Quantitative Aptitude PYQs

Shortcut Trick

When a pyramid is inscribed in a cube sharing the same base and maximum height, its volume is exactly one-third of the cube's volume.

Volume of the inscribed pyramid = (1/3) × (Volume of cube)

⇒ V = (1/3) × a³ = (1/3) × 12³

⇒ V = (1/3) × 1728 = 576 cm³

∴ The correct answer is 576 cm³.

Alternate Method

Given:

Edge length of the cube (a) = 12 cm

The pyramid and the cube share the same base, meaning the base of the pyramid is a square of side 12 cm.

The apex of the pyramid is at the center of the top face of the cube, so the height of the pyramid (h) equals the edge length of the cube = 12 cm.

                       12 cm (Base Edge) h = 12 cm Apex

Formula Used:

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

Calculations:

⇒ Area of base = Side × Side

⇒ Area of base = 12 × 12 = 144 cm²

⇒ Volume of pyramid = (1/3) × 144 × 12

⇒ Volume of pyramid = 144 × 4

⇒ Volume of pyramid = 576 cm³

(Note: The unit specified in the options contains a typographical error by stating cm². Volume is always measured in cubic units, cm³.)

∴ The correct answer is 576 cm³.

Additional Information

Volume of a Prism vs Pyramid

A pyramid's volume is exactly one-third the volume of a prism (like a cube or cuboid) that possesses the exact same base area and height.

Slant Height of Pyramid

The slant height (l) of a square pyramid can be deduced using the right-angled triangle formed by the height, half the base edge, and the slant height itself: l = √(h² + (a/2)²).

Surface Area of Pyramid

Total Surface Area = Base Area + Lateral Surface Area. For a square-based pyramid, the Lateral Surface Area (LSA) equals 2 × a × l.

Shortcut Trick

In a partnership, Profit Ratio = Investment Ratio × Time Ratio.

Directly multiply the respective values for A, B, and C to find their profit shares.

Profit Ratio = (4 × 3) : (5 × 2) : (6 × 1)

Profit Ratio = 12 : 10 : 6 = 6 : 5 : 3

∴ The correct answer is 6:5:3.

Alternate Method

Given:

Investment ratio of A, B, and C = 4 : 5 : 6

Time ratio of A, B, and C = 3 : 2 : 1

Formula Used:

Profit = Investment × Time

Partner A Invest = 4 Time = 3 Profit = 12 Partner B Invest = 5 Time = 2 Profit = 10 Partner C Invest = 6 Time = 1 Profit = 6 Profit Ratio = 12 : 10 : 6 → 6 : 5 : 3

Calculation:

⇒ Let the investments of A, B, and C be 4x, 5x, and 6x respectively.

⇒ Let the time periods for A, B, and C be 3y, 2y, and 1y respectively.

⇒ Profit of A = 4x × 3y = 12xy

⇒ Profit of B = 5x × 2y = 10xy

⇒ Profit of C = 6x × 1y = 6xy

⇒ Ratio of their profits = 12xy : 10xy : 6xy

⇒ Cancelling xy from all terms, Ratio = 12 : 10 : 6

⇒ Simplifying the ratio by dividing by 2, Ratio = 6 : 5 : 3

∴ The correct answer is 6:5:3.

Additional Information

Simple vs Compound Partnership

In a simple partnership, capitals are invested for the same time, and profit is divided solely in the ratio of the capitals. In a compound partnership, capitals are invested for different time periods. Here, Profit Ratio = (C₁ × T₁) : (C₂ × T₂).

Working / Active Partner

An active partner manages the business and receives a fixed salary or percentage of the profit for their work. The remaining profit is then distributed among all partners according to their investment and time.

Joining Mid-way

If a partner joins the business 'n' months after it starts in a given year, their investment time period becomes (12 − n) months for the purpose of calculating their final profit share.

Shortcut Trick

For such questions, the fastest method is to verify the given options.

Condition 1: L × W = 600 | Condition 2: (L + 10) × (W − 2) = 600

Check Option 3: L = 50, W = 12 ⇒ Area = 50 × 12 = 600.

New length = 50 + 10 = 60, New width = 12 − 2 = 10 ⇒ Area = 60 × 10 = 600.

Both conditions are completely satisfied by Option 3.

∴ The correct answer is Length = 50 m, Width = 12 m.

Alternate Method

Given:

Let the original length of the rectangle be L meters and width be W meters.

