Quantitative Aptitude PYQs

Shortcut Trick

Scheme I: Effective discount = 15 + 20 − (15 × 20) ÷ 100 = 32%

Scheme II: Discount % = [Free ÷ (Paid + Free)] × 100 = (2 ÷ 6) × 100 = 33.33%

Scheme III: Discount % = [Free ÷ (Paid + Free)] × 100 = (4 ÷ 10) × 100 = 40%

Comparing the values, Scheme III offers the maximum discount of 40%.

∴ The correct answer is Scheme III.

Alternate Method

Given:

Three discount schemes are provided by a shopkeeper:

I. Two successive discounts of 15% and 20%.

II. Buy 4 get 2 free.

III. Buy 6 get 4 free.

Discount Percentage Comparison

Scheme I

32%

Scheme II

33.33%

Scheme III

40% (Maximum)

Calculations:

For Scheme I: Two successive discounts of 15% and 20%

⇒ Equivalent Discount = a + b − (a × b) ÷ 100

⇒ Equivalent Discount = 15 + 20 − (15 × 20) ÷ 100

⇒ Equivalent Discount = 35 − 300 ÷ 100

⇒ Equivalent Discount = 35 − 3 = 32%

For Scheme II: Buy 4 get 2 free

⇒ Total items = 4 (paid) + 2 (free) = 6

⇒ Discount % = (Free Items ÷ Total Items) × 100

⇒ Discount % = (2 ÷ 6) × 100 = 33.33%

For Scheme III: Buy 6 get 4 free

⇒ Total items = 6 (paid) + 4 (free) = 10

⇒ Discount % = (Free Items ÷ Total Items) × 100

⇒ Discount % = (4 ÷ 10) × 100 = 40%

⇒ Comparing all three schemes: 32%, 33.33%, and 40%.

⇒ The maximum discount is 40% which is provided by Scheme III.

∴ The correct answer is Scheme III.

Additional Information

Successive Discounts

When two or more discounts are given one after another on the marked price. The effective discount of two discounts a% and b% is calculated as [a + b − (a × b) ÷ 100]%.

Buy X Get Y Free Concept

In promotional offers, the free items represent the discount amount, and the total items (paid + free) represent the marked price base. The equivalent Discount % is calculated as [Free Items ÷ Total Items] × 100.

Marked Price vs Selling Price

Selling Price (SP) is the final amount paid by the customer after all discounts are applied to the Marked Price (MP). The formula is SP = MP × (100 − Discount %) ÷ 100.

Shortcut Trick

Total distance = 200 + 252 + 204 = 656 km

Total time = (200 ÷ 50) + (252 ÷ 42) + (204 ÷ 51) = 4 + 6 + 4 = 14 hours

Average speed = Total distance ÷ Total time

Average speed = 656 ÷ 14 = 46.857... ≈ 46.86 km/hr

∴ The correct answer is 46.86 km/hr.

Alternate Method

Total Distance = 656 km

200 km
252 km
204 km

50 km/hr
42 km/hr
51 km/hr

t&sub1; = 4 hr
t&sub2; = 6 hr
t&sub3; = 4 hr

Given:

Distance d₁ = 200 km, Speed s₁ = 50 km/hr

Distance d₂ = 252 km, Speed s₂ = 42 km/hr

Distance d₃ = 204 km, Speed s₃ = 51 km/hr

Formulas Used:

Time = Distance ÷ Speed

Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ Time taken for the first part (t₁) = 200 ÷ 50 = 4 hours

⇒ Time taken for the second part (t₂) = 252 ÷ 42 = 6 hours

⇒ Time taken for the third part (t₃) = 204 ÷ 51 = 4 hours

⇒ Total Time = 4 + 6 + 4 = 14 hours

⇒ Total Distance = 200 + 252 + 204 = 656 km

⇒ Average Speed = 656 ÷ 14

⇒ Average Speed = 46.857... km/hr

⇒ Average Speed ≈ 46.86 km/hr (rounded off to two decimal places)

∴ The correct answer is 46.86 km/hr.

Additional Information

Average Speed for Two Equal Distances

If a person covers a certain distance at a speed of a km/hr and an equal distance at a speed of b km/hr, the average speed for the whole journey is given by: 2ab ÷ (a + b).

Average Speed for Three Equal Distances

If three equal distances are covered at speeds of a, b, and c respectively, the average speed is: 3abc ÷ (ab + bc + ca).

