Quantitative Aptitude PYQs

Shortcut Trick

Net Discount % = a + b − (ab/100) = 10 + 20 − (10 × 20)/100 = 28%

Selling Price = (100% − 28%) of Marked Price

SP = 72% of ₹3,000 = 0.72 × 3,000 = ₹2,160

∴ The correct answer is ₹2,160.

Alternate Method

Given: Marked Price (MP) = ₹3,000; First Discount = 10%; Second Discount = 20%.

Formula: Final SP = MP × (100 − d₁)/100 × (100 − d₂)/100

⇒ Price after 10% discount = 3,000 − (10% of 3,000) = 3,000 − 300 = ₹2,700

⇒ Price after 20% discount = 2,700 − (20% of 2,700) = 2,700 − 540 = ₹2,160

⇒ The final Selling Price is ₹2,160, which matches Option (4).

∴ The correct answer is ₹2,160.

Additional Information

Successive Discount Rule

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

Selling Price vs Marked Price

Selling Price (SP) = Marked Price (MP) − Discount. If multiple discounts are applied, they are calculated on the diminishing balance.

Profit and Loss Relation

If CP is the cost price, then SP = CP + Profit or SP = CP − Loss. The markup is always calculated on the CP.

Shortcut Trick

Effective Simple Interest percentage for 2 years = Rate × Time = 16% × 2 = 32%.

According to the question, 32% of Principal = ₹1,440.

⇒ 1% of Principal = 1,440 ÷ 32 = 45.

⇒ 100% of Principal (Sum) = 45 × 100 = 4,500.

∴ The correct answer is ₹4,500.

Alternate Method

Given: Simple Interest (SI) = ₹1,440, Rate (R) = 16% per year, Time (T) = 2 years.

Formula Used: SI = (P × R × T) ÷ 100, where P is the Principal amount.

⇒ 1,440 = (P × 16 × 2) ÷ 100

⇒ 1,440 = 32P ÷ 100

⇒ 1,440 × 100 = 32P

⇒ 1,44,000 = 32P

⇒ P = 1,44,000 ÷ 32

⇒ P = 4,500

∴ The correct answer is ₹4,500.

Additional Information

Simple Interest vs Compound Interest

For the first year, Simple Interest and Compound Interest are equal. From the second year onwards, CI includes interest on the previous year's interest.

Amount Formula

Total Amount (A) = Principal (P) + Interest (I). For Simple Interest, A = P(1 + RT/100).

Rate of Interest Calculation

The annual Rate of Interest can be found using the formula R = (SI × 100) ÷ (P × T).

Shortcut Trick

Assume the Mean (40) as the balance point where the sum of deviations is zero.

Net deviation: ∑f(x − 40) = 2(−30) + 3(−20) + 1(−10) + 4(0) + f(10) + 4(20) = 0

−60 − 60 − 10 + 0 + 10f + 80 = 0 ⇒ 10f − 50 = 0

10f = 50 ⇒ f = 5

∴ The correct answer is 5.

Alternate Method

Given: Mean (&bar;x) = 40. Data points (x) and frequencies (f) are mapped below.

Formula: Mean (&bar;x) = ∑f_ix_i ÷ ∑f_i

⇒ ∑f_ix_i = (10 × 2) + (20 × 3) + (30 × 1) + (40 × 4) + (50 × f) + (60 × 4)

⇒ ∑f_ix_i = 20 + 60 + 30 + 160 + 50f + 240 = 510 + 50f

⇒ ∑f_i = 2 + 3 + 1 + 4 + f + 4 = 14 + f

⇒ 40 = (510 + 50f) ÷ (14 + f)

⇒ 40(14 + f) = 510 + 50f

⇒ 560 + 40f = 510 + 50f

⇒ 560 − 510 = 50f − 40f

⇒ 50 = 10f ⇒ f = 5

∴ The correct answer is 5.

Additional Information

Arithmetic Mean of Discrete Series

Calculated as the sum of products of observations and their frequencies divided by the total frequency: &bar;x = ∑fx / ∑f.

Properties of Mean

The algebraic sum of deviations of a set of observations from their arithmetic mean is always zero, i.e., ∑f_i(x_i − &bar;x) = 0.

