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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.