Quantitative Aptitude PYQs

Shortcut Trick

List the odd numbers in the given range, excluding those ending in 5: 477, 479, 481, 483, 487, 489, 491.

Eliminate multiples of 3 by checking the sum of digits: 477, 483, and 489 are eliminated.

Remaining numbers to check: 479, 481, 487, 491.

Check divisibility for 481: 481 = 13 × 37, so it is not prime.

The remaining numbers 479, 487, and 491 are not divisible by any prime up to √500 ≈ 22, hence they are prime.

∴ The correct answer is 3.

Alternate Method

Given:

The numbers lie between 476 and 491 (both included).

Concept Used:

A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself.

To check if a number N is prime, we only need to test divisibility by prime numbers up to √N.

Calculation:

⇒ The numbers between 476 and 491 are: 476, 477, 478, 479, 480, 481, 482, 483, 484, 485, 486, 487, 488, 489, 490, 491.

⇒ First, exclude all even numbers and numbers ending in 5, as they cannot be prime.

⇒ The potential prime numbers left are: 477, 479, 481, 483, 487, 489, 491.

⇒ Check 477: 4 + 7 + 7 = 18 (Divisible by 3) → Not prime.

⇒ Check 479: √479 ≈ 21.8. Prime numbers to check ≤ 21 are 2, 3, 5, 7, 11, 13, 17, 19. It is not divisible by any of these → Prime.

⇒ Check 481: 481 ÷ 13 = 37. It is exactly divisible by 13 → Not prime.

⇒ Check 483: 4 + 8 + 3 = 15 (Divisible by 3) → Not prime.

⇒ Check 487: √487 ≈ 22. Not divisible by any prime number ≤ 19 → Prime.

⇒ Check 489: 4 + 8 + 9 = 21 (Divisible by 3) → Not prime.

⇒ Check 491: √491 ≈ 22.1. Not divisible by any prime number ≤ 19 → Prime.

⇒ The prime numbers between 476 and 491 are 479, 487, and 491.

⇒ Total count of prime numbers = 3.

∴ The correct answer is 3.

Additional Information

How to Check for Primality

To check if a number N is prime, calculate the approximate square root of N. Divide N by all prime numbers less than or equal to √N. If none divide N exactly, it is guaranteed to be a prime number.

Divisibility Rule for 3 and 9

A number is divisible by 3 (or 9) if the sum of its digits is divisible by 3 (or 9). This rule is incredibly useful for quickly eliminating large composite numbers during prime number checks.

Co-Prime Numbers

Two numbers are said to be co-prime (or relatively prime) if they do not have any common prime factor other than 1. Their Highest Common Factor (HCF) is always 1.

Shortcut Trick

Since distance is constant, Speed and Time are inversely proportional.

S₁ × T₁ = S₂ × T₂

⇒ 75 × 17 = S₂ × 5

⇒ S₂ = (75 × 17) ÷ 5

⇒ S₂ = 15 × 17 = 255 km/hr

∴ The correct answer is 255 km/hr.

Alternate Method

Total Distance (D) is Constant

Case 1: Speed = 75 km/hr | Time = 17 hrs

Case 2: Speed = ? | Time = 5 hrs

Given:

Initial Speed (S₁) = 75 km/hr

Initial Time (T₁) = 17 hours

New Time (T₂) = 5 hours

Formula Used:

Distance = Speed × Time

Calculation:

⇒ Total Distance = 75 × 17

⇒ Total Distance = 1275 km

Now, to cover the same distance in 5 hours:

⇒ New Speed (S₂) = Total Distance ÷ New Time

⇒ S₂ = 1275 ÷ 5

⇒ S₂ = 255 km/hr

∴ The correct answer is 255 km/hr.

Additional Information

Proportionality Rule

If Distance is constant, Speed and Time are inversely proportional (S ∝ 1 / T). Therefore, the ratio of speeds is the inverse of the ratio of times taken.

Average Speed

When traveling equal distances at two different speeds S₁ and S₂, the average speed is given by (2 × S₁ × S₂) ÷ (S₁ + S₂).

Relative Speed

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

Shortcut Trick

Using the combined exponent rule: a^m ÷ aⁿ × a^p = a^{m − n + p}

Given expression: 48⁻¹⁷ ÷ 48¹² × 48⁻¹³

⇒ Combine the powers since the base (48) is common: −17 − 12 + (−13)

⇒ 48⁻⁴²

∴ The correct answer is 48^-42.

