Quantitative Aptitude PYQs

Shortcut Trick

Extra runs scored above the previous average = 235 − 100 = 135.

This excess is distributed across all 7 innings to find the average increase.

Increase in average = 135 ÷ 7 = 19.285...

Rounding 19.28 to the nearest whole number gives 19.

∴ The correct answer is 19.

Alternate Method

Given: Average of 6 innings = 100 runs, Runs in 7^th inning = 235.

Formula: Average = Total Sum ÷ Number of observations

⇒ Total runs for 6 innings = 6 × 100 = 600

⇒ Total runs for 7 innings = 600 + 235 = 835

⇒ New average for 7 innings = 835 ÷ 7

⇒ New average = 119.2857...

⇒ Increase in average = New average − Old average

⇒ Increase = 119.2857 − 100 = 19.2857

⇒ Rounded average increase = 19

∴ The correct answer is 19.

Additional Information

Concept of Average Change

If a new value is added to a group, the average increases if the new value > old average and decreases if the new value < old average.

Deviation Method for Averages

Net Increase = (New Observation − Old Average) ÷ (Total Number of Observations including the new one).

Rounding Rules

When rounding to the nearest whole number, look at the first decimal place; if it is less than 5, round down (e.g., 19.28 → 19), if 5 or more, round up.

Shortcut Trick

For any quadrilateral PQRS, the angle formed by the bisectors of two adjacent angles (say ∠R and ∠S) is equal to half the sum of the other two angles (∠P and ∠Q):

∠STR = (∠P + ∠Q) ÷ 2

⇒ ∠P + ∠Q = 2 × 44° = 88°

Since the ratio of ∠P : ∠Q is 5 : 6, the total parts are 5 + 6 = 11 parts.

⇒ 11 parts = 88° → 1 part = 8°

The difference between ∠Q and ∠P is (6 − 5) = 1 part = 8°.

∴ The correct answer is 8°.

Alternate Method

Given:

• In quadrilateral PQRS, the bisectors of ∠R and ∠S meet at T.
• ∠STR = 44°
• Ratio of ∠P : ∠Q = 5 : 6

Formula Used:

• Sum of interior angles of a quadrilateral = 360°
• Sum of angles of a triangle = 180°

Calculations:

In triangle STR:

⇒ ∠TSR + ∠TRS + ∠STR = 180°

Since ST and RT are the angle bisectors of ∠S and ∠R respectively:

⇒ ∠TSR = ∠S ÷ 2

⇒ ∠TRS = ∠R ÷ 2

Substituting these values into the triangle angle sum equation:

⇒ (∠S ÷ 2) + (∠R ÷ 2) + 44° = 180°

⇒ (∠S + ∠R) ÷ 2 = 180° − 44°

⇒ (∠S + ∠R) ÷ 2 = 136°

⇒ ∠S + ∠R = 272°

In quadrilateral PQRS, the sum of all interior angles is 360°:

⇒ ∠P + ∠Q + ∠R + ∠S = 360°

⇒ ∠P + ∠Q + 272° = 360°

⇒ ∠P + ∠Q = 360° − 272°

⇒ ∠P + ∠Q = 88°

Let ∠P = 5x and ∠Q = 6x (since ∠P : ∠Q = 5 : 6):

⇒ 5x + 6x = 88°

⇒ 11x = 88°

⇒ x = 8°

The difference between the measures of ∠Q and ∠P is:

⇒ ∠Q − ∠P = 6x − 5x = x

⇒ Difference = 8°

∴ The correct answer is 8°.

Additional Information

Angle Bisector Relation in Quadrilaterals

The angle between the bisectors of two adjacent angles of a quadrilateral is equal to half the sum of the other two angles: ∠STR = (∠P + ∠Q) ÷ 2.

Sum of Interior Angles of a Polygon

The sum of the interior angles of any convex polygon with n sides is (n − 2) × 180°. For a quadrilateral, n = 4, so the sum is 360°.

