G8_Godawari_Kailali_8_2081

1. Universal set \(U = \{1, 2, 3, 4, 5, 6\}\), set \(A = \{1, 3, 5\}\), and set \(B = \{3, 5, 6\}\) are given.
1. Are the sets \(A\) and \(B\) overlapping or disjoint? Write it.[1]
2. Write any two proper subsets from the set \(A\).[1]
3. Show the sets \(A\) and \(B\) in a Venn diagram.[1]
1. Sets \(A\) and \(B\) are overlapping sets.
   Because they share common elements \(3\) and \(5\), i.e., \(A \cap B = \{3, 5\} \neq \emptyset\).
2. Two proper subsets of set \(A = \{1, 3, 5\}\) are
   \(\{1, 3\}\) and \(\{5\}\)
   

   Number of Elements	Proper Subsets	Count
   0	\(\emptyset\)	1
   1	\(\{1\}, \{3\}, \{5\}\)	3
   2	\(\{1, 3\}, \{1, 5\}, \{3, 5\}\)	3
   Total Proper Subsets		7

3. The Venn diagram for sets \(A\) and \(B\) within the universal set \(U = \{1, 2, 3, 4, 5, 6\}\) is described as follows The Venn diagram representing the given sets is shown below:
2. The marked price of a watch is Rs 1500. The shopkeeper sold the watch after a \(10\%\) discount.
1. Write the formula to find the discount percentage.[1]
2. Find the discount amount.[1]
3. If the shopkeeper sold the watch to make a \(8\%\) profit, then find the cost price of the watch.[2]
1. Formula to find the discount percentage:
   Discount Percent = \( \dfrac{\text{Discount}}{\text{Marked Price}} \times 100\% \)
   orDiscount Percent = \( \dfrac{\text{MP} - \text{SP}}{\text{MP}} \times 100\% \)
2. Discount amount:
   Marked Price (MP) = Rs. \(1,500\)
   Discount = \(10\%\) of MP = \( \dfrac{10}{100} \times 1,500 = 150 \)
   So, the discount amount is Rs. \(150\).
3. Cost price when sold at \(5\%\) profit:
   Selling Price (SP) = MP – Discount = \(1,500 - 150 = 1,350\)
   Profit = \(8\%\)
   Thus
   SP=108 % of CP
   or\(1,350 = \dfrac{108}{100} \times \text{CP}\)
   \(\text{CP} = \dfrac{1,350 \times 100}{108} = \dfrac{135,000}{108} = 1,250\)
   So, the cost price of the watch was Rs. \(1,250\).
3. Anu had borrowed a loan of Rs 6,000 from a bank 4 years ago. She paid a total amount of Rs 9,000 and cleared the loan.
   1. Find the simple interest.[2]
   2. What was the rate of interest?[2]
   3. In how many years will the principal and interest be equal?[1]
1. Simple Interest:
   Principal (P) = Rs. 6,000
   Amount (A) = Rs. 9,000
   Thus,
   Interest (I) = A - P = 9,000 - 6,000 = 3,000
   So, the simple interest is Rs. 3,000.
2. Rate of interest:
   Principal (P) = Rs. 6,000
   Time (T) = 4 years
   Interest (I) = Rs. 3,000
   We know,
   Rate (R) = \( \dfrac{I \times 100}{P \times T} = \dfrac{3,000 \times 100}{6,000 \times 4} = 12.5\% \)
   So, the rate of interest was 12.5% per annum.
3. Time for principal and interest to be equal:
   Here, Interest (I) = Principal (P).
   We know, \( T = \dfrac{I \times 100}{P \times R} \)
   Since I = P, the formula becomes \( T = \dfrac{100}{R} \).
   Thus,
   Time (T) = \( \dfrac{100}{12.5} = 8 \) years
   So, the principal and interest will be equal in 8 years.
1. Write the decimal number \(0.000063\) in scientific notation.[1]
2. If the cost of \(10\,\text{kg}\) of oranges is \(\text{Rs.}\,1{,}250\), what will be the cost of \(6\,\text{kg}\) of oranges?[1]
3. Convert \(0.\overline{17}\) into a fraction.[2]
4. Convert the binary number \(10111_2\) into the decimal number system.[1]
5. A circular pond of diameter 14 m is inside a rectangular piece of land of length 40 m and breadth 35 m.
   1. Find the area of the pond. [1]
   2. Find the total area of the land. [1]
   3. Find the area of the land excluding the pond. [1]
   4. Find the cost of fencing the land at the rate of Rs 200 per meter. [2]
1. Area of the pond
   Given that,
   Diameter of pond (\( d \)) = 14 m
   Radius (\( r \)) = \( \frac{14}{2} = 7 \) m
   We know that,
   Area of pond (\( A_p \)) = \( \pi r^2 = \frac{22}{7} \times 7^2 = 154 \) m²
2. Total area of the land
   Given for the rectangular land,
   Length (\( l \)) = 40 m
   Breadth (\( b \)) = 35 m
   Therefore,
   Total Area (\( A_l \)) = \( 40 \times 35 = 1400 \) m²
3. Area of the land excluding the pond
   We know that,
   Remaining Area = Total Area (\( A_l \)) - Area of pond (\( A_p \))
   or Remaining Area = \( 1400 - 154 = 1246 \) m²
4. Cost of fencing the land
   First, calculating the perimeter of the rectangular land,
   Perimeter (\( P \)) = \( 2(l + b) = 2(40 + 35) = 2 \times 75 = 150 \) m
   According to the question, the rate is Rs 200 per meter, so
   Total Cost = \( 150 \times 200 = \) Rs 30,000
1. Simplify by using rules of indices: \(x^{3} \times x^{-3}\)[1]
2. Simplify: \(\dfrac{y^{2}}{y - 2} - \dfrac{4}{y - 2}\)[2]
1. Factorize: \(a^{2} - 64b^{2}\)[2]
2. Solve graphically: \(x + y = 6\) and \(y = x + 4\)[2]
1. Find the H.C.F. of the expressions: \(a^{2} - 25\) and \(a^{2} - 12a + 35\).[2]
2. Solve: \(a^{2} + 7a + 12 = 0\)[2]
9. In the given figure, \(\angle AEB = 60^{\circ}\), and lines are as shown.
1. Find the values of \(y\) and \(z\) from the given figure.[2]
2. Write the formula to find the measure of each exterior angle of a regular polygon.[1]
3. Find the distance between the points \(P(1, 7)\) and \(Q(1, 1)\).[1]
1. Construct a rectangle \(ABCD\) in which \(AB = 7cm\) and \(AD = 4cm\).[3]
2. Verify experimentally that the base angles of an isosceles right-angled triangle are equal to \(45^{\circ}\). (Two figures of different sizes are required.)[2]
1. Draw a net of a cylinder.[1]
2. If \(A(2, 1)\), \(B(5, 1)\), and \(C(4, 4)\) are the vertices of \(\triangle ABC\), find the image \(\triangle A'B'C'\) after reflecting on the \(y\)-axis and show both triangles on the same graph.[3]
3. If the bearing of \(B\) from \(A\) is \(060^{\circ}\), find the bearing of \(A\) from \(B\).[2]
1. Find the mode of the given data: \(30, 40, 60, 80, 30\).[1]
2. Represent the following data in a pie chart:
   

   Gases	Nitrogen	Oxygen	Others
   Percentage	\(78\%\)	\(21\%\)	\(1\%\)

   [2]