Quantitative Aptitude PYQs

Shortcut Trick

CP of 1 pen = 180 ÷ 12 = ₹15

SP of 1 pen = ₹21

Profit = 21 − 15 = ₹6

Profit % = (6 ÷ 15) × 100 = 40%

∴ The correct answer is 40%.

Alternate Method

Given:

CP of 1 dozen pens = ₹180

Number of pens in 1 dozen = 12

SP of 1 pen = ₹21

Formula Used:

Profit = SP − CP

Profit % = (Profit ÷ CP) × 100

CP = ₹15

Profit = ₹6

SP = ₹21
Profit % = (6 ÷ 15) × 100 = 40%

Calculations:

⇒ CP of 1 pen = 180 ÷ 12 = ₹15

⇒ Profit on 1 pen = SP − CP = 21 − 15 = ₹6

⇒ Profit % = (6 ÷ 15) × 100

⇒ Profit % = (2 ÷ 5) × 100 = 40%

∴ The correct answer is 40%.

Additional Information

Loss Percentage

When Cost Price (CP) is greater than Selling Price (SP), there is a loss. Loss % = (Loss ÷ CP) × 100.

Selling Price Formula (with Profit %)

SP = CP × (100 + Profit %) ÷ 100

Marked Price & Discount

Discount is always calculated on Marked Price (MP). SP = MP − Discount.

Shortcut Trick

Apply the BODMAS rule directly left to right based on priority.

⇒ (−9) − [−60 ÷ −10] + [−4 × 7]

⇒ −9 − 6 − 28

⇒ −15 − 28 = −43

∴ The correct answer is −43.

Alternate Method

Given:

(−9) − (−60) ÷ (−10) + (−4) × 7

Concept Used:

BODMAS Rule dictates the order of mathematical operations: Brackets, Orders, Division, Multiplication, Addition, and Subtraction.

Calculations:

⇒ According to the BODMAS rule, perform the division operation first:

⇒ (−60) ÷ (−10) = 6

⇒ Substitute this value back into the expression:

⇒ (−9) − 6 + (−4) × 7

⇒ Next, perform the multiplication operation:

⇒ (−4) × 7 = −28

⇒ Substitute this value back into the expression:

⇒ −9 − 6 − 28

⇒ Finally, perform addition and subtraction from left to right:

⇒ −15 − 28 = −43

∴ The correct answer is −43.

Additional Information

BODMAS Rule

It establishes the order of calculations: Brackets, Orders (powers and roots), Division, Multiplication, Addition, and Subtraction. Division and Multiplication share the same priority (solved left to right), as do Addition and Subtraction.

Sign Rules for Multiplication and Division

When multiplying or dividing two numbers: if the signs are identical (both positive or both negative), the resultant value is positive. If the signs are different, the resultant value is negative.

Shortcut Trick

For Compound Interest, multiplying factor = ^t√(Amount ÷ Principal).

Here, Time (t) = 3 years, so we find the cube root of the ratio.

⇒ ³√(1092727 ÷ 1000000) = 103 ÷ 100

This means ₹100 becomes ₹103 in 1 year.

Rate of interest = (103 − 100) ÷ 100 × 100% = 3%.

∴ The correct answer is 3%.

Alternate Method

Principal (P)
₹10,00,000

Amount (A)
₹10,92,727

Time = 3 Years
Rate = R%

Given:

Principal (P) = ₹10,00,000

Amount (A) = ₹10,92,727

Time (t) = 3 years

Formula Used:

Amount = P × (1 + R ÷ 100)^t

Calculations:

⇒ 1092727 = 1000000 × (1 + R ÷ 100)³

⇒ 1092727 ÷ 1000000 = (1 + R ÷ 100)³

⇒ (103 ÷ 100)³ = (1 + R ÷ 100)³

Taking cube root on both sides:

⇒ 103 ÷ 100 = 1 + R ÷ 100

⇒ R ÷ 100 = 103 ÷ 100 − 1

⇒ R ÷ 100 = (103 − 100) ÷ 100

⇒ R ÷ 100 = 3 ÷ 100

⇒ R = 3

∴ The correct answer is 3%.

Additional Information

Effective Rate Method

For 2 years, the effective CI rate = R + R + (R×R)÷100 %. This is faster for calculating exact compound interest when rates are small numbers.

Difference Between CI and SI

For 2 years: Difference = P × (R÷100)².

For 3 years: Difference = P × (R÷100)² × (3 + R÷100).

