SSC CGL Tier 2 Quant Previous Year Paper (Held On: 8 Aug 2022)

The correct answer is Rs. 700

Key Points

• According to the question, the Cost Price (CP) of the article is Rs. 2800.
• The Profit percentage is given as 20% of the Selling Price (SP). This is different from the usual calculation where profit is a percentage of the Cost Price.
• Let the Selling Price be SP. We know that Profit = SP − CP.
• As per the condition: Profit = 20% of SP = (20/100) × SP = 0.2 × SP.
• Substitute the value of Profit in the basic formula: 0.2 × SP = SP − 2800.
• Rearrange the equation: SP − 0.2 × SP = 2800.
• This simplifies to: 0.8 × SP = 2800.
• Calculate SP: SP = 2800 ÷ 0.8 = Rs. 3500.
• Now, the Actual Profit = SP − CP = 3500 − 2800 = Rs. 700.
• Alternative Method: Let SP be 100 units. Then Profit = 20 units. CP = SP − Profit = 100 − 20 = 80 units. If 80 units = 2800, then 1 unit = 2800 ÷ 80 = 35. Therefore, Profit (20 units) = 20 × 35 = Rs. 700.

Additional Information

• Profit Margin: When profit is calculated as a percentage of the Selling Price, it is often referred to as the Profit Margin.
• Markup: When profit is calculated as a percentage of the Cost Price, it is known as Markup.
• In most standard arithmetic problems, if it is simply mentioned as Profit Percentage, it is assumed to be calculated on the Cost Price unless specified otherwise.
• Formula for Profit Percentage on CP: (Profit ÷ CP) × 100.
• Formula for Profit Percentage on SP: (Profit ÷ SP) × 100.

Shortcut Trick

Base Area = πr² = (22/7) × 7 × 7 = 154 cm².

CSA + TSA = 2 × CSA + 2 × Base Area = 2068.

2 × CSA + 2(154) = 2068 ⇒ 2 × CSA = 1760 ⇒ CSA = 880 cm².

CSA = 2πrh = 880 ⇒ 44 × h = 880 ⇒ h = 20 cm.

Volume = Base Area × height = 154 × 20 = 3080 cm³.

∴ The correct answer is 3080 cm³.

Alternate Method

Given: Radius (r) = 7 cm, CSA + TSA = 2068 cm².

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

⇒ CSA + TSA = 2068

⇒ 2πrh + 2πr(h + r) = 2068

⇒ 2πrh + 2πrh + 2πr² = 2068

⇒ 4πrh + 2πr² = 2068

⇒ 4 × (22/7) × 7 × h + 2 × (22/7) × 7 × 7 = 2068

⇒ 88h + 308 = 2068

⇒ 88h = 2068 − 308 = 1760

⇒ h = 1760 ÷ 88 = 20 cm

⇒ Volume of Cylinder = πr²h

⇒ Volume = (22/7) × 7 × 7 × 20

⇒ Volume = 22 × 7 × 20 = 154 × 20 = 3080 cm³

∴ The correct answer is 3080 cm³.

Additional Information

Curved Surface Area of Cylinder

The area of the curved part of the cylinder is CSA = 2πrh, where r is radius and h is height.

Total Surface Area of Cylinder

The sum of the area of the curved surface and the two circular bases: TSA = 2πr(h + r).

Volume of Cylinder

The capacity or space occupied by the cylinder: Volume = πr²h.

Hollow Cylinder Volume

The volume of material in a hollow cylinder is V = π(R² − r²)h, where R is outer and r is inner radius.

Shortcut Trick Recognize the Pythagorean triplet: 9, 40, 41. Given sin θ = 9/41 = Perpendicular/Hypotenuse. Since the angle is in the first quadrant, all ratios are positive. The Base of the triangle = √(412 − 92) = 40. cot θ = Base ÷ Perpendicular = 40/9. ∴ The correct answer is 40/9.

Alternate Method Given: sin θ = 9/41 and 0° < θ < 90°. Formula Used: sin θ = Perpendicular (P) / Hypotenuse (H); cot θ = Base (B) / Perpendicular (P); H2 = P2 + B2.

⇒ sin θ = 9/41 = P / H ⇒ Let P = 9k and H = 41k ⇒ B2 = H2 − P2 ⇒ B = √((41k)2 − (9k)2) ⇒ B = √(1681k2 − 81k2) = √(1600k2) = 40k ⇒ cot θ = Base / Perpendicular ⇒ cot θ = 40k / 9k = 40/9 ∴ The correct answer is 40/9.

Additional Information Pythagorean Triplets These are sets of three integers (a, b, c) such that a2 + b2 = c2. Standard triples include (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), and (9, 40, 41). Trigonometric Quadrant Signs In the first quadrant (0° to 90°), all trigonometric ratios are positive. In the second, only sin and cosec are positive. Fundamental Identities The key relation between cot and sin is cot2θ = cosec2θ − 1, where cosec θ = 1 ÷ sin θ.

Shortcut Trick

When two equal circles of radius r intersect such that each passes through the other's center, the common chord length is always √3 × r.

Given: Radius (r) = 36 cm.

Length of common chord = 36 × √3 = 36√3 cm.

∴ The correct answer is 36√3 cm.

Alternate Method

Given: Radius of both circles (r) = 36 cm.

Formula: Length of common chord = 2 × √(r² − (d/2)²), where d is the distance between centers.

Since each circle passes through the center of the other, the distance between centers (d) = Radius (r) = 36 cm.

⇒ Half-distance between centers (O₁M) = 36 ÷ 2 = 18 cm.

⇒ Let intersection points be A and B. In right ΔAO₁M:

⇒ AM² = O₁A² − O₁M²

⇒ AM² = 36² − 18²

⇒ AM² = 1296 − 324 = 972

⇒ AM = √972 = 18√3 cm.

⇒ Length of common chord AB = 2 × AM = 2 × 18√3 = 36√3 cm.

∴ The correct answer is 36√3 cm.

Additional Information

Perpendicular Bisector Property

The line joining the centers of two intersecting circles is the perpendicular bisector of the common chord.

Equilateral Triangles in Intersecting Circles

When two circles of equal radius pass through each other's centers, the points of intersection and the two centers form two equilateral triangles.

Area of Common Region

The area of the region common to two such circles is (2/3 × π − √3/2) × r².

Shortcut Trick

For an isosceles right triangle with equal sides 'a', Hypotenuse = a√2.

Inradius (r) = (Sum of legs − Hypotenuse) ÷ 2 = (a + a − a√2) ÷ 2 = a(2 − √2) ÷ 2.

Circumradius (R) = Hypotenuse ÷ 2 = a√2 ÷ 2.

Ratio r : R = [a(2 − √2) ÷ 2] : [a√2 ÷ 2] = (2 − √2) : √2.

∴ The correct answer is 2 − √2 : √2.

Alternate Method

Given:

Triangle ABC is an isosceles right-angled triangle with ∠ABC = 90°.

AB = 12 cm, therefore BC = 12 cm (isosceles property).

Formula Used:

Hypotenuse (h) = √(Base² + Perpendicular²)

Inradius (r) = (Base + Perpendicular − Hypotenuse) ÷ 2

Circumradius (R) = Hypotenuse ÷ 2

Calculations:

⇒ AC = √(12² + 12²) = √(144 + 144) = √288 = 12√2 cm

⇒ Inradius (r) = (12 + 12 − 12√2) ÷ 2

⇒ r = (24 − 12√2) ÷ 2 = 12 − 6√2 cm

⇒ Circumradius (R) = (12√2) ÷ 2 = 6√2 cm

⇒ Ratio r : R = (12 − 6√2) : 6√2

⇒ Ratio r : R = 6(2 − √2) : 6√2

⇒ Ratio r : R = (2 − √2) : √2

∴ The correct answer is 2 − √2 : √2.

Additional Information

Inradius of a Right Triangle

The radius of the incircle is calculated as r = (Perpendicular + Base − Hypotenuse) ÷ 2.

Circumradius of a Right Triangle

The radius of the circumcircle is half the length of the hypotenuse, R = Hypotenuse ÷ 2.

Pythagoras Theorem

In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: c² = a² + b².

Solution:

Profit Ratio = A : B : C = 268 : 236 : 168 = 67 : 59 : 42

Given:

Fractions: $\frac{5}{11}, \frac{7}{12}, \frac{8}{13}, \frac{9}{17}$

Formula used:

Fraction comparison by decimal conversion.

Calculation:

⇒ $\frac{5}{11} \approx 0.454$

⇒ $\frac{7}{12} \approx 0.583$

⇒ $\frac{8}{13} \approx 0.615$

⇒ $\frac{9}{17} \approx 0.529$

⇒ $0.454 < 0.529 < 0.583 < 0.615$

⇒ Least value = $\frac{5}{11}$

∴ The correct answer is option (1).

Shortcut Trick

The number of terms (n) in the series 11, 13, ..., 67 is (67 − 11) ÷ 2 + 1 = 29.

Sum = (99 × n) + (Sum of fractional numerators ÷ 99).

Sum of numerators = n ÷ 2 × (first + last) = 29 ÷ 2 × (11 + 67) = 1131.

Total = (99 × 29) + 1131 ÷ 99 = 2871 + 377 ÷ 33.

Total = (2871 × 33 + 377) ÷ 33 = 95120/33.

∴ The correct answer is 95120/33.

Alternate Method

Given: Series = 99(11/99) + 99(13/99) + 99(15/99) + ... + 99(67/99)

Formula Used: Sum of n terms of an AP (S_n) = n ÷ 2 × (a + l); Number of terms (n) = (l − a) ÷ d + 1; Mixed fraction x(a/b) = x + a/b

⇒ Total terms (n) = (67 − 11) ÷ 2 + 1

⇒ n = 56 ÷ 2 + 1 = 29

⇒ Sum = (99 + 99 + ... 29 times) + (11/99 + 13/99 + ... + 67/99)

⇒ Sum = (99 × 29) + [ (11 + 13 + ... + 67) ÷ 99 ]

⇒ Sum of numerators = 29 ÷ 2 × (11 + 67) = 29 ÷ 2 × 78 = 1131

⇒ Total = 2871 + 1131 ÷ 99

⇒ Total = 2871 + 377 ÷ 33

⇒ Total = (94743 + 377) ÷ 33

⇒ Total = 95120/33

∴ The correct answer is 95120/33.

Additional Information

Mixed Fraction Summation

Mixed fractions follow the distributive property: Σ[A + (b/c)] = (ΣA) + (Σb)/c. This simplifies calculations by separating whole numbers from fractions.

Arithmetic Progression (AP)

For any AP with first term a, last term l, and common difference d, the number of terms n = (l − a)/d + 1 and the sum S_n = n/2 × (a + l).

Fraction Simplification

When adding fractions with a common denominator, simply add the numerators. Always check if the resulting fraction can be reduced by dividing the numerator and denominator by their HCF.

Shortcut Trick

Substitute standard trigonometric values: sin 60° = √3/2 and sin 45° = cos 45° = 1/√2.

Since sin 45° = cos 45°, the last two terms can be combined: (7 + 8) × (1/√2)² = 15/2.

The first term is 5 × (√3/2)² = 15/4.

Total = 15/4 + 15/2 = 15/4 + 30/4 = 45/4.

∴ The correct answer is 45/4.

