Quantitative Aptitude PYQs

Shortcut Trick

The best scheme for the shopkeeper is the one that offers the minimum discount to the customer.

Compare total discounts: A (39+8−3.12=43.88%), B (4/12 × 100=33.33%), C (16%), D (38+34−12.92=59.08%).

Comparing the values 43.88%, 33.33%, 16%, and 59.08%, the minimum discount is 16%.

Therefore, Scheme C is the most profitable for the shopkeeper.

∴ The correct answer is C.

Alternate Method

Formulas: Successive Discount = a + b − (a × b) ÷ 100 | Buy x get y free Discount% = [y ÷ (x + y)] × 100

⇒ Scheme A: 39 + 8 − (39 × 8)/100 = 47 − 3.12 = 43.88%
⇒ Scheme B: (4 ÷ 12) × 100 = 1/3 × 100 = 33.33%
⇒ Scheme C: Single discount = 16%
⇒ Scheme D: 38 + 34 − (38 × 34)/100 = 72 − 12.92 = 59.08%
⇒ For a shopkeeper, the scheme with the lowest discount percentage results in the highest Selling Price (SP) and thus the most profit.

∴ The correct answer is C.

Additional Information

Customer vs Shopkeeper Preference

A customer always looks for the Maximum Discount (Scheme D), while the shopkeeper looks for the Minimum Discount (Scheme C).

Buy x Get y Free Concept

In 'Buy x Get y Free', the total items given to the customer is (x + y). The discount is calculated on the total items: Discount % = [Free Items ÷ Total Items] × 100.

Shortcut Trick

Let the pipes be A, B, and C (outlet).

Combined efficiency (A + B − C) = Total Work ÷ Time = 192 ÷ 48 = 4.

Individual efficiencies: A = 192 ÷ 96 = 2; B = 192 ÷ 64 = 3.

Outlet efficiency (C) = (A + B) − Combined efficiency = (2 + 3) − 4 = 1.

Time taken by pipe C = Total Work ÷ Efficiency = 192 ÷ 1 = 192 min.

∴ The correct answer is 192 min.

Alternate Method

Given: Filling time of Pipe 1 = 96 min, Pipe 2 = 64 min, and Combined time (with outlet) = 48 min.

Calculations:

⇒ Efficiency of Pipe A = 192 ÷ 96 = 2 units/min

⇒ Efficiency of Pipe B = 192 ÷ 64 = 3 units/min

⇒ Combined Efficiency (A + B − C) = 192 ÷ 48 = 4 units/min

⇒ Efficiency of Outlet Pipe C = (Efficiency of A + Efficiency of B) − Combined Efficiency

⇒ Efficiency of Pipe C = (2 + 3) − 4 = 1 unit/min

⇒ Time taken by Pipe C to empty the cistern = Total Work ÷ Efficiency of C

⇒ Time = 192 ÷ 1 = 192 min

∴ The correct answer is 192 min.

Additional Information

Inlet and Outlet Pipes

An inlet pipe fills a tank (positive work), while an outlet pipe empties it (negative work).

Net Efficiency Formula

Net efficiency = ∑(Efficiencies of Inlet pipes) − ∑(Efficiencies of Outlet pipes).

Work and Time Relationship

Time taken to complete a task is inversely proportional to efficiency: Time = Work ÷ Efficiency.

Shortcut Trick

In a quadrilateral, the angle formed by the bisectors of two adjacent angles is equal to half the sum of the other two angles.

∠TOS = (∠M + ∠N) ÷ 2

∠TOS = (63° + 99°) ÷ 2

∠TOS = 162° ÷ 2 = 81°

∴ The correct answer is 81°.

Alternate Method

Given: ∠M = 63°, ∠N = 99°

Formula Used: Sum of interior angles of a quadrilateral = 360°; Sum of angles in a triangle = 180°.

