SSC CGL Tier 2 Quant Previous Year Paper (Held On: 13 September 2019)

Shortcut Trick Ratio of Base to Height for 60° is 1 : √3 and for 30° is √3 : 1. To keep height constant (√3), ratios become: (1 : √3) and (3 : √3). Total distance ratio = 1 + 3 = 4 units. 4 units = 84√3 m ⇒ 1 unit = 21√3 m. Height = √3 units = 21√3 × √3 = 21 × 3 = 63 m. ∴ The correct answer is 63 m.

Alternate Method Given: Angle of elevation from P (θ1) = 60° Angle of elevation from Q (θ2) = 30° Total distance PQ = 84√3 m Formula Used: Height of pole (h) = d ÷ (cot θ1 + cot θ2) [When points are on opposite sides]

Let height of the pole AB be h m. In right ▵ABP: ⇒ cot 60° = BP ÷ AB ⇒ 1 ÷ √3 = BP ÷ h ⇒ BP = h ÷ √3 In right ▵ABQ: ⇒ cot 30° = BQ ÷ AB ⇒ √3 = BQ ÷ h ⇒ BQ = h√3 Given PQ = BP + BQ = 84√3 m: ⇒ (h ÷ √3) + h√3 = 84√3 ⇒ (h + 3h) ÷ √3 = 84√3 ⇒ 4h = 84 × 3 ⇒ h = 21 × 3 ⇒ h = 63 m ∴ The correct answer is 63 m.

Additional Information Distance Formula (Opposite Sides) If two points are on opposite sides of a pole of height h, then the distance between them is d = h(cot α + cot β). Distance Formula (Same Side) If two points are on the same side of a pole, then the distance between them is d = h(cot β − cot α), where α > β. Standard Trigonometric Ratios tan 30° = 1/√3, tan 45° = 1, tan 60° = √3. cot θ = 1/tan θ.

Shortcut Trick

Cyclicity of 3 mod 5 is 4 (3¹=3, 3²=4, 3³=2, 3⁴=1). Since 61284 is divisible by 4, x = 1.

Any power of 4 divided by 6 always leaves a remainder of 4 (4¹=4, 4²=16≡4, 4³=64≡4), so y = 4.

Substitute values in (2x − y) → 2(1) − 4 = −2.

∴ The correct answer is −2.

Alternate Method

Given: x = Remainder (3⁶¹²⁸⁴ ÷ 5) and y = Remainder (4⁹⁶ ÷ 6).

Formula Used: Euler's Totient Theorem and Remainder Properties.

⇒ For x: 5 is prime, so Totient φ(5) = 5 − 1 = 4.

⇒ According to Euler's Theorem, 3⁴ ≡ 1 (mod 5).

⇒ 3⁶¹²⁸⁴ = (3⁴)¹⁵³²¹ ≡ (1)¹⁵³²¹ ≡ 1 (mod 5).

⇒ Therefore, x = 1.

⇒ For y: Check powers of 4 divided by 6.

⇒ 4¹ ÷ 6 leaves Remainder 4.

⇒ 4² = 16 ÷ 6 leaves Remainder 4.

⇒ Since 4ⁿ ≡ 4 (mod 6) for all positive integers n, y = 4.

⇒ Required value = 2x − y = 2(1) − 4 = −2.

∴ The correct answer is −2.

Additional Information

Fermat's Little Theorem

If p is a prime number and a is not divisible by p, then a^p−1 ≡ 1 (mod p).

Powers of 4 divided by 6

The remainder of 4ⁿ when divided by 6 is always 4 for all n ≥ 1, regardless of the exponent value.

Euler's Totient Theorem

For any integer a and n that are coprime, a^φ(n) ≡ 1 (mod n), where φ(n) is the number of integers less than n that are coprime to n.

Shortcut Trick

Put θ = 60° in the given expression.

sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3, cot 60° = 1/√3

Expression = [(√3/2 − 1/2) × (1 + √3 + 1/√3)] ÷ [1 + (√3/2 × 1/2)]

Expression = [((√3−1)/2) × ((4+√3)/√3)] ÷ [(4+√3)/4] = 2 − 2/√3

Checking Option (1): sec 60° − cosec 60° = 2 − 2/√3 (Matches)

∴ The correct answer is secθ − cosecθ.

Alternate Method

Given: Expression = [ (sinθ − cosθ)(1 + tanθ + cotθ) ] ÷ (1 + sinθcosθ)

Formula: tanθ = sinθ/cosθ and cotθ = cosθ/sinθ

⇒ Simplifying the term (1 + tanθ + cotθ):

⇒ 1 + (sinθ/cosθ) + (cosθ/sinθ) = 1 + (sin²θ + cos²θ)/(sinθcosθ)

⇒ 1 + 1/(sinθcosθ) = (sinθcosθ + 1)/(sinθcosθ)

⇒ Substituting this back into the main expression:

⇒ [ (sinθ − cosθ) × (1 + sinθcosθ)/(sinθcosθ) ] ÷ (1 + sinθcosθ)

⇒ Cancelling (1 + sinθcosθ) from numerator and denominator:

⇒ (sinθ − cosθ) ÷ (sinθcosθ)

⇒ sinθ/(sinθcosθ) − cosθ/(sinθcosθ)

⇒ 1/cosθ − 1/sinθ = secθ − cosecθ

∴ The correct answer is secθ − cosecθ.

Additional Information

Trigonometric Identities

Standard Pythagorean identity: sin²θ + cos²θ = 1. Also, sec²θ − tan²θ = 1.

Reciprocal Relations

secθ = 1/cosθ, cosecθ = 1/sinθ, and cotθ = 1/tanθ.

Compound Relation

The expression tanθ + cotθ simplifies directly to 1/(sinθcosθ) or secθcosecθ.

Shortcut Trick

Speed ratio (Usual : New) = 100% : 60% = 5 : 3

Since distance is constant, Time ratio (Usual : New) = 3 : 5

Difference in time ratio = 5 − 3 = 2 units

Delay given = 1 hour 40 minutes = 100 minutes

2 units = 100 minutes ⇒ 1 unit = 50 minutes

Usual time (3 units) = 3 × 50 = 150 minutes = 2 1/2 hours

∴ The correct answer is 2 1/2 hours.

Alternate Method

Given:

New speed = 60% of usual speed

Time delay = 1 hour 40 minutes = 1 + 40÷60 hours = 1 + 2÷3 = 5÷3 hours

Formula Used:

Distance = Speed × Time

When Distance is constant, Speed₁ × Time₁ = Speed₂ × Time₂

Calculations:

Let usual speed be S and usual time be T.

New speed = 0.6S

New time = T + 5÷3

⇒ S × T = 0.6S × (T + 5÷3)

⇒ T = 0.6 × (T + 5÷3)

⇒ T = 0.6T + (0.6 × 5÷3)

⇒ T − 0.6T = 3÷3

⇒ 0.4T = 1

⇒ T = 1 ÷ 0.4

⇒ T = 10 ÷ 4

⇒ T = 2.5 hours

⇒ T = 2 1/2 hours

∴ The correct answer is 2 1/2 hours.

Additional Information

Speed-Time Inversity Rule

For a constant distance, if speed becomes a÷b of its original value, time taken becomes b÷a of the original time.

Average Speed Formula

When a person covers two equal distances at speeds x and y, the average speed is (2 × x × y) ÷ (x + y).

Speed Unit Conversion

To convert kmph to m/s, multiply by 5÷18. To convert m/s to kmph, multiply by 18÷5.

Shortcut Trick

Convert terms using complementary angle rules: sin(90° − θ) = cos θ and cosec(90° − θ) = sec θ.

sin 26° = cos(90° − 26°) = cos 64°, so sin² 64° + cos 64° × cos 64° = sin² 64° + cos² 64° = 1.

cosec 47° = sec(90° − 47°) = sec 43°, so 2 cos 43° × sec 43° = 2(1) = 2.

Total value = 1 + 2 = 3.

∴ The correct answer is 3.

Alternate Method

Given: sin² 64° + cos 64° sin 26° + 2 cos 43° cosec 47°

Formula Used: sin(90° − θ) = cos θ; cosec(90° − θ) = sec θ; sin² θ + cos² θ = 1; cos θ × sec θ = 1.

⇒ Expression = sin² 64° + cos 64° sin(90° − 64°) + 2 cos 43° cosec(90° − 43°)

⇒ Expression = sin² 64° + cos 64° cos 64° + 2 cos 43° sec 43°

⇒ Expression = (sin² 64° + cos² 64°) + 2 × (cos 43° × 1 ÷ cos 43°)

⇒ Expression = 1 + 2 × 1

⇒ Expression = 1 + 2 = 3

∴ The correct answer is 3.

Additional Information

Complementary Angle Identities

Trigonometric ratios of complementary angles: sin(90° − θ) = cos θ, tan(90° − θ) = cot θ, and sec(90° − θ) = cosec θ.

Pythagorean Identities

Fundamental identities include sin² θ + cos² θ = 1, 1 + tan² θ = sec² θ, and 1 + cot² θ = cosec² θ.

Reciprocal Relations

Standard reciprocals: sin θ × cosec θ = 1, cos θ × sec θ = 1, and tan θ × cot θ = 1.

Shortcut Trick

Put θ = 45° as it satisfies the expression constraints.

cot 45° = 1, cosec 45° = √2, sin 45° = 1/√2, cos 45° = 1/√2, sec 45° = √2

Expression = (1 + 1 − √2) × (1 + 1/√2 + 1/√2) × √2

⇒ (2 − √2) × (1 + √2) × √2 = (2 + 2√2 − √2 − 2) × √2

⇒ √2 × √2 = 2

∴ The correct answer is 2.

Alternate Method

Given: (1 + cot θ − cosec θ) (1 + cos θ + sin θ) sec θ

Formula Used: cot θ = cos θ / sin θ | cosec θ = 1 / sin θ | sec θ = 1 / cos θ | (a − b)(a + b) = a² − b²

⇒ (1 + cos θ/sin θ − 1/sin θ) × (1 + cos θ + sin θ) × (1/cos θ)

⇒ [(sin θ + cos θ − 1) / sin θ] × (sin θ + cos θ + 1) × (1/cos θ)

⇒ [(sin θ + cos θ)² − 1²] / (sin θ × cos θ)

⇒ [sin² θ + cos² θ + 2 sin θ cos θ − 1] / (sin θ cos θ)

⇒ [1 + 2 sin θ cos θ − 1] / (sin θ cos θ) [Using sin² θ + cos² θ = 1]

⇒ 2 sin θ cos θ / (sin θ cos θ) = 2

∴ The correct answer is 2.

Additional Information

Pythagorean Identities

Fundamental relations in trigonometry: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Reciprocal Identities

Basic definitions: cosecθ = 1/sinθ, secθ = 1/cosθ, and cotθ = 1/tanθ.

Value Substitution Method

For simplifying trigonometric expressions, substituting standard angles like 45°, 30°, or 60° is often the fastest way to verify identities.

Shortcut Trick

Square of radius r² = (5√13)² = 25 × 13 = 325.

Distance of chord 1 from center d₁ = √(325 − 10²) = √(225) = 15 cm.

Distance of chord 2 from center d₂ = √(325 − 6²) = √(289) = 17 cm.

Distance between chords (same side) = |d₂ − d₁| = 17 − 15 = 2 cm.

∴ The correct answer is 2 cm.

Alternate Method

Given:

Length of chord AB = 20 cm ⇒ half length = 10 cm.

Length of chord CD = 12 cm ⇒ half length = 6 cm.

Radius of the circle (r) = 5√13 cm.

Formula used: Distance of chord from center d = √(r² − (chord/2)²)

⇒ r² = (5√13)² = 325

⇒ Distance of 20 cm chord (d₁) = √(325 − 10²)

⇒ d₁ = √(325 − 100) = √225 = 15 cm

⇒ Distance of 12 cm chord (d₂) = √(325 − 6²)

⇒ d₂ = √(325 − 36) = √289 = 17 cm

⇒ Distance between chords = d₂ − d₁

⇒ Distance = 17 − 15 = 2 cm

∴ The correct answer is 2 cm.

Additional Information

Distance between parallel chords on opposite sides

If two parallel chords are on opposite sides of the center, the distance between them is the sum of their individual distances from the center (d₁ + d₂).

Perpendicular bisector property

The perpendicular drawn from the center of a circle to a chord always bisects the chord. This allows for the application of Pythagoras' theorem: r² = d² + (chord/2)².

Chord length and distance relationship

In a circle, the longer chord is always closer to the center, whereas the shorter chord is further away from the center.

Shortcut Trick

Inner radius (r) = 24 ÷ 2 = 12 cm; Outer radius (R) = 12 + 0.5 = 12.5 cm.

Total Surface Area (TSA) of hollow hemisphere = 3πR² + πr².

TSA = 3π(12.5)² + π(12)².

TSA = 3π × 156.25 + 144π = 468.75π + 144π = 612.75π.

∴ The correct answer is 612.75π cm².

Alternate Method

Given:

Internal diameter = 24 cm ⇒ Internal radius (r) = 12 cm.

Thickness (t) = 0.5 cm ⇒ External radius (R) = 12 + 0.5 = 12.5 cm.

Formula used: TSA of hollow hemisphere = Outer CSA + Inner CSA + Area of the ring.