Original Area = L × W = 600 ⇒ W = 600 ÷ L

New length = L + 10, New width = W − 2

Original Field L W Area = 600 New Field L + 10 W − 2 Area = 600 L × W = (L + 10) × (W − 2)

Formula Used:

Area of a rectangle = Length × Width

Calculations:

⇒ (L + 10) × (W − 2) = 600

⇒ LW − 2L + 10W − 20 = 600

Substitute LW = 600:

⇒ 600 − 2L + 10W − 20 = 600

⇒ −2L + 10W − 20 = 0

Dividing the equation by −2:

⇒ L − 5W + 10 = 0

Substitute W = 600 ÷ L into the equation:

⇒ L − 5 × (600 ÷ L) + 10 = 0

⇒ L − (3000 ÷ L) + 10 = 0

Multiplying the entire equation by L:

⇒ L² + 10L − 3000 = 0

Factorizing the quadratic equation:

⇒ L² + 60L − 50L − 3000 = 0

⇒ L(L + 60) − 50(L + 60) = 0

⇒ (L − 50)(L + 60) = 0

⇒ L = 50 or L = −60

Since the length cannot be negative, L = 50 m.

⇒ W = 600 ÷ 50 = 12 m.

∴ The correct answer is Length = 50 m, Width = 12 m.

Additional Information

Area and Perimeter of Rectangle

The Area is the product of its length and width (A = L × W). The Perimeter is the total length of its boundary (P = 2 × (L + W)).

Quadratic Equations

A quadratic equation is of the form ax² + bx + c = 0. The roots can be found using factorisation or the standard formula x = (−b ± √(b² − 4ac)) ÷ 2a.

Solving Two Variables

When dealing with two unknown variables, you need exactly two independent linear or non-linear equations to determine their unique values.

Shortcut Trick

Total area of regular hexagon = 6 × (√3 / 4) × (side)²

⇒ Area = 1.5 × 1.732 × 20² = 1.5 × 1.732 × 400 = 1039.2 m²

The circular fountain occupies 5% of the garden, leaving 95% empty space.

⇒ Remaining area = 0.95 × 1039.2 = 987.24 m²

∴ The correct answer is 987.24 m².

Alternate Method

Given:

Side of the regular hexagonal garden (a) = 20 m

Area occupied by the circular fountain = 5%

Standard value of √3 ≈ 1.732

Formula Used:

Area of a regular hexagon = (3√3 / 2) × a²

Remaining Area = Total Area − Occupied Area

5% 95% Remaining Garden 20 m

Calculations:

⇒ Total Area of Hexagon = (3√3 / 2) × 20²

⇒ Total Area = (3 × 1.732 / 2) × 400

⇒ Total Area = 1.732 × 600 = 1039.2 m²

⇒ The remaining area will be (100% − 5%) = 95% of the total garden space.

⇒ Remaining Area = (95 / 100) × 1039.2

⇒ Remaining Area = 0.95 × 1039.2 = 987.24 m²

∴ The correct answer is 987.24 m².

Additional Information

Properties of Regular Hexagon

A regular hexagon is naturally composed of 6 congruent equilateral triangles. Its total area can be computed as 6 × (√3 / 4)a², where 'a' is the uniform side length.

Inradius and Circumradius

For any regular hexagon with side 'a', the circumradius (R) exactly equals the side length (R = a), while its inscribed circle radius (r) is (√3 / 2)a.

Sum of Interior Angles

The sum of all interior angles for a regular hexagon is given by (n − 2) × 180° = (6 − 2) × 180° = 720°. This makes each individual interior angle exactly 120°.

Shortcut Trick

The original weight ratio is 7 : 8 : 9 and the new weight ratio is 12 : 13 : 14.

Notice that each individual part has increased uniformly by exactly 5 units (12 − 7 = 5, 13 − 8 = 5, 14 − 9 = 5).

This 5 unit increase directly corresponds to the actual weight gain of 5 kg.

∴ 5 units = 5 kg ⇒ 1 unit = 1 kg.

The original weight of the heaviest friend = 9 units = 9 × 1 = 9 kg.

∴ The correct answer is 9 kg.

Alternate Method

Given:

Initial weight ratio = 7 : 8 : 9

Weight gained by each friend = 5 kg

New weight ratio = 12 : 13 : 14

Friend 1 Friend 2 Heaviest 7 units 8 units 9 units +5 kg +5 kg +5 kg 12 units 13 units 14 units Gain: 12 − 7 = 5 units = 5 kg

Calculation:

⇒ Let the initial weights of the three friends be 7x, 8x, and 9x.

⇒ After gaining 5 kg each, their new weights become (7x + 5), (8x + 5), and (9x + 5).