Alternate Method

Given:

Expression = 48⁻¹⁷ ÷ 48¹² × 48⁻¹³

Formulas Used:

a^m ÷ aⁿ = a^{m − n}

a^m × aⁿ = a^{m + n}

Calculation:

According to BODMAS, multiplication and division have the same priority. We evaluate from left to right.

⇒ (48⁻¹⁷ ÷ 48¹²) × 48⁻¹³

⇒ 48^{(−17 − 12)} × 48⁻¹³

⇒ 48⁻²⁹ × 48⁻¹³

Now apply the multiplication rule:

⇒ 48^{(−29 + (−13))}

⇒ 48^{(−29 − 13)}

⇒ 48⁻⁴²

∴ The correct answer is 48^-42.

Additional Information

Law of Product and Quotient

When multiplying numbers with the same base, add their exponents: a^m × aⁿ = a^{m + n}. When dividing, subtract their exponents: a^m ÷ aⁿ = a^{m − n}.

Power of a Power Rule

To raise a power to another power, multiply the exponents together: (a^m)ⁿ = a^{m × n}.

Negative Exponent Rule

A negative exponent means division. Move the term with the negative exponent to the denominator to make it positive: a⁻ⁿ = 1 ÷ aⁿ.

Zero Exponent Rule

Any non-zero base raised to the power of zero is always equal to 1: a⁰ = 1 (where a ≠ 0).

Shortcut Trick

Since the time period is identical, the Profit Ratio equals the Capital Ratio.

Ratio = 1090 : 1170 : 1340 = 109 : 117 : 134

Total ratio parts = 109 + 117 + 134 = 360

Value of 360 parts = ₹1620 ⇒ 1 part = 1620 ÷ 360 = ₹4.5

Rohan's share = 134 parts × 4.5 = ₹603

∴ The correct answer is ₹603.

Alternate Method

Given:

Sneha's investment = ₹1,090

Yash's investment = ₹1,170

Rohan's investment = ₹1,340

Total Profit = ₹1,620

Formula Used:

When the duration of investment is the same for all partners, their profits are shared directly in the ratio of their investments.

Total Profit = ₹1620 | Total Parts = 360

₹1090
₹1170
₹1340

109x
117x
134x

Sneha
Yash
Rohan

Calculations:

⇒ Profit Ratio (Sneha : Yash : Rohan) = 1090 : 1170 : 1340

⇒ Simplified Ratio = 109 : 117 : 134

⇒ Sum of ratio parts = 109 + 117 + 134 = 360 parts

⇒ Rohan's share = (Rohan's ratio part ÷ Total ratio parts) × Total Profit

⇒ Rohan's share = (134 ÷ 360) × 1620

⇒ Rohan's share = 134 × 4.5

⇒ Rohan's share = ₹603

∴ The correct answer is ₹603.

Additional Information

Simple Partnership

When all partners invest their capitals for the exact same duration, the profit is divided purely in the ratio of their invested amounts.

Compound Partnership

When partners invest for varying durations, the profit ratio is calculated as Capital₁ × Time₁ : Capital₂ × Time₂.

Active and Sleeping Partners

An active partner usually receives a distinct salary or commission for managing operations, which is deducted from the total profit before the remainder is split according to investment shares.

Shortcut Trick

Divisibility rule for 41: Subtract 4 × (last digit) from the remaining truncated number.

58740 − (7 × 4) = 58712

5871 − (2 × 4) = 5863

586 − (3 × 4) = 574

57 − (4 × 4) = 41 (which is clearly divisible by 41)

∴ The correct answer is 41.

Alternate Method

Given number = 587407

We check the divisibility by dividing the number directly by the given options.

⇒ Option 1: 34 is an even number, so it cannot completely divide an odd number like 587407.

⇒ Option 2: 587407 ÷ 43 = 13660 with a remainder of 27.

⇒ Option 3: 587407 ÷ 41 = 14327 with a remainder of 0.

⇒ Option 4: 587407 ÷ 47 = 12498 with a remainder of 1.

⇒ Since division by 41 leaves no remainder, the given number is perfectly divisible by 41.

∴ The correct answer is 41.

Additional Information

Divisibility Rule of 43

Add 13 times the last digit to the remaining truncated number. Repeat the process until a smaller manageable number is obtained.