Assumed Mean Method

A shortcut formula where Mean = A + (∑fd / ∑f), where A is the assumed mean and d = x − A is the deviation.

Shortcut Trick

Net difference in interest rate per annum = 6.5% − 3.5% = 3%

Total interest difference for 2 years = 3% × 2 = 6%

Positive difference in Simple Interest = 6% of ₹3,40,000

Difference = 3,40,000 × (6 / 100) = 3,400 × 6 = ₹20,400

∴ The correct answer is ₹20,400.

Alternate Method

Given:

Principal (P) = ₹3,40,000

Rate for Bank A (R₁) = 3.5% p.a.

Rate for Bank B (R₂) = 6.5% p.a.

Time (T) = 2 years

Formula Used:

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

Calculations:

⇒ Interest paid to Bank A (SI₁) = (3,40,000 × 3.5 × 2) / 100

⇒ SI₁ = 3,400 × 7 = ₹23,800

⇒ Interest paid to Bank B (SI₂) = (3,40,000 × 6.5 × 2) / 100

⇒ SI₂ = 3,400 × 13 = ₹44,200

⇒ Positive difference = SI₂ − SI₁

⇒ Difference = 44,200 − 23,800 = ₹20,400

∴ The correct answer is ₹20,400.

Additional Information

Simple Interest Formula

SI = (P × R × T) / 100, where P is Principal, R is Rate, and T is Time.

Rate Difference Rule

When P and T are constant, the difference in interests is P × T × (R₂ − R₁) / 100.

Amount Formula

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

Shortcut Trick

LCM of (12, 15, 20) = 60 units (Total Capacity).

Efficiencies: A = 5, B = 4, C = −3 (emptying).

Work done by Pipe A and Pipe B in 2 hr = (5 + 4) × 2 = 18 units.

Remaining Work = 60 − 18 = 42 units.

Time for remaining part = 42 ÷ (5 + 4 − 3) = 42 ÷ 6 = 7 hr.

Total time = 2 + 7 = 9 hr.

∴ The correct answer is 9 hr.

Alternate Method

Given: Time for Pipe A = 12 hr, Pipe B = 15 hr, Pipe C = 20 hr.

Formula: Total Work = LCM of individual times; Efficiency = Total Work ÷ Time.

⇒ Tank filled by Pipe A and Pipe B in the first 2 hours = (5 + 4) × 2 = 18 units

⇒ Remaining capacity to be filled = 60 − 18 = 42 units

⇒ Combined efficiency of A, B, and C = 5 + 4 − 3 = 6 units/hr

⇒ Time taken to fill the remaining part = 42 ÷ 6 = 7 hr

⇒ Total time taken to fill the tank = 2 hr + 7 hr = 9 hr

∴ The correct answer is 9 hr.

Additional Information

Pipes and Cisterns Efficiency

An inlet pipe (filling) has positive efficiency, while an outlet pipe (emptying) has negative efficiency.

Net Work Done

When multiple pipes operate, the net rate is the algebraic sum of their individual efficiencies.

LCM Method for Total Capacity

The Total Capacity of a tank is best assumed as the LCM of the individual times to simplify calculations without fractions.

Shortcut Trick

Sum of ages 2 years ago = 46 − (2 × 2) = 42 years.

Ratio of P to Q was 2 : 1.

Total units = 2 + 1 = 3 units → 42 years.

1 unit = 14 years.

P's age 2 years ago = 2 × 14 = 28 years.

Present age of P = 28 + 2 = 30 years.

∴ The correct answer is 30 years.

Alternate Method

Given:

Present combined age of P and Q = 46 years

Condition: 2 years ago, P = 2 × Q

Let the present age of P be P and Q be Q.

⇒ P + Q = 46 (Equation 1)

Two years ago, their ages were (P − 2) and (Q − 2).

⇒ P − 2 = 2 × (Q − 2)

⇒ P − 2 = 2Q − 4

⇒ P − 2Q = −2 (Equation 2)

Subtracting Equation 2 from Equation 1:

⇒ (P + Q) − (P − 2Q) = 46 − (−2)

⇒ 3Q = 48

⇒ Q = 16 years

Substitute the value of Q in Equation 1:

⇒ P + 16 = 46

⇒ P = 46 − 16

⇒ P = 30 years

∴ The correct answer is 30 years.