Alternate Method

Given: 5 sin² 60° + 7 sin² 45° + 8 cos² 45°

Standard Values: sin 60° = √3/2, sin 45° = 1/√2, cos 45° = 1/√2

⇒ 5 × (√3/2)² + 7 × (1/√2)² + 8 × (1/√2)²

⇒ 5 × (3/4) + 7 × (1/2) + 8 × (1/2)

⇒ 15/4 + 7/2 + 8/2

⇒ 15/4 + 15/2

⇒ (15 + 30) / 4

⇒ 45/4

∴ The correct answer is 45/4.

Additional Information

Trigonometric Identities

The fundamental identity is sin²θ + cos²θ = 1, which relates the sine and cosine of the same angle.

Complementary Angles

For any angle θ, sin θ = cos(90° − θ). For example, sin 60° = cos 30° and sin 45° = cos 45°.

Standard Trigonometric Values

sin 30° = 1/2, sin 45° = 1/√2, sin 60° = √3/2, and sin 90° = 1.

Shortcut Trick

Convert speed from m/s to kmph: 15 × (18/5) = 3 × 18 = 54 kmph.

Time = Distance ÷ Speed = 162 ÷ 54 = 3 hours.

∴ The correct answer is 3 hours.

Alternate Method

Given: Distance = 162 km, Speed = 15 m/s.

Formula Used: Time = Distance ÷ Speed; 1 m/s = 18/5 kmph.

⇒ First, convert speed to kmph:

⇒ Speed = 15 × (18/5) = 54 kmph.

⇒ Time = Distance ÷ Speed

⇒ Time = 162 ÷ 54

⇒ Time = 3 hours.

∴ The correct answer is 3 hours.

Additional Information

Unit Conversion Factor

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

Basic Relation

Distance = Speed × Time. Ensure all units are consistent (e.g., km, kmph, hours OR m, m/s, seconds) before calculation.

Average Speed

Average speed for a journey is (Total Distance) ÷ (Total Time taken), not the average of the individual speeds.

Shortcut Trick

Average of first n even positive numbers = n + 1

Average of first n odd positive numbers = n

Average of first 148 even numbers = 148 + 1 = 149

Average of first 129 odd numbers = 129

Required difference = 149 − 129 = 20

∴ The correct answer is 20.

Alternate Method

Given:

Number of even numbers (n₁) = 148

Number of odd numbers (n₂) = 129

Formula Used:

Sum of first n even numbers = n(n + 1)

Sum of first n odd numbers = n²

Average = Sum ÷ Count

Calculations:

⇒ Average of first 148 even numbers = [148 × (148 + 1)] ÷ 148 = 149

⇒ Average of first 129 odd numbers = (129 × 129) ÷ 129 = 129

⇒ Difference = 149 − 129

⇒ Difference = 20

∴ The correct answer is 20.

Additional Information

Sum of first n natural numbers

The sum is calculated as S = n(n + 1) ÷ 2, and the average is (n + 1) ÷ 2.

Sum of squares of first n natural numbers

The sum of squares is given by n(n + 1)(2n + 1) ÷ 6.

Sum of cubes of first n natural numbers

The sum of cubes is [n(n + 1) ÷ 2]², which is the square of the sum of first n natural numbers.

Concept used:

If Ratio of two numbers is given as 1.1 , then the numbers would be written as 11x and 10x where x is HCF of both numbers.

Calculations:

Let the number of males and females in company A in Y1 be 11k and 10k respectively and the number of males and females in company B in Y2 be 9m and 10m respectively.Then,

11k + 9m = 21100 ---------(1)

And 10k + 10m = 20600 ----(2)

On solving (1) and (2), we get:

m = 780

So, The number of males in Company B in Y1 is 9m = 9(780) = 7020

Similarly let is assume the number of males and females in company B in Y3 be 27t and 20t respectively and the number of males and females in company A  in Y3 be 16l and 20l respectively. Then,

27t + 16l = 18025

And 20t + 20l = 16000

On solving (3) and (4), we get:

t = 475 and l = 325

So, Total number of females in company A in Y3 = 20l = 20(325) = 6500

Now, Let the number of males and females in company A in Y5 be 28x and 20x respectively and the number of males and females in company B in Y5 be 17y and 20y respectively.Then,

28x + 17y = 13550 ------------(5)

And 20x + 20y = 11800 ------(6)

On solving equations (5) and (6) , we get :

x = 320 

So, The number of females in company A in Y5

= 20x

= 20(320)

= 6400

Now, required ratio

= 7020 : (6500 + 6400)

= 7020 : 12900

= 117 : 215

Hence,  The Required value is 117 : 215.

Given:

$\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}} \div \frac{\sqrt{14}+\sqrt{10}}{\sqrt{14}-\sqrt{10}}+\frac{\sqrt{10}}{\sqrt{5}}$

Formula used:

$\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$

$\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$

Calculations:

⇒ $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}} \div \frac{\sqrt{2 \times 7}+\sqrt{2 \times 5}}{\sqrt{2 \times 7}-\sqrt{2 \times 5}} + \sqrt{\frac{10}{5}}$

⇒ $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}} \div \frac{\sqrt{2}(\sqrt{7}+\sqrt{5})}{\sqrt{2}(\sqrt{7}-\sqrt{5})} + \sqrt{2}$

⇒ $\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}} \div \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}} + \sqrt{2}$

⇒ 1 + $\sqrt{2}$

∴ The correct answer is option (3).

Shortcut Trick

Initial Water Percentage = (3/5) × 100 = 60%

In each step, 10/100 = 1/10 of the mixture is replaced, leaving (1 − 1/10) = 9/10 of the water.

After 3 operations, Final Water % = 60 × (9/10) × (9/10) × (9/10) = 43.74%

Final Milk Percentage = 100% − 43.74% = 56.26%

∴ The correct answer is 56.26 percent.

Alternate Method

Given: Total volume = 100 L, Initial Milk:Water = 2:3, Process repeated 3 times total.

Formula for quantity of component not added: Final Qty = Initial Qty × (1 − x/V)ⁿ

⇒ Initial Water = (3 / (2 + 3)) × 100 = 60 L

⇒ Final Water = 60 × (1 − 10/100)³

⇒ Final Water = 60 × (0.9)³

⇒ Final Water = 60 × 0.729 = 43.74 L

⇒ Final Milk = Total Volume − Final Water

⇒ Final Milk = 100 − 43.74 = 56.26 L

⇒ Percentage of Milk = (56.26 / 100) × 100 = 56.26%

∴ The correct answer is 56.26 percent.

Additional Information

Successive Replacement Formula

Final Quantity = Initial Quantity × (1 − x/V)ⁿ, where x is volume replaced, V is total volume, and n is number of times.

Mixture Concentration

If multiple replacements are made with different volumes x₁, x₂, ..., the factor becomes (1 − x₁/V)(1 − x₂/V)...

Relative Removal

When mixture is withdrawn, all components are removed in the same ratio as they exist in the mixture at that moment.

Shortcut Trick

Area of equilateral triangle = (√3 / 4) × side² = 36√3

⇒ side² = 36 × 4 = 144 ⇒ side = 12

Direct Formula: Circumradius (R) = side / √3

⇒ R = 12 / √3 = 4√3

∴ The correct answer is 4√3.

Alternate Method

Given:

Area of equilateral triangle ABC = 36√3

Formula Used:

1. Area of equilateral triangle = (√3 / 4) × a² (where a is the side)
2. Circumradius (R) of equilateral triangle = a / √3

Calculations:

⇒ Area = (√3 / 4) × a² = 36√3

⇒ a² = (36√3 × 4) / √3

⇒ a² = 144

⇒ a = √144 = 12

⇒ Circumradius (R) = a / √3

⇒ R = 12 / √3

⇒ R = (4 × 3) / √3

⇒ R = 4√3

∴ The correct answer is 4√3.

Additional Information

Inradius of Equilateral Triangle

The radius of the incircle (r) is given by r = a / (2√3). It is exactly half of the circumradius (R = 2r).

Height of Equilateral Triangle

The altitude or height (h) is given by h = (√3 / 2) × a. Note that h = R + r = 3r.

Area in terms of Height

The area of an equilateral triangle can also be calculated as Area = h² / √3.

Shortcut Trick

In a right-angled triangle, the altitude drawn to the hypotenuse divides the triangle into two similar triangles.

According to the Right Triangle Altitude Theorem (Geometric Mean Theorem):

The square of the altitude (BD) is equal to the product of the segments of the hypotenuse (AD and DC).

BD² = AD × DC

∴ The correct answer is AD × DC.

Alternate Method

Given: ABC is a right-angled triangle where ∠ABC = 90° and BD ⊥ AC.

Formula Used: Similarity of triangles (AA criterion) and ratio of corresponding sides.

⇒ In ▵ABD and ▵BCD:

⇒ ∠ADB = ∠BDC = 90° (Since BD ⊥ AC)

⇒ ∠BAD = 90° − ∠ABD (In ▵ABD)

⇒ ∠CBD = 90° − ∠ABD (Since ∠ABC = 90°)

⇒ ∴ ∠BAD = ∠CBD

⇒ By AA Similarity, ▵ABD ∼ ▵BCD

⇒ Corresponding sides are in proportion: BD/AD = DC/BD

⇒ Cross-multiplying: BD × BD = AD × DC

⇒ BD² = AD × DC

∴ The correct answer is AD × DC.

Additional Information

Leg Rule in Right Triangles

The square of a leg is equal to the product of the hypotenuse and the segment of the hypotenuse adjacent to that leg: AB² = AD × AC and BC² = CD × AC.

Area-Altitude Relationship

The product of the legs of a right triangle is equal to the product of the hypotenuse and the altitude to the hypotenuse: AB × BC = AC × BD.

Reciprocal Property

The reciprocal of the square of the altitude is equal to the sum of the reciprocals of the squares of the legs: 1/BD² = 1/AB² + 1/BC².

Shortcut Trick Convert the range of cube roots to the range of the integer N by cubing the limits. Lower limit for N = 63 = 216 Upper limit for N = 73 = 343 Number of integer values between 216 and 343 = (343 − 216 − 1) = 126 ∴ The correct answer is 126 values.

Alternate Method Given: 3√N lies between 6 and 7, where N is an integer. Formula Used: Count of integers strictly between two numbers x and y = (y − x − 1).

⇒ 6 < 3√N < 7 ⇒ Cubing all terms: 63 < (N1/3)3 < 73 ⇒ 216 < N < 343 ⇒ Since N is an integer, N can take any value in the set {217, 218, 219, ..., 342}. ⇒ Total values of N = 342 − 217 + 1 ⇒ Total values = 125 + 1 = 126 ∴ The correct answer is 126 values.

Additional Information Number of Integers in a Range For any two integers a and b (where a < b), the count of integers between them is (b − a − 1) if both are excluded, and (b − a + 1) if both are included. Properties of Powers and Roots If x > 0 and y > 0, then x < y ⇒ xn < yn for any positive power n. This allows us to cube inequalities to find the range of N. Perfect Cubes Table Common cubes for exams: 53=125, 63=216, 73=343, 83=512, 93=729, 103=1000.

Shortcut Trick

Let the original Cost Price (CP₁) = 100 units.

Original Selling Price (SP₁) = 100 − 25% of 100 = 75 units.

New CP (CP₂) = 2 × 100 = 200 units.

New SP (SP₂) = 200 + 20% profit = 200 + 40 = 240 units.

Difference in SP = 240 − 75 = 165 units → ₹660.

1 unit = 660 ÷ 165 = ₹4. ∴ Original CP = 100 units = 100 × 4 = ₹400.

∴ The correct answer is ₹400.

Alternate Method

Given: Original Loss = 25%, New CP = 2 × Original CP, SP Increase = ₹660, New Profit = 20%.

