Quantitative Aptitude PYQs

Shortcut Trick

For 'Buy X Get Y Free' schemes, the discount percentage is calculated on the total items given.

Discount % = [Free Items ÷ Total Items] × 100

⇒ Discount % = [10 ÷ (30 + 10)] × 100

⇒ Discount % = (10 ÷ 40) × 100 = 25%

∴ The correct answer is 25%.

Alternate Method

Given:

Number of articles bought (Paid) = 30

Number of free articles (Discount) = 10

Total = 40 Items (Marked Price Base)

30 Items Paid
10 Free

Discount
Discount % = (10 ÷ 40) × 100

Calculations:

⇒ Let the Marked Price (MP) of 1 article be 1 unit.

⇒ Total articles taken by the customer = 30 + 10 = 40.

⇒ Total Marked Price (MP) = 40 units.

⇒ Total Selling Price (SP) = 30 units (since the customer pays for 30 articles only).

⇒ Discount = Total MP − Total SP

⇒ Discount = 40 − 30 = 10 units.

⇒ Discount Percentage = (Discount ÷ Total MP) × 100

⇒ Discount Percentage = (10 ÷ 40) × 100

⇒ Discount Percentage = 25%

∴ The correct answer is 25%.

Additional Information

Buy X Get Y Free Rule

In any scheme offering free articles, the discount is always calculated on the total number of articles the customer takes home. Discount % = [Y ÷ (X + Y)] × 100.

Marked Price vs Cost Price

Marked price is the price printed on an article. Discount is always evaluated on the Marked Price (MP) unless specified otherwise. Discount = MP − SP.

Successive Discounts

If two successive discounts of a% and b% are given, the equivalent single discount is given by the formula: [a + b − (a × b ÷ 100)]%.

Shortcut Trick

Expand the brackets directly to cancel out the x² terms:

⇒ 10x² − 35 − 10x² − 16x + 4 = 12

⇒ −16x − 31 = 12

⇒ −16x = 43 → x = −43/16

∴ The correct answer is −(43)/(16).

Alternate Method

Given:

5(2x² − 7) − 2(5x² + 8x − 2) = 12

Calculation:

⇒ Apply the distributive property to expand the terms.

⇒ 5 × 2x² − 5 × 7 − 2 × 5x² − 2 × 8x − 2 × (−2) = 12

⇒ 10x² − 35 − 10x² − 16x + 4 = 12

⇒ Group the like terms together.

⇒ (10x² − 10x²) − 16x − 35 + 4 = 12

⇒ 0 − 16x − 31 = 12

⇒ −16x = 12 + 31

⇒ −16x = 43

⇒ x = −43 / 16

∴ The correct answer is −(43)/(16).

Additional Information

BODMAS Rule

It defines the correct sequence of steps to solve mathematical expressions: Brackets, Orders (powers/roots), Division, Multiplication, Addition, and Subtraction.

Distributive Property

It states that multiplying a sum by a number gives the same result as multiplying each addend by the number and then adding the products: a(b + c) = ab + ac.

Linear Equation

An equation where the highest power of the variable is 1. When the higher powers (like x²) cancel out, the equation reduces to a simple linear form.

Shortcut Trick

Given 3A = 6B = 2C. Find their direct ratio by dividing the LCM of coefficients (6) by each coefficient.

Ratio A : B : C = 6/3 : 6/6 : 6/2 = 2 : 1 : 3.

Total ratio parts = 2 + 1 + 3 = 6 parts.

A's share = (2 ÷ 6) × 12300 = ₹4,100.

∴ The correct answer is ₹4,100.

Alternate Method

Given:

Total amount to be divided = ₹12,300

3 × A's share = 6 × B's share = 2 × C's share

Total Sum = ₹12,300 (6 parts)

A : 2 parts

B : 1 part

C : 3 parts

A's Share = ?

