G8_Nepalganj_8_2081

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  1. The subsets \(A\) and \(B\) of the universal set \(U\) are presented in the Venn diagram.
    1. Identify and write whether the sets \(A\) and \(B\) are overlapping or disjoint.[1]
    2. Write the improper subset that can be made from set \(B\).[1]
    3. How many more or less proper subsets can be made from set \(B\) than from set \(A\)?[1]
    1. Sets \(A\) and \(B\) are overlapping sets.
      Because they share common elements \(4\) and \(5\), i.e., \(A \cap B = \{4, 5\} \neq \emptyset\).
    2. From the Venn diagram, \(B = \{4, 5, 6, 7\}\).
      The improper subset of \(B\) is the set itself, which is
      \(\{4, 5, 6, 7\}\)
    3. We know that, number of proper subsets = \(2^n - 1\), thus
      Proper subsets of \(A = 2^3 - 1 = 8 - 1 = 7\)
      Proper subsets of \(B = 2^4 - 1 = 16 - 1 = 15\)
      Difference = \(15 - 7 = 8\)
      Therefore, 8 more proper subsets can be made from set \(B\) than from set \(A\).
  2. The marked price of a watch is Rs 1500. A shopkeeper sold it at \(10\%\) discount.
    1. Write the formula to calculate discount percentage.[1]
    2. Find the discount amount.[1]
    3. If the shopkeeper earns a profit of \(8\%\) by selling the watch, at what price did he buy the watch?[2]
    1. Formula to calculate discount percentage:
      Discount Percent = \( \dfrac{\text{Discount Amount}}{\text{Marked Price}} \times 100\% \)
      orDiscount Percent = \( \dfrac{MP - SP}{MP} \times 100\% \)
    2. Discount amount.
      Marked Price (MP) = Rs. \(1,500\)
      Discount = \(10\%\)
      Thus
      Discount Amount = \( \dfrac{10}{100} \times 1,500 = 150 \)
      So, the discount amount is Rs. \(150\).
    3. Cost price when the shopkeeper earns an \(8\%\) profit.
      Marked Price (MP) = Rs. \(1,500\)
      Discount = \(10\%\)
      Thus
      Selling Price (SP) = 90% of MP = \( \dfrac{90}{100} \times 1,500 = 1,350 \)
      Now, given that
      Profit = \(8\%\)
      Therefore
      , so SP = \(108\%\) of CP.
      Thus
      SP = 108% of CP
      or\(1,350 = \dfrac{108}{100} \times \text{CP}\)
      or\(\text{CP} = \dfrac{1,350 \times 100}{108} = \dfrac{135,000}{108} = 1,250\)
      So, the shopkeeper bought the watch for Rs. \(1,250\).
  3. If Karuna has deposited Rs 35,000 in a bank at 12% rate of interest per year for 2 years,
    1. Express the simple interest in the form of P, T, and R.[1]
    2. Find the interest.[1]
    3. Divide the interest in the ratio 1:2.[2]
    1. Simple Interest formula:
      Simple Interest (I) = \( \dfrac{P \times T \times R}{100} \)
    2. Find the interest:
      Principal (P) = Rs. 35,000
      Time (T) = 2 years
      Rate (R) = 12%
      Thus,
      Interest (I) = \( \dfrac{35,000 \times 2 \times 12}{100} = 8,400 \)
      So, the interest is Rs. 8,400.
    3. Divide the interest (Ratio 1:2):
      Total Interest = Rs. 8,400
      Sum of ratios = 1 + 2 = 3
      First part = \( \dfrac{1}{3} \times 8,400 = 2,800 \)
      Second part = \( \dfrac{2}{3} \times 8,400 = 5,600 \)
      So, the divided amounts are Rs. 2,800 and Rs. 5,600.
    1. Convert the quinary number \(123_5\) into the decimal number system.[1]
    2. If \(16\) workers can complete a work in \(25\) days, how many workers can complete the same work in \(20\) days?[2]
    3. Write the decimal number \(0.0000045\) in scientific notation.[1]
    4. If \(3, x, 6, 8\) are in proportion, find the value of \(x\).[1]
  5. There is a rectangular plot with a length of 20 m and width of 15 m, and within it there is a circular pond with radius 7 m.
    1. Find the area of the land. [1]
    2. Find the area of the land excluding the pond. [2]
    3. Find the cost of fencing the land at the rate of Rs 175 per meter. [1]
    1. Area of the land
      Given: Length (\( l \)) = 20 m, Width (\( w \)) = 15 m
      Area (\( A_l \)) = \( 20 \times 15 = 300 \) m²
    2. Area excluding the pond
      Area of pond (\( A_p \)) = \( \pi r^2 = \frac{22}{7} \times 7^2 = 154 \) m²
      Remaining Area = Area of land - Area of pond
      Remaining Area = \( 300 - 154 = 146 \) m²
    3. Cost of fencing
      Perimeter of land (\( P \)) = \( 2(l + w) = 2(20 + 15) = 70 \) m
      At the rate of Rs 175 per meter,
      Total Cost = \( 70 \times 175 = \) Rs 12,250
    1. Simplify: \(x^{a-b} \times x^{b-c} \times x^{c-a}\).[1]
    2. Simplify: \(\left(\dfrac{xy}{z}\right)^{-1} \times \left(\dfrac{z}{xy}\right)^{-2}\).[2]
    3. Simplify: \(\dfrac{1}{a + b} + \dfrac{1}{a - b} + \dfrac{2b}{a^2 - b^2}\).[2]
    1. Factorise: \(9x^2 - 25y^2\).[2]
    2. Find the H.C.F. of \(x^2 - 5x + 6\) and \(x^2 - 9\).[2]
    1. Write the expanded form of \((a - 5)^3\).[2]
    2. Solve: \((x - 7)^2 - 64 = 0\).[2]
    1. Find the value of \(x\) in the given figure.[2]
    2. In \(\triangle ABC\), \(\angle B = 90^\circ\), \(AB = 4\,\text{cm}\), and \(BC = 3\,\text{cm}\). Find the length of \(AC\).[2]
    3. What is the measure of each interior angle of a regular pentagon?[1]
    1. Construct a rectangle \(ABCD\) having adjacent sides \(AB = 8cm\), \(BC = 6cm\), and diagonal \(AC = 10cm\).[3]
    2. In the given figure, if \(\triangle PQC \sim \triangle ABC\), find the value of \(QC\).[2]
    1. Find the distance between the points \(P(-2,-4)\) and \(Q(10,1)\).[1]
    2. If the bearing of a place \(S\) from the place \(R\) is \(050^\circ\), what is the bearing of place \(R\) from place \(S\)?[2]
    3. Verify from at least \(2\) experiments that the sum of interior angles of a triangle is \(180^\circ\).[3]
  12. The monthly expenditure of a family is given in the table below:
    13. Title Food Education Clothing Other expense
    Expenditure Amount (Rs.) 3600 2000 1000 600
    1. Represent the above data in a pie chart.[2]
    2. If the mean of \(4, 10, 3, 6, a, 9, 10\) is \(7\), find the value of \(a\).[1]

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