In quadrilateral MNST:

⇒ ∠M + ∠N + ∠S + ∠T = 360°

⇒ 63° + 99° + ∠S + ∠T = 360°

⇒ 162° + ∠S + ∠T = 360°

⇒ ∠S + ∠T = 360° − 162° = 198°

Since TO and SO are angle bisectors of ∠T and ∠S respectively:

⇒ ∠OTS = ∠T ÷ 2 and ∠OST = ∠S ÷ 2

In triangle ΔTOS:

⇒ ∠TOS + ∠OTS + ∠OST = 180°

⇒ ∠TOS + (∠T ÷ 2) + (∠S ÷ 2) = 180°

⇒ ∠TOS + (∠T + ∠S) ÷ 2 = 180°

⇒ ∠TOS + 198° ÷ 2 = 180°

⇒ ∠TOS + 99° = 180°

⇒ ∠TOS = 180° − 99° = 81°

∴ The correct answer is 81°.

Additional Information

Angle Sum Property of Polygons

The sum of all interior angles of an n-sided polygon is (n − 2) × 180°.

Angle Between Opposite Bisectors

The angle between the bisectors of two opposite angles of a quadrilateral is equal to half the difference of the other two angles.

Shortcut Trick

LCM of 10, 15, and 3 is 30.

Divide 539 by 30 to find the remainder: 539 ÷ 30 ⇒ Remainder = 29.

Number to be added = Divisor − Remainder = 30 − 29 = 1.

∴ The correct answer is 1.

Alternate Method

Given: Number = 539, Divisors = 10, 15, 3.

Formula: Least number to be added = LCM(divisors) − (Number % LCM(divisors)).

⇒ LCM of 10, 15, and 3:

⇒ 10 = 2 × 5

⇒ 15 = 3 × 5

⇒ 3 = 3

⇒ LCM = 2 × 3 × 5 = 30.

⇒ Divide 539 by 30:

⇒ 539 = 30 × 17 + 29.

⇒ Remainder = 29.

⇒ Required number to be added = 30 − 29 = 1.

∴ The correct answer is 1.

Additional Information

LCM (Least Common Multiple)

The smallest positive integer that is divisible by each of the given numbers. For divisibility problems involving multiple numbers, always find the LCM first.

Divisibility Rule of 10

A number is divisible by 10 if its last digit is 0. Since 539 ends in 9, the next number ending in 0 is 540, requiring 1 to be added.

Least Number to Subtract

To find the least number to subtract to make a number perfectly divisible, the answer is simply the Remainder obtained after division.

Shortcut Trick

Area of quadrilateral ABCD = Area(ΔBCD) + Area(ΔABD)

In right-angled triangle BCD, Area = 0.5 × 42 × 40 = 840 m²

Hypotenuse BD = √(42² + 40²) = 58 m

In isosceles triangle ABD (sides 36, 36, 58), height h = √(36² − 29²) = √455 ≈ 21.33 m

Area(ΔABD) = 0.5 × 58 × 21.33 ≈ 618.6 m²

Total Area = 840 + 618.6 = 1458.6 m²

∴ The correct answer is 1458.6 m².

Alternate Method

Given: AB = 36 m, BC = 42 m, CD = 40 m, AD = 36 m, and ∠C = 90°.

Formula Used: Area of right triangle = 0.5 × base × height; Pythagoras theorem: a² + b² = c²; Heron's formula: Area = √[s(s−a)(s−b)(s−c)].

A
B
C
D
36 m
42 m
40 m

Shortcut Trick

Remaining distance = 1/3 of 168 = 56 km

Remaining time = (1 − 2/6) of 18 = 4/6 × 18 = 12 hours

Required Speed = 56 ÷ 12 = 4.66...

∴ The correct answer is 4.7 km/hr.

Alternate Method

Given:

Total distance = 168 km

Total time = 18 hours

Calculations:

⇒ Distance covered = 2/3 × 168 = 112 km

⇒ Remaining distance = 168 − 112 = 56 km

⇒ Time taken = 2/6 × 18 = 6 hours

⇒ Remaining time = 18 − 6 = 12 hours

⇒ Required speed = Remaining distance ÷ Remaining time

⇒ Speed = 56 ÷ 12

⇒ Speed = 4.666... ≈ 4.7 km/hr

∴ The correct answer is 4.7 km/hr.