1 part = ₹12,300 ÷ 6 = ₹2,050
A's share = 2 × 2050 = ₹4,100

Calculations:

⇒ Let 3A = 6B = 2C = k

⇒ A = k/3, B = k/6, C = k/2

⇒ Ratio of shares A : B : C = 1/3 : 1/6 : 1/2

⇒ Multiplying by the LCM of denominators (3, 6, 2) which is 6:

⇒ A : B : C = (1/3 × 6) : (1/6 × 6) : (1/2 × 6)

⇒ A : B : C = 2 : 1 : 3

⇒ Sum of the ratio parts = 2 + 1 + 3 = 6 parts

⇒ 6 parts = ₹12,300

⇒ 1 part = 12300 ÷ 6 = ₹2,050

⇒ Share of A = 2 parts = 2 × 2050 = ₹4,100

∴ The correct answer is ₹4,100.

Additional Information

Compound Ratio

When two or more ratios are multiplied term by term, the result is a compound ratio. For a:b and c:d, the compound ratio is ac:bd.

Duplicate & Sub-duplicate Ratio

The duplicate ratio of a:b is a²:b² and the sub-duplicate ratio is √a:√b.

Proportion Rule

If a:b :: c:d, then Product of Extremes = Product of Means, i.e., a × d = b × c.

Shortcut Trick

For expressions with identical bases, multiplication adds the exponents and division subtracts the exponents.

Given expression: 50^-6 ÷ 50¹⁸ × 50^-5

Directly combine the exponents: -6 − 18 + (-5) = -29

Result = 50^-29

∴ The correct answer is 50^-29.

Alternate Method

Given:

50^-6 ÷ 50¹⁸ × 50^-5

Formula Used:

a^m ÷ aⁿ = a^m−n

a^m × aⁿ = a^m+n

Calculation:

⇒ (50^-6 ÷ 50¹⁸) × 50^-5

⇒ 50^{-6 − 18} × 50^-5

⇒ 50^-24 × 50^-5

⇒ 50^{-24 + (-5)}

⇒ 50^-29

∴ The correct answer is 50^-29.

Additional Information

Product Rule for Exponents

When multiplying expressions with the same base, you add the exponents together: a^m × aⁿ = a^m+n.

Quotient Rule for Exponents

When dividing expressions with the same base, you subtract the exponents: a^m ÷ aⁿ = a^m−n.

Power of a Power Rule

When raising a power to another power, you multiply the exponents: (a^m)ⁿ = a^m×n.

Negative Exponent Rule

A negative exponent indicates the reciprocal of the base raised to the positive exponent: a^-n = 1 ÷ aⁿ.

Shortcut Trick

For an odd number of observations, sort the sequence and pick the exact middle value.

Ascending order: 8, 13, 14, 15, 23, 26, 35, 36, 50.

The total terms are 9, so the middle term is the 5^th one.

The 5^th term is 23.

∴ The correct answer is 23.

Alternate Method

Given:

Data set: 26, 13, 14, 15, 8, 36, 23, 35, 50

Total number of terms (n) = 9

Formula Used:

If n is odd, Median = ((n + 1) ÷ 2)^th term of the sorted data.

Sorted Data Set

8

13

14

15

23

26

35

36

50

Median = 5^th Term

Calculations:

⇒ Arranging data in ascending order: 8, 13, 14, 15, 23, 26, 35, 36, 50

⇒ Median position = (n + 1) ÷ 2

⇒ Median position = (9 + 1) ÷ 2 = 10 ÷ 2

⇒ Median position = 5^th term

⇒ From the sorted data set, the 5^th term is 23.

∴ The correct answer is 23.

Additional Information

Mean

The arithmetic average of all observations. Formula: Mean = Sum of observations ÷ Total number of observations.

Mode

The value that occurs most frequently in the given data set.

Empirical Relationship

For a moderately skewed distribution, the relationship is defined as: Mode ≈ 3 × Median − 2 × Mean.

Shortcut Trick

Number of balls (N) = Volume of Cylinder ÷ Volume of one Sphere

N = (π × r² × 8r) ÷ [4/3 × π × (r/2)³]

N = (8 × πr³) ÷ (πr³/6)

N = 8 × 6 = 48

∴ The correct answer is 48.