Formula: SP = CP × (100 ± P/L)%

⇒ Let original CP = x

⇒ Original SP = x × (100 − 25) / 100 = 0.75x

⇒ New CP = 2 × x = 2x

⇒ New SP = 2x × (100 + 20) / 100 = 2x × 1.2 = 2.4x

⇒ Given: New SP − Original SP = 660

⇒ 2.4x − 0.75x = 660

⇒ 1.65x = 660

⇒ x = 660 ÷ 1.65

⇒ x = 66000 ÷ 165 = 400

∴ The correct answer is ₹400.

Additional Information

Profit and Loss Basics

Profit = SP − CP and Loss = CP − SP. Profit% or Loss% is always calculated on CP unless stated otherwise.

Successive Percentage Change

If CP changes by a% and then profit/loss b% is applied, the effective change from original CP is calculated as [a + b + (ab/100)].

Marked Price and Discount

SP = Marked Price × (100 − Discount)% / 100. Relation: CP/MP = (100 − D%)/(100 + P%).

Shortcut Trick

Total sum of ages = 7 × 28 = 196 years.

Sum of first 3 students = 3 × 30 = 90 years.

Age of each of the last 2 students = 30 + 5 = 35 years.

Sum of last 2 students = 2 × 35 = 70 years.

Remaining sum for 4th and 5th student = 196 − (90 + 70) = 36 years.

Average of 4th and 5th student = 36 ÷ 2 = 18 years.

∴ The correct answer is 18 years.

Alternate Method

Given:

Number of students = 7

Average age of 7 students = 28 years

Average age of first 3 students = 30 years

Ages of last 2 students = Average of first 3 + 5

Formula: Average = Sum of observations ÷ Total number of observations

⇒ Total age of 7 students = 7 × 28 = 196 years

⇒ Sum of first 3 students = 3 × 30 = 90 years

⇒ Age of 6th student = Age of 7th student = 30 + 5 = 35 years

⇒ Sum of ages of 6th and 7th student = 35 + 35 = 70 years

⇒ Let age of 5th student be x. Then age of 4th student = x − 4

⇒ Sum of 4th and 5th student = (x − 4) + x = 2x − 4

⇒ 196 = 90 + (2x − 4) + 70

⇒ 196 = 160 + 2x − 4

⇒ 196 = 156 + 2x

⇒ 2x = 196 − 156 = 40

⇒ x = 20 years (Age of 5th student)

⇒ Age of 4th student = 20 − 4 = 16 years

⇒ Average of 4th and 5th student = (16 + 20) ÷ 2 = 36 ÷ 2 = 18 years

∴ The correct answer is 18 years.

Additional Information

Weighted Average

When two or more groups with different averages are combined, the overall average = (n₁×a₁ + n₂×a₂) ÷ (n₁ + n₂).

Sum and Difference Property

If the sum of two numbers is S and their difference is D, then the average of the two numbers is simply S ÷ 2, regardless of the difference.

Effect of Constant Change

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

Shortcut Trick

Effective rate of CI for 2 years = R + R + (R × R) / 100 = 8 + 8 + (8 × 8) / 100 = 16.64%

Effective rate of SI for 2 years = R × T = 8 × 2 = 16%

Given: 16.64% of Sum = Rs. 6656

Required: 16% of Sum = (6656 ÷ 16.64) × 16 = 400 × 16 = 6400

∴ The correct answer is Rs. 6400.

Alternate Method

Given:

Compound Interest (CI) = Rs. 6656

Rate of Interest (R) = 8% per annum

Time (T) = 2 years

Formula:

CI = P[(1 + R/100)^T − 1]

SI = (P × R × T) ÷ 100

Calculations:

⇒ 6656 = P[(1 + 8/100)² − 1]

⇒ 6656 = P[(108/100)² − 1]

⇒ 6656 = P[(27/25)² − 1]

⇒ 6656 = P[729/625 − 1]

⇒ 6656 = P[(729 − 625) ÷ 625]

⇒ 6656 = P × (104 ÷ 625)

⇒ P = (6656 × 625) ÷ 104

⇒ P = 64 × 625

⇒ P = Rs. 40000

Now, calculate SI on the same sum:

⇒ SI = (40000 × 8 × 2) ÷ 100

⇒ SI = 400 × 16

⇒ SI = Rs. 6400

∴ The correct answer is Rs. 6400.

Additional Information

Difference between CI and SI for 2 years

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

Relation between CI and SI for 2 years

CI = SI × (1 + R/200) or SI = CI × [200 ÷ (200 + R)]. For 2 years, CI is always greater than or equal to SI.

Effective Rate of Interest

For Compound Interest, the effective rate for n years can be calculated using successive percentage formula: r_eff = x + y + (xy/100).

Shortcut Trick

Ratio of B : C : D : E = 2 : 3 : 4 : 5. Total = 14 units.

Increased salary of D = 4 × (100 + 40)% = 4 × 1.4 = 5.6 units.

Given 5.6 units = Rs. 560 ⇒ 1 unit = Rs. 100.

Sum of original salaries = 14 units = 14 × 100 = Rs. 1400.

∴ The correct answer is Rs. 1400.

Alternate Method

Given:

Ratio of salaries = 2 : 3 : 4 : 5

Salary increases = 20%, 30%, 40%, 50%

Increased D salary = Rs. 560

Formula Used:

Increased Value = Original Value × (1 + Increase%/100)

Let original salaries be 2x, 3x, 4x, and 5x.

⇒ Increased salary of D = 4x × (140/100)

⇒ 4x × 1.4 = 560

⇒ 5.6x = 560

⇒ x = 560 ÷ 5.6

⇒ x = 100

⇒ Sum of original salaries = 2x + 3x + 4x + 5x = 14x

⇒ Total Sum = 14 × 100 = 1400

∴ The correct answer is Rs. 1400.

Additional Information

Percentage Increase Formula

To find the new value after a percentage increase r%, use: New Value = Old Value × (1 + r/100).

Ratio and Multiplier

If components A, B, and C are in ratio a:b:c, their actual values can be represented as ak, bk, and ck where k is a common multiplier.

Successive Percentage Change

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

Shortcut Trick

Assume original CP = 100 units.

Original SP = 100 − 18 = 82 units.

New CP = 100 − 30% = 70 units.

New SP = 120% of 70 = 84 units.

Difference in SP = 84 − 82 = 2 units → Rs. 144.

1 unit = 72 ∴ Original SP = 82 × 72 = Rs. 5904.

∴ The correct answer is Rs. 5904.

Alternate Method

Given:

Initial loss = 18%

Increase in SP = Rs. 144

Decrease in CP = 30%

Final Profit = 20%

Calculations:

Let the original cost price be 100x.

⇒ Original Selling Price = 100x − 18% of 100x = 82x

⇒ New Cost Price = 100x − 30% of 100x = 70x

⇒ New Selling Price = 70x + 20% of 70x

⇒ New Selling Price = 70x + 14x = 84x

⇒ Difference in Selling Prices = New SP − Original SP

⇒ 84x − 82x = 144

⇒ 2x = 144

⇒ x = 72

⇒ Original Selling Price = 82x = 82 × 72

⇒ Original Selling Price = Rs. 5904

∴ The correct answer is Rs. 5904.

Additional Information

Profit and Loss Basic Formulas

Profit = SP − CP; Loss = CP − SP. Profit% = (Profit/CP) × 100; Loss% = (Loss/CP) × 100.

Marked Price and Discount

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

Selling Price for Profit/Loss

SP = CP × (100 + Profit%)/100 or SP = CP × (100 − Loss%)/100.

Given:

Males(B% - C%) = 3% = 600     (∵ The value 3% is taken from given pie chart)

Females(D% - E%) = 3% = 900

Solution:

Male(1%) = 600/3 = 200 males

Females(1%) = 900/3 = 300 females

Now, 

Female(A% + B% + F%) + Male(D% + E% + A%)

= Female(10% + 14% + 25%) + Male(20% + 20% + 18%)

= Female(49%) + Male(58%) = (49 x 300) + (58 x 200) = 26300

Hence, the sum of the number of females in A, B, and F and the number of males in D, E, and A is 26300.

Shortcut Trick

Standard Circle Data: Radius 7 → Area 154 | Radius 14 → Area 616.

Base Area = 616 cm² ⇒ Radius (r) = 14 cm.

Slant height (l) = 20 cm.

CSA = π × r × l = (22/7) × 14 × 20 = 44 × 20 = 880 cm².

∴ The correct answer is 880 cm².

Alternate Method

Given:

Slant height (l) = 20 cm

Area of base = 616 cm²

Value of π = 22/7

Formula Used:

Area of base of cone = πr²

Curved Surface Area (CSA) of cone = πrl

Calculations:

⇒ Area of base = πr² = 616

⇒ (22/7) × r² = 616

⇒ r² = 616 × (7/22)

⇒ r² = 28 × 7 = 196

⇒ r = √196 = 14 cm

⇒ Curved Surface Area (CSA) = π × r × l

⇒ CSA = (22/7) × 14 × 20

⇒ CSA = 22 × 2 × 20

⇒ CSA = 880 cm²

∴ The correct answer is 880 cm².

Additional Information

Total Surface Area of Cone

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

Volume of a Cone

The space occupied by the cone is given by: Volume = (1/3) × πr²h, where h is the vertical height.

Slant Height Relation

The relationship between slant height (l), vertical height (h), and radius (r) is: l² = r² + h².

Shortcut Trick

Given: Radius (r) = 14 cm.

Height (h) = Radius + 6 = 14 + 6 = 20 cm.

Volume = πr²h = (22/7) × 14 × 14 × 20.

Volume = 44 × 14 × 20 = 12320 cm³.

∴ The correct answer is 12320 cm³.

Alternate Method

Given:

Radius of base (r) = 14 cm

Height of cylinder (h) = r + 6 cm

Formula Used:

Volume of Cylinder (V) = π × r² × h

Calculations:

⇒ Radius (r) = 14 cm

⇒ Height (h) = 14 + 6 = 20 cm

⇒ Volume (V) = (22/7) × 14 × 14 × 20

⇒ V = 22 × (14 ÷ 7) × 14 × 20

⇒ V = 22 × 2 × 14 × 20

⇒ V = 44 × 280

⇒ V = 12320 cm³

∴ The correct answer is 12320 cm³.

Additional Information

Curved Surface Area of Cylinder

The area of the curved part of the cylinder is given by CSA = 2πrh.

Total Surface Area of Cylinder

The sum of the areas of the curved surface and the two circular bases: TSA = 2πr(h + r).

Volume Conversion

1 cm³ = 1 mL and 1000 cm³ = 1 Liter.

Given:

A% = 12%

F% = 28%

Formula used:

360° = 100%

Solution:

1% = 3.6°

Now, the sum of central angle formed by A & F = 12% + 28% = 40%

40% = 40 x 3.6° = 144°

Hence, the sum of central angles formed by the sectors representing tyres of Company F and A is 144 degrees.

Given:

Boys(S6) = Girls(S1) + 0.3[Girls(S1)]

Girls(S1) = 500

Solution:

Boys(S6) = 500 + 0.3(500) = 650 

Girls(S3 + S4) = m x Boys(S6 + S1)

From the graph,

1800 + 400 = m x (650 + 1200)

2200 = m x (1850)

m = 220/185 = 1.1892

m% = 118.92%

Hence, the number of girls in schools S3 and S4 is 118.92% of the number of boys in schools S6 and S1.

Confusion point - If the number of boys in school S6 are 30 percent more than the number of girls in school S1 (Clearly mentioned in the question), S6 is not directly given we have to calculate the value of S6 From S1, and by the given relation between them.