Half-Yearly Compounding

When interest is compounded half-yearly, the rate is halved (R÷2) and the given time period is doubled (2t) before applying the formula.

Shortcut Trick

Sum of the first two numbers = 79 × 2 = 158

Sum of the remaining two numbers = Total sum − (Sum of first two + Third number)

Sum of the remaining two numbers = 655 − (158 + 135) = 362

Average of the remaining two numbers = 362 ÷ 2 = 181

∴ The correct answer is 181.

Alternate Method

Given:

Total sum of 5 numbers = 655

Average of first 2 numbers = 79

Third number = 135

Formula Used:

Sum = Average × Total number of observations

Sum: 158
3^rd: 135
Remaining Sum: 362

Total Sum of 5 numbers = 655

Avg = 79
1 number
Avg = 362 ÷ 2 = 181

Calculations:

⇒ Sum of the first 2 numbers = 79 × 2 = 158

⇒ Sum of the first 3 numbers = 158 + 135 = 293

⇒ Sum of the remaining 2 numbers = Total sum − Sum of first 3 numbers

⇒ Sum of the remaining 2 numbers = 655 − 293 = 362

⇒ Average of the remaining 2 numbers = 362 ÷ 2

⇒ Average of the remaining 2 numbers = 181

∴ The correct answer is 181.

Additional Information

Weighted Average

When dealing with groups of sizes n₁, n₂... having averages A₁, A₂..., the overall combined average is calculated as (n₁A₁ + n₂A₂ + ...) ÷ (n₁ + n₂ + ...).

Deviation Method

Rather than multiplying large datasets, select an assumed average and calculate the net deviation. The actual average is found using: Assumed Average + (Total Deviation ÷ Number of Observations).

Consecutive Series

The average of an arithmetic progression or any sequence of consecutive numbers is exactly the middle term, which can be easily found as (First Term + Last Term) ÷ 2.

Shortcut Trick

Valid votes as a percentage of total registered voters = 80% × 98% = 78.4%.

Winner gets 52% of valid votes, meaning Loser gets 48%. Winning margin = 4% of valid votes.

Margin as a percentage of total voters = 4% of 78.4% = 3.136%.

Given 3.136% → 1568 votes.

Total registered voters (100%) = 1568 ÷ 3.136 × 100 = 50000.

∴ The correct answer is 50000.

Alternate Method

Given:

Votes polled = 80% of registered voters

Invalid votes = 2% of polled votes

Winning candidate votes = 52% of valid votes

Margin of victory = 1568 votes

Formula Used:

Valid Votes = Total Polled Votes − Invalid Votes

Margin = Winner's Votes − Loser's Votes

80% cast vote
98% valid (2% invalid)

Total Voters (100x)

Polled Votes (80x)

Valid Votes (78.4x)

Winner (52%)

Loser (48%)

Margin = 4% of Valid Votes = 1568

Calculations:

⇒ Let the total number of registered voters be 100x.

⇒ Total votes polled = 80% of 100x = 80x

⇒ Invalid votes = 2% of 80x = 1.6x

⇒ Valid votes = 80x − 1.6x = 78.4x

⇒ Winner's votes = 52% of 78.4x

⇒ Loser's votes = (100 − 52)% of 78.4x = 48% of 78.4x

⇒ Margin of victory = Winner's votes − Loser's votes

⇒ Margin = (52% − 48%) of 78.4x = 4% of 78.4x = 3.136x

⇒ Given margin = 1568 votes

⇒ 3.136x = 1568

⇒ x = 1568 ÷ 3.136 = 500

⇒ Total registered voters = 100x = 100 × 500 = 50000

∴ The correct answer is 50000.

Additional Information

Valid and Invalid Votes in Elections

Invalid votes are discarded when determining the results. The winner and loser share percentages are calculated strictly on the base of Valid Votes unless explicitly stated otherwise.

Successive Percentage Change

When multiple percentage reductions occur successively (e.g., voters not voting, then votes becoming invalid), the final quantity can be found by multiplying the remaining fractions: Final Value = Initial Base × (1 − r₁/100) × (1 − r₂/100).

Margin of Victory Formula

The winning margin is directly equivalent to the difference between the percentage obtained by the winner and the loser: Margin = Total Valid Votes × (Winner % − Loser %).

Shortcut Trick

For divisibility by 11, the difference between the sums of alternating digits must be 0 or a multiple of 11.

