SSC CGL Tier 2 Quant Previous Year Paper (Held On: 12 September 2019)

Shortcut Trick

Let Cost Price (CP) = 100.

Profit = 25% when sold at MP → Marked Price (MP) = 125.

New Selling Price (SP) after 9.6% discount = 125 × (100 − 9.6)% = 125 × 90.4%.

New SP = 1.25 × 90.4 = 113.

Profit % = New SP − CP = 113 − 100 = 13%.

∴ The correct answer is 13%.

Alternate Method

Given:

Profit at Marked Price = 25%

Discount offered on Marked Price = 9.6%

Formula Used:

Profit % = [(SP − CP) ÷ CP] × 100

SP = MP − Discount

Calculations:

⇒ Let the Cost Price (CP) of the article be 100.

⇒ Since selling at MP gives 25% profit, MP = 100 + 25% of 100 = 125.

⇒ Discount amount = 9.6% of MP = (9.6 ÷ 100) × 125.

⇒ Discount amount = (9.6 × 1.25) = 12.

⇒ Selling Price (SP) = MP − Discount = 125 − 12 = 113.

⇒ Profit = SP − CP = 113 − 100 = 13.

⇒ Profit % = (13 ÷ 100) × 100 = 13%.

∴ The correct answer is 13%.

Additional Information

Relationship between CP and MP

The ratio of Cost Price to Marked Price is given by CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%).

Successive Discounts

If two successive discounts of a% and b% are allowed, the equivalent single discount is (a + b − ab/100)%.

Effective Profit Percentage

Effective Profit % = Mark up % − Discount % − (Mark up % × Discount % ÷ 100).

Shortcut Trick

According to the Angle Bisector Theorem, the ratio AD : DC is equal to the ratio AB : BC.

Ratio AD : DC = 10 : 11

Total parts in AC = 10 + 11 = 21 units

Value of 21 units = 14 cm

Value of 10 units (AD) = (14 ÷ 21) × 10 = (2/3) × 10 = 20/3 cm

∴ The correct answer is 20/3 cm.

Alternate Method

Given: AB = 10 cm, BC = 11 cm, AC = 14 cm.

In ΔABC, BD is the internal bisector of ∠B.

Formula Used: Angle Bisector Theorem states AD/DC = AB/BC.

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

⇒ AD / DC = AB / BC

⇒ x / (14 − x) = 10 / 11

⇒ 11x = 10 × (14 − x)

⇒ 11x = 140 − 10x

⇒ 11x + 10x = 140

⇒ 21x = 140

⇒ x = 140 / 21

⇒ x = 20 / 3

∴ The correct answer is 20/3 cm.

Additional Information

Length of Internal Angle Bisector

The length of the angle bisector BD can be calculated using the formula: BD² = AB × BC − AD × DC.

External Angle Bisector Theorem

If the bisector of an external angle of a triangle intersects the extended opposite side at a point D', then AD' / CD' = AB / BC.

Incenter Property

The internal angle bisectors of a triangle are concurrent at a point called the incenter, which is equidistant from all sides.

Shortcut Trick

Given: CSA / TSA = 1 / 3

Ratio of Height to (Height + Radius) ⇒ h / (h + r) = 1 / 3

Cross-multiplying: 3h = h + r ⇒ r = 2h

Diameter (d) = 2r = 2 × (2h) = 4h

Ratio of Diameter to Height = d : h = 4h : h = 4 : 1

∴ The correct answer is 4 : 1.

Alternate Method

Given: Curved Surface Area (CSA) of a cylinder = (1/3) × Total Surface Area (TSA)

Formula for CSA = 2πrh

Formula for TSA = 2πr(h + r)

According to the question:

⇒ 2πrh = (1/3) × 2πr(h + r)

⇒ h = (1/3) × (h + r)

⇒ 3h = h + r

⇒ r = 3h − h

⇒ r = 2h

We need to find the ratio of diameter (d) to height (h):

⇒ d = 2r

⇒ d = 2 × (2h) = 4h

⇒ d/h = 4/1

⇒ d : h = 4 : 1

∴ The correct answer is 4 : 1.

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.

Surface Areas of Cone

CSA of a cone = πrl and TSA of a cone = πr(l + r), where l is the slant height (√(r² + h²)).

Surface Area of Sphere

For a sphere, the CSA and TSA are the same, given by 4πr².

Shortcut Trick

Ratio A : B (Last Year) = 4 : 3

A (Present) = 4 × (4 ÷ 3) = 16/3 | B (Present) = 3 × (6 ÷ 5) = 18/5

Ratio A : B (Present) = (16/3) : (18/5) = 80 : 54 = 40 : 27

Total units = 40 + 27 = 67 units → Rs. 8.04 lakh

1 unit = 0.12 lakh ⇒ B Last Year = (5 ÷ 6) × (27 × 0.12) = 2.7 lakh

∴ The correct answer is Rs. 2.7 lakh.

Alternate Method

Given:

Ratio of incomes of A and B (Last Year) = 4 : 3

Individual ratio A (Last : Present) = 3 : 4

Individual ratio B (Last : Present) = 5 : 6

Total Present Income = Rs. 8.04 lakh

Calculations:

⇒ Let last year incomes be A_L = 4x and B_L = 3x.

⇒ A_Present = A_L × (4 ÷ 3) = 4x × 4/3 = 16x/3

⇒ B_Present = B_L × (6 ÷ 5) = 3x × 6/5 = 18x/5

⇒ Total Present Income = 16x/3 + 18x/5 = 8.04

⇒ (80x + 54x) ÷ 15 = 8.04

⇒ 134x ÷ 15 = 8.04

⇒ 134x = 8.04 × 15 = 120.6

⇒ x = 120.6 ÷ 134 = 0.9

⇒ Income of B last year = 3x = 3 × 0.9 = 2.7 lakh

∴ The correct answer is Rs. 2.7 lakh.

Additional Information

Compounded Ratio

For two or more ratios, the compounded ratio is the ratio of the product of the antecedents to the product of the consequents (e.g., for a:b and c:d, it is ac:bd).

Inverse Ratio

The inverse ratio of a:b is b:a. It is often used in speed-time and work-efficiency problems.

Continued Proportion

Three quantities a, b, and c are in continued proportion if a:b = b:c, where b is the mean proportional (b² = ac).

Shortcut Trick

Downstream Speed (v + u) = 18 + 6 = 24 km/h

Upstream Speed (v − u) = 18 − 6 = 12 km/h

Total Time = (Distance ÷ Downstream Speed) + (Distance ÷ Upstream Speed)

Total Time = (90 ÷ 24) + (90 ÷ 12) = 3.75 + 7.5 = 11.25 hours

∴ The correct answer is 11¼ hours.

Alternate Method

Given:

Speed of boat (B) = 18 km/h

Speed of current (C) = 6 km/h

Distance (D) = 90 km

Formula Used:

Time = Distance ÷ Speed

Downstream Speed = Speed of boat + Speed of current

Upstream Speed = Speed of boat − Speed of current

Calculations:

⇒ Downstream speed = 18 + 6 = 24 km/h

⇒ Upstream speed = 18 − 6 = 12 km/h

⇒ Time for downstream trip = 90 ÷ 24 = 3.75 hours

⇒ Time for upstream trip = 90 ÷ 12 = 7.5 hours

⇒ Total time = 3.75 + 7.5 = 11.25 hours

⇒ 11.25 hours = 11 + 0.25 = 11 + (1/4) = 11¼ hours

∴ The correct answer is 11¼ hours.

Additional Information

Speed in Still Water

If downstream speed (D) and upstream speed (U) are known, Speed of boat = (D + U) ÷ 2.

Speed of Current

If downstream speed (D) and upstream speed (U) are known, Speed of current = (D − U) ÷ 2.

Average Speed for Round Trip

Average Speed = (Downstream Speed × Upstream Speed) ÷ Still Water Speed = (D × U) ÷ B.

Shortcut Trick

Net difference in percentage = Gain % − Loss % = 14% − (−16%) = 30%.

Given that 30% of CP = Rs. 33.

Target SP to gain 25% = 125% of CP.

Value of 125% = (33 ÷ 30) × 125 = 1.1 × 125 = 137.5.

∴ The correct answer is Rs. 137.5.

Alternate Method

Given:

Initial Loss = 16%

Potential Gain = 14%

Difference in Selling Price = Rs. 33

Target Gain = 25%

Formula Used:

SP = CP × (100 ± Profit/Loss %)/100

Calculations:

Let the Cost Price (CP) be 100 units.

⇒ Initial Selling Price (SP₁) = 100 − 16 = 84 units

⇒ Second Selling Price (SP₂) = 100 + 14 = 114 units

⇒ Difference in SP = 114 − 84 = 30 units

⇒ According to question, 30 units = Rs. 33

⇒ 1 unit = 33 ÷ 30 = Rs. 1.1

⇒ Required Selling Price (to gain 25%) = 100 + 25 = 125 units

⇒ New Selling Price = 125 × 1.1 = Rs. 137.5

∴ The correct answer is Rs. 137.5.

Additional Information

Profit and Loss Percentage

Profit or Loss is always calculated on the Cost Price (CP) unless stated otherwise. Profit % = (Profit/CP) × 100.

Marked Price and Discount

Discount is always allowed on the Marked Price (MP). SP = MP − Discount.

Relation between CP and MP

CP/MP = (100 − Discount%)/(100 + Profit%). This ratio is useful for solving complex markup and discount problems directly.

Shortcut Trick

Initial average = 72 kg.

Net deviation from students leaving = 10 × (72 − 78) = −60 kg.

Net deviation from students joining = 4 × (80 − 72) = +32 kg.

Total net deviation = −60 + 32 = −28 kg.

New number of students = Net deviation ÷ Change in average = −28 ÷ (−0.7) = 40.

Initial number of students (n) = New count + 10 − 4 = 40 + 6 = 46.

∴ The correct answer is 46 students.

Alternate Method

Given: Initial average = 72 kg | Students leaving = 10 (Avg 78 kg) | Students joining = 4 (Avg 80 kg) | New Average decrease = 0.7 kg.

Let the initial number of students be n.

⇒ Initial total weight = 72 × n

⇒ Weight of students who left = 10 × 78 = 780 kg

⇒ Weight of students who joined = 4 × 80 = 320 kg

⇒ Final number of students = n − 10 + 4 = n − 6

⇒ New Average = 72 − 0.7 = 71.3 kg

⇒ Final Total Weight = (n − 6) × 71.3

⇒ 72n − 780 + 320 = 71.3(n − 6)

⇒ 72n − 460 = 71.3n − 427.8

⇒ 72n − 71.3n = 460 − 427.8

⇒ 0.7n = 32.2

⇒ n = 32.2 ÷ 0.7 = 322 ÷ 7

⇒ n = 46

∴ The correct answer is 46 students.

Additional Information

Deviation Method for Averages

Net Change in Average = (Total Deviation from Mean) ÷ (Total Number of Observations). If the net deviation is negative, the average decreases.

Average weight of a Group

When members leave or join, the new total weight = (Old Average × Old Count) − (Weight of Left) + (Weight of Joined).

Weighted Average Formula

Average_combined = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂), where n is the number of items and x is their respective average.

Shortcut Trick

83 1/3% = 5/6. Let total students = 6 parts (Girls = 5, Boys = 1).

Absent Girls = 100% − 80% = 20% | Absent Boys = 100% − 60% = 40%.

Total absent parts = (5 × 20%) + (1 × 40%) = 100% + 40% = 140%.

Overall absent percentage = 140% ÷ 6 = 70/3% = 23 1/3%.

∴ The correct answer is 23 1/3%.

Alternate Method

Given:

Girls in class = 83 1/3% of total students

Present Boys = 60%, Present Girls = 80%

Formula Used:

Percentage = (Part / Total) × 100

Calculations:

⇒ Let total students = 600

⇒ Number of girls = 83 1/3% × 600 = (250/3)% × 600 = 500

⇒ Number of boys = 600 − 500 = 100

⇒ Absent boys = (100% − 60%) of 100 = 40% of 100 = 40

⇒ Absent girls = (100% − 80%) of 500 = 20% of 500 = 100

⇒ Total absent students = 40 + 100 = 140

⇒ Absent percentage = (140 / 600) × 100

⇒ Absent percentage = 140 / 6 = 70 / 3 = 23 1/3%

∴ The correct answer is 23 1/3%.

Additional Information

Percentage to Fraction Conversion

Standard fractions: 16 2/3% = 1/6, 33 1/3% = 1/3, 66 2/3% = 2/3, 83 1/3% = 5/6. Knowing these saves time in calculation.

Weighted Average Rule

Overall Average = (n₁ × x₁ + n₂ × x₂) ÷ (n₁ + n₂). Here, n represents ratios and x represents the absent percentages.

Base Value Choice

Always choose a base value (like 100 or 600) that is a multiple of the denominator of the given percentage fraction to avoid decimals.

Shortcut Trick

Interest for 2 years (3 yr − 1 yr) = 7504 − 6076 = Rs. 1428

Interest for 1 year = 1428 ÷ 2 = Rs. 714

Principal (Sum) = Amount (1 yr) − Interest (1 yr) = 6076 − 714 = Rs. 5362

Rate % = (Interest ÷ Principal) × 100 = (714 ÷ 5362) × 100 ≈ 13.32%

∴ The correct answer is Rs. 5,362 and 13.32%.

