Quantitative Aptitude PYQs

Shortcut Trick

Total CP of all bulbs = ₹93.

Number of good bulbs remaining = 93 − 19 = 74.

Total SP of remaining bulbs = 74 × ₹1.40 = ₹103.60.

Profit = Total SP − Total CP = 103.60 − 93 = ₹10.60.

∴ The correct answer is ₹10.60.

Alternate Method

Given:

Total bulbs purchased = 93

Total Cost Price (CP) = ₹93

Number of broken bulbs in transit = 19

Selling Price (SP) of one remaining bulb = ₹1.40

Formula Used:

Profit = Total Selling Price − Total Cost Price

Total CP = ₹93

CP Portion = ₹93
Profit = ₹10.60

74 bulbs × ₹1.40 = ₹103.60 (Total SP)

Total SP = ₹103.60

Calculation:

⇒ Number of remaining good bulbs = 93 − 19 = 74

⇒ Total Selling Price = 74 × 1.40

⇒ Total SP = ₹103.60

⇒ Profit = Total SP − Total CP

⇒ Profit = 103.60 − 93 = 10.60

∴ The correct answer is ₹10.60.

Additional Information

Profit and Loss Basics

Profit occurs when the Selling Price (SP) is strictly greater than the Cost Price (CP). Profit = SP − CP.

Loss Scenario

If the total revenue from selling the remaining items is less than the total cost price due to damage or low rates, a loss is incurred. Loss = CP − SP.

Profit / Loss Percentage

Profit or Loss percentage is always evaluated on the total investment or Cost Price unless stated otherwise: Profit % = (Profit ÷ CP) × 100.

Shortcut Trick

Other Number = (HCF × LCM) ÷ Given Number

⇒ Other Number = (6 × 264) ÷ 66

⇒ Other Number = 6 × 4

⇒ Other Number = 24

∴ The correct answer is 24.

Alternate Method

Given:

HCF = 6

LCM = 264

First Number (N₁) = 66

Formula Used:

Product of two numbers = HCF × LCM

⇒ N₁ × N₂ = HCF × LCM

N1 = 66

N2 = ?

×

=

HCF = 6

LCM = 264

×

Calculations:

⇒ 66 × N₂ = 6 × 264

⇒ N₂ = (6 × 264) ÷ 66

⇒ N₂ = 1584 ÷ 66

⇒ N₂ = 24

∴ The correct answer is 24.

Additional Information

Co-prime Numbers

Two numbers are co-prime if their HCF is 1. The LCM of two co-prime numbers is always equal to their product.

HCF and LCM of Fractions

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

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

Relation between HCF and LCM

The HCF of two or more numbers is always a factor of their LCM. That means the LCM is completely divisible by the HCF.

Shortcut Trick

SI is directly proportional to Rate × Time.

Effective rate 1 = 6.8 × (60/7) = 408/7%

Effective rate 2 = 9.7 × (20/3) = 194/3%

Principal = (7344 × 100) ÷ (408/7) = ₹12,600

Amount = 12600 × (100% + 194/3%) = 126 × (494/3) = ₹20,748

∴ The correct answer is ₹20,748.

Alternate Method

Given:

Simple Interest (SI₁) = ₹7,344

Time (T₁) = 8 ⁴/₇ years = 60/7 years

Rate (R₁) = 6.8% per annum

Rate (R₂) = 9.7% per annum

Time (T₂) = 6 ²/₃ years = 20/3 years

Formula Used:

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

Amount = P + SI

Principal = ₹12,600

Case 1
R = 6.8%, T = 60/7 yrs
SI = ₹7,344

Case 2
R = 9.7%, T = 20/3 yrs
SI = ₹8,148
Amount = ₹20,748

Calculations:

⇒ 7344 = (P × 6.8 × (60/7)) ÷ 100

⇒ 7344 = (P × 408) ÷ 700

⇒ P = (7344 × 700) ÷ 408

⇒ P = 18 × 700 = ₹12,600

Now, calculating SI for the second case:

⇒ SI₂ = (P × R₂ × T₂) ÷ 100

⇒ SI₂ = (12600 × 9.7 × (20/3)) ÷ 100

⇒ SI₂ = 126 × 9.7 × (20/3)

⇒ SI₂ = 42 × 9.7 × 20

⇒ SI₂ = 840 × 9.7 = ₹8,148

Finding the Amount payable:

⇒ Amount = P + SI₂

⇒ Amount = 12600 + 8148 = ₹20,748

∴ The correct answer is ₹20,748.