Opposite Angle Bisectors

If the bisectors of two opposite angles of a quadrilateral (say ∠P and ∠R) intersect, the angle between them is equal to half the difference of the other two angles: |∠Q − ∠S| ÷ 2.

Shortcut Trick

Total deduction = ₹(12 + 23 + 43) = ₹78.

Remaining amount = ₹1411 − ₹78 = ₹1333.

Sum of ratio parts = 21 + 16 + 6 = 43 units.

43 units = ₹1333 ⇒ 1 unit = ₹31.

Harsh's reduced share = 16 × 31 = ₹496.

Harsh's actual share = ₹496 + ₹23 = ₹519.

∴ The correct answer is ₹519.

Alternate Method

Given: Total amount = ₹1411; Deductions = ₹12, ₹23, ₹43; Ratio = 21 : 16 : 6.

Formula: Actual Share = (Ratio Share × Value per unit) + Deducted Amount.

⇒ Total Deducted Amount = 12 + 23 + 43 = 78

⇒ New Total Amount = 1411 − 78 = 1333

⇒ Total Ratio = 21 + 16 + 6 = 43

⇒ Value of 1 unit = 1333 ÷ 43 = 31

⇒ Harsh's share in ratio = 16 × 31 = 496

⇒ Harsh's actual share = 496 + 23 = 519

∴ The correct answer is ₹519.

Additional Information

Ratio and Proportion Basics

If a total amount T is divided in the ratio a:b:c, then the share of b is calculated as [b / (a + b + c)] × T.

Concept of Deduction in Ratios

When specific amounts are deducted or added to shares, the ratio applies only to the resultant values. Always adjust the total sum before dividing by the ratio sum.

Shortcut Trick

Use the direct algebraic identity: (x − y)² = (x + y)² − 4xy

Substitute the given values: Sum (x + y) = 20 and Product (xy) = 96

(x − y)² = 20² − 4 × 96

(x − y)² = 400 − 384 = 16

∴ The correct answer is 16.

Alternate Method

Given:

Sum of two numbers (x + y) = 20

Product of two numbers (xy) = 96

Formula Used:

(a − b)² = (a + b)² − 4ab

Calculations:

⇒ (x − y)² = (x + y)² − 4xy

⇒ (x − y)² = (20)² − 4(96)

⇒ (x − y)² = 400 − 384

⇒ (x − y)² = 16

∴ The correct answer is 16.

Additional Information

Fundamental Algebraic Identities

(a + b)² = a² + b² + 2ab and (a − b)² = a² + b² − 2ab

Relationship between Identities

(a + b)² − (a − b)² = 4ab and (a + b)² + (a − b)² = 2(a² + b²)

Difference of Squares

a² − b² = (a + b)(a − b)

Shortcut Trick

Income ratio David : Ashish = 5 : 9 and Expenditure ratio = 2 : 4 (or 1 : 2).

Since savings are equal, make the difference between income and expenditure parts the same.

Multiply Income ratio by (2 − 1) = 1 → 5 : 9 (Diff = 4). Multiply Expenditure ratio by (9 − 5) = 4 → 8 : 16.

This does not align well. Use Cross Multiplication Method: (5×4 − 9×2) units = 39000 × (4 − 2).

2 units = 39000 × 2 ⇒ 1 unit = ₹39,000.

David income = 5 units = 5 × 39,000 = ₹1,95,000.

∴ The correct answer is ₹1,95,000.

Alternate Method

Given: Ratio of incomes = 5 : 9, Ratio of expenditures = 2 : 4, Savings = ₹39,000 each.

Formula Used: Income − Savings = Expenditure

⇒ Let incomes be 5x and 9x.

⇒ (5x − 39000) / (9x − 39000) = 2 / 4

⇒ (5x − 39000) / (9x − 39000) = 1 / 2

⇒ 2(5x − 39000) = 1(9x − 39000)

⇒ 10x − 78000 = 9x − 39000

⇒ 10x − 9x = 78000 − 39000

⇒ x = 39,000

⇒ David monthly income = 5x = 5 × 39,000 = ₹1,95,000.

∴ The correct answer is ₹1,95,000.