Additional Information

Average Speed Formula

Average Speed = Total Distance ÷ Total Time. For two equal distances, Avg Speed = 2xy/(x+y).

Relative Speed

Same direction: S₁ − S₂ | Opposite direction: S₁ + S₂.

Unit Conversion

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

Shortcut Trick

Required percentage = (Marks of B / Marks of A) × 100

Required percentage = (50 / 75) × 100 = (2 / 3) × 100

2/3 of 100 is approximately 66.67%.

∴ The correct answer is 66.67%.

Alternate Method

Given:

Marks obtained by B = 50

Marks obtained by A = 75

Maximum marks = 120 (

Note:

This is redundant for the specific ratio requested)

Formula Used:

Percentage of X relative to Y = (X / Y) × 100

Calculations:

⇒ Required % = (Marks of B / Marks of A) × 100

⇒ Required % = (50 / 75) × 100

⇒ Required % = (2 / 3) × 100

⇒ Required % = 200 / 3

⇒ Required % = 66.666...

⇒ Rounding off to two decimal places, we get 66.67%.

∴ The correct answer is 66.67%.

Additional Information

Percentage Change

Percentage change = (Difference / Original Value) × 100. Used to find how much a value increased or decreased.

Base Value Identification

The value following the word 'than' or 'of' in a comparison is usually taken as the Base Value (denominator).

Fraction to Percentage Conversion

Common fractions: 1/2 = 50%, 1/3 = 33.33%, 2/3 = 66.67%, 1/4 = 25%, 3/4 = 75%.

Shortcut Trick

Ratio of Volumes = n = 8

Ratio of Radii (R/r) = √³(n) = √³(8) = 2

New radius (r) = 14 ÷ 2 = 7 cm

Surface Area = 4 × (22/7) × 7 × 7 = 616 cm²

∴ The correct answer is 616 cm².

Alternate Method

Given:

Radius of large sphere (R) = 14 cm

Number of solid balls (n) = 8

Formula Used:

Volume of Sphere = (4/3) × π × r³

Surface Area of Sphere = 4 × π × r²

Calculations:

⇒ Total Volume of large sphere = 8 × Volume of each small ball

⇒ (4/3) × π × (14)³ = 8 × (4/3) × π × r³

⇒ 14 × 14 × 14 = 8 × r³

⇒ 2744 = 8 × r³

⇒ r³ = 2744 ÷ 8

⇒ r³ = 343

⇒ r = √³(343) = 7 cm

⇒ Surface area of each ball = 4 × π × r²

⇒ Surface area = 4 × (22/7) × 7 × 7

⇒ Surface area = 4 × 22 × 7

⇒ Surface area = 88 × 7 = 616 cm²

∴ The correct answer is 616 cm².

Additional Information

Curved Surface Area of Hemisphere

The CSA of a hemisphere is exactly half of the sphere's surface area, given by 2 × π × r².

Total Surface Area of Hemisphere

The TSA of a solid hemisphere includes the base area, resulting in 3 × π × r².

Volume of Hemisphere

The Volume of a hemisphere is (2/3) × π × r³, which is half the volume of a full sphere.

Shortcut Trick

The Algebraic Identity used is: (Sum)² = (Sum of Squares) + 2 × (Sum of Products).

Substituting the given values: 15² = 83 + 2(ab + bc + ca)

225 = 83 + 2(ab + bc + ca)

2(ab + bc + ca) = 225 − 83 = 142

ab + bc + ca = 142 ÷ 2 = 71

∴ The correct answer is 71.

Alternate Method

Given:

a + b + c = 15

a² + b² + c² = 83

Formula Used:

The standard expansion of the square of a trinomial:

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Calculations:

⇒ (15)² = 83 + 2(ab + bc + ca)

⇒ 225 = 83 + 2(ab + bc + ca)

⇒ 225 − 83 = 2(ab + bc + ca)

⇒ 142 = 2(ab + bc + ca)

⇒ ab + bc + ca = 142 ÷ 2

⇒ ab + bc + ca = 71

∴ The correct answer is 71.

