Quantitative Aptitude PYQs

Shortcut Trick

Let the base value of the goods be 100.

Claimed Selling Price = 100 + 57 = 157

Actual Cost Price (due to using 13% less weight) = 100 − 13 = 87

Profit = Claimed SP − Actual CP = 157 − 87 = 70

Total Profit % = (70 ÷ 87) × 100 ≈ 80.46% ≈ 80%

∴ The correct answer is 80%.

Alternate Method

Given:

Claimed profit percentage = 57%

Dishonest weight used = 13% less than the marked weight

Formula Used:

Profit % = [(Selling Price − Cost Price) ÷ Cost Price] × 100

Dishonest Dealer Profit Calculation

Actual CP: 87

Profit: 70

Selling Price: 157

Total Profit % = (70 ÷ 87) × 100 ≈ 80%

Calculations:

⇒ Let the cost price of 1000 g of vegetables be ₹ 1000.

⇒ The trader claims a profit of 57%, so Selling Price (SP) for 1000 g = 1000 + (57% of 1000) = ₹ 1570.

⇒ However, the trader uses a weight 13% less than 1000 g.

⇒ Actual weight given to the customer = 1000 − (13% of 1000) = 1000 − 130 = 870 g.

⇒ The actual Cost Price (CP) incurred by the trader is for the given 870 g = ₹ 870.

⇒ Profit earned by the trader = SP − CP = 1570 − 870 = ₹ 700.

⇒ Total Profit % = (700 ÷ 870) × 100

⇒ Total Profit % = (70 ÷ 87) × 100 = 7000 ÷ 87 ≈ 80.459%

⇒ Rounding off to the nearest whole number gives 80%.

∴ The correct answer is 80%.

Additional Information

Dishonest Dealer Formula

When a dealer claims to sell at Cost Price but uses false weight, the true profit percentage is given by: [Error ÷ (True Weight − Error)] × 100.

Combined Effect in Dishonest Dealing

When a trader manipulates both the selling price (mark up) and the weight, calculate the effective Cost Price for the actual quantity given and compare it with the charged Selling Price to find the total profit.

Successive Percentage Change

Alternatively, the net effect of two successive changes of x% and y% can be calculated using the formula: Net % = x + y + (x × y) ÷ 100.

Shortcut Trick

The unit digit of the number 861616843 is 3, which is an odd number.

Thus, it cannot be perfectly divided by an even number (36) or a number ending in zero (50).

We test the first remaining option, 43, by splitting the number: 861616843 = 860000000 + 1616843.

860000000 ÷ 43 = 20000000 (divides exactly).

1616843 ÷ 43 = 37601 (divides exactly).

∴ The correct answer is 43.

Alternate Method

Given:

Number = 861616843

Options = 43, 47, 50, 36

Calculation:

⇒ The given number 861616843 ends with an odd digit (3).

⇒ Any number ending with an odd digit cannot be completely divided by an even number or a multiple of 10.

⇒ Therefore, we can immediately eliminate Option 3 (50) and Option 4 (36).

⇒ Let us check divisibility by 43 using standard division.

⇒ First, take the starting digits: 86 ÷ 43 = 2 with a remainder of 0.

⇒ Bring down 161: 161 ÷ 43 = 3 with a remainder of 32.

⇒ Bring down 6 to make 326: 326 ÷ 43 = 7 with a remainder of 25.

⇒ Bring down 8 to make 258: 258 ÷ 43 = 6 with a remainder of 0.

⇒ Bring down 4, which is not divisible by 43, so we add a 0 to the quotient.

⇒ Bring down 3 to make 43: 43 ÷ 43 = 1 with a remainder of 0.

⇒ Combining the quotients gives us exactly 20037601 without any remainder.

⇒ Since the remainder is completely zero, the given number is fully divisible by 43.

∴ The correct answer is 43.

Additional Information

Divisibility Rule of 2, 4, and 8

A number is divisible by 2 if its last digit is even; by 4 if its last two digits form a number divisible by 4; and by 8 if its last three digits form a number divisible by 8.

Divisibility Rule of 3 and 9

A number is divisible by 3 if the sum of its digits is divisible by 3. It is divisible by 9 if the sum of its digits is divisible by 9.