Alternate Method

Given:

Radius of cylinder = r

Height of cylinder (h) = 8r

Radius of each spherical ball (R) = r/2

Formula Used:

Volume of Cylinder = πr²h

Volume of Sphere = (4/3)πR³

Calculations:

⇒ Volume of Cylinder = π × r² × 8r = 8πr³

⇒ Volume of one ball = (4/3) × π × (r/2)³

⇒ Volume of one ball = (4/3) × π × (r³/8) = πr³/6

⇒ Number of balls = (Total Volume) ÷ (Volume of 1 ball)

⇒ Number of balls = (8πr³) ÷ (πr³/6)

⇒ Number of balls = 8 × 6 = 48

∴ The correct answer is 48.

Additional Information

Total Surface Area of Cylinder

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

Surface Area of Sphere

The surface area of a sphere with radius R is given by the formula: A = 4πR².

Conservation of Volume

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

Shortcut Trick

Ratio of Height to Shadow = 30 : 10√3

Dividing by 10√3, we get √3 : 1

In a right-angled triangle, if the ratio of sides is √3 : 1, the angle opposite to √3 is 60°.

∴ The correct answer is 60°.

Alternate Method

Given: Height of the pole (AB) = 30 m, Length of shadow (BC) = 10√3 m.

Formula Used: tan θ = Perpendicular ÷ Base

⇒ Let the angle of elevation be θ.

⇒ tan θ = AB ÷ BC

⇒ tan θ = 30 ÷ 10√3

⇒ tan θ = 3 ÷ √3

⇒ tan θ = √3

⇒ θ = tan⁻¹(√3)

⇒ θ = 60°

∴ The correct answer is 60°.

Additional Information

Standard Trigonometric Ratios

tan 30° = 1/√3, tan 45° = 1, and tan 60° = √3. These are frequently used in height and distance problems.

Angle of Elevation vs Depression

The Angle of Elevation is the angle between the horizontal line and the line of sight when looking up, whereas the Angle of Depression is looking down. Both are numerically equal when measured from alternate interior positions.

Right Triangle Properties

In a 30°-60°-90° triangle, the sides are in the ratio 1 : √3 : 2. Since our ratio of Shadow : Height was 1 : √3, the angle opposite the height must be 60°.

Shortcut Trick

Net rate of filling = (1/6 − 1/9) = 1/18 tank per minute.

Time to fill full tank = 18 minutes.

Time to fill one-half of the tank = 18 ÷ 2 = 9 minutes.

∴ The correct answer is 9 minutes.

Alternate Method

⇒ Total Capacity of tank = LCM(6, 9) = 18 units.

⇒ Combined Efficiency (A + B) = 3 − 2 = 1 unit/min.

⇒ Target Work = 1/2 of 18 units = 9 units.

⇒ Time Required = Target Work ÷ Combined Efficiency = 9 ÷ 1 = 9 minutes.

∴ The correct answer is 9 minutes.

Additional Information

Pipes and Cisterns Concept

Filling pipes contribute positive efficiency, while emptying pipes (leakage) contribute negative efficiency.

Net Work Formula

If pipe A fills in 'x' hours and pipe B empties in 'y' hours, net time = (xy) ÷ (y − x) where y > x.

LCM Method

The Total Work is always assumed as the Least Common Multiple of individual times to simplify efficiency as integers.

Shortcut Trick

Required percentage = (B marks ÷ A marks) × 100

Ratio of B to A = 50 : 75 = 2 : 3

Percentage = (2 ÷ 3) × 100 = 66.666...

Rounding to two decimal places, we get 66.67%.

∴ The correct answer is 66.67%.

Alternate Method

Given:

Marks obtained by A = 75

Marks obtained by B = 50

Maximum Marks = 110

Formula Used:

Percentage of X with respect to Y = (X ÷ Y) × 100

Calculations:

⇒ Required Percentage = (B's marks ÷ A's marks) × 100

⇒ Required Percentage = (50 ÷ 75) × 100

⇒ Required Percentage = (2 ÷ 3) × 100

⇒ Required Percentage = 200 ÷ 3

⇒ Required Percentage = 66.6666...

⇒ Rounding to two decimal places = 66.67%

∴ The correct answer is 66.67%.

Additional Information

Percentage Base Change

When calculating what percent X is of Y, Y is the denominator (base). If the question asked what percent of B's marks was A's marks, the base would be B (50).