Additional Information

Trinomial Square Identity

The formula (a+b+c)² = a²+b²+c²+2(ab+bc+ca) relates the sum of three numbers, the sum of their squares, and the sum of their products taken in pairs.

Cubic Identity

A related identity is a³+b³+c³ − 3abc = (a+b+c)(a²+b²+c² − ab − bc − ca), which is frequently used in algebraic simplifications.

Value of ab+bc+ca

If a+b+c = 0, then a²+b²+c² = −2(ab+bc+ca), implying the sum of squares is twice the negative sum of the cyclic products.

Shortcut Trick

Profit = 15% ⇒ Fraction = 3/20.

Let Cost Price (CP) = 20 units and Profit = 3 units.

Then, Selling Price (SP) = 20 + 3 = 23 units.

Given: 23 units = ₹3,450 ⇒ 1 unit = 3,450 ÷ 23 = ₹150.

CP = 20 units = 20 × 150 = ₹3,000.

∴ The correct answer is ₹3,000.

Alternate Method

Given: Selling Price (SP) = ₹3,450, Profit Percentage = 15%.
Formulas: Profit = SP − CP | SP = CP × (100 + P%) ÷ 100 | CP = (SP × 100) ÷ (100 + P%).

⇒ CP = (3,450 × 100) ÷ (100 + 15)
⇒ CP = (3,450 × 100) ÷ 115
⇒ CP = 30 × 100
⇒ CP = ₹3,000.

∴ The correct answer is ₹3,000.

Additional Information

Cost Price & Selling Price

Cost Price (CP) is the price at which an article is purchased. Selling Price (SP) is the price at which it is sold.

Profit and Loss Percentage

Profit % = (Profit ÷ CP) × 100. Loss % = (Loss ÷ CP) × 100. Both are always calculated on the Cost Price unless stated otherwise.

Shortcut Trick

Let total distance = LCM(7, 5) = 35 units.

Speed of A = 35 ÷ 7 = 5 units/hr, Speed of B = 35 ÷ 5 = 7 units/hr.

Train A travels alone for 2 hours (7:50 a.m. to 9:50 a.m.), covering 5 × 2 = 10 units.

Remaining distance = 35 − 10 = 25 units. Relative speed = 5 + 7 = 12 units/hr.

Time to meet = 25 ÷ 12 hours = 2 hours 5 minutes.

Meeting time = 9:50 a.m. + 2 hours 5 minutes = 11:55 a.m.

∴ The correct answer is 11:55 a.m.

Alternate Method

M
N

P

10 units

25 units

Train A →
Start: 7:50 a.m.
End: 2:50 p.m.

9:50 a.m.

← Train B
Start: 9:50 a.m.
End: 2:50 p.m.

Total Distance = 35 units

Given:

⇒ Train A travel time = 7:50 a.m. to 2:50 p.m. = 7 hours.

⇒ Train B travel time = 9:50 a.m. to 2:50 p.m. = 5 hours.

Calculations:

⇒ Let the total distance between M and N be the LCM of 7 and 5 = 35 km.

⇒ Speed of Train A = Distance ÷ Time = 35 ÷ 7 = 5 km/hr.

⇒ Speed of Train B = Distance ÷ Time = 35 ÷ 5 = 7 km/hr.

⇒ Distance travelled by Train A in 2 hours (from 7:50 a.m. to 9:50 a.m.) = Speed × Time = 5 × 2 = 10 km.

⇒ Remaining distance = 35 − 10 = 25 km.

⇒ Relative speed of both trains when moving towards each other = 5 + 7 = 12 km/hr.

⇒ Time taken to meet after 9:50 a.m. = Remaining distance ÷ Relative speed

⇒ Time taken = 25 ÷ 12 hours = 2 hours and (1/12 × 60) minutes = 2 hours 5 minutes.

⇒ Meeting time = 9:50 a.m. + 2 hours 5 minutes = 11:55 a.m.

∴ The correct answer is 11:55 a.m.