Divisibility Rule of 5 and 10

A number is exactly divisible by 5 if its last digit is either 0 or 5. It is divisible by 10 only if its last digit is 0.

Odd and Even Constraints in Division

An odd number cannot be divided perfectly by any even number. Recognizing this rule allows you to quickly eliminate options in exams without performing extensive calculations.

Shortcut Trick

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

Alternating sums for the digits of N18M05 are (N + 8 + 0) and (1 + M + 5).

⇒ N + 8 = 1 + M + 5

⇒ N + 8 = M + 6

⇒ M − N = 2

∴ The correct answer is M − N = 2.

Alternate Method

Given:

The 6-digit number is N18M05.

The number is perfectly divisible by 11.

Concept Used:

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

Calculation:

⇒ Sum of digits at odd places (from right) = 5 + M + 1 = M + 6

⇒ Sum of digits at even places (from right) = 0 + 8 + N = N + 8

⇒ Difference = (Sum of even places) − (Sum of odd places)

⇒ Difference = (N + 8) − (M + 6)

⇒ Difference = N − M + 2

For the number to be divisible by 11, we equate the difference to 0 (since equating it to 11 or −11 does not yield relations matching the given options for single-digit numbers).

⇒ N − M + 2 = 0

⇒ M − N = 2

∴ The correct answer is M − N = 2.

Additional Information

Divisibility Rule of 3 and 9

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

Divisibility Rule of 4

A number is perfectly divisible by 4 if the number formed by its last two digits is divisible by 4.

Divisibility Rule of 8

A number is perfectly divisible by 8 if the number formed by its last three digits is divisible by 8.

Shortcut Trick
Compare the fractions to a base value like 1/2 to quickly eliminate the smaller ratios.
37/76 and 35/72 are both less than 1/2 (as 37 × 2 < 76 and 35 × 2 < 72).
43/65 and 50/60 are greater than 1/2. Compare them directly via cross-multiplication.
43/65 vs 5/6 ⇒ 43 × 6 = 258 and 65 × 5 = 325.
Since 325 > 258, the ratio 50 : 60 is strictly the greatest among them all.
∴ The correct answer is 50 : 60.
Alternate Method
Given:
Four ratios: 37 : 76, 43 : 65, 35 : 72, and 50 : 60
Method Used:
Convert each ratio into a decimal or percentage value to easily compare their true magnitudes.

37:76

~48.7%
43:65

~66.2%
35:72

~48.6%
50:60

83.3% (Greatest)

0%

50%

100%

Calculations:
⇒ 37 ÷ 76 ≈ 0.4868 or 48.68%
⇒ 43 ÷ 65 ≈ 0.6615 or 66.15%
⇒ 35 ÷ 72 ≈ 0.4861 or 48.61%
⇒ 50 ÷ 60 = 5 ÷ 6 ≈ 0.8333 or 83.33%
⇒ Comparing the evaluated fractions, 83.33% is clearly the highest decimal equivalent.
∴ The correct answer is 50 : 60.
Additional Information
Compound Ratio
For two or more given ratios such as a:b and c:d, their compound ratio is formed by multiplying their respective antecedents and consequents to get ac:bd.
Duplicate & Sub-duplicate Ratio
The duplicate ratio of a:b is calculated as a2:b2, while its sub-duplicate ratio is evaluated as √a:√b.
Proportion Rule
When two respective ratios are exactly equal (a/b = c/d), they form a valid proportion where the product of the extremes directly equals the product of the means (ad = bc).

Shortcut Trick

A rectangle requires all its internal angles to be exactly 90°.

A rhombus requires all its four sides to be equal in length.

A square is a shape that has all 90° angles AND all equal sides.

Since it meets all the defining conditions of both shapes, it inherently belongs to both families.

∴ The correct answer is both a rectangle and a rhombus.

Alternate Method

Given:

Statements comparing a square to a rectangle and a rhombus.

Properties Used:

Definitions of Quadrilaterals (Rectangle, Rhombus, and Square)

Rectangle
Rhombus
Square

=

Angles = 90°

All sides equal

Sides equal &
Angles = 90°

Calculations:

⇒ By definition, a rectangle is a quadrilateral where all internal angles measure exactly 90°.