Calculations:

⇒ Total Surface Area = 2πR² + 2πr² + π(R² − r²)

⇒ Total Surface Area = π[2R² + 2r² + R² − r²]

⇒ Total Surface Area = π[3R² + r²]

⇒ Total Surface Area = π[3 × (12.5)² + (12)²]

⇒ Total Surface Area = π[3 × 156.25 + 144]

⇒ Total Surface Area = π[468.75 + 144]

⇒ Total Surface Area = 612.75π cm².

∴ The correct answer is 612.75π cm².

Additional Information

Total Surface Area of Solid Hemisphere

For a solid hemisphere, the TSA includes the curved surface and the flat base: TSA = 3πr².

Volume of Hollow Hemisphere

The amount of material or capacity of a hollow hemispherical shell: Volume = (2/3)π(R³ − r³).

Curved Surface Area (CSA)

For any hemisphere, the area of the curved part only is CSA = 2πr².

Shortcut Trick

Total time for A = 15 × 3 = 45 days; Total time for B = 18 ÷ 0.75 = 24 days.

Total Work = LCM(45, 24, 36) = 360 units.

Efficiencies: A = 8, B = 15, C = 10.

Work done by B and C in 8 days = (15 + 10) × 8 = 200 units.

Remaining work = 360 − 200 = 160 units.

Time for A to complete remaining work = 160 ÷ 8 = 20 days.

∴ The correct answer is 20 days.

Alternate Method

Given:

A completes 1/3 work in 15 days.

B completes 75% work in 18 days.

C completes full work in 36 days.

Formula Used:

Efficiency = Total Work ÷ Time

Remaining Work = Total Work − Completed Work

⇒ Total days for A = 15 × 3 = 45 days.

⇒ Total days for B = 18 × (100 / 75) = 18 × (4 / 3) = 24 days.

⇒ LCM of 45, 24, and 36 is 360 (Total Work).

⇒ Combined efficiency of B and C = 15 + 10 = 25 units/day.

⇒ Work done by B and C in 8 days = 25 × 8 = 200 units.

⇒ Remaining work = 360 − 200 = 160 units.

⇒ Time taken by A to finish 160 units = 160 ÷ 8 = 20 days.

∴ The correct answer is 20 days.

Additional Information

Work and Efficiency Relation

Efficiency is inversely proportional to the time taken to complete a work. If the ratio of time taken by A and B is x : y, then the ratio of their efficiencies is y : x.

Partial Work Rule

If a person completes 1/n part of a work in 'd' days, then the total time taken to complete the entire work is n × d days.

Percentage to Days Conversion

If p% of work is done in 't' days, total time = (t × 100) ÷ p. For 75% work in 18 days, Total Time = (18 × 100) ÷ 75 = 24 days.

Shortcut Trick

Convert percentage changes to ratios: 20% increase = 1/5 increase (5 → 6).

8 1/3% decrease = 1/12 decrease (12 → 11).

Ratio of Expenditure = (Price Ratio) × (Consumption Ratio).

Initial : Final = (5 × 12) : (6 × 11) = 60 : 66.

Percentage change = (6 / 60) × 100 = 10%.

∴ The correct answer is Increase by 10%.

Alternate Method

Given:

Increase in Price = 20%

Decrease in Consumption = 8 1/3% = 25/3%

Formula: Expenditure = Price × Consumption

⇒ Let original price be 100 and consumption be 100.

⇒ Initial Expenditure = 100 × 100 = 10,000.

⇒ New Price = 100 + 20% of 100 = 120.

⇒ New Consumption = 100 − 25/3% = (300 − 25) / 3 = 275 / 3.

⇒ New Expenditure = 120 × (275 / 3) = 40 × 275 = 11,000.

⇒ Difference in Expenditure = 11,000 − 10,000 = 1,000 (Increase).

⇒ % Increase = (1,000 / 10,000) × 100 = 10%.

∴ The correct answer is Increase by 10%.

Additional Information

Successive Percentage Change Formula

Net % change = a + b + (a × b) / 100, where a and b are percentage changes (use negative for decrease).

Price-Consumption Relation

If expenditure is constant, Price is inversely proportional to Consumption (Price ∝ 1 / Consumption).

Fraction Equivalents

8 1/3% = 1/12, 16 2/3% = 1/6, 12.5% = 1/8. Memorizing these helps in fast ratio calculations.

Shortcut Trick

The circle passing through D, E, and F (where E and F are on extensions of AB and AC) is the A-excircle of ΔABC.

Length of tangent from vertex A to the excircle is equal to the semi-perimeter (s) of the triangle.

s = (AB + AC + BC) ÷ 2 = (10 + 8.6 + 6.4) ÷ 2 = 12.5 cm.

AE = AB + BE = s ⇒ 10 + BE = 12.5.

BE = 12.5 − 10 = 2.5 cm.

∴ The correct answer is 2.5 cm.

Alternate Method

Given: AB = 10 cm, AC = 8.6 cm, BC = 6.4 cm.

In ΔABC, let the circle touch BC at D, AB produced at E, and AC produced at F.

From the properties of tangents from an external point to a circle:

⇒ AE = AF (Tangents from point A)

⇒ BE = BD (Tangents from point B)

⇒ CF = CD (Tangents from point C)

⇒ Perimeter of ΔABC = AB + BC + AC

⇒ Perimeter = AB + (BD + CD) + AC

⇒ Perimeter = AB + (BE + CF) + AC

⇒ Perimeter = (AB + BE) + (AC + CF) = AE + AF

⇒ Since AE = AF, Perimeter = 2 × AE

⇒ 10 + 6.4 + 8.6 = 2 × AE

⇒ 25 = 2 × AE

⇒ AE = 12.5 cm

⇒ BE = AE − AB = 12.5 − 10 = 2.5 cm

∴ The correct answer is 2.5 cm.

Additional Information

Excircle Tangent Property

The distance from a vertex to the points of tangency of the excircle on the extended sides is always equal to the semi-perimeter s = (a + b + c) ÷ 2.

Incircle Tangent Property

The distance from a vertex to the points of contact of the incircle is (s − side opposite to that vertex). For vertex A, the distance is s − a.

Contact Point on Side

For the excircle opposite to vertex A, the distance from vertex B to the contact point D on side BC is BD = s − c and CD = s − b.

Shortcut Trick

Apply Apollonius Theorem: AB² + AD² = 2(AC² + (BD/2)²)

10² + 12² = 2(9² + (BD/2)²)

244 = 2(81 + (BD/2)²) ⇒ 122 = 81 + (BD/2)²

(BD/2)² = 41 ⇒ BD/2 = √41

BD = 2√41 cm

∴ The correct answer is 2√41 cm.

Alternate Method

Given: AB = 10 cm, AD = 12 cm, AC = 9 cm, and C is the midpoint of BD.

Formula: Apollonius Theorem: In ΔABD, if AC is the median to BD, then AB² + AD² = 2(AC² + BC²).

⇒ AB² + AD² = 2(AC² + BC²)

⇒ 10² + 12² = 2(9² + BC²)

⇒ 100 + 144 = 2(81 + BC²)

⇒ 244 = 162 + 2BC²

⇒ 2BC² = 244 − 162

⇒ 2BC² = 82

⇒ BC² = 41

⇒ BC = √41 cm

⇒ BD = 2 × BC = 2√41 cm

∴ The correct answer is 2√41 cm.

Additional Information

Apollonius Theorem

It relates the lengths of the sides of a triangle to the length of its median: Sum of squares of two sides = 2 × (square of median + square of half the third side).

Median and Area

A median of a triangle divides it into two triangles of equal area.

Centroid Property

The centroid of a triangle (intersection of all medians) divides each median in the ratio 2 : 1 from the vertex.

Shortcut Trick

Ratio of Students (Girls : Boys) = 35% : 65% = 7 : 13

Ratio of Average Age (Girls : Boys) = 5 : 6

Weighted Average Units = (7 × 5 + 13 × 6) ÷ (7 + 13) = (35 + 78) ÷ 20 = 113 ÷ 20 = 5.65 units

Given, 5.65 units = 13.56 years ⇒ 1 unit = 13.56 ÷ 5.65 = 2.4 years

Average age of girls = 5 units = 5 × 2.4 = 12 years

∴ The correct answer is 12 years.

Alternate Method

Given:

Total students = 120

Percentage of girls = 35%

Average age of 120 students = 13.56 years

Ratio of average age (Boys : Girls) = 6 : 5

Calculations:

⇒ Number of girls = 35% of 120 = 42

⇒ Number of boys = 120 − 42 = 78

⇒ Let Average age of girls = 5x and Average age of boys = 6x

⇒ Total age = Number of students × Combined Average

⇒ (42 × 5x) + (78 × 6x) = 120 × 13.56

⇒ 210x + 468x = 1627.2

⇒ 678x = 1627.2

⇒ x = 1627.2 ÷ 678 = 2.4

⇒ Average age of girls = 5x = 5 × 2.4 = 12 years

∴ The correct answer is 12 years.

Additional Information

Weighted Average Formula

The combined average of two groups is given by A = (n₁a₁ + n₂a₂) ÷ (n₁ + n₂), where n is the number of items and a is the average.

Ratio Rule in Averages

When calculating a weighted average, you can use the ratio of the quantities (n₁ : n₂) instead of the actual counts to simplify calculations.

Alligation Method

Averages can also be solved using the Alligation cross-method: (Higher Average − Mixed Average) ÷ (Mixed Average − Lower Average) = Quantity Ratio of lower to higher.

Shortcut Trick

Given AD : AB = 3 : 8, let AB = 8 units and AD = 3 units.

BD = AB − AD = 8 − 3 = 5 units.

Since DE || AC, ΔBDE ≈ ΔBAC. The ratio of their areas is the square of the ratio of their corresponding sides.

Area(ΔBDE) : Area(ΔBAC) = (BD)² : (AB)² = 5² : 8² = 25 : 64.

Area(quadrilateral DECA) = Area(ΔBAC) − Area(ΔBDE) = 64 − 25 = 39.

Required Ratio = Area(ΔBDE) : Area(quadrilateral DECA) = 25 : 39.

∴ The correct answer is 25 : 39.

Alternate Method

Given: AD : AB = 3 : 8 and DE || AC.

Formula Used: Ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Since DE || AC, ∠BDE = ∠BAC (Corresponding angles) and ∠BED = ∠BCA.

⇒ ΔBDE ≈ ΔBAC (By AA similarity)

⇒ Area(ΔBDE) / Area(ΔBAC) = (BD / AB)²

⇒ BD = AB − AD

⇒ BD / AB = (AB − AD) / AB = 1 − (AD / AB)

⇒ BD / AB = 1 − (3 / 8) = 5 / 8

⇒ Area(ΔBDE) / Area(ΔBAC) = (5 / 8)² = 25 / 64

⇒ Let Area(ΔBDE) = 25x and Area(ΔBAC) = 64x

⇒ Area(DECA) = Area(ΔBAC) − Area(ΔBDE) = 64x − 25x = 39x

⇒ Area(ΔBDE) : Area(quadrilateral DECA) = 25x : 39x = 25 : 39

∴ The correct answer is 25 : 39.

Additional Information

Area of Similar Triangles

If two triangles are similar, the ratio of their areas is equal to the square of the ratio of any two corresponding sides, altitudes, medians, or angle bisectors.

Basic Proportionality Theorem (Thales Theorem)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio: BD/DA = BE/EC.

Scale Factor

If the scale factor of two similar figures is k (side ratio), then the ratio of their perimeters is k and the ratio of their areas is k².

Shortcut Trick

Simplify x: ⁶√27 = (3³)^1/6 = 3^1/2 = √3 and √(6 3/4) = √(27/4) = 3√3/2.

x = √3 − 1.5√3 = −0.5√3 ⇒ x² = (−√3/2)² = 3/4.

Simplify y: Ratio of coefficients of √5 ⇒ y = (3 + 11 + 7) ÷ (4 + 5) = 21/9 = 7/3.

y² = (7/3)² = 49/9.

x² + y² = 3/4 + 49/9 = (27 + 196) ÷ 36 = 223/36.

∴ The correct answer is 223/36.

Alternate Method

Given:

x = ⁶√27 − √(6 3/4)

y = (√45 + √605 + √245) ÷ (√80 + √125)

Formula Used:

Sum of squares = x² + y²; √(a²b) = a√b; (a/b)ⁿ = aⁿ/bⁿ

Calculations:

⇒ x = (3³)^1/6 − √(27/4)

⇒ x = 3^(3/6) − √(9 × 3) ÷ √4

⇒ x = 3^1/2 − 3√3/2

⇒ x = √3 − 1.5√3 = −0.5√3

⇒ x² = (−√3/2)² = 3/4

⇒ y = (√(9 × 5) + √(121 × 5) + √(49 × 5)) ÷ (√(16 × 5) + √(25 × 5))

⇒ y = (3√5 + 11√5 + 7√5) ÷ (4√5 + 5√5)

⇒ y = (21√5) ÷ (9√5)

⇒ y = 21/9 = 7/3

⇒ y² = (7/3)² = 49/9

⇒ x² + y² = 3/4 + 49/9

⇒ x² + y² = (27 + 196) ÷ 36 = 223/36

∴ The correct answer is 223/36.

Additional Information

Properties of Surds

An n-th root of a number a can be written as a^1/n. Surds can be added or subtracted only if they are like surds (same radicand and index).

Laws of Indices

The rule (a^m)ⁿ = a^m×n is crucial for simplifying higher-order roots into square roots, as seen with ⁶√27 = 3^3/6 = 3^1/2.