Sum of odd-positioned digits = N + 7 + 0 = N + 7

Sum of even-positioned digits = 3 + M + 1 = M + 4

Difference = (N + 7) − (M + 4) = N − M + 3

Equating the difference to 0 gives N − M + 3 = 0 ⇒ M − N = 3.

∴ The correct answer is M − N = 3.

Alternate Method

Given:

The 6-digit number N37M01 is divisible by 11.

Formula Used:

A number is divisible by 11 if the difference of the sum of its digits at odd places and the sum of its digits at even places is 0 or a multiple of 11.

Odd Sum = N + 7 + 0 = N + 7

N

3

7

M

0

1

Even Sum = 3 + M + 1 = M + 4

Calculations:

⇒ Sum of digits at odd places (1^st, 3^rd, 5^th positions) = N + 7 + 0 = N + 7

⇒ Sum of digits at even places (2^nd, 4^th, 6^th positions) = 3 + M + 1 = M + 4

⇒ Difference = (N + 7) − (M + 4) = N − M + 3

⇒ For the number to be perfectly divisible by 11, this difference must be exactly 0, 11, or −11.

⇒ Case 1: Assume Difference = 0

⇒ N − M + 3 = 0 ⇒ M − N = 3

⇒ Case 2: Assume Difference = 11

⇒ N − M + 3 = 11 ⇒ N − M = 8

⇒ Checking the given options, M − N = 3 represents the correctly matched possible relation.

∴ The correct answer is M − N = 3.

Additional Information

Divisibility Rule of 3 & 9

A number is divisible by 3 or 9 if the sum of all its digits is a multiple of 3 or 9, respectively.

Divisibility Rule of 4 & 8

A number is divisible by 4 if its last two digits form a number divisible by 4. It is divisible by 8 if its last three digits form a number divisible by 8.

Combined Divisibility Rule of 7, 11 & 13

A number is divisible by 7, 11, or 13 if the difference between the number formed by its last three digits and the remaining digits is a multiple of 7, 11, or 13.

Shortcut Trick

If sec θ + tan θ = p, then cos θ = 2p ÷ (p² + 1).

Given p = 3/2, substitute directly into the shortcut formula.

cos θ = (2 × 3/2) ÷ ((3/2)² + 1)

cos θ = 3 ÷ (9/4 + 1) = 3 ÷ (13/4) = 12/13

∴ The correct answer is 12/13.

Alternate Method

Given:

sec θ + tan θ = 3/2

Formula Used:

sec² θ − tan² θ = 1

θ
Base = 12
Perpendicular = 5
Hypotenuse = 13

Calculations:

⇒ (sec θ − tan θ) × (sec θ + tan θ) = 1

⇒ (sec θ − tan θ) × (3/2) = 1

⇒ sec θ − tan θ = 2/3

Adding both equations:

⇒ (sec θ + tan θ) + (sec θ − tan θ) = 3/2 + 2/3

⇒ 2 sec θ = (9 + 4) ÷ 6 = 13/6

⇒ sec θ = 13/12

⇒ sec θ = Hypotenuse ÷ Base = 13/12

⇒ By Pythagoras theorem: Perpendicular = √(13² − 12²) = √(169 − 144) = √25 = 5

⇒ cos θ = Base ÷ Hypotenuse = 12/13

∴ The correct answer is 12/13.

Additional Information

Pythagorean Triplets

A Pythagorean triplet consists of three positive integers a, b, and c, such that a² + b² = c². Common exam triplets include (3, 4, 5), (5, 12, 13), and (8, 15, 17).

Trigonometric Identities

The three primary Pythagorean identities are: sin² θ + cos² θ = 1, sec² θ − tan² θ = 1, and cosec² θ − cot² θ = 1.

Secant and Tangent Pair Rule

For any angle θ, if sec θ + tan θ = x, then sec θ − tan θ = 1/x. This specific pairing pattern is derived directly from factoring the difference of squares identity.

Shortcut Trick

Product of two numbers = LCM × HCF = 150.

Given difference between the numbers = 5.

Check the given pairs to find which numbers give a product of 150 and a difference of 5.

The pair 15 and 10 satisfies both conditions (15 × 10 = 150 and 15 − 10 = 5).

∴ The correct answer is 15, 10.

Alternate Method

Given:

Product of LCM and HCF = 150

Difference of the two numbers = 5

Formula Used:

Product of two numbers = LCM × HCF

Calculation:

Let the smaller number be y and the larger number be x.