⇒ A rhombus is defined as a quadrilateral where all four sides are of equal length.

⇒ A square is constructed as a quadrilateral where all four sides are equal AND all four internal angles measure exactly 90°.

⇒ Since a square exhibits opposite sides being equal and all angles as 90°, it completely satisfies the properties of a rectangle.

⇒ Since a square has all four sides equal, it also satisfies the core geometric property of a rhombus.

⇒ Thus, a square inherently functions as a special specialized subset of both figures simultaneously.

∴ The correct answer is both a rectangle and a rhombus.

Additional Information

Parallelogram Hierarchy

A parallelogram is a quadrilateral with opposite sides parallel and equal. Rectangles, rhombuses, and squares are all categorized as special types of parallelograms.

Trapezium

A trapezium (or trapezoid) is a quadrilateral that possesses exactly one pair of opposite parallel sides, setting it apart from the dual-parallel structure of parallelograms.

Diagonals Property

The diagonals of a rectangle are always equal in length. The diagonals of a rhombus always bisect each other at precisely 90°. A square inherits both rules simultaneously.

Shortcut Trick

Using the Alligation Method on Profit/Loss percentages:

Ratio of Cost Prices = (Gain% − Net%) : (Net% − Loss%) = (18 − 0) : (0 − (−9)) = 18 : 9 = 2 : 1

Total parts for both dresses = 2 + 1 = 3 parts.

3 parts = ₹1,200 ⇒ 1 part = ₹400.

Cost price of first dress (2 parts) = 2 × 400 = ₹800.

∴ The correct answer is ₹800.

Alternate Method

Given:

Total Cost Price of two dresses = ₹1,200

Loss on 1^st dress = 9%

Gain on 2^nd dress = 18%

Overall gain or loss = 0%

Formula Used:

Loss = CP × (Loss% ÷ 100)

Gain = CP × (Gain% ÷ 100)

Dress 1 Loss

−9%

Dress 2 Gain

+18%

0%
Net

18

9

CP Ratio = 18 : 9 = 2 : 1
Dress 1 = (2 ÷ 3) × 1200 = ₹800

Calculations:

Let the cost price of the first dress be C₁ and the second dress be C₂.

⇒ C₁ + C₂ = 1200

Since there is neither profit nor loss on the whole transaction:

⇒ Total Loss on 1^st dress = Total Gain on 2^nd dress

⇒ 9% of C₁ = 18% of C₂

⇒ 9 × C₁ = 18 × C₂

⇒ C₁ ÷ C₂ = 18 ÷ 9 = 2 ÷ 1

The cost prices are in the ratio 2 :

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

⇒ 3 parts = ₹1,200

⇒ 1 part = 1200 ÷ 3 = ₹400

⇒ Cost price of the first dress (C₁) = 2 parts = 2 × 400 = ₹800

∴ The correct answer is ₹800.

Additional Information

No Profit No Loss Principle

When a transaction results in neither profit nor loss, the absolute value of the profit made on one item is exactly equal to the absolute value of the loss incurred on the other item. Thus, Profit on Item A = Loss on Item B.

Alligation Method in Profit and Loss

Alligation can be efficiently used to find the ratio of cost prices when two items are sold at different profit/loss percentages and the overall profit/loss percentage is given. The resultant cross-difference ratio strictly represents the ratio of their Cost Prices (CP).

Selling Price Formula

Selling Price (SP) = CP × (100 ± Profit/Loss%) ÷ 100. In a "no profit no loss" case, the sum of the SP of all items equals the sum of the CP of all items.

Shortcut Trick

Effective SI% for Case 1 = 4.2 × 10/3 = 14%

Effective SI% for Case 2 = 6.4 × 12.5 = 80%

Required Amount% = Principal (100%) + SI (80%) = 180%

Since 14% of Principal = ₹5,845

180% of Principal = (5845 ÷ 14) × 180 = 417.5 × 180 = ₹75,150

∴ The correct answer is ₹75,150.