Simplifying Fractions with Surds

When a fraction contains like surds in both numerator and denominator, the surd term can be factored out and cancelled to simplify the expression to a rational number.

Shortcut Trick

In ΔPQR, if PS ⊥ QR and the condition PQ + QS = SR is satisfied, then ∠Q = 2 × ∠R.

Given ∠P = 120°.

Using the angle sum property: ∠P + ∠Q + ∠R = 180°

120° + 2∠R + ∠R = 180° ⇒ 3∠R = 60° ⇒ ∠R = 20°.

Therefore, ∠Q = 2 × 20° = 40°.

∴ The correct answer is 40°.

Alternate Method

Given: ∠P = 120°, PS ⊥ QR, PQ + QS = SR.

Formula Used: Angle sum property of triangle = 180°; SAS Congruence; Exterior angle theorem.

⇒ Let T be a point on SR such that QS = ST.

⇒ In ΔPQS and ΔPTS: PS is common, ∠PSQ = ∠PST = 90°, and QS = ST.

⇒ ∴ ΔPQS ≅ ΔPTS by SAS congruence property.

⇒ This implies PQ = PT and ∠PQS = ∠PTS (Corresponding parts of congruent triangles).

⇒ Given PQ + QS = SR. Since SR = ST + TR and we assumed QS = ST, then PQ + ST = ST + TR ⇒ PQ = TR.

⇒ Since PQ = PT and PQ = TR, then PT = TR.

⇒ In ΔPTR, since PT = TR, ∠TPR = ∠TRP. Let ∠TRP = x.

⇒ Exterior ∠PTS = ∠TPR + ∠TRP = x + x = 2x.

⇒ Since ∠Q = ∠PTS, then ∠Q = 2x.

⇒ In ΔPQR: ∠P + ∠Q + ∠R = 180°

⇒ 120° + 2x + x = 180° ⇒ 3x = 60° ⇒ x = 20°.

⇒ ∠Q = 2x = 2 × 20° = 40°.

∴ The correct answer is 40°.

Additional Information

Triangle Angle Sum Theorem

The sum of interior angles of any triangle is always 180° (∠A + ∠B + ∠C = 180°).

SAS Congruence Criterion

If two sides and the included angle of one triangle are equal to the corresponding parts of another triangle, the triangles are congruent.

Exterior Angle Theorem

An exterior angle of a triangle is equal to the sum of its two opposite interior angles.

Isosceles Triangle Theorem

If two sides of a triangle are equal, then the angles opposite to those sides are also equal.

Given:

25% of $\frac{x}{2}$ = 2.5 × (30% of $\frac{y}{4}$)

Formula used:

Percentage more = $\frac{x - y}{y} \times 100$

Calculations:

$\frac{25}{100} \times \frac{x}{2} = 2.5 \times \frac{30}{100} \times \frac{y}{4}$

⇒ $\frac{1}{4} \times \frac{x}{2} = \frac{5}{2} \times \frac{3}{10} \times \frac{y}{4}$

⇒ $\frac{x}{8} = \frac{15y}{80}$

⇒ $\frac{x}{8} = \frac{3y}{16}$

⇒ $x = \frac{3y \times 8}{16}$

⇒ $x = \frac{3}{2}y$

⇒ x = 1.5y

⇒ Percentage more = (1.5 - 1) × 100

⇒ Percentage more = 50%

∴ The correct answer is option (4).

Shortcut Trick

CP of mixture = 29.90 ÷ 1.15 = ₹26

Using Alligation ratio: (Mean CP − CP₂) : (CP₁ − Mean CP)

Ratio = (26 − 16) : (28 − 26) = 10 : 2 = 5 : 1

Salt B (1 unit) = 39.6 kg

Salt A (5 units) = 5 × 39.6 = 198 kg

∴ The correct answer is 198 kg.

Alternate Method

Given:

CP of Salt A = ₹28/kg

CP of Salt B = ₹16/kg

Quantity of Salt B = 39.6 kg

Selling Price (SP) of mixture = ₹29.90

Profit percentage = 15%

Formula Used:

CP = [SP × 100] ÷ [100 + Profit%]

Alligation Ratio = (Cheaper CP difference) / (Dearer CP difference)

⇒ Mean CP = (29.90 × 100) ÷ (100 + 15)

⇒ Mean CP = 2990 ÷ 115 = ₹26

⇒ Ratio of Quantity A : Quantity B = (26 − 16) : (28 − 26)

⇒ Ratio = 10 : 2 = 5 : 1

⇒ Let quantity of Salt A be x kg.

⇒ x / 39.6 = 5 / 1

⇒ x = 39.6 × 5

⇒ x = 198 kg

∴ The correct answer is 198 kg.

Additional Information

Rule of Alligation

Quantity of Cheaper / Quantity of Dearer = (CP of Dearer − Mean CP) / (Mean CP − CP of Cheaper). All values must be in the same unit (all CP or all SP).

Cost Price from Selling Price

If there is a profit of P%, then CP = [SP ÷ (100 + P)] × 100. If there is a loss of L%, then CP = [SP ÷ (100 − L)] × 100.

Profit and Loss Basics

Profit = SP − CP | Loss = CP − SP | Profit% = (Profit ÷ CP) × 100. Always calculate profit or loss percentage on the Cost Price unless stated otherwise.

Shortcut Trick

For trigonometric expressions that are independent of θ, substitute a convenient value for θ.

Let θ = 0°.

Then, sin 0° = 0 and cos 0° = 1.

Numerator = 2(0⁶ + 1⁶) − 3(0⁴ + 1⁴) = 2(1) − 3(1) = −1.

Denominator = 1⁴ − 0⁴ − 2(1)² = 1 − 2 = −1.

Expression = −1 ÷ −1 = 1.

∴ The correct answer is 1.

Alternate Method

Given: [2(sin⁶θ + cos⁶θ) − 3(sin⁴θ + cos⁴θ)] ÷ [cos⁴θ − sin⁴θ − 2cos²θ]

Formula Used: sin²θ + cos²θ = 1; a³ + b³ = (a + b)(a² − ab + b²); a² − b² = (a + b)(a − b)

⇒ sin⁶θ + cos⁶θ = (sin²θ + cos²θ)(sin⁴θ − sin²θcos²θ + cos⁴θ) = 1(sin⁴θ + cos⁴θ − sin²θcos²θ)

⇒ sin⁴θ + cos⁴θ = (sin²θ + cos²θ)² − 2sin²θcos²θ = 1 − 2sin²θcos²θ

⇒ sin⁶θ + cos⁶θ = (1 − 2sin²θcos²θ) − sin²θcos²θ = 1 − 3sin²θcos²θ

⇒ Numerator = 2(1 − 3sin²θcos²θ) − 3(1 − 2sin²θcos²θ) = 2 − 6sin²θcos²θ − 3 + 6sin²θcos²θ = −1

⇒ Denominator = (cos²θ − sin²θ)(cos²θ + sin²θ) − 2cos²θ = (cos²θ − sin²θ)(1) − 2cos²θ

⇒ Denominator = −cos²θ − sin²θ = −(cos²θ + sin²θ) = −1

⇒ Expression = −1 ÷ −1 = 1

∴ The correct answer is 1.

Additional Information

Standard Trigonometric Identities

The fundamental identity is sin²θ + cos²θ = 1. From this, sin²θ = 1 − cos²θ and cos²θ = 1 − sin²θ.

Higher Order Power Identities

sin⁴θ + cos⁴θ = 1 − 2sin²θcos²θ and sin⁶θ + cos⁶θ = 1 − 3sin²θcos²θ. These are derived using (a+b)² and (a+b)³ expansions.

Difference of Squares

The term cos⁴θ − sin⁴θ is always equal to cos²θ − sin²θ because (cos²θ + sin²θ) = 1.

Shortcut Trick

Identify the algebraic form: Let a = 4.6 and b = 5.4. Then ab = 4.6 × 5.4 = 24.84.

The expression is (a⁴ + b⁴ + a²b²) ÷ (a² + b² + ab).

The identity a⁴ + a²b² + b⁴ = (a² + ab + b²)(a² − ab + b²) simplifies the expression to a² + b² − ab.

Calculation: (4.6)² + (5.4)² − 24.84 = 21.16 + 29.16 − 24.84 = 50.32 − 24.84 = 25.48.

∴ The correct answer is 25.48.

Alternate Method

Given: Expression = [(4.6)⁴ + (5.4)⁴ + (24.84)²] ÷ [(4.6)² + (5.4)² + 24.84]

Formula used: a⁴ + a²b² + b⁴ = (a² + ab + b²)(a² − ab + b²)

⇒ Let a = 4.6 and b = 5.4

⇒ a² = (4.6)² = 21.16

⇒ b² = (5.4)² = 29.16

⇒ ab = 4.6 × 5.4 = 24.84

⇒ Expression = [a⁴ + b⁴ + (ab)²] ÷ [a² + b² + ab]

⇒ Expression = [(a² + b² + ab)(a² + b² − ab)] ÷ (a² + b² + ab)

⇒ Expression = a² + b² − ab

⇒ Expression = 21.16 + 29.16 − 24.84

⇒ Expression = 50.32 − 24.84 = 25.48

∴ The correct answer is 25.48.

Additional Information

Sophie Germain Identity

The polynomial a⁴ + a²b² + b⁴ can always be factored into (a² + ab + b²)(a² − ab + b²).

BODMAS Rule

Always solve expressions in the order: Brackets, Of, Division, Multiplication, Addition, and Subtraction.

Square of a Sum

(a + b)² = a² + 2ab + b²; this helps in quickly finding a² + b² = (a + b)² − 2ab.

Given:

Total male employees = $\frac{5}{8}$

Non-technical males = $\frac{2}{5}$

Technical females = $\frac{2}{3}$

Formula used:

Technical workers = (Males × Technical fraction) + (Females × Technical fraction)

Calculation:

Let total employees = 1

Males = $\frac{5}{8}$

Females = $1 - \frac{5}{8} = \frac{3}{8}$

Technical males = $\frac{5}{8} \times (1 - \frac{2}{5})$

⇒ Technical males = $\frac{5}{8} \times \frac{3}{5} = \frac{3}{8}$

Technical females = $\frac{3}{8} \times \frac{2}{3}$

⇒ Technical females = $\frac{2}{8}$

Total technical workers = $\frac{3}{8} + \frac{2}{8}$

⇒ Total technical workers = $\frac{5}{8}$

∴ The correct answer is option (1).

Shortcut Trick

By Alligation, Ratio of quantities = (Overall % − Loss %) : (Profit % − Overall %)

Ratio = (6 − (−12)) : (8 − 6) = 18 : 2 = 9 : 1

Total units = 9 + 1 = 10 units → 80 articles

1 unit = 8 articles

Number of articles at 8% profit = 9 × 8 = 72

∴ The correct answer is 72.

Alternate Method

Given:

Total articles = 80

Profit on first part (P₁) = +8%

Loss on remaining part (P₂) = −12%

Overall profit (P_avg) = +6%

Formula Used:

Alligation Rule: Ratio = (P_avg − P₂) : (P₁ − P_avg)

Calculations:

⇒ Ratio of articles = (6 − (−12)) : (8 − 6)

⇒ Ratio = 18 : 2 = 9 : 1

⇒ Total article units = 9 + 1 = 10 units

⇒ Value of 10 units = 80 articles

⇒ Value of 1 unit = 80 ÷ 10 = 8 articles

⇒ Articles sold at 8% profit = 9 units × 8 = 72

∴ The correct answer is 72.

Additional Information

Alligation Rule in Profit and Loss

When two parts of a total quantity are sold at different profit or loss percentages, the ratio of the quantities is found by taking the cross-difference between the individual percentages and the overall average percentage.

Weighted Average Profit

Total Profit % = (n₁ × P₁ + n₂ × P₂) ÷ (n₁ + n₂), where n is the quantity and P is the profit/loss percentage.

Treatment of Loss

In alligation or profit formulas, always represent loss as a negative percentage value (e.g., 12% loss = −12%).

Shortcut Trick

Ratio of final Selling Price to Marked Price after 2 discounts = (1 − x/100)²

Selling Price (after 2 discounts) = 1500 − 587.40 = 912.60

Ratio = 912.60/1500 = 1521/2500 ⇒ (1 − x/100) = √(1521/2500) = 39/50

Selling Price after single discount = Marked Price × (39/50) = 1500 × (39/50) = 30 × 39 = 1170

∴ The correct answer is ₹1170.

Alternate Method

Given:

Marked Price (MP) = ₹1500

Total discount after two successive x% discounts = ₹587.40

Formula Used:

Final Selling Price = MP × (1 − x/100)ⁿ

Selling Price (Single Discount) = MP − (x% of MP)

Calculations:

⇒ Final Selling Price = 1500 − 587.40 = 912.60

⇒ 912.60 = 1500 × (1 − x/100)²

⇒ (1 − x/100)² = 912.60 / 1500

⇒ (1 − x/100)² = 9126 / 15000 = 1521 / 2500

⇒ (1 − x/100) = √(1521 / 2500) = 39 / 50

⇒ x/100 = 1 − 39/50 = 11/50

⇒ x = 22%

⇒ SP after single discount of x% = 1500 × (1 − 22/100)

⇒ SP = 1500 × (78/100) = 15 × 78

⇒ SP = 1170

∴ The correct answer is ₹1170.

Additional Information

Successive Discount Formula

For two successive discounts of a% and b%, the effective discount is (a + b − (a × b)/100)%.