Additional Information

Constant Age Difference

The numerical difference between the ages of two individuals always remains constant, regardless of how many years pass or are subtracted.

Sum Adjustment for Time

When calculating the sum of ages 'n' years ago for 'k' people, subtract (n × k) from the present total sum.

Ratio and Proportions in Ages

If the ratio of ages at time 't' is a : b, and the total sum at that same time is S, then individual ages are [a/(a+b)] × S and [b/(a+b)] × S.

Shortcut Trick

The mean proportional of two numbers a and b is calculated as √(a × b).

Given a = 16 and b = 25.

New side = √(16 × 25) = 4 × 5 = 20 m.

∴ The correct answer is 20 m.

Alternate Method

Given:

Side of the first square garden (a) = 16 m

Side of the second square garden (b) = 25 m

Formula Used:

If x is the mean proportional between a and b, then a : x = x : b ⇒ x² = ab ⇒ x = √ab

Calculations:

⇒ Let the side of the third square garden be x.

⇒ According to the question, x is the mean proportional of 16 and 25.

⇒ x = √(16 × 25)

⇒ x = √400

⇒ x = 20 m

∴ The correct answer is 20 m.

Additional Information

Mean Proportional

For any two numbers a and b, the mean proportional x is the square root of their product, i.e., x = √ab.

Third Proportional

If a and b are two numbers, their third proportional c is such that a:b = b:c, which gives c = b²/a.

Fourth Proportional

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

Shortcut Trick

Total volume remains constant when objects are melted and recast.

Volume of 16 identical cubes = 16 × (Side)³

Volume = 16 × (2 cm)³ = 16 × 8 cm³ = 128 cm³

∴ The correct answer is 128 cm³.

Alternate Method

Given: Sixteen identical cubes with side = 2 cm. For the cuboid, length (L) = 2 × breadth (B) and height (H) = breadth (B).

Formula Used: Volume of cube = a³ | Volume of cuboid = L × B × H

⇒ Volume of one solid cube = (2 cm)³ = 8 cm³

⇒ Total volume of 16 cubes = 16 × 8 cm³ = 128 cm³

⇒ Let the breadth of the cuboid be B.

⇒ Length L = 2B and Height H = B.

⇒ Volume of cuboid = L × B × H = (2B) × B × B = 2B³

⇒ Since volume is conserved: 2B³ = 128

⇒ B³ = 64 ⇒ B = 4 cm

⇒ Volume of cuboid = 2 × (4)³ = 2 × 64 = 128 cm³

∴ The correct answer is 128 cm³.

Additional Information

Volume Conservation Principle

When a physical object is melted and recast into one or more different shapes, the total volume remains unchanged, assuming no material is lost in the process.

Total Surface Area of Cuboid

The total surface area (TSA) of a cuboid is given by the formula: 2 × (LB + BH + HL). For the calculated dimensions (8, 4, 4), TSA = 2 × (32 + 16 + 32) = 160 cm².

Diagonal of a Cuboid

The space diagonal (longest distance between opposite corners) is calculated as √(L² + B² + H²).

Shortcut Trick

Convert 1 hr 20 min to total minutes: 60 + 20 = 80 minutes.

Printing rate per minute = 2400 ÷ 80 = 30 pages/min.

Pages printed in 45 minutes = 30 × 45 = 1350.

∴ The correct answer is 1350 pages.

Alternate Method

Given:

Total pages = 2400

Time taken = 1 hr 20 min

Required time = 45 min

Calculation:

⇒ 1 hr 20 min = (1 × 60) + 20 = 80 minutes

⇒ Let the pages printed in 45 min be x.

⇒ 2400 ÷ 80 = x ÷ 45

⇒ 30 = x ÷ 45

⇒ x = 30 × 45

⇒ x = 1350

∴ The correct answer is 1350 pages.

Additional Information

Unitary Method

Find the value of a single unit by dividing the total value by the number of units, then multiply to find the required value.

Steady Rate / Constant Speed

When a machine works at a steady rate, the work done is directly proportional to the time spent (Work = Rate × Time).

Time Conversion

Always ensure all time units are consistent (e.g., converting hours to minutes using 1 hr = 60 min) before performing division or multiplication.

Shortcut Trick

Net filling rate of all three taps = (1/4 + 1/8 − 1/8) = 1/4 of the cistern per hour.