Divisibility Rule of 47

Subtract 14 times the last digit from the remaining truncated number. Repeat the process until a smaller manageable number is obtained.

Divisibility Rule of 37

Subtract 11 times the last digit from the remaining truncated number. Repeat the process until a smaller manageable number is obtained.

Shortcut Trick

Single discount = 40%, and successive discounts 15%, 20% = 15 + 20 − (15 × 20) ÷ 100 = 32%

Difference in discount percentage = 40% − 32% = 8%

8% of Marked Price = ₹400

Marked Price = 400 × (100 ÷ 8) = ₹5,000

∴ The correct answer is ₹5,000.

Alternate Method

Given:

Single discount = 40%

Two successive discounts = 15% and 20%

Difference in selling prices = ₹400

Formula Used:

Equivalent discount for x% and y% = x + y − (x × y) ÷ 100

Marked Price (MP) = 100%

SP1 (60%)
Discount 1 (40%)

SP2 (68%)
Discount 2 (32%)

8% = ₹400
100% = ₹5,000

Calculations:

⇒ Equivalent discount for 15% and 20% = 15 + 20 − (15 × 20) ÷ 100

⇒ Equivalent discount = 35 − 3 = 32%

⇒ Difference between discounts = 40% − 32% = 8%

⇒ We know that difference in selling prices is exactly the difference in discounts computed on the MP.

⇒ 8% of MP = ₹400

⇒ MP = 400 × (100 ÷ 8)

⇒ MP = ₹5,000

∴ The correct answer is ₹5,000.

Additional Information

Successive Discounts

When two discounts x% and y% are given successively, the effective discount is calculated as: [x + y − (x × y) ÷ 100]%.

Marked Price vs Selling Price

Selling Price (SP) is always calculated on the Marked Price (MP) after applying the discount. Formula: SP = MP × (100 − Discount%) ÷ 100.

Three Successive Discounts

For three successive discounts x%, y%, and z%, first find the equivalent of x% and y%, then use that result with z% to find the final equivalent discount.

Shortcut Trick

Volume of a Cylinder = πr²h

Direct substitution: V = 3.14 × 3² × 7

V = 3.14 × 9 × 7 = 3.14 × 63

V = 197.82 cm³

∴ The correct answer is 197.82 cm³.

Alternate Method

Given:

Height of the cylinder (h) = 7 cm

Radius of the cylinder (r) = 3 cm

Value of π = 3.14

Formula Used:

Volume of a Cylinder = π × r² × h

r = 3 cm

h = 7 cm
Volume = π × r × r × h

Calculations:

⇒ Volume = 3.14 × (3)² × 7

⇒ Volume = 3.14 × 9 × 7

⇒ Volume = 3.14 × 63

⇒ Volume = 197.82 cm³

∴ The correct answer is 197.82 cm³.

Additional Information

Curved Surface Area of a Cylinder

The lateral or curved surface area is calculated using the formula CSA = 2πrh.

Total Surface Area of a Cylinder

The total area including both circular bases is calculated as TSA = 2πr(r + h).

Volume of a Cone

A cone with the same base radius and height has exactly one-third the volume of the cylinder: V = (1/3)πr²h.

Shortcut Trick
For a fixed amount, the quantity sold is inversely proportional to the selling price percentage.
Formula: Q1 × SP1% = Q2 × SP2%
⇒ 6 × (100 − 30) = Q2 × (100 + 40)
⇒ 6 × 70 = Q2 × 140
⇒ 420 = 140 × Q2 ⇒ Q2 = 3
∴ The correct answer is 3 articles.

Alternate Method

SP₁ = 70%
6 articles

CP = 100%

SP₂ = 140%
? articles

× 2 (Price multiplier)

÷ 2 (Quantity multiplier)

Given:
Selling Price (SP1) of 6 articles = ₹1
⇒ SP1 per article = ₹ 1/6
Loss percentage = 30%
⇒ SP1 = (100 − 30)% of CP = 70% of CP
⇒ 70% of CP = 1/6
⇒ CP = (1/6) × (100/70) = ₹ 5/21
To find SP for 40% gain:
⇒ New SP (SP2) = (100 + 40)% of CP = 140% of CP
⇒ SP2 = 140% × (5/21)
⇒ SP2 = (140/100) × (5/21) = (7/5) × (5/21) = ₹ 1/3
Since the new selling price is ₹ 1/3 per article,
⇒ Number of articles to be sold for ₹1 = 1 ÷ (1/3) = 3
∴ The correct answer is 3 articles.

