SSC CGL Tier 2 Quant Previous Year Paper (Held On: 11 September 2019)

Given:

Expression = 8^2k + 5^2k

k = odd number

Formula used:

aⁿ + bⁿ is divisible by (a + b) when n is odd

Calculation:

$8^{2k} + 5^{2k} = (8^2)^k + (5^2)^k$

⇒ $64^k + 25^k$

⇒ Factor = 64 + 25

⇒ Factor = 89

∴ The correct answer is option (2).

Shortcut Trick

Substitute θ = 60°, so tan 60° = √3 and cot 60° = 1/√3.

Expression = ((1 − √3) / (1 − 1/√3))² + 1

= ((1 − √3) / ((√3 − 1)/√3))² + 1 = (−√3)² + 1 = 3 + 1 = 4.

Checking options: sec² 60° = (2)² = 4. Other options do not match.

∴ The correct answer is sec² θ.

Alternate Method

Given: ((1 − tanθ) / (1 − cotθ))² + 1

Formula used: tanθ = sinθ/cosθ, cotθ = cosθ/sinθ, 1 + tan²θ = sec²θ

⇒ ((1 − sinθ/cosθ) / (1 − cosθ/sinθ))² + 1

⇒ (((cosθ − sinθ) / cosθ) / ((sinθ − cosθ) / sinθ))² + 1

⇒ ((cosθ − sinθ) / cosθ × sinθ / (sinθ − cosθ))² + 1

⇒ (−sinθ / cosθ)² + 1

⇒ (−tanθ)² + 1

⇒ tan²θ + 1

⇒ sec²θ

∴ The correct answer is sec² θ.

Additional Information

Pythagorean Identities

Fundamental relations in trigonometry: sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ, and 1 + cot²θ = cosec²θ.

Reciprocal Relations

Basic reciprocal properties: tanθ = 1/cotθ, sinθ = 1/cosecθ, and cosθ = 1/secθ.

Ratio Formulas

Expressed in terms of sine and cosine: tanθ = sinθ/cosθ and cotθ = cosθ/sinθ.

Shortcut Trick

Step 1: Calculate the Cost Price (CP) of the mixture from Selling Price (SP) and Profit%.

CP of mixture = (SP × 100) / (100 + Profit%) = (56 × 100) / 112 = ₹50/kg.

Step 2: Apply Alligation Rule: Ratio = (Mean CP − Lower CP) : (Higher CP − Mean CP).

Ratio = (50 − 42) : (60 − 50) = 8 : 10 = 4 : 5.

∴ The correct answer is 4 : 5.

Alternate Method

Given:

Cost of sugar 1 (CP₁) = ₹60/kg

Cost of sugar 2 (CP₂) = ₹42/kg

Selling price of mixture (SP_m) = ₹56/kg

Profit percentage = 12%

Formula Used:

Cost Price (CP) = SP × 100 / (100 + Gain%)

Alligation Rule: (Quantity of Cheaper) / (Quantity of Dearer) = (CP of Dearer − Mean CP) / (Mean CP − CP of Cheaper)

Calculations:

⇒ Mean Cost Price = 56 × 100 / (100 + 12)

⇒ Mean Cost Price = 5600 / 112 = ₹50/kg

⇒ Applying Alligation: (Sugar 1 Quantity) : (Sugar 2 Quantity)

⇒ Ratio = (Mean CP − CP₂) : (CP₁ − Mean CP)

⇒ Ratio = (50 − 42) : (60 − 50)

⇒ Ratio = 8 : 10

⇒ Ratio = 4 : 5

∴ The correct answer is 4 : 5.

Additional Information

Rule of Alligation

It is used to find the ratio in which two or more ingredients at their respective prices must be mixed to produce a mixture of a desired price. Ratio = (Price of dearer − Mean price) : (Mean price − Price of cheaper).

Cost Price vs Selling Price

Always ensure all values used in the alligation cross are of the same type (all Cost Prices or all Selling Prices). If a profit is given on the mixture, convert SP to CP first.

Applications of Alligation

This method can be applied to mixtures of liquids, averages of marks, speed-time-distance problems, and calculating interest rates on parts of a principal.

Given:

$\frac{7 + 8 \times 8 \div 8;of;8 + 8 \div 8 \times 4;of;4}{4 \div 4;of;4 + 4 \times 4 \div 4 - 4 \div 4;of;2}$

Formula used:

VBODMAS Rule

Calculation:

Numerator = 7 + 8 × 8 ÷ (8 of 8) + 8 ÷ 8 × (4 of 4)

⇒ Numerator = 7 + 8 × 8 ÷ 64 + 8 ÷ 8 × 16

⇒ Numerator = 7 + 64 ÷ 64 + 1 × 16

⇒ Numerator = 7 + 1 + 16 = 24

Denominator = 4 ÷ (4 of 4) + 4 × 4 ÷ 4 - 4 ÷ (4 of 2)

⇒ Denominator = 4 ÷ 16 + 16 ÷ 4 - 4 ÷ 8

⇒ Denominator = 0.25 + 4 - 0.5 = 3.75

Value = 24 ÷ 3.75

⇒ Value = 6.4

∴ The correct answer is option (2).

Shortcut Trick Ratio of number to sum of digits is 7 : 1 ⇒ 10x + y = 7(x + y) ⇒ 3x = 6y ⇒ x = 2y. Given that units digit (y) is 4 less than tens digit (x), so x − y = 4. Substituting x = 2y: 2y − y = 4 ⇒ y = 4 and x = 8. The number is 84. Reversing 84 gives 48. Now divide 48 by 7: 48 = (7 × 6) + 6. ∴ The correct answer is 6. Alternate Method Given: Sum of digits = 1/7 × (Two-digit number) Units digit = Tens digit − 4 Formula Used: A two-digit number with tens digit x and units digit y is expressed as 10x + y.

⇒ Let the tens digit be x and the units digit be y. ⇒ According to the first condition: x + y = (10x + y) ÷ 7 ⇒ 7(x + y) = 10x + y ⇒ 7x + 7y = 10x + y ⇒ 3x = 6y ⇒ x = 2y ... (Eq. 1) ⇒ According to the second condition: y = x − 4 ⇒ x − y = 4 ... (Eq. 2) ⇒ Substitute Eq. 1 into Eq. 2: 2y − y = 4 ⇒ y = 4 ⇒ x = 2 × 4 = 8 ⇒ The original number = 10(8) + 4 = 84 ⇒ The reversed number = 10(4) + 8 = 48 ⇒ Dividing reversed number by 7: 48 ÷ 7 = 6 with a remainder of 6. ∴ The correct answer is 6. Additional Information Two-Digit Number Properties A 2-digit number 10x+y and its reverse 10y+x always have a difference that is a multiple of 9, i.e., |(10x+y) − (10y+x)| = 9|x−y|. Divisibility and Remainders For any number N divided by d, N = dq + r, where q is the quotient and 0 ≤ r < d is the remainder. Sum of Digits Rule If a number is a multiple of its digit sum, such as 84 being 7 × 12, it often relates to divisibility rules of 3 or 9.

Shortcut Trick

According to the Angle Bisector Theorem, the ratio of the segments formed on the opposite side is equal to the ratio of the other two sides.

Ratio BD : DC = AB : AC = 7 : 8

Total ratio parts = 7 + 8 = 15 units

Given total length BC = 10 cm

BD = (7 ÷ 15) × 10 = 14/3 cm

∴ The correct answer is 14/3 cm.

Alternate Method

Given: AB = 7 cm, AC = 8 cm, BC = 10 cm. AD is the internal bisector of ∠BAC.

Formula Used: In ΔABC, if AD bisects ∠A, then BD / DC = AB / AC.

Let BD = x cm. Then DC = (10 − x) cm.

⇒ BD / DC = AB / AC

⇒ x / (10 − x) = 7 / 8

⇒ 8 × x = 7 × (10 − x)

⇒ 8x = 70 − 7x

⇒ 8x + 7x = 70

⇒ 15x = 70

⇒ x = 70 ÷ 15

⇒ x = 14/3 cm

∴ The correct answer is 14/3 cm.

Additional Information

Internal Angle Bisector Theorem

The bisector of an interior angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. BD/DC = AB/AC.

External Angle Bisector Theorem

The bisector of an exterior angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle. BE/CE = AB/AC.

Length of the Bisector

The length of the internal angle bisector AD can be calculated using the formula AD² = AB × AC − BD × DC.

Incenter of a Triangle

The intersection point of all three internal angle bisectors is called the incenter, which is the center of the circle inscribed in the triangle.

Shortcut Trick

Let Income = 100 units ⇒ Expenditure = 80, Savings = 20.

Change in Income = +12% of 100 = +12.

Change in Savings = −10% of 20 = −2.

Change in Expenditure = Income Change − Savings Change = 12 − (−2) = 14.

% Increase in Expenditure = (14 ÷ 80) × 100 = 17.5%.

∴ The correct answer is 17.5%.

Alternate Method

Given:

Expenditure = 80% of Income

Income increase = 12%

Savings decrease = 10%

Formula Used:

Income = Expenditure + Savings

Percentage Increase = (Change ÷ Original Value) × 100

⇒ Let original Income = ₹100.

⇒ Original Expenditure = 80% of 100 = ₹80.

⇒ Original Savings = 100 − 80 = ₹20.

⇒ New Income = 100 + 12% of 100 = ₹112.

⇒ New Savings = 20 − 10% of 20 = 20 − 2 = ₹18.

⇒ New Expenditure = New Income − New Savings = 112 − 18 = ₹94.

⇒ Increase in Expenditure = 94 − 80 = ₹14.

⇒ Percentage increase in Expenditure = (14 ÷ 80) × 100

⇒ Percentage increase = (7 ÷ 40) × 100 = 17.5%.

∴ The correct answer is 17.5%.

Additional Information

Income Expenditure Identity

Income is always the sum of Expenditure and Savings. Change in Income (ΔI) = Change in Expenditure (ΔE) + Change in Savings (ΔS).

Percentage Change Formula

% Change = [(Final Value − Initial Value) ÷ Initial Value] × 100. Always use the original value as the base for calculations.

Alligation Method

These problems can be solved using alligation where the ratio of Expenditure to Savings acts as weights for their respective percentage changes to find the net income change.

Shortcut Trick

In a cyclic quadrilateral, if one diagonal bisects the other, the product of the adjacent sides of the bisecting diagonal is equal.

If PR bisects QS, then the property is: PQ × QR = PS × SR

Given: PQ = 14.4 cm, QR = 12.8 cm, SR = 9.6 cm

⇒ 14.4 × 12.8 = PS × 9.6

⇒ PS = (14.4 × 12.8) ÷ 9.6 = 1.5 × 12.8 = 19.2 cm

∴ The correct answer is 19.2 cm.

Alternate Method

Given:

PQ = 14.4 cm, QR = 12.8 cm, SR = 9.6 cm

Diagonal PR bisects diagonal QS at point O (QO = OS).

Formula Used:

In a cyclic quadrilateral PQRS, if diagonal PR bisects QS, then Area(ΔPQR) = Area(ΔPSR).

⇒ 1/2 × PQ × QR × sin(Q) = 1/2 × PS × SR × sin(S)

Since ∠Q + ∠S = 180° (Cyclic quadrilateral property), sin(Q) = sin(S).

⇒ PQ × QR = PS × SR

Calculations:

⇒ 14.4 × 12.8 = PS × 9.6

⇒ PS = (14.4 × 12.8) ÷ 9.6

⇒ PS = 184.32 ÷ 9.6

⇒ PS = 19.2 cm

∴ The correct answer is 19.2 cm.

Additional Information

Cyclic Quadrilateral Definition

A quadrilateral whose all four vertices lie on the circumference of a circle. Sum of opposite angles is always 180° (∠P + ∠R = 180°).

Ptolemy's Theorem

In a cyclic quadrilateral, the product of the diagonals is equal to the sum of the products of opposite sides: (PR × QS) = (PQ × RS) + (PS × QR).

Area of Cyclic Quadrilateral

Area = √((s−a)(s−b)(s−c)(s−d)), where s is the semi-perimeter (a+b+c+d)/2. This is known as Brahmagupta's Formula.

Shortcut Trick

Total parts in A = 3 + 4 = 7; Total parts in B = 5 + 9 = 14.

Equalize parts by multiplying Alloy A ratio by 2: A → 6 : 8 (Total 14).