Additional Information

Compound Interest (CI) Formula

When interest is calculated on the initial principal and the accumulated interest of previous periods, Amount A = P × (1 + R/100)^T.

SI vs CI Difference (for 2 years)

The difference between compound interest and simple interest on a certain sum for exactly 2 years is given by: Difference = P × (R/100)².

Rule of 72

A mathematical rule of thumb to estimate the number of years required to double your money at a given annual compound interest rate (R). Approximate Time ≈ 72 ÷ R.

Shortcut Trick

Simplify the inner fractional term first: 1 + 3/8 = 11/8.

The reciprocal of 11/8 is 8/11.

The given expression simplifies to: 5/2 + 8/11 − 3/5.

Taking the LCM of 2, 11, and 5 as 110, the numerator becomes: (5 × 55) + (8 × 10) − (3 × 22).

Simplifying further: (275 + 80 − 66) / 110 = 289 / 110.

∴ The correct answer is 289/110.

Alternate Method

Given:

Expression: 5/2 + (1 / (1 + 3/8)) − 3/5

Calculations:

⇒ First, we will simplify the term inside the bracket.

⇒ 1 + 3/8 = (8 + 3) / 8 = 11/8

⇒ Dividing 1 by this fraction gives its reciprocal.

⇒ 1 / (11/8) = 8/11

⇒ Now, substitute this back into the original expression:

⇒ 5/2 + 8/11 − 3/5

⇒ To evaluate this, find the Least Common Multiple (LCM) of the denominators 2, 11, and 5.

⇒ LCM(2, 11, 5) = 110

⇒ Convert each fraction to an equivalent fraction with 110 as the denominator:

⇒ (5 × 55)/110 + (8 × 10)/110 − (3 × 22)/110

⇒ 275/110 + 80/110 − 66/110

⇒ (275 + 80 − 66) / 110

⇒ (355 − 66) / 110

⇒ 289 / 110

∴ The correct answer is 289/110.

Additional Information

BODMAS Rule

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

Fraction Addition & Subtraction

To add or subtract unlike fractions, find the LCM of their denominators, convert them to equivalent fractions, and then process the numerators while keeping the common denominator.

Reciprocal of a Fraction

The reciprocal of any fraction a/b is b/a. When 1 is divided by a proper or improper fraction, the result is exactly its reciprocal.

Shortcut Trick

When distance is constant, speed is inversely proportional to time.

Time ratio (T₁ : T₂) = 9 : 12 = 3 : 4

Speed ratio (S₁ : S₂) = 4 : 3

Given 4 units → 48 km/hr

New speed (3 units) → 12 × 3 = 36 km/hr

∴ The correct answer is 36 km/hr.

Alternate Method

Given:

Initial speed (S₁) = 48 km/hr

Initial time (T₁) = 9 hours

New time (T₂) = 12 hours

Formula Used:

Distance = Speed × Time

New Speed = Distance ÷ New Time

Case 1: S = 48 km/hr, T = 9 hours

Case 2: S = ? , T = 12 hours
Distance is Constant

Calculations:

⇒ Total Distance = S₁ × T₁

⇒ Total Distance = 48 × 9 = 432 km

⇒ New Speed (S₂) = Distance ÷ T₂

⇒ S₂ = 432 ÷ 12

⇒ S₂ = 36 km/hr

∴ The correct answer is 36 km/hr.

Additional Information

Average Speed

Average Speed = Total Distance ÷ Total Time. If equal distances are covered at speeds a and b, the average speed becomes 2ab ÷ (a + b).

Relative Speed

When moving in the same direction, Relative Speed = |a − b|. When moving in opposite directions, Relative Speed = a + b.

Proportionality in Speed, Distance and Time

When Distance is constant, Speed is inversely proportional to Time (S ∝ 1 / T). When Time is constant, Distance is directly proportional to Speed (D ∝ S).