Additional Information

Relative Speed

When two bodies travel in opposite directions, their relative speed is the sum of their individual speeds: Relative Speed = S₁ + S₂.

Trains Crossing Each Other

When two trains of lengths L₁ and L₂ cross each other, the total distance covered is L₁ + L₂. Time taken = (L₁ + L₂) ÷ Relative Speed.

Average Speed

For different speeds over equal distances, Average Speed = (2 × S₁ × S₂) ÷ (S₁ + S₂).

Shortcut Trick
Divide the given number 4519 by 30 to find the remainder.
4519 ÷ 30 gives a quotient of 150 and a remainder of 19.
To make the number completely divisible by 30, the required number to be added = Divisor − Remainder.
Required number = 30 − 19 = 11.
∴ The correct answer is 11.

Alternate Method
Given:
The dividend is 4519.
The divisor is 30.
Concept Used:
According to the division algorithm, Dividend = (Divisor × Quotient) + Remainder.
When a number is divided by a divisor D leaving a remainder R, the smallest number that needs to be added to make it completely divisible by D is given by (D − R).
Calculations:
⇒ Let's divide 4519 by 30.
⇒ We know that 30 × 150 = 4500.
⇒ 4519 = 4500 + 19
⇒ 4519 = (30 × 150) + 19
⇒ Here, the remainder (R) is 19.
⇒ The smallest number to be added = Divisor − Remainder
⇒ Number to be added = 30 − 19
⇒ Number to be added = 11
⇒ Verification: 4519 + 11 = 4530, which is perfectly divisible by 30 (4530 ÷ 30 = 151).
∴ The correct answer is 11.

Additional Information
Divisibility Rule of 30
A number is divisible by 30 if it is divisible by both 3 and 10. This means the number must end in a 0, and the sum of its digits must be a multiple of 3.
Number to be Subtracted
If the question asked for the smallest number to be subtracted to make the sum completely divisible by the divisor, the answer would simply be the remainder itself (which is 19 in this case).
Euclid's Division Lemma
For any integer a and positive integer b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b.

Shortcut Trick

Let the ratio of the two numbers be 5 : 1 (since the first is five times the second).

Let the numbers be 5a and a.

Product = 5a × a = 5a² = 2880

⇒ a² = 576 ⇒ a = 24

Sum of the ratio parts = 5 + 1 = 6

Sum of the numbers = 6a = 6 × 24 = 144

∴ The correct answer is 144.

Alternate Method

y
2^nd No.

x = 5y
1^st No.

Sum = x + y = 6y

Product = 5y × y = 2880

Given:

Product of two positive numbers = 2880

First number (x) = 5 × Second number (y)

Formula Used:

Product = x × y

Sum = x + y

Calculations:

⇒ x × y = 2880

Substitute x = 5y in the product equation:

⇒ (5y) × y = 2880

⇒ 5y² = 2880

⇒ y² = 2880 ÷ 5

⇒ y² = 576

Since the numbers are positive, take the positive square root:

⇒ y = √576 = 24

Now, calculate the first number (x):

⇒ x = 5 × 24 = 120

Calculate the sum of the two numbers:

⇒ Sum = x + y = 120 + 24 = 144

∴ The correct answer is 144.

Additional Information

Ratio Multiplier Rule

When solving problems involving the product of ratios, always attach a variable (like a or k) to the parts. The product of a ratio like a : b is (a × b × k²), not just (a × b × k).

Sum and Product Relationship

A useful algebraic identity connecting sum, difference, and product is: (x + y)² = (x − y)² + 4xy. This helps quickly jump between values when two of the three variables are known.

Perfect Squares

Memorizing perfect squares up to 30 speeds up calculation significantly in banking and aptitude exams. For instance, knowing instantly that 24² = 576 skips the need for long division or prime factorization.

Shortcut Trick

When the overall percentage of a group matches the percentage of one of its sub-groups (males), the other sub-group (females) must have the exact same percentage.

Let the female literacy rate be x%.

14% of Total = 14% of Males + x% of Females

14% of (Males + Females) = 14% of Males + x% of Females ⇒ x = 14%

∴ The correct answer is 14%.