Alternate Method

Given:

SI₁ = ₹5,845

R₁ = 4.2%, T₁ = 3 1/3 = 10/3 years

R₂ = 6.4%, T₂ = 12.5 years

Formula Used:

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

Amount = P + SI

0;background:#ffffff;border-radius:10px;border:1px>

Case 1
R = 4.2%, T = 10/3 yrs
SI = ₹5,845
14% of P = 5845

P = ₹41,750

Case 2
R = 6.4%, T = 12.5 yrs
SI = 80% of P
Amount = 180% of P

Total Amount = ₹75,150

Calculations:

⇒ 5845 = (P × 4.2 × 10/3) ÷ 100

⇒ 5845 = (P × 14) ÷ 100

⇒ P = (5845 × 100) ÷ 14

⇒ P = 417.5 × 100 = ₹41,750

Now, finding SI for Case 2:

⇒ SI₂ = (41750 × 6.4 × 12.5) ÷ 100

⇒ SI₂ = (41750 × 80) ÷ 100 = ₹33,400

⇒ Amount = P + SI₂ = 41750 + 33400 = ₹75,150

∴ The correct answer is ₹75,150.

Additional Information

0;font-weight:700;color:#222222;>Effective Interest Rate in SI

In simple interest, the total interest percentage over a given period is the product of Rate (R) and Time (T). Total SI = (R × T)% of Principal.

0;font-weight:700;color:#222222;>Amount Formula

The total amount payable is the sum of the initial principal and the interest earned. Amount = P[1 + (RT ÷ 100)].

0;font-weight:700;color:#222222;>Simple Interest vs Compound Interest

While SI is strictly calculated on the original principal each year, Compound Interest (CI) is calculated on the accumulated amount (Principal + previous Interest). Formula: A = P(1 + R/100)ⁿ.

Shortcut Trick

65% of 65 = 65 × 65 ÷ 100 = 4225 ÷ 100 = 42.25

1/5 of 45 = 9

Difference = 42.25 − 9 = 33.25

∴ The correct answer is 33.25.

Alternate Method

Given:

First value = 65% of 65

Second value = 1/5 of 45

Calculations:

⇒ First value = (65 ÷ 100) × 65 = 4225 ÷ 100

⇒ First value = 42.25

⇒ Second value = (1 ÷ 5) × 45

⇒ Second value = 9

⇒ Required difference = First value − Second value

⇒ Required difference = 42.25 − 9 = 33.25

∴ The correct answer is 33.25.

Additional Information

BODMAS Rule

Always solve bracket first, followed by 'of', division, multiplication, addition, and subtraction in that exact order to simplify any mathematical expression.

Squares Ending in 5

The square of a number ending in 5 (e.g., 65) can be found quickly: multiply the first digit by the next integer (6 × 7 = 42) and simply append 25 to get 4225.

Percentage and Fractions

To convert a percentage into a fraction or decimal, simply divide the number by 100. In mathematics, the word 'of' indicates multiplication between the given values.

Shortcut Trick

Solve brackets from innermost to outermost directly using BODMAS.

⇒ ( 5 − 9 ) = −4

⇒ { 11 + 2 × (−4) } = 11 − 8 = 3

⇒ 459 ÷ 3 = 153

∴ The correct answer is 153.

Alternate Method

Given:

[459 ÷ {11 + 2 × ( 5 − 9 )}]

Formula Used:

BODMAS Rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction)

Calculations:

⇒ 459 ÷ {11 + 2 × ( 5 − 9 )}

⇒ 459 ÷ {11 + 2 × (−4)}

⇒ 459 ÷ {11 − 8}

⇒ 459 ÷ 3

⇒ 153

∴ The correct answer is 153.

Additional Information

BODMAS Rule

BODMAS stands for Brackets, Orders, Division, Multiplication, Addition, and Subtraction. It determines the sequence of operations in a mathematical expression.

Order of Brackets

When solving expressions, evaluate brackets from innermost to outermost: Round brackets ( ) first, then Curly brackets { }, and finally Square brackets [ ].

Divisibility Rule of 3

A number is divisible by 3 if the sum of its digits is a multiple of 3. For 459, the sum is 4 + 5 + 9 = 18, which is divisible by 3, making division easier.

Shortcut Trick

Using the fundamental trigonometric identity:

⇒ cosec² θ = 1 + cot² θ

⇒ cosec θ = √(1 + (x/y)²) = √((y² + x²) / y²)

⇒ cosec θ = (√(x² + y²)) / y

∴ The correct answer is (√(x²+y²))/(y).

