G_8_Bharatpur_8_2081

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  1. Two subsets of universal set \(U = \{a, e, i, o, u\}\) are \(A = \{e, o, u\}\) and \(B = \{a, e, i\}\).
    1. What type of sets are \(A\) and \(B\)—overlapping or disjoint? Write it.[1]
    2. Write one proper and one improper subset of set \(A\).[1]
    3. If the element \(e\) is removed from both sets \(A\) and \(B\), then what type of sets are \(A\) and \(B\)? Write with reason.[1]
    1. Sets \(A\) and \(B\) are overlapping sets.
      Because they share the common element \(e\), i.e., \(A \cap B = \{e\} \neq \emptyset\).
    2. Set \(A = \{e, o, u\}\).
      One proper subset of \(A\): \(\{o, u\}\)
      One improper subset of \(A\): \(\{e, o, u\}\) (the set itself)
      Number of Elements Subsets Count
      0 \(\emptyset\) 1
      1 \(\{e\}, \{o\}, \{u\}\) 3
      2 \(\{e, o\}, \{e, u\}, \{o, u\}\) 3
      3 \(\{e, o, u\}\) 1
      Total Subsets 8
    3. After removing \(e\) from both sets, we get
      New \(A = \{o, u\}\)
      New \(B = \{a, i\}\)
      Now, \(A \cap B = \emptyset\) (no common elements). Therefore, \(A\) and \(B\) become disjoint sets.
  2. In the numeration system, the following sets are given: \(B_{10} = \{0, 1, 2, 3, \ldots, 8, 9\}\), \(B_2 = \{0, 1\}\), \(B_5 = \{0, 1, 2, 3, 4\}\).
    1. Which number system’s digits are used while opening and closing the switch of an electric circuit?[1]
    2. Convert \(35\) into the binary number system.[2]
    3. What is the relation between \(1010_2\) and \(20_5\)? Evaluate.[1]
    4. Convert \(0.\overline{24}\) into a fraction.[1]
  3. Ashok Mahato went to a stationery shop to buy a ball. He saw two balls \(A\) and \(B\) as shown in the figure. The ratio of marked prices of ball \(A\) and ball \(B\) is \(3:2\). The marked price of ball \(A\) is Rs 3750; the marked price of ball \(B\) is Rs 2500, and a \(5\%\) discount is offered.