⇒ According to the given condition, we can equate the ratio of the first two friends:

⇒ (7x + 5) / (8x + 5) = 12 / 13

⇒ 13 × (7x + 5) = 12 × (8x + 5)

⇒ 91x + 65 = 96x + 60

⇒ 96x − 91x = 65 − 60

⇒ 5x = 5

⇒ x = 1

⇒ The original weight of the heaviest friend corresponds to the largest ratio part = 9x.

⇒ Heaviest friend's weight = 9 × 1 = 9 kg.

∴ The correct answer is 9 kg.

Additional Information

Uniform Change in Ratio

When a constant real value is added or subtracted from all quantities, the difference between the initial and final parts of the ratio directly equals that constant value, provided the unit changes are identical across all terms.

Cross-Multiplication Method

Used for solving problems involving two changing quantities. If the initial ratio is a:b and both terms change by x and y respectively to become c:d, then (a ± x) / (b ± y) = c / d.

Compound Ratio

The compound ratio of multiple ratios (like a:b and c:d) is obtained by multiplying their respective antecedents (numerators) and consequents (denominators) together, giving the result as ac:bd.

Shortcut Trick

Sides = 16 m, 24 m, and 32 m (distributed from 2:3:4 ratio and 72 m total perimeter).

Semi-perimeter (s) = 72 ÷ 2 = 36 m.

Area = √(36 × (36−16) × (36−24) × (36−32)) = √(36 × 20 × 12 × 4)

Area = √34560 ≈ 185.9 m² ≈ 186 m².

∴ The correct answer is 186 m².

Alternate Method

Given:

Ratio of sides = 2 : 3 : 4

Perimeter = 72 m

24 m 16 m 32 m

Formula Used:

Heron's Formula: Area = √(s(s − a)(s − b)(s − c))

where s is the semi-perimeter: s = (a + b + c) ÷ 2

Calculations:

⇒ Let the sides of the triangle be 2x, 3x, and 4x.

⇒ Perimeter = 2x + 3x + 4x = 72

⇒ 9x = 72 ⇒ x = 8

⇒ The sides are: a = 2 × 8 = 16 m, b = 3 × 8 = 24 m, and c = 4 × 8 = 32 m.

⇒ Semi-perimeter (s) = 72 ÷ 2 = 36 m.

⇒ Area = √(36 × (36 − 16) × (36 − 24) × (36 − 32))

⇒ Area = √(36 × 20 × 12 × 4)

⇒ Area = √34560 ≈ 185.9 m²

⇒ Rounding to the nearest integer, Area ≈ 186 m².

∴ The correct answer is 186 m².

Additional Information

Right Angled Triangle

The area is calculated as ½ × base × perpendicular height.

Equilateral Triangle

All three sides are equal. Area = (√3 ÷ 4) × a², where 'a' is the side length.

Isosceles Triangle

Two sides are equal. Area = (b ÷ 4) × √(4a² − b²), where 'a' is the equal side and 'b' is the base.

Shortcut Trick

Final Value = Initial Value × (100 − D₁)% × (100 − D₂)% × (100 − D₃)%

Final Value = 10000 × 90% × 80% × 70%

Final Value = 10000 × (9/10) × (8/10) × (7/10)

Final Value = 10 × 504 = 5040

∴ The correct answer is Rs. 5040.

Alternate Method

Given:

Initial Value of the machine = Rs. 10000

Rate of depreciation for 1^st year (R₁) = 10%

Rate of depreciation for 2^nd year (R₂) = 20%

Rate of depreciation for 3^rd year (R₃) = 30%

Formula Used:

Final Value = Initial Value × (1 − R₁/100) × (1 − R₂/100) × (1 − R₃/100)

         10000 9000 7200 5040 −10% −20% −30% Initial Year 1 Year 2 Year 3

Calculations:

⇒ Value after 1^st year = 10000 − (10% of 10000) = 10000 − 1000 = 9000

⇒ Value after 2^nd year = 9000 − (20% of 9000) = 9000 − 1800 = 7200

⇒ Value after 3^rd year = 7200 − (30% of 7200) = 7200 − 2160 = 5040

∴ The correct answer is Rs. 5040.

Additional Information

Successive Percentage Change

When a value undergoes successive percentage changes of a% and b%, the net percentage change is given by (a + b + a×b/100)%. For depreciation, the values of a and b are taken as negative.

Constant Depreciation Formula

If the rate of depreciation is constant at R% per annum, the value after n years is calculated as Value = P × (1 − R/100)ⁿ.

