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Given:

Price increase of honey = 15%

Formula Used:

Let the original price be P and the reduced consumption be C.

New price = P + 0.15P = 1.15P

To keep the expense the same: P × C = 1.15P × (1 - x)

where x is the percentage reduction in consumption.

Calculation:

P × C = 1.15P × (1 - x)

⇒ C = 1.15 × (1 - x)

Since expense is not increased, C = 1

⇒ 1 = 1.15 × (1 - x)

⇒ 1 = 1.15 - 1.15x

⇒ 1.15x = 1.15 - 1

⇒ 1.15x = 0.15

⇒ x = 0.15 / 1.15

⇒ x = 0.1304

⇒ x ≈ 13%

The correct answer is option 2.

Given:

Saving = Rs 8,000

Calculation:

Saving % = 60° =Rs 8,000

Then, 1° = 8000/60 = Rs 800/6

Then, portion spent on food = 120° 

Portion spend on housing = 105° 

Difference in amount spent (in degree) in food and housing =  120° - 105° = 15° 

Then, required difference = 15 × 800/6 = Rs 2,000.

∴ The required difference is Rs 2,000.

Given:

C.P = Rs. 5400

C.P = Cost Price

Calculation:

According to the question,

(5400 × $\dfrac{2}{3}$ × 8%) + (1800 × x%) = (5400 × 18%) 

⇒ (3600 × 8%) + (1800 × x%) = (5400 × 18%)

⇒ 288 + 18x = 54 × 18

⇒ 288 + 18x = 972

⇒ 18x = 972 - 288

⇒ x = 684/18

⇒ x = 38

∴ The correct answer is 38%.

Shortcut Trick

Simplify given eq: x(3 − 2/x) = 3/x ⇒ 3x − 2 = 3/x ⇒ x − 1/x = 2/3

Shortcut: If x − 1/x = k, then x³ − 1/x³ = k³ + 3k

Here k = 2/3, so x³ − 1/x³ = (2/3)³ + 3 × (2/3)

⇒ 8/27 + 2 = (8 + 54) / 27 = 62/27

∴ The correct answer is 62/27.

Alternate Method

Given equation: x(3 − 2/x) = 3/x

Formula used: (a − b)³ = a³ − b³ − 3ab(a − b)

⇒ 3x − 2 = 3/x

⇒ 3x − 3/x = 2

⇒ 3(x − 1/x) = 2

⇒ x − 1/x = 2/3

Cubing both sides of the equation:

⇒ (x − 1/x)³ = (2/3)³

⇒ x³ − 1/x³ − 3 × x × (1/x) × (x − 1/x) = 8/27

⇒ x³ − 1/x³ − 3 × (2/3) = 8/27

⇒ x³ − 1/x³ − 2 = 8/27

⇒ x³ − 1/x³ = 8/27 + 2

⇒ x³ − 1/x³ = (8 + 54) / 27

⇒ x³ − 1/x³ = 62/27

∴ The correct answer is 62/27.

Additional Information

Algebraic Identity for Addition

If x + 1/x = k, then the value of x³ + 1/x³ = k³ − 3k.

Square Identity

If x − 1/x = k, then x² + 1/x² = k² + 2. Similarly, if x + 1/x = k, then x² + 1/x² = k² − 2.

Relation between Sum and Difference

(x + 1/x)² − (x − 1/x)² = 4. This is used to convert between addition and subtraction forms.

Given:

The expression to simplify is √((1191-√(1250-√(599+√676))))

Formula Used:

√a means the square root of a.

Calculation:

Start with the innermost square root: √676

⇒ √676 = 26

Then, 599 + 26 = 625

⇒ √(599 + √676) = √625 = 25

Next, 1250 - 25 = 1225

⇒ √(1250 - √(599 + √676)) = √1225 = 35

Finally, 1191 - 35 = 1156

⇒ √(1191 - √(1250 - √(599 + √676))) = √1156 = 34

∴ The simplified result of the given expression is 34.

Given:

Average of four consecutive odd numbers = 42

Formula used:

Average of n consecutive numbers = Sum of numbers ÷ n

Calculations:

Let the four consecutive odd numbers be (x - 3), (x - 1), (x + 1), and (x + 3).

Average = (Sum of numbers) ÷ 4

⇒ 42 = [(x - 3) + (x - 1) + (x + 1) + (x + 3)] ÷ 4

⇒ 42 = (4x) ÷ 4

⇒ x = 42

Largest number = x + 3 = 42 + 3 = 45

Sum of digits of 45 = 4 + 5 = 9

∴ The sum of the digits of the largest number is 9.

Given:

Time is taken by pipe to fill the tank = 12 hours

Time is taken by both pipe and leak to fill the tank = (12 + 8) = 20 hours

Formula used:

Time is taken = Total work/Efficiency

Calculation:

Part of the tank filled by the pipe in 1 hour = 1/12

Part of the tank filled by pipe and leak in 1 hour = 1/20

Part of the tank empty by a leak in 1 hour = 1/12 - 1/20 = 1/30

Total time is taken by the leak to empty full tank = 30 hours

∴ The time is taken by the leak to empty the full tank in 30 hours

Shortcut Trick

The efficiency of leak = 3 – 5 = - 2     (Here negative sign indicating that it is removing water from the tank)

Time = 60/2 = 30 hours

∴ The time taken by the leak to empty the full tank is 30 hours.

Given:

A table showing the number of trees of 3 different fruits, planted during 2018- 2022.

Formula used:

Percentage increase = (Mango in 2020 - Mango in 2018) / Mango in 2018 × 100

Calculation:

Percentage increase = (52000 - 40000) / 40000 × 100 

⇒ 12000 / 40000 × 100

⇒ 30%

∴ The correct answer is 30%.

Given :

Radius of cone(r) = 11 cm

The slant height of cone (l) = 24 cm

Formula used :

Total surface area of the cone = πr(r + l) 

Calculation:

Total surface area of the cone = πr(r + l) 

⇒  $\dfrac{22}{7}× 11× (11+24)$ 

⇒ ${\dfrac{22}{7}}× 11× 35$

⇒ 22 × 11 × 5 = 1210 cm²

∴ The answer is 1210 cm² .

Shortcut TrickTotal surface area of the cone is the multiple of π. Which is the multiple of 11 and only one option is the multiple of 11 that is 1210.

Given:

Let ABCD be a quadrilateral and AC, BD be diagonals.

The length of BD = 20 cm and the heights of the triangles ABD and BCD are, respectively, 6 cm, 8 cm.

Calculation:

According to the question,

Area of Quadrilateral = Sum of Area (Triangle ABD + Triangle BCD)

⇒ (1/2)× 20 × 6 + (1/2)× 20 × 8

⇒ 60 + 80 

⇒ 140

∴ The area of the quadrilateral ABCD is 140 cm².