Shortcut Trick

Let M and D be Mary's and Dia's present ages.

M − 7D = 9 ⇒ 2M − 14D = 18 ...

(i) 2M − 5D = 27 ...

(ii) Subtracting (i) from (ii): 9D = 9 ⇒ D = 1

M = 7(1) + 9 = 16 ⇒ Difference = 16 − 1 = 15

∴ The correct answer is 15 years.

Alternate Method

Given:

2 times Mary's age is 27 more than 5 times Dia's age.

7 times Dia's age is 9 less than Mary's age.

Mary

16 years

Dia

1 year

Difference = 15 years

Calculations:

Let Mary's present age be M and Dia's present age be D.

⇒ 2M = 5D + 27   ... (Equation 1)

⇒ 7D = M − 9

⇒ M = 7D + 9   ... (Equation 2)

Substitute Equation 2 into Equation 1:

⇒ 2(7D + 9) = 5D + 27

⇒ 14D + 18 = 5D + 27

⇒ 14D − 5D = 27 − 18

⇒ 9D = 9

⇒ D = 1

Substitute the value of D in Equation 2:

⇒ M = 7(1) + 9

⇒ M = 16

Difference between their ages = M − D

⇒ 16 − 1 = 15

∴ The correct answer is 15 years.

Additional Information

Linear Equations in Two Variables

A system of two linear equations can be solved using substitution or elimination methods to logically find the unique values of two unknown variables.

Problems on Ages

If the present age is assumed as x, then the age 'n' years ago is evaluated as (x − n) and the age 'n' years hence will be evaluated as (x + n).

Difference Between Ages

The age difference between two individuals always remains strictly constant throughout their lifetime, regardless of the number of years passed.

Shortcut Trick

Total Distance = 294 + 210 + 330 = 834 km

Total Time = (294 ÷ 42) + (210 ÷ 30) + (330 ÷ 66) = 7 + 7 + 5 = 19 hr

Average Speed = Total Distance ÷ Total Time

Average Speed = 834 ÷ 19 ≈ 43.89 km/hr

∴ The correct answer is 43.89 km/hr.

Alternate Method

Given:

Distance D₁ = 294 km, Speed S₁ = 42 km/hr

Distance D₂ = 210 km, Speed S₂ = 30 km/hr

Distance D₃ = 330 km, Speed S₃ = 66 km/hr

D = 294 km
D = 210 km
D = 330 km

S = 42 km/hr
S = 30 km/hr
S = 66 km/hr

T = 7 hr
T = 7 hr
T = 5 hr

Total = 834 km in 19 hr

Formula Used:

Time = Distance ÷ Speed

Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ T₁ = 294 ÷ 42 = 7 hr

⇒ T₂ = 210 ÷ 30 = 7 hr

⇒ T₃ = 330 ÷ 66 = 5 hr

⇒ Total Time = 7 + 7 + 5 = 19 hr

⇒ Total Distance = 294 + 210 + 330 = 834 km

⇒ Average Speed = 834 ÷ 19 = 43.8947... km/hr

⇒ Average Speed ≈ 43.89 km/hr

∴ The correct answer is 43.89 km/hr.

Additional Information

Average Speed (Different Distances)

It is the ratio of the total distance covered to the total time taken. Formula: Avg Speed = (D₁ + D₂ + ...) ÷ (T₁ + T₂ + ...)

Average Speed (Equal Distances)

If a person covers two equal halves of a journey at speeds x and y, the average speed is (2xy) ÷ (x + y). For three equal parts at speeds x, y, and z, it is (3xyz) ÷ (xy + yz + zx).

Relative Speed

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

Shortcut Trick

When the investment time is equal, Profit Ratio = Capital Ratio.

Ratio of Profit = 1370 : 1500 : 1230 = 137 : 150 : 123

Total ratio parts = 137 + 150 + 123 = 410 parts

Simran's profit share = (123 ÷ 410) × 1010

Simran's profit share = 3 × 101 = ₹303

∴ The correct answer is ₹303.