Compound Growth vs Depreciation

The same successive multiplication principle applies to population growth or compound interest, where rates are added instead of subtracted: Final Value = P × (1 + R/100)ⁿ.

Shortcut Trick

Let the Cost Price (CP) be 100%.

Marked Price (MP) = 100% + 60% = 160%.

Selling Price (SP) after 25% discount = 160% × 0.75 = 120%.

Given the overall profit is 10%, Final SP = 100% + 10% = 110%.

Difference (Cashback) = 120% − 110% = 10% = Rs. 300.

Since 10% = Rs. 300 ⇒ Cost Price (100%) = Rs. 3000.

∴ The correct answer is Rs. 3000.

Alternate Method

Given:

Markup = 60% above CP

Discount = 25%

Cashback = Rs. 300

Overall Profit = 10%

Formula Used:

MP = CP × (1 + Markup% ÷ 100)

SP = MP × (1 − Discount% ÷ 100) − Cashback

      +60% Markup  −25% Discount  −Rs. 300 Cashback  +10% Overall Profit  CP 100%  MP 160%  SP 120%  Final SP 110%

Calculations:

⇒ Let the Cost Price (CP) be x.

⇒ Marked Price (MP) = x × (1 + 60 ÷ 100) = 1.6x

⇒ SP before cashback = 1.6x × (1 − 25 ÷ 100) = 1.6x × 0.75 = 1.2x

⇒ Final SP = 1.2x − 300

⇒ We also know that Final SP = CP × (1 + 10 ÷ 100) = 1.1x

⇒ Equating the two Final SP expressions:

⇒ 1.2x − 300 = 1.1x

⇒ 1.2x − 1.1x = 300

⇒ 0.1x = 300

⇒ x = 300 ÷ 0.1 = 3000

∴ The correct answer is Rs. 3000.

Additional Information

Direct Formula for MP and CP Ratio

When profit and discount percentages are given (without any fixed monetary adjustments), MP ÷ CP = (100 + Profit%) ÷ (100 − Discount%).

Successive Discount

When two consecutive discounts of x% and y% are given on an item, the equivalent single net discount percentage is (x + y − x × y ÷ 100)%.

Dishonest Dealer Concept

If a shopkeeper claims to sell goods at Cost Price but uses a false weight, his profit percentage is given by (True Weight − False Weight) ÷ False Weight × 100%.

Shortcut Trick

Total Work = LCM(10, 50) = 50 units.

Efficiency of (Athar + Baban) = 5 units/day, Chetan = 1 unit/day.

Since Athar = Baban + Chetan, replacing Athar gives: 2 × Baban + Chetan = 5.

2 × Baban + 1 = 5 ⇒ Baban's efficiency = 2 units/day.

Time taken by Baban alone = 50 ÷ 2 = 25 days.

∴ The correct answer is 25 days.

Alternate Method

Given:

⇒ Athar takes the same time as Baban and Chetan together.

⇒ Efficiency of Athar (A) = Efficiency of Baban (B) + Efficiency of Chetan (C)

Calculations:

⇒ A = B + C

⇒ Combined efficiency: A + B = 5

⇒ Substituting A = B + C in the combined efficiency equation:

⇒ (B + C) + B = 5

⇒ 2B + C = 5

⇒ Since C's efficiency is 1:

⇒ 2B + 1 = 5 ⇒ 2B = 4

⇒ B = 2 units/day

⇒ Time taken by Baban alone = Total Work ÷ B's Efficiency = 50 ÷ 2 = 25 days

∴ The correct answer is 25 days.

Additional Information

Total Work using LCM

Total work is generally assumed as the LCM of all the given time periods. This makes the one-day work (efficiency) an integer value, making calculation easier.

Work & Wages

Wages are distributed in the direct proportion of the amount of work done by each individual. If they work for the same duration, wages are split in the ratio of their efficiencies.

Pipes & Cisterns

The core concept remains exactly the same as Time & Work, with the only difference being that a leak or emptying pipe has negative efficiency.

Shortcut Trick

Volume of the rectangular prism = Length × Breadth × Height

⇒ Volume = 50 × 40 × 30 = 60000 cm³

Since 1000 cm³ = 1 liter, the capacity is calculated by dividing the volume by 1000.

⇒ Capacity = 60000 ÷ 1000 = 60 liters

∴ The correct answer is 60 liters.