Rounding Rules

If the third decimal digit is 5 or greater, increase the second decimal digit by 1. Here, 66.666... becomes 66.67.

Percentage Change Formula

Percentage Increase/Decrease = (Difference ÷ Original Value) × 100.

Shortcut Trick

Add both combos: 3 Sandwiches + 3 Juices = ₹80 + ₹70 = ₹150.

Divide by 3 to find 1 Sandwich + 1 Juice = ₹150 ÷ 3 = ₹50.

Subtract this from the first combo: (2 Sandwiches + 1 Juice) − (1 Sandwich + 1 Juice) = ₹80 − ₹50.

Price of one Sandwich = ₹30.

∴ The correct answer is ₹30.

Alternate Method

Given:

Combo 1: 2 Sandwiches and 1 Juice = ₹80

Combo 2: 1 Sandwich and 2 Juices = ₹70

Formula Used:

Method of Substitution or Elimination in Linear Equations.

Let the price of one Sandwich be S and one Juice be J.

⇒ 2S + J = 80 ...

(1) ⇒ S + 2J = 70 ... (2) From (1), J = 80 − 2S. Substitute this in (2):

⇒ S + 2(80 − 2S) = 70

⇒ S + 160 − 4S = 70

⇒ −3S = 70 − 160

⇒ −3S = −90

⇒ S = 30

∴ The correct answer is ₹30.

Additional Information

Linear Equations in Two Variables

An equation of the form ax + by + c = 0 where a, b ≠ 0. It represents a straight line on a graph.

Substitution Method

A method to solve a system of equations by expressing one variable in terms of the other and substituting it into the second equation.

Elimination Method

Scaling equations to make coefficients of one variable equal, then adding or subtracting them to eliminate that variable.

Shortcut Trick

Average speed when equal distances are covered is given by the harmonic mean of the individual speeds.

Speed in first lap (u) = 420 ÷ 140 = 3 m/sec.

Speed in second lap (v) = 420 ÷ 70 = 6 m/sec.

Average speed = (2 × u × v) ÷ (u + v) = (2 × 3 × 6) ÷ (3 + 6) = 36 ÷ 9 = 4 m/sec.

∴ The correct answer is 4 m/sec.

Alternate Method

Given:

Distance of first lap = 420 m, Time taken = 140 sec

Distance of second lap = 420 m, Time taken = 70 sec

Formula Used:

Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ Total Distance = 420 + 420 = 840 m

⇒ Total Time = 140 + 70 = 210 sec

⇒ Average Speed = 840 ÷ 210

⇒ Average Speed = 4 m/sec

∴ The correct answer is 4 m/sec.

Additional Information

Average Speed for Equal Distances

When a journey is split into two equal distances with speeds u and v, the average speed is 2uv ÷ (u + v). For three equal distances with speeds u, v, and w, the formula is 3uvw ÷ (uv + vw + wu).

Average Speed for Equal Time Intervals

If an object travels at speed u for a certain time duration t and at speed v for another equal time duration t, the average speed is the arithmetic mean: (u + v) ÷ 2.

Unit Conversion

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

Shortcut Trick

For two numbers in ratio a:b with HCF = H,

LCM = H × a × b

LCM = 7 × 3 × 4

LCM = 84

∴ The correct answer is 84.

Alternate Method

Given: Ratio = 3 : 4 and HCF = 7

Let the two numbers be 3x and 4x.

⇒ HCF of 3x and 4x is x.

⇒ Given x = 7.

⇒ Number 1 = 3 × 7 = 21

⇒ Number 2 = 4 × 7 = 28

⇒ LCM(21, 28) = 3 × 4 × 7 = 84

∴ The correct answer is 84.

Additional Information

Product Rule

For any two numbers A and B, HCF × LCM = A × B.

Co-prime Numbers

If two numbers are co-prime (HCF = 1), then their LCM is simply their product.

HCF and LCM of Fractions

HCF = HCF of Numerators ÷ LCM of Denominators | LCM = LCM of Numerators ÷ HCF of Denominators.

Shortcut Trick

Subtract the additional years from the person's age and then halve the result.

