G8_Butwal_8_2081

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  1. Two sets \(M\) and \(N\) are presented below: \(M = \{a, p, l, e\}\), \(N = \{p, a, n\}\).
    1. Are the sets \(M\) and \(N\) overlapping sets? Give a reason.[1]
    2. How many subsets can be made from set \(M\)?[1]
    3. Represent the given sets \(M\) and \(N\) in a Venn diagram.[1]
    1. Yes, sets \(M\) and \(N\) are overlapping sets.
      Because they share common elements \(a\) and \(p\), i.e., \(M \cap N = \{a, p\} \neq \emptyset\).
    2. Set \(M = \{a, p, l, e\}\) has \(n = 4\) elements.
      The total number of subsets of a set with \(n\) elements is \(2^n\).
      So, number of subsets of \(M = 2^4 = 16\).
    3. The Venn diagram for sets \(M\) and \(N\) is shown as below.
  2. Kreepa marks a laptop with a price of Rs 120000.
    1. If profit percent, cost price, and selling price are denoted by \(P\%\), \(C.P.\), and \(S.P.\) respectively, write the formula to find cost price (\(C.P.\)).[1]
    2. What will be the price of the laptop if a \(15\%\) discount is given?[2]
    3. Find the cost price if the profit amount is Rs 10000.[1]
    1. Formula to find cost price
      \( SP =(100 + P)\% \times CP \)
      or equivalently,
      \( CP= \dfrac{100}{100 + P} \times SP \)
    2. Price of the laptop after a \(15\%\) discount.
      Marked Price (MP) = Rs. \(120,000\)
      Discount = \(15\%\)
      Thus
      Selling Price = 85% of MP = \( \dfrac{85}{100} \times 120,000 = 102,000 \)
      So, the price after discount is Rs. \(102,000\).
    3. Cost price
      From part (ii), we get that
      Selling Price (SP) = Rs. \(102,000\)
      Profit = Rs. \(10,000\)
      We know that
      CP = SP-profit=\(102,000 - 10,000 = 92,000\)
      So, the cost price was Rs. \(92,000\).
  3. Shriyansh deposited Rs 25,000 in a bank at the rate of 10% per annum simple interest for 2 years.
    1. What do you mean by rate of interest 10% per annum?[1]
    2. Find the interest for 2 years.[1]
    1. Meaning of 10% interest per annum:
      Rate of interest 10% per annum means that Rs. 10 is paid as interest for every Rs. 100 of the principal amount for a period of one year.
    2. Interest for 2 years:
      Principal (P) = Rs. 25,000
      Time (T) = 2 years
      Rate (R) = 10%
      We know,
      Interest (I) = \( \dfrac{P \times T \times R}{100} = \dfrac{25,000 \times 2 \times 10}{100} = 5,000 \)
      So, the interest for 2 years is Rs. 5,000.
    1. Write in scientific notation: 2360000.[1]
    2. If \(2, 6, x, 27\) are in proportion, find the value of \(x\).[2]
    3. Convert \(0.\overline{34}\) into a fraction.[1]
    4. Express in the quinary number system: \((68)_{10}\).[1]
  5. A circular garden having diameter 21 m is made inside a plot which is in the shape of a square of side 40 m.
    1. Write the formula to find the area of a circle. [1]
    2. Find the area of the circular garden. [1]
    3. Find the area of the land excluding the area of the circle. [2]
    4. If the garden were an equilateral triangle having a side of 20 m, how much plot would remain? [1]
    1. Area of circle formula
      We know that,
      Area (\( A \)) = \( \pi r^2 \)
    2. Area of the circular garden
      Given: Diameter (\( d \)) = 21 m, so Radius (\( r \)) = 10.5 m
      Area (\( A_c \)) = \( \frac{22}{7} \times (10.5)^2 = 346.5 \) m²
    3. Area excluding the circle
      Area of square plot (\( A_s \)) = \( 40^2 = 1600 \) m²
      Remaining Area = \( 1600 - 346.5 = 1253.5 \) m²
    4. Remaining plot with equilateral triangle garden
      Area of triangle (\( A_t \)) = \( \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times 20^2 \approx 173.21 \) m²
      Remaining Plot = \( 1600 - 173.21 = 1426.79 \) m²
    1. Express \(a^3 \times a^5\) as a power of \(a\).[1]
    2. Simplify: \(x^{a^2 - b^2} \times x^{b^2 - c^2} \times x^{c^2 - a^2}\).[2]
    1. Solve: \(\dfrac{1}{x} + \dfrac{1}{x+1} = \dfrac{2}{x-1}\).[2]
    2. Find the H.C.F. of \(x^2 - 9\) and \(x^2 - 5x + 6\).[2]
    1. Write the equation of the \(x\)-axis.[1]
    2. Solve graphically: \(x + y = 7\) and \(x - y = 1\).[2]
  9. In the adjoining figure, two parallel lines \(MN\) and \(OP\) are intersected by a straight line \(XY\) at points \(A\) and \(B\) respectively.
    1. Write a pair of corresponding angles.[1]
    2. Find the value of \(x\).[2]
    3. Compare the angles \(\angle CAB\) and \(\angle ABC\).[1]
    1. Construct a parallelogram \(PQRS\) having \(PQ = 6cm\), \(QR = 5cm\), and \(\angle PQR = 75^\circ\).[3]
    2. By which axiom are \(\triangle ABC\) and \(\triangle DEF\) congruent? Also write a pair of corresponding angles.[3]
    1. Draw the net of a cylinder.[1]
    2. Write the bearing of point \(B\) from point \(A\).[1]
    3. \(A(3,4)\), \(B(2,-3)\), and \(C(6,0)\) are the vertices of \(\triangle ABC\). Find the coordinates of the image under reflection on the \(y\)-axis. Also represent \(\triangle ABC\) and its image on graph paper.[3]
  12. The monthly expenditure of a family is given below:
    13. Expenditure Food Health House Rent Education Others
    Amount (Rs.) 8000 6000 4000 12000 6000
    1. Represent the above information in a pie chart.[2]
    2. Find the average expenditure.[1]

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