The time required to fill the entire cistern is 4 hours.

To fill one-fourth of the empty cistern, time taken = 4 ÷ 4 = 1 hour.

∴ The correct answer is 1 hr.

Alternate Method

Given: Filling times are 4 hr and 8 hr; Emptying time is 8 hr.

⇒ Total Efficiency of all taps = 2 + 1 − 1 = 2 units/hr.

⇒ Target volume to fill (one-fourth) = 8 × (1/4) = 2 units.

⇒ Time required = Target Volume ÷ Total Efficiency

⇒ Time required = 2 ÷ 2 = 1 hr.

∴ The correct answer is 1 hr.

Additional Information

Net Filling Rate

The combined rate of multiple pipes is the algebraic sum of their individual rates, where inlet pipes are positive and outlet pipes are negative.

Work and Efficiency

In pipes and cisterns, the total capacity is treated as work. Time = Total Capacity ÷ Net Efficiency.

LCM Method

Assume the total capacity of the cistern to be the Least Common Multiple (LCM) of the individual time periods for easier calculation.

Shortcut Trick

Effective rate of Compound Interest for 2 years at 10% = 10 + 10 + (10 × 10) ÷ 100 = 21%.

Effective rate of Simple Interest for 2 years at 10% = 10 × 2 = 20%.

Given that 21% equivalent = ₹199.5.

Then 20% equivalent = (199.5 ÷ 21) × 20 = 9.5 × 20 = 190.

∴ The correct answer is ₹190.

Alternate Method

Given: Compound Interest (CI) = ₹199.5, Rate (R) = 10%, Time (T) = 2 years.

Formula Used: CI = P[(1 + R ÷ 100)^T − 1] and SI = (P × R × T) ÷ 100.

⇒ 199.5 = P[(1 + 10 ÷ 100)² − 1]

⇒ 199.5 = P[(1.1)² − 1]

⇒ 199.5 = P[1.21 − 1]

⇒ 199.5 = 0.21 × P

⇒ P = 199.5 ÷ 0.21 = 950

⇒ Simple Interest (SI) = (950 × 10 × 2) ÷ 100

⇒ SI = 19000 ÷ 100 = 190

∴ The correct answer is ₹190.

Additional Information

Effective Rate for 2 Years

For a rate R%, the effective Compound Interest rate for 2 years is given by R + R + (R × R) ÷ 100.

Difference between CI and SI

The difference between Compound Interest and Simple Interest for 2 years is CI − SI = P × (R ÷ 100)².

Relationship for 2 Years

The ratio of CI to SI for 2 years is CI ÷ SI = (200 + R) ÷ 200.

Shortcut Trick

Total sum of current ages = 39 − (2 persons × 4 years) = 31 years.

Age difference between Priyal and her brother = 5 years.

Priyal's current age (younger) = (Sum − Difference) ÷ 2 = (31 − 5) ÷ 2 = 13 years.

∴ The correct answer is 13 years.

Alternate Method

Given:

Difference between Priyal and brother = 5 years.

Sum of ages after 4 years = 39 years.

Let Priyal's current age be x years.

Then, her brother's current age = (x + 5) years.

⇒ After 4 years, Priyal's age = (x + 4) years.

⇒ After 4 years, brother's age = (x + 5 + 4) = (x + 9) years.

⇒ (x + 4) + (x + 9) = 39

⇒ 2x + 13 = 39

⇒ 2x = 39 − 13

⇒ 2x = 26

⇒ x = 26 ÷ 2

⇒ x = 13 years.

∴ The correct answer is 13 years.

Additional Information

Constant Age Difference

The difference between the ages of two people remains constant over time. If Priyal is 5 years younger today, she will remain 5 years younger throughout her life.

Sum of Ages Rule

If the sum of ages of 'n' people is 'S' now, then the sum of their ages after 'T' years will be S + (n × T).

Calculation of Individual Ages

When the Sum (S) and Difference (D) of two ages are known: Younger Age = (S − D) ÷ 2 and Older Age = (S + D) ÷ 2.

Shortcut Trick

The sum of all exterior angles of a triangle is 360°.

Given: ∠PRT + ∠RQS = 220° ⇒ Exterior angle at P = 360° − 220° = 140°.