⇒ x − y = 5

⇒ x = y + 5

⇒ x × y = 150

⇒ (y + 5) × y = 150

⇒ y² + 5y − 150 = 0

⇒ y² + 15y − 10y − 150 = 0

⇒ y(y + 15) − 10(y + 15) = 0

⇒ (y − 10)(y + 15) = 0

⇒ The roots are y = 10 and y = −15.

⇒ Since numbers are assumed to be positive in this context, y = 10.

⇒ x = 10 + 5 = 15

∴ The correct answer is 15, 10.

Additional Information

Product Rule of LCM and HCF

The product of two given numbers is exactly equal to the product of their HCF and LCM. Formula: A × B = HCF(A, B) × LCM(A, B). Note that this rule applies only to two numbers, not for three or more.

Co-Prime Numbers

If two numbers are co-prime (having no common factor other than 1), their HCF is exactly 1, and their LCM is simply the product of the two numbers.

HCF and LCM of Fractions

HCF of fractions = (HCF of Numerators) ÷ (LCM of Denominators). LCM of fractions = (LCM of Numerators) ÷ (HCF of Denominators).

Shortcut Trick

Let the original number be 100%.

After a 70% increase, the number becomes 170%.

170% → 3570

100% → (3570 ÷ 170) × 100 = 2100

∴ The correct answer is 2100.

Alternate Method

Original (100%)

• 70%

Total = 170% = 3570

Given:

A number when increased by 70% becomes 3570.

Formula Used:

Final Value = Original Value × (1 + Percentage Increase ÷ 100)

Calculation:

⇒ Let the original number be x.

⇒ Final Value = x + (70 ÷ 100) × x

⇒ 3570 = x + 0.7x

⇒ 1.7x = 3570

⇒ x = 3570 ÷ 1.7

⇒ x = 2100

∴ The correct answer is 2100.

Additional Information

Percentage Change

Percentage change is given by: ((Final Value − Initial Value) ÷ Initial Value) × 100%.

Successive Percentage Change

If a value is changed by a% and then by b%, the net percentage change is (a + b + (a × b) ÷ 100)%.

Base Change Rule

If A is R% more than B, then B is [R ÷ (100 + R) × 100]% less than A.

Shortcut Trick

Let the father's age be F and the son's age be S.

Given equations: 2S + F = 42 and S + 2.2F = 89

Multiply the second equation by 2: 2S + 4.4F = 178

Subtract the first equation: (4.4F + 2S) − (F + 2S) = 178 − 42

⇒ 3.4F = 136 ⇒ F = 40

∴ The correct answer is 40 years.

Alternate Method

Given:

Two times a son's age added to his father's age is 42 years.

2.2 times the father's age added to the son's age is 89 years.

Visual Representation of the Equations

Father

Son

Son

= 42

2.2 × Father

Son

= 89

Calculations:

Let the father's age be F and the son's age be S.

⇒ F + 2S = 42                   --- (1)

⇒ 2.2F + S = 89               --- (2)

From equation (1), express F in terms of S:

⇒ F = 42 − 2S

Substitute this value of F into equation (2):

⇒ 2.2 × (42 − 2S) + S = 89

⇒ 92.4 − 4.4S + S = 89

⇒ 92.4 − 3.4S = 89

⇒ 3.4S = 92.4 − 89

⇒ 3.4S = 3.4

⇒ S = 1 year

Now, substitute S = 1 back into the expression for F:

⇒ F = 42 − 2 × 1

⇒ F = 42 − 2

⇒ F = 40 years

∴ The correct answer is 40 years.

Additional Information

Linear Equations in Two Variables

A system of equations containing two unknown variables, commonly solved using substitution, elimination, or cross-multiplication.

Substitution Method

Express one variable in terms of the other from the first equation and substitute it into the second equation to easily solve for the unknowns.

Elimination Method

Multiply equations by suitable numbers to make the coefficients of one variable equal, then add or subtract the equations to eliminate that variable completely.

Shortcut Trick

When the simple interest earned from two parts over the same time period is equal, the ratio of their principals is inversely proportional to their rates.

Ratio of parts = 1/R₁ : 1/R₂ = 1/14 : 1/4 = 4 : 14 = 2 : 7

Total parts = 2 + 7 = 9 parts

Sum invested at 14% = (2/9) × 21,600 = 4,800

∴ The correct answer is ₹4800.

