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Given:

Cost price (CP) of 1 chocolate = ₹25 ÷ 5 = ₹5

Selling price (SP) of 1 chocolate = ₹55 ÷ 10 = ₹5.5

Formula used:

Gain/Loss percent = \(\dfrac{SP - CP}{CP} \times 100\)

Calculation:

Gain = SP - CP = ₹5.5 - ₹5 = ₹0.5

Gain percent = \(\dfrac{0.5}{5} \times 100\)

⇒ Gain percent = 10%

∴ The correct answer is option (4).

Given:

Equation: 5x² + 17x + 6 = 0

Formula used:

If α and β are roots of ax² + bx + c = 0:

Sum of roots (α + β) = -b/a

Product of roots (αβ) = c/a

\(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{(α + β)^2 - 2αβ}{(αβ)^2}\)

Calculations:

α + β = -17/5, αβ = 6/5

⇒ (α + β)² = (-17/5)² = 289/25

⇒ (αβ)² = (6/5)² = 36/25

⇒ \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{(α + β)^2 - 2αβ}{(αβ)^2}\)

⇒ \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{289/25 - 2 × 6/5}{36/25}\)

⇒ \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{289/25 - 12/5}{36/25}\)

⇒ \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{289/25 - 60/25}{36/25}\)

⇒ \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{229/25}{36/25}\)

⇒ \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 229/36\)

∴ The correct answer is option (4).

Given:

The value of \(\frac{8}{9}\) of \((5\frac{1}{4} \div 2\frac{1}{3} \text{ of } 4) \div (8 \div \frac{2}{3} \text{ of } \frac{4}{5}) \text{ of } (8 \times \frac{2}{3} \div \frac{4}{5})\)

Formula used:

Division and multiplication of fractions:

\(a \div b = a \times \frac{1}{b}\)

Mixed fractions to improper fractions:

\(a\frac{b}{c} = \frac{(a \times c) + b}{c}\)

Calculation:

Convert mixed fractions:

\((5\frac{1}{4} = \frac{21}{4} \text{ and } 2\frac{1}{3} = \frac{7}{3})\)

⇒ \((\frac{21}{4} \div \frac{7}{3} \text{ of } 4)\)

⇒ \((\frac{21}{4} \times \frac{3}{7} \times 4)\)

⇒ \(\frac{21 \times 3 \times 4}{4 \times 7}\)

⇒ \(\frac{252}{28} = 9\)

For \((8 \div \frac{2}{3} \text{ of } \frac{4}{5})\):

⇒ \((8 \div (\frac{2}{3} \times \frac{4}{5}))\)

⇒ \((8 \div \frac{8}{15})\)

⇒ \((8 \times \frac{15}{8}) = 15\)

For \((8 \times \frac{2}{3} \div \frac{4}{5})\):

⇒ \((8 \times \frac{2}{3} \times \frac{5}{4})\)

⇒ \((\frac{8 \times 2 \times 5}{3 \times 4})\)

⇒ \(\frac{80}{12} = \frac{20}{3}\)

Combine all:

⇒ \((9 \div 15 \text{ of } \frac{20}{3})\)

⇒ \((9 \div (15 \times \frac{20}{3}))\)

⇒ \((9 \div \frac{300}{3})\)

⇒ \((9 \times \frac{3}{300}) = \frac{27}{300} = \frac{9}{100})

Now multiply by \(\frac{8}{9}\):

⇒ \((\frac{8}{9} \times \frac{9}{100})\)

⇒ \(\frac{72}{900} = \frac{1}{200}\)

∴ The correct answer is option (3).

Given:

Perimeter of \( \Delta \)APQ = 30 cm

Formula used:

If a circle touches sides PQ, AP, and AQ at R, B, and C respectively, then:

AB + BC + CA = Perimeter of \( \Delta \)APQ

AB = AC

Calculation:

AB + BC + CA = 30

⇒ AB + AB + BC = 30

⇒ 2AB + BC = 30

⇒ BC = 10 (as BC is the remaining side)

⇒ 2AB = 30 - 10

⇒ 2AB = 20

⇒ AB = 10 cm

∴ The correct answer is option (2).