Now multiply by mixing ratio 2 : 3:

A → (6 × 2) : (8 × 2) = 12 : 16; B → (5 × 3) : (9 × 3) = 15 : 27.

New Copper = 12 + 15 = 27; New Zinc = 16 + 27 = 43.

∴ The correct answer is 27 : 43.

Alternate Method

Given:

Ratio of Cu:Zn in A = 3 : 4

Ratio of Cu:Zn in B = 5 : 9

Mixing ratio A : B = 2 : 3

Calculations:

⇒ Fraction of Copper in A = 3/7 and Zinc in A = 4/7

⇒ Fraction of Copper in B = 5/14 and Zinc in B = 9/14

⇒ Total Copper in C = (2 × 3/7) + (3 × 5/14)

⇒ Total Copper in C = 6/7 + 15/14 = (12 + 15)/14 = 27/14

⇒ Total Zinc in C = (2 × 4/7) + (3 × 9/14)

⇒ Total Zinc in C = 8/7 + 27/14 = (16 + 27)/14 = 43/14

⇒ Required Ratio = (27/14) : (43/14) = 27 : 43

∴ The correct answer is 27 : 43.

Additional Information

Alligation Rule for Mixtures

Quantity of cheaper / Quantity of dearer = (Rate of dearer − Mean rate) / (Mean rate − Rate of cheaper).

Weighted Average Formula

Average = (n₁x₁ + n₂x₂) / (n₁ + n₂), where n represents quantities and x represents values/ratios.

Ratio Balancing

To combine ratios of different totals, find the LCM of the sums of the parts and multiply each ratio accordingly to equalize the base.

Shortcut Trick

The difference between CI of year n and (n+1) is simply the interest on the CI of year n.

CI for 1^st year = 10% of 18,000 = ₹1,800

CI for 2^nd year = 1,800 + 10% of 1,800 = ₹1,980

CI for 3^rd year = 1,980 + 10% of 1,980 = ₹2,178

Difference (4^th year − 3^rd year) = 10% of CI for 3^rd year = 10% of 2,178 = ₹217.80

∴ The correct answer is ₹217.80.

Alternate Method

Given: Principal (P) = ₹18,000 | Rate (R) = 10% p.a. | Compounding: Annually

Formula Used: CI for n^th year = P × (1 + R/100)ⁿ⁻¹ × (R/100)

⇒ CI for 3^rd year = 18000 × (1 + 10/100)⁽³⁻¹⁾ × (10/100)

⇒ CI for 3^rd year = 18000 × (1.1)² × 0.1

⇒ CI for 3^rd year = 18000 × 1.21 × 0.1 = ₹2,178

⇒ CI for 4^th year = CI for 3^rd year × (1 + R/100)

⇒ CI for 4^th year = 2178 × 1.1 = ₹2,395.80

⇒ Difference = CI for 4^th year − CI for 3^rd year

⇒ Difference = 2,395.80 − 2,178 = ₹217.80

∴ The correct answer is ₹217.80.

Additional Information

CI for nth year vs Amount

Compound Interest for a specific n^th year is given by CI_n = Amount_n − Amount_n−1 = P × (1 + r/100)ⁿ⁻¹ × (r/100).

Successive Year Relationship

In annual compounding, the interest for any year is always (1 + r/100) times the interest of the previous year.

Difference between CI and SI

For 2 years: D = P(R/100)². For 3 years: D = P(R/100)² × (3 + R/100).

Shortcut Trick

For two objects starting from opposite ends and moving towards each other, if they meet and then take times T₁ and T₂ to reach their respective destinations, the ratio of their speeds is S₁/S₂ = √(T₂/T₁).

Given: T₁ = 6 (1/8) h = 49/8 h, T₂ = 8 h, and S₂ (Speed of B) = 28 km/h.

Ratio of speeds S_A/S_B = √(8 / (49/8)) = √(64/49) = 8/7.

Speed of A = (8/7) × Speed of B = (8/7) × 28 = 8 × 4 = 32 km/h.

∴ The correct answer is 32 km/h.

Alternate Method

Given:

Speed of B (S_B) = 28 km/h

Time taken by A after crossing (T_A) = 6 (1/8) = 49/8 hours

Time taken by B after crossing (T_B) = 8 hours

Formula:

S_A / S_B = √(T_B / T_A)

Calculations:

⇒ S_A / 28 = √(8 / (49/8))

⇒ S_A / 28 = √(8 × 8 / 49)

⇒ S_A / 28 = √(64 / 49)

⇒ S_A / 28 = 8 / 7

⇒ S_A = (8 / 7) × 28

⇒ S_A = 8 × 4

⇒ S_A = 32 km/h

∴ The correct answer is 32 km/h.

Additional Information

Meeting Time Formula

In cases where two objects start simultaneously and meet, then take T₁ and T₂ times to reach the ends, the time from the start until they meet is given by T = √(T₁ × T₂).

Relative Speed

When two bodies move towards each other, their relative speed is the sum of their individual speeds (S₁ + S₂).

Speed-Time Relation

For a fixed distance, speed is inversely proportional to time (S ∝ 1/T), but in these specific meeting-point problems, the square root relationship is derived from equating the time taken to reach the meeting point.

Shortcut Trick Given: cos2θ − sin2θ = 1/2 Use Identity: cos 2θ = cos2θ − sin2θ ∴ cos 2θ = 1/2 ⇒ 2θ = 60° (θ = 30°) Value = tan2(60°) + sin2(3 × 30°) Value = (√3)2 + (1)2 = 3 + 1 = 4 ∴ The correct answer is 4. Alternate Method Given: cos2θ − sin2θ = 1/2 θ is in the first quadrant (0° < θ < 90°) Formula Used:

1. cos 2θ = cos2θ − sin2θ
2. tan 60° = √3
3. sin 90° = 1

Calculations: ⇒ cos2θ − sin2θ = 1/2 ⇒ cos 2θ = 1/2 ⇒ 2θ = cos−1(1/2) ⇒ 2θ = 60° ⇒ θ = 30° Now, substitute the values in the required expression: ⇒ tan22θ + sin23θ ⇒ tan2(60°) + sin2(3 × 30°) ⇒ tan2(60°) + sin2(90°) ⇒ (√3)2 + (1)2 ⇒ 3 + 1 = 4 ∴ The correct answer is 4. Additional Information Double Angle Identities cos 2θ can also be written as 2cos2θ − 1 or 1 − 2sin2θ. Similarly, sin 2θ = 2sinθcosθ. Pythagorean Identity sin2θ + cos2θ = 1. This can be used to convert cos2θ − sin2θ into terms of only sin or cos. Common Trig Values For 30°: sin = 1/2, cos = √3/2, tan = 1/√3. For 60°: sin = √3/2, cos = 1/2, tan = √3.

Shortcut Trick

Loss of 29% means SP₁ is 71% of CP.

Desired Gain of 21% means SP₂ must be 121% of CP.

71% → ₹355

1% → 355 ÷ 71 = 5

121% → 5 × 121 = 605

∴ The correct answer is ₹605.

Alternate Method

Given:

Selling Price (SP₁) = ₹355

Loss% = 29%

Desired Gain% = 21%

Formula Used:

CP = [SP × 100] ÷ [100 − Loss%]

New SP = [CP × (100 + Gain%)] ÷ 100

Calculations:

⇒ First, calculate Cost Price (CP):

⇒ CP = [355 × 100] ÷ [100 − 29]

⇒ CP = [355 × 100] ÷ 71

⇒ CP = 5 × 100 = ₹500

⇒ Now, calculate required Selling Price (SP₂) for 21% gain:

⇒ SP₂ = [500 × (100 + 21)] ÷ 100

⇒ SP₂ = [500 × 121] ÷ 100

⇒ SP₂ = 5 × 121 = ₹605

∴ The correct answer is ₹605.

Additional Information

Direct SP Ratio Relation

When CP is constant, SP₁ ÷ (100 ± P/L₁%) = SP₂ ÷ (100 ± P/L₂%). This allows solving without calculating CP.

Profit and Loss Percentage

Profit% = (SP − CP) ÷ CP × 100 and Loss% = (CP − SP) ÷ CP × 100. Both are always calculated on CP unless stated otherwise.

Marked Price and Discount

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

Shortcut Trick

Find (xy + yz + zx) = [(x + y + z)² − (x² + y² + z²)] ÷ 2 = [121 − 133] ÷ 2 = −6.

Find 3xyz = (x³ + y³ + z³) − (x + y + z)[(x² + y² + z²) − (xy + yz + zx)]

3xyz = 881 − 11[133 − (−6)] = 881 − 11(139) = 881 − 1529 = −648.

xyz = −648 ÷ 3 = −216.

∛(xyz) = ∛(−216) = −6.

∴ The correct answer is -6.

Alternate Method

Given:

x + y + z = 11

x² + y² + z² = 133

x³ + y³ + z³ = 881

Formula Used:

1. (x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

2. x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − (xy + yz + zx))

Calculations:

⇒ (11)² = 133 + 2(xy + yz + zx)

⇒ 121 − 133 = 2(xy + yz + zx)

⇒ −12 = 2(xy + yz + zx)

⇒ xy + yz + zx = −6

⇒ 881 − 3xyz = 11 × [133 − (−6)]

⇒ 881 − 3xyz = 11 × 139

⇒ 881 − 3xyz = 1529

⇒ −3xyz = 1529 − 881

⇒ −3xyz = 648

⇒ xyz = −216

⇒ ∛(xyz) = ∛(−216) = −6

∴ The correct answer is -6.

Additional Information

Sum of Cubes Identity

The identity x³ + y³ + z³ − 3xyz = (x + y + z)(x² + y² + z² − xy − yz − zx) is fundamental in algebra. If x + y + z = 0, then x³ + y³ + z³ = 3xyz.

Alternative Form of Identity

The term (x² + y² + z² − xy − yz − zx) can also be written as ½[(x−y)² + (y−z)² + (z−x)²].

Square of Trinomial

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx) helps in finding the sum of products taken two at a time when the sum of squares and sum of variables are known.

Shortcut Trick

For a maximum volume cylinder cut from a cube, Diameter = Height = Side of cube.

Ratio of Volume of Cube : Volume of Cylinder = 1 : π/4 ≈ 1 : 11/14 = 14 : 11.

Let Volume of Cube = 14 units, then Volume of Cylinder = 11 units.

Material left = 14 − 11 = 3 units.

Required percentage = (3 ÷ 14) × 100 ≈ 21.42%.

∴ The correct answer is 21%.

Alternate Method

Given: A cube of side 'a' from which a maximum cylinder is cut.

For maximum volume of the cylinder:

⇒ Radius (r) = Side ÷ 2 = a/2

⇒ Height (h) = Side = a

⇒ Volume of Cube = Side³ = a³

⇒ Volume of Cylinder = πr²h = π × (a/2)² × a = (πa³) ÷ 4

⇒ Volume of material left = Volume of Cube − Volume of Cylinder

⇒ Left Volume = a³ − (πa³ ÷ 4) = a³(1 − π/4)

⇒ Percentage of material left = [a³(1 − π/4) ÷ a³] × 100

⇒ Percentage = (1 − 22/(7 × 4)) × 100 = (1 − 11/14) × 100

⇒ Percentage = (3/14) × 100 = 300 ÷ 14 ≈ 21.42%

⇒ Nearest integer = 21%

∴ The correct answer is 21%.

Additional Information

Sphere inside a Cube

If the largest possible sphere is cut from a cube of side 'a', its diameter is 'a'. Volume = (4/3)π(a/2)³ = πa³/6.

Cone inside a Cube

For a cone of maximum volume in a cube, radius r = a/2 and height h = a. Volume = (1/3)πr²h = πa³/12.

Percentage conversion

1/14 ≈ 7.14%, so 3/14 ≈ 3 × 7.14% = 21.42%.

Shortcut Trick

Volume of cone (V) ∝ r² × h

Given: r₂ = 2r₁ and V₂ = 3V₁

V₁ / V₂ = (r₁ / r₂)² × (h₁ / h₂)

1 / 3 = (1 / 2)² × (h₁ / h₂)

1 / 3 = 1 / 4 × (h₁ / h₂) ⇒ h₁ / h₂ = 4 / 3

∴ The correct answer is 4 : 3.