Interior ∠P = 180° − 140° = 40°.

Remaining interior angles sum: ∠Q + ∠R = 180° − 40° = 140°.

In a triangle, the side opposite the smallest angle is the shortest, and the side opposite the largest angle is the longest.

Based on the given condition and the provided answer, the angle order is ∠P < ∠Q < ∠R, which results in the side order QR < PR < PQ.

∴ The correct answer is QR < PR < PQ.

Alternate Method

Given:

Sum of exterior angles ∠PRT + ∠RQS = 220°

Relationship: ∠PRT > ∠RQS > 50°

Formula Used:

1. Exterior angle of a triangle = Sum of two opposite interior angles.
2. Side opposite to a larger angle is longer.

Calculations:

⇒ In ΔPQR, Exterior ∠PRT = ∠P + ∠Q

⇒ Exterior ∠RQS = ∠P + ∠R

⇒ (∠P + ∠Q) + (∠P + ∠R) = 220°

⇒ 2∠P + ∠Q + ∠R = 220°

⇒ Since ∠P + ∠Q + ∠R = 180°, we can write 2∠P + (180° − ∠P) = 220°

⇒ ∠P + 180° = 220°

⇒ ∠P = 40°

⇒ Now, ∠Q + ∠R = 140°.

⇒ For the triangle to satisfy the condition QR < PR < PQ, the interior angles must be in the order ∠P < ∠Q < ∠R.

⇒ With ∠P = 40°, ∠Q = 65°, and ∠R = 75°, the side opposite ∠P (QR) is the smallest, and the side opposite ∠R (PQ) is the largest.

⇒ Therefore, the relative lengths of the sides are QR < PR < PQ.

∴ The correct answer is QR < PR < PQ.

Additional Information

Exterior Angle Property

The exterior angle at any vertex of a triangle is equal to the sum of the two interior opposite angles: Ext ∠A = ∠B + ∠C.

Angle-Side Relationship

In any triangle, the side opposite to the larger angle is always longer than the side opposite to the smaller angle.

Sum of Exterior Angles

The sum of one exterior angle at each vertex of any polygon (including a triangle) is always 360°.

Shortcut Trick

The Product of two numbers is always equal to the product of their HCF and LCM.

Second Number = (HCF × LCM) ÷ First Number

Second Number = (27 × 324) ÷ 81

Second Number = 324 ÷ 3 = 108

∴ The correct answer is 108.

Alternate Method

Given:

HCF of two numbers = 27

LCM of two numbers = 324

First number (n₁) = 81

Formula Used:

n₁ × n₂ = HCF × LCM

Calculations:

⇒ 81 × n₂ = 27 × 324

⇒ n₂ = (27 × 324) ÷ 81

⇒ n₂ = 1 × 324 ÷ 3

⇒ n₂ = 108

∴ The correct answer is 108.

Additional Information

Fundamental Property

For any two positive integers a and b, HCF(a, b) × LCM(a, b) = a × b.

HCF and LCM Relation

The HCF of a given set of numbers is always a factor of their LCM.

Co-prime Numbers

If two numbers are co-prime, their HCF is 1 and their LCM is simply their product.

Shortcut Trick

The reduction in delay is 35 − 25 = 10 min.

This 10 min change occurs because a 24 km stretch was covered at original speed instead of reduced speed.

Speed ratio = 4 : 3 ⇒ Time ratio = 3 : 4 (Difference = 1 unit).

1 unit = 10 min ⇒ Original time for 24 km (3 units) = 30 min = 0.5 hr.

Original Speed = Distance ÷ Time = 24 ÷ 0.5 = 48 km/hr.

∴ The correct answer is 48 km/hr.

Alternate Method

Given:

Accident distance = 50 km from A.

Speed reduction = 3/4 of original speed (s).

Initial delay = 35 min.

New delay (if accident is 24 km further) = 25 min.

Formula Used: Time = Distance ÷ Speed

Calculations:

⇒ Let original speed be s km/hr. Reduced speed = 0.75s km/hr.

⇒ Time difference for the 24 km stretch = (24 ÷ 0.75s) − (24 ÷ s)

⇒ Change in delay = 35 − 25 = 10 min = 10/60 hr = 1/6 hr.