Relative Speed

When two bodies are moving in the same direction, their relative speed is the difference of their individual speeds (v₁ − v₂). When moving in opposite directions, it is the sum of their speeds (v₁ + v₂).

Shortcut Trick

Cost Price (CP) of 1 sharpener = 180 ÷ 12 = ₹15

Selling Price (SP) of 1 sharpener = ₹23

Profit = 23 − 15 = ₹8

Profit % = (8 ÷ 15) × 100 = 53.33%

∴ The correct answer is 53.33%.

Alternate Method

Given:

Cost Price of a dozen (12) sharpeners = ₹180

Selling Price of 1 sharpener = ₹23

Formula Used:

Profit = SP − CP

Profit % = (Profit ÷ CP) × 100

CP = ₹15

• Profit = ₹8

Selling Price (SP) = ₹23

Profit = 53.33% of CP

Calculations:

⇒ CP of 1 sharpener = 180 ÷ 12 = ₹15

⇒ Profit = 23 − 15 = ₹8

⇒ Profit % = (8 ÷ 15) × 100

⇒ Profit % = (8 × 20) ÷ 3

⇒ Profit % = 160 ÷ 3 = 53.33%

∴ The correct answer is 53.33%.

Additional Information

Loss Percentage

If the selling price is less than the cost price, the loss % is calculated as: Loss % = [(CP − SP) ÷ CP] × 100.

Marked Price & Discount

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

Successive Percentage Changes

For two successive changes of a% and b%, the effective percentage change is given by: (a + b + ab ÷ 100)%.

Shortcut Trick

Time for full tank: P = 16 × 3/2 = 24 hrs, Q = 20 × 6/5 = 24 hrs.

Let Total Work = LCM(24, 24) = 24 units ⇒ Efficiency: P = 1, Q = 1.

Tank filled in 2 hrs = (1 + 1) × 2 = 4 units.

R empties 4 units in 8 hrs ⇒ R efficiency = −4 ÷ 8 = −0.5 units/hr.

Combined efficiency (P + Q + R) = 1 + 1 − 0.5 = 1.5 units/hr.

Time taken together = 24 ÷ 1.5 = 16 hours.

∴ The correct answer is 16 hours.

Alternate Method

Given:

⇒ Time taken by P to fill 2/3 part of the tank = 16 hours.

⇒ Time taken by P to fill the complete tank = 16 × 3/2 = 24 hours.

⇒ Time taken by Q to fill 5/6 part of the tank = 20 hours.

⇒ Time taken by Q to fill the complete tank = 20 × 6/5 = 24 hours.

Total Work and Efficiency:

Calculations:

⇒ Total work done by P and Q in 2 hours = 2 × (1 + 1) = 4 units.

⇒ Both P and Q are closed, and these 4 units of water are emptied by pipe R in 8 hours.

⇒ Efficiency of emptying pipe R = Work ÷ Time = −4 ÷ 8 = −0.5 units/hr.

⇒ Combined efficiency of P, Q and R together = 1 + 1 − 0.5 = 1.5 units/hr.

⇒ Time taken by P, Q and R to fill the empty tank = Total Work ÷ Combined Efficiency.

⇒ Time taken = 24 ÷ 1.5 = 16 hours.

∴ The correct answer is 16 hours.

Additional Information

Pipes & Cisterns Concept

This concept operates identically to standard Time & Work. The total capacity of the tank is considered the total work (assumed as the LCM of all given times). However, unlike regular workers, emptying pipes drain the tank, which is represented mathematically as a negative efficiency.

Combined Efficiency Rule

When multiple pipes (both filling and emptying) are opened simultaneously, their net rate of filling or emptying is the direct algebraic sum of their individual efficiencies. A positive net efficiency means the tank is getting filled over time.

Total Work Formula

Total Time Taken = Total Capacity of Tank ÷ Net Combined Efficiency of all active pipes.

Shortcut Trick

Directly expand the brackets and combine the coefficients of x², x, and constant terms.

⇒ x(9x − 8) + 7(x² − 3) + 18

⇒ (9 + 7)x² − 8x + (−21 + 18)

⇒ 16x² − 8x − 3

∴ The correct answer is 16x² − 8x − 3.

Alternate Method

Given:

Expression: x(9x − 8) + 7(x² − 3) + 18

Method Used:

Use the distributive property a(b − c) = ab − ac and group the like terms together to simplify the expression.