Alternate Method

Given:

Amount after 1 year (A₁) = Rs. 6076

Amount after 3 years (A₃) = Rs. 7504

Formula Used:

Simple Interest (SI) = Principal × Rate × Time ÷ 100

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

Calculations:

⇒ SI for 2 years (3 − 1 years) = A₃ − A₁

⇒ SI for 2 years = Rs. 7504 − Rs. 6076 = Rs. 1428

⇒ SI for 1 year = 1428 ÷ 2 = Rs. 714

⇒ Principal (Sum) = Amount after 1 year − SI for 1 year

⇒ Principal = 6076 − 714 = Rs. 5362

⇒ Rate of Interest (R) = (SI × 100) ÷ (P × T)

⇒ R = (714 × 100) ÷ (5362 × 1)

⇒ R ≈ 13.3159% → 13.32%

∴ The correct answer is Rs. 5,362 and 13.32%.

Additional Information

Nature of Simple Interest

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

Amount and Principal Relation

Amount after T years is A = P + (P × R × T) ÷ 100. If we know the amounts at different years, the difference in amounts gives the interest for that specific time gap.

Effective Rate of Interest

For SI, the effective rate for T years is simply R × T. In Compound Interest, the effective rate is calculated using the formula [ (1 + R/100)^T − 1 ] × 100.

Shortcut Trick

Combine ratios: A : B = 6 : 7 and B : C = 3 : 2.

A : B : C = (6 × 3) : (7 × 3) : (7 × 2) = 18 : 21 : 14.

Difference between A and C parts = 18 − 14 = 4 units.

Given 4 units = ₹540 → 1 unit = ₹135.

Total sum x = 18 + 21 + 14 = 53 units = 53 × 135 = ₹7155.

∴ The correct answer is ₹7155.

Alternate Method

Given: Ratio A : B = 6 : 7 and B : C = 3 : 2. Difference between share of A and C = ₹540.

Formula Used: To combine ratios A : B and B : C, equate the value of the common term B.

⇒ Share of A : Share of B = 6 : 7

⇒ Share of B : Share of C = 3 : 2

⇒ To make B common, multiply first ratio by 3 and second ratio by 7:

⇒ A : B = (6 × 3) : (7 × 3) = 18 : 21

⇒ B : C = (3 × 7) : (2 × 7) = 21 : 14

⇒ A : B : C = 18 : 21 : 14

⇒ Difference between shares of A and C = 18 units − 14 units = 4 units

⇒ 4 units = ₹540

⇒ 1 unit = ₹540 ÷ 4 = ₹135

⇒ Total sum x = (18 + 21 + 14) units = 53 units

⇒ x = 53 × ₹135 = ₹7155

∴ The correct answer is ₹7155.

Additional Information

Combining Ratios

If A : B = a : b and B : C = c : d, then A : B : C = (a × c) : (b × c) : (b × d).

Compound Ratio

The compound ratio of (a : b) and (c : d) is (ac : bd), which is the product of the individual ratios.

Proportional Division

If a sum S is divided into parts A, B, and C in the ratio p : q : r, then Share of A = S × p / (p + q + r).

Shortcut Trick

Profit ratio = (Investment Ratio × Time Ratio)

Profit Ratio A : B : C = (2 × 4) : (3 × 2) : (5 × 3) = 8 : 6 : 15

Difference between C and A = 15 − 8 = 7 units

Given, 7 units = ₹1,47,000 → 1 unit = ₹21,000

B’s share = 6 units = 6 × 21,000 = ₹1,26,000

∴ The correct answer is ₹1,26,000.

Alternate Method

Given: Capital Ratio = 2 : 3 : 5 and Time Ratio = 4 : 2 : 3

Formula: Profit = Capital × Time

⇒ Let Profit shares of A, B and C be 8x, 6x and 15x respectively.

⇒ Difference between C and A = 15x − 8x = 7x

⇒ 7x = 1,47,000

⇒ x = 1,47,000 ÷ 7 = 21,000

⇒ B’s share = 6x = 6 × 21,000

⇒ B’s share = 1,26,000

∴ The correct answer is ₹1,26,000.

Additional Information

Partnership Profit Division

Profit of a partner is directly proportional to both capital invested and the duration of investment: P = C × T.

Ratio of Capitals

If profits and times are known, capital ratio is calculated as: C₁ : C₂ = (P₁/T₁) : (P₂/T₂).

Ratio of Time Periods

If profits and capitals are known, time ratio is calculated as: T₁ : T₂ = (P₁/C₁) : (P₂/C₂).

Working vs Sleeping Partner

A working partner manages the business and may receive a fixed salary or extra percentage of profit before the remaining profit is divided.

Given:

9 × 6 ÷ 24 + 8 ÷ 2 of 5 – 30 ÷ 4 of 4 + 27 × 5 ÷ 9 = ?

Formula used:

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

Calculation:

9 × 6 ÷ 24 + 8 ÷ 2 of 5 – 30 ÷ 4 of 4 + 27 × 5 ÷ 9

⇒ 9 × 6 ÷ 24 + 8 ÷ 10 – 30 ÷ 16 + 27 × 5 ÷ 9

⇒ 9 × $\frac{1}{4}$ + $\frac{4}{5}$ – $\frac{15}{8}$ + 3 × 5

⇒ $\frac{9}{4} + \frac{4}{5} - \frac{15}{8} + 15$

⇒ $\frac{90 + 32 - 75 + 600}{40}$

⇒ $\frac{647}{40}$

∴ The correct answer is option (4).

Shortcut Trick

Tangents from an external point are equal: DR = DS = 6 cm.

AS = AD − DS = 24 − 6 = 18 cm, and AP = AS = 18 cm.

PB = AB − AP = 27 − 18 = 9 cm.

Since angle B = 90⁰, the radius r of the inscribed circle is equal to PB = 9 cm.

Circumference = 2 × π × r = 2 × π × 9 = 18π.

∴ The correct answer is 18π cm.

Alternate Method

Given:

∠B = 90°, AD = 24 cm, AB = 27 cm, DR = 6 cm

Circle touches AB at P, BC at Q, CD at R, and AD at S.

Formula Used:

Tangents drawn from an external point to a circle are equal.

AP = AS, BP = BQ, CR = CQ, DR = DS

Since ∠B = 90°, the radius of the circle is equal to BP = BQ.

Calculation:

DS = DR = 6 cm

⇒ AS = AD − DS

⇒ AS = 24 − 6 = 18 cm

⇒ AP = AS = 18 cm

⇒ BP = AB − AP

⇒ BP = 27 − 18 = 9 cm

Since ∠B = 90°,

⇒ r = BP = BQ = 9 cm

⇒ Circumference = 2πr

⇒ Circumference = 2 × π × 9

⇒ Circumference = 18π cm

∴ The circumference of the circle is 18π cm.

Additional Information

Tangents from External Points

The lengths of two tangents drawn from an external point to a circle are always equal (e.g., $AP = AS$).

Tangential Quadrilateral (Pitot Theorem)

For any quadrilateral circumscribed around a circle, the sum of opposite sides is equal: $AB + CD = BC + DA$.

Radius at Perpendicular Tangents

If two tangents are perpendicular to each other, the radii to the points of contact and the tangents form a square of side length $r$.

Shortcut Trick

For triangles sharing the same base, the ratio of their areas is equal to the ratio of their heights.

Area(ΔABC) ÷ Area(ΔDBC) = Height(AO) ÷ Height(DO)

x ÷ Area(ΔDBC) = a ÷ b

Area(ΔDBC) = (b ÷ a) × x = bx/a

∴ The correct answer is bx/a cm².

Alternate Method

Given: AO = a cm, DO = b cm, Area(ΔABC) = x cm². AD ⊥ BC at O.

Formula Used: Area of triangle = 1/2 × base × height.

⇒ Area(ΔABC) = 1/2 × BC × AO

⇒ x = 1/2 × BC × a

⇒ BC = 2x / a

⇒ Area(ΔDBC) = 1/2 × BC × DO

⇒ Area(ΔDBC) = 1/2 × (2x / a) × b

⇒ Area(ΔDBC) = (x × b) / a

⇒ Area(ΔDBC) = bx/a

∴ The correct answer is bx/a cm².

Additional Information

Triangle Area Ratios

If two triangles have the same base, the ratio of their areas is equal to the ratio of their altitudes: Area₁/Area₂ = h₁/h₂.

Triangles between Parallels

Two triangles having the same base and lying between the same parallel lines have equal areas.

Median Property

A median of a triangle divides it into two triangles of equal area because they share the same height and have equal bases.

Given: x > y > z $\frac{z}{x} = \frac{9}{16}$ $\frac{z}{x} = y + 0.0625$ $x + y + z = 1\frac{13}{24}$

Formula used: Basic fraction arithmetic

Calculation: ⇒ $\frac{9}{16} = y + 0.0625$ ⇒ $\frac{9}{16} = y + \frac{1}{16}$ ⇒ $y = \frac{9}{16} - \frac{1}{16} = \frac{8}{16} = \frac{1}{2}$ ⇒ $x + y + z = \frac{37}{24}$ ⇒ $x + \frac{1}{2} + z = \frac{37}{24}$ ⇒ $x + z = \frac{37}{24} - \frac{1}{2}$ ⇒ $x + z = \frac{37 - 12}{24} = \frac{25}{24}$

∴ The correct answer is option (4).

Shortcut Trick

If x = (√a − √b) / (√a + √b) and y = 1/x, then x + y = 2(a + b) / (a − b).

Substituting a = 5 and b = 3: x + y = 2(5 + 3) / (5 − 3) = 2 × 8 / 2 = 8.

Since y is the reciprocal of x, xy = 1.

Value of x³ + y³ = (x + y)³ − 3xy(x + y) = 8³ − 3(1)(8).

Final Result = 512 − 24 = 488.

∴ The correct answer is 488.

Alternate Method

Given: x = (√5 − √3) / (√5 + √3) and y = 1/x

Formula: x³ + y³ = (x + y)³ − 3xy(x + y)

⇒ Rationalizing x: x = [(√5 − √3) × (√5 − √3)] / [(√5 + √3) × (√5 − √3)]

⇒ x = (5 + 3 − 2√15) / (5 − 3) = (8 − 2√15) / 2 = 4 − √15

⇒ Similarly, y = 1/x = 4 + √15

⇒ x + y = (4 − √15) + (4 + √15) = 8

⇒ xy = (4 − √15)(4 + √15) = 16 − 15 = 1

⇒ x³ + y³ = 8³ − 3 × 1 × 8

⇒ x³ + y³ = 512 − 24 = 488

∴ The correct answer is 488.

Additional Information

Algebraic Identity for Cubes

The sum of cubes is given by a³ + b³ = (a + b)(a² − ab + b²) or a³ + b³ = (a + b)³ − 3ab(a + b).

Rationalizing the Denominator

To rationalize a denominator like (√a + √b), multiply both numerator and denominator by its conjugate (√a − √b).

Reciprocal Sum Property

If x = a − √b and y = a + √b where a² − b = 1, then x and y are reciprocals (xy = 1).

Shortcut Trick

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

Here, (2x)³ − (3y)³ = (2x − 3y)(4x² + 6xy + 9y²)

Comparing coefficients: A = 2, B = −3, C = 4, −D = 9 ⇒ D = −9

Required value: A + B + C − D = 2 − 3 + 4 − (−9) = 12

∴ The correct answer is 12.

Alternate Method

Given Equation: 8x³ − 27y³ = (Ax + By) (Cx² − Dy² + 6xy)

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

⇒ 8x³ − 27y³ = (2x)³ − (3y)³

⇒ (2x − 3y) ((2x)² + (2x)(3y) + (3y)²)

⇒ (2x − 3y) (4x² + 6xy + 9y²)

Now, comparing this with the given RHS: (Ax + By) (Cx² − Dy² + 6xy)

⇒ Ax + By = 2x − 3y ⇒ A = 2, B = −3

⇒ Cx² − Dy² + 6xy = 4x² + 9y² + 6xy

⇒ C = 4 and −D = 9 ⇒ D = −9

⇒ Value of (A + B + C − D) = 2 + (−3) + 4 − (−9)

⇒ 2 − 3 + 4 + 9 = 12

∴ The correct answer is 12.

Additional Information

Sum of Cubes Identity

The identity for the sum of two cubes is a³ + b³ = (a + b)(a² − ab + b²).

Whole Cube Expansion

(a − b)³ = a³ − b³ − 3ab(a − b), which is different from the difference of two cubes identity.

Coefficient Comparison

When two algebraic expressions are identical for all values of variables, the coefficients of corresponding terms (like x, y, x², xy) must be equal.

Shortcut Trick

Third proportional of a and b = b² ÷ a

⇒ Third proportional (P₁) = (0.8 × 0.8) ÷ 0.4 = 0.64 ÷ 0.4 = 1.6

Mean proportional of x and y = √(x × y)

⇒ Mean proportional (P₂) = √(13.5 × 0.24) = √3.24 = 1.8

Required Ratio = 1.6 : 1.8 = 16 : 18 = 8 : 9

∴ The correct answer is 8 : 9.