Sibling's age = (Person's age − 5) ÷ 2

Sibling's age = (27 − 5) ÷ 2 = 22 ÷ 2 = 11

∴ The correct answer is 11 years.

Alternate Method

Given:

Person's Age = 27 years

Let the Sibling's Age be x.

Equation:

According to the question: 2x + 5 = 27

Calculations:

⇒ 2x = 27 − 5

⇒ 2x = 22

⇒ x = 22 ÷ 2

⇒ x = 11

∴ The correct answer is 11 years.

Additional Information

Linear Equations in One Variable

Equations where the highest power of the variable is 1, used to find unknown ages or values by translating verbal statements into mathematical expressions.

Age Relation Rules

If person A is n years older than B, A = B + n. If A is n times as old as B, A = n × B. These can be combined as A = (n × B) + m.

Verification Method

Always plug the answer back into the original condition: 2 × 11 + 5 = 22 + 5 = 27. This confirms the solution is correct.

Shortcut Trick

Difference in Profit Percentage = 17% − 7% = 10%.

According to the question, 10% of Cost Price = ₹1,280.

100% of Cost Price (CP) = 1,280 × (100 ÷ 10).

Cost Price = ₹12,800.

∴ The correct answer is ₹12,800.

Alternate Method

Given: Initial Profit = 7%, New Profit = 17%, Price difference = ₹1,280.

Formulas: SP = CP × (100 + P%) ÷ 100 | Difference in SP = CP × (P₂% − P₁%) ÷ 100.

⇒ Let the Cost Price be x.

⇒ First Selling Price (SP₁) = 1.07x.

⇒ Second Selling Price (SP₂) = 1.17x.

⇒ Difference: SP₂ − SP₁ = 1.17x − 1.07x = 0.10x.

⇒ 0.10x = 1,280 ⇒ x = 1,280 ÷ 0.10 = 12,800.

∴ The correct answer is ₹12,800.

Additional Information

Profit and Loss Variations

When there is a shift from loss to profit, the percentage gap is the sum of the loss% and profit%. For profit to profit, take the difference.

Formula for Price Change

CP = (Difference in Price × 100) ÷ (Difference in Profit/Loss Percentages).

Shortcut Trick

Total ceiling area = 2500 ft².

Unoccupied ceiling area = 2500 − (4 × 25) = 2400 ft².

Wall area = 4 × side × height = 4 × 50 × 10 = 2000 ft².

Cubicle wall area reduction = 4 × (2 × 5 × 10) = 400 ft².

Total painted area = 2400 + 2000 − 400 = 4000 ft².

∴ The correct answer is 4000 ft².

Alternate Method

Given:

⇒ Volume of the room = 25,000 ft³

⇒ Height of the room (h) = 10 ft

⇒ Base area of each cubicle = 25 ft²

Formula Used:

⇒ Volume = Base Area × Height

⇒ Base Area of square room = a² (where a is the side length)

⇒ Total painted area = (Unoccupied Ceiling Area) + (Painted Wall Area)

Calculations:

⇒ Base Area of the room = 25,000 ÷ 10 = 2500 ft²

⇒ Side of the room (a) = √2500 = 50 ft

⇒ Side of each cubicle (c) = √25 = 5 ft

⇒ Unoccupied ceiling area = Total Base Area − (4 × Cubicle Base Area)

⇒ Unoccupied ceiling area = 2500 − (4 × 25) = 2400 ft²

⇒ Painted wall area: Since the 4 corners contain cubicles of 5 ft × 5 ft, the walls of the room inside these cubicles are not painted, but the outer partition walls of the cubicles facing the room are painted.

⇒ For each corner, the room walls of length 5 ft + 5 ft = 10 ft are not painted, but the cubicle partition walls of length 5 ft + 5 ft = 10 ft are painted.

⇒ Therefore, the total perimeter of the walls to be painted remains the same as the perimeter of the room = 4 × 50 = 200 ft.

⇒ But the partition walls of the cubicles block the corner room walls, maintaining an equivalent surface area. Thus, the total wall area to be painted is equal to the total original wall area of the room minus the blocked parts plus the new partition walls, which sums up to: 4 × 50 × 10 − 4 × (5 + 5) × 10 + 4 × (5 + 5) × 10 = 2000 ft².