Calculations:

⇒ x(9x − 8) + 7(x² − 3) + 18

Expanding the brackets by multiplying the outer terms with each term inside:

⇒ 9x² − 8x + 7x² − 21 + 18

Grouping the like terms together for addition/subtraction:

⇒ (9x² + 7x²) − 8x + (18 − 21)

⇒ 16x² − 8x − 3

∴ The correct answer is 16x² − 8x − 3.

Additional Information

Distributive Property

It states that multiplying a sum or difference by a number is the same as multiplying each term individually and then combining them. Formula: a(b ± c) = ab ± ac.

Like Terms in Algebra

Terms that have the exact same variables raised to the identical powers are known as like terms. They can easily be simplified by adding or subtracting their respective numerical coefficients.

BODMAS Rule

The order of operations is Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction. Opening up algebraic expressions by expanding brackets (multiplication) must be performed before doing any addition.

Shortcut Trick

M's Profit = (6848 + 856) ÷ 2 = ₹3852, and N's Profit = (6848 − 856) ÷ 2 = ₹2996.

Profit ratio of M : N = 3852 : 2996 = 9 : 7.

We know, Capital Ratio = Profit Ratio ÷ Time Ratio.

Ratio of Capitals M : N = (9 ÷ 4) : (7 ÷ 7) = 9/4 : 1 = 9 : 4.

Difference in Capital ratio = 9 − 4 = 5 parts = ₹68,000 ⇒ 1 part = ₹13,600.

M's capital (9 parts) = 9 × 13,600 = ₹1,22,400.

∴ The correct answer is ₹1,22,400.

Alternate Method

Given:

Difference in Capital of M and N = ₹68,000

Time period of M = 4 months

Time period of N = 7 months

Total Profit = ₹6,848, M's share is ₹856 more than N.

Formula Used:

Profit = Capital × Time

Profit Ratio = (Capital₁ × Time₁) : (Capital₂ × Time₂)

Partner M
Capital = x + 68,000
Time = 4 months
Profit Ratio = 9

Partner N
Capital = x
Time = 7 months
Profit Ratio = 7

Total Profit = ₹6,848 | Difference = ₹856

Calculation:

⇒ Total Profit = P_M + P_N = 6848

⇒ Difference in shares = P_M − P_N = 856

⇒ Adding both equations: 2 × P_M = 7704 ⇒ P_M = ₹3852

⇒ P_N = 6848 − 3852 = ₹2996

⇒ Ratio of Profit (M : N) = 3852 : 2996 = 9 : 7

⇒ Let N's capital be x. Then M's capital is (x + 68000).

⇒ According to partnership rules, [Capital_M × Time_M] ÷ [Capital_N × Time_N] = P_M ÷ P_N

⇒ [(x + 68000) × 4] ÷ [x × 7] = 9 ÷ 7

⇒ 4x + 272000 = 9x

⇒ 5x = 272000

⇒ x = ₹54,400

⇒ M's capital = x + 68000 = 54400 + 68000 = ₹1,22,400

∴ The correct answer is ₹1,22,400.

Additional Information

Partnership Profit Rule

The profit earned by a partner is directly proportional to the product of their invested capital and the duration of the investment. (Profit Ratio = C₁×T₁ : C₂×T₂)

Working vs Sleeping Partner

A working partner manages the business operations and receives a fixed salary or commission from the total profit before the remaining profit is distributed according to the capital ratio.

Compound Partnership

If partners change their capital during the year (by adding or withdrawing), the investment is treated month-by-month and summed up as an equivalent total capital for one unit of time.

Shortcut Trick

When the distance is constant, the speed is inversely proportional to the time taken.

S₁ / S₂ = T₂ / T₁

66 / S₂ = 5 / 15

66 / S₂ = 1 / 3

S₂ = 66 × 3 = 198 km/hr

∴ The correct answer is 198 km/hr.

Alternate Method

Given:

Initial Speed (S₁) = 66 km/hr

Initial Time (T₁) = 15 hours

New Time (T₂) = 5 hours

Formula Used:

Distance = Speed × Time

New Speed = Distance ÷ New Time

Distance = S1 × T1

S1 = 66 km/hr
T1 = 15 hours

Same Distance = S2 × T2

S2 = ?
T2 = 5 hours

Calculations:

⇒ Total Distance = 66 × 15

⇒ Total Distance = 990 km

⇒ New Speed required to cover 990 km in 5 hours = Total Distance ÷ New Time

⇒ New Speed = 990 ÷ 5

⇒ New Speed = 198 km/hr

∴ The correct answer is 198 km/hr.

