If two sets A = apple, orange, banana and B = banana, grape, lychee are given: a. Is the set a finite or infinite set? Give one reason. b. Form any two subsets of the given set. c. Is C = 5, 6, 7, 8 equivalent to set A? Give one reason.

Step 1: Solve Question 1. *a) Set $A = \{apple, orange, banana\}$. Set A is a finite set. Reason: It has a definite, countable number of elements (3 elements).

*b) Two subsets of set A are: • $\{apple\}$ • $\{orange, banana\}$

*c) Set $C = \{5, 6, 7, 8\}$. Set $A = \{apple, orange, banana\}$. Set C has 4 elements. Set A has 3 elements. Set C is not equivalent to set A. Reason: Equivalent sets must have the same number of elements. Set C has 4 elements, while Set A has 3 elements.

Step 2: Solve Question 2. *a) Find the square root of 225 using the Prime Factorization Method. $225 = 3 \times 75$ $225 = 3 \times 3 \times 25$ $225 = 3 \times 3 \times 5 \times 5$ To find the square root, group the prime factors in pairs: $\sqrt{225} = \sqrt{(3 \times 3) \times (5 \times 5)}$ $\sqrt{225} = 3 \times 5$ $\sqrt{225} = 15$ The square root of 225 is $\boxed{15}$.

*b) Find the greatest number that can exactly divide 24, 28, and 40 using the Common Division Method (HCF). Divide the numbers by common prime factors: $2 | \underline{24, 28, 40}$ $2 | \underline{12, 14, 20}$ $- | \underline{6, 7, 10}$ There are no more common prime factors for 6, 7, and 10. The HCF is the product of the common prime factors: $HCF = 2 \times 2 = 4$ The greatest number that can exactly divide 24, 28, and 40 is $\boxed{4}$.

Step 3: Solve Question 3. Shyam's income fractions spent: Food: $\frac{2}{9}$ Education: $\frac{2}{5}$ Clothes: $\frac{2}{6} = \frac{1}{3}$

*a) Compare spending on food ($\frac{2}{9}$) and education ($\frac{2}{5}$). To compare, find a common denominator (LCM of 9 and 5 is 45): Food: $\frac{2}{9} = \frac{2 \times 5}{9 \times 5} = \frac{10}{45}$ Education: $\frac{2}{5} = \frac{2 \times 9}{5 \times 9} = \frac{18}{45}$ Since $\frac{18}{45} > \frac{10}{45}$, Shyam spends more on education.

*b) If his monthly income is Rs. 90,000, how much does he save? Total fraction spent: $\frac{2}{9} + \frac{2}{5} + \frac{1}{3}$ Find the LCM of 9, 5, and 3, which is 45. $\frac{2 \times 5}{9 \times 5} + \frac{2 \times 9}{5 \times 9} + \frac{1 \times 15}{3 \times 15}$ $\frac{10}{45} + \frac{18}{45} + \frac{15}{45}$ $\frac{10 + 18 + 15}{45} = \frac{43}{45}$ Fraction saved = $1 - \text{Total fraction spent}$ $1 - \frac{43}{45} = \frac{45}{45} - \frac{43}{45} = \frac{2}{45}$ Amount saved = Fraction saved $\times$ Total income $\frac{2}{45} \times 90,000$ $2 \times \frac{90,000}{45}$ $2 \times 2,000 = 4,000$ Shyam saves $\boxed{Rs. 4,000}$.

Step 4: Solve Question 4. *a) Find the square of $(4x + 5y)$. Using the identity $(a+b)^2 = a^2 + 2ab + b^2$: $(4x + 5y)^2 = (4x)^2 + 2(4x)(5y) + (5y)^2$ $= 16x^2 + 40xy + 25y^2$ The square of $(4x + 5y)$ is $\boxed{16x^2 + 40xy + 25y^2}$.

*b) Simplify: $\frac{x^5 x^{-2} y^3 xy^2}{x^4 xy^6}$ Combine terms with the same base by adding exponents in the numerator and denominator separately. Numerator: $x^{5 + (-2) + 1} y^{3 + 2} = x^{4} y^{5}$ Denominator: $x^{4 + 1} y^{6} = x^{5} y^{6}$ Now divide: $\frac{x^4 y^5}{x^5 y^6} = x^{4-5} y^{5-6}$ $= x^{-1} y^{-1}$ $= \frac{1}{xy}$ The simplified expression is $\boxed{\frac{1}{xy}}$.

*c) Factorize: $x^2 - y^2$. Using the difference of squares identity $a^2 - b^2 = (a-b)(a+b)$: $x^2 - y^2 = (x - y)(x + y)$ The factorization of $x^2 - y^2$ is $\boxed{(x - y)(x + y)}$.

Step 5: Solve Question 5. Points are $A(4,5)$ and $B(2,0)$. To reflect a point $(x, y)$ across the x-axis, the new point is $(x, -y)$. Reflecting point A: $A(4,5) \to A'(4, -5)$ Reflecting point B: $B(2,0) \to B'(2, -0) = B'(2, 0)$ The reflected points are $\boxed{A'(4, -5) and B'(2, 0)}$.

Step 6: Solve Question 6. Construct triangle ABC with $AB = 7 \text{ cm}$, $BC = 6 \text{ cm}$, and $CA = 4.5 \text{ cm}$. Steps for construction:

1. Draw a line segment $AB = 7 \text{ cm}$.
2. With A as the center, draw an arc with a radius of $4.5 \text{ cm}$ (length of CA).
3. With B as the center, draw an arc with a radius of $6 \text{ cm}$ (length of BC).
4. The intersection point of the two arcs is point C.
5. Join AC and BC to form triangle ABC.

Step 7: Solve Question 7. Given: Selling Price (S.P.) = Rs 400, Loss percent = 20%. We need to find the Cost Price (C.P.). Formula for S.P. with loss: $S.P. = C.P. \times \left(1 - \frac{Loss \%}{100}\right)$ $400 = C.P. \times \left(1 - \frac{20}{100}\right)$ $400 = C.P. \times \left(1 - 0.20\right)$ $400 = C.P. \times 0.80$ To find C.P.: $C.P. = \frac{400}{0.80}$ $C.P. = \frac{400}{\frac{80}{100}}$ $C.P. = \frac{400 \times 100}{80}$ $C.P. = \frac{40000}{80}$ $C.P. = 500$ The Cost Price (C.P.) is $\boxed{Rs. 500}$.

Step 8: Solve Question 8. Given numbers: $7, 12, 6, 10, p$. Number of terms = 5. Average = 8. Formula for average: $\text{Average} = \frac{Sum of numbers}{Number of terms}$ $8 = \frac{7 + 12 + 6 + 10 + p}{5}$ $8 \times 5 = 7 + 12 + 6 + 10 + p$ $40 = 35 + p$ To find $p$: $p = 40 - 35$ $p = 5$ The value of $p$ is $\boxed{5}$.