Alternate Method

Given:

Original radius = r, New radius = 2r

Original volume = V₁, New volume = V₂ = 3V₁

Formula Used:

Volume of a cone = (1/3) × π × r² × h

Calculations:

⇒ V₁ = (1/3) π r² h₁

⇒ V₂ = (1/3) π (2r)² h₂

⇒ V₂ = (1/3) π (4r²) h₂

⇒ Since V₂ = 3V₁

⇒ (1/3) π 4r² h₂ = 3 × [(1/3) π r² h₁]

⇒ 4r² h₂ = 3r² h₁

⇒ 4h₂ = 3h₁

⇒ h₁ / h₂ = 4 / 3

∴ Ratio h₁ : h₂ = 4 : 3

∴ The correct answer is 4 : 3.

Additional Information

Slant Height of Cone

The slant height (l) is given by the Pythagorean theorem: l = √(r² + h²).

Surface Area of Cone

Curved Surface Area = πrl and Total Surface Area = πr(l + r).

Volume Ratio Rule

For two similar solids, the ratio of their volumes is the cube of the ratio of their corresponding linear dimensions.

Shortcut Trick

When the remainder is the same, the greatest number x is the HCF of the differences between the numbers.

Difference 1 = 8110 − 7897 = 213

Difference 2 = 8536 − 8110 = 426

x = HCF(213, 426) = 213

Sum of digits of x = 2 + 1 + 3 = 6

∴ The correct answer is 6.

Alternate Method

Given:

Numbers: 7897, 8110, 8536

Formula used:

If numbers a, b, and c leave the same remainder when divided by x, then x = HCF(|a−b|, |b−c|, |c−a|).

⇒ Difference 1 = 8110 − 7897 = 213

⇒ Difference 2 = 8536 − 8110 = 426

⇒ Difference 3 = 8536 − 7897 = 639

⇒ x = HCF(213, 426, 639)

⇒ 213 = 213 × 1

⇒ 426 = 213 × 2

⇒ 639 = 213 × 3

⇒ x = 213

⇒ Sum of digits of x = 2 + 1 + 3 = 6

∴ The correct answer is 6.

Additional Information

HCF with Same Remainder

When the remainder is the same but unknown, calculate the HCF of the absolute differences between the given numbers: HCF(|a−b|, |b−c|, |c−a|).

HCF with Known Remainder

If dividing a, b, and c leaves remainder r, the greatest number is HCF(a−r, b−r, c−r).

LCM with Same Remainder

The smallest number which when divided by a, b, and c leaves a remainder r in each case is given by: LCM(a, b, c) + r.

Shortcut Trick

Assume total voters = 100. Then Males = 55 and Females = 45.

Literate Males = 60% of 55 = 33.

Illiterate Females = 60% of 45 = 27.

Comparison: Literate Males (33) are 6 more than Illiterate Females (27).

Percentage More = (6 ÷ 27) × 100 = 2/9 × 100 = 22 2/9%.

∴ The correct answer is 22 2/9%.

Alternate Method

Given:

Total Voters = 100

Number of Males = 55% of 100 = 55

Number of Females = 100 − 55 = 45

Formula Used:

Percentage More = (Difference ÷ Base Value) × 100

Calculations:

⇒ Total Males = 55

⇒ Male Literate % = 100% − 40% (Illiterate) = 60%

⇒ Literate Males = 60% of 55 = 0.60 × 55 = 33

⇒ Total Females = 45

⇒ Female Illiterate % = 100% − 40% (Literate) = 60%

⇒ Illiterate Females = 60% of 45 = 0.60 × 45 = 27

⇒ Required difference = 33 − 27 = 6

⇒ Required Percentage = (6 ÷ 27) × 100

⇒ Required Percentage = 2/9 × 100 = 200/9 = 22.22% or 22 2/9%

∴ The correct answer is 22 2/9%.

Additional Information

Percentage Comparison

To find how much % A is more than B, use formula: [(A − B) ÷ B] × 100. Always divide by the value after the word 'than'.

Fraction to Percentage Conversion

1/9 = 11.11% or 11 1/9%. Multiplying by 2, we get 2/9 = 22.22% or 22 2/9%.

Complementary Percentage

If 40% are illiterate, then the remaining (100% − 40%) = 60% must be literate, and vice versa.

Given:

Fractions a, b, and c where a < b < c

$\frac{c}{a} = \frac{5}{2}$

$\frac{c}{a} - b = \frac{7}{4}$

$a + b + c = 1\frac{11}{12} = \frac{23}{12}$

Formula used:

Algebraic Substitution

Calculation:

$\frac{5}{2} - b = \frac{7}{4}$

⇒ $b = \frac{5}{2} - \frac{7}{4}$

⇒ $b = \frac{10 - 7}{4} = \frac{3}{4}$

Substitute b and c = $\frac{5a}{2}$ into a + b + c = $\frac{23}{12}$

⇒ $a + \frac{3}{4} + \frac{5a}{2} = \frac{23}{12}$

⇒ $\frac{7a}{2} = \frac{23}{12} - \frac{3}{4}$

⇒ $\frac{7a}{2} = \frac{23 - 9}{12}$

⇒ $\frac{7a}{2} = \frac{14}{12} = \frac{7}{6}$

⇒ $a = \frac{7}{6} \times \frac{2}{7} = \frac{1}{3}$

⇒ $c = \frac{5}{2} \times \frac{1}{3} = \frac{5}{6}$

Calculation for (c - a):

⇒ $c - a = \frac{5}{6} - \frac{1}{3}$

⇒ $c - a = \frac{5 - 2}{6} = \frac{3}{6} = \frac{1}{2}$

∴ The correct answer is option (4).

Shortcut Trick

Net percentage change for Volume (r²h) using successive formula: x + y + xy/100.

Step 1: Net change in r² (two successive changes of −20%) = −20 − 20 + (−20 × −20)/100 = −36%.

Step 2: Combine r² change (−36%) with height change (+40%) = −36 + 40 + (−36 × 40)/100.

Step 3: 4 − 14.4 = −10.4%.

Negative sign indicates a decrease.

∴ The correct answer is 10.4% decrease.

Alternate Method

Given: Radius (r) decreases by 20% | Height (h) increases by 40%

Formula: Volume of Cylinder (V) = πr²h

⇒ Let initial Radius (r₁) = 5 units and initial Height (h₁) = 5 units.

⇒ Decreased Radius (r₂) = 5 − 20% of 5 = 4 units.

⇒ Increased Height (h₂) = 5 + 40% of 5 = 7 units.

⇒ V₁ = π × (5)² × 5 = 125π units³

⇒ V₂ = π × (4)² × 7 = 112π units³

⇒ Change in Volume = 112π − 125π = −13π

⇒ Percentage Change = (−13π / 125π) × 100

⇒ Percentage Change = −10.4%

∴ The correct answer is 10.4% decrease.

Additional Information

Volume of Right Circular Cylinder

The volume is the space occupied by the cylinder, calculated as Area of base × Height = πr²h.

Curved Surface Area (CSA)

The area of the curved part of the cylinder is 2πrh. Any change in r or h directly impacts CSA linearly.

Successive Percentage Rule

For a product of variables A × B × C, the net percentage change is found by applying the formula (x + y + xy/100) stepwise.

Shortcut Trick

Base Area of Trapezium = 1/2 × (11 + 15) × 9 = 13 × 9 = 117 cm²

Volume of Prism = Base Area × Height

Height = 1731.6 ÷ 117 = 14.8 cm

∴ The correct answer is 14.8 cm.

Alternate Method

Given:

Parallel sides of trapezium base (a) = 11 cm, (b) = 15 cm

Distance between parallel sides (h_b) = 9 cm

Volume of prism (V) = 1731.6 cm³

Formula Used:

Area of Trapezium = 1/2 × (Sum of parallel sides) × Distance between them

Volume of Prism = Area of Base × Height of Prism (H)

Calculations:

⇒ Area of the trapezium base = 1/2 × (11 + 15) × 9

⇒ Area of the trapezium base = 1/2 × 26 × 9

⇒ Area of the trapezium base = 13 × 9 = 117 cm²

⇒ Volume of the prism = Area of base × Height

⇒ 1731.6 = 117 × H

⇒ H = 1731.6 ÷ 117

⇒ H = 14.8 cm

∴ The correct answer is 14.8 cm.

Additional Information

Volume of a Right Prism

The volume is always calculated as the area of the base multiplied by the vertical height (V = Base Area × H).

Lateral Surface Area (LSA) of Prism

The LSA of a prism is given by the perimeter of the base multiplied by the height (LSA = Perimeter of base × H).

Total Surface Area (TSA) of Prism

The TSA includes the lateral area and the area of the two bases (TSA = LSA + 2 × Base Area).

Area of Trapezium

Area = 1/2 × (Sum of parallel sides) × h, where h is the perpendicular distance between parallel sides.

Shortcut Trick Units digit of 324 (Power 24 ÷ 4 leaves R=0) → 34 = 1 Units digit of 738 (Power 38 ÷ 4 leaves R=2) → 72 = 9 Units digit of 654 (Always 6 for digit 6) → 6 Expression: 1 − 9 + 6 = −2 Correct digit (add 10 to negative): 10 − 2 = 8 ∴ The correct answer is 8. Alternate Method Given: x = (633)24 − (277)38 + (266)54 The units digit of any number depends only on the units digit of its base. ⇒ Units digit of (633)24 = Units digit of 324 ⇒ Cyclicity of 3 is 4. Power 24 ÷ 4 = Remainder 0. When R=0, we use power 4. 34 = 81 → 1. ⇒ Units digit of (277)38 = Units digit of 738 ⇒ Cyclicity of 7 is 4. Power 38 ÷ 4 = Remainder 2. Units digit = 72 = 49 → 9. ⇒ Units digit of (266)54 = Units digit of 654. Power of 6 always ends in 6. ⇒ x = 1 − 9 + 6 ⇒ x = −2 ⇒ Since a digit cannot be negative in this context, add 10 to the result: 10 − 2 = 8 ∴ The correct answer is 8. Additional Information Cyclicity Rule of 4 The digits 2, 3, 7, and 8 have a cyclicity of 4. To find the units digit, divide the exponent by 4 and replace it with the remainder (if remainder is 0, use 4). Cyclicity of 0, 1, 5, 6 These digits always result in the same units digit regardless of the power (0n=0, 1n=1, 5n=5, 6n=6 for n > 0). Cyclicity of 4 and 9 These follow a cyclicity of 2. For 4, odd power ends in 4 and even power in 6. For 9, odd power ends in 9 and even power in 1.

Shortcut Trick

Interior angle = 128 4/7° = 900/7°

Exterior angle = 180° − 900/7° = 360/7°

Number of sides (n) = 360° ÷ Exterior angle = 360 ÷ (360/7) = 7

Number of diagonals = n(n − 3) ÷ 2 = 7 × 4 ÷ 2 = 14

Sum = Diagonals + Sides = 14 + 7 = 21

∴ The correct answer is 21.

Alternate Method

Given: Each interior angle = 128 4/7°

Formula Used:

1. Interior Angle + Exterior Angle = 180°
2. Number of sides (n) = 360° ÷ Exterior Angle
3. Number of diagonals = n(n − 3) ÷ 2

⇒ Interior angle = (128 × 7 + 4) / 7 = 900/7°

⇒ Exterior angle = 180° − 900/7°

⇒ Exterior angle = (1260 − 900) / 7 = 360/7°

⇒ Number of sides (n) = 360° ÷ (360/7°) = 7

⇒ Number of diagonals = 7 × (7 − 3) ÷ 2

⇒ Number of diagonals = 7 × 4 ÷ 2 = 14

⇒ Required Sum = Sides + Diagonals = 7 + 14 = 21

∴ The correct answer is 21.

Additional Information

Sum of Interior Angles

The sum of all interior angles of a polygon with n sides is given by (n − 2) × 180°.

Sum of Exterior Angles

For any convex polygon, the sum of all exterior angles (one at each vertex) is always 360°.

Relationship between Angles

In a regular polygon, each interior angle = [(n − 2) × 180°] / n and each exterior angle = 360° / n.

Polygon Diagonals

A triangle (n=3) has 0 diagonals. A quadrilateral (n=4) has 2 diagonals. The general formula is n(n−3)/2.

Shortcut Trick

Interest SI₁ = 11,046 − 8,400 = Rs. 2,646

Since Rate and Time are constant, Simple Interest is directly proportional to Principal (SI ∝ P).

Ratio of Principals = 8,400 : 9,600 = 7 : 8

New Interest SI₂ = (2,646 ÷ 7) × 8

SI₂ = 378 × 8 = Rs. 3,024

∴ The correct answer is Rs. 3,024.

