G8_Janakpurdham_8_2081

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  1. The elements of sets \(A\) and \(B\) are shown in the adjoining Venn diagram.
    1. Define proper subset.[1]
    2. Write the improper subset of set \(A\).[1]
    3. If \(e, i, o, u\) are the only elements of set \(B\), what type of relation exists between sets \(A\) and \(B\)? Write with reason.[1]
    1. A proper subset of a set \(A\) is a subset that contains some, but not all, elements of given set \(A\), or is the empty set.
      Formally, \(B\) is a proper subset of \(A\) if \(B \subseteq A\) and \(B \neq A\).
    2. From the Venn diagram, set \(A = \{a, b, c\}\).
      The improper subset of \(A\) is the set \(A\) itself, which is
      \(\{a, b, c\}\)
    3. If \(B = \{e, i, o, u\}\), then \(A = \{a, b, c\}\) and \(B = \{e, i, o, u\}\) share no common elements.
      Therefore, sets \(A\) and \(B\) are disjoint sets.
  2. The marked price of a book is Rs 3900.
    1. If marked price and discount are represented by \(M\) and \(D\) respectively, write the formula to find the discount percent.[1]
    2. If the discount percent is \(7\%\), how much discount is given?[1]
    3. If the shopkeeper earned a \(17\%\) profit after selling it at \(7\%\) discount, find the cost price of the book.[2]
    1. Formula to find the discount percent:
      Discount Percent = \( \dfrac{D}{M} \times 100\% \)
      orDiscount Percent = \( \dfrac{M - SP}{M} \times 100\% \)
    2. Discount amount
      Marked Price (\(M\)) = Rs. \(3,900\)
      Discount Percent = \(7\%\)
      Thus
      Discount = \( \dfrac{7}{100} \times 3,900 = 273 \)
      So, the discount given is Rs. \(273\).
    3. Cost price
      Marked Price (\(M\)) = Rs. \(3,900\)
      Discount = \(7\%\)
      Thus
      Selling Price (SP) = 93% of MP = \( \dfrac{93}{100} \times 3,900 = 3,627 \)
      Given that
      Profit = \(17\%\)
      Thus
      SP = 117% of CP
      or\(3,627 = \dfrac{117}{100} \times \text{CP}\)
      or\(\text{CP} = \dfrac{3,627 \times 100}{117} = \dfrac{362,700}{117} = 3,100\)
      So, the cost price of the book was Rs. \(3,100\).
  3. Rs 6,000 is lent out at the rate of \(12\dfrac{1}{2}\%\) per annum for 10 months.
    1. How much interest does Sangita earn in 10 months?[1]
    2. How much amount will Sangita receive after 2 years at the same rate?[2]
    3. The ages of Sangita’s elder and younger brothers are 12 years and 8 years respectively. If she divides Rs 200,000 between them in the ratio of their ages, how much more will the elder brother receive?[2]
    1. Interest for 10 months:
      Principal (P) = Rs. 6,000
      Rate (R) = \(12.5\%\)
      Time (T) = 10 months = \( \dfrac{10}{12} \) years
      Thus,
      Interest (I) = \( \dfrac{6,000 \times \frac{10}{12} \times 12.5}{100} = 625 \)
      So, Sangita earns Rs. 625 interest.
    2. Amount after 2 years:
      Principal (P) = Rs. 6,000
      Time (T) = 2 years
      Rate (R) = 12.5%
      Interest (I) = \( \dfrac{6,000 \times 2 \times 12.5}{100} = 1,500 \)
      Amount (A) = P + I = 6,000 + 1,500 = 7,500
      So, Sangita will receive Rs. 7,500.
    3. Distribution between brothers:
      Ratio of ages (Elder : Younger) = 12 : 8 = 3 : 2
      Sum of ratios = 3 + 2 = 5
      Total amount = Rs. 200,000
      Elder brother's share = \( \dfrac{3}{5} \times 200,000 = 120,000 \)
      Younger brother's share = \( \dfrac{2}{5} \times 200,000 = 80,000 \)
      Difference = 120,000 - 80,000 = 40,000
      So, the elder brother will receive Rs. 40,000 more.
