G8_Kushma_Parbat_2081

1. Study the given Venn diagram.
1. Write the type of sets \(A\) and \(B\)—overlapping or disjoint.[1]
2. Construct all the possible subsets using the members of set \(A\).[2]
1. Sets \(A\) and \(B\) are overlapping sets.
   Because they share the common element \(5\), i.e., \(A \cap B = \{5\} \neq \emptyset\).
2. From the Venn diagram, set \(A = \{1, 4, 5\}\).
   All possible subsets of \(A\) are listed below .

   Number of Elements	Subsets	Count
   0	\(\emptyset\)	1
   1	\(\{1\}, \{4\}, \{5\}\)	3
   2	\(\{1, 4\}, \{1, 5\}, \{4, 5\}\)	3
   3	\(\{1, 4, 5\}\)	1
   Total Subsets		8

   (Since \(A\) has \(n = 3\) elements, total subsets = \(2^3 = 8\).)

2. Ram fixed the marked price of a laptop at Rs 63000 and sold it with a \(10\%\) discount.
1. Write the formula to find the discount percentage.[1]
2. At what price did Ram sell the laptop? Find it.[2]
3. How much discount amount was given if Ram had fixed the price at Rs 57000 and sold it at the above selling price?[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. Selling price of the laptop:
   Marked Price (MP) = Rs. \(63,000\)
   Discount = \(10\%\)
   Thus
   SP=90% of MP = \( \dfrac{90}{100} \times 63,000 = 56,700 \)
   So, Ram sold the laptop at Rs. \(56,700\).
3. Discount amount if marked price was Rs. 57,000 but sold at the same SP Rs. 56,700:
   New Marked Price = Rs. 57,000
   Selling Price = Rs. 56,700
   Thus
   Discount = MP – SP = 57,000 - 56,700 = 300
   So, the discount amount would be Rs. 300.
3. Ruby took a loan for 2 years at the simple interest rate of 10% per annum. If the interest for that period was Rs 2,000,
   1. Define interest.[1]
   2. How much loan did she take? Find it.[2]
   3. Find the ratio between interest and principal.[1]
1. Definition of Interest:
   Interest is the extra money paid by a borrower to a lender or bank for the use of the borrowed principal amount.
2. Loan amount (Principal):
   Time (T) = 2 years
   Rate (R) = 10%
   Interest (I) = Rs. 2,000
   We know,
   Principal (P) = \( \dfrac{I \times 100}{T \times R} = \dfrac{2,000 \times 100}{2 \times 10} = 10,000 \)
   So, she took a loan of Rs. 10,000.
3. Ratio between Interest and Principal:
   Interest = Rs. 2,000
   Principal = Rs. 10,000
   Thus,
   Ratio = \( \dfrac{2,000}{10,000} = \dfrac{1}{5} \)
   So, the ratio between interest and principal is 1:5.
1. Two numbers are in the ratio \(3:4\). If their sum is \(133\), find the numbers.[2]
2. If the price of \(10\) pens is \(\text{Rs.}\,200\), what will be the price of \(3\) dozen pens?[1]
3. The distance between Earth and the Sun is \(149{,}600{,}000{,}000\,\text{m}\). Write this number in scientific notation.[1]
4. Convert the binary number \(110111_2\) into the decimal number system.[1]
5. Flowers are planted in a trapezium-shaped garden inside a parallelogram-shaped land.
   
   1. Write the formula to find the area of a parallelogram. [1]
   2. Calculate the area of the trapezium-shaped garden. [2]
   3. Find the area of the parallelogram-shaped land excluding the trapezium garden. [2]
   4. How much will it cost to fence once around the parallelogram-shaped land at the rate of Rs 500 per meter? [2]
1. Area of a parallelogram formula
   We know that,
   Area (\( A \)) = base (\( b \)) \(\times\) height (\( h \))
2. Area of the trapezium-shaped garden
   From the figure, for the garden:
   Parallel sides (\( p_1 \)) = 6 m and (\( p_2 \)) = 9 m
   Height (\( h \)) = 4 m
   Now using the formula, we get
   Area of garden (\( A_g \)) = \( \frac{1}{2} \times (p_1 + p_2) \times h \)
   or \( A_g = \frac{1}{2} \times (6 + 9) \times 4 = 15 \times 2 = 30 \) m²
3. Area of the land excluding the garden
   First, calculating the total area of the land,
   Base (\( b \)) = 40 m and Height (\( h \)) = 20 m
   Total Area (\( A_l \)) = \( 40 \times 20 = 800 \) m²
   Now to find the area excluding the garden,
   Remaining Area = \( A_l - A_g = 800 - 30 = 770 \) m²
4. Cost of fencing the land
   Calculating the perimeter of the parallelogram,
   Perimeter (\( P \)) = \( 2(a + b) = 2(25 + 40) = 2 \times 65 = 130 \) m
   According to the question, the rate is Rs 500 per meter, so
   Total Cost = \( 130 \times 500 = \) Rs 65,000
1. Find the value of \((xyz)^0\).[1]
2. Using the laws of indices, prove that: \[ \dfrac{x^{p - q + 1} \cdot x^{q - r + 1} \cdot x^{r - p + 1}}{x^{3}} = 1 \][2]
1. Find the H.C.F. of the algebraic expressions: \(x^{2} - 7x + 12\) and \(3x^{2} - 27\).[2]
2. Simplify: \(\dfrac{x^{2} - 5x + 6}{x^{2} - 4}\)[2]
8. Two equations are given: \(x + y = 2\) and \(2x + y = 7\).
1. What is the system of the given equations called?[1]
2. Solve the equations using the graphical method and find the values of \(x\) and \(y\).[2]
9. In the given figure,
1. Write one pair of alternate angles.[1]
2. Find the value of \(q\).[2]
3. Find an exterior angle of a regular polygon with \(6\) sides.[1]
1. Construct a parallelogram \(PQRS\) with \(PQ = 6cm\), \(QR = 5cm\), and \(\angle PQR = 75^{\circ}\).[3]
2. In the adjoining figure, \(BC \parallel PQ\). Show that \(\triangle ABC \sim \triangle APQ\).[2]
1. Define regular tessellation.[1]
2. Reflect \(\triangle ABC\) with vertices \(A(1,1)\), \(B(4,1)\), and \(C(4,6)\) on the \(x\)-axis. Show both triangles on a graph and write the coordinates of the image.[3]
3. Find the distance between points \(A(1,1)\) and \(C(4,6)\).[2]
12. The ages (in years) of \(10\) students of Grade VIII are: \(13, 14, 15, 14, 16, 18, 14, 13, 13, 14\).
1. Find the mode age of the students.[1]
2. Find the average age of the students.[1]
3. The total weekly expenditure of a family is \(\text{Rs.}\,6{,}000\), distributed as shown in the pie chart:
   Food: \(120^{\circ}\), Education: \(80^{\circ}\), Cloth: \(90^{\circ}\), Others: \(70^{\circ}\).
   How much does the family spend on food per week? Find it.[1]