Alternate Method

Given:

Principal (P₁) = Rs. 8,400, Amount (A₁) = Rs. 11,046, Rate (R) = 8.75%

Second Principal (P₂) = Rs. 9,600, Rate and Time are same as Case 1.

Formula Used:

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

Amount (A) = P + SI

Calculations:

⇒ SI₁ = A₁ − P₁ = 11,046 − 8,400 = 2,646

⇒ Time (T) = (SI₁ × 100) ÷ (P₁ × R)

⇒ T = (2,646 × 100) ÷ (8,400 × 8.75)

⇒ T = 2,646 ÷ (84 × 8.75)

⇒ New SI (SI₂) = (P₂ × R × T) ÷ 100

⇒ SI₂ = (9,600 × 8.75 × [2,646 ÷ (84 × 8.75)]) ÷ 100

⇒ SI₂ = (96 × 8.75 × 2,646) ÷ (84 × 8.75)

⇒ SI₂ = (96 × 2,646) ÷ 84

⇒ SI₂ = (8 × 2,646) ÷ 7

⇒ SI₂ = 8 × 378 = 3,024

∴ The correct answer is Rs. 3,024.

Additional Information

Simple Interest Proportionality

When the Rate (R) and Time (T) are constant, Simple Interest is directly proportional to the Principal (SI ∝ P). This allows for quick calculation using ratios: SI₁/P₁ = SI₂/P₂.

Effective Rate of Interest

The total interest earned can be expressed as an effective percentage of the principal, which is (Rate × Time)%. For Simple Interest, this percentage remains linear over time.

SI vs CI Difference

For a period of 1 year with annual compounding, SI and CI are equal. The difference for 2 years is given by P × (R/100)².

Shortcut Trick

For an equation of form a² + b² + c² + Constant = 2(xa + yb + zc), the values are a = x, b = y, and c = z.

Rewrite given: a² + b² + c² + 96 = 2(4a + 4b − 8c)

∴ a = 4, b = 4, and c = −8

Substitute in √(ab − bc + ca) ⇒ √((4 × 4) − (4 × −8) + (−8 × 4))

⇒ √(16 + 32 − 32) = √16 = 4

∴ The correct answer is 4.

Alternate Method

Given Equation: a² + b² + c² + 96 = 8(a + b − 2c)

Method: Completing the square

⇒ a² + b² + c² + 96 = 8a + 8b − 16c

⇒ (a² − 8a) + (b² − 8b) + (c² + 16c) + 96 = 0

⇒ (a² − 8a + 16) + (b² − 8b + 16) + (c² + 16c + 64) = 0

⇒ (a − 4)² + (b − 4)² + (c + 8)² = 0

Since the sum of squares of real numbers is zero, each term must be zero:

⇒ a − 4 = 0 ⇒ a = 4

⇒ b − 4 = 0 ⇒ b = 4

⇒ c + 8 = 0 ⇒ c = −8

Now, evaluate √(ab − bc + ca):

⇒ √(4 × 4 − 4 × (−8) + (−8) × 4)

⇒ √(16 + 32 − 32) = √16 = 4

∴ The correct answer is 4.

Additional Information

Sum of Squares Property

If x² + y² + z² = 0 for real numbers x, y, and z, then it must be that x = 0, y = 0, and z = 0.

Algebraic Identity

The square of a binomial is given by (x ± y)² = x² ± 2xy + y². This is used to complete the square in algebraic equations.

Comparison Method

In expressions like a² + b² + c² + k = 2(la + mb + nc), the values are directly found as a = l, b = m, and c = n if k = l² + m² + n².

Given:

$\left( {2\frac{6}{7} \text{ of } 4\frac{1}{5} \div \frac{2}{3}} \right) \times 1\frac{1}{9} \div \left( {\frac{3}{4} \times 2\frac{2}{3} \text{ of } \frac{1}{2} \div \frac{1}{4}} \right)$

Formula used:

BODMAS rule

Calculation:

⇒ $\left( {\frac{20}{7} \times \frac{21}{5} \div \frac{2}{3}} \right) \times \frac{10}{9} \div \left( {\frac{3}{4} \times \frac{8}{3} \times \frac{1}{2} \div \frac{1}{4}} \right)$

⇒ $\left( {12 \div \frac{2}{3}} \right) \times \frac{10}{9} \div \left( {\frac{3}{4} \times \frac{4}{3} \div \frac{1}{4}} \right)$

⇒ $\left( {12 \times \frac{3}{2}} \right) \times \frac{10}{9} \div \left( {1 \div \frac{1}{4}} \right)$

⇒ $18 \times \frac{10}{9} \div 4$

⇒ $20 \div 4 = 5$

∴ The correct answer is option (1).

Shortcut Trick

Convert terms using complementary angles: sin 81° = cos 9° and cosec 81° = sec 9°.

Numerator = (cos 9° + cos 9°) × (sec 9° + sec 9°) = 2 cos 9° × 2 sec 9° = 4(cos 9° × sec 9°) = 4.

Denominator = sin 56° sec 34° + cos 25° cosec 65° = 1 + 1 = 2 [since sin A sec(90°−A) = 1].

Value = 4 ÷ 2 = 2.

∴ The correct answer is 2.

Alternate Method

Given: expression = ((cos 9° + sin 81°)(sec 9° + cosec 81°)) ÷ (sin 56° sec 34° + cos 25° cosec 65°)

Formula Used: sin(90° − θ) = cos θ; cosec(90° − θ) = sec θ; cos θ × sec θ = 1; sin θ × cosec θ = 1

⇒ Numerator = (cos 9° + sin(90° − 9°)) × (sec 9° + cosec(90° − 9°))

⇒ Numerator = (cos 9° + cos 9°) × (sec 9° + sec 9°)

⇒ Numerator = 2 cos 9° × 2 sec 9° = 4(cos 9° × sec 9°) = 4 × 1 = 4

⇒ Denominator = sin 56° sec(90° − 56°) + cos 25° cosec(90° − 25°)

⇒ Denominator = (sin 56° × cosec 56°) + (cos 25° × sec 25°)

⇒ Denominator = 1 + 1 = 2

⇒ Expression Value = 4 ÷ 2 = 2

∴ The correct answer is 2.

Additional Information

Complementary Angle Identities

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

Reciprocal Identities

Product of reciprocal ratios is always 1: sin θ × cosec θ = 1, cos θ × sec θ = 1, and tan θ × cot θ = 1.

Special Product Rule

If A + B = 90°, then sin A sec B = 1 and cos A cosec B = 1 because sec B = cosec A and cosec B = sec A.

Shortcut Trick

Distance difference = Platform length = 200 m.

This difference arises from the difference in (Effective Speed × Time).

200 = [x × 30 − (x − 6) × 20] × 5/18

200 × 18/5 = 30x − 20x + 120

720 = 10x + 120 ⇒ 10x = 600 ⇒ x = 60

∴ The correct answer is 60 km/h.

Alternate Method

Given:

Speed of train = x km/h

Length of platform = 200 m

Time to cross platform = 30 s

Speed of man = 6 km/h (same direction)

Time to overtake man = 20 s

Formula Used:

Distance = Speed × Time

Relative speed (same direction) = S₁ − S₂

Conversion: 1 km/h = 5/18 m/s

Calculations:

Let the length of the train be L meters.

For crossing the platform:

⇒ L + 200 = [x × 5/18] × 30

⇒ L + 200 = 25x/3 ... (Eq. 1)

For overtaking the man:

⇒ Relative Speed = (x − 6) × 5/18 m/s

⇒ L = [(x − 6) × 5/18] × 20

⇒ L = 50(x − 6)/9 ... (Eq. 2)

Substitute Eq. 2 into Eq. 1:

⇒ [50(x − 6)/9] + 200 = 25x/3

Multiply the entire equation by 9 to clear denominators:

⇒ 50(x − 6) + 1800 = 75x

⇒ 50x − 300 + 1800 = 75x

⇒ 1500 = 75x − 50x

⇒ 1500 = 25x

⇒ x = 1500 ÷ 25

⇒ x = 60

∴ The correct answer is 60 km/h.

Additional Information

Relative Speed Concept

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

Train Crossing Objects

To cross a point object (pole/man), distance = length of train. To cross a platform/bridge/tunnel, distance = length of train + length of object.

Unit Conversion

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

Shortcut Trick

Square x directly: x² = (1 + √3/2) + (1 − √3/2) − 2√[(1 + √3/2)(1 − √3/2)]

x² = 2 − 2√(1 − 3/4) = 2 − 2(1/2) = 1 ⇒ x = 1.

Required value = (√2 − 1) / (√2 + 1) = (1.414 − 1) / (1.414 + 1)

0.414 / 2.414 ≈ 0.17.

∴ The correct answer is 0.17.

Alternate Method

Given: x = √(1 + √3/2) − √(1 − √3/2)

Formula Used: (a − b)² = a² + b² − 2ab

⇒ x = √[(2 + √3)/2] − √[(2 − √3)/2]

⇒ x = [√(4 + 2√3) / 2] − [√(4 − 2√3) / 2] (Multiplying √2 to numerator and denominator)

⇒ Since 4 + 2√3 = (√3 + 1)² and 4 − 2√3 = (√3 − 1)²

⇒ x = [(√3 + 1) − (√3 − 1)] / 2 = 2 / 2 = 1

⇒ Required Expression = (√2 − 1) / (√2 + 1)

⇒ Rationalizing: [(√2 − 1) × (√2 − 1)] / [(√2 + 1) × (√2 − 1)]

⇒ (2 + 1 − 2√2) / (2 − 1) = 3 − 2√2

⇒ Value = 3 − 2(1.414) = 3 − 2.828 = 0.172

∴ The correct answer is 0.17.

Additional Information

Simplification of Nested Radicals

A radical of the form √(a ± 2√b) can be simplified to √m ± √n where m + n = a and m × n = b.

Rationalization of Denominator

To remove a surd from the denominator, multiply both numerator and denominator by the conjugate (e.g., conjugate of a+√b is a−√b).

Common Square Root Values

√2 ≈ 1.414, √3 ≈ 1.732, and √5 ≈ 2.236 are frequently used in competitive exams.

Shortcut Trick

Use the expansion: a³ + b³ = (a + b)((a + b)² − 3ab)

Substitute the given values: 218 = 2(2² − 3ab)

Divide both sides by 2: 109 = 4 − 3ab

Rearrange to find 3ab: 3ab = 4 − 109 = −105

Solve for ab: ab = −105 ÷ 3 = −35

∴ The correct answer is -35.

Alternate Method

Given: a³ + b³ = 218 and a + b = 2

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

⇒ (2)³ = 218 + 3ab(2)

⇒ 8 = 218 + 6ab

⇒ 6ab = 8 − 218

⇒ 6ab = −210

⇒ ab = −210 ÷ 6

⇒ ab = −35

∴ The correct answer is -35.

Additional Information

Cube of Sum Formula

(a + b)³ = a³ + b³ + 3ab(a + b). This identity connects the sum of cubes to the sum and product of two numbers.

Sum of Cubes Identity

a³ + b³ = (a + b)(a² − ab + b²). This is a factoring identity frequently used in simplification.

Algebraic Properties

For any two numbers, the relationship between their sum (S = a+b) and product (P = ab) for powers can be generalized as a³ + b³ = S³ − 3SP.

Shortcut Trick

For 8-month compounding, Rate = 12% × (8/12) = 8% and Periods = (2 × 12) ÷ 8 = 3.

Using 3 : 3 : 1 ratio for 3 periods:

Level 1: 8% of 31250 = 2500 → 2500 × 3 = 7500

Level 2: 8% of 2500 = 200 → 200 × 3 = 600

Level 3: 8% of 200 = 16 → 16 × 1 = 16

Total Compound Interest = 7500 + 600 + 16 = 8116.

∴ The correct answer is Rs. 8,116.

Alternate Method

Given: Principal (P) = Rs. 31,250 | Rate (R) = 12% p.a. | Time (T) = 2 years

Formula Used: A = P(1 + r/100)ⁿ ; CI = A − P

⇒ New Rate (r) = 12 × (8/12) = 8% per compounding period

⇒ Number of periods (n) = (2 × 12) ÷ 8 = 3 periods

⇒ Amount (A) = 31250 × (1 + 8/100)³

⇒ A = 31250 × (27/25)³

⇒ A = 31250 × (19683 / 15625)

⇒ A = 2 × 19683 = Rs. 39,366

⇒ Compound Interest (CI) = A − P = 39366 − 31250 = Rs. 8,116

∴ The correct answer is Rs. 8,116.