Alternate Method

Given:

Total population = 351000

Number of males = 164000

Overall literacy rate = 14%

Male literacy rate = 14%

Total Population: 351,000
14% Literate

Males: 164,000
14% Literate

Females: 187,000
?% Literate

Literate Females = Total Literate − Literate Males

Calculations:

⇒ Number of females = Total population − Number of males

⇒ Number of females = 351000 − 164000 = 187000

⇒ Total literate population = 14% of 351000 = (14 ÷ 100) × 351000 = 49140

⇒ Literate males = 14% of 164000 = (14 ÷ 100) × 164000 = 22960

⇒ Literate females = Total literate − Literate males

⇒ Literate females = 49140 − 22960 = 26180

⇒ Percentage of literate females = (Literate females ÷ Total females) × 100

⇒ Percentage of literate females = (26180 ÷ 187000) × 100 = 14%

∴ The correct answer is 14%.

Additional Information

Weighted Average Formula

When a group is divided into multiple parts, the overall average is calculated as: A_w = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂)

Alligation Method

It is a quick method used to determine the ratio in which two components are combined to form a mixture: (Quantity 1 ÷ Quantity 2) = (A₂ − A_w) ÷ (A_w − A₁)

Base Independence

If two distinct parts possess the exact same percentage value (e.g., 14%), the combined percentage will also remain unchanged at that value (14%), completely irrespective of the absolute numbers in each group.

Shortcut Trick
Product of the two numbers = HCF × LCM = 6 × 36 = 216.
Check the options to find the pair with a product of 216 and a difference of 30.
12 × 42 = 504 | 24 × 54 = 1296 | 18 × 48 = 864
Option 3 gives 6 × 36 = 216 and a difference of 36 − 6 = 30.
∴ The correct answer is 6 and 36.

Alternate Method
Given:
HCF = 6
LCM = 36
Difference = 30
Formula Used:
Numbers = HCF × x and HCF × y (where x and y are co-prime)
Calculations:
⇒ Let the two numbers be 6x and 6y (where x > y).
⇒ LCM = 6xy = 36
⇒ xy = 6
⇒ Given difference = 6x − 6y = 30
⇒ x − y = 5
⇒ The only co-prime factors of 6 with a difference of 5 are 6 and 1.
⇒ So, x = 6 and y = 1.
⇒ The numbers are 6 × 6 = 36 and 6 × 1 = 6.
∴ The correct answer is 6 and 36.

Additional Information
Product Rule of HCF and LCM
For any two numbers A and B, the product of the numbers is equal to the product of their HCF and LCM: A × B = HCF × LCM.
Co-prime Factors Rule
If the HCF of two numbers is H, they can be represented as Hx and Hy, where x and y are co-prime integers.
LCM of Hx and Hy
The LCM of two numbers represented as Hx and Hy is given by the formula: LCM = H × x × y.

Shortcut Trick

When Simple Interest and time are identical, the ratio of Principals is inversely proportional to their Rates.

⇒ P₁ : P₂ = R₂ : R₁ = 9 : 10

⇒ Total investment ratio parts = 9 + 10 = 19 parts ≡ ₹20,900

⇒ 1 part = 20,900 ÷ 19 = ₹1,100

⇒ Sum invested at 10% (P₁) = 9 parts = 9 × 1,100 = ₹9,900

∴ The correct answer is ₹9,900.

Alternate Method

Given:

Total invested sum = ₹20,900

Rate for 1^st part (R₁) = 10% and Rate for 2^nd part (R₂) = 9%

Time (T) = 5 years for both parts.

Simple interest earned from both parts is equal (SI₁ = SI₂).

Formula Used:

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

Total Sum = ₹20,900

Part 1 (P1)
10% for 5 years

Part 2 (P2)
9% for 5 years

SI1 = SI2

Calculations:

⇒ Let the sum invested at 10% be P₁ and at 9% be P₂.

