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Shortcut Trick

Identify Total Distance = 90 + 120 + 150 = 360 km.

Calculate Total Time = (90 ÷ 45) + (120 ÷ 60) + (150 ÷ 75) = 2 + 2 + 2 = 6 hours.

Average Speed = Total Distance ÷ Total Time = 360 ÷ 6 = 60 km/h.

∴ The correct answer is 60 km/h.

Alternate Method

Given:

Distance₁ (d₁) = 90 km, Speed₁ (s₁) = 45 km/h

Distance₂ (d₂) = 120 km, Speed₂ (s₂) = 60 km/h

Distance₃ (d₃) = 150 km, Speed₃ (s₃) = 75 km/h

Formula Used:

Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ Time taken for first part (t₁) = 90 ÷ 45 = 2 hours

⇒ Time taken for second part (t₂) = 120 ÷ 60 = 2 hours

⇒ Time taken for third part (t₃) = 150 ÷ 75 = 2 hours

⇒ Total Distance = 90 + 120 + 150 = 360 km

⇒ Total Time = 2 + 2 + 2 = 6 hours

⇒ Average Speed = 360 ÷ 6 = 60 km/h

∴ The correct answer is 60 km/h.

Additional Information

Average Speed for Equal Distances

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

Unit Conversion

To convert speed from km/h to m/s, multiply by 5/18. To convert from m/s to km/h, multiply by 18/5.

Relative Speed

When two objects move in the same direction, relative speed is s₁ − s₂. In opposite directions, it is s₁ + s₂.

Shortcut Trick

Sum both equations to find the cost of 8 chairs and 8 tables.

(5C + 3T) + (3C + 5T) = 12,500 + 15,500 ⇒ 8C + 8T = ₹28,000.

To find the cost of 2 chairs and 2 tables, divide the total sum by 4.

Cost = 28,000 ÷ 4 = 7,000.

∴ The correct answer is ₹7,000.

Alternate Method

Given:

Cost of 5 chairs and 3 tables = ₹12,500

Cost of 3 chairs and 5 tables = ₹15,500

Formula Used:

Total Cost = (Cost per unit × Quantity)

Calculations:

Let the cost of one chair be C and one table be T.

⇒ 5C + 3T = 12,500 — (Equation 1)

⇒ 3C + 5T = 15,500 — (Equation 2)

Adding Equation 1 and Equation 2:

⇒ (5C + 3C) + (3T + 5T) = 12,500 + 15,500

⇒ 8C + 8T = 28,000

⇒ 8 × (C + T) = 28,000

⇒ C + T = 28,000 ÷ 8

⇒ C + T = 3,500

We need the cost of 2 chairs and 2 tables:

⇒ 2 × (C + T) = 2 × 3,500

⇒ 2C + 2T = 7,000

∴ The correct answer is ₹7,000.

Additional Information

Linear Equations in Two Variables

Equations in the form ax + by = c can be solved using elimination, substitution, or cross-multiplication methods.

Symmetric Equation Property

When coefficients are swapped (ax + by and bx + ay), always add and subtract the equations to simplify the system quickly.

Unitary Concept

Finding the combined value of one unit of each item (C + T) often allows solving for any multiple (nC + nT) without finding individual prices.

Given:

(((16/21))) / (((16/4))) ÷ ((4/5 × 10/12) + 4/7) + 4/8 ÷ (26/14 of 14/8)

Formula used:

The VBODMAS rule: Brackets, Of, Division, Multiplication, Addition, Subtraction.

Calculations:

Simplify the first complex fraction term:

⇒ (16/21) / (16/4) = 16/21 × 4/16 = 4/21

Simplify the expression inside the brackets:

⇒ (4/5 × 10/12 + 4/7) = (2/3 + 4/7) = (14 + 12) / 21 = 26/21

Solve the Of operation (which takes priority over division):

⇒ 26/14 of 14/8 = (26/14) × (14/8) = 26/8 = 13/4

Now, substitute these simplified values back into the full expression:

⇒ (4/21 ÷ 26/21) + (4/8 ÷ 13/4)

⇒ (4/21 × 21/26) + (4/8 × 4/13)

⇒ 4/26 + 16/104

⇒ 2/13 + 2/13

⇒ 4/13

∴ The correct answer is option (4).

Shortcut Trick

The numbers are in continued proportion, so the ratio 16 : 24 must equal the ratio 24 : third number.

Simplify the first ratio: 16 : 24 = 2 : 3.

This means the second ratio must also be 2 : 3.

If 2 units = 24, then 1 unit = 12.

Therefore, 3 units = 12 × 3 = 36.

∴ The correct answer is 36.