Alternate Method

Given:

Numbers for third proportional: a = 0.4, b = 0.8

Numbers for mean proportional: x = 13.5, y = 0.24

Formula Used:

Third Proportional (c) of a and b is given by a : b = b : c ⇒ c = b²/a

Mean Proportional (m) of x and y is given by x : m = m : y ⇒ m = √(xy)

Calculations:

⇒ Let the third proportional be P₁.

⇒ P₁ = (0.8 × 0.8) ÷ 0.4

⇒ P₁ = 0.64 ÷ 0.4 = 1.6

⇒ Let the mean proportional be P₂.

⇒ P₂ = √(13.5 × 0.24)

⇒ P₂ = √(3.24)

⇒ P₂ = 1.8

⇒ Required Ratio = P₁ : P₂

⇒ Required Ratio = 1.6 : 1.8

⇒ Multiply both by 10: 16 : 18

⇒ Divide both by 2: 8 : 9

∴ The correct answer is 8 : 9.

Additional Information

Mean Proportional

For two numbers a and b, the mean proportional x satisfies a : x = x : b, giving x = √(ab).

Third Proportional

For two numbers a and b, the third proportional c satisfies a : b = b : c, giving c = b²/a.

Fourth Proportional

For three numbers a, b, and c, the fourth proportional d satisfies a : b = c : d, giving d = (b × c)/a.

Continued Proportion

Three quantities a, b, c are in continued proportion if a : b = b : c, which implies b² = ac.

Shortcut Trick

A's investment is 33 1/3% (or 1/3) of the total capital.

Since the time period for all partners is the same, the profit sharing ratio is identical to the investment ratio.

A's share of profit = 33 1/3% of total profit → (1/3) × 1,62,000.

A's share = Rs. 54,000.

∴ The correct answer is Rs. 54,000.

Alternate Method

Given:

Total Profit = Rs. 1,62,000

A's investment = 33 1/3% of total capital

B's investment = 25% of the remaining capital

Formula Used:

Profit Share = Investment × Time period

Calculations:

⇒ Let total capital = 1 unit

⇒ A's investment = 33 1/3% = 1/3

⇒ Remaining capital = 1 − 1/3 = 2/3

⇒ B's investment = 25% of 2/3 = 1/4 × 2/3 = 1/6

⇒ C's investment = 2/3 − 1/6 = (4 − 1)/6 = 3/6 = 1/2

⇒ Investment ratio A : B : C = 1/3 : 1/6 : 1/2 = 2 : 1 : 3

⇒ Total ratio parts = 2 + 1 + 3 = 6 units

⇒ 6 units = Rs. 1,62,000

⇒ 1 unit = 1,62,000 ÷ 6 = Rs. 27,000

⇒ A's share (2 units) = 2 × 27,000 = Rs. 54,000

∴ The correct answer is Rs. 54,000.

Additional Information

Profit Sharing Rule

In any partnership, if the time period of investment is the same for all partners, the profit is distributed in the ratio of their capitals: P₁ : P₂ = C₁ : C₂.

Compound Partnership

If partners invest for different time periods, the profit is divided in the ratio of the product of capital and time: P₁ : P₂ = (C₁ × T₁) : (C₂ × T₂).

Active and Sleeping Partners

An active partner manages the business and may receive a salary or fixed percentage of profit before the remaining profit is distributed among all partners in their investment ratio.

Shortcut Trick

Solve equations: 3α − 5β = 19 and −7α + 3β = −1

Multiply Eq(1) × 3 ⇒ 9α − 15β = 57

Multiply Eq(2) × 5 ⇒ −35α + 15β = −5

Adding both: −26α = 52 ⇒ α = −2

Substitute α = −2 in Eq(1): 3(−2) − 5β = 19 ⇒ −5β = 25 ⇒ β = −5

Required: 3α − β = 3(−2) − (−5) = −6 + 5 = −1

∴ The correct answer is -1.

Alternate Method

Given Equations:

Line 1: 3x − 5y = 19 ... (i)

Line 2: 3y − 7x + 1 = 0 ⇒ −7x + 3y = −1 ... (ii)

To find the point of intersection P(α, β), solve (i) and (ii) simultaneously:

⇒ Multiply (i) by 3: 9x − 15y = 57

⇒ Multiply (ii) by 5: −35x + 15y = −5

⇒ Add the two resulting equations:

⇒ (9x − 35x) + (−15y + 15y) = 57 − 5

⇒ −26x = 52

⇒ x = 52 ÷ (−26) = −2

⇒ ∴ α = −2

⇒ Substitute x = −2 in (i):

⇒ 3(−2) − 5y = 19

⇒ −6 − 5y = 19

⇒ −5y = 19 + 6 = 25

⇒ y = 25 ÷ (−5) = −5

⇒ ∴ β = −5

⇒ Now, calculate the value of (3α − β):

⇒ 3(−2) − (−5)

⇒ −6 + 5 = −1

∴ The correct answer is -1.

Additional Information

Condition for Unique Solution

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

Cramer's Rule

Linear equations can be solved using determinants where x = D_x/D and y = D_y/D, provided the main determinant D ≠ 0.

Consistency of Equations

A system is called consistent if it has at least one solution (intersecting or coincident lines) and inconsistent if it has no solution (parallel lines).

Shortcut Trick

Divide the equation x⁴ − 83x² + 1 = 0 by x² to get x² + 1/x² = 83.

Find x − 1/x using the relation: x − 1/x = √(x² + 1/x² − 2) = √(83 − 2) = 9.

Using the cube shortcut: If x − 1/x = k, then x³ − 1/x³ = k³ + 3k.

Putting k = 9: x³ − 1/x³ = 9³ + 3(9) = 729 + 27 = 756.

∴ The correct answer is 756.

Alternate Method

Given: x⁴ − 83x² + 1 = 0

Formula Used: (a − b)³ = a³ − b³ − 3ab(a − b)

⇒ x⁴ − 83x² + 1 = 0

⇒ (x⁴ + 1) / x² = 83x² / x²

⇒ x² + 1/x² = 83

⇒ (x − 1/x)² + 2 = 83

⇒ (x − 1/x)² = 81

⇒ x − 1/x = 9

⇒ (x − 1/x)³ = x³ − 1/x³ − 3(x)(1/x)(x − 1/x)

⇒ 9³ = x³ − 1/x³ − 3(9)

⇒ 729 = x³ − 1/x³ − 27

⇒ x³ − 1/x³ = 729 + 27

⇒ x³ − 1/x³ = 756

∴ The correct answer is 756.

Additional Information

Reciprocal Identities for Squares

If x + 1/x = k, then x² + 1/x² = k² − 2. Conversely, if x − 1/x = k, then x² + 1/x² = k² + 2.

Reciprocal Identities for Cubes

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

Higher Power Reductions

Equations of the form ax⁴ + bx² + a = 0 are symmetric and can be solved by dividing the entire equation by x² to create reciprocal terms.

Shortcut Trick

For a number to be divisible by 72, it must be divisible by its coprime factors 8 and 9.

Divisibility by 8: Last three digits 67y ÷ 8 ⇒ 672 ÷ 8 = 84, so y = 2.

Divisibility by 9: Sum of digits (5+6+7+8+x+4+3+2+6+7+y) = 48 + x + y must be divisible by 9.

For y = 2, sum = 50 + x ⇒ Nearest multiple of 9 is 54, so x = 4.

Required value = √(5 × 4 + 8 × 2) = √(20 + 16) = √36 = 6.

∴ The correct answer is 6.

Alternate Method

Given:

Number: 5678x43267y

Divisibility: Divisible by 72

Formula Used:

Divisibility Rule of 8: Last three digits of the number must be divisible by 8.

Divisibility Rule of 9: Sum of all digits must be divisible by 9.

Calculations:

⇒ Divisibility by 8: Last three digits = 67y

⇒ 67y ÷ 8 ⇒ 672 ÷ 8 = 84 (Exact division)

⇒ Therefore, y = 2

⇒ Divisibility by 9: Sum of digits = 5+6+7+8+x+4+3+2+6+7+2

⇒ Sum = 50 + x

⇒ For (50 + x) to be divisible by 9, the next multiple is 54

⇒ x = 54 − 50 = 4

⇒ Value to find = √(5x + 8y)

⇒ √(5 × 4 + 8 × 2) = √(20 + 16)

⇒ √36 = 6

∴ The correct answer is 6.

Additional Information

Divisibility by 72

A number is divisible by 72 if and only if it is divisible by both 8 and 9, as 8 and 9 are coprime factors (HCF = 1).

Divisibility by 11

The difference between the sum of digits at odd places and the sum of digits at even places must be 0 or a multiple of 11.

Coprime Factor Rule

If a number N is divisible by two coprime numbers a and b, then N is also divisible by the product a × b.

Shortcut Trick

Volume ratio method: Number of bottles (n) = Volume of bowl ÷ Volume of bottle

n = [(2/3) × R³] ÷ [r² × h] (where π cancels out)

n = (2 × 18 × 18 × 18) ÷ (3 × 3 × 3 × 12)

n = (2 × 6 × 6 × 1.5) × 1.5? Let us simplify: n = (2 × 5832) ÷ (3 × 9 × 12) = 11664 ÷ 324 = 36

∴ The correct answer is 36 bottles.

Alternate Method

Given:

Diameter of bowl = 36 cm ⇒ Radius (R) = 18 cm

Radius of bottle (r) = 3 cm

Height of bottle (h) = 12 cm

Formula Used:

Volume of hemispherical bowl = (2/3)πR³

Volume of cylindrical bottle = πr²h

Calculations:

⇒ Volume of liquid in bowl = (2/3) × π × (18)³

⇒ Volume of liquid in bowl = (2/3) × π × 5832 = 3888π cm³

⇒ Volume of one bottle = π × (3)² × 12

⇒ Volume of one bottle = π × 9 × 12 = 108π cm³

⇒ Number of bottles (n) = (3888π) ÷ (108π)

⇒ n = 3888 ÷ 108

⇒ n = 36

∴ The correct answer is 36 bottles.

Additional Information

Volume of Sphere

The volume of a complete sphere is given by (4/3)πr³, which is exactly double the volume of a hemisphere of the same radius.

Surface Area of Hemisphere

Curved Surface Area (CSA) = 2πr² and Total Surface Area (TSA) = 3πr² (including the circular base).

Conservation of Volume

When a liquid is transferred from one container to multiple smaller containers, the total volume remains constant: V_total = n × V_small.

Shortcut Trick

AB is diameter ⇒ ∠ACB = 90° (angle in semicircle)

∠ACB is exterior angle of ▵BCE ⇒ ∠CBE = 90° − ∠AEB = 90° − 68° = 22°

∠COD = 2 × ∠CBE (central angle = 2 × inscribed angle on same arc)

∠COD = 2 × 22° = 44°

∴ The correct answer is 44°.

Alternate Method

Given:

∠AEB = 68°

Formula used:

Angle in a semicircle = 90°

Exterior angle of a triangle = Sum of the two opposite interior angles

Angle at the centre = 2 × Angle at the circumference standing on the same arc

Calculation:

AB is the diameter of the circle.

⇒ ∠ACB = 90°

∠ACB is the exterior angle of △BCE.

⇒ ∠ACB = ∠BEC + ∠CBE

⇒ 90° = 68° + ∠CBE

⇒ ∠CBE = 90° − 68°

⇒ ∠CBE = 22°

⇒ ∠COD = 2 × ∠CBE

⇒ ∠COD = 2 × 22°

⇒ ∠COD = 44°

∴ ∠COD = 44°.

Additional Information

Angle in a Semicircle

Any angle subtended by the diameter at the circumference of the circle is always 90°.

Central Angle Theorem

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

External Intersection of Chords

If two chords AC and BD intersect at point E outside the circle, then ∠AEB = 1/2 × (major arc − minor arc).

Shortcut Trick

LCM(15, 18, 20, 27) = 540.

x = 540k + 10 must be divisible by 31.

Testing k: (13k + 10) ÷ 31 ⇒ 13(4) + 10 = 62, which is 31 × 2.

x = 540 × 4 + 10 = 2170.

Next perfect square after 2170 is 47² = 2209.

Number to be added = 2209 − 2170 = 39.

∴ The correct answer is 39.

Alternate Method

Given: Divisors are 15, 18, 20, 27 with remainder 10. x is a multiple of 31.

Step 1: Find LCM of 15, 18, 20, and 27.

⇒ 15 = 3 × 5

⇒ 18 = 2 × 3²

⇒ 20 = 2² × 5

⇒ 27 = 3³

⇒ LCM = 2² × 3³ × 5 = 4 × 27 × 5 = 540

Step 2: Find the least multiple of 31 in the form 540k + 10.

⇒ (540k + 10) ÷ 31

⇒ (31 × 17k + 13k + 10) ÷ 31

⇒ 13k + 10 should be a multiple of 31.

⇒ For k = 4: 13 × 4 + 10 = 52 + 10 = 62 (Divisible by 31)

⇒ x = 540 × 4 + 10 = 2170

Step 3: Find the next perfect square.