Shortcut Trick

Area of base = Volume ÷ Height

Area of base = 4800 ÷ 20

Area of base = 240 cm²

∴ The correct answer is 240 cm².

Alternate Method

Given:

Volume of the cuboid = 4800 cm³

Height of the cuboid = 20 cm

Formula Used:

Volume of a cuboid = Length × Breadth × Height

Volume of a cuboid = Area of base × Height

Calculations:

⇒ Volume = Area of base × Height

⇒ 4800 = Area of base × 20

⇒ Area of base = 4800 ÷ 20

⇒ Area of base = 240 cm²

∴ The correct answer is 240 cm².

Additional Information

Total Surface Area of Cuboid

The total area of all six faces is given by 2(lb + bh + hl), where l, b, and h are length, breadth, and height.

Lateral Surface Area of Cuboid

The area of the four side walls (excluding top and bottom) is 2h(l + b).

Diagonal of a Cuboid

The longest rod that can be placed inside a cuboid is its diagonal, calculated as √(l² + b² + h²).

Shortcut Trick

If a large sphere of radius R is melted into n equal small spheres of radius r, then r = R ÷ n^1/3.

Here n = 8, so r = 14 ÷ 8^1/3 = 14 ÷ 2 = 7 cm.

Sum of Surface Areas of n small spheres = n × 4πr².

Sum = 8 × 4 × (22/7) × 7 × 7 = 32 × 154 = 4928 cm².

∴ The correct answer is 4928 cm².

Alternate Method

Given:

Radius of large sphere (R) = 14 cm

Number of small spheres (n) = 8

Formula Used:

Volume of Sphere = (4/3)πr³

Total Surface Area (TSA) of Sphere = 4πr²

Calculations:

⇒ Volume of large sphere = 8 × Volume of one small sphere

⇒ (4/3) × π × (14)³ = 8 × (4/3) × π × r³

⇒ 14 × 14 × 14 = 8 × r³

⇒ r³ = (14 × 14 × 14) ÷ 8

⇒ r³ = 2744 ÷ 8 = 343

⇒ r = √³(343) = 7 cm

⇒ TSA of one small sphere = 4 × (22/7) × 7 × 7

⇒ TSA of one small sphere = 4 × 22 × 7 = 616 cm²

⇒ Sum of TSA of all 8 small spheres = 8 × 616

⇒ Sum of TSA = 4928 cm²

∴ The correct answer is 4928 cm².

Additional Information

Volume and Surface Area Scaling

If the radius of a sphere is scaled by a factor k, the volume scales by k³ and the surface area scales by k².

Hemisphere Formulas

For a solid hemisphere of radius r, Volume = (2/3)πr³, Curved Surface Area = 2πr², and Total Surface Area = 3πr².

Mass and Density Relation

When an object is melted and recast, its volume remains constant (provided there is no material loss), such that V_initial = V_final.

Shortcut Trick

Convert terms into sine and cosine: sec A − cos A = (1/cos A) − cos A

Take the LCM: (1 − cos² A) / cos A

Using identity sin² A + cos² A = 1: sin² A / cos A

Split the term: (sin A / cos A) × sin A = tan A × sin A

∴ The correct answer is tan A.sin A.

Alternate Method

Given expression: sec A − cos A

Formula used: sec A = 1/cos A; tan A = sin A/cos A; sin² A + cos² A = 1

⇒ sec A − cos A = (H/B) − (B/H)

⇒ sec A − cos A = (H² − B²) / (BH)

⇒ In right triangle, H² − B² = P²

⇒ sec A − cos A = P² / (BH)

⇒ sec A − cos A = (P/B) × (P/H)

⇒ sec A − cos A = tan A × sin A

∴ The correct answer is tan A.sin A.

Additional Information

Fundamental Trigonometric Identities

The three main Pythagorean identities are: sin² A + cos² A = 1, 1 + tan² A = sec² A, and 1 + cot² A = cosec² A.

Reciprocal Relations

Trigonometric ratios are related as: sin A = 1/cosec A, cos A = 1/sec A, and tan A = 1/cot A.

Quotient Identities

The tangent and cotangent can be expressed as: tan A = sin A / cos A and cot A = cos A / sin A.

Shortcut Trick

Sides 6, 8, and 10 form a Pythagorean triple because 6² + 8² = 10² (36 + 64 = 100).

Thus, both ΔABC and ΔPQR are right-angled triangles with the longest side (10 cm) as the hypotenuse.

In ΔABC, BC is the hypotenuse ⇒ ∠A = 90°. So, ∠ABC + ∠ACB = 90°.

Since the triangles are congruent (SSS rule), ∠PRQ = ∠ACB.

Given ∠ABC = x, then ∠ACB = 90° − x.

∴ The correct answer is (90 − x) degree.

Alternate Method

Given: AB = PQ = 6 cm, BC = QR = 10 cm, AC = PR = 8 cm, and ∠ABC = x°.

Formula: In a right-angled triangle, the two acute angles are complementary (sum to 90°).

⇒ In ΔABC, sides are 6, 8, 10. Check: 6² + 8² = 36 + 64 = 100 = 10².

⇒ Since BC² = AB² + AC², ∠BAC = 90°.

⇒ In ΔABC, ∠ABC + ∠ACB + ∠BAC = 180°

⇒ x + ∠ACB + 90° = 180° ⇒ ∠ACB = 90° − x.

⇒ Since AB = PQ, BC = QR, and AC = PR, then ΔABC ≅ ΔPQR (By SSS Congruence Rule).

⇒ Corresponding parts of congruent triangles (CPCT) are equal, so ∠PRQ = ∠ACB.

⇒ ∠PRQ = 90° − x.

∴ The correct answer is (90 − x) degree.

Additional Information

SSS Congruence Rule

If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

Pythagorean Triplets

Sets of three positive integers (a, b, c) such that a² + b² = c², which always form a right-angled triangle (e.g., 3-4-5, 6-8-10, 5-12-13).

Complementary Angles

In any right-angled triangle, the two non-right angles always add up to 90°, making them complementary to each other.

Shortcut Trick

Area of a quadrilateral given one diagonal and perpendiculars from opposite vertices is calculated as:

Area = 1/2 × Diagonal × (Sum of perpendiculars)

Area = 1/2 × 24 × 10

Area = 12 × 10 = 120 cm²

∴ The correct answer is 120 cm².

Alternate Method

Given:

Length of diagonal (AC) = 24 cm

Sum of perpendiculars (h₁ + h₂) = 10 cm

Formula used:

Area of Quadrilateral = Area of ΔABC + Area of ΔADC

Area = (1/2 × AC × h₁) + (1/2 × AC × h₂) = 1/2 × AC × (h₁ + h₂)

Calculations:

⇒ Area = 1/2 × AC × (h₁ + h₂)

⇒ Area = 1/2 × 24 × 10

⇒ Area = 12 × 10

⇒ Area = 120 cm²

∴ The correct answer is 120 cm².

Additional Information

Area of Rhombus

If d₁ and d₂ are the lengths of the two diagonals of a rhombus, then Area = 1/2 × d₁ × d₂.

Area of Trapezium

Area = 1/2 × (Sum of parallel sides) × Height = 1/2 × (a + b) × h.

Area of Parallelogram

Area = Base × Corresponding Height = b × h.

Shortcut Trick

Income Ratio = 4 : 5 | Savings Ratio = 14 : 19 | Constant Expenditure = Rs. 1200

Using Cross-Product Method: (4 × 19) − (5 × 14) = 76 − 70 = 6 units

Equating to Expenditure Change: 1200 × (19 − 14) = 1200 × 5 = Rs. 6000

6 units = Rs. 6000 ⇒ 1 unit = Rs. 1000

Difference in Monthly Incomes (5 − 4) = 1 unit = Rs. 1000

∴ The correct answer is Rs. 1000.

Alternate Method

Given: Monthly Income Ratio A : B = 4 : 5 | Savings Ratio A : B = 14 : 19 | Expenditure A = Expenditure B = Rs. 1200.

Formula: Savings = Income − Expenditure

⇒ Let Monthly Incomes of A and B be 4x and 5x respectively.

⇒ Savings of A = 4x − 1200

⇒ Savings of B = 5x − 1200

⇒ According to the ratio: (4x − 1200) ÷ (5x − 1200) = 14 ÷ 19

⇒ 19 × (4x − 1200) = 14 × (5x − 1200)

⇒ 76x − 22800 = 70x − 16800

⇒ 76x − 70x = 22800 − 16800

⇒ 6x = 6000

⇒ x = 1000

⇒ Difference between Incomes = 5x − 4x = x

⇒ Difference = Rs. 1000

∴ The correct answer is Rs. 1000.

Additional Information

Income and Expenditure Formula

Income = Expenditure + Savings. This basic identity is the foundation for solving problems related to earnings and savings.

Cross Multiplication Trick

For two ratios a:b and c:d where a fixed value K is subtracted/added to both, then 1 unit = |K × (c − d)| ÷ |(a × d) − (b × c)|.

Balanced Ratio Method

If the difference between income and savings units is the same for both individuals, that difference directly represents the expenditure.

Shortcut Trick

If (x + 1/x)² = 3, then x + 1/x = √3.

Standard algebraic property: If x + 1/x = √3, then x⁶ = −1.

We need to find the value of x⁶ + x⁻⁶, which is x⁶ + 1/x⁶.

Substituting x⁶ = −1, we get −1 + 1/(−1) = −1 − 1 = −2.

∴ The correct answer is −2.

Alternate Method

Given: (x + 1/x)² = 3

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

⇒ (x + 1/x)² = 3

⇒ x + 1/x = √3

⇒ Cubing both sides: (x + 1/x)³ = (√3)³

⇒ x³ + 1/x³ + 3 × x × (1/x) × (x + 1/x) = 3√3

⇒ x³ + 1/x³ + 3 × (√3) = 3√3

⇒ x³ + 1/x³ = 3√3 − 3√3 = 0

⇒ (x⁶ + 1) / x³ = 0

⇒ x⁶ + 1 = 0 ⇒ x⁶ = −1

⇒ Now, x⁶ + x⁻⁶ = x⁶ + 1/x⁶

⇒ (−1) + 1/(−1) = −1 − 1 = −2

∴ The correct answer is −2.

Additional Information

Special Case for x + 1/x = √3

Whenever x + 1/x = √3 or (x + 1/x)² = 3, the expression x⁶ always equals −1.

Algebraic Identity for Cubes

If x + 1/x = k, then x³ + 1/x³ = k³ − 3k. Substituting k = √3 gives (√3)³ − 3(√3) = 0.

Negative Exponents

The term x⁻ⁿ is mathematically defined as the reciprocal 1/xⁿ.

Standard Result for x + 1/x = 1

If x + 1/x = 1, then x² − x + 1 = 0, which leads to the standard result x³ = −1.

Shortcut Trick

Use the substitution method by assuming x = 1, y = 1, and z = 1.

Numerator = (1 + 1 + 1)(1 × 1 + 1 × 1 + 1 × 1) − (1 × 1 × 1) = 3 × 3 − 1 = 8.

Denominator = (1 + 1)(1 + 1)(1 + 1) = 2 × 2 × 2 = 8.

Simplified Value = 8 ÷ 8 = 1.

∴ The correct answer is 1.