Additional Information

Income-Expenditure Relationship

Basic identity: Income = Expenditure + Savings. For two people with same savings, the difference in income units equals the difference in expenditure units if scaled correctly.

Ratio and Proportion

If a : b = c : d, then ad = bc. This cross-multiplication property is fundamental in solving ratio-based linear equations.

Cross Multiplication Trick

For equations like (ax + k)/(bx + k) = m/n, the value of x = [k × (m − n)] / [an − bm]. This is the fastest way to solve income-expenditure problems.

Shortcut Trick

Net change formula: a + b + (a × b) ÷ 100

Here, markup a = +40% and discount b = −20%.

Profit% = 40 − 20 + (40 × −20) ÷ 100

Profit% = 20 − 8 = 12%.

∴ The correct answer is 12%.

Alternate Method

Given: Markup = 40%, Discount = 20%.
Formulas: MP = CP × (100 + Markup%) ÷ 100 | SP = MP × (100 − Discount%) ÷ 100 | Profit% = (SP − CP) ÷ CP × 100

⇒ Let Cost Price (CP) = 100.
⇒ Marked Price (MP) = 100 + 40% of 100 = 140.
⇒ Selling Price (SP) = 140 − 20% of 140 = 140 − 28 = 112.
⇒ Profit = SP − CP = 112 − 100 = 12.
⇒ Profit% = (12 ÷ 100) × 100 = 12%.

∴ The correct answer is 12%.

Additional Information

Relationship between CP and MP

The ratio of MP to CP can be expressed as: MP ÷ CP = (100 + Profit%) ÷ (100 − Discount%).

Marked Price (MP)

It is the price displayed on the label of an article. Discount is always calculated on the Marked Price.

Shortcut Trick

The Curved Surface Area (CSA) of a cylinder is given by π × diameter (d) × height (h).

924 = (22 ÷ 7) × d × 14

924 = 22 × d × 2 ⇒ 924 = 44 × d

d = 924 ÷ 44 = 21 cm

∴ The correct answer is 21 cm.

Alternate Method

Given:

Curved Surface Area (CSA) = 924 cm²

Height (h) = 14 cm

Formula Used:

Curved Surface Area of Cylinder = 2 × π × r × h

Diameter (d) = 2 × r

Calculations:

⇒ CSA = 2 × π × r × h

⇒ 924 = 2 × (22 ÷ 7) × r × 14

⇒ 924 = 2 × 22 × r × 2

⇒ 924 = 88 × r

⇒ r = 924 ÷ 88

⇒ r = 10.5 cm

⇒ Diameter (d) = 2 × r = 2 × 10.5

⇒ d = 21 cm

∴ The correct answer is 21 cm.

Additional Information

Total Surface Area (TSA)

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

Volume of Cylinder

The volume measures the space inside the cylinder: Volume = πr²h.

Base Area and Perimeter

Base Area = πr² and the circumference (perimeter) of the base = 2πr.

Shortcut Trick

Net percentage interest for 2 years at 15% compound interest = x + y + (xy/100)

⇒ 15 + 15 + (15 × 15)/100 = 30 + 2.25 = 32.25%

Total Amount = 100% (Principal) + 32.25% (Interest) = 132.25% of Principal

⇒ Amount = 1200 × 1.3225 = ₹1,587

∴ The correct answer is ₹1,587.

Alternate Method

Given:

Principal (P) = ₹1,200

Rate of Interest (R) = 15% per annum

Time (n) = 2 years

Formula Used:

Amount (A) = P(1 + R/100)ⁿ

Calculations:

⇒ A = 1200 × (1 + 15/100)²

⇒ A = 1200 × (1 + 0.15)²

⇒ A = 1200 × (1.15) × (1.15)

⇒ A = 1200 × 1.3225

⇒ A = ₹1,587

∴ The correct answer is ₹1,587.

Additional Information

Compound Interest Formula

The interest earned is calculated as CI = A − P, where A = P(1 + r/100)ⁿ.

Difference between CI and SI for 2 years

The difference between compound interest and simple interest for two years is given by D = P(R/100)².