Alternate Method

Given:

Priya's investment = ₹1370

Deepak's investment = ₹1500

Simran's investment = ₹1230

Total profit at the end of the year = ₹1010

Formula Used:

When the time of investment is the same for all partners, the profit is distributed in proportion to their initial capital investments.

Capital Investment and Profit Ratio

Priya

₹1370 (137 parts)

Deepak

₹1500 (150 parts)

Simran

₹1230 (123 parts)

Total Profit
₹1010

Calculations:

⇒ Profit Ratio (Priya : Deepak : Simran) = 1370 : 1500 : 1230

⇒ Simplified Profit Ratio = 137 : 150 : 123

⇒ Sum of the profit ratios = 137 + 150 + 123 = 410 units

⇒ Simran's share = (Simran's ratio ÷ Total ratio) × Total profit

⇒ Simran's share = (123 ÷ 410) × 1010

⇒ Simran's share = 123 × (1010 ÷ 410)

⇒ Simran's share = 123 × (101 ÷ 41)

⇒ Simran's share = 3 × 101 = 303

∴ The correct answer is ₹303.

Additional Information

Partnership Profit Ratio

Profit ratio is basically calculated as the product of the capital invested and the respective time period of investment. (P₁ : P₂ = C₁×T₁ : C₂×T₂)

Active and Sleeping Partners

An active partner actively works to manage the business and receives a fixed salary or a specific percentage of profit as management fee. The remaining profit is distributed according to invested capitals. A sleeping partner only invests capital.

Joining Mid-way

If a partner joins the business after 'n' months in a 1-year operational business, their investment time period is considered as (12 − n) months while evaluating their profit share.

Shortcut Trick

Let Cost Price (CP) = 100 units ⇒ List Price (MP) = 125 units.

Selling Price (SP) after 16% discount = 125 × 0.84 = 105 units.

Profit = SP − CP = 105 − 100 = 5 units.

Given, 5 units = ₹140 ⇒ 1 unit = ₹28.

List Price (MP) = 125 × 28 = 3500.

∴ The correct answer is ₹3500.

Alternate Method

Given:

Markup Percentage = 25%

Discount Percentage = 16%

Profit = ₹140

CP = 100u

SP = 105u

MP = 125u

Markup: +25%

Discount
−16%

Profit = 5u = ₹140

Calculations:

⇒ Let the Cost Price (CP) be 100 units.

⇒ List Price (MP) = CP + 25% of CP = 100 + 25 = 125 units.

⇒ Discount is always calculated on the List Price.

⇒ Selling Price (SP) = MP − Discount

⇒ SP = 125 − (16% of 125)

⇒ SP = 125 − 20 = 105 units.

⇒ Profit = SP − CP = 105 − 100 = 5 units.

⇒ According to the question, Profit = ₹140

⇒ 5 units = 140

⇒ 1 unit = 140 ÷ 5 = 28

⇒ List Price (MP) = 125 units = 125 × 28 = 3500.

∴ The correct answer is ₹3500.

Additional Information

Marked Price and Discount

Discount is always calculated on the Marked Price (or List Price) of an article. The Selling Price is derived as: SP = MP × (1 − Discount%).

Profit and Loss Calculation

Profit and Loss percentages are always computed on the Cost Price (CP). Profit = SP − CP, and Profit% = (Profit ÷ CP) × 100.

Direct Relation Formula

The relationship between Cost Price and Marked Price can be directly established using the effective formula: CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%).

Shortcut Trick

Since the order of successive discounts does not change the final price, calculate the 10% discount first.

Intermediate Price after 10% discount = 39250 − (10% of 39250) = 39250 − 3925 = ₹35,325

The second discount reduces the price from ₹35,325 to ₹32,499.

Discount amount = 35325 − 32499 = ₹2,826

Second discount x = (2826 ÷ 35325) × 100 = 8%

∴ The correct answer is 8%.