Alternate Method

Given expression: ((x + y + z)(xy + yz + zx) − xyz) ÷ ((x + y)(y + z)(z + x))

Expand the Numerator:

⇒ (x + y + z)(xy + yz + zx) − xyz

⇒ x(xy + yz + zx) + y(xy + yz + zx) + z(xy + yz + zx) − xyz

⇒ (x²y + xyz + zx²) + (xy² + y²z + xyz) + (xyz + yz² + z²x) − xyz

⇒ x²y + xy² + y²z + yz² + z²x + zx² + 2xyz

Now, expand the Denominator:

⇒ (x + y)(y + z)(z + x) = (xy + xz + y² + yz)(z + x)

⇒ z(xy + xz + y² + yz) + x(xy + xz + y² + yz)

⇒ (xyz + xz² + y²z + yz²) + (x²y + x²z + xy² + xyz)

⇒ x²y + xy² + y²z + yz² + z²x + zx² + 2xyz

Dividing Numerator by Denominator:

⇒ Result = (x²y + xy² + y²z + yz² + z²x + zx² + 2xyz) ÷ (x²y + xy² + y²z + yz² + z²x + zx² + 2xyz) = 1

∴ The correct answer is 1.

Additional Information

Algebraic Identity for Product of Sums

(x + y)(y + z)(z + x) = (x + y + z)(xy + yz + zx) − xyz. This identity is a standard result in cyclic algebraic expressions.

Special Case for x + y + z = 0

If x + y + z = 0, the expression simplifies such that (x + y)(y + z)(z + x) = −xyz. Also, x³ + y³ + z³ = 3xyz.

Square of a Trinomial

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx). This is used to find the sum of products from the sum of variables and the sum of their squares.

Shortcut Trick

The given expression is cot θ + tan θ.

Standard trigonometric identity: tan θ + cot θ = sec θ × cosec θ.

By cross-multiplying: (sin²θ + cos²θ) ÷ (sin θ cos θ) = 1 ÷ (sin θ cos θ).

Since 1/sin θ = cosec θ and 1/cos θ = sec θ, the result is cosec θ.sec θ.

∴ The correct answer is cosec θ.sec θ.

Alternate Method

Given:

Expression = (cos θ ÷ sin θ) + (sin θ ÷ cos θ)

Formula Used:

1. sin²θ + cos²θ = 1
2. sec θ = 1 ÷ cos θ
3. cosec θ = 1 ÷ sin θ

⇒ [(cos θ × cos θ) + (sin θ × sin θ)] ÷ (sin θ × cos θ)

⇒ (cos²θ + sin²θ) ÷ (sin θ cos θ)

⇒ 1 ÷ (sin θ cos θ) [Using sin²θ + cos²θ = 1]

⇒ (1 ÷ sin θ) × (1 ÷ cos θ)

⇒ cosec θ × sec θ

∴ The correct answer is cosec θ.sec θ.

Additional Information

Pythagorean Identities

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

Reciprocal Identities

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

Ratio Identities

tan θ = sin θ/cos θ and cot θ = cos θ/sin θ.

Shortcut Trick For any triangle with sides x, y, and z (where z is the longest side): If x2 + y2 = z2 → Right Angled Triangle If x2 + y2 < z2 → Obtuse Angled Triangle If x2 + y2 > z2 → Acute Angled Triangle Given: x2 + y2 > z2, therefore the triangle must be acute. ∴ The correct answer is Acute angled triangle. Alternate Method Given: Sides of triangle are x, y, and z. z is the largest side. Condition: x2 + y2 > z2. Formula Used: Law of Cosines: cos Z = (x2 + y2 − z2) ÷ (2xy), where Z is the angle opposite to side z.

⇒ According to the Cosine Rule for the largest side z and its opposite angle Z: ⇒ cos Z = (x2 + y2 − z2) ÷ (2xy) ⇒ Since x2 + y2 > z2, then (x2 + y2 − z2) > 0. ⇒ As x and y are lengths of sides, 2xy is always positive. ⇒ ∴ cos Z > 0, which implies that angle Z is less than 90° (0 < Z < 90°). ⇒ Since z is the largest side, Z is the largest angle. If the largest angle is acute, then all angles must be acute. ∴ The correct answer is Acute angled triangle. Additional Information Right Angled Triangle Property A triangle is right-angled if the square of the longest side is exactly equal to the sum of the squares of the other two sides (z2 = x2 + y2). Obtuse Angled Triangle Property A triangle is obtuse-angled if the square of the longest side is greater than the sum of the squares of the other two sides (z2 > x2 + y2). Triangle Inequality Theorem For any triangle to exist, the sum of any two sides must be strictly greater than the third side (x + y > z, y + z > x, and z + x > y).

Shortcut Trick

Find two numbers such that their sum is 6 and product is 5.

By inspection, the numbers are 5 and 1 (as 5 + 1 = 6 and 5 × 1 = 5).

Substituting these values: a³ + b³ = 5³ + 1³

a³ + b³ = 125 + 1 = 126.

∴ The correct answer is 126.

Alternate Method

Given: a + b = 6 and ab = 5

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

⇒ a³ + b³ = (6)³ − 3 × 5 × 6

⇒ a³ + b³ = 216 − 30 × 3

⇒ a³ + b³ = 216 − 90

⇒ a³ + b³ = 126

∴ The correct answer is 126.

Additional Information

Sum of Cubes Identity

a³ + b³ can also be expressed as (a + b)(a² − ab + b²).

Difference of Cubes Identity

a³ − b³ = (a − b)(a² + ab + b²) or (a − b)³ + 3ab(a − b).

Square of Sum and Difference

(a + b)² − (a − b)² = 4ab, which helps in finding a − b if sum and product are known.

GIVEN:

Selling price and discount percentage of seven different articles from A₁ to A₇ 

FORMULA USED:

 M.P. = 100 S.P. / ( 100  - d) 

Where, S.P. = Selling price , M.P. = Marked price and d = discount%

CALCULATION:

Here, we have 

S.P. of A1  = Rs420 and discount%  = 30 

⇒ M.P. = 100 S.P. / ( 100 -d)  

⇒ M.P. = 100 × 420 / ( 100 - 30) 

⇒ M.P. = 100 × 420/70  = Rs600 

Now, S.P. of A2 = Rs600 and d% = 25 

⇒ M.P. =  100 × 600 / 75 =  Rs800 

Now, S.P.  of A3 = Rs816  and d = 32%

⇒ M.P. = 100 × 816 / 68 = Rs1200 

Now, S.P. of A4  = Rs825 , d = 45% 

⇒ M.P. = 100 × 825/55 = Rs1500 

Now, S.P. of A5 =  Rs425 ,d = 15%

⇒ M.P. = 100 × 425 /85 = Rs500

Now, S.P. of A6 = Rs800 , d = 20%

⇒ M.P. = 100 × 800/80 = Rs1,000

Now, S.P. of A7 = Rs840 ,d = 40%

⇒ M.P. = 100 × 840/60 = Rs1400 

Total Marked Price from A1 to A7 = ( 600 + 800 + 1200 + 1500 + 500 + 1000 + 1400 ) = Rs7,000 

Hence, sum of the marked price of the seven articles is Rs7,000

Shortcut Trick

Substitute standard values: sin 30° = 0.5, cos 60° = 0.5, sec 45° = √2

Expression = 3 × (0.5)² + 0.6 × (0.5)² − 2 × (√2)²

Expression = 3 × 0.25 + 0.6 × 0.25 − 2 × 2 = 0.75 + 0.15 − 4

Expression = 0.9 − 4 = −3.1

Convert decimal to fraction: −3.1 = −31/10

∴ The correct answer is −31/10.

Alternate Method

Given: 3 sin² 30° + 3/5 cos² 60° − 2 sec² 45°

Values Used:

sin 30° = 1/2

cos 60° = 1/2

sec 45° = √2

Calculations:

⇒ 3 × (1/2)² + 3/5 × (1/2)² − 2 × (√2)²

⇒ 3 × (1/4) + 3/5 × (1/4) − 2 × 2

⇒ 3/4 + 3/20 − 4

⇒ (3 × 5)/20 + 3/20 − (4 × 20)/20

⇒ 15/20 + 3/20 − 80/20

⇒ (15 + 3 − 80) / 20

⇒ −62 / 20

⇒ −31 / 10

∴ The correct answer is −31/10.

Additional Information

Standard Trigonometric Ratios

sin 30° = cos 60° = 1/2 and sin 60° = cos 30° = √3/2.

Reciprocal Ratios

sec θ = 1/cos θ. Since cos 45° = 1/√2, then sec 45° = √2.

Pythagorean Identity

sin² θ + cos² θ = 1. This can be used to simplify expressions when the angle θ is the same.

BODMAS Rule

Always solve exponents (squares) first, then multiplication/division, and finally addition/subtraction from left to right.

Shortcut Trick

Identify the ratio between terms using √a / √b = √(a/b).

Term 1: √(29.16 / 1.1664) = √(291600 / 11664) = √25 = 5

Term 2: √(0.2916 / 116.64) = √(2916 / 1166400) = √(1/400) = 1/20

Term 3: √(0.0036 / 0.36) = √(36 / 3600) = √(1/100) = 1/10 = 2/20

Total Sum = 5 + 1/20 + 2/20 = 5 + 3/20 = 103/20.

∴ The correct answer is 103/20.

Alternate Method

Given: Expression = (√29.16 / √1.1664) + (√0.2916 / √116.64) + (√0.0036 / √0.36)

We know that √2916 = 54 and √11664 = 108.

⇒ √29.16 = 5.4

⇒ √1.1664 = 1.08

⇒ √0.2916 = 0.54

⇒ √116.64 = 10.8

⇒ √0.0036 = 0.06

⇒ √0.36 = 0.6

Substitute these values into the expression:

⇒ First term = 5.4 / 1.08 = 540 / 108 = 5

⇒ Second term = 0.54 / 10.8 = 54 / 1080 = 1 / 20

⇒ Third term = 0.06 / 0.6 = 6 / 60 = 1 / 10

⇒ Total = 5 + 1/20 + 1/10

⇒ Total = (100 + 1 + 2) / 20 = 103/20

∴ The correct answer is 103/20.

Additional Information

Square Root of Decimals

When finding the square root of a decimal, the number of decimal places in the result is always half the number of decimal places in the original number.

Ratio Property of Surds

The rule √a / √b = √(a / b) allows for simplifying calculations by dividing the numbers before extracting the square root.

BODMAS Rule

For complex expressions, prioritize Brackets, then Orders (roots/powers), then Division and Multiplication, followed by Addition and Subtraction.

Shortcut Trick

Difference between a two-digit number and its reverse is always 9 × (Difference of digits).

9 × (x − y) = 36 ⇒ x − y = 4. (Statement I is correct)

Possible numbers (x − y = 4): 40, 51, 62, 73, 84, 95.

Since 84 is in the list, Statement II is correct. Since 73 is prime, Statement III is incorrect.

∴ The correct answer is I and II.

Alternate Method

Given: A two-digit number decreases by 36 when digits are reversed.

Let the tens digit be x and the units digit be y. Number = 10x + y.

Reversed number = 10y + x.

⇒ (10x + y) − (10y + x) = 36

⇒ 9x − 9y = 36

⇒ 9(x − y) = 36

⇒ x − y = 4 (This confirms Statement I)

⇒ Possible values for (x, y) where x − y = 4 are: (4,0), (5,1), (6,2), (7,3), (8,4), (9,5).

⇒ Possible numbers: 40, 51, 62, 73, 84, 95.

⇒ Number 84 satisfies the condition (84 − 48 = 36). (This confirms Statement II)

⇒ 73 is a Prime number, so the number is not always composite. (Statement III is False)

∴ The correct answer is I and II.