Alternate Method

Given:

Length (l) = 50 cm

Breadth (b) = 40 cm

Height (h) = 30 cm

Formula Used:

Volume of a rectangular prism = l × b × h

1 liter = 1000 cm³

   50 cm 30 cm 40 cm Storage Box (Internal Dimensions)

Calculations:

⇒ Volume = 50 × 40 × 30

⇒ Volume = 60000 cm³

⇒ Capacity in liters = 60000 ÷ 1000

⇒ Capacity = 60 liters

∴ The correct answer is 60 liters.

Additional Information

Total Surface Area of Cuboid

The total surface area is the sum of the areas of its six faces, given by the formula: TSA = 2(lb + bh + hl).

Diagonal of a Cuboid

The longest rod that can be placed inside a cuboidal box is its diagonal. Length of diagonal = √(l² + b² + h²).

Important Capacity Conversions

1 m³ = 1000 liters and 1 liter = 1000 cm³.

Shortcut Trick

Using the deviation method, assume an arbitrary average of Rs. 45.

Sum of deviations = (−5 × 2) + (−3 × 3) + (0 × 4) + (+3 × 5) + (+5 × 6) = −10 − 9 + 0 + 15 + 30 = +26.

Total weight ratio parts = 2 + 3 + 4 + 5 + 6 = 20.

Average deviation = +26 ÷ 20 = +1.3.

Actual average = 45 + 1.3 = 46.3.

∴ The correct answer is Rs. 46.3.

Alternate Method

Given:

Ratio of weights = 2 : 3 : 4 : 5 : 6

Prices per kg = Rs. 40, Rs. 42, Rs. 45, Rs. 48, Rs. 50

Weighted Average of Mixture 2k 3k 4k 5k 6k Rs.40 Rs.42 Rs.45 Rs.48 Rs.50 Total Weight = 20k Average = Rs. 46.3 / kg

Formula Used:

Average Cost = Total Cost ÷ Total Quantity

Calculations:

⇒ Let the weights be 2 kg, 3 kg, 4 kg, 5 kg, and 6 kg respectively.

⇒ Total Weight = 2 + 3 + 4 + 5 + 6 = 20 kg

⇒ Total Cost = (2 × 40) + (3 × 42) + (4 × 45) + (5 × 48) + (6 × 50)

⇒ Total Cost = 80 + 126 + 180 + 240 + 300

⇒ Total Cost = Rs. 926

⇒ Average Cost = 926 ÷ 20

⇒ Average Cost = 46.3

∴ The correct answer is Rs. 46.3.

Additional Information

Weighted Average

When quantities (w₁, w₂, ..., w_n) have different values (x₁, x₂, ..., x_n), the weighted average = (w₁x₁ + w₂x₂ + ... + w_nx_n) ÷ (w₁ + w₂ + ... + w_n).

Deviation Method for Averages

You can assume an arbitrary average (A) closer to the middle values to simplify calculations. Actual Average = A + (Sum of Deviations ÷ Total Elements).

Multiple Mixtures

When mixing more than two ingredients, the alligation cross-method becomes complicated. The weighted average formula is the most efficient and standard approach for such questions.

Shortcut Trick

Using the Alligation rule directly on the fraction of Tin in the alloys:

Fraction of Tin: Alloy 1 = 5/8, Alloy 2 = 3/4, Mixture = 4/6 = 2/3

Ratio of quantities = (3/4 − 2/3) : (2/3 − 5/8) = 1/12 : 1/24 = 2 : 1

Given 1 part (Alloy 2) corresponds to 20 kg.

So, 2 parts (Alloy 1) = 2 × 20 = 40 kg.

∴ The correct answer is 40 kg.

Alternate Method

Given:

Ratio of Tin to Copper in Alloy 1 = 5 : 3

Ratio of Tin to Copper in Alloy 2 = 3 : 1

Target Ratio of Tin to Copper in New Alloy = 4 : 2 = 2 : 1

Quantity of Alloy 2 = 20 kg

Formula Used:

Alligation method applied to fractions of a single component (Tin) in the mixtures.

  Alloy 1 (5/8)  Alloy 2 (3/4)  Mix (2/3)  1/12  1/24 Ratio = 1/12 : 1/24 = 2 : 1

Calculations:

⇒ Fraction of Tin in Alloy 1 = 5 ÷ (5 + 3) = 5/8

⇒ Fraction of Tin in Alloy 2 = 3 ÷ (3 + 1) = 3/4

⇒ Fraction of Tin in New Alloy = 4 ÷ (4 + 2) = 4/6 = 2/3

Applying the rule of alligation to find the required ratio of quantities:

⇒ Ratio of Alloy 1 to Alloy 2 = (3/4 − 2/3) : (2/3 − 5/8)

⇒ Ratio = (9/12 − 8/12) : (16/24 − 15/24)

⇒ Ratio = 1/12 : 1/24

⇒ Required Ratio = 2 : 1

We are given that the quantity of Alloy 2 used is 20 kg (which represents 1 unit in the ratio):

⇒ Quantity of Alloy 1 = 2 × 20 kg

⇒ Quantity of Alloy 1 = 40 kg

∴ The correct answer is 40 kg.