⇒ √2170 ≈ 46.58

⇒ Next integer = 47

⇒ 47² = 2209

⇒ Difference = 2209 − 2170 = 39

∴ The correct answer is 39.

Additional Information

General Form of Numbers with Constant Remainder

Any number x which when divided by a, b, c leaves remainder R is given by x = LCM(a, b, c) × k + R.

Perfect Square Estimation

To find the number to be added, calculate √x, take the next integer n, and find n² − x.

Divisibility Check for Variables

When solving (Ak + B) ÷ D, reduce A by taking A mod D to simplify the testing of k values.

Shortcut Trick

Ratio of efficiencies X : Y = 5 : 4

Total efficiency of (X + Y) = 5 + 4 = 9 units/day

Total work = Combined efficiency × Days = 9 × 10 = 90 units

Work done by Y in 5 days = 4 × 5 = 20 units

Remaining work = 90 − 20 = 70 units

Time taken by X alone = 70 ÷ 5 = 14 days

∴ The correct answer is 14 days.

Alternate Method

Given:

Efficiency ratio of X to Y = 5 : 4

Time taken by X and Y together = 10 days

Time for which Y worked alone = 5 days

Formula Used:

Total Work = Total Efficiency × Total Time

Time = Work ÷ Efficiency

Calculations:

⇒ Let efficiency of X = 5 units/day and efficiency of Y = 4 units/day

⇒ Combined efficiency of X and Y = 5 + 4 = 9 units/day

⇒ Total work = 9 × 10 = 90 units

⇒ Work completed by Y in 5 days = 4 × 5 = 20 units

⇒ Remaining work = Total work − Work done by Y

⇒ Remaining work = 90 − 20 = 70 units

⇒ Time taken by X to complete remaining work = Remaining work ÷ Efficiency of X

⇒ Time taken by X = 70 ÷ 5 = 14 days

∴ The correct answer is 14 days.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a work. If the efficiency ratio is a:b, the time ratio will be b:a.

Work and Wages

Wages are always distributed in the ratio of the actual work done by each person, which is Efficiency × Time.

Combined Work Formula

If A takes 'x' days and B takes 'y' days, together they take (x × y) ÷ (x + y) days to complete the work.

Shortcut Trick

Units digit of 4^{odd power} is always 4.

Units digit of 3³³⁷ → 337 ÷ 4 gives remainder 1 → 3¹ = 3.

Units digit of 7⁷²⁶ → 726 ÷ 4 gives remainder 2 → 7² = 49 → 9.

Expression units digit = 4 + 3 − 9 = −2.

Since the digit cannot be negative, add 10 → 10 − 2 = 8.

∴ The correct answer is 8.

Alternate Method

Given: x = (164)¹⁶⁹ + (333)³³⁷ − (727)⁷²⁶

To find the units digit of x, we calculate the units digit of each term individually:

⇒ Term 1: (164)¹⁶⁹ → last digit of base is 4. Power 169 is odd. 4^odd always ends in 4.

⇒ Term 2: (333)³³⁷ → last digit of base is 3. Divide power 337 by 4 → 337 = (4 × 84) + 1. Remainder = 1. Units digit = 3¹ = 3.

⇒ Term 3: (727)⁷²⁶ → last digit of base is 7. Divide power 726 by 4 → 726 = (4 × 181) + 2. Remainder = 2. Units digit = 7² = 49 → 9.

⇒ Units digit of x = (Units digit of Term 1) + (Units digit of Term 2) − (Units digit of Term 3)

⇒ Units digit of x = 4 + 3 − 9

⇒ Units digit of x = 7 − 9 = −2

⇒ If the result is negative, add 10: −2 + 10 = 8.

∴ The correct answer is 8.

Additional Information

Cyclicity of Digits

The digits 2, 3, 7, and 8 follow a cycle of 4. For any power n, the units digit is the same as the base raised to (n mod 4). If the remainder is 0, use 4 as the power.

Parity Rule for 4 and 9

For base 4: 4^odd ends in 4 and 4^even ends in 6. For base 9: 9^odd ends in 9 and 9^even ends in 1.

Handling Negative Results

In units digit arithmetic, if the subtraction results in a negative value, add 10 to obtain the correct positive units digit representing the borrow from the tens place.

Shortcut Trick

Total time A and B worked = 10:30 AM to 8:30 PM = 10 hours.

Total time C worked = 10:30 AM to 2:30 PM = 4 hours.

Fraction of tank filled by A and B in 10 hours = 10/16 + 10/24 = 5/8 + 5/12 = (15 + 10)/24 = 25/24.

Excess work to be emptied by C in 4 hours = 25/24 − 1 = 1/24 of the tank.

If C empties 1/24 of the tank in 4 hours, then C empties full tank in = 4 × 24 = 96 hours.

∴ The correct answer is 96 hours.

Alternate Method

Given:

Time taken by A = 16 hours

Time taken by B = 24 hours

Time taken by C = x hours (Outlet pipe)

Let the Total Work (Capacity) be the LCM of 16 and 24 = 48 units.

Calculations:

⇒ Total duration A and B were open (10:30 AM to 8:30 PM) = 10 hours

⇒ Total duration C was open (10:30 AM to 2:30 PM) = 4 hours

⇒ Total work done by A and B in 10 hours = 10 × (3 + 2) = 50 units

⇒ Since the tank capacity is 48 units, the extra 2 units were emptied by C.

⇒ Work emptied by C in 4 hours = 50 − 48 = 2 units

⇒ Efficiency of Pipe C = 2 ÷ 4 = 0.5 units/hr

⇒ Time taken by C to empty the full tank (x) = Total Capacity ÷ Efficiency of C

⇒ x = 48 ÷ 0.5 = 96 hours

∴ The correct answer is 96 hours.

Additional Information

Negative Efficiency

In Pipes and Cisterns, an inlet pipe has positive efficiency as it adds water, while an outlet pipe (drain pipe) has negative efficiency as it removes water.

Combined Rate Formula

If Pipe A fills in 'a' hours and Pipe B empties in 'b' hours, the net work done in 1 hour is (1/a − 1/b). If the result is positive, the tank fills; if negative, it empties.

LCM Method for Capacity

Taking the LCM of individual times as the total capacity of the tank simplifies calculations by converting fractions into whole numbers for efficiency.

Shortcut Trick

Let the initial expenditure be 100.

New Price = 100 + 19 = 119; Target Expenditure = 100 + 12 = 112.

Quantity reduction % = (New Price − Target Expenditure) ÷ New Price × 100

⇒ (119 − 112) ÷ 119 × 100 = 700 ÷ 119 ≈ 5.88%.

∴ The correct answer is 6%.

Alternate Method

Given: Price increase = 19%, Intended expenditure increase = 12%

Formula: Expenditure = Price × Quantity

⇒ Let original Price = 100 and original Quantity = 100.

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

⇒ New Price = 100 + 19 = 119.

⇒ New Expenditure = 10,000 + 12% of 10,000 = 11,200.

⇒ New Quantity = 11,200 ÷ 119 ≈ 94.117.

⇒ Percentage reduction in quantity = 100 − 94.117 = 5.883%.

⇒ Nearest integer = 6%.

∴ The correct answer is 6%.

Additional Information

Price-Consumption-Expenditure Rule

When the price of a commodity increases by P% and the expenditure increases by E%, the new consumption is (100+E)/(100+P) of the original consumption.

Constant Expenditure Case

If the expenditure remains constant (0% increase), the percentage reduction in consumption is given by [r/(100+r)] × 100, where r is the percentage increase in price.

Successive Percentage Formula

Net % change in expenditure can also be calculated as [a + b + (ab/100)]%, where a is % change in price and b is % change in consumption.

The correct answer is 8

Key Points

• Let the two-digit number be represented as 10x + y, where x is the tens digit and y is the units digit.
• The sum of the digits is (x + y).
• According to the first condition: When the number is multiplied by the sum of its digits, the product is 424.
  

(10x + y)(x + y) = 424 −−− (Equation 1)

• The number obtained by interchanging its digits is 10y + x.
• According to the second condition: When this new number is multiplied by the sum of the digits, the result is 280.
  

(10y + x)(x + y) = 280 −−− (Equation 2)

• To find the sum of the digits, add Equation 1 and Equation 2:
  

(10x + y)(x + y) + (10y + x)(x + y) = 424 + 280
(x + y) [(10x + y) + (10y + x)] = 704
(x + y) [11x + 11y] = 704
(x + y) * 11(x + y) = 704

• Simplify the equation:
  

11(x + y)² = 704
(x + y)² = 704 ÷ 11
(x + y)² = 64

• Taking the square root of both sides:
  

x + y = √64
x + y = 8

• The sum of the digits of the given number is 8. Therefore, Option 4 is the correct answer.

Additional Information

• A two-digit number can always be expressed in the form 10t + u, where t is the tens place and u is the units place.
• If we need to find the specific number in this problem:
  

We know x + y = 8.
From Equation 1: (10x + y)(8) = 424, which means 10x + y = 53.
By comparing, the tens digit x = 5 and the units digit y = 3.
The original number is 53, and the interchanged number is 35.

• Check: 35 * 8 = 280, which matches the second condition given in the question.
• The sum of digits (5 + 3) is 8.

Shortcut Trick

Volume of Sphere = 4851 cm³

⇒ r³ = 4851 × 3/4 × 7/22 = 441 × 21/8 = (21/2)³ ⇒ r = 10.5 cm

Surface Area = 4 π r² = 4 × 22/7 × 10.5 × 10.5

Surface Area = 88/7 × 110.25 = 1386 cm²

∴ The correct answer is 1386 cm².

Alternate Method

Given:

Volume (V) = 4851 cm³

π = 22/7

Formula Used:

Volume of sphere = (4/3) × π × r³

Surface area of sphere = 4 × π × r²

Calculations:

⇒ (4/3) × (22/7) × r³ = 4851

⇒ r³ = (4851 × 3 × 7) ÷ (4 × 22)

⇒ r³ = (441 × 21) ÷ 8

⇒ r³ = (21 × 21 × 21) ÷ (2 × 2 × 2)

⇒ r = 21/2 = 10.5 cm

⇒ Surface Area = 4 × (22/7) × (21/2) × (21/2)

⇒ Surface Area = (4 × 22 × 21 × 21) ÷ (7 × 2 × 2)

⇒ Surface Area = (88 × 441) ÷ 28

⇒ Surface Area = 22 × 63 = 1386 cm²

∴ The correct answer is 1386 cm².

Additional Information

Properties of Sphere and Hemisphere

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

Hemisphere Formulas

For a hemisphere: Volume = (2/3)πr³, Curved Surface Area = 2πr², and Total Surface Area = 3πr².

Ratio of Volume to Surface Area

For a sphere, the ratio of Volume to Surface Area is r/3.

Shortcut Trick

Break the surds using the rule: √(x ± 2√y) = √m ± √n where m+n=x and m×n=y.

√(10 − 2√21) = √7 − √3 (since 7+3=10 and 7×3=21)

√(8 + 2√15) = √5 + √3 (since 5+3=8 and 5×3=15)

Sum = (√7 − √3) + (√5 + √3) = √7 + √5

Comparing with √a + √b, we get a = 7 and b = 5.

√(ab) = √(7 × 5) = √35 ≈ 5.9

∴ The correct answer is 5.9.

Alternate Method

Given Expression: √(10 − 2√21) + √(8 + 2√15) = √a + √b

Step 1: Simplify √(10 − 2√21)

⇒ √(10 − 2√21) = √[7 + 3 − 2√(7 × 3)]

⇒ √[(√7)² + (√3)² − 2√7√3]

⇒ √[(√7 − √3)²] = √7 − √3

Step 2: Simplify √(8 + 2√15)

⇒ √(8 + 2√15) = √[5 + 3 + 2√(5 × 3)]

⇒ √[(√5)² + (√3)² + 2√5√3]

⇒ √[(√5 + √3)²] = √5 + √3

Step 3: Combine the simplified terms

⇒ (√7 − √3) + (√5 + √3) = √7 + √5

⇒ √a + √b = √7 + √5

⇒ a = 7, b = 5 (as a and b are positive integers)

Step 4: Calculate √(ab)

⇒ √(ab) = √(7 × 5) = √35

⇒ Since 5.9² = 34.81 and 6.0² = 36

⇒ √35 ≈ 5.916

∴ The correct answer is 5.9.

Additional Information

Square Root of a Surd

A surd of the form √(x ± 2√y) can be simplified to √m ± √n only when x = m + n and y = mn. This is based on the identity (a ± b)² = a² + b² ± 2ab.

Root Approximations

Common square root values helpful in exams: √2 ≈ 1.414, √3 ≈ 1.732, √5 ≈ 2.236, √6 ≈ 2.449, √7 ≈ 2.645.

Comparing Surds

To compare √a + √b with √c + √d, if a+b = c+d, the one with the larger product ab is larger.