⇒ P₁ + P₂ = 20,900

Since the simple interest earned from both parts is equal:

⇒ (P₁ × 10 × 5) ÷ 100 = (P₂ × 9 × 5) ÷ 100

⇒ 50 × P₁ = 45 × P₂

⇒ 10 × P₁ = 9 × P₂

⇒ P₁ / P₂ = 9 / 10

⇒ Ratio of sums invested (P₁ : P₂) = 9 : 10

⇒ Total ratio parts = 9 + 10 = 19

⇒ 19 parts = ₹20,900

⇒ 1 part = 20,900 ÷ 19 = ₹1,100

We need the sum invested at 10% (P₁), which corresponds to 9 parts.

⇒ P₁ = 9 × 1,100 = ₹9,900

∴ The correct answer is ₹9,900.

Additional Information

Equal Simple Interests Rule

If parts of a sum are invested such that their simple interests are equal, the ratio of the invested parts is P₁ : P₂ : P₃ = 1/(R₁×T₁) : 1/(R₂×T₂) : 1/(R₃×T₃).

Equal Final Amounts Rule

If parts are invested such that their final return amounts (Principal + SI) are equal, the principal ratio becomes P₁ : P₂ : P₃ = 1/(100 + R₁×T₁) : 1/(100 + R₂×T₂) : 1/(100 + R₃×T₃).

Difference Between SI and CI

For 2 years, the difference between Compound Interest (CI) and Simple Interest (SI) on a certain sum is calculated directly by the formula D = P × (R/100)².

Shortcut Trick

Solve the components mentally by separating the variables and constants.

x² terms: 3 × (4/3) − 4 = 4 − 4 = 0

x terms: 3 × (−30) − 4 × 8 = −90 − 32 = −122

Constant terms: 3 × 18 − 4 × (−14) = 54 + 56 = 110

Combine all parts: −122x + 110

∴ The correct answer is −122x + 110.

Alternate Method

Given:

Expression: 3((4/3)x² − 30x + 18) − 4(x² + 8x − 14)

Calculation:

⇒ Distribute 3 into the first bracket:

⇒ 3 × (4/3)x² − 3 × 30x + 3 × 18 = 4x² − 90x + 54

⇒ Distribute −4 into the second bracket:

⇒ −4 × x² − 4 × 8x − 4 × (−14) = −4x² − 32x + 56

⇒ Combine both expanded expressions:

⇒ (4x² − 90x + 54) + (−4x² − 32x + 56)

⇒ Group the like terms together:

⇒ (4 − 4)x² + (−90 − 32)x + (54 + 56)

⇒ 0x² − 122x + 110

⇒ −122x + 110

∴ The correct answer is −122x + 110.

Additional Information

BODMAS Rule

It specifies the order of operations: Brackets, Orders, Division/Multiplication, Addition/Subtraction. Always solve the expressions inside the brackets first.

Distributive Property

Multiplication distributes over addition and subtraction: a(b ± c) = ab ± ac. This removes the brackets and makes simplification easier.

Like Terms Combination

Only terms with identical variables and powers can be added or subtracted (e.g., ax² and bx²). Different variable terms remain separate.

Shortcut Trick

Convert the ratios to approximate percentages for a quick comparison:

40 : 51 → 40 ÷ 51 ≈ 40 ÷ 50 = 80%

44 : 74 → 44 ÷ 74 ≈ 44 ÷ 75 ≈ 58.6%

42 : 70 → 42 ÷ 70 = 60%

22 : 76 → 22 ÷ 76 ≈ 22 ÷ 75 ≈ 29.3%

∴ The correct answer is 40 : 51.

Alternate Method

Given:

Ratios: 40 : 51, 44 : 74, 42 : 70, 22 : 76

Method Used:

Convert each ratio into a decimal value by dividing the antecedent (first term) by the consequent (second term).

40 : 51
44 : 74
42 : 70
22 : 76

~78.4%

~59.4%

60.0%

~28.9%

Comparison of Ratio Percentages

Calculations:

⇒ 40 ÷ 51 ≈ 0.784

⇒ 44 ÷ 74 ≈ 0.594

⇒ 42 ÷ 70 = 0.600

⇒ 22 ÷ 76 ≈ 0.289

⇒ Comparing the decimal values: 0.784 > 0.600 > 0.594 > 0.289

⇒ The greatest decimal value is 0.784, which matches the ratio 40 : 51.

