² and Volume = (4/3)πr³.

Ratio of TSA to Volume for Hemisphere

The ratio of TSA to Volume for a solid hemisphere is (3πr²) : ((2/3)πr³) = 9 : 2r.