Additional Information

Price and Quantity Relationship
When the total expenditure or selling amount remains constant, the price per item is inversely proportional to the quantity of items (Price × Quantity = Constant).

Basic Profit and Loss Rules
Profit occurs when SP > CP (Profit = SP − CP) and Loss occurs when CP > SP (Loss = CP − SP). Both profit and loss percentages are always calculated on the Cost Price (CP).

Selling Price Formula
SP can be directly found from CP using the multiplier: SP = CP × (100 + Profit%) ÷ 100 or SP = CP × (100 − Loss%) ÷ 100.

Shortcut Trick

The mode of a data set is simply the number that occurs most often.

Looking at the given marks: 19, 14, 11, 20, 18, 17, 14, 10, 11, 11.

The number 11 is repeated 3 times, while 14 is repeated twice and all other numbers appear only once.

Thus, 11 has the highest frequency in the given set.

∴ The correct answer is 11.

Alternate Method

Given:

The marks scored by 10 students: 19, 14, 11, 20, 18, 17, 14, 10, 11, 11

Concept Used:

The mode is the specific value that appears most frequently in a given set of data observations.

1
2
3

10

11

14

17

18

19

20
Frequency Chart of Marks
Highest

Calculations:

⇒ Let us arrange the given marks in ascending order to count the occurrences systematically.

⇒ Ascending order: 10, 11, 11, 11, 14, 14, 17, 18, 19, 20

⇒ From the arranged data, we can observe the frequency of each distinct mark:

⇒ 10 → appears 1 time

⇒ 11 → appears 3 times

⇒ 14 → appears 2 times

⇒ 17, 18, 19, 20 → appear 1 time each

⇒ The mark 11 has the highest frequency of 3 among all the numbers.

⇒ Therefore, 11 is the true mode of the given set of marks.

∴ The correct answer is 11.

Additional Information

Mean

The mean is the basic arithmetic average of all the numbers in a data set. Formula: Sum of all observations ÷ Total number of observations.

Median

The middle value when data is sorted in order. If the number of observations (n) is odd, it is the ((n+1)÷2)^th term. If even, it is the exact average of the (n÷2)^th and (n÷2 + 1)^th terms.

Empirical Relationship

For moderately skewed normal distributions, the Mean, Median, and Mode are fundamentally connected by the relation: Mode ≈ 3 × Median − 2 × Mean.

Range

The total overall difference between the absolute maximum and minimum values in a statistical data set. Formula: Range = Maximum value − Minimum value.

Shortcut Trick

Net percentage increase = 4 + 25 + (4 × 25)/100 = 30%.

This means the current sales are 130% of the initial sales.

Initial sales = Current sales × (100 ÷ 130)

Initial sales = 140400 × (10 ÷ 13) = 108000

∴ The correct answer is ₹1,08,000.

Alternate Method

Given:

Current sales = ₹1,40,400

First year increase = 4%

Second year increase = 25%

Formula Used:

Final Value = Initial Value × (1 + R₁/100) × (1 + R₂/100)

Initial (x)
Year 1
₹1,40,400

+4%
+25%

× 1.04
× 1.25

Successive Percentage Change

Calculations:

⇒ Let the sales two years ago be x.

⇒ x × (1 + 4/100) × (1 + 25/100) = 1,40,400

⇒ x × (104/100) × (125/100) = 1,40,400

⇒ x × (26/25) × (5/4) = 1,40,400

⇒ x × (130/100) = 1,40,400

⇒ x × 1.3 = 1,40,400

⇒ x = 1,40,400 ÷ 1.3

⇒ x = 1,08,000

∴ The correct answer is ₹1,08,000.

Additional Information

Successive Percentage Change

When a value is increased successively by a% and b%, the net percentage change is given by the formula: Net Change = a + b + (a × b)/100.

Multiplier Method

To increase a number by r%, multiply it directly by its multiplier (1 + r/100). For successive changes across multiple periods, simply multiply all factors sequentially.

Reversing a Percentage Increase

If the final value is known after a total r% increase, the initial value can be found by multiplying the final value by 100 ÷ (100 + r).