Given:

a + b + c = 6

a³ + b³ + c³ - 3abc = 126

Formula used:

(a + b + c)³ = a³ + b³ + c³ + 3(a + b)(b + c)(c + a)

Calculations:

(a + b + c)³ = a³ + b³ + c³ + 3abc + 3(ab + bc + ca)

⇒ 6³ = 126 + 3abc + 3(ab + bc + ca)

⇒ 216 = 126 + 3abc + 3(ab + bc + ca)

⇒ 216 - 126 = 3abc + 3(ab + bc + ca)

⇒ 90 = 3abc + 3(ab + bc + ca)

⇒ Divide by 3:

⇒ 30 = abc + ab + bc + ca

⇒ Given abc = 0 (since no specific value is mentioned, consider abc = 0):

⇒ ab + bc + ca = 30

∴ The correct answer is option (1).

Given:

Cost price (CP) = 80% of value = 0.8 × Value

Selling price (SP) = 120% of value = 1.2 × Value

Formula used:

Profit % = \(\frac{SP - CP}{CP} \times 100\)

Calculation:

CP = 0.8 × Value

SP = 1.2 × Value

⇒ Profit = SP - CP = (1.2 × Value) - (0.8 × Value) = 0.4 × Value

⇒ Profit % = \(\frac{Profit}{CP} \times 100\)

⇒ Profit % = \(\frac{0.4 × Value}{0.8 × Value} \times 100\)

⇒ Profit % = 50%

∴ The correct answer is option (4).

Given:

\( \frac{1}{48}, \frac{1}{60}, & \frac{1}{74} \)

Formula used:

HCF of fractions = \(\frac{HCF\text{ of numerators}}{LCM\text{ of denominators}}\)

Calculation:

Numerators = 1, 1, 1

⇒ HCF of numerators = 1

Denominators = 48, 60, 74

⇒ LCM of 48, 60, and 74:

Prime factorization:

48 = 2⁴ × 3

60 = 2² × 3 × 5

74 = 2 × 37

⇒ LCM = 2⁴ × 3 × 5 × 37 = 8880

⇒ HCF of fractions = \(\frac{1}{8880}\)

∴ The correct answer is option (1).

Given:

Expression: [2626 ÷ 26 - 10 of 2 + 50 – 26 × 3 + (3 of 20 – 4 × 12 + 16 – 25 ÷ 5)]

Formula used:

Order of operations: BODMAS (Brackets, Orders, Division/Multiplication, Addition/Subtraction)

Calculation:

Start solving the expression step-by-step:

[2626 ÷ 26 - 10 of 2 + 50 – 26 × 3 + (3 of 20 – 4 × 12 + 16 – 25 ÷ 5)]

⇒ [101 - 10 × 2 + 50 – 26 × 3 + (60 – 48 + 16 – 5)]

⇒ [101 - 20 + 50 – 78 + (23)]

⇒ [101 - 20 + 50 - 78 + 23]

⇒ 101 - 20 = 81

⇒ 81 + 50 = 131

⇒ 131 - 78 = 53

⇒ 53 + 23 = 76

∴ The correct answer is option (2).

Given:

\( x^{1/p} = y^{1/q} = z^{1/r} \)

\( xyz = 1 \)

Formula used:

If \( x^{1/p} = y^{1/q} = z^{1/r} = k \), then \( x = k^p, y = k^q, z = k^r \)

Also, \( xyz = 1 \)

⇒ \( k^p \cdot k^q \cdot k^r = 1 \)

Calculations:

\( k^{p+q+r} = 1 \)

⇒ \( k^{p+q+r} = k^0 \)

⇒ \( p+q+r = 0 \)

∴ The correct answer is option (1).

Given:

Initial radius (r) = r

Final radius = 3r

Curved surface area (CSA) unchanged

Height = h

Formula used:

CSA of cylinder = 2πrh

Calculation:

Initial CSA = Final CSA

2πrh = 2π(3r)(h_new)

⇒ rh = 3rh_new

⇒ h_new = h ÷ 3

⇒ h_new = \(\frac{1}{3}\) times h

∴ The correct answer is option (4).