⇒ However, the walls enclosed within the cubicles (which are 2 walls of the room per cubicle) are not painted. The new partition walls of the cubicles facing the inside of the room are painted. This means the perimeter being painted is exactly equal to the boundary of the unoccupied region.

⇒ Perimeter of unoccupied region = 4 × (50 − 2 × 5) + 4 × (2 × 5) = 4 × 40 + 40 = 200 ft.

⇒ Painted Wall Area = Perimeter × Height = 200 × 10 = 2000 ft².

⇒ Total Painted Area = 2400 + 2000 = 4400 ft². Since the outer walls of the cubicles are part of the cubicle structures and only the room's walls are to be painted, we deduct the cubicle wall overlap: 2000 (Room walls) − 400 (covered by cubicles) = 1600 ft².

⇒ Total Painted Area = 2400 (Ceiling) + 1600 (Remaining Room Walls) = 4000 ft².

∴ The correct answer is 4000 ft².

Additional Information

Total Surface Area of a Cuboid

The total surface area of a cuboid is given by 2(lb + bh + hl), where l, b, and h are length, breadth, and height respectively.

Lateral Surface Area of a Room

The area of the four walls of a room is equal to its perimeter multiplied by the height: 2(l + b) × h.

2

Shortcut Trick
Net percentage change = a + b + (a × b) ÷ 100
Here, markup percentage a = +44% and discount percentage b = −75%.
Net change = 44 − 75 + (44 × −75) ÷ 100
Net change = −31 − (3300 ÷ 100) = −31 − 33 = −64%
Negative sign indicates a loss.
∴ The correct answer is 64% loss.

Alternate Method
Given → Markup % = 44%, Discount % = 75%.
Formulas: MP = CP × (100 + Markup%) ÷ 100 | SP = MP × (100 − Discount%) ÷ 100 | Loss% = (CP − SP) ÷ CP × 100

⇒ Let the Cost Price (CP) be 100.
⇒ Marked Price (MP) = 100 + 44% of 100 = 144.
⇒ Selling Price (SP) = 144 − 75% of 144.
⇒ SP = 144 − (0.75 × 144) = 144 − 108 = 36.
⇒ Since CP (100) > SP (36), it is a Loss.
⇒ Loss% = (100 − 36) ÷ 100 × 100 = 64%.
∴ The correct answer is 64% loss.

Additional Information
Marked Price & Discount
The Marked Price is the price printed on the label. Discount is always calculated on the Marked Price to arrive at the Selling Price.
Successive Percentage Formula
When an item is marked up and then discounted, the overall Gain/Loss% can be found using x + y + (xy/100), where x is markup and y is discount (negative).

Shortcut Trick

Convert all terms to base 2

4 = 2², 64 = 2⁶, 8 = 2³

2^2(x+1) = 2⁶ × 2^3x = 2^6+3x

Equate powers ⇒ 2x + 2 = 6 + 3x ⇒ x = −4

∴ Answer = −4

Alternate Method

Given:

4^{(x + 1)} = 64 × 8^x

Formula used:

a^m × aⁿ = a^m+n, (a^m)ⁿ = a^mn

Calculation:

⇒ 4^(x+1) = (2²)^(x+1) = 2^2x+2

⇒ 64 × 8^x = 2⁶ × (2³)^x = 2^6+3x

⇒ 2^2x+2 = 2^6+3x

⇒ 2x + 2 = 6 + 3x

⇒ 2 − 6 = 3x − 2x

⇒ −4 = x

∴ The correct answer is −4.

Additional Information

Exponent Rule

a^m × aⁿ = a^m+n

Power of Power

(a^m)ⁿ = a^mn

Equal Bases

If bases are equal, equate exponents directly

Standard Bases

4 = 2², 8 = 2³, 64 = 2⁶

Shortcut Trick

Total amount = ₹1,964. Total deduction = 50 + 17 + 10 = ₹77.

Remaining amount = 1,964 − 77 = ₹1,887.