Shortcut Trick
Compare to 1/2 (or 50%): 34/77 and 39/89 are less than 1/2, so eliminate them directly.
The remaining fractions 46/75 and 36/69 (simplified to 12/23) are both greater than 1/2.
Cross multiply the remaining fractions: 46/75 and 12/23.
46 × 23 = 1058 and 75 × 12 = 900.
Since 1058 > 900, 46/75 is the greatest ratio.
∴ The correct answer is 46 : 75.

Alternate Method
Given:
Four ratios: 46 : 75, 34 : 77, 39 : 89, 36 : 69
Method Used:
Convert each ratio into a percentage by multiplying by 100 to easily compare them on a uniform scale.

46 : 75
34 : 77
39 : 89
36 : 69

61.33%
44.16%
43.82%
52.17%

Visual Comparison of Ratios as Percentages

Calculations:
⇒ 1^st ratio = (46 ÷ 75) × 100 = 61.33%
⇒ 2^nd ratio = (34 ÷ 77) × 100 ≈ 44.16%
⇒ 3^rd ratio = (39 ÷ 89) × 100 ≈ 43.82%
⇒ 4^th ratio = (36 ÷ 69) × 100 ≈ 52.17%
⇒ Comparing all percentages: 61.33% > 52.17% > 44.16% > 43.82%
⇒ The greatest percentage is 61.33%, which corresponds to the first ratio.
∴ The correct answer is 46 : 75.

Additional Information
Cross Multiplication Method
To compare two fractions a/b and c/d, compute the cross products a × d and b × c. If ad > bc, then a/b > c/d.
Percentage Method
Converting ratios into percentages (by multiplying the fraction by 100) provides a uniform scale to compare multiple values effortlessly.
Denominator Equalization
Another fundamental way to compare fractions or ratios is by calculating the LCM of their denominators and expressing each fraction with this common base.

Shortcut Trick
Break 7 cos2 θ into 3 cos2 θ + 4 cos2 θ to simplify the equation instantly.
⇒ 3(sin2 θ + cos2 θ) + 4 cos2 θ = 4
Since sin2 θ + cos2 θ = 1, the equation becomes:
⇒ 3(1) + 4 cos2 θ = 4 ⇒ 4 cos2 θ = 1
⇒ cos θ = 1/2. Since 0° < θ < 90°, θ must be 60°.
∴ The correct answer is 60°.

Alternate Method
Given:
3 sin2 θ + 7 cos2 θ = 4
0° < θ < 90°
Formula Used:
sin2 θ + cos2 θ = 1
sin2 θ = 1 − cos2 θ

60°
Base = 1
Perp = √3
Hyp = 2
cos θ = 1/2 ⇒ θ = 60°

Calculations:
⇒ 3 sin2 θ + 7 cos2 θ = 4
Substituting sin2 θ = 1 − cos2 θ into the main equation:
⇒ 3(1 − cos2 θ) + 7 cos2 θ = 4
⇒ 3 − 3 cos2 θ + 7 cos2 θ = 4
⇒ 3 + 4 cos2 θ = 4
⇒ 4 cos2 θ = 4 − 3
⇒ 4 cos2 θ = 1
⇒ cos2 θ = 1/4
⇒ cos θ = 1/2 (Taking the positive root since 0° < θ < 90°)
We know that cos 60° = 1/2.
⇒ θ = 60°
∴ The correct answer is 60°.

Additional Information
Trigonometric Identities
The fundamental Pythagorean identities are: sin2 θ + cos2 θ = 1, 1 + tan2 θ = sec2 θ, and 1 + cot2 θ = cosec2 θ.
Trigonometric Values at 60°
For the standard angle 60°: sin 60° = √3/2, cos 60° = 1/2, and tan 60° = √3.
Quadrant Rule (ASTC)
In the first quadrant (angles between 0° and 90°), all trigonometric ratios yield positive values.

Shortcut Trick
Given: 20 × SP = 24 × CP ⇒ SP / CP = 24 / 20 = 6 / 5
Profit = SP − CP = 6 − 5 = 1 unit
Gain % = (1 / 5) × 100 = 20%
∴ The correct answer is 20% gain.