Alternate Method

Given:

Marked Price (MP) = ₹39,250

Selling Price (SP) = ₹32,499

Successive discounts = x% and 10%

Formula Used:

SP = MP × (1 − x/100) × (1 − 10/100)

Marked Price
₹39,250

−10%

New Price
₹35,325

−x%

Selling Price
₹32,499

Discount % = (35325 − 32499) ÷ 35325 × 100 = 8%

Calculations:

⇒ 32499 = 39250 × (1 − x/100) × (1 − 10/100)

⇒ 32499 = 39250 × (1 − x/100) × (90/100)

⇒ 32499 = 35325 × (1 − x/100)

⇒ 32499 ÷ 35325 = 1 − x/100

⇒ 0.92 = 1 − x/100

⇒ x/100 = 1 − 0.92 = 0.08

⇒ x = 0.08 × 100 = 8

∴ The correct answer is 8%.

Additional Information

Successive Discounts Formula

When two successive discounts of a% and b% are applied, the equivalent single discount is given by: (a + b − ab/100)%.

Marked Price vs Cost Price Relation

The relationship linking Marked Price (MP), Cost Price (CP), Profit % (P) and Discount % (D) is: MP/CP = (100 + P) ÷ (100 − D).

Effective Multiplier Concept

The final Selling Price after any number of successive discounts can be found using multipliers directly: SP = MP × (100 − d₁)/100 × (100 − d₂)/100.

Shortcut Trick

According to the rule of weighted averages, Total % = (Male % × Male Pop + Female % × Female Pop) ÷ Total Pop

If the overall percentage of a population exactly matches the percentage of one of its sub-groups, the remaining sub-group must also share that exact same percentage to maintain the overall average.

Since both the total population and the male population have a literacy rate of exactly 62%, the female literacy rate must also inherently be 62%.

∴ The correct answer is 62%.

Alternate Method

Given:

Total Population = 368000

Male Population = 220000

Total Literate = 62%

Literate Males = 62%

Total Pop: 368,000
Literate: 62%

Males: 220,000
Literate: 62%

Females: 148,000
Literate: ? %

Since Total % = Male %, Female % must be 62%

Calculations:

⇒ Female population = Total Population − Male Population

⇒ Female population = 368000 − 220000 = 148000

⇒ Number of total literate people = 62% of 368000

⇒ Number of literate males = 62% of 220000

⇒ Number of literate females = Total literate − Literate males

⇒ Literate females = (62% of 368000) − (62% of 220000)

⇒ Literate females = 62% × (368000 − 220000)

⇒ Literate females = 62% of 148000

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

⇒ Percentage of literate females = [(62% of 148000) ÷ 148000] × 100

⇒ Percentage of literate females = 62%

∴ The correct answer is 62%.

Additional Information

Weighted Average Rule

When a population is divided into distinct subgroups, the total percentage is the weighted average of the sub-group percentages: Overall % = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂).

Constant Percentage Property

If the overall percentage of a mixture or group matches the percentage of one of its components exactly, all other components must inherently have that exact same percentage.

Base Independence

In questions where sub-group percentages are perfectly equal to the total group percentage, actual population or quantity numbers become redundant and do not impact the final percentage value.

Shortcut Trick

Convert the percentage to a fraction: 20% = 1/5.

Equate the parts: (1/5) × First Number = (4/5) × Second Number.

The denominators (5) cancel out directly from both sides.

First Number = 4 × Second Number.

Ratio of First Number to Second Number = 4 :

1. ∴ The correct answer is 4 : 1.

Alternate Method

20% of 1^st

=

4/5 of 2^nd

1/5 × 1^st = 4/5 × 2^nd

Ratio = 4 : 1

Given:

20% of the first number is equal to four-fifths of the second number.

Calculations:

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

⇒ 20% of x = (4/5) of y

⇒ (20/100) × x = (4/5) × y

⇒ (1/5) × x = (4/5) × y

Multiplying both sides by 5 to eliminate the fractions:

⇒ x = 4 × y

To find the ratio of the first number to the second, rearrange the equation:

⇒ x/y = 4/1

⇒ x : y = 4 : 1

∴ The correct answer is 4 : 1.

Additional Information

Ratio & Proportion

A ratio compares two quantities by division, expressed as a : b. A proportion is an equation stating that two ratios are equivalent to each other.

Percentage to Fraction Conversion

Percentages can be directly converted into standard fractions by dividing by 100. For example, 20% = 20/100 = 1/5. Memorizing base fractional values significantly reduces calculation time.