Additional Information

Sum of Digits Property

The sum of a two-digit number and the number formed by reversing its digits is always divisible by 11. Sum = 11(x + y).

Difference of Digits Property

The difference between a two-digit number and its reverse is always divisible by 9. Difference = 9|x − y|.

Prime vs Composite Numbers

A prime number has exactly two factors (1 and itself). A composite number has more than two factors. For example, 73 is prime, whereas 51 (3 × 17) is composite.

Shortcut Trick

For complementary angles (A + B = 90°), the ratios sin A / cos B = 1 and cosec A / sec B = 1.

In the expression, 50° + 40° = 90° and 80° + 10° = 90°.

⇒ cos 50° / sin 40° = 1

⇒ 3 cosec 80° / sec 10° = 3 × 1 = 3

⇒ 2 cos 50° cosec 40° = 2 × sin 40° × cosec 40° = 2 × 1 = 2

⇒ Total = 1 + 3 − 2 = 2

∴ The correct answer is 2.

Alternate Method

Given:

Expression: (cos 50° / sin 40°) + (3 cosec 80° / sec 10°) − 2 cos 50° cosec 40°

Formula Used:

1. cos θ = sin(90° − θ)
2. cosec θ = sec(90° − θ)
3. cosec θ = 1 / sin θ

Calculations:

⇒ cos 50° = sin(90° − 50°) = sin 40°

⇒ cosec 80° = sec(90° − 80°) = sec 10°

⇒ Expression = (sin 40° / sin 40°) + (3 sec 10° / sec 10°) − 2 cos 50° cosec 40°

⇒ Expression = 1 + 3 − 2 (sin 40° × (1 / sin 40°))

⇒ Expression = 4 − 2(1)

⇒ Expression = 2

∴ The correct answer is 2.

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.

Reciprocal Identities

Basic reciprocal relations: sin θ × cosec θ = 1, cos θ × sec θ = 1, and tan θ × cot θ = 1.

Pythagorean Identities

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

Shortcut Trick

√3 tan θ = 3 sin θ ⇒ √3 (sin θ / cos θ) = 3 sin θ

Canceling sin θ on both sides: √3 / cos θ = 3 ⇒ cos θ = √3 / 3 = 1 / √3

Then, cos² θ = 1 / 3

Expression: sin² θ − cos² θ = (1 − cos² θ) − cos² θ = 1 − 2 cos² θ

Substituting value: 1 − 2(1 / 3) = 1 − 2/3 = 1/3

∴ The correct answer is 1/3.

Alternate Method

Given: √3 tan θ = 3 sin θ

Formula Used: tan θ = sin θ / cos θ; sin² θ + cos² θ = 1

⇒ √3 (sin θ / cos θ) = 3 sin θ

⇒ √3 / cos θ = 3

⇒ cos θ = √3 / 3 = 1 / √3

⇒ cos² θ = (1 / √3)² = 1/3

⇒ From identity, sin² θ = 1 − cos² θ

⇒ sin² θ = 1 − 1/3 = 2/3

⇒ Required value = sin² θ − cos² θ

⇒ Required value = 2/3 − 1/3 = 1/3

∴ The correct answer is 1/3.

Additional Information

Fundamental Trigonometric Identity

The primary identity relates sine and cosine as sin²θ + cos²θ = 1 for any angle θ.

Double Angle Identity for Cosine

cos 2θ = cos²θ − sin²θ. The expression in the question, sin²θ − cos²θ, is equivalent to −cos 2θ.

Tangent Property

The tangent ratio is defined as the quotient of sine and cosine: tan θ = sin θ / cos θ.

Pythagoras Theorem in Trigonometry

For a right-angled triangle, Perpendicular² + Base² = Hypotenuse², which leads to the ratio identities.

Shortcut Trick

Base area of right-angled triangle = (1/2) × product of smaller sides

Base area = (1/2) × 9 × 12 = 54 cm²

Height of prism = Volume ÷ Base area

Height = 648 ÷ 54 = 12 cm

∴ The correct answer is 12 cm.

Alternate Method

Given:

Sides of the triangular base = 9 cm, 12 cm and 15 cm

Volume of the prism = 648 cm³

Formula Used:

Area of right-angled triangle = 1/2 × base × height

Volume of prism = Area of base × Height of prism

Calculations:

⇒ Area of base = (1/2) × 9 × 12

⇒ Area of base = 9 × 6 = 54 cm²

⇒ Volume = Area of base × Height

⇒ 648 = 54 × Height

⇒ Height = 648 ÷ 54

⇒ Height = 12 cm

∴ The correct answer is 12 cm.

Additional Information

Volume of Prism

The volume of any prism is given by the product of its base area and its vertical height (V = A_base × h).

Lateral Surface Area (LSA) of Prism

LSA = Perimeter of base × Height. For this prism, LSA = (9 + 12 + 15) × 12 = 36 × 12 = 432 cm².

Total Surface Area (TSA) of Prism

TSA = Lateral Surface Area + 2 × Area of base. TSA = 432 + 2 × 54 = 540 cm².

Shortcut Trick

Weight ratio = 5 : 9 → Total units = 14

Cost is proportional to square of weight, so let Original Cost = 14² = 196 units

Cost after breakage = (5² + 9²) = 25 + 81 = 106 units

Loss in value = 196 − 106 = 90 units

Loss % = (Loss ÷ Original Cost) × 100 = (90 ÷ 196) × 100 ≈ 45.92%

∴ The correct answer is 45.92 percent.

Alternate Method

Given:

Total weight of diamond = 14 gm

Cost of 14 gm diamond = Rs. 2560

Breakage ratio = 5 : 9

Formula Used:

Cost (C) ∝ Weight (W)² ⇒ C = k × W²

Calculations:

⇒ Let the weights of the two pieces be 5x and 9x.

⇒ Total weight = 5x + 9x = 14x = 14 gm ⇒ x = 1 gm.

⇒ Weights of pieces = 5 gm and 9 gm.

⇒ Since C ∝ W², we can work directly with units as the constant k cancels out in percentage calculation.

⇒ Original Cost equivalent = 14² = 196

⇒ New Cost equivalent = 5² + 9² = 25 + 81 = 106

⇒ Loss = 196 − 106 = 90

⇒ Loss % = (90 ÷ 196) × 100

⇒ Loss % = (45 ÷ 98) × 100

⇒ Loss % ≈ 45.918...

∴ The correct answer is 45.92 percent.

Additional Information

Direct Proportion of Squares

When a value Y is directly proportional to the square of X (Y ∝ X²), doubling X results in a 4-fold increase in Y, and tripling X results in a 9-fold increase in Y.

Percentage Change in Ratios

Percentage change calculations are independent of the actual values if the entire system scales linearly; here, the Rs. 2560 value was not required to calculate the loss percentage.

Maximum Loss Condition

In breakage problems where C ∝ W², the maximum loss occurs when the diamond breaks into the maximum number of equal pieces, while minimum loss (zero) occurs if it doesn't break at all.

GIVEN:

Pie chart shows number of males  in six department 

Number of males in S = 1080 

CALCULATION:

As, number of males in S = 1080

⇒ 54^{$^\circ$} = 1080 

⇒ 1^{$^\circ$} = 20 

Now, 

Number of males in T department is  90^{$^\circ$} =  90 × 20 = 1800 

Number of males in P department is 30^{$^\circ$} = 30 × 20 = 600 

Number of males in Q department is 90^{$^\circ$} = 90 × 20 = 1800 

As, J = Total number of males in T and P department = ( 1800 + 600 ) = 2400 

Also, K = Difference of number of males in P and Q department = ( 1800 - 600) = 1200 

Now, J + K = ( 2400 + 1200 ) = 3600

Hence, the required value is 3600.

Shortcut Trick

A regular hexagon is composed of 6 identical equilateral triangles that meet at the center O.

Triangle AOB is one such equilateral triangle where the side is equal to the side of the hexagon.

Given side (a) = 36 cm.

Area of equilateral triangle = (√3/4) × a² = (√3/4) × 36 × 36.

Area = √3 × 9 × 36 = 324√3 cm².

∴ The correct answer is 324√3 cm².

Alternate Method

Given:

Side of regular hexagon ABCDEF = 36 cm

Point O is the center of the regular hexagon.

Formula Used:

In a regular hexagon, the distance from the center to any vertex is equal to the side of the hexagon.

Area of an equilateral triangle with side 'a' = (√3/4) × a²

Calculations:

⇒ In regular hexagon ABCDEF, OA = OB = AB = 36 cm.

⇒ Therefore, ▵AOB is an equilateral triangle with side a = 36 cm.

⇒ Area of ▵AOB = (√3 / 4) × (36)²

⇒ Area = (√3 / 4) × 1296

⇒ Area = 324 × √3

⇒ Area = 324√3 cm²

∴ The correct answer is 324√3 cm².

Additional Information

Area of Regular Hexagon

The total area of a regular hexagon is 6 × Area of one equilateral triangle = (3√3 / 2) × a².

Angles of a Regular Hexagon

Each interior angle is 120° and the sum of all interior angles is 720°. Each exterior angle is 60°.

Diagonals of a Regular Hexagon

The longest diagonal (e.g., AD) is 2a, and the shorter diagonal (e.g., AC) is a√3.

Shortcut Trick

Since both articles have the same selling price, each SP = 48640 ÷ 2 = Rs. 24320.

For article A (28% profit): CP_A = SP × (100/128) = 24320 × (25/32) = Rs. 19000.

For article B (24% loss): CP_B = SP × (100/76) = 24320 × (25/19) = Rs. 32000.

The respective cost prices are Rs. 19000 and Rs. 32000.

∴ The correct answer is Rs. 19000, Rs. 32000.

Alternate Method

Given:

Total Selling Price (SP_A + SP_B) = Rs. 48640

Condition: SP_A = SP_B

Profit on A = 28% | Loss on B = 24%

Formula Used:

CP = (SP × 100) ÷ (100 ± P/L%)

Calculations:

⇒ Individual Selling Price (SP) = 48640 ÷ 2 = Rs. 24320

⇒ For Article A: CP_A = (24320 × 100) ÷ (100 + 28)

⇒ CP_A = 2432000 ÷ 128 = Rs. 19000

⇒ For Article B: CP_B = (24320 × 100) ÷ (100 − 24)

⇒ CP_B = 2432000 ÷ 76 = Rs. 32000

∴ The correct answer is Rs. 19000, Rs. 32000.

Additional Information

Direct CP-SP Relation

When SP is constant, the Cost Price is inversely proportional to the (100 ± Gain/Loss%). CP₁ × (100 + G₁%) = CP₂ × (100 − L₂%).

Net Profit or Loss Percentage

Total Profit/Loss % = (Total SP − Total CP) ÷ Total CP × 100. In this case, Total CP = 19000 + 32000 = 51000, Net Loss = 51000 − 48640 = 2360.

Same SP, Same % Profit and Loss

If two items are sold at the same price, one at x% profit and another at x% loss, the transaction always results in a loss of (x/10)² %.

Shortcut Trick

Common difference for S1 (d₁) = 9 − 2 = 7 and for S2 (d₂) = 11 − 7 = 4.

First common term (a) = 23 (by inspection: 2, 9, 16, 23... and 7, 11, 15, 19, 23...).

Common difference of the series of common terms (D) = LCM(7, 4) = 28.

Number of common terms (n): 23 + (n − 1) × 28 ≤ 632 ⇒ (n − 1) ≤ 21.75 ⇒ n = 22.

Sum = (n ÷ 2) × [2a + (n − 1)D] = 11 × [46 + (21 × 28)] = 11 × 634 = 6974.