Alternate Method
Given:
Selling price of 20 books = Cost price of 24 books
Formula Used:
Profit % = (Profit ÷ Cost Price) × 100

Profit Calculation using Unitary Method

CP of 20 books = ₹20

SP of 20 books = ₹24

Profit (₹4)

Calculations:
Let the cost price of 1 book be ₹1.
⇒ Cost price of 20 books = 20 × 1 = ₹20
⇒ Selling price of 20 books = Cost price of 24 books = 24 × 1 = ₹24
⇒ Since Selling Price > Cost Price, a profit is made.
⇒ Profit earned on selling 20 books = SP − CP = 24 − 20 = ₹4
⇒ Gain % = (Profit ÷ CP) × 100
⇒ Gain % = (4 ÷ 20) × 100 = 20%
∴ The correct answer is 20% gain.

Additional Information
Profit and Loss Basics
Profit occurs when the Selling Price (SP) is greater than the Cost Price (CP). Profit = SP − CP.
Profit Percentage Formula
Profit % is always calculated on the Cost Price unless stated otherwise. Formula: Profit % = (Profit ÷ CP) × 100.
Dishonest Dealer Concept
When a dealer uses a false weight, Profit % = (Error ÷ True Value − Error) × 100.

Shortcut Trick

Total Work = LCM(22, 11, 4) = 44 units

Efficiency: A = 44 ÷ 22 = 2, B = 44 ÷ 11 = 4, A+B+C = 44 ÷ 4 = 11

Efficiency of C = 11 − (2 + 4) = 5 units/day

Time taken by C = 44 ÷ 5 = 8 ⁴/₅ days

∴ The correct answer is 8 ⁴/₅ days.

Alternate Method

Given:

Time taken by A alone = 22 days

Time taken by B alone = 11 days

Time taken by A, B and C together = 4 days

Calculations:

⇒ Combined efficiency of A, B and C = 11

⇒ Efficiency of A + Efficiency of B + Efficiency of C = 11

⇒ 2 + 4 + Efficiency of C = 11

⇒ 6 + Efficiency of C = 11

⇒ Efficiency of C = 11 − 6 = 5 units/day

⇒ Time taken by C alone = Total Work ÷ Efficiency of C

⇒ Time taken by C alone = 44 ÷ 5

⇒ Time taken by C alone = 8 ⁴/₅ days

∴ The correct answer is 8 ⁴/₅ days.

Additional Information

Work & Wages

Wages are distributed in direct proportion to the work done by each person. If all people work for the exact same number of days, the wages will be distributed in the ratio of their efficiencies.

Pipes & Cisterns

It operates on the same logic as Time and Work. An inlet pipe is considered positive work as it fills the tank, whereas an outlet pipe (or leak) is treated as negative efficiency since it empties the tank.

Alternate Day Work

When individuals work on alternate days, find the total work and combine their efficiencies for one complete cycle (e.g., A works day 1, B works day 2), then scale this cycle up to approach the total work.

Shortcut Trick

The largest circle inscribed completely inside a rectangle has its diameter equal to the shorter side (breadth) of the rectangle.

Diameter of circle (d) = Breadth of rectangle = 168 m

Circumference = π × d = 22/7 × 168

⇒ Circumference = 22 × 24 = 528 m

∴ The correct answer is 528 m.

Alternate Method

Given:

Length of the rectangle (l) = 249 m

Breadth of the rectangle (b) = 168 m

d = 168 m
Length = 249 m
168 m

Formula Used:

Circumference of a circle = 2 × π × r

Calculations:

⇒ To fit completely inside the rectangle without crossing the boundaries, the circle's diameter cannot exceed the rectangle's smallest dimension.

⇒ Maximum possible diameter (d) = Breadth = 168 m

⇒ Radius (r) = 168 ÷ 2 = 84 m

⇒ Circumference = 2 × (22/7) × 84

⇒ Circumference = 2 × 22 × 12

⇒ Circumference = 44 × 12 = 528 m

∴ The correct answer is 528 m.

Additional Information

Largest Circle in a Rectangle

The maximum diameter is always bounded by the shorter side of the rectangle. Diameter = breadth.

Largest Square in a Circle

When a square is perfectly inscribed inside a circle, the diagonal of the square is equal to the diameter of the circle. Side of square = d ÷ √2.

Circumcircle of a Rectangle

For a circle circumscribing a rectangle (touching all four corners), the diagonal of the rectangle serves as the diameter of the circle. Diameter = √(l² + b²).

Shortcut Trick

In right-angled △BDC, ∠C = 90° − ∠DBC = 90° − 76° = 14°.