Alternate Method

θ
Base = x
Perpendicular = y
Hypotenuse = √(x² + y²)

Given:

⇒ cot θ = x / y

Formula Used:

⇒ cot θ = Base / Perpendicular

⇒ cosec θ = Hypotenuse / Perpendicular

Calculations:

⇒ Let Base = x and Perpendicular = y.

⇒ According to Pythagoras theorem:

⇒ Hypotenuse² = Base² + Perpendicular²

⇒ Hypotenuse = √(x² + y²)

⇒ Substituting the computed values into the formula for cosec θ:

⇒ cosec θ = Hypotenuse / Perpendicular

⇒ cosec θ = (√(x² + y²)) / y

∴ The correct answer is (√(x²+y²))/(y).

Additional Information

Trigonometric Ratios

In a right-angled triangle, sin θ = Perpendicular / Hypotenuse, cos θ = Base / Hypotenuse, and tan θ = Perpendicular / Base. Their corresponding reciprocals are cosec θ, sec θ, and cot θ.

Pythagoras Theorem

It states that in any right-angled triangle, the square of the hypotenuse is exactly equal to the sum of the squares of the perpendicular and the base: Hypotenuse² = Perpendicular² + Base².

Fundamental Trigonometric Identities

The three core identities universally used in trigonometry are: sin² θ + cos² θ = 1, sec² θ − tan² θ = 1, and cosec² θ − cot² θ = 1.

Shortcut Trick

Time to fill full tank: P = 27 × 4/3 = 36 hours, Q = 12 × 3/2 = 18 hours.

In 6 hours, part filled by P and Q = 6 × (1/36 + 1/18) = 1/2 tank.

Pipe R empties this 1/2 tank in 8 hours ⇒ Full tank empty time for R = 16 hours.

Let total capacity = LCM(36, 18, 16) = 144 units.

Net efficiency of (P + Q − R) = (144 ÷ 36) + (144 ÷ 18) − (144 ÷ 16) = 4 + 8 − 9 = 3 units/hour.

∴ The correct answer is 48 hours.

Alternate Method

Given:

Time taken by P to fill 3/4 part of the tank = 27 hours.

⇒ Total time taken by P to fill the empty tank = 27 × 4/3 = 36 hours.

Time taken by Q to fill 2/3 part of the tank = 12 hours.

⇒ Total time taken by Q to fill the empty tank = 12 × 3/2 = 18 hours.

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

Pipe R empties this 1/2 tank in 8 hours.

⇒ Total time taken by R alone to empty the full tank = 8 × 2 = 16 hours.

Calculations:

⇒ Net efficiency of P, Q, and R together = 4 + 8 − 9 = 3 units/hour.

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

⇒ Time taken = 144 ÷ 3 = 48 hours.

∴ The correct answer is 48 hours.

Additional Information

Inlet and Outlet Pipes

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

LCM Method for Time & Work

Total work is assumed as the LCM of all the given individual times. Efficiency is calculated as Total Work ÷ Time taken. The total time for combined work = Total Work ÷ Net Efficiency.

Part of Tank Filled or Emptied

The fraction of the tank filled or emptied in a given time period is equal to the Time taken × Rate of work (Combined Efficiency).

Shortcut Trick

The area of a triangle can be directly calculated by multiplying the base and height and dividing by 2.

Area = (Base × Height) ÷ 2

Area = (12 × 8) ÷ 2 = 96 ÷ 2 = 48 cm².

∴ The correct answer is 48 cm².

Alternate Method

Given:

⇒ Base of the triangle (b) = 12 cm

⇒ Height of the triangle (h) = 8 cm

Base = 12 cm
Height = 8 cm

Formula Used:

⇒ Area of a Triangle = (1 ÷ 2) × Base × Height

Calculations:

⇒ Area = (1 ÷ 2) × 12 × 8

⇒ Area = 6 × 8

⇒ Area = 48

∴ The correct answer is 48 cm².

Additional Information

Heron's Formula

When the lengths of all three sides (a, b, c) are known, Area = √[s × (s−a) × (s−b) × (s−c)], where s is the semi-perimeter: s = (a + b + c) ÷ 2.