    4. ⚽🏀

    1. Write the marked prices of balls \(A\) and \(B\) in proportion.[1]
    2. If Ashok Mahato decided to buy ball \(B\), what amount should he pay for it? Find it.[1]
    3. If the shopkeeper earned \(25\%\) profit by selling ball \(B\) after the discount, at what price did he buy ball \(B\)?[2]
    1. Marked prices of balls in proportion:
      Marked Price of \(A\) : Marked Price of \(B\) = \(3750 : 2500\)
      or\(3750 :1250 = 3: 2\)
    2. Amount Ashok should pay for ball \(B\).
      Marked Price of ball \(B\) = Rs. \(2,500\)
      Discount = \(5\%\)
      Thus
      Selling Price = 95% of MP = \( \dfrac{95}{100} \times 2,500 = 2,375 \)
      So, Ashok should pay Rs. \(2,375\) for ball \(B\).
    3. Cost price of ball .
      Selling Price = Rs. \(2,375\)
      Profit = \(25\%\)
      Thus
      SP = 1250% of CP
      or\(2,375 = \dfrac{125}{100} \times \text{CP}\)
      or\(\text{CP} = \dfrac{2,375 \times 100}{125} = \dfrac{237,500}{125} =1900\)
      So, the shopkeeper bought ball \(B\) for Rs. 1900.
  4. Sarita invested Rs 5,000 as a loan and lent it to Mansur for 3 years at 10% simple interest per annum.
    1. How much interest is obtained by Sarita?[1]
    2. How long should Sarita wait to get double the invested sum?[2]
    3. What will be the interest on Rs 12,000 at the same rate and for the same time?[1]
    1. Interest obtained:
      Principal (P) = Rs. 5,000
      Time (T) = 3 years
      Rate (R) = 10%
      We know,
      Interest (I) = \( \dfrac{P \times T \times R}{100} = \dfrac{5,000 \times 3 \times 10}{100} = 1,500 \)
      So, Sarita obtains Rs. 1,500 as interest.
    2. Time to double the sum:
      To double the sum, the Interest (I) must be equal to the Principal (P).
      Here, I = P = Rs. 5,000
      Rate (R) = 10%
      Thus,
      Time (T) = \( \dfrac{I \times 100}{P \times R} = \dfrac{5,000 \times 100}{5,000 \times 10} = 10 \) years
      So, Sarita should wait 10 years to double her investment.
    3. Interest on Rs. 12,000:
      New Principal (P) = Rs. 12,000
      Time (T) = 3 years
      Rate (R) = 10%
      Thus,
      Interest (I) = \( \dfrac{12,000 \times 3 \times 10}{100} = 3,600 \)
      So, the interest on Rs. 12,000 will be Rs. 3,600.
  5. In the figure, ABCD is a kite-shaped plot where diagonals AC = 6 m and BD = 5 m. Inside it, there is a circular well of radius 50 cm.
    1. Write the formula to find the area of the kite. [1]
    2. Calculate the area of the kite-shaped land. [1]
    3. Find the area and circumference of the circular well. [2]
    4. What is the difference in area between the circular well and the kite-shaped plot? Find it by calculation. [1]
    1. Area of kite formula
      We know that,
      Area (\( A \)) = \( \frac{1}{2} \times d_1 \times d_2 \)
    2. Area of the kite-shaped land
      Given: \( d_1 = 6 \text{ m} \), \( d_2 = 5 \text{ m} \)
      Area (\( A_k \)) = \( \frac{1}{2} \times 6 \times 5 = 15 \text{ m}^2 \)
    3. Area and circumference of the well
      Radius (\( r \)) = \( 50 \text{ cm} = 0.5 \text{ m} \)
      Area (\( A_w \)) = \( \pi r^2 = 3.14 \times (0.5)^2 = 0.785 \text{ m}^2 \)
      Circumference (\( C \)) = \( 2\pi r = 2 \times 3.14 \times 0.5 = 3.14 \text{ m} \)
    4. Difference in area
      Difference = Area of kite - Area of well
      Difference = \( 15 - 0.785 = 14.215 \text{ m}^2 \)
    1. Which geometrical figure's area is represented by \(x^2\)?[1]
    2. Simplify: \(\left(\dfrac{x^3 y^2}{x^4 y}\right)^2\).[2]
  7. Two algebraic expressions are given: \(x^2 - 16\) and \(x^2 - 9x + 20\).
    1. Find the Highest Common Factor (H.C.F.) of the given expressions.[2]
    2. For what value of \(x\) does the expression \(x^2 - 16\) become zero?[2]
    1. What type of equation is \(ax + by + c = 0\)?[1]
    2. Find the quotient when \(\dfrac{m^2 - n^2}{n^2}\) is divided by \(\dfrac{m^2 + mn}{mn}\).[2]
  9. In the given figure, \(RS \parallel MN\), \(\angle PQM = 55^\circ\), \(\angle QMR = y\), \(\angle MRS = (2x + 3)^\circ\), and \(\angle RMN = 73^\circ\).
    1. If \(y = 55^\circ\), write the relation between line segments \(PQ\) and \(MR\).[1]
    2. Find the value of \(x\).[1]
    3. Experimentally verify that the base angles of an isosceles triangle are equal. (Two figures of different sizes are necessary.)[3]
  10. The given figure \(ABCDEF\) is a regular polygon. \(ACDF\) is a rectangle where \(AC = 5.5cm\) and \(AF = 3.6cm\). Side \(FE\) is produced to a point \(G\) such that \(\angle GED = m\).
    1. Find the sum of interior angles of the given regular polygon.[2]
    2. Construct another rectangle having the same dimensions as rectangle \(ACDF\).[3]
    1. What type of tessellation is shown in the given figure?[1]
    2. A man walks \(3\,\text{m}\) north and then turns east and walks \(4\,\text{m}\). What is the shortest distance between his starting and ending points? Calculate it.[1]
    3. \(A(2,2)\), \(B(4,6)\), and \(C(6,3)\) are the vertices of \(\triangle ABC\). Draw \(\triangle ABC\) on graph paper and also plot its image after reflection on the \(x\)-axis.[3]
  12. The adjoining pie chart presents the number of animals in Devchuli Animal Husbandry:
    1. Write the names of animals that represent the mode value.[1]
    2. Find the average number of animals.[2]

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