Alternate Method

Given: First number = 16, Second number = 24.

Formula Used: For three numbers a, b, and c in continued proportion, a ÷ b = b ÷ c.

Calculations:

⇒ Let the third number be c.

⇒ 16 ÷ 24 = 24 ÷ c

⇒ (16 × c) = (24 × 24)

⇒ 16c = 576

⇒ c = 576 ÷ 16

⇒ c = 36

∴ The correct answer is 36.

Additional Information

Mean Proportional

If a, b, and c are in continued proportion, then b is called the mean proportional between a and c, where b = √(a × c).

Third Proportional

In the proportion a : b = b : c, the value c is known as the third proportional to a and b, calculated as c = b² ÷ a.

Fourth Proportional

For four numbers a, b, c, and d in proportion (a : b = c : d), d is the fourth proportional, where d = (b × c) ÷ a.

Shortcut Trick

The Mean Proportional between a and b is √(a × b).

Given Mean Proportional = 12, numbers are x and 36.

Square of Mean = Product of numbers ⇒ 12² = x × 36.

144 = 36 × x ⇒ x = 144 ÷ 36 = 4.

∴ The correct answer is 4.

Alternate Method

Given:

First term = x

Mean proportional (Middle term) = 12

Third term = 36

Formula Used:

In a Continued Proportion (a : b :: b : c), the relation is b² = a × c.

Calculations:

⇒ x / 12 = 12 / 36

⇒ x × 36 = 12 × 12

⇒ 36x = 144

⇒ x = 144 / 36

⇒ x = 4

∴ The correct answer is 4.

Additional Information

Mean Proportional

For any two numbers a and c, the mean proportional b is calculated as b = √(ac).

Third Proportional

In a continued proportion a : b = b : c, the term c is the third proportional, where c = b² / a.

Fourth Proportional

For four quantities a, b, c, d in proportion, a/b = c/d, the fourth proportional is d = (b × c) / a.

Shortcut Trick

Let the Original Price be 100.

Step 1: 20% increase → 100 + 20 = 120.

Step 3: Final price becomes 100 after a 10% increase of the Original Price (+10).

This means the price before Step 3 was 100 − 10 = 90.

Step 2: Value of x is the percentage decrease from 120 to 90 ⇒ (30 ÷ 120) × 100 = 25.

∴ The correct answer is 25.

Alternate Method

Given:

Initial Price = 100

First Increase = 20%

Final Increase = 10% of Original Price

Final Price = Original Price = 100

Calculations:

⇒ Let original price = 100

⇒ Price after 20% increase = 100 + 20 = 120

⇒ Price after x% decrease = 120 − (x/100 × 120) = 120 − 1.2x

⇒ Final price after 10% of original price increase = (120 − 1.2x) + 10

⇒ Given final price = 100

⇒ 130 − 1.2x = 100

⇒ 1.2x = 30

⇒ x = 30 ÷ 1.2 = 25

∴ The correct answer is 25.

Additional Information

Successive Percentage Change

When a value is changed by a% and then by b%, the net effective change is given by the formula: [a + b + (ab ÷ 100)]%.

Base Value Concept

Percentage calculations always depend on the base value. In this problem, x% was calculated on the new price (120), while the 10% increase was calculated on the original price (100).

Percentage to Fraction Conversion

Common fractions help in fast calculation: 25% = 1/4, 20% = 1/5, 10% = 1/10.

Shortcut Trick

Assume the car travelled all 9.5 hr at the slower speed of 40 km/h.

Hypothetical distance = 40 × 9.5 = 380 km.

Shortfall in distance = 560 − 380 = 180 km.

Difference in speeds = 160 − 40 = 120 km/h.

Time spent at 160 km/h = 180 ÷ 120 = 1.5 hr.

Distance covered at 160 km/h = 1.5 × 160 = 240 km.

∴ The correct answer is 240 km.

Alternate Method

Given: Total distance = 560 km, Total time = 9.5 hr, Speeds = 40 km/h and 160 km/h.

Formula Used: Time = Distance ÷ Speed

Let the distance travelled at 160 km/h be d km.

The distance travelled at 40 km/h will be (560 − d) km.

⇒ Total Time = Time₁ + Time₂

⇒ 9.5 = (560 − d) ÷ 40 + d ÷ 160

⇒ 9.5 = [4(560 − d) + d] ÷ 160

⇒ 9.5 × 160 = 2240 − 4d + d

⇒ 1520 = 2240 − 3d

⇒ 3d = 2240 − 1520

⇒ 3d = 720

⇒ d = 240 km

∴ The correct answer is 240 km.