Half-Yearly Compounding

If interest is compounded half-yearly, the rate is halved (R/2) and the time is doubled (2n).

Shortcut Trick

Convert all dimensions to cm: Wall = 2000 × 990 × 30, Brick = 25 × 11 × 8.

Since 10% is cement, 90% of the wall volume is occupied by bricks.

Number of bricks = (0.9 × 2000 × 990 × 30) ÷ (25 × 11 × 8)

Number of bricks = 53,460,000 ÷ 2,200 = 24,300.

∴ The correct answer is 24,300 bricks.

Alternate Method

Given:

Wall length = 20 m = 2000 cm

Wall height = 9.9 m = 990 cm

Wall thickness = 30 cm

Brick dimensions = 25 cm × 11 cm × 8 cm

Cement volume = 10% of wall volume

Formula Used:

Volume of Cuboid = Length × Height × Thickness

Number of bricks = (Net volume of wall for bricks) ÷ (Volume of one brick)

⇒ Total volume of the wall = 2000 × 990 × 30 = 59,400,000 cm³

⇒ Volume occupied by cement = 10% of 59,400,000 = 5,940,000 cm³

⇒ Volume occupied by bricks = 59,400,000 − 5,940,000 = 53,460,000 cm³

⇒ Volume of one brick = 25 × 11 × 8 = 2,200 cm³

⇒ Total number of bricks = 53,460,000 ÷ 2,200 = 24,300

∴ The correct answer is 24,300 bricks.

Additional Information

Volume of a Cuboid

The volume of any cuboidal structure is calculated as Length × Width × Height. Ensure all units are consistent (e.g., all in cm or all in m) before calculation.

Net Volume Calculation

When a percentage of volume is occupied by mortar or cement, the net volume for bricks is (100 − Mortar%) ÷ 100 × Total Volume.

Unit Conversion

1 meter = 100 centimeters | 1 m³ = 1,000,000 cm³. Converting larger units to smaller units (m to cm) usually makes manual calculation easier by avoiding decimals.

Shortcut Trick

The relation between Cost Price (CP) and Marked Price (MP) is given by the ratio:

CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%)

CP ÷ 62,000 = (100 − 73) ÷ (100 + 20) = 27 ÷ 120

CP = 62,000 × (27 ÷ 120) = 62,000 × (9 ÷ 40)

CP = 1,550 × 9 = 13,950

∴ The correct answer is ₹13,950.

Alternate Method

Given: Marked Price (MP) = ₹62,000, Discount = 73%, Profit = 20%.

Formulas: SP = MP × (100 − Discount%) ÷ 100 | CP = SP × 100 ÷ (100 + Profit%)

⇒ Selling Price (SP) = 62,000 × (100 − 73) ÷ 100

⇒ SP = 62,000 × 27 ÷ 100 = 620 × 27 = ₹16,740

⇒ Cost Price (CP) = SP × 100 ÷ (100 + 20)

⇒ CP = 16,740 × 100 ÷ 120 = 16,740 × 5 ÷ 6

⇒ CP = 2,790 × 5 = ₹13,950

∴ The correct answer is ₹13,950.

Additional Information

Marked Price & Discount

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

CP-MP Relationship

This direct ratio (100−D):(100+P) is extremely useful for exams to bypass calculating the Selling Price.

Shortcut Trick

Use the Inverse Proportion formula: M₁ × D₁ = M₂ × D₂

Initial men (M₁) = 1209, Days (D₁) = 24

New men (M₂) = 1209 + 403 = 1612

1209 × 24 = 1612 × D₂

D₂ = (1209 × 24) ÷ 1612 = 18 days

∴ The correct answer is 18 days.

Alternate Method

Given:

Initial number of men (M₁) = 1209

Initial duration of provisions (D₁) = 24 days

Increase in men = 403

Formula Used:

Total Work (Provisions) = Men × Days

Calculations:

⇒ Total M₂ = 1209 + 403 = 1612

⇒ According to the work equivalence rule: M₁D₁ = M₂D₂

⇒ 1209 × 24 = 1612 × D₂

⇒ D₂ = (1209 × 24) ÷ 1612

⇒ D₂ = 29016 ÷ 1612

⇒ D₂ = 18 days

∴ The correct answer is 18 days.