Divide ₹1,887 in ratio 20 : 4 : 13. Sum of parts = 37 units.

37 units = ₹1,887 → 1 unit = ₹51.

Lalita's reduced share = 4 × 51 = ₹204. Original share = 204 + 17 = ₹221.

∴ The correct answer is ₹221.

Alternate Method

Given: Total amount = ₹1,964. Ratio after deductions = 20 : 4 : 13. Deductions = ₹50, ₹17, ₹10.

Formula: Actual Share = (Ratio Part × Value of 1 unit) + Deduction

⇒ Total deduction = 50 + 17 + 10 = ₹77

⇒ Remaining amount to be divided = 1,964 − 77 = ₹1,887

⇒ Sum of ratio parts = 20 + 4 + 13 = 37 units

⇒ 37 units = ₹1,887

⇒ 1 unit = 1,887 ÷ 37 = ₹51

⇒ Lalita's share after deduction = 4 units × ₹51 = ₹204

⇒ Lalita's original share = 204 + 17 = ₹221

∴ The correct answer is ₹221.

Additional Information

Ratio and Proportion

When values are subtracted from original shares, always subtract the total deduction from the total sum before applying the ratio parts.

Compound Ratio

The compound ratio of (a : b) and (c : d) is given by (ac : bd).

Invertendo and Alternendo

If a/b = c/d, then b/a = d/c (Invertendo) and a/c = b/d (Alternendo).

Shortcut Trick

Total work in man-hours:

Mohan = 6 days × 5 hours = 30 hours

Rohan = 8 days × 5 hours = 40 hours (assuming same hours/day initially)

Combined time in hours = (30 × 40) ÷ (30 + 40) = 1200 ÷ 70 = 120/7 hours

Days required at 8 hours/day = (120/7) ÷ 8 = 15/7 = 2(1/7) days

∴ The correct answer is 2(1/7) days.

Alternate Method

Given:

Mohan takes 6 days at 5 hours/day = 30 hours

Rohan takes 8 days at 5 hours/day = 40 hours

Calculations:

⇒ Combined Efficiency = 4 + 3 = 7 units/hour

⇒ Total hours required = 120 ÷ 7 hours

⇒ Number of days at 8 hours per day = (120 ÷ 7) ÷ 8

⇒ Number of days = 15/7 days

⇒ Number of days = 2(1/7) days

∴ The correct answer is 2(1/7) days.

Additional Information

Work-Hours Relation

Total Work = Persons × Days × Hours per day. This is often solved using the M₁D₁H₁/W₁ = M₂D₂H₂/W₂ formula.

Combined Efficiency

When multiple people work together, their efficiencies (work done per unit time) are added to find the total time taken for the task.

Shortcut Trick

Apply the algebraic identity: (a − b)² + 4ab = (a + b)².

In this expression, let a = 4x and b = 5y. Then 4ab = 4(4x)(5y) = 80xy.

The numerator (4x − 5y)² + 80xy simplifies exactly to (4x + 5y)².

The entire expression becomes (4x + 5y)² ÷ (4x + 5y)² = 1.

∴ The correct answer is 1.

Alternate Method

Given: [(4x − 5y)² + 80xy] ÷ [(4x + 5y)²]

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

⇒ Expand the numerator: (4x)² − 2(4x)(5y) + (5y)² + 80xy

⇒ Numerator = 16x² − 40xy + 25y² + 80xy

⇒ Numerator = 16x² + 40xy + 25y²

⇒ Now, expand the denominator: (4x + 5y)² = (4x)² + 2(4x)(5y) + (5y)²

⇒ Denominator = 16x² + 40xy + 25y²

⇒ Expression = (16x² + 40xy + 25y²) ÷ (16x² + 40xy + 25y²)

⇒ Simplified Value = 1

∴ The correct answer is 1.

Additional Information

Binomial Square Identities

The expansion of (a ± b)² is a core concept in algebra, given by a² ± 2ab + b².

Relationship between Identities

The identities are linked such that (a + b)² − (a − b)² = 4ab and (a + b)² + (a − b)² = 2(a² + b²).

Simplification Rule

Any non-zero expression divided by itself, such as k ÷ k, always results in 1.