Additional Information

Alligation Method

It is a rule used to quickly find the ratio in which two or more ingredients must be mixed to produce a specific desired average or concentration. It can be applied on percentages, prices, and component fractions.

Replacement Method

If a container holds x units of pure liquid and y units are taken out and replaced by water n times, the remaining quantity of pure liquid is given by x × (1 − y/x)ⁿ.

Multiple Mixtures

When combining multiple mixtures containing differing ratios of the same components, always evaluate the quantities by converting the component ratios into absolute fractional parts out of the whole before combining them.

Shortcut Trick

For any angle A, the expression (a×sinA − b×cosA)² + (b×sinA + a×cosA)² simplifies to a² + b².

In the given expression, a = 3 and b = 4.

Required value = 3² + 4² = 9 + 16 = 25.

The given value sinA = 2/3 is extra information and is not needed.

∴ The correct answer is 25.

Alternate Method

Given:

sin A = 2/3 (Extraneous data)

Formula Used:

(x − y)² = x² − 2xy + y²

(x + y)² = x² + 2xy + y²

sin²A + cos²A = 1

Calculation:

⇒ Let E = (3sinA − 4cosA)² + (4sinA + 3cosA)²

⇒ Expand both terms using the algebraic formulas.

⇒ (3sinA − 4cosA)² = 9sin²A − 24sinA×cosA + 16cos²A

⇒ (4sinA + 3cosA)² = 16sin²A + 24sinA×cosA + 9cos²A

⇒ E = (9sin²A − 24sinA×cosA + 16cos²A) + (16sin²A + 24sinA×cosA + 9cos²A)

⇒ The middle terms (−24sinA×cosA and +24sinA×cosA) cancel each other out.

⇒ E = 9sin²A + 16cos²A + 16sin²A + 9cos²A

⇒ E = 25sin²A + 25cos²A

⇒ E = 25 × (sin²A + cos²A)

⇒ Since sin²A + cos²A = 1, we get:

⇒ E = 25 × 1 = 25

∴ The correct answer is 25.

Additional Information

Pythagorean Trigonometric Identities

The fundamental identity is sin²A + cos²A = 1. Dividing this by cos²A gives 1 + tan²A = sec²A, and dividing by sin²A gives 1 + cot²A = cosec²A.

Standard Algebraic Identity in Trigonometry

The expression (a×sinA − b×cosA)² + (b×sinA + a×cosA)² always simplifies to a² + b², completely independent of the angle A.

Extraneous Information in Exams

Exam questions often provide unnecessary data (like sin A = 2/3 in this problem) to tempt candidates into lengthy calculations. Always simplify the main expression algebraically first before substituting given values.

Shortcut Trick

If sin A + cos A = √2, then A = 45° is the only valid angle in the primary interval.

Substitute A = 45° into sin³ A + cos³ A.

sin³ 45° + cos³ 45° = (1/√2)³ + (1/√2)³

⇒ 1/(2√2) + 1/(2√2) = 2/(2√2)

⇒ 1/√2 = √2 / 2

∴ The correct answer is √2 / 2.

Alternate Method

Given:

sin A + cos A = √2

Formula Used:

(a + b)² = a² + b² + 2ab

a³ + b³ = (a + b)(a² + b² − ab)

sin² A + cos² A = 1

Calculations:

⇒ Squaring both sides of the given equation:

⇒ (sin A + cos A)² = (√2)²

⇒ sin² A + cos² A + 2 sin A cos A = 2

⇒ 1 + 2 sin A cos A = 2

⇒ 2 sin A cos A = 1

⇒ sin A cos A = 1/2

⇒ Now, we need to find the value of sin³ A + cos³ A

⇒ Using the algebraic identity for the sum of cubes:

⇒ sin³ A + cos³ A = (sin A + cos A)(sin² A + cos² A − sin A cos A)

⇒ Substitute the known values into the identity:

⇒ sin³ A + cos³ A = (√2) × (1 − 1/2)

⇒ sin³ A + cos³ A = √2 × (1/2)

⇒ sin³ A + cos³ A = √2 / 2

∴ The correct answer is √2 / 2.