Alternate Method

Given:

Total Investment = ₹21,600

Rate 1 (R₁) = 14% per annum, Rate 2 (R₂) = 4% per annum

Time (T) = 6 years for both

Simple Interest 1 (SI₁) = Simple Interest 2 (SI₂)

Formula Used:

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

Total: ₹21,600

Part 1 (x)
R = 14%, T = 6 yr
SI₁

Part 2 (y)
R = 4%, T = 6 yr
SI₂
SI₁ = SI₂

x : y = 4 : 14 = 2 : 7

Calculations:

Let the amount invested at 14% be x and at 4% be y.

⇒ Total Investment = x + y = 21,600

According to the question, SI on both investments are equal.

⇒ (x × 14 × 6) ÷ 100 = (y × 4 × 6) ÷ 100

⇒ 84x = 24y

⇒ x / y = 24 / 84 = 2 / 7

Thus, x : y = 2 : 7

⇒ Sum of ratio parts = 2 + 7 = 9 parts

⇒ 9 parts = 21,600

⇒ 1 part = 21,600 ÷ 9 = 2,400

⇒ Sum invested at 14% (x) = 2 parts = 2 × 2,400 = 4,800

∴ The correct answer is ₹4800.

Additional Information

Equal Simple Interest Rule

When a sum is divided into parts such that their Simple Interests are equal, the parts are in the ratio of 1/(R₁T₁) : 1/(R₂T₂) : 1/(R₃T₃).

Equal Amount Rule

When a sum is divided into parts such that the Amounts obtained from them are equal, the parts are in the ratio of 1/(100 + R₁T₁) : 1/(100 + R₂T₂).

Simple Interest Principle

Simple Interest = (Principal × Rate × Time) ÷ 100. The principal remains constant throughout the entire period, making interest accumulation linear over time.

Shortcut Trick
Total capacity = LCM(33, 22, 18) = 198 units.
Efficiency of filling Pipe A = 198 ÷ 33 = +6 units/min.
Efficiency of filling Pipe B = 198 ÷ 22 = +9 units/min.
Combined Efficiency (A + B − C) = 198 ÷ 18 = +11 units/min.
Efficiency of emptying Pipe C = (6 + 9) − 11 = 4 units/min.
Time taken by Pipe C to empty = 198 ÷ 4 = 49.5 minutes.
∴ The correct answer is 49.5 minutes.
Alternate Method
Given:
⇒ Time taken by Pipe A to fill the cistern = 33 minutes
⇒ Time taken by Pipe B to fill the cistern = 22 minutes
⇒ Time taken by all three pipes (A, B and emptying Pipe C) to fill = 18 minutes
Formula Used:
⇒ Total Work = Efficiency × Time taken

Pipe
Time (min)
Total Work
Efficiency

Pipe A
33
198
+6

Pipe B
22
+9

A + B − C
18
+11

Calculations:
⇒ Let the efficiency of the emptying pipe C be x.
⇒ Total efficiency when all pipes are open = Efficiency of A + Efficiency of B − Efficiency of C
⇒ 11 = 6 + 9 − x
⇒ 11 = 15 − x
⇒ x = 4 units/min.
⇒ Time taken by Pipe C to empty the completely filled cistern = Total Work ÷ Efficiency of C
⇒ Time taken = 198 ÷ 4
⇒ Time taken = 49.5 minutes.
∴ The correct answer is 49.5 minutes.
Additional Information
Pipes and Cisterns Sign Convention
In pipe and cistern problems, the work done by a filling pipe is considered positive (+), whereas the work done by an emptying pipe or leak is taken as negative (−).
Efficiency & Time Relation
Efficiency is inversely proportional to the time taken. For any setup, Total Capacity = Efficiency × Time.
Combined Work Rule
If pipe A can fill a tank in x hours and pipe B can empty it in y hours (where y > x), the net part filled in 1 hour is given by (1/x) − (1/y).

Shortcut Trick

Directly expand the given expression and observe that the quadratic terms cancel out:

10x² − 10 − 10x² − 45x + 25 = 18

⇒ −45x + 15 = 18

⇒ −45x = 3 ⇒ x = −1/15

∴ The correct answer is −1/15.

Alternate Method

Given:

2(5x² − 5) − 5(2x² + 9x − 5) = 18

Calculations:

⇒ Apply the distributive property to expand the terms:

⇒ 10x² − 10 − 10x² − 45x + 25 = 18

⇒ Combine like terms (notice that the 10x² terms cancel each other out):

⇒ −45x + 15 = 18

⇒ Isolate x on one side by subtracting 15 from both sides:

⇒ −45x = 18 − 15

⇒ −45x = 3

⇒ Divide both sides by −45:

⇒ x = 3 / (−45)

⇒ x = −1/15

∴ The correct answer is −1/15.