  4. The capacity of a water tank to supply water to a village is \(37{,}200\) liters.
    1. Write the capacity of the water tank in scientific notation.[1]
    2. If a family consumes \(8{,}520\) liters in a month and the cost per liter is \(\text{Rs.}\,0.40\), how much must they pay for one month?[2]
    3. Convert \(0.\overline{34}\) into a fraction.[1]
    4. Convert the binary number \(11100_2\) into decimal.[1]
  5. A trapezium-shaped land belongs to Bimala. A circular well of diameter 1.4 m is dug inside it. The parallel sides of the land are 45 m and 55 m, and the height is 30 m.
    1. Write the formula to find the area of the circular well. [1]
    2. Find the area of the top of the well. (\(\pi = \frac{22}{7}\)) [1]
    3. Find the area of the land excluding the well. [2]
    4. Bimala wants to fence the land three times. How much wire is needed? [1]
    1. Formula for the area of the well
      Since the top of the well is circular,
      Area (\( A \)) = \( \pi r^2 \) or \( \frac{\pi d^2}{4} \)
    2. Area of the well
      Given: Diameter (\( d \)) = 1.4 m, so Radius (\( r \)) = 0.7 m.
      Area (\( A_w \)) = \( \frac{22}{7} \times (0.7)^2 = 1.54 \) m²
    3. Area excluding the well
      Area of trapezium plot:
      Area (\( A_t \)) = \( \frac{1}{2} \times (p_1 + p_2) \times h = \frac{1}{2} \times (45 + 55) \times 30 = 1500 \) m²
      Remaining Area = \( 1500 - 1.54 = 1498.46 \) m²
    4. Wire needed for fencing
      First, find the perimeter. Using the figure and Pythagorean theorem, the non-parallel side length is \( \sqrt{30^2 + (55-45)^2} \approx 31.62 \) m.
      Perimeter (\( P \)) = \( 45 + 55 + 30 + 31.62 = 161.62 \) m.
      Wire needed for 3 rounds = \( 3 \times 161.62 = 484.86 \) m.
    1. Express \(\dfrac{x^m}{x^n}\) as a power of \(x\).[1]
    2. Simplify: \(\dfrac{x}{x - y} + \dfrac{y}{y - x}\).[2]
  7. Two equations are given: \(x + y = 4\) and \(3x - 2y = -13\).
    1. What is this system of equations called?[1]
    2. Solve the equations using a graph.[2]
    1. Find the H.C.F. of \(x^2 - 1\) and \(x^2 - 2x + 1\).[2]
    2. For what value of \(x\) does the expression \(x^2 - 5x + 6\) become zero?[2]
  9. In the adjoining figure, line \(EF\) intersects straight lines \(AB\) and \(CD\) at points \(G\) and \(H\) respectively.
    1. Write a pair of co-interior angles.[1]
    2. Find the value of \(x\).[2]
    3. For what value of \(\angle BGK\) will lines \(AB\) and \(CD\) be parallel?[1]
    1. Construct a rectangle \(ABCD\) with \(AB = 8cm\) and \(BC = 6cm\).[3]
    1. Which type of triangle is used to form a regular tessellation?[1]
    2. In \(\triangle ABC\), the bearing of point \(C\) from point \(B\) is \(090^\circ\). Find the bearing of point \(B\) from point \(C\).[2]
    3. Rotate \(\triangle ABC\) through \(+90^\circ\) about the origin \(O(0,0)\) and write the coordinates of the image \(\triangle A'B'C'\).[3]
  12. The monthly expenditure of a family is given below:
    13. Month Baishakh Jestha Ashar Shrawan
    Expenditure (Rs.) 25000 19000 28000 18000
    1. What is the average monthly expenditure of the family?[1]
    2. Present the family’s expenditure in a pie chart.[2]

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