Additional Information

Inverse Variation

When the number of men increases, the days for which provisions last will decrease. This is a case of inverse proportion where x × y = constant.

Chain Rule

The general formula is (M₁ × D₁ × H₁) ÷ W₁ = (M₂ × D₂ × H₂) ÷ W₂, where M=Men, D=Days, H=Hours, and W=Work.

Shortcut Trick

Let the total capacity of the cistern be 36 units (LCM of 9 and 36).

Combined efficiency of P and Q = 36 ÷ 9 = 4 units/hour.

Efficiency of P alone = 36 ÷ 36 = 1 unit/hour.

Efficiency of Q alone = 4 − 1 = 3 units/hour.

Time taken by Q to fill one-fourth of the cistern = (36 × 1/4) ÷ 3 = 9 ÷ 3 = 3 hours.

∴ The correct answer is 3 hours.

Alternate Method

Given:

Time taken by P and Q together to fill the cistern = 9 hours

Time taken by P alone to fill the cistern = 36 hours

Formula Used:

1/T_Q = 1/T_P+Q − 1/T_P

Calculations:

⇒ Efficiency of Q = Efficiency of (P + Q) − Efficiency of P

⇒ Efficiency of Q = 4 − 1 = 3 units/hour

⇒ One-fourth of the total cistern capacity = 36 ÷ 4 = 9 units

⇒ Time taken by Q to fill 9 units = 9 ÷ 3 = 3 hours

∴ The correct answer is 3 hours.

Additional Information

Total Work LCM Method

Taking the LCM of individual times as the total work simplifies calculations by avoiding fractions during efficiency determination.

Work and Efficiency Relation

Time taken is inversely proportional to efficiency. Time = Total Work ÷ Efficiency.

Shortcut Trick

Net discount = a + b − (a × b) ÷ 100

Net discount = 20 + 10 − (20 × 10) ÷ 100 = 30 − 2 = 28%

Selling Price (SP) = 100% − 28% = 72% of Marked Price (MP)

Price paid = 1620 × 0.72 = 1166.40

∴ The correct answer is ₹1,166.40.

Alternate Method

Given: Marked Price (MP) = ₹1,620, Discount 1 = 20%, Discount 2 = 10%.
Formulas: SP = MP × (100 − D₁%) ÷ 100 × (100 − D₂%) ÷ 100

⇒ Price after 1^st discount = 1620 × (80 ÷ 100) = 1620 × 0.8 = 1296
⇒ Price after 2^nd discount = 1296 × (90 ÷ 100) = 1296 × 0.9
⇒ Final Price = 1166.40

∴ The correct answer is ₹1,166.40.

Additional Information

Marked Price & Discount

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

Successive Discount

When two or more discounts are applied one after another, the net discount is less than the sum of individual discounts. Formula for two: a + b − ab/100.

Shortcut Trick

Net discount for first two = 22 + 8 − (22 × 8) ÷ 100 = 30 − 1.76 = 28.24%.

Net discount with third = 28.24 + 17 − (28.24 × 17) ÷ 100.

⇒ 45.24 − (480.08 ÷ 100) = 45.24 − 4.8008 = 40.4392%.

∴ The correct answer is 40.44%.

Alternate Method

Given → Discounts are 22%, 8%, and 17%.
Formulas: Successive discount = a + b − (ab ÷ 100) | SP = MP × (100 − D₁)% × (100 − D₂)% × (100 − D₃)%

⇒ Let the Marked Price (MP) be 100.

⇒ Price after 22% discount = 100 × (78 ÷ 100) = 78.

⇒ Price after 8% discount = 78 × (92 ÷ 100) = 71.76.

⇒ Price after 17% discount = 71.76 × (83 ÷ 100) = 59.5608.