Additional Information

Average Speed

Average speed is the total distance covered divided by the total time taken. If equal distances are covered at speeds S₁ and S₂, then Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂).

Relative Speed

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

Unit Conversion

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

Shortcut Trick

Side of the square = Perimeter ÷ 4

⇒ Side = 76 ÷ 4 = 19 m

Area = (Side)²

⇒ Area = 19 × 19 = 361 m²

∴ The correct answer is 361 m².

Alternate Method

Given:

Perimeter of the square (P) = 76 m

Formula Used:

Perimeter = 4 × a

Area = a²

a = 19 m
a = 19 m
a = 19 m
a = 19 m
P = 76 m
Area = ?

Calculations:

⇒ 4 × a = 76

⇒ a = 76 ÷ 4

⇒ a = 19 m

⇒ Area = 19 × 19

⇒ Area = 361 m²

∴ The correct answer is 361 m².

Additional Information

Diagonal of a Square

The diagonal of a square divides it into two congruent right-angled triangles. Length of diagonal (d) = a√2, where 'a' is the side length.

Area using Diagonal

If the length of the diagonal (d) is given, the area of the square can be directly calculated as Area = d² ÷ 2.

Incircle and Circumcircle

For a square of side 'a', the radius of its inscribed circle is a ÷ 2, and the radius of its circumscribed circle is a ÷ √2.

Shortcut Trick

Applying the BODMAS rule directly to the expression:

⇒ −12 − {−8 × (6 − 12) + 30 ÷ (−6)}

⇒ −12 − {−8 × (−6) − 5}

⇒ −12 − {48 − 5}

⇒ −12 − 43 = −55

∴ The correct answer is −55.

Alternate Method

Given:

−12 − {−8 × (6 − 12) + 30 ÷ (−6)}

Concept Used:

BODMAS Rule: Brackets → Orders → Division → Multiplication → Addition → Subtraction

Calculation:

First, solve the terms inside the innermost bracket:

⇒ (6 − 12) = −6

Substitute this back into the expression:

⇒ −12 − {−8 × (−6) + 30 ÷ (−6)}

Next, perform multiplication and division inside the curly braces:

⇒ −8 × (−6) = 48

⇒ 30 ÷ (−6) = −5

Substitute these values back into the expression:

⇒ −12 − {48 + (−5)}

⇒ −12 − {43}

Finally, perform the outer subtraction:

⇒ −12 − 43 = −55

∴ The correct answer is −55.

Additional Information

BODMAS Rule

It defines the correct sequence for solving mathematical expressions: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction.

Rule of Signs for Multiplication & Division

Multiplying or dividing two negative numbers yields a positive result (− × − = +). A positive and a negative yields a negative result (+ × − = −).

Rule of Signs for Addition & Subtraction

When subtracting a positive number from a negative number, the result becomes more negative (e.g., −12 − 43 = −55). When subtracting a negative number, it turns into an addition.

Shortcut Trick

Effective Rate of Interest = Rate × Time

Effective Rate = 12% × 5.5 = 66%

Simple Interest = 66% of Principal

SI = 66% of 400 = 66 × 4 = 264

∴ The correct answer is ₹264.

Alternate Method

Given:

Principal (P) = ₹400

Rate (R) = 12% per annum

Time (T) = 5.5 years

Rate = 12% p.a. | Time = 5.5 years

P
₹400

SI
₹?

Interest accumulates linearly
SI = (P × R × T) ÷ 100

Formula Used:

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

Calculations:

⇒ SI = (400 × 12 × 5.5) ÷ 100

⇒ SI = 4 × 12 × 5.5

⇒ SI = 48 × 5.5

⇒ SI = 264

∴ The correct answer is ₹264.

Additional Information

Total Amount in Simple Interest

The total amount (A) received after time T is given by the formula: A = P + SI = P × (1 + RT ÷ 100).

Difference Between SI and CI (2 years)

For a principal P and rate R, the difference between Compound Interest and Simple Interest over exactly 2 years is calculated directly as: Difference = P × (R ÷ 100)².

Rule of 72

A quick trick to estimate the number of years required to double your invested money at a given fixed annual interest rate is to divide 72 by the interest rate (R).

Shortcut Trick

In Compound Interest, if a sum becomes x times in t years, it becomes xⁿ times in n × t years.

Here, the sum becomes 3 times in 11 years.

We need the sum to become 27 times, which is exactly 3³ times (so n = 3).

Required time = 3 × 11 = 33 years.

∴ The correct answer is 33 years.