Area of an Equilateral Triangle

If all sides of a triangle are equal (side = a), the formula for the area is directly given as (√3 ÷ 4) × a².

Area of a Right-Angled Triangle

For a right-angled triangle, the area is simply (1 ÷ 2) × Perpendicular × Base, since the perpendicular side acts as the height.

Shortcut Trick

SP = 50000 − 18000 = 32000

Ratio of SP to MP = 32000 / 50000 = 16 / 25

For 2 successive discounts, multiplying factor = √(SP / MP) = √(16 / 25) = 4 / 5

Discount % (x) = (1 − 4/5) × 100% = 20%

∴ The correct answer is 20%.

Alternate Method

Given:

Marked Price (MP) = ₹50,000

Total Discount = ₹18,000

Formula Used:

SP = MP × (1 − x/100)²

MP = ₹50,000

SP = ₹32,000

2 discounts of x%
Discount = ₹18,000

Calculation:

⇒ Selling Price (SP) = MP − Total Discount

⇒ SP = 50000 − 18000 = 32000

⇒ SP = MP × (1 − x/100)²

⇒ 32000 = 50000 × (1 − x/100)²

⇒ (1 − x/100)² = 32000 / 50000

⇒ (1 − x/100)² = 16 / 25

⇒ 1 − x/100 = √(16 / 25)

⇒ 1 − x/100 = 4 / 5

⇒ x/100 = 1 − 4/5 = 1/5

⇒ x = 100 / 5 = 20

∴ The correct answer is 20%.

Additional Information

Successive Discount

For two successive discounts of a% and b%, the effective single discount is given by (a + b − ab/100)%.

Relation between MP and SP

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

Multiplier Concept

If an article is given n successive equal discounts of x%, then SP = MP × (1 − x/100)ⁿ.

Shortcut Trick

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

Here, the total number of terms, n = 160.

Average = 160 + 1 = 161.

∴ The correct answer is 161.

Alternate Method

Given:

Sequence of the first 160 even numbers: 2, 4, 6, ..., up to 160 terms.

Formula Used:

Average of an Arithmetic Progression (AP) = (First term + Last term) ÷ 2

2
161
320

First Term
(1^st even)

Last Term
(160^th even)

Average = (2 + 320) ÷ 2

...
...

Average of an AP is its midpoint

Calculations:

⇒ First even number (First term) = 2 × 1 = 2

⇒ 160^th even number (Last term) = 2 × 160 = 320

⇒ Average = (2 + 320) ÷ 2

⇒ Average = 322 ÷ 2

⇒ Average = 161

∴ The correct answer is 161.

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' consecutive odd numbers is simply equal to n.

Average of squares of first 'n' Natural Numbers

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

Average of an Arithmetic Progression

For any sequence in an arithmetic progression, the average is always equal to the midpoint of the series, i.e., (First term + Last term) ÷ 2.

Shortcut Trick

Total Capacity of tank = LCM(18, 21, 16) = 1008 units.

Combined Efficiency = (1008 ÷ 18) + (1008 ÷ 21) + (1008 ÷ 16) = 56 + 48 + 63 = 167 units/hr.

Time to fill together = Total Capacity ÷ Combined Efficiency = 1008 ÷ 167 = 6 ⁶/₁₆₇ hours.

∴ The correct answer is 6 ⁶/₁₆₇ hours.

Alternate Method

Given:

Time taken by Pipe A = 18 hours

Time taken by Pipe B = 21 hours

Time taken by Pipe C = 16 hours

Formula Used:

Total Work = LCM of given time periods

Efficiency = Total Work ÷ Time Taken

Time Taken Together = Total Work ÷ Combined Efficiency

Calculations:

⇒ Total combined efficiency = 56 + 48 + 63 = 167 units/hr

⇒ Time taken to fill the tank together = Total Work ÷ Total Efficiency

⇒ Time taken = 1008 ÷ 167 = 6 ⁶/₁₆₇ hours

∴ The correct answer is 6 ⁶/₁₆₇ hours.

Additional Information

Pipes & Cisterns Concept

If a pipe can fill a tank in x hours, the part of the tank filled in 1 hour is 1/x. This unit-work approach is equivalent to finding fractional efficiency.