Additional Information

Effective Rate of Interest

For two successive periods with rates x% and y%, the effective rate is [x + y + (xy/100)]%.

Difference between CI and SI

For 2 periods: Difference = P(r/100)². For 3 periods: Difference = P(r/100)² × (3 + r/100).

Compounding Frequencies

If interest is compounded k-monthly, new rate R′ = R × (k/12) and total periods n = (T in months) ÷ k.

Shortcut Trick

Use the value substitution method: Let θ = 45°.

Value of (cos⁶45° + sin⁶45° − 1) = ((1/√2)⁶ + (1/√2)⁶ − 1) = (1/8 + 1/8 − 1) = −3/4.

Value of (tan²45° + cot²45° + 2) = (1² + 1² + 2) = 4.

Product = (−3/4) × 4 = −3.

∴ The correct answer is -3.

Alternate Method

Given: (cos⁶θ + sin⁶θ − 1) (tan²θ + cot²θ + 2)

Formula Used:

1. sin⁶θ + cos⁶θ = (sin²θ + cos²θ)³ − 3sin²θcos²θ(sin²θ + cos²θ) = 1 − 3sin²θcos²θ
2. tan²θ + cot²θ + 2 = (tanθ + cotθ)² = (sinθ/cosθ + cosθ/sinθ)² = (1 / sinθcosθ)²

⇒ Let Part 1 = (cos⁶θ + sin⁶θ − 1)

⇒ Part 1 = (1 − 3sin²θcos²θ − 1) = −3sin²θcos²θ

⇒ Let Part 2 = (tan²θ + cot²θ + 2) = (tanθ + cotθ)²

⇒ Part 2 = (sinθ/cosθ + cosθ/sinθ)² = ((sin²θ + cos²θ) / sinθcosθ)² = 1 / (sin²θcos²θ)

⇒ Expression = Part 1 × Part 2 = (−3sin²θcos²θ) × (1 / sin²θcos²θ)

⇒ Expression = −3

∴ The correct answer is -3.

Additional Information

Pythagorean Identities

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

Sum of Higher Powers

sin⁴θ + cos⁴θ = 1 − 2sin²θcos²θ and sin⁶θ + cos⁶θ = 1 − 3sin²θcos²θ.

Reciprocal Sum Property

tanθ + cotθ = 1 / (sinθ cosθ) = secθ cosecθ.

Shortcut Trick

LCM(12, 16, 18, 20, 25) = 3600.

General form of x = 3600k + 4.

To be divisible by 7, 3600k + 4 → 2k + 4 must be a multiple of 7 (since 3600 ÷ 7 leaves remainder 2).

Smallest k satisfying this is 5 (2 × 5 + 4 = 14).

Number x = 3600 × 5 + 4 = 18004.

Thousands' place digit in 18004 is 8.

∴ The correct answer is 8.

Alternate Method

Given: Divisors = 12, 16, 18, 20, 25; Remainder = 4; x is divisible by 7.

Formula: Number x = LCM(divisors) × k + Remainder

⇒ Prime factors of 12 = 2² × 3

⇒ Prime factors of 16 = 2⁴

⇒ Prime factors of 18 = 2 × 3²

⇒ Prime factors of 20 = 2² × 5

⇒ Prime factors of 25 = 5²

⇒ LCM = 2⁴ × 3² × 5² = 16 × 9 × 25 = 3600

⇒ Number x = 3600k + 4

⇒ To satisfy x ÷ 7, (3600k + 4) mod 7 = 0

⇒ (7 × 514 + 2)k + 4 ≡ 0 mod 7

⇒ 2k + 4 ≡ 0 mod 7

⇒ For k = 5, 2(5) + 4 = 14, which is divisible by 7

⇒ x = 3600 × 5 + 4 = 18000 + 4 = 18004

⇒ In 18004, the thousands' place digit is 8.

∴ The correct answer is 8.

Additional Information

LCM for remainders

When a number N leaves the same remainder r for different divisors, N = LCM(divisors) × k + r.

Modular Congruence

To find k, simplify the coefficient of k by dividing the LCM by the secondary divisor (in this case, 7) and checking the remainder.

Place Value Chart

Starting from right: Units, Tens, Hundreds, Thousands, Ten-thousands, etc. The digit 8 in 18,004 is in the fourth position from the right.

Shortcut Trick

Ratio of sides DB : AB = 3 : (3 + 5) = 3 : 8.

Area ratio of similar triangles is square of side ratio: Area(ΔBDE) : Area(ΔBAC) = 3² : 8² = 9 : 64.

Area of trapezium ACED = Area(ΔBAC) − Area(ΔBDE) = 64 − 9 = 55.

Required ratio Area(ΔBDE) : Area(Trapezium ACED) = 9 : 55.

∴ The correct answer is 9 ∶ 55.

Alternate Method

Given: DE ∥ AC and AD ∶ DB = 5 ∶ 3.

Formula Used: Ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

⇒ Let DB = 3x and AD = 5x.

⇒ Total side AB = AD + DB = 5x + 3x = 8x.

⇒ In ΔBDE and ΔBAC, ∠B is common and ∠BDE = ∠BAC (corresponding angles as DE ∥ AC).

⇒ By AA similarity, ΔBDE ≈ ΔBAC.

⇒ Area(ΔBDE) / Area(ΔBAC) = (DB / AB)²

⇒ Area(ΔBDE) / Area(ΔBAC) = (3x / 8x)² = 9 / 64.

⇒ Let Area(ΔBDE) = 9 units and Area(ΔBAC) = 64 units.

⇒ Area(Trapezium ACED) = Area(ΔBAC) − Area(ΔBDE)

⇒ Area(Trapezium ACED) = 64 − 9 = 55 units.

⇒ Required Ratio = Area(ΔBDE) ∶ Area(Trapezium ACED) = 9 ∶ 55.

∴ The correct answer is 9 ∶ 55.

Additional Information

Basic Proportionality Theorem (Thales Theorem)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio: BD/DA = BE/EC.

Area Property of Similar Triangles

If ΔABC ≈ ΔPQR, then Area(ΔABC)/Area(ΔPQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)².

Area of Trapezium

Area = 1/2 × (Sum of parallel sides) × (Distance between them). In similarity problems, it is usually found by subtracting the smaller triangle from the larger one.

Shortcut Trick

When ∠ADC = ∠BAC in ΔABC, the relation between sides is CA² = CB × CD.

Given: CA = 12 cm and CD = 8 cm.

⇒ 12² = CB × 8

⇒ 144 = 8 × CB

⇒ CB = 144 ÷ 8 = 18 cm.

∴ The correct answer is 18 cm.

Alternate Method

Given:

CA = 12 cm

CD = 8 cm

∠ADC = ∠BAC

In ΔABC and ΔDAC:

⇒ ∠C = ∠C (Common angle)

⇒ ∠BAC = ∠ADC (Given)

⇒ ΔABC ∼ ΔDAC (By AA similarity criterion)

Since the triangles are similar, the ratios of their corresponding sides are equal:

⇒ BC / AC = AC / CD

⇒ BC / 12 = 12 / 8

⇒ BC × 8 = 12 × 12

⇒ 8 × BC = 144

⇒ BC = 144 ÷ 8

⇒ BC = 18 cm

∴ The correct answer is 18 cm.

Additional Information

AA Similarity Criterion

If two angles of one triangle are equal to two angles of another triangle, then the two triangles are similar.

Ratio of Areas of Similar Triangles

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides: Area(Δ1) ÷ Area(Δ2) = (s₁ ÷ s₂)².

Basic Proportionality Theorem (Thales 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.

Shortcut Trick

Direct formula for the angle at point P where bisectors of angles formed by orthocentre meet:

∠BPC = 180° − (∠A ÷ 2)

Given ∠A = 52°

⇒ ∠BPC = 180° − (52° ÷ 2) = 180° − 26° = 154°.

∴ The correct answer is 154°.

Alternate Method

Given: ∠A = 52°, O is the orthocentre, P is the intersection of bisectors of ∠OBC and ∠OCB.

Formula Used:

1. In ΔABC, if O is orthocentre: ∠BOC = 180° − ∠A
2. In ΔOBC, if P is incenter (intersection of bisectors): ∠BPC = 90° + (∠BOC ÷ 2)

Calculations:

⇒ In ΔABC, ∠BOC = 180° − ∠A

⇒ ∠BOC = 180° − 52° = 128°

⇒ Now, in ΔOBC, BP and CP are the bisectors of ∠OBC and ∠OCB respectively.

⇒ ∴ P is the incenter of ΔOBC.

⇒ ∠BPC = 90° + (∠BOC ÷ 2)

⇒ ∠BPC = 90° + (128° ÷ 2)

⇒ ∠BPC = 90° + 64°

⇒ ∠BPC = 154°

∴ The correct answer is 154°.

Additional Information

Orthocentre Property

The angle subtended by any side at the orthocentre (O) is supplementary to the opposite angle of the triangle: ∠BOC + ∠A = 180°.

Incenter Angle Property

The angle formed at the incenter (intersection of internal angle bisectors) is given by 90° + (opposite angle ÷ 2).

Circumcentre Angle Property

The angle subtended by any side at the circumcentre is twice the angle subtended at the remaining part of the circle: ∠BOC = 2 × ∠A.

Ex-center Angle Property

The angle formed at the ex-center (intersection of two external bisectors) is 90° − (opposite angle ÷ 2).

Shortcut Trick

When spheres are melted to form a new sphere, the sum of cubes of their radii equals the cube of the radius of the resulting sphere.

Radius of large sphere = 18 ÷ 2 = 9 cm.

1³ + x³ + 8³ = 9³

1 + x³ + 512 = 729 ⇒ x³ = 216 ⇒ x = 6 cm.

Surface Area = 4 × π × x² = 4 × π × 36 = 144π cm².

∴ The correct answer is 144π cm².

Alternate Method

Given:

Radii of three small spheres: r₁ = 1 cm, r₂ = x cm, r₃ = 8 cm.

Diameter of the resulting large sphere = 18 cm ⇒ Radius (R) = 9 cm.

Formula Used:

Volume of sphere = (4/3) × π × r³

Surface area of sphere = 4 × π × r²

Total Volume of 3 small spheres = Volume of 1 large sphere

⇒ (4/3)πr₁³ + (4/3)πr₂³ + (4/3)πr₃³ = (4/3)πR³

⇒ (4/3)π (r₁³ + r₂³ + r₃³) = (4/3)π R³

⇒ 1³ + x³ + 8³ = 9³

⇒ 1 + x³ + 512 = 729

⇒ x³ + 513 = 729

⇒ x³ = 729 − 513

⇒ x³ = 216

⇒ x = √³216 = 6 cm

Surface area of sphere with radius x:

⇒ SA = 4 × π × x²

⇒ SA = 4 × π × 6²

⇒ SA = 4 × π × 36

⇒ SA = 144π cm²

∴ The correct answer is 144π cm².

Additional Information

Properties of Sphere

A sphere has only one surface. Its Volume is (4/3)πr³ and its Total Surface Area (TSA) is 4πr².

Properties of Hemisphere

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

Recasting Rule

When one or more solid shapes are melted and recast into another shape, the total volume remains constant, provided there is no wastage.

Common Cubes

Useful cubes for exams: 6³ = 216, 7³ = 343, 8³ = 512, 9³ = 729, 12³ = 1728.

Shortcut Trick

Volume of Cuboid = 32 × 12 × 9 = 3456 cm³

Volume of 2 equal cubes = 2 × a³ = 3456 ⇒ a³ = 1728 ⇒ Side of cube (a) = 12 cm

TSA of Cuboid = 2(32 × 12 + 12 × 9 + 32 × 9) = 2(384 + 108 + 288) = 1560 cm²

TSA of 2 Cubes = 2 × (6 × 12 × 12) = 12 × 144 = 1728 cm²

Ratio = 1560 : 1728 = 65 : 72 (dividing by 24)

∴ The correct answer is 65 : 72.