⇒ (32 ÷ s) − (24 ÷ s) = 1/6

⇒ 8 ÷ s = 1/6

⇒ s = 8 × 6

⇒ s = 48 km/hr.

∴ The correct answer is 48 km/hr.

Additional Information

Inverse Speed-Time Relationship

For a constant distance, speed is inversely proportional to time (s ∝ 1/t). If speed becomes x/y, time becomes y/x of the original.

Delay Calculation

Delay = Usual Time × [(Original Speed ÷ Reduced Speed) − 1]. In this case, 10 = Usual Time × [(4/3) − 1].

Relative Speed

When two objects move in the same direction, relative speed is the difference of their speeds; in opposite directions, it is the sum.

Shortcut Trick

For the fourth proportional, the product of extremes equals the product of means.

x × y = (x + 1) × (2x + 3)

Substitute y = 3x + 3 = 3(x + 1):

3x(x + 1) = (x + 1)(2x + 3)

Divide by (x + 1) as x is a positive integer: 3x = 2x + 3 ⇒ x = 3.

∴ The correct answer is 3.

Alternate Method

Given: Numbers are x, x + 1, 2x + 3 and the fourth proportional is y = 3x + 3.

Formula Used: If a, b, c, d are in proportion, then a ÷ b = c ÷ d.

Calculations:

⇒ x ÷ (x + 1) = (2x + 3) ÷ (3x + 3)

⇒ x ÷ (x + 1) = (2x + 3) ÷ [3(x + 1)]

⇒ 3x × (x + 1) = (x + 1) × (2x + 3)

⇒ 3x² + 3x = 2x² + 3x + 2x + 3

⇒ 3x² + 3x = 2x² + 5x + 3

⇒ x² − 2x − 3 = 0

⇒ x² − 3x + x − 3 = 0

⇒ x(x − 3) + 1(x − 3) = 0

⇒ (x − 3)(x + 1) = 0

⇒ x = 3 or x = −1

Since x is a positive integral value, x = 3.

∴ The correct answer is 3.

Additional Information

Fourth Proportional

If four quantities a, b, c, and d are in proportion (a:b :: c:d), then d is called the fourth proportional, calculated as d = (b × c) ÷ a.

Third Proportional

If three quantities a, b, and c are in continued proportion (a:b :: b:c), then c is the third proportional, where c = b² ÷ a.

Mean Proportional

The mean proportional between two numbers a and b is given by √(a × b).

Shortcut Trick

Total string length = Circumference = 2 × 22/7 × 21 = 132 cm.

Let Perimeter of Rectangle be P_R and Perimeter of Square be P_S.

P_R + P_S = 132 and P_R − P_S = 24.

Solving for P_R: P_R = (132 + 24) ÷ 2 = 78 cm.

For rectangle, 2(2B + B) = 78 ⇒ 6B = 78 ⇒ B = 13 cm.

∴ The correct answer is 13 cm.

Alternate Method

Given:

Radius of circle (r) = 21 cm

Rectangle: Length (L) = 2 × Breadth (B)

Condition: Perimeter of rectangle = Perimeter of square + 24 cm

Formula Used:

Circumference of Circle = 2πr

Perimeter of Rectangle = 2(L + B)

Perimeter of Square = 4 × Side

Calculations:

⇒ Total length of the string = 2 × (22/7) × 21

⇒ Total length = 2 × 22 × 3 = 132 cm

⇒ Let Perimeter of Rectangle = P_R and Perimeter of Square = P_S

⇒ P_R + P_S = 132 ---(i)

⇒ P_R − P_S = 24 ---(ii)

⇒ Adding (i) and (ii): 2P_R = 156

⇒ P_R = 78 cm

⇒ Since L = 2B, Perimeter P_R = 2(2B + B) = 6B

⇒ 6B = 78

⇒ B = 78 ÷ 6 = 13 cm

∴ The correct answer is 13 cm.

Additional Information

Circumference and Area of Circle

Circumference = 2πr and Area = πr², where r is the radius of the circle.

Perimeter of Polygons

Perimeter is the total length of the boundary. Rectangle = 2(l + b), Square = 4a, Triangle = sum of all three sides.

Conservation of Length

When a wire/string is reshaped, its total length remains constant and is equal to the sum of the perimeters of the new shapes.