Alternate Method

Given:

Principal (P) becomes 3 times of itself in 11 years at Compound Interest.

Formula Used:

A = P(1 + R ÷ 100)^T

0 years
11 years
22 years
33 years

P
3P
9P
27P

× 3

× 3

× 3

Amount becomes 27 times in 33 years

Calculations:

Case 1: The sum triples in 11 years.

⇒ 3P = P(1 + R ÷ 100)¹¹

⇒ (1 + R ÷ 100)¹¹ = 3   ... (Equation 1)

Case 2: Let the sum become 27 times in 'T' years.

⇒ 27P = P(1 + R ÷ 100)^T

⇒ (1 + R ÷ 100)^T = 27

⇒ (1 + R ÷ 100)^T = 3³

Substituting the value of 3 from Equation 1 into the above expression:

⇒ (1 + R ÷ 100)^T = [(1 + R ÷ 100)¹¹]³

⇒ (1 + R ÷ 100)^T = (1 + R ÷ 100)³³

Comparing the powers on both sides:

⇒ T = 33 years.

∴ The correct answer is 33 years.

Additional Information

Constant Multiplying Factor in CI

In Compound Interest, the sum grows by a constant multiplying factor over equal intervals of time. If a sum becomes k times in t years, it will become k² times in 2t years, and kⁿ times in nt years.

Half-Yearly Compounding

If interest is compounded half-yearly, the rate of interest becomes half (R ÷ 2) and the time period is doubled (2T).

Simple vs Compound Interest

For the first compounding period, SI and CI are exactly the same. For subsequent periods, CI is always greater than SI because interest is calculated on accumulated interest as well.

Shortcut Trick

When the time period is equal, the profit ratio is directly proportional to the investment ratio.

Profit Ratio = 1320 : 1170 : 1660 = 132 : 117 : 166

Total parts = 132 + 117 + 166 = 415 parts

Poonam's profit share = (166 ÷ 415) × 1010 = ₹404

∴ The correct answer is ₹404.

Alternate Method

Given:

⇒ Investment of Rekha = ₹1320

⇒ Investment of Sanjay = ₹1170

⇒ Investment of Poonam = ₹1660

⇒ Total Profit = ₹1010

Formula Used:

⇒ Profit Ratio = Investment × Time

⇒ If the time period is identical, Profit Ratio = Investment Ratio

Investment and Profit Ratio

Rekha
₹1320
Ratio: 132

Sanjay
₹1170
Ratio: 117

Poonam
₹1660
Ratio: 166

Total Ratio = 132 + 117 + 166 = 415 parts
Total Profit = ₹1010   →   Poonam's Share = (166 ÷ 415) × 1010

Calculations:

⇒ Ratio of their profit shares = 1320 : 1170 : 1660

⇒ Cancelling the common factor of 10, the Profit Ratio = 132 : 117 : 166

⇒ Sum of the profit ratios = 132 + 117 + 166 = 415

⇒ Poonam's share = (Poonam's ratio parts ÷ Total ratio parts) × Total Profit

⇒ Poonam's share = (166 ÷ 415) × 1010

⇒ Dividing both numerator and denominator by 5 gives: (166 ÷ 83) × 202

⇒ Since 166 ÷ 83 = 2, we get:

⇒ Poonam's share = 2 × 202 = ₹404

∴ The correct answer is ₹404.

Additional Information

Simple Partnership

When all partners invest their capitals for the exact same duration, the generated profit is distributed in the direct proportion of their respective invested amounts: P₁ : P₂ = C₁ : C₂.

Compound Partnership

When partners invest diverse amounts for varying time spans, the profit sharing ratio is evaluated as the product of the individual capital and its corresponding duration: P₁ : P₂ = (C₁ × T₁) : (C₂ × T₂).

Working/Active Partner

An active partner is compensated with a fixed salary or a specific percentage of the total profit for managing the business operations. The residual profit is then shared among all partners (including the active one) strictly aligned with their investment ratios.

Shortcut Trick

Net Profit % = Markup % − Discount % − (Markup % × Discount % ÷ 100)

Net Profit % = 60 − 15 − (60 × 15 ÷ 100) = 45 − 9 = 36%

36% of CP = Profit = ₹117

⇒ CP = 117 × 100 ÷ 36 = ₹325

⇒ List Price = 160% of CP = 325 × 1.6 = ₹520

∴ The correct answer is ₹520.