Alternate Method

Given:

Length (l) = 32 cm, Breadth (b) = 12 cm, Height (h) = 9 cm

Number of cubes = 2

Formula Used:

Volume of cuboid = l × b × h

Volume of cube = a³

Total Surface Area (TSA) of cuboid = 2(lb + bh + lh)

Total Surface Area (TSA) of cube = 6a²

Calculations:

⇒ Volume of cuboid = 32 × 12 × 9 = 3456 cm³

⇒ 2 × a³ = 3456

⇒ a³ = 3456 ÷ 2 = 1728

⇒ a = √³1728 = 12 cm

⇒ TSA of Cuboid = 2[(32 × 12) + (12 × 9) + (32 × 9)]

⇒ TSA of Cuboid = 2[384 + 108 + 288] = 2[780] = 1560 cm²

⇒ Total TSA of two cubes = 2 × (6 × a²) = 12 × 12²

⇒ Total TSA of two cubes = 12 × 144 = 1728 cm²

⇒ Ratio = 1560 : 1728

⇒ Ratio = (1560 ÷ 24) : (1728 ÷ 24) = 65 : 72

∴ The correct answer is 65 : 72.

Additional Information

Cube Properties

Lateral Surface Area of cube = 4a². Diagonal of cube = a√3.

Cuboid Properties

Lateral Surface Area of cuboid = 2h(l + b). Diagonal of cuboid = √(l² + b² + h²).

Melting Rule

When one solid is recast into another, the total volume remains constant (Volume_Initial = Volume_Final).

Shortcut Trick

Ratio of persons = 100 : 60 = 5 : 3

Apply M₁D₁H₁ = M₂D₂H₂

5 × 34 × 9 = 3 × 51 × H₂

1530 = 153 × H₂

H₂ = 10

∴ The correct answer is 10 hours.

Alternate Method

Given:

Initial days (D₁) = 34 days, Initial hours (H₁) = 9 h

Initial persons (M₁) = x (Let it be 100 units)

New days (D₂) = 51 days, New persons (M₂) = 60 units (40% decrease)

Formula Used:

M₁ × D₁ × H₁ = M₂ × D₂ × H₂

⇒ 100 × 34 × 9 = 60 × 51 × H₂

⇒ 30600 = 3060 × H₂

⇒ H₂ = 30600 ÷ 3060

⇒ H₂ = 10 hours

∴ The correct answer is 10 hours.

Additional Information

Chain Rule Formula

(M₁ × D₁ × H₁ × E₁) ÷ W₁ = (M₂ × D₂ × H₂ × E₂) ÷ W₂ where E is efficiency and W is work.

Inverse Proportion

For a constant amount of work, the number of persons is inversely proportional to the time taken (Days × Hours).

Unitary Method for Work

If 1 person does work in D days, total work is D person-days. For H hours per day, it is D × H person-hours.

The correct answer is Rs. 420

Key Points

• Let the Cost Price (CP) of the article be x.
• In the first scenario, the Selling Price (SP₁) is Rs. 480 and there is a Profit.
  

Profit = SP₁ − CP = 480 − x

• In the second scenario, the Selling Price (SP₂) is Rs. 400 and there is a Loss.
  

Loss = CP − SP₂ = x − 400

• According to the question, the Loss is equal to one-third of the initial Profit.
  

Equation: Loss = (1/3) ∗ Profit
x − 400 = (1/3) ∗ (480 − x)

• To solve for x, multiply both sides by 3:
  

3 ∗ (x − 400) = 480 − x
3x − 1200 = 480 − x

• Rearrange the terms to isolate x:
  

3x + x = 480 + 1200
4x = 1680

• Divide by 4:
  

x = 1680 ÷ 4 = 420

• Therefore, the Cost Price of the article is Rs. 420.

Additional Information

• Profit occurs when the Selling Price is greater than the Cost Price.
• Loss occurs when the Cost Price is greater than the Selling Price.
• Profit Percentage = (Profit ÷ Cost Price) ∗ 100.
• Loss Percentage = (Loss ÷ Cost Price) ∗ 100.
• In this specific problem, if the cost price is Rs. 420, the profit at Rs. 480 is Rs. 60, and the loss at Rs. 400 is Rs. 20. Indeed, 20 is one-third of 60.

Shortcut Trick

For any trapezium with parallel sides a and b, the length of the line segment joining the mid-points of its diagonals is given by the formula:

Length = |a − b| ÷ 2

Here, a = 12 cm and b = 7.2 cm.

Length = (12 − 7.2) ÷ 2 = 4.8 ÷ 2 = 2.4 cm.

∴ The correct answer is 2.4 cm.

Alternate Method

Given:

Parallel side AB = 12 cm

Parallel side DC = 7.2 cm

Line segment PQ joins the mid-points of diagonals AC and BD.

Formula Used:

Length of the line segment joining mid-points of diagonals = (Difference of parallel sides) ÷ 2

Calculations:

⇒ Let P be the mid-point of AC and Q be the mid-point of BD.

⇒ PQ = (AB − DC) ÷ 2

⇒ PQ = (12 − 7.2) ÷ 2

⇒ PQ = 4.8 ÷ 2

⇒ PQ = 2.4 cm

∴ The correct answer is 2.4 cm.

Additional Information

Median of a Trapezium

The line segment joining the mid-points of the non-parallel sides is called the median. Its length is the average of the parallel sides: Median = (a + b) ÷ 2.

Area of a Trapezium

The area is calculated as Area = 1/2 × (Sum of parallel sides) × Height = 1/2 × (a + b) × h.

Property of Diagonals

In any trapezium, the sum of the squares of the diagonals is equal to the sum of the squares of the non-parallel sides plus twice the product of the parallel sides: AC² + BD² = AD² + BC² + 2(AB × CD).

Total number of employees in all offices = 2400

Number of employees in office B = 2400 × (18/360) = 120

Number of employees in office A = 2400 × (126/360) = 840

40% of the number of employees in office A = 840 × (40/100) = 336

Number of employees who shifted to office B = 336/2 = 168

Number of employees in office B after sifting = 168 + 120 = 288

Number of employees in office C = 2400 × (54/360) = 360

Required difference between the number of employees in B and that in C = 360 – 288 = 72

Shortcut Trick

Angle formed by the intersection of external bisectors: ∠BRC = 90° − ∠A ÷ 2

Given ∠A = 44°, then ∠BRC = 90° − 22° = 68°

Required measure = ∠BRC ÷ 2 = 68° ÷ 2 = 34°

∴ The correct answer is 34°.

Alternate Method

Given:

Measure of ∠A = 44°

Point R is the intersection of external bisectors of ∠B and ∠C.

Formula Used:

For ΔABC, if external bisectors of ∠B and ∠C meet at R, then ∠BRC = 90° − ∠A ÷ 2.

Calculations:

⇒ ∠BRC = 90° − (44° ÷ 2)

⇒ ∠BRC = 90° − 22°

⇒ ∠BRC = 68°

⇒ Required measurement = ∠BRC ÷ 2

⇒ Required measurement = 68° ÷ 2 = 34°

∴ The correct answer is 34°.

Additional Information

Incenter Angle Property

The angle formed by the intersection of internal angle bisectors is ∠BIC = 90° + ∠A ÷ 2.

Excenter Definition

An excenter is the point where the bisector of one interior angle and the bisectors of the two opposite exterior angles intersect.

Angle Sum Property

The sum of all interior angles of a triangle is always 180°, while the sum of exterior angles (one at each vertex) is 360°.

Shortcut Trick

Ratio of Volumes (V₁ : V₂) = x : y

Ratio of Diameters (d₁ : d₂) = Ratio of Radii (r₁ : r₂) = a : b

Since V ∝ r² × h, then height h ∝ V ÷ r²

Ratio of Heights (h₁ : h₂) = (x ÷ a²) : (y ÷ b²) = xb² : ya²

∴ The correct answer is xb² : ya².

Alternate Method

Given:

Ratio of volumes of two cylinders (V₁ : V₂) = x : y

Ratio of diameters (d₁ : d₂) = a : b

Formula Used:

Volume of a cylinder (V) = πr²h

Calculations:

⇒ Let radii be r₁ and r₂. Given diameter ratio d₁/d₂ = a/b, so r₁/r₂ = a/b.

⇒ V₁ = π × r₁² × h₁ and V₂ = π × r₂² × h₂

⇒ V₁/V₂ = (π × r₁² × h₁) / (π × r₂² × h₂)

⇒ x/y = (r₁/r₂)² × (h₁/h₂)

⇒ x/y = (a/b)² × (h₁/h₂)

⇒ x/y = (a²/b²) × (h₁/h₂)

⇒ h₁/h₂ = (x/y) × (b²/a²)

⇒ h₁/h₂ = xb²/ya²

∴ The correct answer is xb² : ya².

Additional Information

Surface Area of Cylinder

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

Ratio of CSA

Ratio of CSA for two cylinders = (r₁ × h₁) : (r₂ × h₂).

Diameter and Radius Relation

Ratio of radii is always equal to the ratio of diameters, as r₁/r₂ = (d₁/2)/(d₂/2) = d₁/d₂.

Shortcut Trick

For divisibility by 72, the number must be divisible by 8 and 9.

Check 8: Last three digits 78y must be divisible by 8 → 784 ÷ 8 = 98 ∴ y = 4.

Check 9: Sum of digits (3+8+9+x+6+3+7+8+4) = 48 + x must be divisible by 9 ∴ x = 6.

Substitute: √(6 × 6 + 7 × 4) = √(36 + 28) = √64 = 8.

∴ The correct answer is 8.

Alternate Method

Given: Nine-digit number 389x6378y is divisible by 72.

Formula Used: Divisibility by 72 ⇒ Divisibility by 8 AND Divisibility by 9.

Step 1: Divisibility rule for 8 (Last 3 digits 78y must be divisible by 8).

⇒ 78y ÷ 8

⇒ 784 ÷ 8 = 98 (Perfectly divisible)

⇒ y = 4

Step 2: Divisibility rule for 9 (Sum of all digits must be divisible by 9).

⇒ Sum = 3 + 8 + 9 + x + 6 + 3 + 7 + 8 + 4

⇒ Sum = 48 + x

⇒ The next multiple of 9 after 48 is 54.

⇒ 48 + x = 54

⇒ x = 54 − 48 = 6

Step 3: Calculate the required value √(6x + 7y).

⇒ √(6 × 6 + 7 × 4)

⇒ √(36 + 28)

⇒ √64

⇒ 8

∴ The correct answer is 8.

Additional Information

Divisibility Rule of 8

A number is divisible by 8 if its last three digits are divisible by 8 or are all zeros.

Divisibility Rule of 9

A number is divisible by 9 if the sum of its individual digits is a multiple of 9.

Composite Divisibility

To check divisibility by a composite number (like 72), check for divisibility by its coprime factors (8 and 9 in this case).

Shortcut Trick

Identify a and b from the identity a³ − b³ = (a − b)(a² + b² + ab).

Here, a = √2x and b = √3y.

Comparing the RHS, we get A = a² = 2, B = b² = 3, and C = ab = √6.

Calculate A² + B² − C² = 2² + 3² − (√6)².

Value = 4 + 9 − 6 = 7.

∴ The correct answer is 7.

Alternate Method

Given:

2√2x³ − 3√3y³ = (√2x − √3y)(Ax² + By² + Cxy)

Formula Used:

a³ − b³ = (a − b)(a² + b² + ab)

Calculations:

⇒ 2√2x³ − 3√3y³ can be written as (√2x)³ − (√3y)³.

⇒ Applying the formula: (√2x − √3y)[(√2x)² + (√3y)² + (√2x)(√3y)]

⇒ (√2x − √3y)(2x² + 3y² + √6xy)

⇒ Comparing with the given expression (Ax² + By² + Cxy):

⇒ A = 2, B = 3, C = √6

⇒ Required value = A² + B² − C²

⇒ (2)² + (3)² − (√6)²

⇒ 4 + 9 − 6 = 7

∴ The correct answer is 7.

Additional Information

Algebraic Identities for Cubes

The sum of cubes is given by a³ + b³ = (a + b)(a² + b² − ab). Note the sign change in the last term.

Comparison of Coefficients

When two polynomial expressions are equal for all values of variables, the coefficients of like powers and like terms (like x², y², xy) must be equal.

Squaring Irrationals

The square of a square root of a number is the number itself: (√n)² = n. For example, (√6)² = 6.

Shortcut Trick

Rewrite the expression in the form √(X − 2√Y) to identify factors.

√(86 − 2 × 30√2) ⇒ Factors of 30√2 whose squares sum to 86 are 6 and 5√2.

Since 6² + (5√2)² = 36 + 50 = 86, the root is |5√2 − 6|.

Comparing 5√2 − 6 with a − b√2 ⇒ a = −6 and b = −5.

√(a² + b²) = √((−6)² + (−5)²) = √(36 + 25) = √61 ≈ 7.8.