Additional Information

BODMAS Rule

It dictates the required order of mathematical operations: Brackets, Orders, Division, Multiplication, Addition, and Subtraction. Expressions inside brackets must be simplified or expanded first.

Distributive Property

Multiplying a sum by a given number yields the same result as multiplying each addend by that number and then adding the products together: a(b + c) = ab + ac.

Linear Equations

Equations of the first degree involving one variable. They can be solved directly by bringing all constant values to one side and variable terms to the opposite side to isolate the variable.

Shortcut Trick

Using the BODMAS rule, evaluate the expression from innermost brackets to outermost:

⇒ [640 ÷ { 22 + 6 × ( 4 − 5 )}] = [640 ÷ { 22 + 6 × (−1) }]

⇒ [640 ÷ { 22 − 6 }] = [640 ÷ 16]

⇒ 40

∴ The correct answer is 40.

Alternate Method

Given:

The expression is [640 ÷ { 22 + 6 × ( 4 − 5 )}].

Concept Used:

BODMAS Rule: Brackets, Orders, Division, Multiplication, Addition, Subtraction.

Calculations:

⇒ First, simplify the terms inside the innermost parentheses ():

⇒ (4 − 5) = −1

⇒ Substitute this back into the expression:

⇒ [640 ÷ { 22 + 6 × (−1) }]

⇒ Next, solve the multiplication inside the curly brackets {}:

⇒ 6 × (−1) = −6

⇒ Now, solve the subtraction inside the curly brackets:

⇒ { 22 − 6 } = 16

⇒ Finally, solve the division inside the square brackets []:

⇒ 640 ÷ 16 = 40

∴ The correct answer is 40.

Additional Information

BODMAS Rule

It dictates the order of mathematical operations: Brackets, Of/Orders, Division, Multiplication, Addition, and Subtraction. Always evaluate expressions in this sequence.

Order of Brackets

When multiple brackets are present, solve them from the innermost to outermost: first parenthesis (), then curly braces {}, and finally square brackets [].

Sign Rules in Multiplication

Multiplying a positive number by a negative number results in a negative product. Example: (+) × (−) = (−).

Shortcut Trick

We can use the successive percentage change formula: Overall % = a + b + (a × b) ÷ 100

Here, the markup percentage (a) = +40% and the discount percentage (b) = −45%

Overall Profit/Loss % = 40 − 45 − (40 × 45) ÷ 100

Overall Profit/Loss % = −5 − 1800 ÷ 100 = −5 − 18 = −23%

The negative sign indicates a net loss of 23%.

∴ The correct answer is 23%.

Alternate Method

Given:

Markup = 40% above the cost price

Discount = 45% on the marked price

Formula Used:

Marked Price (MP) = CP + (Markup% of CP)

Selling Price (SP) = MP − (Discount% of MP)

Loss % = ((CP − SP) ÷ CP) × 100

SP (77)
CP (100)
MP (140)

+40% (Markup)
−45% (Discount)
23% Loss

Calculations:

⇒ Let the Cost Price (CP) of the goods be 100.

⇒ The dealer marks his goods 40% above the CP.

⇒ Marked Price (MP) = 100 + 40% of 100 = 140

⇒ He allows a discount of 45% on the MP.

⇒ Discount Amount = 45% of 140 = (45 ÷ 100) × 140 = 63

⇒ Selling Price (SP) = MP − Discount = 140 − 63 = 77

⇒ Since the SP (77) is less than the CP (100), there is a loss.

⇒ Loss = CP − SP = 100 − 77 = 23

⇒ Loss % = (23 ÷ 100) × 100 = 23%

∴ The correct answer is 23%.

Additional Information

Direct Relation between CP and MP

The ratio of Cost Price to Marked Price can be directly calculated using the rule: CP ÷ MP = (100 − Discount%) ÷ (100 + Profit% or − Loss%).

Successive Percentage Change

When a value is successfully changed by a% and then b%, the net effective change is calculated as (a + b + (a × b) ÷ 100)%. Use positive signs for increases/markups and negative signs for decreases/discounts.

Same Markup and Discount

If a product's price is marked up by x% and then a discount of x% is given on it, the seller always incurs a net loss which is exactly equal to (x² ÷ 100)%.

Shortcut Trick

Let the original CP = 100a. Since the initial profit is 54%, the original SP = 154a.

After changes, New CP = 100a + 75 and New SP = 154a − 66.

The new profit is 40.8%, which means the ratio of New SP to New CP is 140.8/100 = 176/125.