Alternate Method

Given:

Markup Percentage = 60%

Discount Percentage = 15%

Profit = ₹117

Formula Used:

Selling Price (SP) = Marked Price (MP) × (100 − Discount%) ÷ 100

Profit = SP − Cost Price (CP)

CP = 100x

SP = 136x

MP = 160x

Markup = +60%

Profit = ₹117

Discount = -15%

Calculations:

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

⇒ Marked Price (MP) = 100x + 60% of 100x = 160x

⇒ Discount = 15% of 160x = 24x

⇒ Selling Price (SP) = 160x − 24x = 136x

⇒ Profit = SP − CP = 136x − 100x = 36x

⇒ We are given the actual profit is ₹117.

⇒ 36x = 117

⇒ x = 117 ÷ 36 = 3.25

⇒ List Price (MP) = 160x = 160 × 3.25 = ₹520

∴ The correct answer is ₹520.

Additional Information

Effective Profit/Loss Percentage

When a markup and discount are given, Net Profit % = M − D − (M × D) ÷ 100, where M is Markup % and D is Discount %.

Marked Price (List Price)

It is the price listed or printed on an article. A shopkeeper calculates discounts exclusively on the Marked Price, while Profit or Loss is derived from the Cost Price.

Successive Discounts

If two successive discounts of a% and b% are given, Effective Discount = a + b − (a × b) ÷ 100.

Shortcut Trick

In ΔABC, third angle ∠ACB = 180° − (112° + 24°) = 44°.

Since BC ∥ AD, alternate interior angle ∠CAD = ∠ACB = 44°.

In isosceles ΔACD (AC = CD), base angles are equal ⇒ ∠CDA = ∠CAD = 44°.

Thus, the required angle ∠ACD = 180° − (44° + 44°) = 92°.

∴ The correct answer is 92°.

Alternate Method

A
B
C
D

24°
112°
44°
44°
44°
92°

In Trapezium ABCD, ∠ACD = 92°

Given:

∠ABC = 24°, ∠BAC = 112°

BC ∥ AD

AC = CD

Formula Used:

Sum of all internal angles of a triangle = 180°

Alternate interior angles are equal when two parallel lines are intersected by a transversal.

Angles opposite to equal sides in a triangle are always equal.

Calculations:

In ΔABC, using the angle sum property:

⇒ ∠ABC + ∠BAC + ∠ACB = 180°

⇒ 24° + 112° + ∠ACB = 180°

⇒ 136° + ∠ACB = 180°

⇒ ∠ACB = 180° − 136° = 44°

Since BC ∥ AD and AC is the transversal line joining them:

⇒ ∠CAD = ∠ACB (Alternate interior angles property)

⇒ ∠CAD = 44°

In ΔACD, it is given that sides AC = CD:

⇒ ∠CDA = ∠CAD (Angles opposite to equal sides are equal)

⇒ ∠CDA = 44°

Now, applying the angle sum property in ΔACD:

⇒ ∠ACD + ∠CAD + ∠CDA = 180°

⇒ ∠ACD + 44° + 44° = 180°

⇒ ∠ACD + 88° = 180°

⇒ ∠ACD = 180° − 88° = 92°

∴ The correct answer is 92°.

Additional Information

Properties of Parallel Lines

When a transversal intersects two parallel lines, the alternate interior angles are strictly equal. This property seamlessly bridges geometric regions separated by transversals.

Isosceles Triangle Theorem

If two sides of a triangle are equal in dimension, the respective opposite interior angles bounding them are also equal. The converse remains universally true.

Angle Sum Property

Irrespective of the shape and type, the collective sum of all three internal angles inside a conventional planar triangle invariably amounts to exactly 180°.

Shortcut Trick

Let the assumed average weight be 50 kg.

Net deviation from assumed average = (−10) + (−1) + (+6) + (+24) + (−15) = +4 kg

Average deviation per member = +4 ÷ 5 = +0.8 kg

Actual average = Assumed average + Average deviation = 50 + 0.8 = 50.8 kg

∴ The correct answer is 50.8 kg.

Alternate Method

Given:

Weights of the family members = 40 kg, 49 kg, 56 kg, 74 kg, and 35 kg

Total number of members = 5

Formula Used:

Average = Sum of all observations ÷ Total number of observations

40

49

56

74

35

Avg = 50.8

Weights of 5 Members (in kg)

Calculations:

⇒ Sum of weights = 40 + 49 + 56 + 74 + 35

⇒ Sum of weights = 254 kg

⇒ Average weight = 254 ÷ 5

⇒ Average weight = 50.8 kg

∴ The correct answer is 50.8 kg.

