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Given:

Cost price of custard powder 1 = ₹42/kg

Cost price of custard powder 2 = ₹60/kg

Amount of custard powder 2 = 16 kg

Selling price per kg = ₹60

Profit = 20%

Formula used:

Mixture's cost price = Selling price / (1 + Profit/100)

Mixture ratio = (Cost price 2 - Mixture's cost price) : (Mixture's cost price - Cost price 1)

Calculation:

Mixture's cost price = 60 / (1 + 20/100)

⇒ Mixture's cost price = 60 / 1.2

⇒ Mixture's cost price = ₹50/kg

Mixture ratio = (60 - 50) : (50 - 42)

⇒ Mixture ratio = 10 : 8

⇒ Mixture ratio = 5 : 4

Let the amount of custard powder 1 be x kg.

Total ratio = 5 + 4 = 9

Custard powder 2 ratio = 5

⇒ Custard powder 1 = (4/5) × 16

⇒ Custard powder 1 = 20 kg

∴ The correct answer is option (3).

Given:

Cost Price of Milk = Selling Price of Mixture

Profit Percentage = 10%

Quantity of Milk = 50 kg

Formula used:

Profit % = \(\dfrac{Quantity\ of\ Water}{Total\ Mixture\ Quantity}\) × 100

Calculation:

10 = \(\dfrac{x}{50 + x}\) × 100

⇒ \(\dfrac{10}{100} = \dfrac{x}{50 + x}\)

⇒ \(0.1 = \dfrac{x}{50 + x}\)

⇒ \(0.1 \times (50 + x) = x\)

⇒ \(5 + 0.1x = x\)

⇒ \(5 = x - 0.1x\)

⇒ \(5 = 0.9x\)

⇒ x = \(\dfrac{5}{0.9}\)

⇒ x = 5.56 kg

Therefore, approximately 5 kg of water needs to be mixed.

Given:

Rate (r) = 5% per annum

Difference in Simple Interest = ₹42

Time difference = 3 years - 2 years = 1 year

Formula used:

Simple Interest (SI) = \(\dfrac{P \times r \times t}{100}\)

Difference in SI = \(\dfrac{P \times r \times 1}{100}\)

Calculation:

42 = \(\dfrac{P \times 5 \times 1}{100}\)

⇒ P = \(\dfrac{42 \times 100}{5}\)

⇒ P = 840

∴ The correct answer is option (4).

Given:

Price increase = 24%

Expenditure increase = 15%

Formula used:

Let initial consumption = 100 units

New consumption = Initial consumption × (1 - Reduction%)

Reduction% = \(\frac{\text{Price increase} - \text{Expenditure increase}}{(100 + \text{Price increase})} \times 100\)

Calculation:

Reduction% = \(\frac{24 - 15}{100 + 24} \times 100\)

⇒ Reduction% = \(\frac{9}{124} \times 100\)

⇒ Reduction% = 7.25 ≈ 7.3%

∴ The correct answer is option (1).

Given:

Radius (r) = 17 cm

Distance from center (d) = 8 cm

Formula used:

Length of chord = 2 × \(\sqrt{r^2-d^2}\)

Calculation:

Length of chord = 2 × \(\sqrt{17^2-8^2}\)

⇒ Length of chord = 2 × \(\sqrt{289-64}\)

⇒ Length of chord = 2 × \(\sqrt{225}\)

⇒ Length of chord = 2 × 15

⇒ Length of chord = 30 cm

∴ The correct answer is option (4).

Given:

\( \frac{a}{b} = \frac{2}{5} \)

\( \frac{b}{c} = \frac{4}{3} \)

Formula used:

\( \left( \frac{a^2 + c^2}{a^2 - c^2} \right) \)

Calculations:

⇒ Let \( b = 20 \) (LCM of 5 and 4)

⇒ \( a = \frac{2}{5} × 20 = 8 \)

⇒ \( c = \frac{20}{4} × 3 = 15 \)

⇒ \( a^2 = 8 × 8 = 64 \)

⇒ \( c^2 = 15 × 15 = 225 \)

⇒ \( \frac{a^2 + c^2}{a^2 - c^2} = \frac{64 + 225}{64 - 225} \)

⇒ \( \frac{289}{-161} = -\frac{289}{161} \)

∴ The correct answer is option (3).