Additional Information

Pythagorean Trigonometric Identities

The primary foundational identity is sin² A + cos² A = 1. Other related identities include 1 + tan² A = sec² A and 1 + cot² A = cosec² A.

Algebraic Identities for Cubes

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

Maximum Value of Sine and Cosine Expressions

For any general trigonometric expression of the form a sin x + b cos x, the maximum value it can attain is √(a² + b²). In this specific case, 1 sin A + 1 cos A has a theoretical maximum of √(1² + 1²) = √2, which perfectly matches the given equation and confirms A = 45°.

Shortcut Trick

Compare the given line equations with the standard slope-intercept form y = mx + c.

Slope of the first line (m₁) = 2

Slope of the second line (m₂) = −1/2

Product of slopes = 2 × (−1/2) = −1

Two lines are perpendicular if the product of their slopes is −1, which is exactly the case here.

∴ The correct answer is Yes.

Alternate Method

Given:

Equation of first line: y = 2x + 3

Equation of second line: y = −1/2x − 1

Formula Used:

Condition for two intersecting lines to be perpendicular: m₁ × m₂ = −1 (where m represents the slope).

O

y = 2x + 3 y = -1/2x - 1

(−1.6, −0.2)

90°

Calculations:

⇒ The standard slope-intercept form of a linear equation is y = mx + c.

⇒ From the first equation y = 2x + 3, the slope m₁ = 2.

⇒ From the second equation y = −1/2x − 1, the slope m₂ = −1/2.

⇒ Now, multiplying the slopes of both lines:

⇒ m₁ × m₂ = 2 × (−1/2)

⇒ m₁ × m₂ = −1

⇒ Since the product evaluates exactly to −1, the lines intersect at right angles (90°).

∴ The correct answer is Yes.

Additional Information

Parallel Lines Condition

Two straight lines are parallel if they have exactly the same slope. They never intersect, and the condition is represented as m₁ = m₂.

Slope from General Equation

For a line given in the general algebraic form ax + by + c = 0, the slope can be directly determined using the formula m = −a ÷ b.

Angle Between Intersecting Lines

The acute angle θ formed between two intersecting lines with slopes m₁ and m₂ is found using the formula: tan θ = |(m₁ − m₂) ÷ (1 + m₁ × m₂)|.

Shortcut Trick

When sin A = cos B, their angles are complementary, meaning A + B = 90°.

⇒ x + 5x − 60° = 90°

⇒ 6x = 150°

⇒ x = 25°

∴ The correct answer is 25°.

Alternate Method

Given:

sin x = cos(5x − 60°)

Formula Used:

sin θ = cos(90° − θ)

Calculations:

⇒ cos(90° − x) = cos(5x − 60°)

Equating the angles on both sides:

⇒ 90° − x = 5x − 60°

⇒ 5x + x = 90° + 60°

⇒ 6x = 150°

⇒ x = 150° ÷ 6

⇒ x = 25°

∴ The correct answer is 25°.

Additional Information

Complementary Angle Identities

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

Supplementary Angle Identities

For any angle θ, sin(180° − θ) = sin θ, cos(180° − θ) = −cos θ, and tan(180° − θ) = −tan θ.

Important Trigonometric Values

sin 30° = cos 60° = 1 ÷ 2 | sin 45° = cos 45° = 1 ÷ √2 | sin 60° = cos 30° = √3 ÷ 2.

Shortcut Trick

Complementary angles always sum up to 90°.

We can check the given options for the smaller angle directly.

If the smaller angle is 20°, the larger angle must be 90° − 20° = 70°.

Now verify the given condition: 3 × 20° + 10° = 60° + 10° = 70°.

Since this perfectly matches the larger angle, our assumption is correct.

∴ The correct answer is 20°.

Alternate Method

Given:

The two angles are complementary (their sum is 90°).

The measure of one angle is 10° more than three times the other angle.

       90°  x 3x + 10°  Complementary Angles: Sum = 90°

Calculation:

Let the measure of the smaller angle be x.

Then, the measure of the larger angle = 3x + 10°

Since the angles are complementary, their sum must be equal to 90°.

⇒ x + (3x + 10°) = 90°

⇒ 4x + 10° = 90°

⇒ 4x = 90° − 10°

⇒ 4x = 80°

⇒ x = 80° ÷ 4

⇒ x = 20°

∴ The correct answer is 20°.