Shortcut Trick Successive percentage change formula: Net % = a + b + (a × b)/100 Here, markup percentage (a) = +36% and discount percentage (b) = −40% Net change = 36 − 40 + (36 × −40)/100 Net change = −4 − 14.4 = −18.4% Negative sign indicates a loss of 18.4%. ∴ The correct answer is 18.4%. Alternate Method Given: Markup % = 36% Discount % = 40% Formula Used: Marked Price (MP) = CP × (100 + Markup%)/100 Selling Price (SP) = MP × (100 − Discount%)/100 Loss % = [(CP − SP) / CP] × 100

Calculations: ⇒ Let Cost Price (CP) = 100 ⇒ Marked Price (MP) = 100 + 36% of 100 = 136 ⇒ Discount = 40% of 136 = (40 ÷ 100) × 136 = 54.4 ⇒ Selling Price (SP) = 136 − 54.4 = 81.6 ⇒ Loss = CP − SP = 100 − 81.6 = 18.4 ⇒ Loss Percentage = (Loss / CP) × 100 ⇒ Loss Percentage = (18.4 / 100) × 100 = 18.4% ∴ The correct answer is 18.4%. Additional Information Marked Price and Markup Marked price is the price displayed on the product. Markup % = [(MP − CP) / CP] × 100. Successive Percentage Rule When a value is changed by x% and then by y%, the net change is [x + y + (xy/100)]%. Profit and Loss Relation Profit occurs when SP > CP, and Loss occurs when CP > SP. Both are always calculated on CP unless stated otherwise.

Shortcut Trick

Assume Total Employees = 120 units (LCM of 3, 8, 5, 15).

Males = 2/3 × 120 = 80 units; Females = 1/3 × 120 = 40 units.

Permanent: Male = 5/8 × 80 = 50; Female = 3/5 × 40 = 24. Total = 74 units.

Given 74 units = 740 ⇒ 1 unit = 10.

Requirement: (7/15 × 120) − (Temp Female: 2/5 × 40) = 56 − 16 = 40 units.

∴ The correct answer is 400.

Alternate Method

Given:

Ratio of Male to Female = 2 : 1

Temporary Male = 3/8 of total males | Temporary Female = 2/5 of total females

Total Permanent employees = 740

Calculations:

⇒ Let total employees = 120x

⇒ Males = 2/3 × 120x = 80x; Females = 1/3 × 120x = 40x

⇒ Permanent Males = (1 − 3/8) × 80x = 5/8 × 80x = 50x

⇒ Permanent Females = (1 − 2/5) × 40x = 3/5 × 40x = 24x

⇒ Total Permanent = 50x + 24x = 74x = 740 ⇒ x = 10

⇒ Total employees = 120 × 10 = 1200

⇒ 7/15 of total = 7/15 × 1200 = 560

⇒ Temporary females = 2/5 × (40 × 10) = 2/5 × 400 = 160

⇒ Difference = 560 − 160 = 400

∴ The correct answer is 400.

Additional Information

Fractional Parts and Ratios

When multiple fractions like 2/3, 3/8, and 2/5 are involved, assuming the total as the LCM of denominators (3, 8, 5) simplifies the calculation by avoiding decimals.

Complementary Fractions

If x/y are temporary, then (1 − x/y) are permanent. For example, if 3/8 are temporary, then 5/8 are permanent.

Base of Percentage/Fraction

Always identify the base: 3/8 is of males, whereas 7/15 is of the total employees. Multiplying a fraction by the wrong base is a common error.

Shortcut Trick

Multiply the given equation by 4 to get 4x + 1/(4x) = 12.

Use the standard algebraic identity: If a + 1/a = n, then a³ + 1/a³ = n³ − 3n.

Substituting n = 12, we get 64x³ + 1/64x³ = 12³ − 3(12) = 1728 − 36 = 1692.

The required expression 16x³ + 1/256x³ is exactly 1/4 of (64x³ + 1/64x³).

Value = 1692 ÷ 4 = 423.

∴ The correct answer is 423.

Alternate Method

Given: x + 1/16x = 3

To find: 16x³ + 1/256x³

Formula used: (a + b)³ = a³ + b³ + 3ab(a + b)

⇒ x + 1/16x = 3

Multiplying both sides by 4:

⇒ 4(x + 1/16x) = 4 × 3

⇒ 4x + 1/4x = 12

Now, cubing both sides of the equation:

⇒ (4x + 1/4x)³ = 12³

⇒ (4x)³ + (1/4x)³ + 3 × 4x × (1/4x) × (4x + 1/4x) = 1728

⇒ 64x³ + 1/64x³ + 3 × 1 × 12 = 1728

⇒ 64x³ + 1/64x³ + 36 = 1728

⇒ 64x³ + 1/64x³ = 1728 − 36 = 1692

Dividing the entire equation by 4 to get the required form:

⇒ (64x³ + 1/64x³) ÷ 4 = 1692 ÷ 4

⇒ 16x³ + 1/256x³ = 423

∴ The correct answer is 423.

Additional Information

Square Identity

If k + 1/k = n, then k² + 1/k² = n² − 2. This is derived from (k + 1/k)² = k² + 1/k² + 2.

Difference Cube Identity

If k − 1/k = n, then k³ − 1/k³ = n³ + 3n. This is derived from (k − 1/k)³ = k³ − 1/k³ − 3(k − 1/k).

Higher Power Relation

To find k⁴ + 1/k⁴, square the result of k² + 1/k² and subtract 2, i.e., (k² + 1/k²)² − 2.

Shortcut Trick

Effective rate for first 2 years = 20 + 20 + (20 × 20)/100 = 44%

Rate for remaining 2/5 year = 20 × (2/5) = 8%

Total effective rate = 44 + 8 + (44 × 8)/100 = 52 + 3.52 = 55.52%

Compound Interest = 7200 × 55.52% = 3997.44 ≈ Rs. 3,997

∴ The correct answer is Rs. 3,997.

Alternate Method

Given: Principal (P) = Rs. 7200, Rate (R) = 20% p.a., Time (n) = 2 (2/5) years

Formula: Amount = P × (1 + R/100)^{full years} × (1 + (R × fraction)/100)

⇒ Amount (A) = 7200 × (1 + 20/100)² × (1 + (20 × 2/5)/100)

⇒ A = 7200 × (1.2)² × (1 + 8/100)

⇒ A = 7200 × 1.44 × 1.08

⇒ A = 10368 × 1.08 = 11197.44

⇒ Compound Interest (CI) = Amount − Principal

⇒ CI = 11197.44 − 7200 = 3997.44

⇒ Nearest Integer = Rs. 3,997

∴ The correct answer is Rs. 3,997.

Additional Information

Effective Interest Rate for Fractional Years

When the time is in fraction form (T + a/b), the rate for the fractional part is calculated as R × (a/b)% and then compounded with the previous periods.

Successive Percentage Change

For two successive changes of x% and y%, the net change is (x + y + (x × y)/100)%. This is highly effective for calculating CI for 2 years.

CI vs SI for 2 Years

The difference between CI and SI for 2 years is P(R/100)². For the 1st year, CI and SI are always equal.

Shortcut Trick

Use the direct cubic identity: x³ + y³ + z³ − 3xyz = (x + y + z)[(x + y + z)² − 3(xy + yz + zx)]

Substitute values: x³ + y³ + z³ − 3(−12) = 2[2² − 3(−11)]

⇒ x³ + y³ + z³ + 36 = 2[4 + 33] = 2 × 37 = 74

⇒ x³ + y³ + z³ = 74 − 36 = 38

⇒ Required value = √(38 − 2) = √36 = 6

∴ The correct answer is 6.

Alternate Method

Given: x + y + z = 2, xy + yz + zx = −11, xyz = −12

Formula used: (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

⇒ 2² = x² + y² + z² + 2(−11)

⇒ 4 = x² + y² + z² − 22

⇒ x² + y² + z² = 26

⇒ x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − (xy + yz + zx))

⇒ x³ + y³ + z³ − 3(−12) = 2(26 − (−11))

⇒ x³ + y³ + z³ + 36 = 2(26 + 11)

⇒ x³ + y³ + z³ + 36 = 2 × 37 = 74

⇒ x³ + y³ + z³ = 38

⇒ Expression = √(x³ + y³ + z³ − 2)

⇒ Expression = √(38 − 2) = √36 = 6

∴ The correct answer is 6.

Additional Information

Algebraic Identity for Cubes

The formula a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca) is a fundamental identity used to solve symmetric equations.

Special Case: a + b + c = 0

If the sum of three variables is zero (a + b + c = 0), then the sum of their cubes equals three times their product: a³ + b³ + c³ = 3abc.

Square of a Trinomial

The square of a sum of three terms is given by (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

Shortcut Trick

Net deviation from the original average (74):

First 17 numbers: 17 × (72.8 − 74) = 17 × (−1.2) = −20.4

Last 17 numbers: 17 × (77.2 − 74) = 17 × (+3.2) = +54.4

Excess value of 17^th number = −20.4 + 54.4 = 34

Average of remaining 32 numbers = 74 − (Excess ÷ Remaining Count) = 74 − (34 ÷ 32) = 74 − 1.0625 = 72.9375

∴ The correct answer is 72.9.

Alternate Method

Given:

Total numbers (n) = 33 | Average = 74

First 17 numbers average = 72.8 | Last 17 numbers average = 77.2

Formula used: Sum = Average × Number of items

⇒ Sum of all 33 numbers = 33 × 74 = 2442

⇒ Sum of first 17 numbers = 17 × 72.8 = 1237.6

⇒ Sum of last 17 numbers = 17 × 77.2 = 1312.4

⇒ Sum of (first 17 + last 17) numbers = 1237.6 + 1312.4 = 2550

⇒ 17^th number = 2550 − 2442 = 108

⇒ Sum of remaining 32 numbers = 2442 − 108 = 2334

⇒ New Average = 2334 ÷ 32 = 72.9375

⇒ New Average (rounded) ≈ 72.9

∴ The correct answer is 72.9.

Additional Information

Net Deviation Method

In any group, the sum of deviations of all observations from their mean is always zero (∑(x_i − x_avg) = 0).

Effect of Exclusion

If a number greater than the average is excluded, the new average decreases. If a number smaller than the average is excluded, the new average increases.

New Average Formula

New Average = (Old Sum − Excluded Value) ÷ (Original Count − 1)

Shortcut Trick

Base Area = (√3/4) × 8 × 8 = 16√3 cm²

Volume = (1/3) × Base Area × Height

Volume = (1/3) × 16√3 × 24√3 = 16 × 8 × 3 = 384

∴ The correct answer is 384 cm³.

Alternate Method

Given:

Side of equilateral triangle base (s) = 8 cm

Height of the pyramid (h) = 24√3 cm

Formula Used:

Area of equilateral triangle = (√3/4) × s²

Volume of pyramid = (1/3) × Base Area × Height

Calculations:

⇒ Base Area = (√3 / 4) × 8 × 8

⇒ Base Area = (√3 / 4) × 64

⇒ Base Area = 16√3 cm²

⇒ Volume = (1/3) × 16√3 × 24√3

⇒ Volume = (1/3) × 16 × 24 × (√3 × √3)

⇒ Volume = (1/3) × 16 × 24 × 3

⇒ Volume = 16 × 24

⇒ Volume = 384 cm³

∴ The correct answer is 384 cm³.

Additional Information

Slant Height and Slant Edge

Slant height (l) is √(h² + r²) where r is the inradius of the base. Slant edge (e) is √(h² + R²) where R is the circumradius.

Surface Area of Pyramid

Lateral Surface Area = (1/2) × Perimeter of base × Slant height. Total Surface Area = LSA + Base Area.

Prism vs Pyramid Volume

The volume of a pyramid is exactly one-third of the volume of a prism with the same base area and height (V = Base Area × h).

Shortcut Trick

Profit is given in terms of the Cost Price (CP) of 3 articles.

Number of articles sold = 2.

Profit % = (Gain in articles / Number of articles sold) × 100

Profit % = (3 / 2) × 100 = 1.5 × 100 = 150%.

∴ The correct answer is 150%.

Alternate Method

Given:

Selling Price (SP) of 2 articles = ₹800

Gain = CP of 3 articles

Let the Cost Price (CP) of 1 article be ₹x.

⇒ Total CP of 2 articles = ₹2x

⇒ Gain = CP of 3 articles = ₹3x

⇒ Formula for Profit % = (Profit ÷ CP) × 100

⇒ Profit % = (3x ÷ 2x) × 100

⇒ Profit % = 3/2 × 100

⇒ Profit % = 1.5 × 100

⇒ Profit % = 150%

∴ The correct answer is 150%.

Additional Information

Profit and Loss on Cost Price

Unless otherwise stated, profit or loss percentage is always calculated on the Cost Price (CP). Profit % = (SP − CP)/CP × 100.

Selling Price Formula

If there is a gain of G%, then SP = CP × (100 + G) ÷ 100. If there is a loss of L%, then SP = CP × (100 − L) ÷ 100.

Marked Price and Discount

Discount is the reduction offered on the Marked Price (MP) to arrive at the Selling Price (SP). SP = MP − Discount.