Marked Price vs Selling Price

Selling Price (SP) = Marked Price (MP) − Discount. Discount is always calculated on the Marked Price unless stated otherwise.

Equal Successive Discounts

If n successive discounts of x% are given, then SP = MP × (1 − x/100)ⁿ. This is similar to the concept of depreciation.

Shortcut Trick

Ratio A:B = 11:10 and B:C = 6:5 (Since 20% = 1/5)

Combined Ratio A:B:C = (11×6) : (10×6) : (10×5) = 66 : 60 : 50

Difference (A − C) = 66 − 50 = 16 units

Given 16 units = 272 → 1 unit = 272 ÷ 16 = 17

Marks of B = 60 units = 60 × 17 = 1020

∴ The correct answer is 1020 marks.

Alternate Method

Given:

A = B + 10% of B

B = C + 20% of C

A − C = 272

Formula Used:

Percentage Increase: Result = Base × (100 + Increase %)/100

⇒ Let marks obtained by C = 100x

⇒ Marks obtained by B = 100x + 20% of 100x = 120x

⇒ Marks obtained by A = 120x + 10% of 120x = 132x

⇒ Difference A − C = 132x − 100x = 32x

⇒ 32x = 272

⇒ x = 272 ÷ 32 = 8.5

⇒ Marks obtained by B = 120 × 8.5 = 1020 marks

∴ The correct answer is 1020 marks.

Additional Information

Successive Percentage Change

When a value is increased by x% and then by y%, the net percentage change is given by [x + y + (xy ÷ 100)]%.

Percentage to Ratio Conversion

To simplify calculations, convert percentages to fractions: 10% = 1/10 (multiplier 11/10) and 20% = 1/5 (multiplier 6/5).

Base Shifting Rule

If A is x% more than B, B is [x ÷ (100 + x)] × 100% less than A. Always identify the correct base for percentage calculations.

Shortcut Trick

For any odd positive integer n, the expression (aⁿ + bⁿ) is always divisible by (a + b).

Here, a = 127, b = 97, and n = 97 (which is an odd number).

Sum = 127 + 97 = 224.

Check if 224 is divisible by 32: 224 ÷ 32 = 7.

Since the divisor 32 is a factor of 224, the remainder must be 0.

∴ The correct answer is 0.

Alternate Method

Given: Find the remainder when (127⁹⁷ + 97⁹⁷) is divided by 32.

⇒ Identity: aⁿ + bⁿ = (a + b)(aⁿ⁻¹ − aⁿ⁻²b + aⁿ⁻³b² − ... + bⁿ⁻¹) for odd n.

⇒ Here n = 97, which is odd ∴ (127 + 97) is a factor of (127⁹⁷ + 97⁹⁷).

⇒ a + b = 127 + 97 = 224.

⇒ We need to divide the total expression by 32.

⇒ Since 224 = 32 × 7, the number 224 is perfectly divisible by 32.

⇒ Therefore, any multiple of 224 is also perfectly divisible by 32.

⇒ Remainder = 0.

∴ The correct answer is 0.

Additional Information

Divisibility of aⁿ − bⁿ

(aⁿ − bⁿ) is always divisible by (a − b) for all natural numbers n.

Even Power Property

(aⁿ − bⁿ) is divisible by (a + b) if and only if n is an even number.

Remainder Theorem Property

If a number N is divisible by k, then it is also divisible by every factor of k. Here, k = 224 and 32 is a factor of 224.

Shortcut Trick Difference in percentage = 14% (Profit) − (−8% (Loss)) = 22% 22% of CP = ₹121 1% of CP = 121 ÷ 22 = ₹5.5 Cost Price (100%) = 5.5 × 100 = ₹550 Loss % = [(550 − 536.25) × 100] ÷ 550 = 13.75 ÷ 5.5 = 2.5% ∴ The correct answer is Loss, 2.5%. Alternate Method Given: Initial Profit = 14% | Second Case Loss = 8% | Price Decrease = ₹121 | New Selling Price = ₹536.25

Calculations: ⇒ Let the Cost Price (CP) be 100 units. ⇒ First Selling Price (SP1) = 100 + 14 = 114 units. ⇒ Second Selling Price (SP2) = 100 − 8 = 92 units. ⇒ Difference = SP1 − SP2 = 114 − 92 = 22 units. ⇒ Given: 22 units = ₹121. ⇒ 1 unit = 121 ÷ 22 = ₹5.5. ⇒ CP = 100 units = 100 × 5.5 = ₹550. ⇒ New SP = ₹536.25. ⇒ Since New SP < CP, it is a loss. ⇒ Loss Amount = 550 − 536.25 = ₹13.75. ⇒ Loss % = (Loss ÷ CP) × 100 ⇒ Loss % = (13.75 ÷ 550) × 100 = 2.5%. ∴ The correct answer is Loss, 2.5%. Additional Information Profit and Loss Percentage Profit % = (Profit ÷ CP) × 100 and Loss % = (Loss ÷ CP) × 100. Always calculate on Cost Price unless stated otherwise. Marked Price and Discount Selling Price (SP) = Marked Price (MP) × (100 − Discount%) ÷ 100. Discount is always calculated on the Marked Price. CP and MP Relationship CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%). This is a direct shortcut to find the ratio between CP and MP.

Shortcut Trick

Difference (D) = 2001, Quotient (Q) = 9, Remainder (R) = 41.

Smaller number (S) = (D − R) ÷ (Q − 1)

S = (2001 − 41) ÷ (9 − 1) = 1960 ÷ 8 = 245.

Larger number (L) = S + D = 245 + 2001 = 2246.

Sum of digits of L = 2 + 2 + 4 + 6 = 14.

∴ The correct answer is 14.

Alternate Method

Let the larger number be L and the smaller number be S.

Given:

L − S = 2001

L = 9 × S + 41

⇒ Substitute L in the difference equation:

⇒ (9S + 41) − S = 2001

⇒ 8S + 41 = 2001

⇒ 8S = 2001 − 41

⇒ 8S = 1960

⇒ S = 1960 ÷ 8 = 245

⇒ L = S + 2001

⇒ L = 245 + 2001 = 2246

⇒ Sum of digits of 2246 = 2 + 2 + 4 + 6 = 14

∴ The correct answer is 14.

Additional Information

Euclid's Division Lemma

For any two positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.

Digital Sum Property

The digital sum of a number is the sum of all its digits. It is often used to check divisibility by 3 and 9 (if the sum is divisible by 9, the number is divisible by 9).

Remainder Rule

In any division process, the remainder must always be strictly less than the divisor. In this case, 41 < 245, which satisfies the condition.

Shortcut Trick

Base side (a) = √(4 × Area / √3) = √(4 × 16√3 / √3) = 8 cm.

Slant height (l) = (2 × Lateral Face Area) ÷ side = (2 × 30) ÷ 8 = 7.5 cm.

In-radius of base (r) = a / (2√3) = 8 / (2√3) = 4/√3 cm.

Height (h) = √(l² − r²) = √((15/2)² − (4/√3)²).

h = √(225/4 − 16/3) = √((675 − 64)/12) = √(611/12) cm.

∴ The correct answer is √(611/12) cm.

Alternate Method

Given:

Area of equilateral base = 16√3 cm²

Area of one lateral face = 30 cm²

Formula Used:

Area of equilateral triangle = (√3/4) × side²

Area of lateral face = 1/2 × base × slant height (l)

Relationship: l² = h² + r² (where r is in-radius of the base)

⇒ Area of base = (√3/4) × side² = 16√3

⇒ side² = 16 × 4 = 64 ⇒ side = 8 cm

⇒ Area of lateral face = 1/2 × side × slant height = 30

⇒ 1/2 × 8 × l = 30 ⇒ 4l = 30 ⇒ l = 7.5 cm or 15/2 cm

⇒ In-radius of equilateral triangle (r) = side / (2√3)

⇒ r = 8 / (2√3) = 4/√3 cm

⇒ Using Pythagoras theorem: h² = l² − r²

⇒ h² = (15/2)² − (4/√3)²

⇒ h² = 225/4 − 16/3

⇒ h² = (675 − 64) / 12 = 611 / 12

⇒ h = √(611/12) cm

∴ The correct answer is √(611/12) cm.

Additional Information

Volume of a Pyramid

The volume of any pyramid is given by (1/3) × Base Area × Vertical Height (h).

Slant Height vs Slant Edge

Slant height (l) relates to height (h) and in-radius (r) as l² = h² + r². Slant edge (e) relates to height (h) and circum-radius (R) as e² = h² + R².

Lateral Surface Area

For a right pyramid, the lateral surface area is (1/2) × Perimeter of base × Slant height.

Shortcut Trick

Identify the sides: 12, 35, 37.

These form a Pythagorean triplet since 12² + 35² = 144 + 1225 = 1369 = 37².

For any right-angled triangle, the circumradius (R) is equal to half of the hypotenuse.

The longest side is the hypotenuse = 37 cm.

R = 37 ÷ 2 = 18.5 cm.

∴ The correct answer is 18.5 cm.

Alternate Method

Given:

Sides of the triangle: a = 12 cm, b = 35 cm, c = 37 cm.

Formula Used:

In a right-angled triangle, Circumradius (R) = Hypotenuse ÷ 2

Calculations:

⇒ First, verify if the triangle is right-angled using Pythagoras theorem.

⇒ a² + b² = 12² + 35²

⇒ a² + b² = 144 + 1225 = 1369

⇒ c² = 37² = 1369

⇒ Since a² + b² = c², the triangle is a right-angled triangle where the longest side (37 cm) is the hypotenuse.

⇒ Circumradius (R) = Hypotenuse ÷ 2

⇒ R = 37 ÷ 2

⇒ R = 18.5 cm

∴ The correct answer is 18.5 cm.

Additional Information

Inradius of a Right-Angled Triangle

The inradius (r) of a right-angled triangle can be calculated as r = (Perpendicular + Base − Hypotenuse) ÷ 2.

Circumcenter Location

In a right-angled triangle, the circumcenter always lies at the midpoint of the hypotenuse and is equidistant from all three vertices.

General Circumradius Formula

For any triangle with sides a, b, c and area A, the circumradius R = (a × b × c) ÷ (4 × A).

Pythagorean Triplets

Common triplets include (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), and (12, 35, 37).

Shortcut Trick

Total numbers after addition = 18 + 6 = 24

Total increase in sum = Total numbers × Increase in average = 24 × 1 = 24

Contribution of 6 numbers to the increase = 24

Increase per added number over original average = 24 ÷ 6 = 4

Value of x = Old Average + Increase per added number = 37.5 + 4 = 41.5

∴ The correct answer is 41.5.

Alternate Method

Given:

Initial count of numbers = 18

Initial average = 37.5

Added count of numbers = 6

Average of added numbers = x

Increase in average = 1

Formula Used:

Average = Total Sum ÷ Total Count

Total Sum = Average × Count

Calculations:

⇒ Total sum of 18 numbers = 18 × 37.5 = 675

⇒ New total count of numbers = 18 + 6 = 24

⇒ New average = 37.5 + 1 = 38.5

⇒ New total sum = 24 × 38.5 = 924

⇒ Sum of the 6 added numbers = 924 − 675 = 249

⇒ x = Sum of 6 numbers ÷ 6

⇒ x = 249 ÷ 6

⇒ x = 41.5

∴ The correct answer is 41.5.

Additional Information

Weighted Average Formula

When two groups of sizes n₁ and n₂ with averages A₁ and A₂ are combined, the combined average is (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Net Deviation Principle

The sum of deviations of all observations from the mean is always zero. This can be used to find missing values quickly by comparing old and new averages.

Arithmetic Mean Property

If each number in a set is increased or decreased by a constant k, the average also increases or decreases by the same constant k.

Shortcut Trick

Test standard angles: For θ = 30°, 2cos²30° + 3sin 30° = 2 × (3/4) + 3 × (1/2) = 1.5 + 1.5 = 3. (Satisfied)

Expression = sin²60° + cos²30° + tan²60° + cosec²60°

Expression = (√3/2)² + (√3/2)² + (√3)² + (2/√3)²

Expression = 3/4 + 3/4 + 3 + 4/3 = 3/2 + 3 + 4/3 = (9 + 18 + 8) ÷ 6 = 35/6.

∴ The correct answer is 35/6.

Alternate Method

Given: 2cos²θ + 3sin θ = 3, 0° < θ < 90°

Formula: cos²θ = 1 − sin²θ

⇒ 2(1 − sin²θ) + 3sin θ = 3

⇒ 2 − 2sin²θ + 3sin θ = 3

⇒ 2sin²θ − 3sin θ + 1 = 0

⇒ (2sin θ − 1)(sin θ − 1) = 0

⇒ sin θ = 1/2 or sin θ = 1

⇒ Since 0° < θ < 90°, sin θ = 1/2 ⇒ θ = 30°

⇒ Expression = sin²(2 × 30°) + cos²30° + tan²(2 × 30°) + cosec²(2 × 30°)

⇒ Expression = sin²60° + cos²30° + tan²60° + cosec²60°

⇒ Expression = (√3/2)² + (√3/2)² + (√3)² + (2/√3)²

⇒ Expression = 3/4 + 3/4 + 3 + 4/3

⇒ Expression = 6/4 + 3 + 4/3 = 3/2 + 3 + 4/3

⇒ Expression = (9 + 18 + 8) ÷ 6 = 35/6

∴ The correct answer is 35/6.