Cross Multiplication Rule

When two fractions are proportional (a/b = c/d), applying cross multiplication yields a × d = b × c. When denominators are identical on both sides of an equation, they can be directly cancelled out.

Shortcut Trick

The relation between Volume (V), Curved Surface Area (CSA), and radius (r) of a cylinder is directly given by the formula:

Volume = (CSA × r) ÷ 2

⇒ Volume = (5280 × 21) ÷ 2

⇒ Volume = 2640 × 21

⇒ Volume = 55440

∴ The correct answer is 55,440 cm³.

Alternate Method

Given:

Curved Surface Area (CSA) = 5280 cm²

Radius (r) = 21 cm

Formulas Used:

CSA = 2πrh

Volume = πr²h

r = 21 cm

h = ?

Curved Surface Area = 5280 sq cm

Calculations:

⇒ 2 × (22 ÷ 7) × 21 × h = 5280

⇒ 2 × 22 × 3 × h = 5280

⇒ 132 × h = 5280

⇒ h = 5280 ÷ 132 = 40 cm

Now, substitute the value of h to find the volume:

⇒ Volume = π × r² × h

⇒ Volume = (22 ÷ 7) × 21 × 21 × 40

⇒ Volume = 22 × 3 × 21 × 40

⇒ Volume = 66 × 840

⇒ Volume = 55440 cm³

∴ The correct answer is 55,440 cm³.

Additional Information

Total Surface Area of a Cylinder

The TSA of a solid cylinder includes the curved surface area plus the area of the two circular bases. Formula: TSA = 2πr(r + h).

Volume of a Cone

A cone with the same base radius and height as a cylinder has exactly one-third of the cylinder's volume. Formula: V = (1 ÷ 3)πr²h.

Hollow Cylinder

For a hollow tube with outer radius R and inner radius r, the volume of material used is given by V = πh(R² − r²).

Shortcut Trick

Put θ = 45° to quickly evaluate the expression.

⇒ sec 45° = √2, sin 45° = 1/√2, cos 45° = 1/√2

⇒ (1 − sec²45°) ÷ sin²45° × cos²45°

⇒ (1 − 2) ÷ (1/2) × (1/2)

⇒ −1 × 2 × (1/2) = −1

∴ The correct answer is −1.

Alternate Method

Given:

[(1 − sec²θ) / sin²θ] × cos²θ

Formula Used:

sec²θ − tan²θ = 1

tanθ = sinθ / cosθ

Calculation:

⇒ [(1 − sec²θ) / sin²θ] × cos²θ

⇒ [−tan²θ / sin²θ] × cos²θ

⇒ [(−sin²θ / cos²θ) / sin²θ] × cos²θ

⇒ (−1 / cos²θ) × cos²θ

⇒ −1

∴ The correct answer is −1.

Additional Information

Trigonometric Identities

sin²θ + cos²θ = 1 | sec²θ − tan²θ = 1 | cosec²θ − cot²θ = 1

Reciprocal Relations

sinθ = 1 / cosecθ | cosθ = 1 / secθ | tanθ = 1 / cotθ

Value Substitution Method

For expressions independent of θ, substituting standard values like 30°, 45°, or 60° gives the answer quickly without solving algebraically.

Shortcut Trick

Use the LCM method to quickly find the total capacity and combined efficiency.

Total Capacity (LCM of 16, 28, and 8) = 112 units.

Combined Efficiency = (112 ÷ 16) + (112 ÷ 28) + (112 ÷ 8) = 7 + 4 + 14 = 25 units/hour.

Time Taken = Total Capacity ÷ Combined Efficiency = 112 ÷ 25 = 4 ¹²/₂₅ hours.

∴ The correct answer is 4 ¹²/₂₅ hours.

Alternate Method

Given:

Time taken by Pipe A = 16 hours

Time taken by Pipe B = 28 hours

Time taken by Pipe C = 8 hours

Calculations:

⇒ Total combined efficiency of all pipes = 7 + 4 + 14 = 25 units/hour

⇒ Total time taken when all pipes operate together = Total Work ÷ Total Efficiency

⇒ Time taken = 112 ÷ 25

⇒ Time taken = 4 ¹²/₂₅ hours

∴ The correct answer is 4 ¹²/₂₅ hours.

