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Calculation:

Here, from the table, we can see, the number of Chocobars is what percent more or less with respect to Cornettos is given

So, we can say, Arun has 20% more Chocobars than Cornettos 

Let, the total number of Cornettos be 100

So, the ratio of Chocobars and Cornettos = (100 + 20) ∶ 100

⇒ 6 ∶ 5

Similarly, we can say the same for others,

So, put those data in a table, we get,

 

The total number of Cornettos all of them have = (100 + 200 + 200 + 80 + 100) = 680

The total number of Chocobars all of them have = (120 + 140 + 180 + 100 + 160) = 700

So, the difference = (700 - 680) = 20

∴ The required difference is 20

Shortcut Trick

Initial Average age = 20 years.

New students' age = 7.5 years each. Deficit per new student = 20 − 7.5 = 12.5 years.

Total deficit for 4 students = 4 × 12.5 = 50 years.

This deficit caused the total average to decrease by 5.

Total number of students (x + 4) = Total deficit ÷ decrease in average = 50 ÷ 5 = 10.

x = 10 − 4 = 6.

∴ The correct answer is 6.

Alternate Method

Given: Initial students = x, Initial average = 20, New students = 4, Age of each = 7.5, New average = 20 − 5 = 15.

Formula Used: Total Age = Number of students × Average Age

⇒ Total initial age = 20 × x

⇒ Age of 4 new students = 4 × 7.5 = 30

⇒ New total students = x + 4

⇒ New total age = 15 × (x + 4)

⇒ 20x + 30 = 15(x + 4)

⇒ 20x + 30 = 15x + 60

⇒ 5x = 30

⇒ x = 6

∴ The correct answer is 6.

Additional Information

Weighted Average Formula

When two groups are combined, the new average is A = (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Net Deviation Principle

In any average problem, the sum of deviations of all observations from the average is always zero.

Shortcut Trick

Side of large cube (A) = √³(3³ + 4³ + 5³) = √³(27 + 64 + 125) = √³(216) = 6 cm.

Sum of Total Surface Areas (TSA) of small cubes = 6 × (3² + 4² + 5²) = 6 × (9 + 16 + 25) = 6 × 50 = 300 cm².

TSA of large cube = 6 × 6² = 6 × 36 = 216 cm².

Required Ratio = 300 : 216 = 25 : 18.

∴ The correct answer is 25 : 18.

Alternate Method

Given: Sides of three cubes are 3 cm, 4 cm, and 5 cm.

Formula Used: Volume of cube = a³; Total Surface Area = 6a².

⇒ Total Volume = 3³ + 4³ + 5³ = 27 + 64 + 125 = 216 cm³

⇒ Side of large cube (A) = √³(216) = 6 cm

⇒ Sum of TSAs of small cubes = 6(3²) + 6(4²) + 6(5²) = 6(9 + 16 + 25) = 6 × 50 = 300 cm²

⇒ TSA of large cube = 6 × 6² = 216 cm²

⇒ Ratio = 300 : 216

⇒ Dividing by 12: 25 : 18

∴ The correct answer is 25 : 18.

Additional Information

Cube Mensuration

Lateral Surface Area of a cube = 4a² and Diagonal of a cube = a√3.

Conservation of Volume

When objects are melted and recast, the total Volume remains constant (V₁ + V₂ + ... = V_final).

Surface Area Change

Upon melting smaller shapes to form a larger one, the Total Surface Area usually decreases because internal faces are lost.

Shortcut Trick

Speed of Bus = 1330 ÷ 19 = 70 km/h

Speed of Bike = 50% of 70 = 35 km/h

Speed of Cycle = 35 − 15 = 20 km/h

Time Ratio (Cycle : Bike) for same distance = Speed of Bike : Speed of Cycle = 35 : 20 = 7 : 4

∴ The correct answer is 7 : 4.

Alternate Method

Given: Distance = 1330 km, Time = 19 hours, Fixed Distance = 280 km

Formula Used: Speed = Distance ÷ Time; Time = Distance ÷ Speed

⇒ Speed of Bus = 1330 ÷ 19 = 70 km/h

⇒ Speed of Bike = 70 − (50% of 70) = 70 − 35 = 35 km/h

⇒ Speed of Cycle = 35 − 15 = 20 km/h

⇒ Time taken by Cycle = 280 ÷ 20 = 14 hours

⇒ Time taken by Bike = 280 ÷ 35 = 8 hours

⇒ Required Ratio (Cycle : Bike) = 14 : 8 = 7 : 4

∴ The correct answer is 7 : 4.

Additional Information

Inverse Relationship of Speed and Time

When the distance is constant, the time taken is inversely proportional to the speed. Ratio of Speed S₁:S₂ = Ratio of Time T₂:T₁.

Percentage Decrease

If a value is x% less than another value V, the new value is V × (100 − x)/100.

Relative Speed

In the same direction, relative speed is the difference (S₁ − S₂). In opposite directions, it is the sum (S₁ + S₂).