Additional Information

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1 | 1 + tan²θ = sec²θ | 1 + cot²θ = cosec²θ.

Standard Trig Values at 30° and 60°

sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3. | sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3.

Quadratic Roots in Trig

When solving asax² + bx + c = 0 where x = sin θ, ensure the root lies between −1 and 1.

Shortcut Trick

LCM of 72, 90, and 120 (2 hours) = 360 units (Total Capacity).

Efficiencies: A = 5, B = 4, and Combined (A + B + C) = 3.

Efficiency of C = 3 − (5 + 4) = −6 (Emptying pipe).

Water filled by A and B in 12 min = (5 + 4) × 12 = 108 units.

Time for C to empty = 108 ÷ 6 = 18 minutes.

∴ The correct answer is 18 minutes.

Alternate Method

Given: A fills in 72 min, B fills in 90 min, and (A + B + C) fill in 120 min.

Total work = LCM(72, 90, 120) = 360 units.

⇒ Total efficiency of A and B = 5 + 4 = 9 units/min.

⇒ Efficiency of Pipe C = (Efficiency of A+B+C) − (Efficiency of A+B)

⇒ Efficiency of Pipe C = 3 − 9 = −6 units/min.

⇒ Amount of water filled by A and B in 12 minutes = 9 × 12 = 108 units.

⇒ Time taken by C to empty 108 units = 108 ÷ 6 = 18 minutes.

∴ The correct answer is 18 minutes.

Additional Information

Emptying Pipe Efficiency

An emptying pipe (outlet) is always assigned a negative efficiency value because it reduces the total volume of the tank over time.

Time and Efficiency Relation

Time taken to fill or empty is inversely proportional to efficiency: Time = Total Work ÷ Net Efficiency.

Pipes and Cisterns Logic

If multiple pipes work together, their net efficiency is the algebraic sum of their individual efficiencies (Inlet = positive, Outlet = negative).

Shortcut Trick

Vertices on x-axis (where y = 0): x = 3 and x = 12/2 = 6.

Base of triangle on x-axis = |6 − 3| = 3 units.

Intersection height (y): Multiply first eq by 2 and subtract from second: (2x+5y) − (2x+2y) = 12 − 6 ⇒ 3y = 6 ⇒ y = 2.

Area = 1/2 × Base × Height = 1/2 × 3 × 2 = 3 square units.

∴ The correct answer is 3 square units.

Alternate Method

Given Equations: x + y = 3, 2x + 5y = 12, and y = 0 (x-axis).

Formula: Area of Triangle = 1/2 × |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)| or 1/2 × Base × Height.

Vertex A: Intersection of x + y = 3 and y = 0

⇒ x + 0 = 3 ⇒ x = 3 ⇒ A = (3, 0)

Vertex B: Intersection of 2x + 5y = 12 and y = 0

⇒ 2x + 0 = 12 ⇒ x = 6 ⇒ B = (6, 0)

Vertex C: Intersection of x + y = 3 and 2x + 5y = 12

⇒ From first eq: x = 3 − y

⇒ Substitute in second: 2(3 − y) + 5y = 12

⇒ 6 − 2y + 5y = 12 ⇒ 3y = 6 ⇒ y = 2

⇒ x = 3 − 2 = 1 ⇒ C = (1, 2)

Calculation of Area:

⇒ Base (on x-axis) = |x_B − x_A| = |6 − 3| = 3 units

⇒ Height = y-coordinate of C = 2 units

⇒ Area = 1/2 × 3 × 2 = 3 square units

∴ The correct answer is 3 square units.

Additional Information

Area using Coordinates

For vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃), Area = 1/2 |x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂)|.

X and Y Intercepts

X-intercept is found by putting y = 0, and Y-intercept is found by putting x = 0 in the linear equation.

Triangular Region bounded by Axis

When bounded by x-axis, the height is the y-coordinate of the intersection point of the two lines.

Shortcut Trick

Given relation: 3A = 2B = 4C

Ratio A : B : C = (1/3) : (1/2) : (1/4)

Multiplying by LCM(3, 2, 4) = 12, we get Ratio = 4 : 6 : 3

C's share (3 units) = ₹4,863

A's share (4 units) = (4,863 ÷ 3) × 4 = 1,621 × 4 = ₹6,484

∴ The correct answer is ₹6,484.

Alternate Method

Given:

3 × Investment of A = 2 × Investment of B = 4 × Investment of C

C's share in profit = ₹4,863

Formula Used:

Profit Share Ratio = (Investment₁ × Time₁) : (Investment₂ × Time₂) : (Investment₃ × Time₃)

Since time is constant for all, Profit Ratio = Investment Ratio.

Calculations:

⇒ Let 3A = 2B = 4C = k

⇒ A = k/3, B = k/2, C = k/4

⇒ A : B : C = (k/3) : (k/2) : (k/4)

⇒ A : B : C = (1/3 × 12) : (1/2 × 12) : (1/4 × 12)

⇒ A : B : C = 4 : 6 : 3

⇒ Given, Profit share of C = 3 units = ₹4,863

⇒ Value of 1 unit = 4,863 ÷ 3

⇒ Value of 1 unit = ₹1,621

⇒ Profit share of A = 4 units = 4 × 1,621

⇒ Profit share of A = ₹6,484

∴ The correct answer is ₹6,484.

Additional Information

Basic Partnership Law

Profit is always distributed in the ratio of the product of Capital and Time (P = C × T). If time is same, Profit ∝ Capital.

Ratio of Reciprocals

If aA = bB = cC, then the ratio A : B : C = (1/a) : (1/b) : (1/c), which can be simplified by multiplying with the LCM of a, b, and c.

Active vs Sleeping Partner

An active partner manages the business and may receive a salary or extra commission before the remaining profit is divided in the investment ratio.

Shortcut Trick

According to the Angle Bisector Theorem, the ratio of the segments of the base is equal to the ratio of the other two sides.

Ratio BD : DC = AB : AC = 15 : 13.

Total ratio parts = 15 + 13 = 28 units.

Given total length BC = 14 cm, so 28 units = 14 cm ⇒ 1 unit = 0.5 cm.

DC = 13 units = 13 × 0.5 cm = 6.5 cm.

∴ The correct answer is 6.5 cm.

Alternate Method

Given: AB = 15 cm, AC = 13 cm, BC = 14 cm.

Formula Used: Internal Angle Bisector Theorem states that in ΔABC, if AD bisects ∠A, then AB/AC = BD/DC.

⇒ Let DC = x cm, then BD = (BC − DC) = (14 − x) cm.

⇒ According to the theorem: AB / AC = BD / DC

⇒ 15 / 13 = (14 − x) / x

⇒ 15x = 13 × (14 − x)

⇒ 15x = 182 − 13x

⇒ 15x + 13x = 182

⇒ 28x = 182

⇒ x = 182 ÷ 28 = 6.5 cm

∴ The correct answer is 6.5 cm.

Additional Information

Internal Angle Bisector Theorem

The bisector of an internal angle of a triangle divides the opposite side into segments whose lengths are in the same ratio as the lengths of the other two sides: AB/AC = BD/DC.

External Angle Bisector Theorem

The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle: AB/AC = BE/CE (where E is the external point on extended side BC).

Incenter and Bisectors

The point where all three internal angle bisectors of a triangle intersect is called the incenter. The distance from the incenter to the vertices can be found using the ratio (AB+AC)/BC = AI/ID.

Shortcut Trick

Remaining solution volume = 32 − 12 = 20 litres.

In remaining solution, Ratio Acid : Water = 5 : 3 (Total = 8 units).

8 units = 20 litres ⇒ 1 unit = 2.5 litres.

Acid = 5 × 2.5 = 12.5 litres | Water = (3 × 2.5) + 7.5 = 15 litres.

New Ratio = 12.5 : 15 = 5 : 6.

∴ The correct answer is 5 : 6.

Alternate Method

Given:

Initial Volume = 32 litres

Initial Ratio (Acid : Water) = 5 : 3

Quantity taken out = 12 litres

Water added = 7.5 litres

Calculations:

⇒ Volume of remaining solution = 32 − 12 = 20 litres

⇒ In 20 litres, quantity of Acid = 20 × (5 ÷ (5 + 3))

⇒ Acid = 20 × (5 ÷ 8) = 12.5 litres

⇒ In 20 litres, quantity of Water = 20 × (3 ÷ 8) = 7.5 litres

⇒ New quantity of Water = 7.5 (existing) + 7.5 (added) = 15 litres

⇒ New Ratio (Acid : Water) = 12.5 : 15

⇒ New Ratio = 125 : 150 (Multiplying by 10)

⇒ New Ratio = 5 : 6 (Dividing by 25)

∴ The correct answer is 5 : 6.

Additional Information

Ratio Conservation in Mixture Removal

When a portion of a mixture is removed, the concentration and ratio of the components in the remaining mixture remains exactly the same as the original.

Component Addition Rule

If a pure component is added to a mixture, only the quantity of that specific component increases while the quantity of other components remains constant.

Unitary Method for Mixtures

Total quantity = Sum of ratio parts × Unit value. Component quantity = Specific ratio part × Unit value.

Shortcut Trick

Volume of Cylinder = π × 12² × 15 = 2160π cm³

Volume of one toy = Cone vol. + Hemisphere vol. = ((1/3) × π × 3² × 9) + ((2/3) × π × 3³)

Volume of one toy = 27π + 18π = 45π cm³

n = Total Volume ÷ Single Toy Volume = 2160π ÷ 45π = 48

∴ The correct answer is 48.

Alternate Method

Given:

Radius of cylinder (R) = 12 cm, Height of cylinder (H) = 15 cm

Radius of toy (r) = 3 cm, Height of cone part (h) = 9 cm

Formula Used:

Volume of cylinder = πR²H

Volume of cone = (1/3)πr²h

Volume of hemisphere = (2/3)πr³

⇒ Volume of Cylinder = π × 12² × 15 = 2160π cm³

⇒ Volume of one toy = Volume of cone + Volume of hemisphere

⇒ Volume of one toy = (1/3) × π × 3² × 9 + (2/3) × π × 3³

⇒ Volume of one toy = 27π + 18π = 45π cm³

⇒ Number of toys (n) = Total Volume ÷ Volume of one toy

⇒ n = 2160π ÷ 45π

⇒ n = 48

∴ The correct answer is 48.

Additional Information

Volume of Solids

For any solid, Volume = Base Area × Height (if uniform). For a sphere, Volume = (4/3)πr³ and for a hemisphere, Volume = (2/3)πr³.

Recasting Rule

When a 3D object is melted and recast into other shapes, the total volume of material remains constant: Volume_old = Volume_new1 + Volume_new2 + ...

Surface Area of Cylinder

Curved Surface Area = 2πrh; Total Surface Area = 2πr(h + r). This differs from volume which measures the space occupied inside.

Shortcut Trick

Equate the volume of earth removed from the cylindrical well to the volume of the rectangular platform.

Volume of well = Volume of platform

πr²h = L × b × H

(22/7) × 2.8 × 2.8 × 15 = L × 8 × 1.5

24.64 × 15 = 12 × L

L = 369.6 ÷ 12 = 30.8 m

∴ The correct answer is 30.8 m.

Alternate Method

Given:

Radius of the cylindrical well (r) = 2.8 m

Depth of the cylindrical well (h) = 15 m

Breadth of the rectangular platform (b) = 8 m

Height of the rectangular platform (H) = 1.5 m

Formula Used:

Volume of Cylinder = πr²h

Volume of Cuboid = Length × Breadth × Height

Calculations:

⇒ Volume of earth taken out from well = π × (2.8)² × 15

⇒ Volume = (22/7) × 2.8 × 2.8 × 15

⇒ Volume = 22 × 0.4 × 2.8 × 15

⇒ Volume = 8.8 × 42

⇒ Volume = 369.6 m³

⇒ Volume of platform = Length × Breadth × Height

⇒ 369.6 = L × 8 × 1.5

⇒ 369.6 = L × 12

⇒ L = 369.6 ÷ 12

⇒ L = 30.8 m

∴ The correct answer is 30.8 m.

Additional Information

Cylinder Surface Area

Curved Surface Area (CSA) = 2πrh | Total Surface Area (TSA) = 2πr(h + r).

Cuboid Properties

Diagonal of a cuboid = √(L² + b² + H²) | Total Surface Area = 2(Lb + bH + HL).

Conservation of Volume

When a solid is reshaped or earth is transferred, the total volume remains constant unless stated otherwise.

Shortcut Trick

Identify the expression as the sum of cubes identity: a³ + b³ = (a + b)(a² − ab + b²).

Here, a = √5x and b = √2y.

By inspection, A = √5, B = 5, and C = −√10 (from the −ab term).

Calculate: A² + B² − C² = (√5)² + 5² − (−√10)² = 5 + 25 − 10 = 20.

∴ The correct answer is 20.

Alternate Method

Given: 5√5 x³ + 2√2 y³ = (Ax + √2y)(Bx² + 2y² + Cxy)

Formula Used: a³ + b³ = (a + b)(a² − ab + b²)

⇒ 5√5 x³ + 2√2 y³ = (√5x)³ + (√2y)³

⇒ (√5x + √2y)[(√5x)² − (√5x)(√2y) + (√2y)²]

⇒ (√5x + √2y)(5x² − √10xy + 2y²)

Comparing this with the given expression (Ax + √2y)(Bx² + 2y² + Cxy):

⇒ A = √5

⇒ B = 5

⇒ C = −√10

Now, calculate the value of (A² + B² − C²):

⇒ (√5)² + (5)² − (−√10)²

⇒ 5 + 25 − 10

⇒ 30 − 10 = 20

∴ The correct answer is 20.