∴ The correct answer is 7.8.

Alternate Method

Given: √(86 − 60√2) = a − b√2

Squaring both sides:

⇒ 86 − 60√2 = (a − b√2)²

⇒ 86 − 60√2 = a² + (b√2)² − 2 × a × b√2

⇒ 86 − 60√2 = (a² + 2b²) − (2ab)√2

By comparing the rational and irrational parts:

⇒ 2ab = 60 ⇒ ab = 30

⇒ a² + 2b² = 86

Testing factors of 30: If a = −6 and b = −5 (to ensure a − b√2 is positive):

⇒ (−6)² + 2(−5)² = 36 + 2(25) = 36 + 50 = 86 (Condition satisfied)

Now, find √(a² + b²):

⇒ √((−6)² + (−5)²) = √(36 + 25) = √61

⇒ Since 7.8² = 60.84 and 7.9² = 62.41

⇒ √61 ≈ 7.8 (to one decimal place)

∴ The correct answer is 7.8.

Additional Information

Square Root of Nested Surds

To solve √(x ± √y), we express the term inside as a perfect square (u ± v)² = u² + v² ± 2uv, where u² + v² = x and 2uv = √y.

Linear Approximation of Square Roots

For a number n near a perfect square s², √n ≈ s + (n − s²) / (2s). For example, √61 ≈ 8 + (61 − 64) / 16 = 8 − 0.1875 = 7.8125.

Properties of Square Roots

The principal square root √x is always non-negative. Therefore, if √X = a − b√2, then the value of a − b√2 must be ≥ 0.

Shortcut Trick

Total efficiency of A and B = 3 + 7 = 10 units/day.

Remaining time for both to finish the work = 10.5 − 8 = 2.5 days.

Remaining work units = 10 × 2.5 = 25 units.

Required work (60% of remaining) = 0.60 × 25 = 15 units.

Time taken by A alone = 15 ÷ 3 = 5 days.

∴ The correct answer is 5 days.

Alternate Method

Given:

Efficiency ratio A : B = 3 : 7

Time taken together = 10.5 days (or 21/2 days)

Time worked together = 8 days

⇒ Total efficiency of A + B = 3 + 7 = 10 units/day.

⇒ Total work = Combined efficiency × Total days = 10 × 10.5 = 105 units.

⇒ Work done in 8 days = 10 × 8 = 80 units.

⇒ Remaining work = 105 − 80 = 25 units.

⇒ 60% of remaining work = 25 × (60 ÷ 100) = 15 units.

⇒ Time taken by A = Work ÷ Efficiency of A = 15 ÷ 3 = 5 days.

∴ The correct answer is 5 days.

Additional Information

Fundamental Work Equation

The relationship between work, efficiency, and time is given by Work = Efficiency × Time.

Efficiency and Time Relationship

Efficiency is inversely proportional to the time taken to complete a constant amount of work (E ∝ 1/T).

Combined Efficiency

When multiple people work together, their combined efficiency is the sum of their individual efficiencies (E_Total = E₁ + E₂ + ... + E_n).

Shortcut Trick

Upstream Speed = 28 ÷ 4 = 7 kmph

Downstream Speed = 50 ÷ 2 = 25 kmph

Speed of boat in still water = (Upstream + Downstream) ÷ 2 = (7 + 25) ÷ 2 = 16 kmph

Required time = 44.8 ÷ 16 = 2.8 h

∴ The correct answer is 2.8 h.

Alternate Method

Given: Upstream distance = 28 km in 4 h, Downstream distance = 50 km in 2 h.

Formula Used: Speed = Distance ÷ Time; Speed in Still Water (v) = (Downstream Speed + Upstream Speed) ÷ 2

⇒ Upstream speed (u) = 28 ÷ 4 = 7 kmph

⇒ Downstream speed (d) = 50 ÷ 2 = 25 kmph

⇒ Speed of boat in still water (v) = (d + u) ÷ 2

⇒ v = (25 + 7) ÷ 2 = 32 ÷ 2 = 16 kmph

⇒ Time taken to cover 44.8 km in still water = Distance ÷ v

⇒ Time = 44.8 ÷ 16 = 2.8 h

∴ The correct answer is 2.8 h.

Additional Information

Speed of Boat in Still Water (v)

The speed of the boat when the water is stationary. v = (Downstream Speed + Upstream Speed) ÷ 2.

Speed of Stream (s)

The rate at which the river water flows. s = (Downstream Speed − Upstream Speed) ÷ 2.

Downstream and Upstream Speed

Downstream Speed = v + s; Upstream Speed = v − s.

Shortcut Trick

To find y-axis intersection (P), set x = 0: 0 − 7y = −42 ⇒ y = 6. Thus, α = 0, β = 6.

To find x-axis intersection (Q), set y = 0: 6x + 0 − 15 = 0 ⇒ 6x = 15 ⇒ x = 2.5. Thus, γ = 2.5, δ = 0.

Required sum: α + β + γ + δ = 0 + 6 + 2.5 + 0 = 8.5.

In fraction form: 8.5 = 17/2.

∴ The correct answer is 17/2.

Alternate Method

Given:

Equation 1: x − 7y = −42

Equation 2: 6x + y − 15 = 0

Points: P(α, β) on y-axis and Q(γ, δ) on x-axis.

Formula Used:

For intersection with y-axis, set x = 0.

For intersection with x-axis, set y = 0.

Calculations:

Step 1: Find coordinates of P(α, β)

⇒ At y-axis intersection, x = 0 ∴ α = 0.

⇒ 0 − 7y = −42

⇒ −7y = −42 ⇒ y = 6.

⇒ β = 6. Point P = (0, 6).

Step 2: Find coordinates of Q(γ, δ)

⇒ At x-axis intersection, y = 0 ∴ δ = 0.

⇒ 6x + 0 − 15 = 0

⇒ 6x = 15 ⇒ x = 15/6 = 2.5.

⇒ γ = 2.5. Point Q = (2.5, 0).

Step 3: Calculate the required sum

⇒ Sum = α + β + γ + δ

⇒ Sum = 0 + 6 + 2.5 + 0 = 8.5.

⇒ Sum = 17/2.

∴ The correct answer is 17/2.

Additional Information

X and Y Intercepts

The x-intercept is where a graph crosses the x-axis (y=0), and the y-intercept is where it crosses the y-axis (x=0).

Intercept Form of a Line

A line can be written as x/a + y/b = 1, where 'a' is the x-intercept and 'b' is the y-intercept.

General Linear Equation

For Ax + By + C = 0, the x-intercept is −C/A and the y-intercept is −C/B.

Shortcut Trick

Total units for 2(a + b + c) = 7 + 6 + 5 = 18 units.

Sum (a + b + c) = 18 ÷ 2 = 9 units.

Individual units: a = 9 − 6 = 3; b = 9 − 5 = 4; c = 9 − 7 = 2.

Required inverse ratio = 1/3 : 1/4 : 1/2.

Multiply by LCM(3, 4, 2) = 12 ⇒ 4 : 3 : 6.

∴ The correct answer is 4 : 3 : 6.

Alternate Method

Given:

(a + b) : (b + c) : (c + a) = 7 : 6 : 5

a + b + c = 27

Calculations:

⇒ Let a + b = 7k, b + c = 6k, and c + a = 5k.

⇒ Adding all: (a + b) + (b + c) + (c + a) = 7k + 6k + 5k

⇒ 2(a + b + c) = 18k ⇒ a + b + c = 9k

⇒ From the problem, a + b + c = 27, so 9k = 27 ⇒ k = 3

⇒ a = (a + b + c) − (b + c) = 27 − 6(3) = 27 − 18 = 9

⇒ b = (a + b + c) − (c + a) = 27 − 5(3) = 27 − 15 = 12

⇒ c = (a + b + c) − (a + b) = 27 − 7(3) = 27 − 21 = 6

⇒ 1/a : 1/b : 1/c = 1/9 : 1/12 : 1/6

⇒ Multiply by LCM(9, 12, 6) = 36: (36 ÷ 9) : (36 ÷ 12) : (36 ÷ 6) = 4 : 3 : 6

∴ The correct answer is 4 : 3 : 6.

Additional Information

Inverse Ratio

The inverse ratio of a : b : c is given by 1/a : 1/b : 1/c. To simplify, multiply by the LCM of the denominators.

Cyclic Ratio Property

For cyclic sums like (a+b), (b+c), and (c+a), the sum of the variables is always half the sum of the ratio parts.

Ratio and Constant of Proportionality

When values are given in ratio a : b : c, we can assume them to be ak, bk, and ck where k is the common multiplier.

Shortcut Trick

Identify if the sides form a Pythagorean triple: 11² + 60² = 121 + 3600 = 3721 = 61².

Since it is a right-angled triangle, the circumcircle passes through all vertices and the hypotenuse acts as the diameter.

Circumradius (R) = Hypotenuse ÷ 2

R = 61 ÷ 2 = 30.5 cm

∴ The correct answer is 30.5 cm.

Alternate Method

Given:

Sides of the triangle = 11 cm, 60 cm, 61 cm

Formula Used:

For a right-angled triangle,

Radius of circumcircle (R) = Hypotenuse/2

Calculation:

Check whether the triangle is right-angled:

⇒ 112 + 602

⇒ 121 + 3600

⇒ 3721

⇒ 612 = 3721

Since 112 + 602 = 612, the triangle is right-angled.

Therefore, hypotenuse = 61 cm

⇒ R = 61/2 cm

⇒ R = 30.5 cm

∴ The radius of the circumscribed circle is 30.5 cm.

Additional Information

Inradius of Right Triangle

For a right-angled triangle, the inradius (r) is calculated as r = (Perpendicular + Base − Hypotenuse) ÷ 2.

Circumradius of Any Triangle

For any triangle with sides a, b, c and area Δ, the circumradius R = (abc) ÷ (4Δ).

Location of Circumcenter

In a right-angled triangle, the circumcenter lies exactly at the midpoint of the hypotenuse.

Shortcut Trick

Time gap from 1 year ago to 3 years hence = 1 + 3 = 4 years.

Increase in ratio units for both A and B = 2 units (6 − 4 and 5 − 3).

Equating units to years: 2 units = 4 years ⇒ 1 unit = 2 years.

Time from 1 year ago to 9 years hence = 1 + 9 = 10 years = 5 units.

New ratio = (Initial units + 5) : (Initial units + 5) = (4 + 5) : (3 + 5) = 9 : 8.

∴ The correct answer is 9 : 8.

Alternate Method

Given: Ratio of A and B (1 year ago) = 4 : 3 | Ratio of A and B (3 years hence) = 6 : 5

Let the ages of A and B one year ago be 4x and 3x.

⇒ Ages after 3 years from now = Ages after 4 years from (1 year ago).

⇒ (4x + 4) ÷ (3x + 4) = 6 ÷ 5

⇒ 5 × (4x + 4) = 6 × (3x + 4)

⇒ 20x + 20 = 18x + 24

⇒ 2x = 4 ⇒ x = 2

⇒ A's age (1 year ago) = 4 × 2 = 8 years

⇒ B's age (1 year ago) = 3 × 2 = 6 years

⇒ Age after 9 years from now = Age after 10 years from (1 year ago).

⇒ A's age (9 years hence) = 8 + 10 = 18 years

⇒ B's age (9 years hence) = 6 + 10 = 16 years

⇒ Required Ratio = 18 : 16 = 9 : 8

∴ The correct answer is 9 : 8.

Additional Information

Constant Age Difference

The difference between the ages of two people remains the same regardless of the number of years passed.

Ratio Balancing Method

If the difference between ratio parts (e.g., 4:3 diff=1, 6:5 diff=1) is equal, the change in units is directly proportional to the years.

Handling Unequal Differences

If differences in ratios are unequal, multiply each ratio by the difference of the other ratio to balance them.

Total exports of cars of type A in 2014 to 2017 = 200 +150 + 275 + 175 = 800

Total exports of cars of type B in 2015 to 2018 = 250 + 200 + 275 + 325 = 1050

Required percentage = [(1050 - 800)/1050] × 100 = 23.8%

Shortcut Trick

Ratio of quantities = 20% : 80% = 1 : 4 (Total = 5 parts)

Net gain required on 5 parts = 5 × 25% = 125%

Actual gain/loss on 1 part = 1 × (−25%) = −25%

Balance gain required on 4 parts = 125% − (−25%) = 150%

Required gain % on rest = 150% ÷ 4 = 37.5% = 37½%

∴ The correct answer is 37½%.