Additional Information

Pipes & Cisterns Concept

The rules for solving pipe filling problems are identical to Time and Work questions. If a pipe fills a tank in x hours, its 1-hour work is 1/x.

Emptying Pipe (Leakage)

If an outlet pipe empties a full tank in y hours, its 1-hour work is taken as negative (−1/y).

Volume / Capacity of Tank

To avoid complex fraction calculations, the total capacity of the tank is generally assumed to be the LCM of the given time periods.

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 = 155.

⇒ Average = 155 + 1 = 156

∴ The correct answer is 156.

Alternate Method

Given:

The sequence consists of the first 155 even numbers.

Formula Used:

For an Arithmetic Progression (A.P.), Average = (First term + Last term) ÷ 2

2
1^st Term

310
155^th Term

156
Average = (2+310)÷2

4

6

308

306

Number Line showing A.P. of Even Numbers

Calculations:

⇒ The series of the first 155 even numbers is 2, 4, 6, 8, ..., up to 155 terms.

⇒ First term (a) = 2

⇒ Common difference (d) = 2

⇒ The 155^th term (Last term) = a + (n − 1)d

⇒ Last term = 2 + (155 − 1) × 2 = 2 + 154 × 2 = 2 + 308 = 310

⇒ Average = (First term + Last term) ÷ 2

⇒ Average = (2 + 310) ÷ 2

⇒ Average = 312 ÷ 2 = 156

∴ The correct answer is 156.

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 Numbers in A.P.

For any sequence of numbers in an Arithmetic Progression, the average is exactly the middle term, calculated as (First term + Last term) ÷ 2.

Shortcut Trick

Expand the brackets directly to cancel the x² terms:

⇒ 6x² − 12 − 6x² − 24x + 15 = 14

⇒ −24x + 3 = 14

⇒ −24x = 11 ⇒ x = −11/24

∴ The correct answer is −11/24.

Alternate Method

Given:

2(3x² − 6) − 3(2x² + 8x − 5) = 14

Calculations:

Expand the terms using the distributive property:

⇒ (2 × 3x²) − (2 × 6) − (3 × 2x²) − (3 × 8x) − (3 × −5) = 14

⇒ 6x² − 12 − 6x² − 24x + 15 = 14

Group the like terms together:

⇒ (6x² − 6x²) − 24x + (−12 + 15) = 14

⇒ 0 − 24x + 3 = 14

⇒ −24x = 14 − 3

⇒ −24x = 11

⇒ x = −11/24

∴ The correct answer is −11/24.

Additional Information

Linear Equation in One Variable

An equation representing a straight line that can be written in the form ax + b = 0, where a and b are constants and a ≠ 0.

Distributive Property

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

BODMAS Rule

The sequence of operations used to simplify mathematical expressions: Brackets, Orders, Division/Multiplication, Addition/Subtraction.

Shortcut Trick

Work equivalency: 6M × 9 = 6W × 6 ⇒ 54M = 36W ⇒ M ÷ W = 2 ÷ 3

Total work = 54 × 2 = 108 units

Combined efficiency of 3M and 2W = (3 × 2) + (2 × 3) = 6 + 6 = 12 units/day

Required days = 108 ÷ 12 = 9 days

∴ The correct answer is 9 days.

Alternate Method

Given:

Time taken by 6 men = 9 days ⇒ Time taken by 1 man = 6 × 9 = 54 days

Time taken by 6 women = 6 days ⇒ Time taken by 1 woman = 6 × 6 = 36 days

Calculations:

⇒ Efficiency of 3 men = 3 × 2 = 6 units/day

⇒ Efficiency of 2 women = 2 × 3 = 6 units/day

⇒ Combined efficiency of 3 men and 2 women = 6 + 6 = 12 units/day

⇒ Total days required = Total Work ÷ Combined Efficiency

⇒ Days = 108 ÷ 12 = 9 days

∴ The correct answer is 9 days.