Additional Information

Sum of Cubes Identity

The factorization follows the form a³ + b³ = (a + b)(a² − ab + b²), where the middle term of the quadratic factor has a negative sign.

Difference of Cubes Identity

The factorization follows the form a³ − b³ = (a − b)(a² + ab + b²), where the linear factor has a negative sign and the quadratic factor has all positive signs.

Comparing Polynomial Coefficients

If two polynomials are equal for all x and y, then the coefficients of like terms such as x, y, x², y², and xy must be identical on both sides.

Shortcut Trick

Each exterior angle of a regular polygon = 360° ÷ n

Given: Exterior angle = 51 3/7° = 360/7° ⇒ n = 7

Ratio of Diagonals to Sides = [n(n − 3) / 2] : n = (n − 3) : 2

For n = 7, Ratio = (7 − 3) : 2 = 4 : 2 = 2 : 1

∴ The correct answer is 2 : 1.

Alternate Method

Given:

Measure of each exterior angle = 51 3/7°

Formula Used:

1. Number of sides (n) = 360° ÷ Exterior Angle
2. Number of diagonals = n(n − 3) ÷ 2

Calculations:

⇒ Each exterior angle = (51 × 7 + 3) / 7 = 360/7°

⇒ Number of sides (n) = 360 ÷ (360/7) = 360 × (7/360) = 7

⇒ Number of diagonals = 7 × (7 − 3) ÷ 2

⇒ Number of diagonals = 7 × 4 ÷ 2 = 14

⇒ Required Ratio = Diagonals : Sides = 14 : 7

⇒ Ratio = 2 : 1

∴ The correct answer is 2 : 1.

Additional Information

Sum of Interior Angles

The sum of all interior angles of a polygon with n sides is given by (n − 2) × 180°.

Sum of Exterior Angles

For any convex polygon, the sum of all exterior angles is always 360°.

Interior and Exterior Relationship

At any vertex of a polygon, the Interior Angle + Exterior Angle = 180°.

Diagonal Property

A triangle (n=3) has 0 diagonals, while a quadrilateral (n=4) has 2 diagonals as per the formula n(n−3)/2.

Shortcut Trick

Assume total distance is 100 km.

First part (40%) = 40 km at 8 km/h → Time = 40 ÷ 8 = 5 hours.

Second part (40% of 60 km) = 24 km at 9 km/h → Time = 24 ÷ 9 = 8/3 hours.

Third part (Remaining) = 36 km at 12 km/h → Time = 36 ÷ 12 = 3 hours.

Average Speed = Total Distance ÷ Total Time = 100 ÷ (5 + 3 + 8/3) = 100 ÷ (32/3) = 300/32 = 75/8 km/h.

∴ The correct answer is 9 3/8 km/h.

Alternate Method

Given:

Speed 1 (v₁) = 8 km/h, Speed 2 (v₂) = 9 km/h, Speed 3 (v₃) = 12 km/h.

Distance 1 (d₁) = 40% of total.

Distance 2 (d₂) = 40% of remaining distance.

Formula:

Average Speed = Total Distance ÷ Total Time

Total Time = (d₁/v₁) + (d₂/v₂) + (d₃/v₃)

Let Total Distance = 100 km

⇒ d₁ = 40% of 100 = 40 km

⇒ Remaining distance = 100 − 40 = 60 km

⇒ d₂ = 40% of 60 = 24 km

⇒ d₃ = 100 − (40 + 24) = 36 km

⇒ t₁ = 40 ÷ 8 = 5 hours

⇒ t₂ = 24 ÷ 9 = 8/3 hours

⇒ t₃ = 36 ÷ 12 = 3 hours

⇒ Total Time = 5 + 3 + 8/3 = 8 + 8/3 = 32/3 hours

⇒ Average Speed = 100 ÷ (32/3) = (100 × 3) ÷ 32 = 300 ÷ 32 = 75/8 km/h

⇒ Average Speed = 9 3/8 km/h

∴ The correct answer is 9 3/8 km/h.

Additional Information

General Average Speed Formula

Average Speed = Total Distance ÷ Total Time. This applies regardless of the number of segments or variation in speeds.

Average Speed for Equal Distances

If two equal distances are covered at speeds x and y, Average Speed = 2xy ÷ (x + y). For three equal distances at speeds x, y, and z, Average Speed = 3xyz ÷ (xy + yz + zx).

Unit Conversion

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

Shortcut Trick

From AD × AC = AB × AE ⇒ AD/AB = AE/AC.

With ∠A common, ΔADE ∼ ΔABC by SAS similarity.

Thus, ∠ADE = ∠ABC = 78°.

Given ∠ADE = ∠ACB + 30° ⇒ 78° = ∠ACB + 30° ⇒ ∠ACB = 48°.

In ΔABC, ∠A = 180° − (78° + 48°) = 54°.

∴ The correct answer is 54°.

Alternate Method

Given: AD × AC = AB × AE, ∠ABC = 78°, and ∠ADE = ∠ACB + 30°.

⇒ In ΔADE and ΔABC:

⇒ AD / AB = AE / AC (Rearranging the given product)

⇒ ∠DAE = ∠BAC (Common angle)

⇒ ∴ ΔADE ∼ ΔABC (By SAS Similarity Rule)

⇒ Corresponding angles are equal: ∠ADE = ∠ABC

⇒ ∠ADE = 78°

⇒ From question: ∠ADE = ∠ACB + 30°

⇒ 78° = ∠ACB + 30° ⇒ ∠ACB = 48°

⇒ In ΔABC: ∠A + ∠B + ∠C = 180°

⇒ ∠A + 78° + 48° = 180°

⇒ ∠A + 126° = 180° ⇒ ∠A = 54°

∴ The correct answer is 54°.

Additional Information

SAS Similarity Criterion

If two triangles have one pair of corresponding angles equal and the sides including these angles are in the same ratio, the triangles are similar.

Properties of Similar Triangles

When ΔADE ∼ ΔABC, then ∠ADE = ∠ABC, ∠AED = ∠ACB, and AD/AB = AE/AC = DE/BC.

Angle Sum Property

The sum of the interior angles of any triangle is exactly 180°.

Shortcut Trick

Using the value substitution method, let θ = 45°.

First term = (sin 45° + cos 45° − 1) ÷ (sin 45° − cos 45° + 1) = (1/√2 + 1/√2 − 1) ÷ (1) = √2 − 1.

Numerator of second term = tan² 45° (cosec² 45° − 1) = 1² × (2 − 1) = 1.

Denominator of second term = sec 45° − tan 45° = √2 − 1.

Total Expression = (√2 − 1) × [1 ÷ (√2 − 1)] = 1.

∴ The correct answer is 1.

Alternate Method

Given: [(sin θ + cos θ − 1) ÷ (sin θ − cos θ + 1)] × [tan² θ (cosec² θ − 1) ÷ (sec θ − tan θ)]

Formula used: cosec² θ − 1 = cot² θ; tan θ × cot θ = 1; (sin θ + cos θ − 1) ÷ (sin θ − cos θ + 1) = (1 − sin θ) ÷ cos θ

⇒ Let Part A = (sin θ + cos θ − 1) ÷ (sin θ − cos θ + 1)

⇒ Rationalizing Part A with [sin θ + (cos θ − 1)] gives: Part A = (1 − sin θ) ÷ cos θ = sec θ − tan θ

⇒ Let Part B = tan² θ (cosec² θ − 1) = tan² θ × cot² θ = 1

⇒ Final Expression = (sec θ − tan θ) × [1 ÷ (sec θ − tan θ)]

⇒ Final Expression = 1

∴ The correct answer is 1.

Additional Information

Pythagorean Identities

sin² θ + cos² θ = 1, 1 + tan² θ = sec² θ, and 1 + cot² θ = cosec² θ.

Fundamental Reciprocals

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

Rationalization Property

The expression (sin θ + cos θ − 1) ÷ (sin θ − cos θ + 1) is equivalent to (1 − sin θ) ÷ cos θ or (cos θ) ÷ (1 + sin θ).

Shortcut Trick

Rate for 1^st year = 15% | Rate for 2^nd year = 15%

Rate for remaining 2/5 year = 15% × 2/5 = 6%

Net Multiplier = 1.15 × 1.15 × 1.06 = 1.40185

Principal (P) = Amount ÷ 1.40185 = 4205.55 ÷ 1.40185 = 3000

∴ The correct answer is Rs. 3,000.

Alternate Method

Given:

Amount (A) = Rs. 4,205.55

Rate (R) = 15% per annum

Time (T) = 2(2/5) years

Formula Used:

When time is in fraction like n(a/b) years, Amount A = P × (1 + R/100)ⁿ × (1 + (R × a/b)/100)

Calculations:

⇒ R for first 2 years = 15%

⇒ R for next 2/5 year = 15% × 2/5 = 6%

⇒ A = P × (1 + 15/100)² × (1 + 6/100)

⇒ 4205.55 = P × (115/100) × (115/100) × (106/100)

⇒ 4205.55 = P × (23/20) × (23/20) × (53/50)

⇒ 4205.55 = P × (529/400) × (53/50)

⇒ 4205.55 = P × (28037 / 20000)

⇒ P = (4205.55 × 20000) ÷ 28037

⇒ P = 84111000 ÷ 28037

⇒ P = 3000

∴ The correct answer is Rs. 3,000.

Additional Information

Compound Interest for Fractional Time

For a time period of T = T₁ + f years (where f is a fraction), Amount = P(1 + R/100)^T₁ × (1 + Rf/100).

Effective Rate of Interest

The effective rate for successive rates r₁, r₂ is (r₁ + r₂ + r₁r₂/100)%. This can be extended to three or more years.

Difference between CI and SI for 2 Years

The difference (D) between CI and SI for 2 years is given by D = P(R/100)².

Shortcut Trick

Initial ratio = 3 : 5, Final ratio = 10 : 21. Value change in both = −13.

Using the cross-multiplication method: Difference of ratio products = (21 × 3) − (10 × 5) = 63 − 50 = 13 units.

Difference of value products = (21 × 13) − (10 × 13) = 13 × (21 − 10) = 143.

13 units = 143 ⇒ 1 unit = 11. The initial numbers are (3 × 11) = 33 and (5 × 11) = 55.

New numbers after adding 15 are (33 + 15) = 48 and (55 + 15) = 70.

∴ The correct answer is 24 : 35.

Alternate Method

Given:

The initial numbers are in the ratio 3 : 5.

When 13 is subtracted from each, the new ratio becomes 10 : 21.

Formula Used:

If a ratio is a : b, the values can be taken as ax and bx.

First Number Second Number

3x

5x

−13

−13

10 parts

21 parts Cross-multiply: 21(3x − 13) = 10(5x − 13)

Calculations:

⇒ Let the original numbers be 3x and 5x.

⇒ According to the question, (3x − 13) / (5x − 13) = 10 / 21

⇒ 21 × (3x − 13) = 10 × (5x − 13)

⇒ 63x − 273 = 50x − 130

⇒ 63x − 50x = 273 − 130

⇒ 13x = 143

⇒ x = 11

⇒ The original numbers are 3x = 3 × 11 = 33 and 5x = 5 × 11 = 55.

⇒ If 15 is added to each number, the new numbers will be:

⇒ First number = 33 + 15 = 48

⇒ Second number = 55 + 15 = 70

⇒ Required ratio = 48 : 70

⇒ Simplifying by dividing both by 2, we get 24 : 35

∴ The correct answer is 24 : 35.

Additional Information

Compound Ratio

The compound ratio of two individual ratios a : b and c : d is calculated by multiplying the corresponding parts, given by (a × c) : (b × d).

Duplicate and Sub-duplicate Ratios

The duplicate ratio of a : b is given by a² : b², while its sub-duplicate ratio is calculated as √a : √b.

Cross-Multiplication Method

When terms of a given ratio are increased or decreased by equal amounts, equating the difference in the cross products of the ratio terms to the difference in cross products of the numeric values gives the exact equivalent change in units.

Shortcut Trick

Interest for 19/5 years = 13,825 − 10,500 = ₹3,325.

Interest for 1 year at original rate = 3,325 × (5/19) = ₹875.

The new condition is double the rate for 5 years, which is equivalent to 10 years at the original rate (2 × 5 = 10).

New Simple Interest = Interest for 1 year × 10 = 875 × 10 = ₹8,750.

∴ The correct answer is ₹8,750.

Alternate Method

Given: Principal (P) = ₹10,500, Amount (A) = ₹13,825, Time (T₁) = 3 4/5 = 3.8 years.

Formula: Simple Interest (SI) = (P × R × T) ÷ 100 | Amount = P + SI

⇒ SI₁ = 13,825 − 10,500 = ₹3,325

⇒ Rate (R₁) = (SI₁ × 100) ÷ (P × T₁)

⇒ R₁ = (3325 × 100) ÷ (10500 × 3.8)

⇒ R₁ = 332500 ÷ 39900 ≈ 8.333%

⇒ New Rate (R₂) = 2 × R₁ = 2 × [332500 ÷ 39900]

⇒ New Time (T₂) = 5 years

⇒ New SI = (10500 × R₂ × 5) ÷ 100

⇒ New SI = [10500 × 2 × (332500 ÷ 39900) × 5] ÷ 100

⇒ New SI = (10500 × 10 × 3325) ÷ 39900

⇒ New SI = 349125000 ÷ 39900 = ₹8,750

∴ The correct answer is ₹8,750.