∴ The correct answer is 6974.

Alternate Method

Given:

Series S1: 2, 9, 16, ..., 632 (Arithmetic Progression with a₁ = 2, d₁ = 7)

Series S2: 7, 11, 15, ..., 743 (Arithmetic Progression with a₂ = 7, d₂ = 4)

Formula Used:

n^th term of AP: T_n = a + (n − 1)d

Sum of AP: S_n = (n ÷ 2) × (a + l), where l is the last term.

Calculations:

⇒ For the common series, the first term (a) is the first number that appears in both series. Listing S1: 2, 9, 16, 23, 30... and S2: 7, 11, 15, 19, 23... so a = 23.

⇒ The common difference (D) of common terms is LCM(d₁, d₂) = LCM(7, 4) = 28.

⇒ The last term (L) of the common series must be ≤ min(632, 743), so L ≤ 632.

⇒ Let the number of common terms be n. Then, T_n = 23 + (n − 1) × 28 ≤ 632.

⇒ 28(n − 1) ≤ 632 − 23 ⇒ 28(n − 1) ≤ 609.

⇒ n − 1 ≤ 609 ÷ 28 ⇒ n − 1 ≤ 21.75.

⇒ n − 1 = 21 ⇒ n = 22.

⇒ Last common term (L) = 23 + (21 × 28) = 23 + 588 = 611.

⇒ Sum = (n ÷ 2) × (a + L) = (22 ÷ 2) × (23 + 611).

⇒ Sum = 11 × 634 = 6974.

∴ The correct answer is 6974.

Additional Information

Common Difference of Intersecting APs

If two arithmetic progressions have common differences d₁ and d₂, the series formed by their common terms is also an AP with a common difference equal to LCM(d₁, d₂).

Sum of First n Natural Numbers

The sum of the first n natural numbers is given by Σn = [n × (n + 1)] ÷ 2.

General Term of an AP

The n^th term T_n of an AP is calculated as a + (n − 1)d, where a is the first term and d is the common difference.

Shortcut Trick

Time for A = 12 × 5 ÷ 2 = 30 days.

B is 25% faster than A ⇒ B's time = 30 ÷ 1.25 = 24 days.

C's time = 24 − 12 = 12 days; D is 25% slower than C ⇒ D's time = 12 ÷ 0.75 = 16 days.

Total Work = LCM(30, 24, 12, 16) = 240 units.

Combined Efficiency = 8 (A) + 10 (B) + 20 (C) + 15 (D) = 53 units/day.

∴ The correct answer is 240/53 days.

Alternate Method

Given:

A completes 2/5 work in 12 days.

B's efficiency = 1.25 × A's efficiency.

C's time = B's time − 12 days.

D's efficiency = 0.75 × C's efficiency.

Formula Used:

Total Work = Time Taken × Efficiency

Efficiency = 1 ÷ Time Taken (when work is constant)

⇒ Time taken by A for full work = 12 × (5 ÷ 2) = 30 days.

⇒ Since B is 25% more efficient, B's time = 30 ÷ 1.25 = 24 days.

⇒ C's time = 24 − 12 = 12 days.

⇒ Since D is 25% less efficient than C, D's time = 12 ÷ 0.75 = 16 days.

⇒ Let Total Work be LCM(30, 24, 12, 16) = 240 units.

⇒ Total Efficiency (A + B + C + D) = 8 + 10 + 20 + 15 = 53 units/day.

⇒ Time taken together = Total Work ÷ Total Efficiency = 240 / 53 days.

∴ The correct answer is 240/53 days.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a work. If the ratio of efficiency is x:y, the ratio of time taken is y:x.

Work and Wages

Wages are generally distributed in the ratio of the work done by each individual. If they work for the same duration, wages are divided in the ratio of their efficiencies.

Combined Work Formula

If A can do a work in 'a' days and B in 'b' days, together they complete it in (a × b) ÷ (a + b) days.

Given: Value = $\sqrt[3]{12}$ Formula used: If a > b, then an > bn Calculation: Target value = 121/3 Comparing with Option (2) = 332141/12 Raise both to the power of 12: ⇒ (121/3)12 = 124 ⇒ 124 = 12 × 12 × 12 × 12 = 20736 ⇒ (332141/12)12 = 33214 ⇒ 33214 > 20736 Checking other options: ⇒ Option (1): $\sqrt[6]{121}$ = (112)1/6 = 111/3 < 121/3 ⇒ Option (4): $\sqrt[9]{1500}$ < $\sqrt[9]{1728}$ = (123)1/9 = 121/3 ⇒ Option (3): 121/3 = 2488321/15 and 601/5 = 2160001/15 ∴ The correct answer is option (2).

Given:

A = $\frac{\sqrt{0.0004} \times \sqrt[3]{0.000008}}{\sqrt[4]{16000} \times \sqrt[3]{125000} \times \sqrt[4]{810}}$

B = $\frac{\sqrt[3]{0.729} \times \sqrt[4]{0.0016}}{\sqrt{0.16}}$

Formula used:

$\sqrt[n]{x} \times \sqrt[n]{y} = \sqrt[n]{xy}$

Calculation:

A = $\frac{0.02 \times 0.02}{\sqrt[4]{16000 \times 810} \times 50}$

⇒ A = $\frac{0.0004}{\sqrt[4]{12960000} \times 50}$

⇒ A = $\frac{0.0004}{60 \times 50}$

⇒ A = $\frac{4 \times 10^{-4}}{3000}$ = $\frac{4}{3}$ × 10^-7

B = $\frac{0.9 \times 0.2}{0.4}$

⇒ B = $\frac{0.18}{0.4}$ = 0.45

A × B = $\frac{4}{3}$ × 10^-7 × 0.45

⇒ A × B = $\frac{1.8}{3}$ × 10^-7

⇒ A × B = 0.6 × 10^-7

⇒ A × B = 6 × 10^-8

∴ The correct answer is option (3).

Shortcut Trick

Assume θ = 45°.

Expression = 1/tan 45° + tan 45° = 1/1 + 1 = 2.

Check Option (1): sec 45° × cosec 45° = √2 × √2 = 2.

Matches the value of the expression.

∴ The correct answer is sec θ.cosec θ.

Alternate Method

Given: (1 / tan θ) + tan θ

Formula Used: tan θ = sin θ / cos θ, cot θ = cos θ / sin θ, and sin² θ + cos² θ = 1.

⇒ 1/tan θ + tan θ = cot θ + tan θ

⇒ (cos θ / sin θ) + (sin θ / cos θ)

⇒ Take LCM of denominators: sin θ × cos θ

⇒ (cos² θ + sin² θ) / (sin θ × cos θ)

⇒ 1 / (sin θ × cos θ) [Since sin² θ + cos² θ = 1]

⇒ (1 / sin θ) × (1 / cos θ)

⇒ cosec θ × sec θ

∴ The correct answer is sec θ.cosec θ.

Additional Information

Fundamental Trigonometric Identities

There are three primary identities: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Reciprocal Relations

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

Complementary Angles

Trigonometric ratios of (90° − θ): sin(90°−θ) = cosθ, tan(90°−θ) = cotθ, and sec(90°−θ) = cosecθ.

Shortcut Trick

Profit Ratio = (Investment × Time)

A : B : C = (100,000 × 12) : (140,000 × 10) : (200,000 × 3)

A : B : C = (10 × 12) : (14 × 10) : (20 × 3) = 120 : 140 : 60 = 6 : 7 : 3

Given: C's share = 3 units = ₹1155 ⇒ 1 unit = ₹385

A + B total share = (6 + 7) × 385 = 13 × 385 = ₹5005

∴ The correct answer is ₹5005.

Alternate Method

Given:

Investment of A = ₹100,000; Duration = 12 months

Investment of B = ₹140,000; Duration = 10 months (3 + 7)

Investment of C = ₹200,000; Duration = 3 months

Profit share of C = ₹1155

Formula: Profit Ratio = (I₁ × T₁) : (I₂ × T₂) : (I₃ × T₃)

Calculations:

⇒ Ratio of Profit of A, B and C = (100000 × 12) : (140000 × 10) : (200000 × 3)

⇒ Ratio = 1,200,000 : 1,400,000 : 600,000

⇒ Ratio = 12 : 14 : 6 = 6 : 7 : 3

⇒ Let common factor be x.

⇒ Profit share of C = 3x = 1155

⇒ x = 1155 ÷ 3 = 385

⇒ Total share of A and B = 6x + 7x = 13x

⇒ Total share = 13 × 385 = ₹5005

∴ The correct answer is ₹5005.

Additional Information

Partnership Principle

Profit is distributed in the ratio of (Capital × Time). If capitals are C₁, C₂ for times T₁, T₂, then Profit Ratio P₁:P₂ = C₁T₁ : C₂T₂.

Working Partner

A partner who manages the business is a working partner. They often receive a fixed percentage of profit or salary before the remaining profit is shared.

Time Calculation

Always ensure all time units are consistent (either all in months or all in years) before calculating the ratio.

Shortcut Trick

Effective interest (%) for P₁ = 10% × 3 = 30%

Effective interest (%) for P₂ = 15% × 3 = 45%

Total SI = (30% of 20000) + (45% of 40000)

Total SI = 6000 + 18000 = 24000

∴ The correct answer is ₹24000.

Alternate Method

Given:

Principal 1 (P₁) = ₹20000, Rate 1 (R₁) = 10% p.a., Time (T) = 3 years

Principal 2 (P₂) = ₹40000, Rate 2 (R₂) = 15% p.a., Time (T) = 3 years

Formula Used:

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

Calculations:

⇒ SI₁ = (20000 × 10 × 3) ÷ 100

⇒ SI₁ = 200 × 10 × 3 = ₹6000

⇒ SI₂ = (40000 × 15 × 3) ÷ 100

⇒ SI₂ = 400 × 15 × 3 = ₹18000

⇒ Total SI = SI₁ + SI₂

⇒ Total SI = 6000 + 18000

⇒ Total SI = ₹24000

∴ The correct answer is ₹24000.

Additional Information

Total Amount Formula

The total amount (A) in Simple Interest is given by A = P + SI = P(1 + RT/100).

Difference between SI and CI

The difference between Compound Interest and Simple Interest for 2 years is P(R/100)².

Effective Rate of Interest

Effective annual rate for SI remains constant each year, whereas for CI it increases due to compounding.

Shortcut Trick

Substitute the identity 1 = sec² θ − tan² θ only in the numerator.

Expression = [(tan θ + sec θ) − (sec² θ − tan² θ)] ÷ (tan θ − sec θ + 1).

Factor out (sec θ + tan θ) to get: (sec θ + tan θ) [1 − (sec θ − tan θ)] ÷ (tan θ − sec θ + 1).

The term [1 − sec θ + tan θ] cancels out perfectly with the denominator.

Remaining result = sec θ + tan θ = 1/cos θ + sin θ/cos θ.

∴ The correct answer is (1 + sin θ) / cos θ.

Alternate Method

Given: Expression = (tan θ + sec θ − 1) ÷ (tan θ − sec θ + 1)

Formula Used: sec² θ − tan² θ = 1

⇒ Expression = [ (tan θ + sec θ) − (sec² θ − tan² θ) ] ÷ (tan θ − sec θ + 1)

⇒ Using a² − b² = (a − b)(a + b) in the numerator:

⇒ Numerator = (tan θ + sec θ) − [(sec θ − tan θ)(sec θ + tan θ)]

⇒ Factorize (tan θ + sec θ): (tan θ + sec θ) [1 − (sec θ − tan θ)]

⇒ Numerator = (tan θ + sec θ) (1 − sec θ + tan θ)

⇒ Substituting back: [(tan θ + sec θ)(1 − sec θ + tan θ)] ÷ (tan θ − sec θ + 1)

⇒ Since (1 − sec θ + tan θ) = (tan θ − sec θ + 1), the terms cancel out.