In △AEC, the exterior angle at vertex E is equal to the sum of its two interior opposite angles.

⇒ ∠AEB = ∠CAE + ∠C

⇒ ∠AEB = 74° + 14° = 88°.

∴ The correct answer is 88°.

Alternate Method

A
B
C
D
E

76°
74°
14°
88°

Given:

BD ⊥ AC at point D, implying ∠BDC = 90°

∠DBC = 76°

∠CAE = 74°

Calculations:

Consider the right-angled triangle BDC formed by altitude BD:

⇒ ∠C + ∠DBC + ∠BDC = 180°

⇒ ∠C + 76° + 90° = 180°

⇒ ∠C = 180° − 166°

⇒ ∠C = 14°

Now, observe the triangle AEC. The point E lies on the segment BC. Hence, the angle ∠AEB forms an exterior angle to △AEC at vertex E.

According to the Exterior Angle Theorem, an exterior angle is strictly equal to the sum of the two opposite interior angles of the triangle:

⇒ ∠AEB = ∠CAE + ∠ACE

⇒ ∠AEB = 74° + ∠C

⇒ ∠AEB = 74° + 14°

⇒ ∠AEB = 88°

∴ The correct answer is 88°.

Additional Information

Exterior Angle Theorem

The measure of an exterior angle of a triangle equals the sum of the measures of its two remote interior angles. Formula: ∠Exterior = ∠Opposite₁ + ∠Opposite₂

Triangle Angle Sum Property

The total measure of the internal angles of any valid triangle is strictly 180°. In right-angled triangles, the two non-right (acute) angles must complement each other (sum to exactly 90°).

Linear Pair Axiom

When two adjacent angles form a straight continuous line, they are called a linear pair and are supplementary. Therefore, their sum is always equal to 180°.

Shortcut Trick

Use option verification for the fastest result.

Test Option 2: Let Q's present age = 63 years.

From statement 1: 4P = 63 + 41 = 104 ⇒ P = 26 years.

Ages after 10 years: Q = 73 years, P = 36 years.

Check statement 2: 3 × 73 = 219 and (7 × 36) − 33 = 252 − 33 = 219. (Condition satisfies)

∴ The correct answer is 63 years.

Alternate Method

Present Ages
P
Q

After 10 Years
P + 10
Q + 10

+10 years

Eq 1: 4P − Q = 41
Eq 2: 3(Q + 10) = 7(P + 10) − 33

Given:

⇒ Present age condition: 4P − Q = 41  ... (Equation 1)

⇒ Future age condition: 3(Q + 10) = 7(P + 10) − 33  ... (Equation 2)

Calculations:

⇒ From Equation 1, we get: Q = 4P − 41

⇒ Expanding Equation 2:

⇒ 3Q + 30 = 7P + 70 − 33

⇒ 3Q + 30 = 7P + 37

⇒ 3Q = 7P + 7

⇒ Substitute Q = 4P − 41 into the simplified Equation 2:

⇒ 3(4P − 41) = 7P + 7

⇒ 12P − 123 = 7P + 7

⇒ 12P − 7P = 123 + 7

⇒ 5P = 130

⇒ P = 26 years

⇒ Now, substitute P = 26 into Equation 1:

⇒ Q = 4(26) − 41

⇒ Q = 104 − 41 = 63 years

∴ The correct answer is 63 years.

Additional Information

Difference of Ages

The age difference between two individuals always remains constant regardless of the time period (past, present, or future).

Ratio of Ages

If the current age ratio is a : b, their ages are taken as ax and bx. After n years, their ages become (ax + n) and (bx + n).

Average Age Property

If the average age of n people is A, then after k years, the new average age of the same group will naturally become A + k.

Shortcut Trick

Under a 'Buy 4 Get 2 Free' scheme, the customer takes home exactly 6 shirts but only pays the price of 4 shirts.

Total amount to pay for 6 shirts = Price of 4 shirts

⇒ Amount to pay = 4 × ₹1,250 = ₹5,000

∴ The correct answer is ₹5,000.

Alternate Method

Given:

Marked price of a shirt = ₹1,250

Scheme offered = 'buy four get two free'

Total shirts required by customer = 6

Buy 4 Get 2 Free Scheme

₹1250

₹1250

₹1250

₹1250

FREE

FREE

Paid (4 Shirts) = ₹5000

Free (2 Shirts)

Calculations:

⇒ In a 'buy 4 get 2 free' scheme, the customer gets a total of 6 shirts (4 paid + 2 free).