Additional Information

Simple Interest Relationships

Simple Interest is directly proportional to Principal (P), Rate (R), and Time (T). If the rate is doubled, the interest for the same period also doubles.

Constant SI Property

Unlike Compound Interest, Simple Interest remains constant for every year if the principal and rate are unchanged.

Total Amount Formula

Amount (A) = P + SI = P(1 + RT/100). This formula helps find the final value of an investment over time.

Shortcut Trick

Multiply Eq(1) by 2: 10x − 4y = −2. Eq(2) is: −3x + 4y = −5.

Add the equations: (10x − 3x) + (−4y + 4y) = −2 − 5 ⇒ 7x = −7 ⇒ x = −1.

Substitute x = −1 in Eq(1): 5(−1) − 2y + 1 = 0 ⇒ −4 = 2y ⇒ y = −2.

The intersection point (α, β) is (−1, −2).

Value = 2α − 3β = 2(−1) − 3(−2) = −2 + 6 = 4.

∴ The correct answer is 4.

Alternate Method

Given:

Line 1: 5x − 2y + 1 = 0

Line 2: 4y − 3x + 5 = 0

Calculations:

⇒ Equation 1: 5x − 2y = −1

⇒ Equation 2: −3x + 4y = −5

⇒ Multiply Equation 1 by 2: 10x − 4y = −2

⇒ Add this to Equation 2: (10x − 3x) + (−4y + 4y) = −2 + (−5)

⇒ 7x = −7 ⇒ x = −1

⇒ Put x = −1 in Equation 1: 5(−1) − 2y = −1

⇒ −5 − 2y = −1 ⇒ −2y = 4 ⇒ y = −2

⇒ Point P(α, β) = (−1, −2) ⇒ α = −1, β = −2

⇒ Value = 2α − 3β = 2(−1) − 3(−2) = −2 + 6 = 4

∴ The correct answer is 4.

Additional Information

Intersection of Linear Equations

The point of intersection of two linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 represents the unique solution (α, β) that satisfies both equations simultaneously.

Nature of Solutions

A system of two linear equations has a unique solution if a₁/a₂ ≠ b₁/b₂. If a₁/a₂ = b₁/b₂ ≠ c₁/c₂, the lines are parallel (no solution). If a₁/a₂ = b₁/b₂ = c₁/c₂, the lines are coincident (infinite solutions).

Point-Slope Form

Any line can be expressed as y = mx + c, where m is the slope and c is the y-intercept. For 5x − 2y + 1 = 0, the slope is 5/2 = 2.5.

Shortcut Trick

Rewrite the expression: [3(x² + 1) / 3x] − [7x / 3x] = 6.

Simplify to get: x + 1/x − 7/3 = 6 ⇒ x + 1/x = 25/3.

We know that (√x + 1/√x) = √(x + 1/x + 2).

Substitute the value: √(25/3 + 2) = √(31/3).

∴ The correct answer is √(31/3).

Alternate Method

Given: [3(x² + 1) − 7x] / 3x = 6

Formula Used: (a + b)² = a² + b² + 2ab

⇒ 3(x² + 1) − 7x = 18x

⇒ 3x² + 3 − 7x = 18x

⇒ 3x² + 3 = 25x

⇒ Divide both sides by 3x:

⇒ (3x² / 3x) + (3 / 3x) = 25x / 3x

⇒ x + 1/x = 25/3

⇒ Now, consider (√x + 1/√x)² = (√x)² + (1/√x)² + 2 × √x × (1/√x)

⇒ (√x + 1/√x)² = x + 1/x + 2

⇒ (√x + 1/√x)² = 25/3 + 2

⇒ (√x + 1/√x)² = (25 + 6) / 3 = 31/3

⇒ √x + 1/√x = √(31/3)

∴ The correct answer is √(31/3).

Additional Information

Relationship between x + 1/x and √x + 1/√x

If x + 1/x = k, then √x + 1/√x = √(k + 2) and √x − 1/√x = √(k − 2).

Algebraic Identities

The fundamental identity (a + b)² = a² + b² + 2ab is used to relate a sum to the sum of squares.

Reciprocal Sums

For any x ≠ 0, (x + 1/x)² = x² + 1/x² + 2 and (x − 1/x)² = x² + 1/x² − 2.

Shortcut Trick

Total interest for 1 year = 4125 ÷ 2 = ₹2,062.5.

Effective average interest rate = (2062.5 ÷ 25000) × 100 = 8.25% p.a.

Ratio of sums = 1/4 : 3/4 = 1 : 3.

Net deviation for the whole sum from the base rate = (3 ÷ 4) × 4% = 3%.

Base rate of the first sum = 8.25% − 3% = 5.25%.

Rate of the second sum = 5.25% + 4% = 9.25%.

∴ The correct answer is 9.25%.

Alternate Method

Given:

Total Principal (P) = ₹25,000 | Time (T) = 2 years | Total SI = ₹4,125.

P₁ = 1/4 × 25,000 = ₹6,250 | P₂ = 3/4 × 25,000 = ₹18,750.

Formula Used:

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

Let the interest rate for the first part be R%.

⇒ [(6250 × R × 2) ÷ 100] + [(18750 × (R + 4) × 2) ÷ 100] = 4125

⇒ 125R + 375 × (R + 4) = 4125

⇒ 125R + 375R + 1500 = 4125

⇒ 500R = 4125 − 1500

⇒ 500R = 2625

⇒ R = 2625 ÷ 500 = 5.25%

⇒ Rate of the second sum = R + 4 = 5.25 + 4 = 9.25%

∴ The correct answer is 9.25%.

Additional Information

Weighted Average Rate

When a sum is split into parts P₁ and P₂ with rates R₁ and R₂, the average rate R_avg = (P₁R₁ + P₂R₂) ÷ (P₁ + P₂).

Simple Interest Properties

Simple interest is calculated only on the principal amount. For the same principal and rate, the interest grows linearly with time.

Effect of Rate Change

If the rate increases by X%, the extra interest earned for T years is (P × X × T) ÷ 100.

Shortcut Trick

Use substitution: Let a = 1, b = 2, and c = 3.

Expression: 1×2(1 − 2) + 2×3(2 − 3) + 3×1(3 − 1) ⇒ −2 − 6 + 6 = −2.

Option 1: (1 + 2)(2 − 3)(3 − 1) = 3 × (−1) × 2 = −6.

Option 2: (2 − 1)(2 − 3)(3 − 1) = 1 × (−1) × 2 = −2.

Since the values match, Option 2 is the correct factorization.

∴ The correct answer is (b − a)(b − c)(c − a).

Alternate Method

Given: ab(a − b) + bc(b − c) + ca(c − a)

Formula Used: Factorization of cyclic expressions by grouping terms.

⇒ Expansion: a²b − ab² + b²c − bc² + c²a − ca²

⇒ Rearranging to group by (b − c): a²(b − c) − a(b² − c²) + bc(b − c)

⇒ (b − c) [a² − a(b + c) + bc]

⇒ (b − c) [a² − ab − ac + bc]

⇒ (b − c) [a(a − b) − c(a − b)]

⇒ (b − c)(a − b)(a − c)

⇒ (a − b)(b − c)[−(c − a)]

⇒ [−(a − b)](b − c)(c − a)

⇒ (b − a)(b − c)(c − a)

∴ The correct answer is (b − a)(b − c)(c − a).

Additional Information

Cyclic Expressions

An expression is cyclic if it remains unchanged when variables are replaced in a cycle (a → b, b → c, c → a). For example, ab(a−b) + bc(b−c) + ca(c−a).

Factor Theorem in Cyclic Polynomials

If substituting a = b makes the expression zero, then (a − b) is a factor. By symmetry, (b − c) and (c − a) will also be factors.

Standard Result

a²(b − c) + b²(c − a) + c²(a − b) = −(a − b)(b − c)(c − a).

Symmetric vs Alternating

An alternating expression changes sign when two variables are swapped, whereas a symmetric expression remains identical.

Shortcut Trick

Successive Percentage Increase Formula = a + b + (a × b / 100)

Working hours increase (a) = 15%

Wages per hour increase (b) = 20%

Total increase = 15 + 20 + (15 × 20 / 100)

Total increase = 35 + 3 = 38%

∴ The correct answer is 38%.

Alternate Method

Given:

Increase in working hours = 15%

Increase in wages per hour = 20%

Formula Used:

Daily Earnings = (Working Hours per day) × (Wages per hour)

Calculations:

Let initial working hours = 100 units and initial wages = 100 units.

⇒ Initial daily earnings = 100 × 100 = 10,000 units

⇒ New working hours = 100 + 15% of 100 = 115 units

⇒ New wages per hour = 100 + 20% of 100 = 120 units

⇒ New daily earnings = 115 × 120 = 13,800 units

⇒ Increase in earnings = 13,800 − 10,000 = 3,800 units

⇒ Percentage increase = (3,800 ÷ 10,000) × 100

⇒ Percentage increase = 38%

∴ The correct answer is 38%.

Additional Information

Successive Percentage Change

When a value is increased by x% and then by y%, the net change is given by [x + y + (xy/100)]%.

Product Stability and Area

This logic applies to any product relation (A = B × C), such as Area = Length × Breadth or Revenue = Price × Consumption.

Percentage Decrease Case

If a value decreases, use a negative sign in the formula. For example, a 10% decrease means x = −10.

Shortcut Trick

Break each term into (a ± b)² form where 2ab is the irrational part.

√(28 + 10√3) = √(25 + 3 + 2 × 5 × √3) = 5 + √3

√(7 − 4√3) = √(4 + 3 − 2 × 2 × √3) = 2 − √3

Net value = (5 + √3) − (2 − √3) = 3 + 2√3

Approximate value = 3 + 2(1.732) = 3 + 3.464 = 6.464 ≈ 6.5

∴ The correct answer is 6.5.

Alternate Method

Given: √(28 + 10√3) − √(7 − 4√3)

Formula Used: (a + b)² = a² + b² + 2ab and (a − b)² = a² + b² − 2ab

⇒ √(28 + 10√3) = √(25 + 3 + 2 × 5 × √3)

⇒ √(5² + (√3)² + 2 × 5 × √3) = 5 + √3

⇒ √(7 − 4√3) = √(4 + 3 − 2 × 2 × √3)

⇒ √(2² + (√3)² − 2 × 2 × √3) = 2 − √3

⇒ Original Expression = (5 + √3) − (2 − √3)

⇒ 5 + √3 − 2 + √3

⇒ 3 + 2√3

⇒ Taking √3 ≈ 1.732

⇒ 3 + 2 × 1.732 = 3 + 3.464 = 6.464

⇒ Value ≈ 6.5

∴ The correct answer is 6.5.

Additional Information

Square Root of Surds

To find √(x ± √y), we try to express (x ± √y) as (a ± b)². Here, a² + b² = x and 2ab = √y.

Approximation of Roots

Common values used in competitive exams: √2 ≈ 1.414, √3 ≈ 1.732, and √5 ≈ 2.236.

Conjugate Surds

For a surd (a + √b), the conjugate is (a − √b). Their product (a + √b)(a − √b) = a² − b is always a rational number.

Shortcut Trick

Profit is given in terms of Cost Price (CP).

Profit = CP of 30 kg rice

Base Cost = CP of 80 kg rice

Profit % = (30 ÷ 80) × 100

Profit % = 3/8 × 100 = 37.5%

∴ The correct answer is 37½%.

Alternate Method

Given:

Quantity bought = 80 kg

Profit = Cost Price of 30 kg

Formula:

Profit % = (Profit ÷ Cost Price) × 100

Let the Cost Price (CP) of 1 kg of rice = ₹1

⇒ CP of 80 kg of rice = ₹80

⇒ Profit = CP of 30 kg of rice = ₹30

⇒ Profit % = (30 ÷ 80) × 100

⇒ Profit % = 300 ÷ 8 = 37.5%

⇒ 37.5% = 37½%

∴ The correct answer is 37½%.

Additional Information

Profit when SP is given

If profit is equal to the SP of 'n' items, Profit % = [n ÷ (Total Items − n)] × 100.

Loss when CP is given

If loss is equal to the CP of 'n' items, Loss % = (n ÷ Total Items) × 100.

Marked Price & Discount

Discount is always calculated on Marked Price (MP). SP = MP × (100 − Discount%) ÷ 100.

Given:

$24 \times 2 \div 12 + 12 \div 6 \text{ of } 2 \div (15 \div 8 \times 4) \text{ of } (28 \div 7 \text{ of } 5)$

Formula used:

BODMAS Rule (Brackets, Of, Division, Multiplication, Addition, Subtraction)

Calculation:

$24 \times 2 \div 12 + 12 \div 6 \text{ of } 2 \div (15 \div 8 \times 4) \text{ of } (28 \div 7 \text{ of } 5)$

⇒ $48 \div 12 + 12 \div 12 \div (\frac{15}{2}) \text{ of } (\frac{4}{5})$

⇒ $4 + 1 \div (\frac{15}{2} \times \frac{4}{5})$

⇒ $4 + 1 \div 6$

⇒ $4\frac{1}{6}$

∴ The correct answer is option (4).

Shortcut Trick

Lateral Surface Area (LSA) of a right prism = Perimeter of the base × Height.

Perimeter of triangular base = 5 + 8 + 12 = 25 cm.

LSA = 25 × 18 = 450 cm².