The correct answer is 17.8

Key Points

• Assume the Initial Cost Price (CP) of the goods is 100 units.
• Since the man sells the goods at a 20% profit, the Initial Selling Price (SP) is calculated as: 100 + (20% of 100) = 120 units.
• The problem states that the New Cost Price (CP) increases by 10%. Thus, the New CP = 100 + (10% of 100) = 110 units.
• The man then sells them at an 8% higher price than the initial selling price. The New Selling Price (SP) = 120 + (8% of 120).
• Calculating 8% of 120: (8 ÷ 100) * 120 = 9.6. So, the New SP = 120 + 9.6 = 129.6 units.
• The New Profit earned is the difference between the New SP and the New CP: 129.6 − 110 = 19.6 units.
• To find the Profit Percentage, we use the formula: (Profit ÷ New CP) * 100.
• Calculation: (19.6 ÷ 110) * 100 = 1960 ÷ 110 = 17.818...%
• When rounded to one decimal place, the profit percent is 17.8%.

Additional Information

• Profit is defined as the surplus amount gained when the Selling Price (SP) exceeds the Cost Price (CP). Formula: Profit = SP − CP.
• The Profit Percentage is always calculated on the Cost Price unless specifically mentioned otherwise in the problem. Formula: Profit % = (Profit ÷ CP) * 100.
• When there is a Percentage Increase in a value, the new value can be found using: New Value = Original Value * (1 + Rate ÷ 100). For an 8% increase on 120: 120 * (1 + 8 ÷ 100) = 120 * 1.08 = 129.6.

Shortcut Trick

Term 1: tan 29° cot 61° − cosec² 61° = cot² 61° − cosec² 61° = −1

Term 2: cot² 54° − sec² 36° = cot² 54° − cosec² 54° = −1

Series: sin² 1° + sin² 3° + ... + sin² 89° has n = [(89 − 1) ÷ 2] + 1 = 45 terms.

Value of series = (44 ÷ 2) + sin² 45° = 22 + 1/2 = 22.5.

Final Value = −1 − 1 + 22.5 = 20.5.

∴ The correct answer is 20 1/2.

Alternate Method

Given: (tan 29° cot 61° − cosec² 61°) + cot² 54° − sec² 36° + (sin² 1° + sin² 3° + ... + sin² 89°)

Formula Used: tan A = cot(90 − A); sec A = cosec(90 − A); sin² A + sin²(90 − A) = 1; cot² θ − cosec² θ = −1

⇒ Part 1: tan 29° cot 61° − cosec² 61° = cot 61° × cot 61° − cosec² 61° = cot² 61° − cosec² 61° = −1

⇒ Part 2: cot² 54° − sec² 36° = cot² 54° − cosec² 54° = −1

⇒ Part 3 (Series): sin² 1° + sin² 3° + ... + sin² 89°

⇒ Angle AP: a = 1, d = 2, l = 89. Number of terms n = [(89 − 1) ÷ 2] + 1 = 45.

⇒ Number of complementary pairs (A + B = 90°) = 44 ÷ 2 = 22.

⇒ Sum = (22 × 1) + sin² 45° = 22 + (1/√2)² = 22 + 1/2 = 22.5.

⇒ Total = (−1) + (−1) + 22.5 = 20.5.

∴ The correct answer is 20 1/2.

Additional Information

Complementary Trigonometric Ratios

When A + B = 90°, sin² A + sin² B = 1 and cos² A + cos² B = 1. Also, tan A = cot B and sec A = cosec B.

Fundamental Trigonometric Identities

sin² θ + cos² θ = 1; 1 + tan² θ = sec² θ; 1 + cot² θ = cosec² θ.

Terms in a Series

For an arithmetic progression, the number of terms n = [(Last term − First term) ÷ Common difference] + 1.

Shortcut Trick

Solve 2x + 3y = 11 and x − 2y = −12 simultaneously by multiplying second equation by 2: 2x − 4y = −24.

Subtracting both: (2x + 3y) − (2x − 4y) = 11 − (−24) ⇒ 7y = 35 ⇒ y₁ = 5 and x₁ = −2.

For point Q on x-axis in x − 2y + 12 = 0, put y = 0 ⇒ x₂ = −12 and y₂ = 0.

Required value = x₁ − x₂ + y₁ + y₂ = −2 − (−12) + 5 + 0 = 15.

∴ The correct answer is 15.

Alternate Method

Given:

Line 1: 2x + 3y = 11

Line 2: x − 2y + 12 = 0

P(x₁, y₁) is the intersection of Line 1 and Line 2.

Q(x₂, y₂) is the intersection of Line 2 and the x-axis.

Step 1: Finding coordinates of P(x₁, y₁)

Multiply x − 2y = −12 by 2:

⇒ 2x − 4y = −24 --- (i)

Given first equation:

⇒ 2x + 3y = 11 --- (ii)

Subtract (i) from (ii):

⇒ (2x + 3y) − (2x − 4y) = 11 − (−24)

⇒ 7y = 35 ⇒ y₁ = 5

Substitute y = 5 in Line 2:

⇒ x − 2(5) + 12 = 0

⇒ x − 10 + 12 = 0 ⇒ x₁ = −2

⇒ P(x₁, y₁) = (−2, 5)

Step 2: Finding coordinates of Q(x₂, y₂)

Since Q lies on the x-axis, y-coordinate y₂ = 0.

Substitute y = 0 in equation of Line 2: x − 2y + 12 = 0

⇒ x − 2(0) + 12 = 0

⇒ x + 12 = 0 ⇒ x₂ = −12

⇒ Q(x₂, y₂) = (−12, 0)

Step 3: Calculating the required value

⇒ (x₁ − x₂ + y₁ + y₂)

⇒ (−2 − (−12) + 5 + 0)

⇒ −2 + 12 + 5 + 0 = 15

∴ The correct answer is 15.

Additional Information

Intersection of Two Lines

The point of intersection of two linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 is found by solving them simultaneously as a system of equations.

Intercepts on Axes

To find the x-intercept of a line, set y = 0. To find the y-intercept, set x = 0. A point on the x-axis always has the form (x, 0).

Condition for Parallel Lines

Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel if their slopes are equal, i.e., a₁/a₂ = b₁/b₂ ≠ c₁/c₂.

Shortcut Trick

Relative speed of trains = Distance ÷ Time = 396 ÷ 5.5 = 72 km/h.

Since trains move towards each other, Sum of speeds (Y + X) = 72 km/h.

Given Difference of speeds (Y − X) = 10 km/h.

Speed of faster train (Y) = (Sum + Difference) ÷ 2 = (72 + 10) ÷ 2 = 41 km/h.

∴ The correct answer is 41 km/h.

Alternate Method

Given:

Distance (D) = 396 km

Time (t) = 5.5 hours

Speed of train Y = Speed of train X + 10 km/h

Formula:

Relative Speed = Total Distance ÷ Time taken to meet

⇒ Let speed of Train X be S km/h.

⇒ Speed of Train Y = (S + 10) km/h.

⇒ Relative Speed = S + (S + 10) = 2S + 10 km/h.

⇒ 2S + 10 = 396 ÷ 5.5

⇒ 2S + 10 = 72

⇒ 2S = 72 − 10 = 62

⇒ S = 31 km/h (Speed of Train X)

⇒ Speed of Train Y = 31 + 10 = 41 km/h.

∴ The correct answer is 41 km/h.

Additional Information

Relative Speed in Opposite Directions

When two objects move towards each other, their relative speed is the sum of their individual speeds (S₁ + S₂).

Relative Speed in Same Direction

When two objects move in the same direction, their relative speed is the difference between their individual speeds (S₁ − S₂).

Distance Formula

Distance = Relative Speed × Time. In meeting point problems, the time taken by both to meet is the same if they start simultaneously.

Shortcut Trick

Relative speed in opposite direction = 50 + 58 = 108 km/h

Relative speed in m/s = 108 × 5/18 = 30 m/s

Total distance (L_A + L_B) = Speed × Time = 30 × 15 = 450 m

Given L_B = 3L_A, so total length = L_A + 3L_A = 4L_A

4L_A = 450 ⇒ L_A = 450 ÷ 4 = 112.5 m

∴ The correct answer is 112.5 m.

Alternate Method

Given:

Speed of Train A (S_A) = 50 km/h

Speed of Train B (S_B) = 58 km/h

Time to cross each other (t) = 15 seconds

Length of Train B (L_B) = 3 × Length of Train A (L_A)

Formula Used:

Relative Speed (Opposite Direction) = S_A + S_B

Total Distance = Length of Train A + Length of Train B

Distance = Speed × Time

Calculations:

⇒ Relative speed = 50 + 58 = 108 km/h

⇒ Converting speed to m/s: 108 × (5/18) = 6 × 5 = 30 m/s

⇒ Total distance = (L_A + L_B)

⇒ L_A + 3L_A = 30 × 15

⇒ 4L_A = 450

⇒ L_A = 450 ÷ 4

⇒ L_A = 112.5 m

∴ The correct answer is 112.5 m.

Additional Information

Relative Speed Rules

When two objects move in opposite directions, relative speed = S₁ + S₂. When they move in the same direction, relative speed = |S₁ − S₂|.

Unit Conversion

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

Crossing Length

When a train crosses another train, the total distance covered is always the sum of the lengths of both trains, regardless of the direction.

Shortcut Trick

Put φ = 45° to evaluate the expression quickly.

tan 45° = 1, cot 45° = 1, sec 45° = √2, cosec 45° = √2

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

Expression = 1 + 1 − 2 × 2 = 2 − 4 = −2

∴ The correct answer is -2.

Alternate Method

Given: tan²φ + cot²φ − sec²φ cosec²φ

Formula Used:

1. sec²φ = 1 + tan²φ
2. cosec²φ = 1 + cot²φ
3. sec²φ cosec²φ = sec²φ + cosec²φ

Calculations:

⇒ We know that sec²φ cosec²φ = (1/cos²φ) × (1/sin²φ)

⇒ sec²φ cosec²φ = (sin²φ + cos²φ) / (sin²φ cos²φ)

⇒ sec²φ cosec²φ = (sin²φ / sin²φ cos²φ) + (cos²φ / sin²φ cos²φ)

⇒ sec²φ cosec²φ = (1 / cos²φ) + (1 / sin²φ) = sec²φ + cosec²φ

⇒ Substitute this in the original expression: tan²φ + cot²φ − (sec²φ + cosec²φ)

⇒ Using identities sec²φ = 1 + tan²φ and cosec²φ = 1 + cot²φ:

⇒ tan²φ + cot²φ − (1 + tan²φ + 1 + cot²φ)

⇒ tan²φ + cot²φ − (2 + tan²φ + cot²φ)

⇒ tan²φ + cot²φ − 2 − tan²φ − cot²φ = −2

∴ The correct answer is -2.

Additional Information

Trigonometric Identities

Standard Pythagorean identities: sin²θ + cos²θ = 1, sec²θ − tan²θ = 1, and cosec²θ − cot²θ = 1.

Product to Sum Property

For secant and cosecant: sec²θ × cosec²θ = sec²θ + cosec²θ. Similarly for tan and cot: tan²θ × cot²θ = 1.

Complementary Angles

sin(90° − θ) = cos θ, tan(90° − θ) = cot θ, and sec(90° − θ) = cosec θ.

Shortcut Trick

Total time: A = 12 ÷ 0.4 = 30 days, B = 15 ÷ 0.6 = 25 days.

Let Total Work = LCM(30, 25) = 150 units. Efficiencies: A = 5, B = 6.

(A+B) in 10 days = (5 + 6) × 10 = 110. Rem. work = 150 − 110 = 40.

C's Efficiency = 40 ÷ 4 = 10. Total Efficiency (A+B+C) = 5 + 6 + 10 = 21.

Time for 28% work = (150 × 0.28) ÷ 21 = 42 ÷ 21 = 2 days.

∴ The correct answer is 2 days.

Alternate Method

Given:

A completes 40% in 12 days ⇒ 100% in (12 × 100) ÷ 40 = 30 days.

B completes 60% in 15 days ⇒ 100% in (15 × 100) ÷ 60 = 25 days.

⇒ Work done by A and B in 10 days = (5 + 6) × 10 = 110 units.

⇒ Remaining work left for C = 150 − 110 = 40 units.

⇒ Efficiency of C = Remaining Work ÷ Days = 40 ÷ 4 = 10 units/day.

⇒ Combined efficiency of A, B and C = 5 + 6 + 10 = 21 units/day.

⇒ 28% of total work = 150 × (28 ÷ 100) = 42 units.

⇒ Time taken by A, B and C = 42 ÷ 21 = 2 days.

∴ The correct answer is 2 days.

Additional Information

Efficiency and Time Relation

The time taken to complete a work is inversely proportional to the efficiency. Work = Efficiency × Time.

Combined Efficiency

When multiple people work together, their combined efficiency is the sum of their individual efficiencies (Total Efficiency = E₁ + E₂ + ...).

Percentage Work calculation

To find the time for a partial work, multiply the total work by the percentage fraction (e.g., 28% = 28/100) before dividing by efficiency.

Shortcut Trick

The expression is in the form √[a + b + c + 2(√ab + √bc + √ca)], which is √(a + b + c)².

Compare the signs of the products: 2(√6 − √10 − √15).

Positive term √6 comes from √2 × √3. Negative terms −√10 and −√15 share −√5 as a common factor.

Thus, the terms are √3, √2, and −√5.

Sum of squares = 3 + 2 + 5 = 10, which matches the constant term.

∴ The correct answer is √3 + √2 − √5.