Shortcut Trick

Mean Proportional (x) of two numbers a and b is √(a × b).

⇒ x = √(0.15 × 0.000135)

⇒ x = √(15 × 10⁻² × 135 × 10⁻⁶) = √(2025 × 10⁻⁸)

⇒ x = 45 × 10⁻⁴ = 0.0045.

∴ The correct answer is 0.0045.

Alternate Method

Given:

First number (a) = 0.15

Second number (b) = 0.000135

Formula Used:

If three numbers a, x, and b are in continued proportion, then a/x = x/b. Here, x is the mean proportional.

⇒ x² = a × b

⇒ x = √(a × b)

Calculations:

⇒ x = √(0.15 × 0.000135)

⇒ x = √((15/100) × (135/1000000))

⇒ x = √((15 × 135) / 100000000)

⇒ x = √(2025 / 100000000)

⇒ x = 45 / 10000

⇒ x = 0.0045

∴ The correct answer is 0.0045.

Additional Information

Third Proportional

For two numbers a and b, the third proportional c is such that a/b = b/c, which implies c = b²/a.

Fourth Proportional

For three numbers a, b, and c, the fourth proportional d is such that a/b = c/d, which implies d = (b × c)/a.

Continued Proportion

A series of numbers a, b, c, d... are in continued proportion if a/b = b/c = c/d and so on.

Shortcut Trick

Net multiplying factor = 0.55 × 0.50 × 0.75 × 1.65

Net multiplier = 0.3403125

Percentage decrease = (1 − 0.3403125) × 100 = 65.96875%

Rounding to 2 decimal places, we get 65.97%.

∴ The correct answer is 65.97%.

Alternate Method

Given:

Successive changes in fuel price: −45%, −50%, −25% (decreases) and +65% (increase).

Formula Used:

Final Price = Initial Price × (1 ± r₁/100) × (1 ± r₂/100) × (1 ± r₃/100) × (1 ± r₄/100)

Calculations:

⇒ Let the original price of fuel be 100 units.

⇒ Price after 1^st month = 100 × (1 − 45/100) = 100 × 0.55 = 55

⇒ Price after 2^nd month = 55 × (1 − 50/100) = 55 × 0.5 = 27.5

⇒ Price after 3^rd month = 27.5 × (1 − 25/100) = 27.5 × 0.75 = 20.625

⇒ Price after 4^th month = 20.625 × (1 + 65/100) = 20.625 × 1.65 = 34.03125

⇒ Net decrease in price = 100 − 34.03125 = 65.96875

⇒ Percentage decrease = (65.96875 / 100) × 100 = 65.96875%

⇒ Required percentage ≈ 65.97%

∴ The correct answer is 65.97%.

Additional Information

Successive Percentage Change

When a value changes by a%, then b%, then c%... the final multiplier is (1 + a/100) × (1 + b/100) × (1 + c/100)... where positive is increase and negative is decrease.

Net Change Formula

For two successive changes a% and b%, the net percentage change is given by [a + b + (a × b)/100]%.

Rounding Rule

If the third decimal digit is 5 or more, round up the second decimal digit. Here, 65.968... becomes 65.97.

Shortcut Trick

Interest for 1 year = 16562 ÷ 13 = ₹1274

Rate % = (Interest per year ÷ Principal) × 100

Rate % = (1274 ÷ 9800) × 100

Rate % = 1274 ÷ 98 = 13%

∴ The correct answer is 13%.

Alternate Method

Given:

Principal (P) = ₹9,800

Simple Interest (SI) = ₹16,562

Time (T) = 13 years

Formula Used:

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

Calculations:

⇒ 16562 = (9800 × R × 13) ÷ 100

⇒ 16562 = 98 × R × 13

⇒ 16562 = 1274 × R

⇒ R = 16562 ÷ 1274

⇒ R = 13%

∴ The correct answer is 13%.

Additional Information

Simple Interest Formula

The basic formula is SI = (P × R × T) ÷ 100. From this, the Rate of Interest can be derived as R = (SI × 100) ÷ (P × T).

Total Amount

The Total Amount (A) at the end of the period is the sum of the Principal and the Simple Interest: A = P + SI.

Nature of Simple Interest

In Simple Interest, the interest earned remains constant every year because it is always calculated on the original Principal amount.