⇒ Since the customer takes home exactly 6 shirts, they only have to pay for the 4 shirts.

⇒ Total amount to be paid = Number of paid shirts × Marked price of one shirt

⇒ Total amount to be paid = 4 × 1250

⇒ Total amount to be paid = ₹5,000

∴ The correct answer is ₹5,000.

Additional Information

Buy X Get Y Free Concept

The effective discount percentage in such promotional schemes is calculated as: Discount % = (Free Items ÷ Total Items) × 100.

Marked Price and Selling Price

Selling Price (SP) = Marked Price (MP) × (1 − Discount/100). The discount is always offered on the marked price of the article.

Successive Discounts

When two consecutive discounts of a% and b% are applied on an item, the single equivalent discount = (a + b − (a × b)/100)%.

Shortcut Trick

Time for P to fill the tank = 14 × 9/7 = 18 hours.

Time for Q to fill the tank = 20 × 9/5 = 36 hours.

Work done by P and Q in 2 hours = 2 × (1/18 + 1/36) = 1/6 of the tank.

R empties 1/6 of the tank in 6 hours, so R's total time to empty the tank = 36 hours.

Combined efficiency of P, Q, and R = 1/18 + 1/36 − 1/36 = 1/18 per hour.

∴ The correct answer is 18 hours.

Alternate Method

Given:

Time for pipe P to fill 7/9 part = 14 hours.

Time for pipe Q to fill 5/9 part = 20 hours.

Time for pipe R to empty the filled part = 6 hours.

Formula Used:

Total Work = Efficiency × Time

Calculations:

⇒ Time for P to fill the complete tank = 14 × 9/7 = 18 hours.

⇒ Time for Q to fill the complete tank = 20 × 9/5 = 36 hours.

⇒ Part of the tank filled by P in 1 hour = 1/18

⇒ Part of the tank filled by Q in 1 hour = 1/36

⇒ Part of the tank filled by P and Q in 2 hours = 2 × (1/18 + 1/36) = 2 × (3/36) = 1/6.

⇒ Pipe R empties this 1/6 filled part in 6 hours.

⇒ Time taken by R to empty the complete tank = 6 × 6 = 36 hours.

⇒ Let Total Work = LCM of (18, 36, 36) = 36 units.

⇒ Combined efficiency of P, Q, and R = 2 + 1 − 1 = 2 units/hr.

⇒ Time taken by P, Q, and R together to fill the tank = Total Work ÷ Combined Efficiency

⇒ Time = 36 ÷ 2 = 18 hours.

∴ The correct answer is 18 hours.

Additional Information

Pipes & Cisterns Concept

If a pipe fills a tank in x hours, its efficiency is positive (1/x). If a pipe empties the tank in y hours, its efficiency is taken as negative (−1/y).

Combined Work Rule

When multiple pipes are opened simultaneously, their combined efficiency is simply the algebraic sum of their individual efficiencies.

Fractional Work

If a/b part of a task is completed in D days (or hours), then the entire task will be completed in D × (b/a) days.

Shortcut Trick

Let the original number be 100%.

After a 70% increase, the new number becomes 100% + 70% = 170%.

Given that 170% = 3230.

So, 1% = 3230 ÷ 170 = 19.

Original number (100%) = 19 × 100 = 1900.

∴ The correct answer is 1900.

Alternate Method

Given:

Resulting number = 3230

Percentage increase = 70%

Formula Used:

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

100% (Original)

+70% Increase

Total: 170% = 3230

Calculations:

Let the original number be x.

⇒ Final Value = x × (1 + 70 ÷ 100)

⇒ 3230 = x × (1 + 0.70)

⇒ 3230 = x × 1.70

⇒ x = 3230 ÷ 1.70

⇒ x = 32300 ÷ 17

⇒ x = 1900

∴ The correct answer is 1900.

Additional Information

Percentage Decrease

If a number is decreased by R%, the final value is calculated as Initial Value × (1 − R ÷ 100).

Successive Percentage Change

If a number undergoes successive percentage changes of A% and B%, the effective overall percentage change is given by A + B + (A × B) ÷ 100.

Base for Percentage

Percentage change is always calculated on the initial value (the base) unless a different base is explicitly mentioned in the problem.