Alternate Method

Given: √[10 + 2(√6 − √10 − √15)]

Formula Used: (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

Let x = √3, y = √2, and z = −√5.

⇒ x² + y² + z² = (√3)² + (√2)² + (−√5)² = 3 + 2 + 5 = 10

⇒ 2(xy + yz + zx) = 2[(√3 × √2) + (√2 × −√5) + (−√5 × √3)]

⇒ 2(xy + yz + zx) = 2(√6 − √10 − √15)

Substituting these into the given expression:

⇒ √[10 + 2(√6 − √10 − √15)] = √[(√3 + √2 − √5)²]

⇒ √3 + √2 − √5

∴ The correct answer is √3 + √2 − √5.

Additional Information

Square of a Trinomial

The formula (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca helps in simplifying nested radicals with three irrational terms.

Simplification rule for Surds

When simplifying √(x ± 2√y), look for two numbers whose sum is x and product is y, then the result is √a ± √b.

BODMAS Rule

Always solve terms inside the innermost brackets first, followed by orders (roots/powers), then division, multiplication, addition, and subtraction.

Shortcut Trick

Check for Pythagorean triplet: 56² + 90² = 3136 + 8100 = 11236 = 106².

Since the triangle is right-angled, the circumradius R = Hypotenuse ÷ 2 = 106 ÷ 2 = 53 cm.

Circumference of circumcircle = 2 × π × R = 2 × π × 53 = 106π cm.

∴ The correct answer is 106π cm.

Alternate Method

Given:

Sides of the triangle: a = 56 cm, b = 90 cm, c = 106 cm.

Formula Used:

For a right-angled triangle, the circumradius (R) = Hypotenuse ÷ 2.

Circumference of a circle = 2 × π × R

Calculations:

⇒ First, we check if the triangle is right-angled using Pythagoras theorem:

⇒ a² + b² = 56² + 90² = 3136 + 8100 = 11236

⇒ c² = 106² = 11236

⇒ Since a² + b² = c², the triangle is right-angled at the vertex opposite to the 106 cm side.

⇒ In a right-angled triangle, the circumcircle passes through all vertices, and the hypotenuse is the diameter.

⇒ Diameter (d) = Hypotenuse = 106 cm

⇒ Circumradius (R) = 106 ÷ 2 = 53 cm

⇒ Circumference = 2 × π × R = π × d

⇒ Circumference = 106π cm

∴ The correct answer is 106π cm.

Additional Information

Circumradius of any Triangle

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

Inradius of a Right-angled Triangle

The inradius (r) of a right-angled triangle is given by r = (Perpendicular + Base − Hypotenuse) ÷ 2.

Pythagorean Triplets

Common triplets include (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29), and (28, 45, 53).

Shortcut Trick

Prime factorize 3600: 3600 = 2⁴ × 3² × 5².

For N = p^a × q^b × r^c, Total Factors = (a+1)(b+1)(c+1).

Number of factors = (4+1) × (2+1) × (2+1) = 5 × 3 × 3 = 45.

∴ The correct answer is 45.

Alternate Method

Given Number: 3600

Formula: If a number N = p₁^a × p₂^b × p₃^c..., the total number of factors = (a+1) × (b+1) × (c+1)...

⇒ 3600 = 36 × 100

⇒ 3600 = (2² × 3²) × (2² × 5²)

⇒ 3600 = 2²⁺² × 3² × 5²

⇒ 3600 = 2⁴ × 3² × 5²

⇒ Exponents: a = 4, b = 2, c = 2

⇒ Total Factors = (4 + 1) × (2 + 1) × (2 + 1)

⇒ Total Factors = 5 × 3 × 3 = 45

∴ The correct answer is 45.

Additional Information

Sum of Factors

The sum of all factors is calculated as Σ = [ (p^a+1 − 1)/(p − 1) ] × [ (q^b+1 − 1)/(q − 1) ] × ...

Product of Factors

The product of all factors of a number N is given by N^{(Total Factors / 2)}.

Even and Odd Factors

For N = 2^a × q^b × r^c, Odd factors = (b+1)(c+1) and Even factors = a × (b+1)(c+1).

Perfect Square Factors

The number of factors that are perfect squares is calculated as ([a/2]+1) × ([b/2]+1) × ...

Shortcut Trick

Apply Componendo and Dividendo to (1 + sin ϕ) / (1 − sin ϕ) = p² / q².

1 / sin ϕ = (p² + q²) / (p² − q²) ⇒ sin ϕ = (p² − q²) / (p² + q²).

Recognize the Pythagorean triplet: Perpendicular = p² − q², Hypotenuse = p² + q², Base = 2pq.

sec ϕ = Hypotenuse / Base = (p² + q²) / 2pq.

Splitting the terms: sec ϕ = p²/2pq + q²/2pq = 1/2 (p/q + q/p).

∴ The correct answer is 1/2 (p/q + q/p).

Alternate Method

Given: (1 + sin ϕ) / (1 − sin ϕ) = p² / q²

Formula Used: Componendo and Dividendo: If a/b = c/d, then (a + b) / (a − b) = (c + d) / (c − d). Also, sec ϕ = 1 / cos ϕ.

⇒ Applying Componendo and Dividendo:

⇒ [(1 + sin ϕ) + (1 − sin ϕ)] / [(1 + sin ϕ) − (1 − sin ϕ)] = (p² + q²) / (p² − q²)

⇒ 2 / 2sin ϕ = (p² + q²) / (p² − q²)

⇒ 1 / sin ϕ = (p² + q²) / (p² − q²) ⇒ sin ϕ = (p² − q²) / (p² + q²)

⇒ Since sin ϕ = Perpendicular / Hypotenuse, we have P = p² − q² and H = p² + q².

⇒ Base (B) = √[H² − P²] = √[(p² + q²)² − (p² − q²)²]

⇒ B = √[4p²q²] = 2pq

⇒ sec ϕ = Hypotenuse / Base = (p² + q²) / 2pq

⇒ sec ϕ = p² / 2pq + q² / 2pq = p/2q + q/2p = 1/2 (p/q + q/p)

∴ The correct answer is 1/2 (p/q + q/p).

Additional Information

Componendo and Dividendo Rule

It is a theorem on proportions that allows for the simplification of equations of the form (a+b)/(a−b) = k.

Pythagorean Triplet for Algebra

The set of values (p² − q²), 2pq, and (p² + q²) always forms a right-angled triangle triplet.

Reciprocal Trigonometric Ratios

The secant function (sec ϕ) is the reciprocal of the cosine function (cos ϕ = Base/Hypotenuse), so sec ϕ = Hypotenuse/Base.

Shortcut Trick

The three medians of a triangle divide it into 6 smaller triangles of equal area.

Area of ΔAOE = 1 part (since it is one of the 6 small triangles).

Area of ΔABD = 3 parts (since the median AD divides the triangle into two equal halves, i.e., 6/2 = 3 parts).

Ratio = Area of ΔABD : Area of ΔAOE = 3 : 1.

∴ The correct answer is 3 : 1.

Alternate Method

Given: In ΔABC, AD, BE, and CF are medians intersecting at centroid O.

Formula: A median divides a triangle into two triangles of equal area. The centroid divides a triangle into six triangles of equal area.

Let the total area of ΔABC be 6 units.

⇒ Area of ΔABD = 1/2 × Area of ΔABC (Property of median AD)

⇒ Area of ΔABD = 1/2 × 6 = 3 units.

⇒ The six triangles ΔAOF, ΔBOF, ΔBOD, ΔCOD, ΔCOE, and ΔAOE have equal area.

⇒ Area of ΔAOE = 1/6 × Area of ΔABC

⇒ Area of ΔAOE = 1/6 × 6 = 1 unit.

⇒ Ratio = Area(ΔABD) / Area(ΔAOE)

⇒ Ratio = 3 / 1

⇒ Ratio = 3 : 1.

∴ The correct answer is 3 : 1.

Additional Information

Centroid Property

The centroid (O) divides each median in the ratio 2 : 1. For example, AO : OD = 2 : 1, BO : OE = 2 : 1, and CO : OF = 2 : 1.

Area of Median Triangles

Area of ΔAOB = Area of ΔBOC = Area of ΔCOA = 1/3 × Area of ΔABC.

Apollonius Theorem

In ΔABC with median AD, the theorem states: AB² + AC² = 2(AD² + BD²).

Shortcut Trick

Difference in given percentages: 60% − 20% = 40%

Value of 40% = 120 ⇒ 1% = 120 ÷ 40 = 3

Required gap: 33¹/₃% − 28% = 5¹/₃%

Target Value = 3 × 5¹/₃ = 3 × (16/3) = 16

∴ The correct answer is 16.

Alternate Method

Given: Difference between 60% and 20% of a number is 120.

Formula Used: Value = (Percentage ÷ 100) × Number

⇒ Let the number be x.

⇒ (60/100) × x − (20/100) × x = 120

⇒ (40/100) × x = 120

⇒ x = (120 × 100) ÷ 40 = 300

⇒ 33¹/₃% of 300 = (1/3) × 300 = 100

⇒ 28% of 300 = (28/100) × 300 = 84

⇒ Difference = 100 − 84 = 16

∴ The correct answer is 16.

Additional Information

Fraction to Percentage Conversion

Standard fractions simplify calculations: 1/3 = 33¹/₃%, 1/4 = 25%, 1/5 = 20%.

Percentage Base Rule

When the base number is the same, percentage differences can be added or subtracted directly before calculating the final value.

Percentage Change Formula

% Change = [(New Value − Old Value) ÷ Old Value] × 100.

Shortcut Trick

Identify complementary angles: Since (78° + θ) + (12° − θ) = 90°, then sin(78° + θ) = cos(12° − θ), making their difference 0.

Simplify numerator identity: tan²70° − cosec²20° = tan²70° − sec²70° = −1 (as sec²A − tan²A = 1).

Simplify denominator: sin 25° cos 65° + cos 25° sin 65° = sin(25° + 65°) = sin 90° = 1.

Final calculation: (0 + (−1)) ÷ 1 = −1.

∴ The correct answer is −1.

Alternate Method

Given:

Expression = (sin(78° + θ) − cos(12° − θ) + (tan²70° − cosec²20°)) ÷ (sin 25° cos 65° + cos 25° sin 65°)

Formula Used:

1. sin A = cos(90° − A)
2. sec²A − tan²A = 1 ⇒ tan²A − sec²A = −1
3. sin A cos B + cos A sin B = sin(A + B)

Calculations:

⇒ Numerator Term 1: sin(78° + θ) − cos(12° − θ)

⇒ sin(78° + θ) = cos(90° − (78° + θ)) = cos(12° − θ)

⇒ cos(12° − θ) − cos(12° − θ) = 0

⇒ Numerator Term 2: tan²70° − cosec²20°

⇒ cosec 20° = sec(90° − 20°) = sec 70°

⇒ tan²70° − sec²70° = −(sec²70° − tan²70°) = −1

⇒ Denominator: sin 25° cos 65° + cos 25° sin 65°

⇒ sin(25° + 65°) = sin 90° = 1

⇒ Final Value = (0 − 1) ÷ 1 = −1

∴ The correct answer is −1.

Additional Information

Complementary Angle Ratios

For any angle θ, sin(90° − θ) = cos θ, tan(90° − θ) = cot θ, and sec(90° − θ) = cosec θ.

Trigonometric Identities

Fundamental identities include sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Compound Angle Formulas

sin(A ± B) = sin A cos B ± cos A sin B and cos(A ± B) = cos A cos B ∓ sin A sin B.

Shortcut Trick

Sides (20, 21, 29) form a right-angled triangle as 20² + 21² = 29².

Base Area = 1/2 × 20 × 21 = 210 cm².

Height of prism = Volume ÷ Base Area = 7560 ÷ 210 = 36 cm.

Lateral Surface Area = Perimeter of base × Height = (20 + 21 + 29) × 36.

LSA = 70 × 36 = 2520 cm².

∴ The correct answer is 2520 cm².

Alternate Method

Given:

Sides of the triangular base = 20 cm, 21 cm, 29 cm

Volume of the prism = 7560 cm³

Formula Used:

Volume of Prism = Area of Base × Height

Lateral Surface Area (LSA) = Perimeter of Base × Height

Calculations:

⇒ Base is a right triangle as 20² + 21² = 400 + 441 = 841 = 29².

⇒ Area of Base = 1/2 × Base × Height = 1/2 × 21 × 20 = 210 cm².

⇒ Height (h) of the prism = Volume ÷ Area of Base.

⇒ h = 7560 ÷ 210 = 36 cm.

⇒ Perimeter of triangular base = 20 + 21 + 29 = 70 cm.

⇒ Lateral Surface Area = Perimeter × h = 70 × 36 = 2520 cm².

∴ The correct answer is 2520 cm².

Additional Information

Total Surface Area of Prism

TSA = Lateral Surface Area + 2 × Area of Base.

Right Prism Properties

In a right prism, the side faces are rectangles and are perpendicular to the bases.

Pythagorean Triplets

Common triplets like (3,4,5), (5,12,13), (8,15,17), and (20,21,29) help identify right-angled bases quickly.