∴ The correct answer is 40 : 51.

Additional Information

Cross-Multiplication Method

To rapidly compare two fractions a/b and c/d without decimals, simply multiply diagonally. If a × d > b × c, then the fraction a/b is greater.

Percentage Conversion

Any fraction can be efficiently analyzed by multiplying it by 100 to get a percentage, creating an easily comparable base of 100 for all fractions.

Decimal Approximation

In competitive exams, slightly adjusting the denominator to a nearby round number (like changing 51 to 50) saves crucial time while maintaining relative accuracy.

Shortcut Trick

Total CP = ₹456 + ₹20 = ₹476

Total SP = ₹612

Profit = ₹612 − ₹476 = ₹136

Profit % = (136 ÷ 476) × 100 = 200/7%

∴ The correct answer is 200/7% profit.

Alternate Method

Total CP
₹476

Selling Price
₹612

Profit: ₹136

Base CP: ₹456

• Decor: ₹20

Profit % = (136 ÷ 476) × 100 = 200/7%

Given:

Base Cost Price (CP) = ₹456

Overhead expenses (Decoration) = ₹20

Selling Price (SP) = ₹612

Formula Used:

Total CP = Base CP + Overhead expenses

Profit = SP − Total CP

Profit % = (Profit ÷ Total CP) × 100

Calculation:

⇒ Total CP = 456 + 20 = 476

⇒ Profit = 612 − 476 = 136

⇒ Profit % = (136 ÷ 476) × 100

⇒ Profit % = (2 ÷ 7) × 100 = 200/7%

∴ The correct answer is 200/7% profit.

Additional Information

Marked Price & Discount

Discount is always calculated on the Marked Price (MP). Formula: SP = MP − Discount, or SP = MP × (100 − D%) ÷ 100.

Successive Discounts

When two discounts a% and b% are given, the equivalent single discount is [a + b − (a × b) ÷ 100]%.

Dishonest Dealer

If a dealer sells goods at CP but uses a false weight, Profit % = (Error ÷ True Value − Error) × 100.

Shortcut Trick

Let the Principal sum be 100%.

Interest for the 1^st year = 20%

Interest for the 2^nd year = 20% + (20% of 20%) = 20% + 4% = 24%

Given that 24% of the sum = ₹6,786

Sum (100%) = (6786 ÷ 24) × 100 = 282.75 × 100 = ₹28,275

∴ The correct answer is ₹28,275.

Alternate Method

Given:

Interest earned during the 2^nd year = ₹6,786

Rate of interest (R) = 20% per annum

Formula Used:

Amount after 1^st year (A₁) = P × (1 + R/100)

Compound Interest for the 2^nd year = A₁ × (R/100)

  Year 0 Year 1 Year 2 P 1.2 P 1.44 P  1<sup>st</sup> Yr CI = 0.2P  2<sup>nd</sup> Yr CI = 0.24P 0.24 P = ₹6,786

Calculations:

⇒ Let the principal sum be P.

⇒ Amount after 1^st year = P × (1 + 20/100) = 1.2P

⇒ Compound interest for 2^nd year = 1.2P × (20/100) = 0.24P

⇒ Given, 0.24P = 6786

⇒ P = 6786 ÷ 0.24

⇒ P = 678600 ÷ 24

⇒ P = 28275

∴ The correct answer is ₹28,275.

Additional Information

Compound Interest Formula

The total amount A after t years is given by A = P(1 + R/100)^t, where P is the principal and R is the rate of interest.

Effective Rate of Interest

For 2 years at a rate of R%, the effective compound interest rate is calculated as [R + R + (R×R)÷100]%.

Interest for the n^th Year

The compound interest for a specific n^th year is calculated by applying the interest rate directly on the total amount accumulated at the end of the (n−1)^th year.