⇒ Single Equivalent Discount = MP − Final SP = 100 − 59.5608 = 40.4392%.

∴ The correct answer is 40.44%.

Additional Information

Successive Discounts

When two or more discounts are applied one after another, they are called successive discounts. The order of application does not change the final amount.

Effective Discount Formula

For three discounts d₁, d₂, and d₃, the equivalent discount is 100 − [100 × (1 − d₁/100) × (1 − d₂/100) × (1 − d₃/100)].

Shortcut Trick

Assume R = 2x. Then P = x, S = x − 10, and T = 2x + 20.

Total sum = 2x + x + (x − 10) + (2x + 20) = 6x + 10.

6x + 10 = 130 ⇒ 6x = 120 ⇒ x = 20.

R = 2x = 2 × 20 = 40 years.

∴ The correct answer is 40 years.

Alternate Method

Given:

T = R + 20

S = P − 10

P = R ÷ 2

Sum: T + S + P + R = 130

Substitute all variables in terms of R:

⇒ (R + 20) + (R/2 − 10) + (R/2) + R = 130

⇒ R + R/2 + R/2 + R + 20 − 10 = 130

⇒ 3R + 10 = 130

⇒ 3R = 120

⇒ R = 40 years.

∴ The correct answer is 40 years.

Additional Information

Linear Equations in Ages

Expressing all unknown ages in terms of a single variable (usually the youngest person or the person linked to all others) simplifies the calculation.

Ratio and Proportion in Ages

If the ratio of ages of A and B is a:b, their ages can be assumed as ax and bx years respectively.

Age Difference Constant

The difference between the ages of two persons remains the same at any point in time.

Shortcut Trick

Convert all bases to 3: 9 = 3², 27 = 3³

Expression becomes: 3^(4+x) × 3^2(x−4) ÷ (3^3(x−3) × 3³)

Combine powers: Numerator = 3^(4+x+2x−8) = 3^(3x−4)

Denominator = 3^(3x−9+3) = 3^(3x−6)

Final = 3^{(3x−4 − (3x−6))} = 3² = 9

∴ Answer = 9

Alternate Method

Given:

(3^(4+x) × 9^(x−4)) ÷ (27^(x−3) × 3³)

Concept Used:

Index laws: a^m × aⁿ = a^m+n, (a^m)ⁿ = a^mn, a^m ÷ aⁿ = a^m−n

Calculation:

⇒ (3^(4+x) × (3²)^(x−4)) ÷ ((3³)^(x−3) × 3³)

⇒ (3^(4+x) × 3^(2x−8)) ÷ (3^(3x−9) × 3³)

⇒ 3^(4+x+2x−8) ÷ 3^(3x−9+3)

⇒ 3^(3x−4) ÷ 3^(3x−6)

⇒ 3^{((3x−4) − (3x−6))}

⇒ 3²

⇒ 9

∴ The correct answer is 9.

Additional Information

Laws of Exponents

a^m × aⁿ = a^m+n, a^m ÷ aⁿ = a^m−n

Power of a Power

(a^m)ⁿ = a^mn

Base Conversion

Convert numbers like 9 and 27 into powers of 3 for easy simplification.

Simplification Trick

Combine powers early to reduce expression quickly.

Shortcut Trick

Assume initial population is 100.

After 40% left: 100 − 40 = 60.

After 10% of remaining left: 60 − (10% of 60) = 60 − 6 = 54.

Ratio of initial to present = 100 : 54 = 50 : 27.

∴ The correct answer is 50:27.

Alternate Method

Given:

Percentage who left due to water crisis = 40%

Percentage of remaining who left for jobs = 10%

Present population = 1080

Calculations:

⇒ Let Initial Population be P.

⇒ Remaining after water crisis = P × (100 − 40)/100 = 0.6P

⇒ Present Population = 0.6P × (100 − 10)/100

⇒ Present Population = 0.6P × 0.9 = 0.54P

⇒ Ratio = Initial / Present = P / 0.54P

⇒ Ratio = 1 / 0.54 = 100 / 54

⇒ Ratio = 50 / 27

∴ The correct answer is 50:27.