⇒ Result = sec θ + tan θ

⇒ Convert to sine and cosine: 1/cos θ + sin θ/cos θ = (1 + sin θ) / cos θ

∴ The correct answer is (1 + sin θ) / cos θ.

Additional Information

Trigonometric Identities

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

Specific Result Pattern

For expressions of the form (tan θ + sec θ − 1)/(tan θ − sec θ + 1), the result is always (sec θ + tan θ). Similarly, for cot and cosec, (cot θ + cosec θ − 1)/(cot θ − cosec θ + 1) = cosec θ + cot θ.

Quotient Relations

tan θ = sin θ ÷ cos θ and sec θ = 1 ÷ cos θ are used to convert expressions into primary sine/cosine terms.

Shortcut Trick

The sum of a series where each term is the product of two consecutive integers n(n + 1) up to k terms is given by the formula: S_k = [k × (k + 1) × (k + 2)] ÷ 3.

Here, the number of terms k = 20.

Sum = [20 × (20 + 1) × (20 + 2)] ÷ 3 = [20 × 21 × 22] ÷ 3.

Sum = 20 × 7 × 22 = 140 × 22 = 3080.

∴ The correct answer is 3080.

Alternate Method

Given:

Series: (1 × 2) + (2 × 3) + (3 × 4) + ... up to 20 terms.

The n^th term (a_n) of the series is n(n + 1) = n² + n.

Formula Used:

Sum of n terms (S_n) = ∑ a_n = ∑ n² + ∑ n

Where, ∑ n² = [n(n + 1)(2n + 1)] ÷ 6 and ∑ n = [n(n + 1)] ÷ 2

Calculations:

⇒ S₂₀ = ∑_n=1²⁰ n² + ∑_n=1²⁰ n

⇒ S₂₀ = [20 × (20 + 1) × (2 × 20 + 1)] ÷ 6 + [20 × (20 + 1)] ÷ 2

⇒ S₂₀ = [20 × 21 × 41] ÷ 6 + [20 × 21] ÷ 2

⇒ S₂₀ = [10 × 7 × 41] + [10 × 21]

⇒ S₂₀ = 2870 + 210

⇒ S₂₀ = 3080

∴ The correct answer is 3080.

Additional Information

Sum of first n natural numbers

The sum of the first n natural numbers is given by ∑ n = n(n + 1) ÷ 2.

Sum of squares of first n natural numbers

The sum of squares of the first n natural numbers is given by ∑ n² = n(n + 1)(2n + 1) ÷ 6.

General product series formula

For a series of product of r consecutive integers, ∑ n(n+1)...(n+r-1) = [n(n+1)...(n+r)] ÷ (r+1).

Sum of cubes of first n natural numbers

The sum of cubes of the first n natural numbers is given by ∑ n³ = [n(n + 1) ÷ 2]².

Shortcut Trick

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

MP ÷ CP = (100 + 14) ÷ (100 − 5) = 114 ÷ 95 = 6 ÷ 5

Let CP = 500 and MP = 600. New Discount = 11% of 600 = 66.

New SP = 600 − 66 = 534. Profit = 534 − 500 = 34.

Profit% = (34 ÷ 500) × 100 = 6.8%.

∴ The correct answer is 6.8 percent.

Alternate Method

Given:

Initial Discount (D₁) = 5%

Initial Profit (P₁) = 14%

New Discount (D₂) = 11%

Formula Used:

MP ÷ CP = (100 + P%) ÷ (100 − D%)

SP = MP × (100 − D%) ÷ 100

Calculations:

⇒ MP ÷ CP = (100 + 14) ÷ (100 − 5)

⇒ MP ÷ CP = 114 ÷ 95 = 1.2

⇒ Let CP = 100, then MP = 1.2 × 100 = 120

⇒ New SP = 120 × (100 − 11) ÷ 100

⇒ New SP = 120 × 89 ÷ 100 = 1.2 × 89 = 106.8

⇒ New Profit = SP − CP = 106.8 − 100 = 6.8

⇒ Profit% = (6.8 ÷ 100) × 100 = 6.8%

∴ The correct answer is 6.8 percent.

Additional Information

Marked Price and Cost Price Relation

The ratio between Marked Price (MP) and Cost Price (CP) is given by MP ÷ CP = (100 + Profit%) ÷ (100 − Discount%). If there is a loss, Profit% is replaced by −Loss%.

Successive Discounts

When two successive discounts d₁% and d₂% are given, the equivalent single discount is (d₁ + d₂ − d₁d₂ ÷ 100)%.

Net Profit with Discount

Net Profit % = [Markup% − Discount% − (Markup% × Discount%) ÷ 100], where Markup is calculated over CP.

Shortcut Trick

In a circle, the ratio of Circumference to Radius is 2π : 1, which is approximately 44/7 : 1.

The difference (Circumference − Radius) = 44/7 − 1 = 37/7 units.

According to the question: 37/7 units = 37 cm ⇒ 1 unit = 7 cm.

So, the radius (r) = 7 cm.

Volume = πr²h = 22/7 × 7 × 7 × 35 = 154 × 35 = 5390 cm³.

∴ The correct answer is 5390 cm³.

Alternate Method

Given:

Height of cylinder (h) = 35 cm

Circumference of base − Radius = 37 cm

Formula used:

Circumference of base = 2πr

Volume of cylinder = πr²h

Step 1: Find the radius (r) using the base property.

⇒ 2πr − r = 37

⇒ r(2π − 1) = 37

⇒ r(2 × 22/7 − 1) = 37

⇒ r(44/7 − 1) = 37

⇒ r(37/7) = 37

⇒ r = 37 × 7/37

⇒ r = 7 cm

Step 2: Calculate the volume of the cylinder.

⇒ Volume = π × r² × h

⇒ Volume = 22/7 × 7 × 7 × 35

⇒ Volume = 22 × 7 × 35

⇒ Volume = 154 × 35

⇒ Volume = 5390 cm³

∴ The correct answer is 5390 cm³.

Additional Information

Surface Area of Cylinder

The Curved Surface Area (CSA) is 2πrh and the Total Surface Area (TSA) is 2πr(h + r).

Volume of a Cone

The volume of a cone with the same base radius and height is exactly one-third of the cylinder: V = 1/3 × πr²h.

Circle Constants

For a circle with radius 7, the circumference is 44 and the area is 154. These are useful benchmarks for fast calculations.

Given:

Concept used:

Average = Sum of all the value/Total values

Calculations:

The average amount invested in purchasing raw materials during 2012 to 2017

⇒ A = (140 + 250 + 350 + 325 + 525 + 450)/6

⇒ A = 2040/6 = 340

The average value of sales of finished goods during the same period 2012 to 2017

⇒ A = (220 + 300 + 475 + 400 + 600 + 525)/6

⇒ A = 2520/6 = 420

The difference between the average = 420 - 340 = 80 lakhs

∴ The difference between the average of both purchasing and finished goods is 80 lakhs.

Shortcut Trick

Net effective compound interest rate for 3 years at 20% = 72.8%

Net effective compound interest rate for 3 years at 30% = 119.7%

CI of Vipul = 72.8% of 15000 = 10920

CI of Manish = 119.7% of 20000 = 23940

Total Compound Interest = 10920 + 23940 = 34860

∴ The correct answer is ₹34860.

Alternate Method

Given:

Vipul: Principal (P₁) = ₹15000, Rate (R₁) = 20%, Time (T) = 3 years

Manish: Principal (P₂) = ₹20000, Rate (R₂) = 30%, Time (T) = 3 years

Formula Used:

Compound Interest (CI) = P[(1 + R/100)^T − 1]

Calculations:

⇒ CI_Vipul = 15000 × [(1 + 20/100)³ − 1]

⇒ CI_Vipul = 15000 × [(6/5)³ − 1]

⇒ CI_Vipul = 15000 × [216/125 − 1] = 15000 × [91/125]

⇒ CI_Vipul = 120 × 91 = ₹10920

⇒ CI_Manish = 20000 × [(1 + 30/100)³ − 1]

⇒ CI_Manish = 20000 × [(1.3)³ − 1]

⇒ CI_Manish = 20000 × [2.197 − 1] = 20000 × 1.197

⇒ CI_Manish = ₹23940

⇒ Total CI = 10920 + 23940 = 34860

∴ The correct answer is ₹34860.

Additional Information

Successive Percentage Method

For compound interest, net rate for 2 years = x + y + (xy/100). For 3 years, apply this twice.

CI vs SI for 3 Years

Difference (D) = P(R/100)² × (3 + R/100).

Effective Rates for 3 Years

10% p.a. = 33.1% | 20% p.a. = 72.8% | 5% p.a. = 15.7625%.

Shortcut Trick

Assume Total Income = 100 units.

Spent on Transport + Food = 10 + 20 = 30 units → Remaining = 70 units.

Spent on Clothing = 30% of 70 = 21 units.

Total Savings = 70 − 21 = 49 units.

Given: 49 units = Rs. 26460 ⇒ 1 unit = Rs. 540.

Expenditure on Food + Clothing = 20 + 21 = 41 units = 41 × 540 = Rs. 22140.

∴ The correct answer is Rs. 22140.

Alternate Method

Given: Transport = 10%, Food = 20%, Clothing = 30% of remaining, Savings = Rs. 26460.

Let the total income be Rs. x.

Income spent on Transport and Food = 10% + 20% = 30% of x.

⇒ Remaining Income = x − 30% of x = 70% of x.

Expenditure on Clothing = 30% of (70% of x)

⇒ Clothing Expenditure = 0.30 × 0.70x = 0.21x (or 21% of x).

Total Savings = Remaining − Clothing

⇒ Savings = 70% of x − 21% of x = 49% of x.

⇒ 49% of x = 26460

⇒ 0.49x = 26460

⇒ x = 26460 ÷ 0.49 = 54000.

Now, calculate expenditure on Food and Clothing together:

⇒ Food Expenditure = 20% of 54000 = Rs. 10800.

⇒ Clothing Expenditure = 21% of 54000 = Rs. 11340.

⇒ Total = 10800 + 11340 = Rs. 22140.

∴ The correct answer is Rs. 22140.

Additional Information

Percentage of Remaining Value

When a percentage is spent from the "remaining" amount, it is calculated as: (Percentage ÷ 100) × Remaining Value. Total % = Direct % + [% of (100 − Direct %)].

Successive Percentage Change

If values change by a% and then b% of the remainder, the final value = Initial × (100−a)/100 × (100−b)/100.

Income-Expenditure Rule

Income = Total Expenditure + Savings. Always ensure the base for percentages is clearly identified as "Total Income" or "Remaining Income".

Shortcut Trick

Total Work = LCM of (14, 28, 56) = 56 units.

Efficiencies: A = 4, B = 2, C = 1; Total efficiency = 7 units/day.

Adjustment: Add work of those who left early → A (3 days) and B (2 days).

New Total Work = 56 + (4 × 3) + (2 × 2) = 56 + 12 + 4 = 72 units.

Total time = Adjusted Work ÷ Total Efficiency = 72/7 days.

∴ The correct answer is 72/7 days.

Alternate Method

Given:

Time taken by A = 14 days, B = 28 days, C = 56 days.

Condition: B left 2 days before and A left 3 days before completion.