G8_SiddarthaNagar_8_2081

1. Two sets are given: \(A = \{1, 2, 3\}\) and \(B = \{2, 3, 4\}\).
1. Are sets \(A\) and \(B\) disjoint sets? Give a reason.[1]
2. Write the improper subsets of set \(A\).[1]
3. Write the proper subsets formed from set \(B\).[1]
1. No, sets \(A\) and \(B\) are not disjoint sets.
   Because they have common elements \(2\) and \(3\), i.e., \(A \cap B = \{2, 3\} \neq \emptyset\).
2. The improper subsets of set \(A\) is
   \(\{1, 2, 3\}\) (the set itself)
   
3. The proper subsets of set \(B = \{2, 3, 4\}\) are all subsets except \(B\) itself. They are
   \(\{2\}, \{3\}, \{4\}, \{2, 3\}, \{2, 4\}, \{3, 4\}, \emptyset\)

   Number of Elements	Subsets	Count
   0	\(\emptyset\)	1
   1	\(\{2\}, \{3\}, \{4\}\)	3
   2	\(\{2, 3\}, \{2, 4\}, \{3, 4\}\)	3
   Total Proper Subsets		7

2. The distance between Bhairahawa and Kathmandu is 275000 m.
1. Write the distance in scientific notation.[1]
2. Convert 1050 into the quinary number system.[2]
3. The distance between Bhairahawa and Nepalgunj is 255000 m. Find the ratio of the distances between the two pairs of cities.[2]
3. Pranavi bought a mobile phone of marked price Rs 12000 at a \(20\%\) discount.
1. Write the formula to calculate the discount amount if the discount rate is \(D\%\) and the marked price is MP.[1]
2. Find the discount amount on the mobile phone.[1]
3. Find the selling price of the mobile phone.[1]
4. If the mobile phone is sold at a \(4\%\) loss, find the cost price.[1]
1. Formula to calculate the discount amount is
   Discount Amount = \( \dfrac{D}{100} \times \text{MP} \)
2. Finding the discount amount
   Marked Price (MP) = Rs. \(12,000\)
   Discount Rate = \(20\%\)
   Therefore,
   Discount = \( \dfrac{20}{100} \times 12,000 = 2,400 \)
   So, the discount amount is Rs. \(2,400\).
3. Finding the selling price (SP)
   SP = MP – Discount = \(12,000 - 2,400 = 9,600\)
   So, the selling price of the mobile phone is Rs. \(9,600\).
4. Finding the cost price (CP) when sold at a \(10\%\) loss
   Selling Price (SP) = Rs. \(9,600\)
   Loss = \(4\%\)
   Thus,
   \( SP= 96 \% \times \text{CP} \)
   or\( 9600 = \dfrac{96}{100} \times \text{CP} \)
   or\( \text{CP} = \dfrac{9600 \times 100}{96} =10000 \)
   So, the cost price of the mobile phone was Rs. \(10,000\).
4. Anita deposited Rs 3,50,000 in a bank at the rate of 12% per annum for 2 years.
   1. Write the formula to calculate the amount.[1]
   2. How much interest does Anita get in 2 years?[2]
   3. 18 men can complete a work in 30 days. How many men can finish the same work in 36 days?[2]
1. Formula to calculate amount:
   Amount (A) = Principal (P) + Interest (I)
   or,
   Amount (A) = \( P \left( 1 + \dfrac{T \times R}{100} \right) \)
2. Interest for 2 years:
   Principal (P) = Rs. 3,50,000
   Time (T) = 2 years
   Rate (R) = 12%
   Thus,
   Interest (I) = \( \dfrac{P \times T \times R}{100} = \dfrac{3,50,000 \times 2 \times 12}{100} = 84,000 \)
   So, Anita gets Rs. 84,000 interest.
3. Work and Men calculation:
   To finish work in 30 days, men required = 18
   To finish work in 1 day, men required = \( 18 \times 30 \)
   Thus,
   To finish work in 36 days, men required = \( \dfrac{18 \times 30}{36} = 15 \)
   So, 15 men can finish the work in 36 days.
5. Manisha has a plot of land in the shape of a parallelogram. She wishes to construct a circular pond of radius 14 m inside it. The base of the parallelogram is 150 m and the height is 18 m.
   
   1. Write the formula to calculate the area of a parallelogram and a circle. [1]
   2. Find the area of Manisha's plot. [1]
   3. What is the area of the plot excluding the pond? [2]
   4. How many meters of wire netting are needed to fence around the pond? [1]
1. Area formulas
   We know that,
   Area of parallelogram (\( A \)) = base (\( b \)) \(\times\) height (\( h \))
   Area of circle (\( A \)) = \( \pi r^2 \)
2. Area of Manisha's plot
   Given that,
   Base (\( b \)) = 150 m
   Height (\( h \)) = 18 m
   Now using the formula, we get
   Area of plot (\( A_{p} \)) = \( 150 \times 18 = \) 2700 m²
3. Area of the plot excluding the pond
   We know that,
   Radius of pond (\( r \)) = 14 m
   Calculating the area of the circular pond,
   Area of pond (\( A_{c} \)) = \( \pi r^2 = \frac{22}{7} \times 14^2 = 616 \) m²
   Therefore,
   Area excluding pond = \( A_{p} - A_{c} = 2700 - 616 = \) 2084 m²
4. Wire netting needed for the pond
   We know that,
   Circumference (\( C \)) = \( 2\pi r \)
   According to the question,
   Wire needed = \( 2 \times \frac{22}{7} \times 14 = \) 88 m
1. Find the H.C.F. of: \(3x^{2} - 27\) and \(2x - 6\).[2]
2. Simplify: \(\dfrac{x^{2}}{x + y} - \dfrac{y^{2}}{x + y}\)[2]
1. What is the value of \((9 + x)^{0}\)? Write it.[1]
2. At what value(s) of \(x\) is the expression \(x^{2} - 3x + 2\) equal to zero?[2]
8. Two equations are given: \(x + y = 6\) and \(x - y = 2\).
1. What are these equations called?[1]
2. Solve the given equations using the graphical method.[2]
9. In the given figure, lines \(KL\) and \(MN\) are intersected by transversal \(DE\) at points \(A\) and \(C\) respectively. \(\angle KAC = 60^{\circ}\).
1. Write one pair of alternate angles.[1]
2. Find the value of \(x\).[2]
3. Verify experimentally that the base angles of an isosceles triangle are equal. (Two figures of different sizes are required.)[3]
1. Find the coordinates of the image of the line segment joining \(P(5, 3)\) and \(Q(1, 6)\) after reflection in the \(y\)-axis.[1]
2. Find the distance between points \(P\) and \(Q\).[2]
3. The bearing of \(P\) from \(Q\) is \(060^{\circ}\). Find the bearing of \(Q\) from \(P\).[1]
1. Construct a rectangle \(ABCD\) with \(AB = 5cm\) and \(BC = 4cm\).[3]
2. In the given figure, \(\angle RPQ = \angle ABC\), \(PR = BC\), and \(PQ = AB\). Prove that \(\triangle PQR \cong \triangle ABC\).[2]
12. The marks obtained by a Grade VIII student (out of \(50\)) in four subjects are given below:

Subject	English	Mathematics	Science	Social Studies
Obtained Marks	\(45\)	\(30\)	\(42\)	\(30\)

1. Find the mode of the given data.[1]
2. Represent the given data in a pie chart.[2]