Inlet vs Outlet Pipes

Inlet pipes add water to the tank, so their efficiency is taken as positive. Outlet (leak) pipes remove water, so their efficiency is taken as negative when computing combined net work.

Efficiency & Time Relationship

Efficiency is inversely proportional to time taken (Efficiency = Work ÷ Time). A pipe that takes fewer hours to fill a tank has a higher individual filling capacity.

Shortcut Trick

The last digit of the number 814242541 is 1 (an odd number).

Numbers ending in an odd digit cannot be divided by even numbers, so 8 and 12 are eliminated.

Numbers divisible by 15 must end in 0 or 5, so 15 is also eliminated.

By the method of elimination, the given number must be divisible by 7.

∴ The correct answer is 7.

Alternate Method

Given:

Number = 814242541

Calculations:

Let us check the divisibility rule for 7 directly.

Rule for 7: Form groups of 3 digits starting from the rightmost digit. The difference between the sum of odd-positioned groups and even-positioned groups must be divisible by 7.

⇒ Groups of three digits from right to left: 541, 424, 814

⇒ Sum of odd-positioned groups = 541 + 814 = 1355

⇒ Sum of even-positioned groups = 424

⇒ Difference = 1355 − 424 = 931

⇒ Checking divisibility of 931 by 7: 931 ÷ 7 = 133

⇒ Since 931 is exactly divisible by 7, the entire number 814242541 is also divisible by 7.

∴ The correct answer is 7.

Additional Information

Divisibility Rule of 7, 11, and 13

Form groups of 3 digits from right to left. Find the difference between the sum of odd-placed groups and even-placed groups. If the difference is divisible by 7, 11, or 13, then the original number is divisible by 7, 11, or 13 respectively.

Divisibility Rule of 8

A number is divisible by 8 if the number formed by its last three digits is a multiple of 8.

Divisibility Rule of 12

A number is divisible by 12 if it is divisible by both 3 and 4.

Divisibility Rule of 15

A number is divisible by 15 if it is divisible by both 3 and 5. This means it must end in 0 or 5 and the sum of its digits must be a multiple of 3.

Shortcut Trick

Speed ratio of Anmol = 1 : 2

Time ratio of Anmol = 2 : 1 (Difference = 1 unit)

Total time saved by doubling speed = 7 hrs (late) + 8 hrs (early) = 15 hrs

1 unit = 15 hrs ⇒ Original time (2 units) = 30 hrs

Anmol's speed = Distance ÷ Time = 30 ÷ 30 = 1 km/hr

∴ The correct answer is 1 km/hr.

Alternate Method

Given:

Total Distance = 30 km

Anmol takes 7 hours more than Nikhil initially.

If Anmol doubles his speed, he takes 8 hours less than Nikhil.

Distance = 30 kmNormal speed time → t + 7 hrsDouble speed time → t − 8 hrsAnmol's total time difference = 15 hrs

Formula Used:

Speed = Distance ÷ Time

Calculations:

Let the original time taken by Anmol be t_a and by Nikhil be t_n.

⇒ t_a − t_n = 7   ...

(i) When Anmol's speed is doubled, his new time becomes t_a ÷ 2.

⇒ t_n − (t_a ÷ 2) = 8   ...

(ii) Adding equations (i) and (ii):

⇒ t_a − (t_a ÷ 2) = 7 + 8

⇒ t_a ÷ 2 = 15

⇒ t_a = 30 hours

Anmol's original speed = Distance ÷ Time

⇒ Speed = 30 ÷ 30 = 1 km/hr

∴ The correct answer is 1 km/hr.

Additional Information

Speed and Time Inverse Relation

When the distance is constant, speed is inversely proportional to time. If speed becomes n times, the required time becomes 1/n times.

Average Speed

Average speed is the ratio of total distance traveled to total time taken. For two equal distances covered at speeds x and y, Average Speed = (2 × x × y) ÷ (x + y).

Relative Speed

When two objects move in the same direction, relative speed is the difference of their speeds. In opposite directions, it is the sum of their speeds.

Shortcut Trick

Net Population = Initial Base × Successive Multiplying Factors

Final Population = 12500 × (100 + 52)% × (100 + 30)% × (100 − 15)%

Final Population = 12500 × 1.52 × 1.30 × 0.85

Final Population = 19000 × 1.30 × 0.85 = 24700 × 0.85 = 20995

∴ The correct answer is 20,995.