∴ The correct answer is 450 cm².

Alternate Method

Given:

Sides of the triangular base (a, b, c) = 5 cm, 8 cm, 12 cm

Height of the prism (h) = 18 cm

Formula: Lateral Surface Area = (a + b + c) × h

⇒ Perimeter of the base = 5 + 8 + 12

⇒ Perimeter of the base = 25 cm

⇒ Lateral Surface Area = Perimeter × Height

⇒ Lateral Surface Area = 25 × 18

⇒ Lateral Surface Area = 450 cm²

∴ The correct answer is 450 cm².

Additional Information

Total Surface Area of a Prism

Total Surface Area (TSA) = Lateral Surface Area + 2 × Area of Base.

Volume of a Prism

Volume = Area of the base × Height.

Heron's Formula for Base Area

Area of triangle = √(s(s − a)(s − b)(s − c)), where s is the semi-perimeter (a + b + c) ÷ 2.

Shortcut Trick

Direct relation between Volume (V), Curved Surface Area (CSA), and Radius (r):

Volume (V) = (CSA × r) ÷ 2

Given: CSA = 60π cm² and r = 3 cm

V = (60π × 3) ÷ 2 = 180π ÷ 2 = 90π cm³

∴ The correct answer is 90π cm³.

Alternate Method

Given: Radius (r) = 3 cm, Curved Surface Area (CSA) = 60π cm²

Formula used: Curved Surface Area (CSA) = 2πrh | Volume (V) = πr²h

⇒ 2 × π × r × h = 60π

⇒ 2 × 3 × h = 60

⇒ 6h = 60

⇒ h = 10 cm

⇒ Volume = π × r² × h

⇒ Volume = π × 3² × 10

⇒ Volume = π × 9 × 10

⇒ Volume = 90π cm³

∴ The correct answer is 90π cm³.

Additional Information

Total Surface Area (TSA) of a Cylinder

The total surface area includes the curved surface and the two circular ends: TSA = 2πrh + 2πr² = 2πr(h + r).

Base Area of a Cylinder

The area of the base (or top) of a right circular cylinder is the area of a circle: Area = πr².

Relationship between CSA and Volume

Since CSA = 2πrh and V = πr²h, the Volume can be written as V = (CSA × r) ÷ 2. This is useful when the height is not directly required.

Shortcut Trick

Upstream Speed = 900 m ÷ 12 min = 75 m/min

Downstream Speed = 900 m ÷ 9 min = 100 m/min

Speed in still water = (100 + 75) ÷ 2 = 87.5 m/min

To km/h: 87.5 × (60 ÷ 1000) = 5.25 km/h

5.25 km/h = 5 ¹/₄ km/h

∴ The correct answer is 5 1/4 km/h.

Alternate Method

Given: Distance = 900 m, Upstream time = 12 min, Downstream time = 9 min.

Formula Used: Speed = Distance ÷ Time; Speed in still water = (Downstream speed + Upstream speed) ÷ 2.

⇒ Upstream speed = 0.9 ÷ (12/60) = 0.9 × 5 = 4.5 km/h

⇒ Downstream speed = 0.9 ÷ (9/60) = 0.9 × (60/9) = 6 km/h

⇒ Speed in still water = (6 + 4.5) ÷ 2

⇒ Speed = 10.5 ÷ 2 = 5.25 km/h

⇒ 5.25 = 5 ²⁵/₁₀₀ = 5 ¹/₄ km/h

∴ The correct answer is 5 1/4 km/h.

Additional Information

Downstream and Upstream Speed

If the boat speed is u and stream speed is v, Downstream speed = u + v and Upstream speed = u − v.

Speed of Stream

Speed of current/stream (v) = (Downstream speed − Upstream speed) ÷ 2.

Unit Conversion

To convert m/min to km/h, multiply the value by 60 and divide by 1000 (or multiply by 0.06).

Shortcut Trick

Rate half-yearly = 8% ÷ 2 = 4% → 1/25 in fraction.

Ratio (Principal : Instalment) for 1st half-year = 25 : 26.

Ratio for 2nd half-year = 25² : 26² = 625 : 676.

Make instalments equal: 1st ratio × 26 → (25 × 26) : (26 × 26) = 650 : 676.

Total Principal = 650 + 625 = 1275 units. Total Amount paid = 676 × 2 = 1352 units.

Interest = 1352 − 1275 = 77 units. Since 676 units = ₹6,760, then 1 unit = ₹10.

Interest = 77 × 10 = ₹770.

∴ The correct answer is ₹770.

Alternate Method

Given:

Instalment (I) = ₹6,760

Annual Rate (R) = 8% p.a. → Half-yearly Rate (r) = 4%

Number of instalments (n) = 2

Formula Used:

Principal (P) = [I ÷ (1 + r/100)¹] + [I ÷ (1 + r/100)²]

Total Interest = (n × I) − P

Calculations:

⇒ P = [6760 ÷ (1 + 4/100)] + [6760 ÷ (1 + 4/100)²]

⇒ P = [6760 ÷ (1.04)] + [6760 ÷ (1.04)²]

⇒ P = [6760 ÷ (26/25)] + [6760 ÷ (676/625)]

⇒ P = [6760 × (25/26)] + [6760 × (625/676)]

⇒ P = [260 × 25] + [10 × 625]

⇒ P = 6500 + 6250 = ₹12,750

⇒ Total Amount paid = 2 × 6,760 = ₹13,520

⇒ Total Interest Paid = 13520 − 12750 = ₹770

∴ The correct answer is ₹770.

Additional Information

Compound Interest Installment Formula

For n equal installments, P = I × [(xⁿ − 1) ÷ (r × xⁿ)] where x = (1 + r/100).

Half-yearly Compounding

When interest is compounded half-yearly, the annual rate is halved (R/2) and the time period is doubled (2n).

Effective Rate of Interest

If interest is compounded half-yearly at r% per half-year, the effective annual rate is [ (1 + r/100)² − 1 ] × 100%.

Shortcut Trick

Ratio of r : h is 5 : 12. By Pythagorean triplets (5, 12, 13), the slant height l ratio is 13.

Let r = 5x, h = 12x, and l = 13x.

CSA = πrl ⇒ 3.14 × 5x × 13x = 816.4 ⇒ 204.1x² = 816.4 ⇒ x² = 4 ⇒ x = 2.

Volume = 1/3 × πr²h = 1/3 × 3.14 × (10)² × 24 = 314 × 8 = 2512 cm³.

∴ The correct answer is 2512 cm³.

Alternate Method

Given:

Ratio of radius (r) to height (h) = 5 : 12

Curved Surface Area (CSA) = 816.4 cm²

π = 3.14

Formula Used:

Slant height (l) = √(r² + h²)

CSA of cone = πrl

Volume of cone (V) = (1/3) πr²h

Calculations:

Let the radius r = 5x and height h = 12x.

⇒ Slant height l = √((5x)² + (12x)²)

⇒ l = √(25x² + 144x²) = √(169x²) = 13x

⇒ CSA = πrl = 3.14 × 5x × 13x = 816.4

⇒ 3.14 × 65x² = 816.4

⇒ 204.1x² = 816.4

⇒ x² = 816.4 ÷ 204.1 = 4

⇒ x = 2

⇒ Radius (r) = 5 × 2 = 10 cm

⇒ Height (h) = 12 × 2 = 24 cm

⇒ Volume (V) = (1/3) × 3.14 × (10)² × 24

⇒ V = 3.14 × 100 × 8

⇒ V = 314 × 8 = 2512 cm³

∴ The correct answer is 2512 cm³.

Additional Information

Total Surface Area of a Cone

The total surface area includes the curved surface and the circular base: TSA = πr(l + r).

Relationship between Cone and Cylinder

If a cone and a cylinder have the same radius and height, the volume of the cone is exactly 1/3 the volume of the cylinder: V_cone = (1/3) × V_cylinder.

Slant Height of a Cone

The slant height l is the distance from the apex to any point on the base circumference, calculated as l = √(h² + r²).

Shortcut Trick

In ΔOBC, P is the incenter. ∴ ∠BPC = 90° + ∠BOC/2

⇒ 148° = 90° + ∠BOC/2 → ∠BOC = 2 × (148° − 90°) = 116°

Since O is the orthocenter of ΔABC, ∠A + ∠BOC = 180°

⇒ ∠A = 180° − 116° = 64°

∴ The correct answer is 64°.

Alternate Method

Given: ∠BPC = 148°, BE ⊥ AC, CD ⊥ AB, P is the intersection of angle bisectors of ∠OBC and ∠OCB.

Formula Used:

1. Angle at incenter: ∠BIC = 90° + ∠A/2
2. In a quadrilateral ADOE, sum of angles = 360°

⇒ In ΔOBC, BP and CP are bisectors of ∠OBC and ∠OCB respectively.

⇒ ∠BPC = 90° + ∠BOC / 2

⇒ 148° = 90° + ∠BOC / 2

⇒ ∠BOC / 2 = 148° − 90° = 58°

⇒ ∠BOC = 116°

⇒ In quadrilateral ADOE, ∠ADO = 90° and ∠AEO = 90° (Given BE ⊥ AC, CD ⊥ AB)

⇒ ∠DOE = ∠BOC = 116° (Vertically opposite angles)

⇒ ∠A + ∠ADO + ∠DOE + ∠AEO = 360°

⇒ ∠A + 90° + 116° + 90° = 360°

⇒ ∠A + 296° = 360°

⇒ ∠A = 360° − 296° = 64°

∴ The correct answer is 64°.

Additional Information

Orthocenter Properties

The angle formed by the orthocenter (O) at the side, ∠BOC, and the opposite vertex angle ∠A are supplementary: ∠BOC + ∠A = 180°.

Incenter Properties

The angle at the incenter (I) is given by ∠BIC = 90° + ∠A/2. This applies to any triangle where bisectors meet at a point.

Altitudes and Cyclicity

In an acute triangle, the quadrilateral formed by a vertex and the feet of altitudes from other vertices (like ADOE) is always cyclic if the opposite angles sum to 180°.

Shortcut Trick

Savings % of A, B, and C are 20% (100−80), 15% (100−85), and 25% (100−75) respectively.

Income Ratio = (Savings ÷ Savings %) ⇒ A : B : C = 8/20% : 9/15% : 20/25%

Income Ratio = 40 : 60 : 80 → 2 : 3 : 4

Difference (C − A) = 4 − 2 = 2 units → ₹18,000 ∴ 1 unit = ₹9,000

Income of B = 3 units × ₹9,000 = ₹27,000.

∴ The correct answer is ₹27,000.

Alternate Method

Given:

Expenditure: A = 80%, B = 85%, C = 75%

Savings Ratio A : B : C = 8 : 9 : 20

Difference (Income C − Income A) = ₹18,000

Calculations:

Savings % = 100% − Expenditure %

⇒ Savings % of A = 100% − 80% = 20%

⇒ Savings % of B = 100% − 85% = 15%

⇒ Savings % of C = 100% − 75% = 25%

Let Savings be 8k, 9k, and 20k.

⇒ Income of A = 8k ÷ 20% = 8k × 100/20 = 40k

⇒ Income of B = 9k ÷ 15% = 9k × 100/15 = 60k

⇒ Income of C = 20k ÷ 25% = 20k × 100/25 = 80k

Income Ratio A : B : C = 40 : 60 : 80 = 2 : 3 : 4

⇒ Let Income A = 2x, Income B = 3x, Income C = 4x

⇒ Income C − Income A = 4x − 2x = 2x

⇒ 2x = 18,000

⇒ x = 9,000

⇒ Income of B = 3x = 3 × 9,000 = ₹27,000

∴ The correct answer is ₹27,000.

Additional Information

Income, Expenditure and Savings Relationship

Income = Expenditure + Savings. If spending is x%, then savings is (100−x)%.

Ratio Scaling

To convert savings ratio to income ratio, divide each ratio part by its respective fractional savings rate.

Percentage to Fraction Conversion

20% = 1/5 | 15% = 3/20 | 25% = 1/4. Using these fractions can often simplify income calculations.

Shortcut Trick

Total work = 25 persons × 60 days = 1500 units.

If 15 persons (25 − 10) worked for the entire 80 days, they would complete: 15 × 80 = 1200 units.

The remaining 300 units (1500 − 1200) were completed by the 10 persons who left.

Number of days those 10 persons worked (x) = 300 ÷ 10 = 30 days.

∴ The correct answer is 30 days.

Alternate Method

Given: Initial persons = 25, Initial time = 60 days, Persons left = 10, Total time taken = 80 days.

Formula Used: Total Work = Persons × Days

⇒ Total work units = 25 × 60 = 1500 units.

⇒ Work done in first phase (x days) by 25 persons = 25 × x = 25x.

⇒ Work done in second phase (80 − x days) by 15 persons = 15 × (80 − x).

⇒ 25x + 15(80 − x) = 1500

⇒ 25x + 1200 − 15x = 1500

⇒ 10x = 1500 − 1200

⇒ 10x = 300

⇒ x = 30 days.

∴ The correct answer is 30 days.

Additional Information

Man-Days Concept

Total work is the product of the number of workers and the time they take. If efficiency is constant, W = M × D.

Chain Rule Formula

For different groups working on the same task: M₁ × D₁ × H₁ = M₂ × D₂ × H₂, where H represents hours per day.

Work and Efficiency

Efficiency of a person is inversely proportional to the time taken to complete the work. Efficiency = Total Work ÷ Time.