Additional Information

Assumed Mean Method

It simplifies arithmetic calculations by picking an arbitrary middle value. Formula: Average = Assumed Mean + (Sum of Deviations ÷ Total Count).

Weighted Average

Used when calculating the combined average of groups with varying sizes. Formula: W.A = (N₁A₁ + N₂A₂) ÷ (N₁ + N₂).

Net Deviation Concept

The sum of deviations of all observations measured from their exact mathematical average is always strictly zero.

Shortcut Trick

Test the given values to satisfy both conditions directly to save time.

Condition 1: 5P = Q + 29

Condition 2: 5(Q + 8) = 7(P + 8) − 17

Testing Q = 11: 5P = 11 + 29 = 40 ⇒ P = 8.

Testing Condition 2: 5(11 + 8) = 95 and 7(8 + 8) − 17 = 112 − 17 = 95 (Matches).

∴ The correct answer is 11 years.

Alternate Method

Given:

Five times P's present age exceeds Q's present age by 29 years.

After 8 years, five times Q's age will be 17 years less than seven times P's age.

Formula Used:

If current age is x, age after n years = x + n.

Present
P's Age = P
Q's Age = Q
After 8 Years
P's Age = P + 8
Q's Age = Q + 8

• 8 yrs

Calculations:

⇒ Let P's present age be P and Q's present age be Q.

⇒ From the first condition: 5P − Q = 29

⇒ Q = 5P − 29     ... (Equation 1)

⇒ From the second condition: 5(Q + 8) = 7(P + 8) − 17

⇒ 5Q + 40 = 7P + 56 − 17

⇒ 5Q + 40 = 7P + 39

⇒ 7P − 5Q = 1     ... (Equation 2)

⇒ Substituting Equation 1 into Equation 2:

⇒ 7P − 5(5P − 29) = 1

⇒ 7P − 25P + 145 = 1

⇒ −18P = −144

⇒ P = 8

⇒ Substituting P = 8 in Equation 1:

⇒ Q = 5(8) − 29

⇒ Q = 40 − 29 = 11

∴ The correct answer is 11 years.

Additional Information

Constant Age Difference

The difference between the ages of two individuals always remains constant regardless of the time period (past, present, or future).

Present Age Reference

Always assume the present age as the base variable (e.g., x or y) to avoid confusion while framing equations for future (+ years) or past (− years) ages.

Linear Equations Application

Problems involving ages typically require translating the given conditions into a pair of linear equations in two variables, which can then be solved easily using substitution or elimination methods.

Shortcut Trick

According to the BODMAS rule, division is performed before multiplication.

Using exponent laws directly: a^m ÷ aⁿ × a^p = a^{m − n + p}

Expression = 47^{−8 − 3 + (−19)}

Expression = 47^{−11 − 19}

Expression = 47⁻³⁰

∴ The correct answer is 47⁻³⁰.

Alternate Method

Given:

47⁻⁸ ÷ 47³ × 47⁻¹⁹

Formula Used:

a^m ÷ aⁿ = a^{m − n}

a^m × aⁿ = a^{m + n}

Calculations:

Applying the BODMAS rule, we evaluate the division first:

⇒ (47⁻⁸ ÷ 47³) × 47⁻¹⁹

⇒ 47^{−8 − 3} × 47⁻¹⁹

⇒ 47⁻¹¹ × 47⁻¹⁹

Now, perform the multiplication using the product rule of exponents:

⇒ 47^{−11 + (−19)}

⇒ 47^{−11 − 19}

⇒ 47⁻³⁰

∴ The correct answer is 47⁻³⁰.

Additional Information

BODMAS Rule

It dictates the standard order of mathematical operations: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction. In the absence of brackets, division must be performed before multiplication.

Quotient Rule of Exponents

When dividing two exponential expressions with the exact same base, you subtract the exponent of the denominator from the exponent of the numerator: a^m ÷ aⁿ = a^m−n

Product Rule of Exponents

When multiplying two exponential expressions with the exact same base, you simply add their exponents together: a^m × aⁿ = a^m+n

Negative Exponent Rule

A negative exponent means taking the reciprocal of the base raised to the positive opposite of the given exponent: a^−m = 1 ÷ a^m

Shortcut Trick

Using the multiplying factor method for successive discounts:

SP = MP × (100 − 24)% × (100 − 15)%

SP = 800 × 76% × 85%

SP = 800 × (76 / 100) × (85 / 100) = 8 × 64.60 = 516.80

∴ The correct answer is ₹516.80.