Alternate Method

Given:

Initial Population (P) = 12,500

Successive changes over three years: +52%, +30%, and −15%

Formula Used:

Final Value = P × (1 ± R₁/100) × (1 ± R₂/100) × (1 ± R₃/100)

Initial
End of Yr 1
End of Yr 2
End of Yr 3

12,500
19,000
24,700
20,995

+52%

+30%

−15%

Calculations:

⇒ Population at the end of Year 1 = 12500 × (1 + 52/100)

⇒ Population at the end of Year 1 = 12500 × 152/100 = 19000

⇒ Population at the end of Year 2 = 19000 × (1 + 30/100)

⇒ Population at the end of Year 2 = 19000 × 130/100 = 24700

⇒ Population at the end of Year 3 = 24700 × (1 − 15/100)

⇒ Population at the end of Year 3 = 24700 × 85/100 = 20995

∴ The correct answer is 20,995.

Additional Information

Successive Percentage Change

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

Final Value Formula

For successive changes over multiple periods, the final value is calculated as: Final Value = Initial Value × (1 ± r₁/100) × (1 ± r₂/100) × ...

Population Depreciation

If a population or machine value decreases by a constant rate of r% every year, the value after n years is calculated using P(1 − r/100)ⁿ.

Shortcut Trick

The product of two consecutive numbers x and (x + 1) is approximately equal to x².

We know the squares of nearby numbers: 26² = 676 and 27² = 729.

The given product 702 lies perfectly between 676 and 729, meaning the numbers are 26 and 27.

Checking the product directly: 26 × 27 = 702.

Hence, the greater of the two numbers is 27.

∴ The correct answer is 27.

Alternate Method

Given:

The product of two consecutive natural numbers = 702

Calculation:

Let the smaller natural number be n.

Then, the greater consecutive natural number will be (n + 1).

⇒ n × (n + 1) = 702

⇒ n² + n = 702

⇒ n² + n − 702 = 0

⇒ n² + 27n − 26n − 702 = 0

⇒ n(n + 27) − 26(n + 27) = 0

⇒ (n − 26)(n + 27) = 0

⇒ n = 26 or n = −27

Since n must be a natural number, it cannot be a negative value.

⇒ n = 26

The greater number = n + 1

⇒ 26 + 1 = 27

∴ The correct answer is 27.

Additional Information

Consecutive Numbers

Numbers that follow each other continuously in order from smallest to largest, with a common difference of 1. They are generally represented as n, n+1, n+2, etc.

Consecutive Even or Odd Numbers

These numbers skip one integer, meaning they always maintain a difference of 2. They are represented as 2n, 2n+2 for even numbers, and 2n+1, 2n+3 for odd numbers.

Quadratic Equation Roots

For a quadratic equation in the form ax² + bx + c = 0, the sum of roots is given by (−b ÷ a) and the product of roots is given by (c ÷ a).

Shortcut Trick

Make the number of decimal places equal: 12.800 and 0.004.

Treat the numbers as integers: 12800 and 4.

Find the LCM of the integers: LCM of 12800 and 4 is 12800.

Place the decimal point back 3 places from the right: 12.800

∴ The correct answer is 12.8.

Alternate Method

Given:

Numbers: 12.8 and 0.004

Formula Used:

LCM of fractions = LCM of numerators ÷ HCF of denominators

Calculation:

⇒ Convert 12.8 to its simplest fraction: 128/10 = 64/5

⇒ Convert 0.004 to its simplest fraction: 4/1000 = 1/250

⇒ Find the LCM of the numerators (64 and 1) = 64

⇒ Find the HCF of the denominators (5 and 250) = 5

⇒ LCM of the fractions = 64 ÷ 5

⇒ LCM = 12.8

∴ The correct answer is 12.8.

Additional Information

HCF of Decimals

Make the decimal places equal, find the HCF of the numbers as integers, and place the decimal point back.

HCF of Fractions

HCF = HCF of numerators ÷ LCM of denominators.

Product of Two Numbers

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

Simplification of Fractions

Always convert decimal numbers to their simplest fractional forms before applying the LCM or HCF formulas to avoid calculation errors.