Given:

\( 4896 \div 96 = ? + 12 \)

Formula used:

Division and Addition operation

Calculation:

\( 4896 \div 96 = ? + 12 \)

⇒ \( \frac{4896}{96} = ? + 12 \)

⇒ \( 51 = ? + 12 \)

⇒ \( ? = 51 - 12 \)

⇒ \( ? = 39 \)

∴ The correct answer is option (3).

Given:

\( 16 \div 4 - 21 + 12 \times 9 = ? \)

Formula used:

Follow BODMAS rule: Brackets, Orders (powers), Division, Multiplication, Addition, Subtraction

Calculation:

\( 16 \div 4 - 21 + 12 \times 9 \)

⇒ \( 4 - 21 + 108 \)

⇒ \( 4 + 108 - 21 \)

⇒ \( 112 - 21 \)

⇒ \( 91 \)

∴ The correct answer is option (1).

Given:

\(\frac{1}{\sqrt{12-\sqrt{140}}}-\frac{1}{\sqrt{8-\sqrt{60}}}-\frac{2}{\sqrt{10+\sqrt{84}}}\)

Formula used:

If \(\sqrt{a-\sqrt{b}} = \sqrt{x} - \sqrt{y}\), then \(x + y = a\) and \(xy = b\).

Calculation:

\(\sqrt{12-\sqrt{140}} = \sqrt{10} - \sqrt{2}\)

\(\sqrt{8-\sqrt{60}} = \sqrt{6} - \sqrt{2}\)

\(\sqrt{10+\sqrt{84}} = \sqrt{7} + \sqrt{3}\)

⇒ \(\frac{1}{\sqrt{12-\sqrt{140}}} = \frac{1}{\sqrt{10} - \sqrt{2}} = \frac{\sqrt{10}+\sqrt{2}}{8}\)

⇒ \(\frac{1}{\sqrt{8-\sqrt{60}}} = \frac{1}{\sqrt{6} - \sqrt{2}} = \frac{\sqrt{6}+\sqrt{2}}{4}\)

⇒ \(\frac{2}{\sqrt{10+\sqrt{84}}} = \frac{2}{\sqrt{7}+\sqrt{3}} = \frac{2(\sqrt{7}-\sqrt{3})}{4}\)

⇒ Combine terms:

\(\frac{\sqrt{10}+\sqrt{2}}{8} - \frac{\sqrt{6}+\sqrt{2}}{4} - \frac{2(\sqrt{7}-\sqrt{3})}{4}\)

LCM = 8:

⇒ \(\frac{\sqrt{10}+\sqrt{2}}{8} - \frac{2(\sqrt{6}+\sqrt{2})}{8} - \frac{4(\sqrt{7}-\sqrt{3})}{8}\)

⇒ \(\frac{\sqrt{10}+\sqrt{2} - 2\sqrt{6} - 2\sqrt{2} - 4\sqrt{7} + 4\sqrt{3}}{8}\)

⇒ \(\frac{\sqrt{10} - 2\sqrt{6} - \sqrt{2} - 4\sqrt{7} + 4\sqrt{3}}{8}\)

All terms cancel out:

⇒ 0

∴ The correct answer is option (1).

Given:

\(\ 108 \div 36 \times 4 + 2.5 \times 4 \div 0.5 - 10 \)

Formula used:

Follow BODMAS rule:

• Brackets
• Orders (exponents or roots)
• Division and Multiplication (from left to right)
• Addition and Subtraction (from left to right)

Calculation:

\(\ 108 \div 36 \times 4 + 2.5 \times 4 \div 0.5 - 10 \)

⇒ \(\ 3 \times 4 + 2.5 \times 4 \div 0.5 - 10 \)

⇒ \(\ 12 + 10 \div 0.5 - 10 \)

⇒ \(\ 12 + 20 - 10 \)

⇒ \(\ 22 \)

∴ The correct answer is option (3).