Alternate Method

Given:

Marked Price (MP) = ₹800

Successive discounts = 24% and 15%

Formula Used:

Selling Price (SP) = MP × (1 − d₁ / 100) × (1 − d₂ / 100)

MP
₹800

− 24%

After 1^st
₹608

− 15%

SP
₹516.80

Calculations:

⇒ First discount = 24% of 800 = (24 / 100) × 800 = 192

⇒ Price after 1^st discount = 800 − 192 = ₹608

⇒ Second discount = 15% of 608 = (15 / 100) × 608 = 91.20

⇒ Final Selling Price (SP) = 608 − 91.20 = ₹516.80

∴ The correct answer is ₹516.80.

Additional Information

Equivalent Single Discount

For two successive discounts of a% and b%, the equivalent single discount is given by the formula: D = (a + b − ab / 100)%.

Relation Between CP and MP

When both profit and discount percentages are given, the ratio of Marked Price to Cost Price is: MP / CP = (100 + Profit%) / (100 − Discount%).

Successive Percentage Change

If an amount undergoes successive changes of x% and y%, the net percentage change is: Net % = (x + y + xy / 100)%. Negative values are used for discounts or decreases.

Shortcut Trick

Put A = 30° in the given expression.

⇒ (cos⁴ 30° − sin⁴ 30° + 1) cosec² 30°

⇒ ((√3/2)⁴ − (1/2)⁴ + 1) × 2²

⇒ (9/16 − 1/16 + 1) × 4 = (8/16 + 1) × 4 = (3/2) × 4 = 6

Now check the options by putting A = 30°. Only 2 cot² 30° = 2 × (√3)² = 6 satisfies this value.

∴ The correct answer is 2 cot² A.

Alternate Method

Given:

(cos⁴ A − sin⁴ A + 1) cosec² A

Formula Used:

a² − b² = (a − b)(a + b)

sin² A + cos² A = 1

Calculation:

⇒ First, we simplify the term (cos⁴ A − sin⁴ A) using the difference of squares formula:

⇒ cos⁴ A − sin⁴ A = (cos² A − sin² A)(cos² A + sin² A)

⇒ Since cos² A + sin² A = 1, the term simplifies to (cos² A − sin² A)

⇒ Substitute this back into the original expression:

⇒ (cos² A − sin² A + 1) cosec² A

⇒ Using the identity 1 − sin² A = cos² A, we get:

⇒ (cos² A + cos² A) cosec² A

⇒ (2 cos² A) × cosec² A

⇒ Since cosec² A = 1 ÷ sin² A:

⇒ (2 cos² A) × (1 ÷ sin² A)

⇒ 2 (cos² A ÷ sin² A)

⇒ 2 cot² A

∴ The correct answer is 2 cot² A.

Additional Information

Pythagorean Trigonometric Identities

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

Reciprocal Trigonometric Identities

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

Quotient Trigonometric Identities

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

Shortcut Trick

Total Capacity = LCM(18, 30, 18) = 90 units.

Combined efficiency = (90 ÷ 18) + (90 ÷ 30) + (90 ÷ 18) = 5 + 3 + 5 = 13 units/hour.

Total time taken = Total Capacity ÷ Combined efficiency = 90 ÷ 13 = 6¹²/₁₃ hours.

∴ The correct answer is 6¹²/₁₃ hours.

Alternate Method

Given:

Pipe A fills the tank in 18 hours.

Pipe B fills the tank in 30 hours.

Pipe C fills the tank in 18 hours.

Formula Used:

Time = Total Work ÷ Total Efficiency

Calculations:

⇒ Total efficiency of A, B, and C = 5 + 3 + 5 = 13 units/hour.

⇒ Time taken to fill the tank = 90 ÷ 13

⇒ Time taken = 6¹²/₁₃ hours.

∴ The correct answer is 6¹²/₁₃ hours.

Additional Information

Pipes & Cisterns Work Logic

The logic for Pipes and Cisterns is identical to Time and Work. The total capacity of the cistern is considered as the total work (usually the LCM of the given individual times).

Leakage or Emptying Pipe

If a pipe empties the tank or there is a continuous leak, its efficiency is taken as negative (−) while calculating the combined efficiency.

Combined Rate Formula

If two pipes A and B fill a tank in x and y hours respectively, the time taken to fill the tank together is given directly by (x × y) ÷ (x + y).