Additional Information

Distance, Speed, and Time Relationship

The fundamental relation is Distance = Speed × Time. If speed is constant, distance is directly proportional to time.

Average Speed

Average Speed = Total Distance ÷ Total Time. It is not the simple arithmetic mean of the individual speeds unless the time spent at each speed is equal.

Unit Conversion

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

Alligation Rule for Speed

When alligation is applied to speeds, the resulting ratio is the ratio of the time taken at those respective speeds.

Shortcut Trick

The total surface area of three spheres is given by the formula: 4π(r₁² + r₂² + r₃²).

Sum of squares of radii = 2² + 3² + 6² = 4 + 9 + 36 = 49 cm².

Total Area = 4 × (22/7) × 49 = 88 × 7 = 616 cm².

∴ The correct answer is 616 cm².

Alternate Method

Given:

Radius of first sphere (r₁) = 2 cm

Radius of second sphere (r₂) = 3 cm

Radius of third sphere (r₃) = 6 cm

Value of π = 22/7

Formula Used:

Surface area of a sphere = 4πr²

Calculations:

⇒ Surface area of first sphere = 4 × (22/7) × 2² = 4 × (22/7) × 4 = 352/7 cm²

⇒ Surface area of second sphere = 4 × (22/7) × 3² = 4 × (22/7) × 9 = 792/7 cm²

⇒ Surface area of third sphere = 4 × (22/7) × 6² = 4 × (22/7) × 36 = 3168/7 cm²

⇒ Sum of surface areas = (352/7) + (792/7) + (3168/7)

⇒ Sum of surface areas = (352 + 792 + 3168) ÷ 7

⇒ Sum of surface areas = 4312 ÷ 7

⇒ Sum of surface areas = 616 cm²

∴ The correct answer is 616 cm².

Additional Information

Volume of a Sphere

The volume of a sphere with radius r is given by the formula V = (4/3) × π × r³.

Surface Area of Hemisphere

Curved Surface Area (CSA) = 2πr² and Total Surface Area (TSA) = 3πr².

Volume of a Hemisphere

The volume of a hemisphere is exactly half of the sphere, given by V = (2/3) × π × r³.

Given:

55 1/8 + 55 3/8 + 55 5/8 + 55 7/8

Formula used:

A Mixed Fraction is the sum of its Whole Number part and its Fractional Part:

x a/b = x + a/b

Calculations:

55 1/8 + 55 3/8 + 55 5/8 + 55 7/8

⇒ (55 + 1/8) + (55 + 3/8) + (55 + 5/8) + (55 + 7/8)

⇒ (55 + 55 + 55 + 55) + (1/8 + 3/8 + 5/8 + 7/8)

⇒ (55 × 4) + (1 + 3 + 5 + 7) / 8

⇒ 220 + 16 / 8

⇒ 220 + 2

⇒ 222

∴ The correct answer is option (1).

Shortcut Trick

The highest frequency is 13, which corresponds to the modal class 50−55.

Using the mode formula: Mode = l + [ (f₁ − f₀) ÷ (2f₁ − f₀ − f₂) ] × h

Here, l = 50, f₁ = 13, f₀ = 4, f₂ = 5, and h = 5.

Mode = 50 + [ (13 − 4) ÷ (26 − 9) ] × 5 = 50 + (9 ÷ 17) × 5 = 50 + 2.65 = 52.65.

∴ The correct answer is 52.65.

Alternate Method

Given Data:

Modal Class (Class with highest frequency) = 50−55

Lower limit of the modal class (l) = 50

Frequency of the modal class (f₁) = 13

Frequency of the class preceding the modal class (f₀) = 4

Frequency of the class succeeding the modal class (f₂) = 5

Class interval width (h) = 5

Calculations:

⇒ Mode = l + [ (f₁ − f₀) ÷ (2 × f₁ − f₀ − f₂) ] × h

⇒ Mode = 50 + [ (13 − 4) ÷ (2 × 13 − 4 − 5) ] × 5

⇒ Mode = 50 + [ 9 ÷ (26 − 9) ] × 5

⇒ Mode = 50 + [ 9 ÷ 17 ] × 5

⇒ Mode = 50 + (45 ÷ 17)

⇒ Mode = 50 + 2.64705...

⇒ Mode ≈ 52.65

∴ The correct answer is 52.65.

Additional Information

Empirical Relationship

For a moderately skewed distribution, the relationship is: Mode = 3 × Median − 2 × Mean.

Median for Grouped Data

Median = l + [ (N÷2 − cf) ÷ f ] × h, where cf is the cumulative frequency of the preceding class.

Modal Class Definition

It is the class interval which has the maximum frequency in a grouped frequency distribution.