Additional Information

Total Work Formula

Total Work = Number of workers × Efficiency per worker × Number of days.

MDH Rule

M₁ × D₁ × H₁ ÷ W₁ = M₂ × D₂ × H₂ ÷ W₂, where M is the number of workers, D is days, H is hours, and W is the work done.

Work and Wages

Wages are always distributed in the ratio of the total work done by the individuals, which directly correlates to their working efficiencies if they work for the exact same number of days.

Shortcut Trick

Between 430 and 438, odd numbers are 431, 433, 435, 437.

435 is divisible by 5, so it is clearly not prime.

Check primes up to √438 ≈ 20 (i.e., 2, 3, 5, 7, 11, 13, 17, 19) for the remaining candidates.

431 and 433 are not divisible by any of these primes, so they are prime numbers.

437 is exactly divisible by 19 (19 × 23), so it is not prime.

∴ The correct answer is 2 prime numbers.

Alternate Method

Given:

The range of numbers is from 430 to 438 (both included).

Concept Used:

A prime number is only divisible by 1 and itself.

To check if a number N is prime, we must divide it by all prime numbers up to √N.

Calculation:

⇒ Numbers in the given range: 430, 431, 432, 433, 434, 435, 436, 437, 438

⇒ Eliminate all even numbers as they are divisible by 2.

⇒ Remaining numbers: 431, 433, 435, 437

⇒ Eliminate 435 since it ends in 5 and is divisible by 5.

⇒ Remaining candidates to check: 431, 433, 437

⇒ For these numbers, √N ≈ √438 ≈ 20.9. Prime numbers to check ≤ 20 are 2, 3, 5, 7, 11, 13, 17, 19.

⇒ Checking 431: Not divisible by any prime ≤ 19. Hence, 431 is prime.

⇒ Checking 433: Not divisible by any prime ≤ 19. Hence, 433 is prime.

⇒ Checking 437: Divisible by 19 (437 ÷ 19 = 23). Hence, 437 is a composite number.

⇒ The prime numbers between 430 and 438 are exactly 431 and 433.

⇒ Total count of prime numbers = 2

∴ The correct answer is 2 prime numbers.

Additional Information

Prime Numbers

A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself (e.g., 2, 3, 5, 7, 11).

Primality Test

To check if a number N is prime, it is sufficient to test its divisibility by all prime numbers less than or equal to √N. If none divide perfectly, N is prime.

Twin Primes

Two prime numbers that differ by exactly 2 are called twin primes. For example, 431 and 433 are twin primes since 433 − 431 = 2.

Divisibility Rule for 19

To check if a number is divisible by 19, multiply its last digit by 2 and add it to the remaining truncated number. If the final result is a multiple of 19, the number is divisible by 19.

Shortcut Trick

The mode of a dataset is simply the value that appears most frequently.

By visually observing the given marks, the value 13 appears 3 times, which is the highest repetition.

All other values appear only 1 or 2 times.

∴ The correct answer is 13 marks.

Alternate Method

Given:

Marks scored by 10 students: 13, 20, 15, 13, 19, 12, 12, 11, 13, 10

Formula Used:

Mode = Observation with the maximum frequency in the data set.

0
1
2
3
Freq

1
10

1
11

2
12

3
13

1
15

1
19

1
20

Marks (Values)

Calculations:

⇒ Let us arrange the data in ascending order to easily count the frequency of each mark:

⇒ 10, 11, 12, 12, 13, 13, 13, 15, 19, 20

⇒ Count of 10 = 1

⇒ Count of 11 = 1

⇒ Count of 12 = 2

⇒ Count of 13 = 3

⇒ Count of 15, 19, and 20 = 1 each

⇒ Since 13 has the maximum frequency (it occurs 3 times), it is the mode of the given data.

∴ The correct answer is 13 marks.

Additional Information

Mean

The mean is the average of a given set of numbers, calculated as the sum of all observations divided by the total number of observations.

Median

The median is the exact middle value when the dataset is arranged in ascending or descending order. If the count is even, it is the average of the two middle values.

Empirical Relationship

For a moderately skewed distribution, the relationship between central tendencies is given by: Mode = 3 × Median − 2 × Mean.