Shortcut Trick

Let Upstream speed be U and Downstream speed be D.

Set 1: 52/U + 42/D = 10 | Set 2: 60/U + 66/D = 13

By inspection or HCF method for D (42, 66), try D = 6 or 14. If D = 14, 42/14 = 3 hours, then 52/U = 7 → U = 52/7 (not integer). If D = 6, not possible for downstream.

Try D = 14 and U = 6. Set 1: 52/6.5 + 42/14 = 8 + 3 = 11 (close). Using Elimination: U = 6.5 km/hr, D = 14 km/hr (approx) or solving: U = 8 km/hr, D = 12 km/hr.

Checking U = 8, D = 12: 52/8 + 42/12 = 6.5 + 3.5 = 10 (Matches!). 60/8 + 66/12 = 7.5 + 5.5 = 13 (Matches!).

Speed in still water = (D + U) ÷ 2 = (12 + 8) ÷ 2 = 10 km/hr.

∴ The correct answer is 10 km/hr.

Alternate Method

Let speed of man be u and speed of current be v. Upstream speed U = u − v, Downstream speed D = u + v.

From given data:

⇒ 52/U + 42/D = 10 ... (i)

⇒ 60/U + 66/D = 13 ... (ii)

Multiply (i) by 15 and (ii) by 13 to equate 1/U terms:

⇒ 780/U + 630/D = 150

⇒ 780/U + 858/D = 169

Subtracting the equations:

⇒ 228/D = 19

⇒ D = 228 ÷ 19 = 12 km/hr

Substitute D = 12 in (i):

⇒ 52/U + 42/12 = 10 ⇒ 52/U + 3.5 = 10 ⇒ 52/U = 6.5

⇒ U = 52 ÷ 6.5 = 8 km/hr

Speed of man in still water = (D + U) ÷ 2

⇒ (12 + 8) ÷ 2 = 20 ÷ 2 = 10 km/hr

∴ The correct answer is 10 km/hr.

Additional Information

Upstream and Downstream formulas

Downstream speed (D) = Speed of boat (u) + Speed of stream (v). Upstream speed (U) = u − v.

Speed of Still Water

Speed of boat in still water = (Downstream speed + Upstream speed) ÷ 2.

Speed of Current

Speed of stream or current = (Downstream speed − Upstream speed) ÷ 2.

Follow BODMAS rule to solve this question, as per the order given below, 

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and follow BODMAS rule in the bracket,

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step - 4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value,

So, we can write these values to their nearest integers.

Given expression is

25.96% of 250.05 + 75.96% of 75.05 + √80.99 = ?

⇒ 26% of 250 + 76% of 75 + √81 = ? 

⇒ 65 + 57 + 9 = ?

⇒ ? = 131

Follow BODMAS rule to solve this question, as per the order given below,

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and follow BODMAS rule in the bracket,

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step - 4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value,

So, we can write these values to their nearest integers.

Given expression is

2.93 × 4.85 - ? - 6.97 × (2.02 + 0.99) = ∛511.9

3 × 5 - x - 7 × (2 + 1) = ∛512

⇒ 3 × 5 - x - 7 × 3 = 8 

⇒ 15 - x - 21 = 8

⇒ -x - 6 = 8

∴ x = -14

Follow BODMAS rule to solve this question, as per the order given below,

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and follow BODMAS rule in the bracket,

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step - 4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value,

So, we can write these values to their nearest integers.

Given expression is

599.99 × 3/2 of 19.92% of (109.99 – 9.99) = ?

= 600 × 3/2 of 20% of 100 = ?

= 600 × 3/2 of 20/100 of 100 = ?

= 600 × 3/2 of 20 = ?

= 600 × 3/2 × 20 = ?

= 600 × 30 = ?

= 18000 = ?

Follow BODMAS rule to solve this question, as per the order given below,

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and follow BODMAS rule in the bracket,

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step - 4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value,

So, we can write these values to their nearest integers.

Given expression is

(2.992 × 3.992× 4.99) ÷ 35.99 = (?)2 – 79.99

⇒ (32 × 42× 5) ÷ 36 = (?)2 – 80

⇒ (?)2 = ((9 × 16 × 5) ÷ 36) + 80

⇒ (?)2 = (720 ÷ 36) + 80

⇒ (?)2 = 20 + 80 = 100

⇒ (?) = √100 = 10

Follow BODMAS rule to solve this question, as per the order given below,

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and follow BODMAS rule in the bracket,

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next,

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step - 4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value,

So, we can write these values to their nearest integers.

Given expression is

324.95 – [(11.99)2 – 74.9] = (?)2 – 67.99

⇒ 325 – [(12)2 – 75] = (?)2 – 68

⇒ 325 – [144 – 75] + 68 = (?)2

⇒ 325 – 69 + 68 = (?)2

⇒ 324 = (?)2

⇒ 18 = ?