Additional Information

Successive Percentage Change

When a value changes by x% and then the remaining changes by y%, the net effective multiplier is (1 ± x/100) × (1 ± y/100).

Ratio of Change

If a population decreases by R%, the ratio of original to new population is 100 : (100 − R).

Percentage to Fraction

Converting percentages to fractions (40% = 2/5, 10% = 1/10) can often simplify complex population calculations.

Shortcut Trick

Relative Speed = Total distance ÷ Total time

Total distance = 164 + 136 = 300 m; Time = 10 s

Relative Speed = 300 ÷ 10 = 30 m/s = 30 × (18/5) = 108 km/hr

Speed of second train = 108 − 82 = 26 km/hr

∴ The correct answer is 26 km/hr.

Alternate Method

Given: Length₁ = 164 m, Length₂ = 136 m, Time = 10 s, Speed₁ = 82 km/hr

Formula Used: Relative Speed (opposite) = (L₁ + L₂) ÷ Time

⇒ Total Distance = 164 + 136 = 300 m

⇒ Relative Speed (m/s) = 300 ÷ 10 = 30 m/s

⇒ Relative Speed (km/hr) = 30 × (18/5) = 108 km/hr

⇒ Relative Speed = Speed₁ + Speed₂

⇒ 108 = 82 + Speed₂

⇒ Speed₂ = 108 − 82 = 26 km/hr

∴ The correct answer is 26 km/hr.

Additional Information

Relative Speed in Opposite Direction

When two objects move in opposite directions, their relative speed is the sum of their individual speeds: S_rel = S₁ + S₂.

Unit Conversion

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

Crossing Distance

When a train crosses another train (stationary or moving), the total distance covered is always the sum of their lengths (L₁ + L₂).

Shortcut Trick

To find the larger number when the sum and difference of two numbers are known:

Larger Number = (Sum + Difference) ÷ 2

⇒ Larger Number = (45 + 5) ÷ 2 = 50 ÷ 2 = 25

∴ The correct answer is 25.

Alternate Method

Given:

Sum of two numbers = 45

Difference between the numbers = 5

Method: Linear Equation in one variable

Let the smaller number be x.

Then the larger number = x + 5

According to the question:

⇒ x + (x + 5) = 45

⇒ 2x + 5 = 45

⇒ 2x = 45 − 5

⇒ 2x = 40

⇒ x = 20

The larger number is x + 5:

⇒ 20 + 5 = 25

∴ The correct answer is 25.

Additional Information

Finding Smaller Number

To calculate the smaller value directly from sum (S) and difference (D), use the formula: Smaller Number = (S − D) ÷ 2.

System of Linear Equations

If x + y = S and x − y = D, then adding both equations eliminates y to find x, and subtracting them eliminates x to find y.

Average of Two Numbers

The average of two numbers is exactly halfway between them. Average = Sum ÷ 2. Here, 45 ÷ 2 = 22.5. The numbers are 22.5 ± (Difference ÷ 2), which is 22.5 ± 2.5.

Shortcut Trick

Total Distance = 187 + 273 = 460 km

Total Time = 1 + 1 = 2 hours

Average Speed = 460 ÷ 2 = 230 km/hr

∴ The correct answer is 230 km/hr.

Alternate Method

Given:

Distance in 1^st hour (d₁) = 187 km

Distance in 2^nd hour (d₂) = 273 km

Formula Used:

Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ Total Distance = d₁ + d₂ = 187 + 273

⇒ Total Distance = 460 km

⇒ Total Time = 1 + 1 = 2 hours

⇒ Average Speed = 460 ÷ 2

⇒ Average Speed = 230 km/hr

∴ The correct answer is 230 km/hr.

Additional Information

Average Speed (Equal Distances)

If a journey is divided into two equal distances with speeds x and y, then Average Speed = 2xy ÷ (x + y).

Average Speed (Equal Time)

If a body travels at different speeds x and y for equal time intervals, Average Speed = (x + y) ÷ 2.

Unit Conversion

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