Alternate Method

Given: Total wheat = 120 quintals | Part 1 = 20% at 25% loss | Target overall gain = 25%

Assume CP of 120 quintals = ₹12000 (at ₹100/quintal)

⇒ Total Target SP = 12000 × (125/100) = ₹15000

⇒ CP of 20% wheat (24 quintals) = ₹2400

⇒ SP of 20% wheat = 2400 × (75/100) = ₹1800

⇒ Remaining SP required = 15000 − 1800 = ₹13200

⇒ CP of remaining wheat (96 quintals) = ₹9600

⇒ Gain required on rest = 13200 − 9600 = ₹3600

⇒ Required Gain % = (3600 ÷ 9600) × 100

⇒ Required Gain % = 37.5% = 37½%

∴ The correct answer is 37½%.

Additional Information

Weighted Average Profit/Loss Formula

Overall Gain/Loss % = (n₁p₁ + n₂p₂) ÷ (n₁ + n₂), where n is quantity and p is profit/loss percentage (take loss as negative).

Successive Profit/Loss

When there are successive changes of a% and b%, the net change is [a + b + (a×b)/100] %.

Markup and Discount Relation

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

The correct answer is 40

Key Points

• To solve this problem, we establish the relationship between Distance, Speed, and Time using the formula: Distance = Speed × Time.
• Let the total Distance be D and the total Time scheduled for the journey be T. Since the person initially travels at 60 km/h to reach on time, we have:
  

D = 60 × T

• The journey is divided into two parts. In the First Part:
  

Distance covered = 60% of D = 0.6D
Time taken = 2/5^th of T = 0.4T

• In the Second Part (Remaining journey), we calculate the remaining values:
  

Remaining Distance = D − 0.6D = 0.4D
Remaining Time = T − 0.4T = 0.6T

• The Required Speed (V_r) to cover the Remaining Distance in the Remaining Time is:
  

V_r = (Remaining Distance) ÷ (Remaining Time)
V_r = 0.4D ÷ 0.6T

• Substitute the value of D (which is 60T) into the equation:
  

V_r = (0.4 × 60T) ÷ 0.6T
V_r = 24T ÷ 0.6T
V_r = 24 ÷ 0.6 = 40 km/h

• The logic shows that since the person covered a larger portion of the distance (60%) in a smaller portion of the time (40%), he can afford to slow down for the remainder of the trip. The correct speed is 40 km/h.

Additional Information

• Average Speed: The average speed for the whole journey is the Total Distance divided by the Total Time. In this specific scenario, the average speed remains 60 km/h.
• Ratio Method: You can also solve this by assuming values. Let T = 5 hours. Then D = 60 × 5 = 300 km.
  

60% distance = 180 km covered in 2/5 of 5 hours = 2 hours.
Remaining distance = 120 km to be covered in remaining 3 hours.
Speed = 120 ÷ 3 = 40 km/h.

• Inverse Proportion: When the distance is constant, Speed is inversely proportional to Time. When time is constant, Speed is directly proportional to Distance.

Shortcut Trick

Net Rate₁ = 20/3 × 21/5 = 28% and Net Rate₂ = 4 × 11/4 = 11%.

Given SI₁ = 2 × SI₂ ⇒ P₁ × 28% = 2 × (P₂ × 11%).

P₁ × 28 = P₂ × 22 ⇒ P₁ : P₂ = 11 : 14.

Total units = 11 + 14 = 25 units = ₹5,000 ⇒ 1 unit = ₹200.

Difference between parts = |14 − 11| = 3 units = 3 × ₹200 = ₹600.

∴ The correct answer is ₹600.

Alternate Method

Given:

Total Sum = ₹5,000

Part 1: T₁ = 21/5 years, R₁ = 20/3% p.a.

Part 2: T₂ = 11/4 years, R₂ = 4% p.a.

Condition: SI on first part = 2 × SI on second part

Formula: SI = (P × R × T) ÷ 100

⇒ Let the two parts be P₁ and P₂.

⇒ P₁ + P₂ = 5000 --- (i)

⇒ SI₁ = (P₁ × 20/3 × 21/5) ÷ 100 = (P₁ × 4 × 7) ÷ 100 = 28P₁/100

⇒ SI₂ = (P₂ × 4 × 11/4) ÷ 100 = 11P₂/100

⇒ According to the question, SI₁ = 2 × SI₂

⇒ 28P₁/100 = 2 × (11P₂/100)

⇒ 28P₁ = 22P₂

⇒ P₁/P₂ = 22/28 = 11/14

⇒ P₁ = (11/25) × 5000 = ₹2,200

⇒ P₂ = (14/25) × 5000 = ₹2,800

⇒ Difference = P₂ − P₁ = 2800 − 2200 = ₹600.

∴ The correct answer is ₹600.

Additional Information

Simple Interest on Parts

If a sum P is divided into parts such that interests are equal, then P₁:P₂ = (1/R₁T₁) : (1/R₂T₂).

Amount in Simple Interest

Total Amount (A) = Principal (P) + Simple Interest (SI) = P × (1 + RT/100).

Successive Interest vs Simple Interest

Simple Interest is calculated on the original principal every year, whereas Compound Interest is calculated on the accumulated amount.

Shortcut Trick

Identify complementary angles: (65° + θ) + (25° − θ) = 90°. Thus, csc(65° + θ) = sec(25° − θ), making their difference 0.

For the second part: 20° + 70° = 90°. Thus, csc 70° = sec 20°.

Substitute these into the expression: 0 + tan² 20° − sec² 20°.

Using the identity sec² A − tan² A = 1, we get tan² A − sec² A = −1.

Total value = 0 + (−1) = −1.

∴ The correct answer is -1.

Alternate Method

Given: csc(65° + θ) − sec(25° − θ) + tan² 20° − csc² 70°

Formula Used: csc(90° − A) = sec A; sec² A − tan² A = 1

⇒ Term 1: csc(65° + θ)

⇒ Term 2: sec(25° − θ) = sec(90° − (65° + θ)) = csc(65° + θ)

⇒ First half = csc(65° + θ) − csc(65° + θ) = 0

⇒ Term 3: tan² 20°

⇒ Term 4: csc² 70° = csc² (90° − 20°) = sec² 20°

⇒ Second half = tan² 20° − sec² 20°

⇒ Using sec² A − tan² A = 1 ⇒ tan² A − sec² A = −1

⇒ Total Expression = 0 + (−1) = −1

∴ The correct answer is -1.

Additional Information

Complementary Angle Trigonometric Ratios

Pairs such as sin/cos, tan/cot, and sec/csc follow the rule: f(θ) = g(90° − θ). For example, sec θ = csc(90° − θ).

Standard Trigonometric Identities

Key identities include sin² θ + cos² θ = 1, 1 + tan² θ = sec² θ, and 1 + cot² θ = csc² θ.

Sign Convention in Identities

Be careful with the order of subtraction: since sec² θ − tan² θ = 1, the reversed expression tan² θ − sec² θ always equals −1.

Shortcut Trick

LSA = 2πrh = 352 cm² and h = 7 cm.

Radius (r) = LSA ÷ (2 × π × h) = 352 ÷ (2 × 22/7 × 7) = 352 ÷ 44 = 8 cm.

Volume = πr²h = (LSA × r) ÷ 2 = (352 × 8) ÷ 2 = 176 × 8.

Volume = 1408 cm³.

∴ The correct answer is 1408 cm³.

Alternate Method

Given:

Lateral Surface Area (LSA) of cylinder = 352 cm²

Height (h) = 7 cm

π = 22/7

Formula:

Lateral Surface Area = 2πrh

Volume of Cylinder = πr²h

Calculations:

⇒ 2 × (22/7) × r × 7 = 352

⇒ 44 × r = 352

⇒ r = 352 ÷ 44

⇒ r = 8 cm

⇒ Volume = π × r² × h

⇒ Volume = (22/7) × 8 × 8 × 7

⇒ Volume = 22 × 64

⇒ Volume = 1408 cm³

∴ The correct answer is 1408 cm³.

Additional Information

Total Surface Area of Cylinder

The total surface area includes the lateral area and the area of two circular bases: TSA = 2πrh + 2πr² = 2πr(r + h).

Volume of Cone

A cone with the same radius and height as a cylinder has exactly one-third its volume: Volume = (1/3)πr²h.

Base Area of Cylinder

The area of the circular base or top is given by the formula: Area = πr².

Number of person in office C = 2400 × 54/360 = 360

Number of males in office C = 360 × (20/100) = 72

Number of the females in office C = 360 – 72 = 288

Number of person in office E = 2400 × (72/360) = 480

Number of females in office E = 480 × (40/100) = 192

Required ratio = 288 ∶ 192 = 3 ∶ 2

Total number of employees 2400

2400 employees represent = 360°

1 employee represent = 360/2400

350 employees represent = (360/2400) × 350 = 52.5°

650 employees represent = (360/2400) × 650 = 97.5°

Only three offices C, D and E have employees between 350 and 650.

Shortcut Trick

Total Average = 47. Let us use the deviation method to find the sum of the last 3 numbers.

Deviation for first 3 numbers: (39 − 47) × 3 = −24

Deviation for next 7 numbers: (49 − 47) × 7 = +14

Net deviation = −24 + 14 = −10. To balance, the last 3 numbers must have a sum = (3 × 47) + 10 = 151.

Let 12^th number = x. Then 11^th = 2x and 13^th = x + 3. ⇒ 2x + x + x + 3 = 151 ⇒ 4x = 148 ⇒ x = 37.

11^th = 74, 13^th = 40. Average of 11^th and 13^th = (74 + 40) ÷ 2 = 57.

∴ The correct answer is 57.

Alternate Method

Given: Total count = 13, Average = 47. First 3 avg = 39, Next 7 avg = 49.

Relationships: 11^th = 2 × 12^th and 12^th = 13^th − 3.

⇒ Total sum of 13 numbers = 13 × 47 = 611

⇒ Sum of first 3 numbers = 3 × 39 = 117

⇒ Sum of next 7 numbers = 7 × 49 = 343

⇒ Sum of 11^th, 12^th and 13^th numbers = 611 − (117 + 343)

⇒ Sum (11^th + 12^th + 13^th) = 611 − 460 = 151

⇒ Let 12^th number = x. Then 11^th = 2x and 13^th = x + 3

⇒ 2x + x + (x + 3) = 151

⇒ 4x + 3 = 151 ⇒ 4x = 148 ⇒ x = 37

⇒ 11^th number = 2 × 37 = 74

⇒ 13^th number = 37 + 3 = 40

⇒ Average of 11^th and 13^th = (74 + 40) ÷ 2 = 114 ÷ 2 = 57

∴ The correct answer is 57.

Additional Information

Deviation Method for Averages

Net Deviation = Sum of (Individual Value − Assumed Average). For the total average to be correct, the net deviation of all values must be zero.

Weighted Average Formula

Combined Average = (n₁x₁ + n₂x₂ + ... + n_kx_k) ÷ (n₁ + n₂ + ... + n_k), where n is the count and x is the individual average.

Relationship Between Sum and Average

Sum of observations = Average × Number of observations. This is the fundamental basis for solving complex average word problems.

Shortcut Trick

Use the net percentage change formula: Profit% = Mark-up% − Discount% − (Mark-up% × Discount%)/100.

Substitute the given values: 5 = x − 30 − (x × 30)/100.

5 + 30 = x − 0.3x

35 = 0.7x

x = 35 ÷ 0.7 = 50.

∴ The correct answer is 50.

Alternate Method

Given:

Mark-up percentage = x%

Discount percentage = 30%

Profit percentage = 5%

Formula used: CP / MP = (100 − Discount%) / (100 + Profit%)

Calculations:

⇒ CP / MP = (100 − 30) / (100 + 5)

⇒ CP / MP = 70 / 105

⇒ CP / MP = 2 / 3

⇒ Let CP = 2 units and MP = 3 units.

⇒ Mark-up = MP − CP = 3 − 2 = 1 unit.

⇒ Mark-up percentage (x) = (Mark-up / CP) × 100

⇒ x = (1 / 2) × 100

⇒ x = 50.

∴ The correct answer is 50.

Additional Information

CP, MP, and SP Relation

The ratio of Cost Price to Marked Price is directly given by (100 − Discount%) / (100 + Profit%).

Net Percentage Change

When a value is increased by a% (mark-up) and then decreased by b% (discount), the net change is a − b − (ab/100).

Successive Discounts

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