Equating ratios: (154a − 66) ÷ (100a + 75) = 176 ÷ 125

⇒ 11(14a − 6) ÷ 25(4a + 3) = 176 ÷ 125 ⇒ 5(14a − 6) = 16(4a + 3)

⇒ 70a − 30 = 64a + 48 ⇒ 6a = 78 ⇒ a = 13. Original CP = 100a = 100 × 13 = 1300.

∴ The correct answer is ₹1300.

Alternate Method

Given:

Original Profit = 54%

Let the original Cost Price (CP) be x.

⇒ Original Selling Price (SP) = x + 54% of x = 1.54x

If CP is increased by ₹75 and SP is reduced by ₹66:

⇒ New CP = x + 75

⇒ New SP = 1.54x − 66

New Profit = 40.8%

Formula Used:

Selling Price = Cost Price × (1 + Profit/100)

Calculations:

⇒ 1.54x − 66 = (x + 75) × (1 + 40.8/100)

⇒ 1.54x − 66 = (x + 75) × 1.408

⇒ 1.54x − 66 = 1.408x + (75 × 1.408)

⇒ 1.54x − 66 = 1.408x + 105.6

⇒ 1.54x − 1.408x = 105.6 + 66

⇒ 0.132x = 171.6

⇒ x = 171.6 ÷ 0.132

⇒ x = 171600 ÷ 132 = 1300

∴ The correct answer is ₹1300.

Additional Information

Profit/Loss Percentage

Profit and Loss percentages are always calculated on the base Cost Price (CP) unless specifically mentioned otherwise. Formula: Profit % = [(SP − CP) ÷ CP] × 100.

Successive Absolute Changes

When both CP and SP change by different absolute numerical values, the profit ratio changes entirely. Applying the standard relation New SP = New CP × (1 + Profit/100) helps in framing a direct linear equation to solve such problems effortlessly.

Multiplying Factor Method

Instead of calculating the absolute profit and adding it to the CP, you can directly find the SP by multiplying the CP by a combined decimal factor. For instance, for a 40.8% profit margin, the directly used multiplying factor is 1.408 (i.e., SP = CP × 1.408).

Shortcut Trick

Convert the fraction directly into a percentage.

4/5 = (4 ÷ 5) × 100% = 80%

Given: 69% of First Number = 80% of Second Number

⇒ First Number / Second Number = 80 / 69

∴ The correct answer is 80 : 69.

Alternate Method

Given:

69% of the first number is equal to 4/5 of the second number.

69% of I

=

4/5 of II (80%)

I : II = 80 : 69

Calculations:

Let the first number be x and the second number be y.

⇒ (69 / 100) × x = (4 / 5) × y

⇒ x / y = (4 / 5) ÷ (69 / 100)

⇒ x / y = (4 / 5) × (100 / 69)

⇒ x / y = (4 × 100) / (5 × 69)

⇒ x / y = (4 × 20) / 69

⇒ x / y = 80 / 69

∴ The correct answer is 80 : 69.

Additional Information

Ratio & Proportion

A ratio compares two quantities using division (a/b or a:b). By equating percentages or fractions, we deduce the proportional relationship between the elements.

Fraction to Percentage Conversion

To convert a fraction into a percentage, directly multiply it by 100. For example, 4/5 × 100% = 80%.

Inverse Relationship

If A % of x = B % of y, then the ratio of x to y is always inversely proportional to their percentages, giving x : y = B : A.

Shortcut Trick

For symmetric travel times, use the direct linear relationship:

Time for walking both ways = Time for (Walk + Drive) + Time gained by driving

⇒ Total Time = 4 hours 15 minutes + 4 hours

⇒ Total Time = 8 hours 15 minutes

∴ The correct answer is 8 hours 15 minutes.

Alternate Method

Given:

Time taken to walk one way and drive back = 4 hours 15 minutes = 255 minutes.

Time gained by driving both ways = 4 hours = 240 minutes.

Walk + Drive

Walk (W)
D
4h 15m

Drive + Drive

D
D

Gained 4h
15m

Walk + Walk

Walk (W)
Walk (W)
8h 15m

Calculations:

Let the time taken to walk one way be W and drive one way be D.

⇒ W + D = 255 minutes

Time gained by driving both ways is 240 minutes, which means 2D takes 240 minutes less than walking and driving.