Shortcut Trick

The angle formed by the intersection of two exterior angle bisectors (Excenter) is given by: ∠QUR = 90° − (∠P ÷ 2).

Given ∠QUR = 79°.

79° = 90° − (∠P ÷ 2)

∠P ÷ 2 = 90° − 79° = 11°

∠P = 11° × 2 = 22°

∴ The correct answer is 22°.

Alternate Method

Given:

In ΔPQR, ∠QUR = 79° (where U is the intersection of exterior bisectors of ∠Q and ∠R).

Formula Used:

Exterior Angle Bisector Theorem: ∠QUR = 90° − ∠P/2

Calculations:

⇒ In ΔPQR, the bisectors of external angles ∠SQR and ∠TRQ meet at point U.

⇒ The relationship between the vertex angle ∠P and the excenter angle ∠QUR is:

⇒ ∠QUR = 90° − ∠P/2

⇒ 79° = 90° − ∠P/2

⇒ ∠P/2 = 90° − 79°

⇒ ∠P/2 = 11°

⇒ ∠P = 11° × 2

⇒ ∠P = 22°

∴ The correct answer is 22°.

Additional Information

Incenter Angle Formula

When the internal bisectors of ∠B and ∠C of ΔABC meet at I, the angle formed is ∠BIC = 90° + ∠A/2.

Internal and External Bisector Intersection

If the internal bisector of ∠B and the external bisector of ∠C meet at point E, then ∠BEC = ∠A/2.

Sum of Angles Property

The sum of all interior angles of a triangle is always 180°. (∠P + ∠Q + ∠R = 180°).

Excenters

A triangle has three excenters, each formed by the intersection of one internal angle bisector and two external angle bisectors.

Shortcut Trick

Identity: If a³ + b³ + c³ = 3abc, then a + b + c = 0.

Here, a = 5x + 1, b = x − 3, and c = 2(3x − 4) = 6x − 8.

Applying the condition: (5x + 1) + (x − 3) + (6x − 8) = 0.

12x − 10 = 0 ⇒ 12x = 10 ⇒ x = 5/6.

∴ The correct answer is 5/6.

Alternate Method

Given Equation: (5x + 1)³ + (x − 3)³ + 8(3x − 4)³ = 6(5x + 1)(x − 3)(3x − 4)

Formula Used: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)

If a³ + b³ + c³ = 3abc, then a + b + c = 0 (for non-equal a, b, c).

Let a = (5x + 1)

Let b = (x − 3)

Let c = 2(3x − 4) = (6x − 8)

Now, checking the RHS: 3abc = 3 × (5x + 1) × (x − 3) × 2(3x − 4) = 6(5x + 1)(x − 3)(3x − 4)

This matches the given equation structure a³ + b³ + c³ = 3abc.

⇒ a + b + c = 0

⇒ (5x + 1) + (x − 3) + (6x − 8) = 0

⇒ 5x + x + 6x + 1 − 3 − 8 = 0

⇒ 12x − 10 = 0

⇒ 12x = 10

⇒ x = 10/12

⇒ x = 5/6

∴ The correct answer is 5/6.

Additional Information

Cubic Algebraic Identity

The expression a³ + b³ + c³ − 3abc equals 0 if either a + b + c = 0 or a = b = c.

Sum and Product of Roots

In quadratic equations ax² + bx + c = 0, the sum of roots is −b/a and the product of roots is c/a.

BODMAS Rule

Always solve mathematical expressions in the order: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction.

Shortcut Trick Let the two numbers be 21a and 21b, where a and b are co-prime. LCM = HCF × a × b ⇒ 221 × 21 = 21 × ab ⇒ ab = 221. Factors of 221 are 13 and 17 (since 13 × 17 = 221). First number = 21 × 13 = 273 (lies between 200 and 300). Other number = 21 × 17 = 357. Sum of digits = 3 + 5 + 7 = 15. ∴ The correct answer is 15.

Alternate Method Given: HCF (H) = 21 LCM (L) = 221 × 21 One number (n1) lies between 200 and 300. Formula Used: Product of two numbers = HCF × LCM LCM = HCF × (Product of co-prime factors)

Calculations: ⇒ Let the numbers be x and y. ⇒ x = 21a and y = 21b, where HCF(a, b) = 1. ⇒ LCM = 21 × a × b ⇒ 221 × 21 = 21 × a × b ⇒ ab = 221 ⇒ Prime factors of 221: 13 × 17 = 221. ⇒ So, {a, b} = {13, 17}. ⇒ x = 21 × 13 = 273 ⇒ y = 21 × 17 = 357 ⇒ Since 200 < 273 < 300, the other number is 357. ⇒ Sum of digits of 357 = 3 + 5 + 7 = 15. ∴ The correct answer is 15.

Additional Information Fundamental HCF-LCM Relation For any two numbers A and B: A × B = HCF(A, B) × LCM(A, B). LCM of Fractions LCM of fractions = (LCM of numerators) ÷ (HCF of denominators). Co-prime Property If two numbers are expressed as H×a and H×b (where H is HCF), then a and b must be co-prime, meaning HCF(a, b) = 1.

Shortcut Trick

Using the identity: cot² ϕ − cos² ϕ = cot² ϕ × cos² ϕ

3(cot² ϕ × cos² ϕ) = cos² ϕ ⇒ 3 cot² ϕ = 1

⇒ cot ϕ = 1/√3 ⇒ ϕ = 60°

Value = tan² 60° + cosec² 60° + sin² 60° = 3 + 4/3 + 3/4

Value = (36 + 16 + 9) / 12 = 61/12

∴ The correct answer is 61/12.

Alternate Method

Given: 3(cot² ϕ − cos² ϕ) = cos² ϕ where 0° < ϕ < 90°.

Formula: cot ϕ = cos ϕ / sin ϕ | tan ϕ = 1 / cot ϕ | cosec ϕ = 1 / sin ϕ

⇒ 3(cot² ϕ − cos² ϕ) = cos² ϕ

⇒ 3(cos² ϕ / sin² ϕ − cos² ϕ) = cos² ϕ

⇒ 3 cos² ϕ (1 / sin² ϕ − 1) = cos² ϕ

⇒ 3 (1 / sin² ϕ − 1) = 1 (Since cos ϕ ≠ 0)

⇒ 3 / sin² ϕ − 3 = 1 ⇒ 3 / sin² ϕ = 4

⇒ sin² ϕ = 3/4 ⇒ sin ϕ = √3/2 ⇒ ϕ = 60°

⇒ Required value = tan² 60° + cosec² 60° + sin² 60°

⇒ (√3)² + (2/√3)² + (√3/2)²

⇒ 3 + 4/3 + 3/4

⇒ (36 + 16 + 9) / 12 = 61/12

∴ The correct answer is 61/12.

Additional Information

Trigonometric Identities

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

Product Rule for Cot and Cos

A useful simplification in trigonometry is cot² θ − cos² θ = cot² θ × cos² θ.

Specific Angle Values

At 60°: sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3, cot 60° = 1/√3, and cosec 60° = 2/√3.

Shortcut Trick

In orthocenter P, the angle between altitudes at the vertex is supplementary to the vertex angle: ∠C + ∠EPD = 180°.

⇒ ∠C = 180° − 116° = 64°.

In incenter Q, the angle formed by bisectors of ∠A and ∠B is: ∠AQB = 90° + ∠C/2.

⇒ ∠AQB = 90° + 64°/2 = 90° + 32° = 122°.

∴ The correct answer is 122°.

Alternate Method

Given: ∠EPD = 116°, P is the orthocenter, and Q is the incenter.

Formula Used: Sum of angles in quadrilateral = 360°; Incenter angle = 90° + (Third angle / 2).

⇒ In quadrilateral CEPD, ∠CEP = 90° and ∠CDP = 90° (Perpendiculars).

⇒ Sum of angles: ∠C + ∠CEP + ∠EPD + ∠CDP = 360°.

⇒ ∠C + 90° + 116° + 90° = 360°.

⇒ ∠C + 296° = 360°.

⇒ ∠C = 360° − 296° = 64°.

⇒ Since Q is the meeting point of angle bisectors of ∠A and ∠B, Q is the Incenter of ΔABC.

⇒ Property of Incenter: ∠AQB = 90° + (∠C / 2).

⇒ ∠AQB = 90° + (64° / 2).

⇒ ∠AQB = 90° + 32° = 122°.

∴ The correct answer is 122°.

Additional Information

Orthocenter Angle Property

The angle formed by the orthocenter P at any side is supplementary to the opposite vertex angle: ∠BPC = 180° − ∠A.

Incenter Angle Property

The angle formed by the incenter I is always obtuse: ∠BIC = 90° + ∠A/2, ∠AIC = 90° + ∠B/2, and ∠AIB = 90° + ∠C/2.

Circumcenter Angle Property

The angle formed by the circumcenter O at the center is twice the angle at the vertex: ∠BOC = 2 × ∠A.

Ex-center Angle Property

The angle formed by the external bisectors of ∠B and ∠C at ex-center I_a is: ∠BI_aC = 90° − ∠A/2.

Shortcut Trick

Angle at the in-centre ∠QIR = 90° + ∠P / 2

107° = 90° + ∠P / 2

∠P / 2 = 107° − 90° = 17°

∠P = 17° × 2 = 34°

∴ The correct answer is 34°.

Alternate Method

Given:

In ΔPQR, I is the in-centre.

Measure of ∠QIR = 107°

Formula Used:

The angle formed by the internal bisectors of two angles of a triangle at the in-centre is given by: ∠In-centre = 90° + (1/2) × Opposite Vertex Angle.

For ΔPQR: ∠QIR = 90° + ∠P / 2

Calculations:

⇒ ∠QIR = 90° + ∠P / 2

⇒ 107° = 90° + ∠P / 2

⇒ 107° − 90° = ∠P / 2

⇒ 17° = ∠P / 2

⇒ ∠P = 17° × 2

⇒ ∠P = 34°

∴ The correct answer is 34°.

Additional Information

Ex-centre Angle

The angle subtended by external bisectors of ∠Q and ∠R at ex-centre E is ∠QER = 90° − ∠P / 2.

Circumcentre Angle

The angle subtended by sides at the circumcentre O is exactly double the angle at the vertex: ∠QOR = 2 × ∠P.

Orthocentre Angle

The angle subtended by sides at the orthocentre H is supplementary to the vertex angle: ∠QHR = 180° − ∠P.

In-centre Definition

The in-centre is the intersection point of all three internal angle bisectors and is equidistant from all sides of the triangle.

Shortcut Trick

Sum of interior angles = (n − 2) × 180° = 1260° → n − 2 = 7 → n = 9 sides.

Exterior angle (E) = 360° ÷ 9 = 40°.

Interior angle (I) = 180° − 40° = 140°.

Difference = 140° − 40° = 100°.

∴ The correct answer is 100°.

Alternate Method

Given: Sum of interior angles = 1260°

Formula: Sum of interior angles = (n − 2) × 180°

⇒ (n − 2) × 180° = 1260°

⇒ n − 2 = 1260 ÷ 180

⇒ n − 2 = 7

⇒ n = 9

⇒ Interior angle = 1260° ÷ 9 = 140°

⇒ Exterior angle = 360° ÷ 9 = 40°

⇒ Difference = 140° − 40° = 100°

∴ The correct answer is 100°.

Additional Information

Sum of Exterior Angles

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

Relationship between Interior and Exterior Angles

At any vertex of a polygon, the interior angle and exterior angle are supplementary: Interior + Exterior = 180°.

Number of Diagonals

The number of diagonals in a polygon with n sides is given by the formula: [n × (n − 3)] ÷ 2.

Shortcut Trick

Apply the external secant segment theorem: PB × PA = PC × PD.

Here, PA = PB + AB = 6.3 + 7.7 = 14 cm.

Substitute: 6.3 × 14 = 5.6 × (5.6 + CD).

88.2 = 31.36 + 5.6 × CD ⇒ 5.6 × CD = 56.84.

CD = 56.84 ÷ 5.6 = 10.15 cm.

∴ The correct answer is 10.15 cm.

Alternate Method

Given:

Length of chord segment AB = 7.7 cm

External segment PB = 6.3 cm

External segment PC = 5.6 cm

Formula Used:

Intersecting Chords Theorem (External): PA × PB = PD × PC

Calculations:

⇒ PA = AB + PB

⇒ PA = 7.7 + 6.3 = 14 cm

⇒ According to the theorem, PA × PB = PD × PC

⇒ 14 × 6.3 = PD × 5.6

⇒ 88.2 = PD × 5.6

⇒ PD = 88.2 ÷ 5.6

⇒ PD = 15.75 cm

⇒ Now, PD = PC + CD

⇒ 15.75 = 5.6 + CD

⇒ CD = 15.75 − 5.6

⇒ CD = 10.15 cm

∴ The correct answer is 10.15 cm.

Additional Information

Intersecting Chords Theorem (Internal)

When two chords AB and CD intersect inside a circle at point E, then AE × EB = CE × ED.

Tangent-Secant Theorem

If a tangent PT and a secant PAB are drawn from an external point P to a circle, then PT² = PA × PB.

External Secant Segments

The theorem states that for two secant lines, the product of the length of the entire secant segment and its external part is constant: PA × PB = PC × PD.