Additional Information

Supplementary Angles

Two angles are considered supplementary if the sum of their measures is exactly 180°.

Adjacent Angles

Two angles are adjacent when they have a common side and a common vertex (corner point), and do not overlap.

Linear Pair of Angles

A linear pair is a pair of adjacent angles formed by intersecting lines. The sum of a linear pair is always equal to 180°.

Vertically Opposite Angles

These are the angles opposite each other when two straight lines cross. Vertically opposite angles are always equal to each other.

Shortcut Trick

For any triangle where the internal angle bisectors meet at incenter I:

∠BIC = 90° + ∠A ÷ 2

⇒ 110° = 90° + ∠A ÷ 2

⇒ ∠A ÷ 2 = 20° ⇒ ∠A = 40°

∴ The correct answer is 40°.

Alternate Method

    A B C I  110°  ?

Given:

In ΔABC, angle bisectors of ∠B and ∠C intersect at point I.

∠BIC = 110°

Formula Used:

When the internal angle bisectors of two base angles of a triangle meet, the angle formed is:

∠BIC = 90° + ∠A ÷ 2

Calculations:

⇒ ∠BIC = 90° + ∠A ÷ 2

⇒ 110° = 90° + ∠A ÷ 2

⇒ ∠A ÷ 2 = 110° − 90°

⇒ ∠A ÷ 2 = 20°

⇒ ∠A = 20° × 2

⇒ ∠A = 40°

∴ The correct answer is 40°.

Additional Information

Incenter Properties

Formed by the intersection of internal angle bisectors. Subtended angle ∠BIC = 90° + ∠A ÷ 2.

Excenter Properties

Formed by the intersection of external angle bisectors. Subtended angle ∠BEC = 90° − ∠A ÷ 2.

Circumcenter Properties

Formed by the intersection of perpendicular bisectors. Central angle ∠BOC = 2 × ∠A.

Orthocenter Properties

Formed by the intersection of altitudes. Subtended angle ∠BHC = 180° − ∠A.

Shortcut Trick

For an algebraic expression of the form (√a + √b)², the direct result is (a + b) + 2√(ab).

Here, a = 3 and b = 2.

⇒ (3 + 2) + 2√(3 × 2)

⇒ 5 + 2√6

∴ The correct answer is 5 + 2√6.

Alternate Method

Given:

(√3 + √2)²

Formula Used:

(a + b)² = a² + 2ab + b²

Calculations:

⇒ (√3 + √2)² = (√3)² + 2 × √3 × √2 + (√2)²

⇒ (√3)² = 3

⇒ (√2)² = 2

⇒ 2 × √3 × √2 = 2√6

⇒ 3 + 2√6 + 2

⇒ 5 + 2√6

∴ The correct answer is 5 + 2√6.

Additional Information

Square of a Difference

(a − b)² = a² − 2ab + b². This rule is used for expanding terms like (√3 − √2)².

Difference of Squares

(a + b)(a − b) = a² − b². This identity is most often used to rationalize denominators involving square roots.

Product of Square Roots

√a × √b = √(ab), provided both a and b are non-negative numbers.

Shortcut Trick

According to the Basic Proportionality Theorem (BPT), since ST is parallel to QR, the sides are divided in the same ratio.

⇒ PS / SQ = PT / TR

⇒ 5 / 10 = 6 / TR

⇒ 1 / 2 = 6 / TR

⇒ TR = 12 cm

∴ The correct answer is 12 cm.

Alternate Method

Given:

In triangle PQR, line ST is parallel to the base QR.

Length of PS = 5 cm

Length of SQ = 10 cm

Length of PT = 6 cm

P Q R S T

5 cm 10 cm 6 cm TR = ?

Formula Used:

Basic Proportionality Theorem (Thales's Theorem): If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

⇒ PS / SQ = PT / TR

Calculation:

⇒ 5 / 10 = 6 / TR

⇒ 1 / 2 = 6 / TR

⇒ TR = 6 × 2

⇒ TR = 12 cm

∴ The correct answer is 12 cm.

Additional Information

Basic Proportionality Theorem (BPT)

Also known as Thales Theorem, it states that a line drawn parallel to one side of a triangle divides the other two sides proportionally.

Converse of BPT

If a line divides any two sides of a triangle in the same ratio, then the line must be strictly parallel to the third side.

Similar Triangles Property

When ST is parallel to QR, ΔPST is directly similar to ΔPQR. The ratio of their areas equals the square of their corresponding side ratio: Area(ΔPST) ÷ Area(ΔPQR) = (PS/PQ)².