⇒ 2D = 255 − 240 = 15 minutes

⇒ D = 7.5 minutes

Substitute the value of D in the first equation:

⇒ W + 7.5 = 255

⇒ W = 255 − 7.5 = 247.5 minutes

Time taken to walk both ways = 2W

⇒ 2W = 2 × 247.5 = 495 minutes

⇒ 495 minutes = 8 hours 15 minutes

∴ The correct answer is 8 hours 15 minutes.

Additional Information

Relative Speed

When two objects move in opposite directions, relative speed = S₁ + S₂. When moving in the same direction, relative speed = |S₁ − S₂|.

Average Speed

If a person covers equal distances at speeds S₁ and S₂, the average speed for the entire journey is calculated as (2 × S₁ × S₂) ÷ (S₁ + S₂).

Linear Equations in Time & Distance

For symmetric travel times, 2W = (W + D) + (W − D) and 2D = (W + D) − (W − D). This directly bypasses calculating single variable speeds or distances.

Shortcut Trick

The average of the first n even natural numbers is given directly by the formula: (n + 1).

Here, the number of terms (n) = 168.

Average = 168 + 1

Average = 169

∴ The correct answer is 169.

Alternate Method

Given:

Number of terms (n) = 168

The series of first 168 even numbers is: 2, 4, 6, 8, ... up to 168 terms.

Formula Used:

For an Arithmetic Progression, Average = (First Term + Last Term) ÷ 2

Last term (l) = a + (n − 1)d

Average of an Arithmetic Progression

2
(1^st term)

4
6

169
Average

334

336
(168^th term)

Avg = (First + Last) ÷ 2

Calculations:

⇒ First term (a) = 2

⇒ Common difference (d) = 2

⇒ Last term (l) = 2 + (168 − 1) × 2

⇒ l = 2 + 167 × 2

⇒ l = 2 + 334 = 336

⇒ Average = (2 + 336) ÷ 2

⇒ Average = 338 ÷ 2

⇒ Average = 169

∴ The correct answer is 169.

Additional Information

Average of first n natural numbers

The average of the first n natural numbers is given by the formula (n + 1) ÷ 2.

Average of first n odd numbers

The average of the first n odd natural numbers is exactly equal to n.

Average of squares of first n natural numbers

The average of the squares of the first n natural numbers is (n + 1)(2n + 1) ÷ 6.

Average of an Arithmetic Progression

If a series is an Arithmetic Progression (having a constant difference between consecutive terms), its average is always the middle term, which can be found using (First Term + Last Term) ÷ 2.

Shortcut Trick

Height of an isosceles triangle (h) = √(a² − (b/2)²)

Where a = equal side = 85 cm, b = base = 72 cm

h = √(85² − 36²) = √((85 − 36) × (85 + 36)) = √(49 × 121) = 7 × 11 = 77 cm

Area = 1/2 × base × height = 1/2 × 72 × 77 = 36 × 77 = 2772 cm²

∴ The correct answer is 2772 cm².

Alternate Method

Given:

Length of equal sides of the triangle (a) = 85 cm

Base of the triangle (b) = 72 cm

Formula Used:

Area of triangle = 1/2 × base × height

Height of an isosceles triangle = √(a² − (b/2)²)

A
B
C
D
85 cm
85 cm
h = 77 cm
36 cm
36 cm

72 cm

Calculations:

⇒ The altitude drawn to the base of an isosceles triangle bisects the base into two equal halves.

⇒ Base of the right-angled triangle formed (BD) = 72 ÷ 2 = 36 cm

⇒ Using Pythagoras theorem in right-angled triangle ABD: (Height)² + (Base)² = (Hypotenuse)²

⇒ h² + 36² = 85²

⇒ h² = 85² − 36²

⇒ Using a² − b² = (a − b)(a + b), we get:

⇒ h = √((85 − 36) × (85 + 36))

⇒ h = √(49 × 121) = 7 × 11 = 77 cm

⇒ Area of the triangle = 1/2 × Base × Height

⇒ Area = 1/2 × 72 × 77 = 36 × 77

⇒ Area = 2772 cm²

∴ The correct answer is 2772 cm².

Additional Information

Heron's Formula

The area of any triangle can also be found using Area = √(s(s − a)(s − b)(s − c)), where s is the semi-perimeter (a + b + c)/2.

Properties of Isosceles Triangle

In an isosceles triangle, the perpendicular dropped from the apex to the unequal base also acts as an angle bisector and a median, dividing the triangle into two congruent right-angled triangles.

Equilateral Triangle Area

If all three sides of a triangle are equal (a), Area = (√3 / 4) × a² and Height = (√3 / 2) × a.