G8_Dharan_8_2081

1. Study the given Venn diagram and answer the following questions.
1. Write the set notation for the shaded region.[1]
2. Write all the possible subsets formed from set \(B\).[1]
3. Which elements must be removed from both sets to make them disjoint sets?[1]
1. The shaded region described as
   Shaded region = \(A \cap B = \{1, 2\}\)
   
2. From the Venn diagram, set \(B = \{1, 2, 6\}\)
   All possible subsets of \(B\) are listed below:

   Number of Elements	Subsets	Count
   0	\(\emptyset\)	1
   1	\(\{1\}, \{2\}, \{6\}\)	3
   2	\(\{1,2\}, \{1,6\}, \{2,6\}\)	3
   3	\(\{1, 2, 6\}\)	1
   Total Subsets		8

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

3. The elements \(1\) and \(2\) from both sets must be removed so that new sets \(A = \{4, 8\}\) and \(B = \{6\}\) will be disjoint.
2. The cost of 16 kg apples is Rs 4800.
1. Write the cost of 16 kg apples in scientific notation.[1]
2. What is the cost of 1 kg apple in the quinary number system?[1]
3. How many kg of apples can be bought for Rs 5100? Find it.[1]
3. The cost of a mobile is Rs 12800. However, the mobile's marked price is \(20\%\) above the cost price. The shopkeeper sold the mobile, allowing a \(10\%\) discount.
1. Find the marked price of the mobile.[2]
2. Find the selling price of the mobile.[2]
3. What is the shopkeeper's profit or loss percent from the mobile? Find it.[1]
1. Marked price of the mobile.
   Cost Price (CP) = Rs. \(12,800\)
   Marked Price is \(20\%\) above CP.
   Thus
   MP = 120% of CP = \( \dfrac{120}{100} \times 12,800 = 15,360 \)
   So, the marked price is Rs. \(15,360\).
2. Selling price of the mobile.
   Marked Price (MP) = Rs. \(15,360\)
   Discount = \(10\%\)
   Thus
   SP = 90% of MP = \( \dfrac{90}{100} \times 15,360 = 13,824 \)
   So, the selling price is Rs. \(13,824\).
3. Profit or loss percent.
   Cost Price (CP) = Rs. \(12,800\)
   Selling Price (SP) = Rs. \(13,824\)
   Since SP > CP, there is a profit.
   Profit = \(13,824 - 12,800 = 1,024\)
   Thus
   Profit Percent = \( \dfrac{1,024}{12,800} \times 100\% = 8\% \)
   So, the shopkeeper made a profit of \(8\%\).
4. Ramesh gave Suresh a loan of Rs 24,500 at a fixed simple interest rate for 2 years. After the term ended, Suresh repaid a total amount of Rs 30,380.
   1. Write the formula to find simple interest when principal (P), time (T), and rate of interest (R) are given.[1]
   2. Find the interest rate paid by Suresh.[1]
   3. How much amount is returned by Suresh to Ramesh, if he paid at the end of 3 years?[2]
   4. If Ramesh distributes the total amount he received in two years between Ganesh and Mahesh in the ratio 3:7, who will receive how much more?[2]
1. Simple Interest formula:
   Simple Interest (I) = \( \dfrac{P \times T \times R}{100} \)
2. Rate of interest:
   Principal (P) = Rs. 24,500
   Amount (A) = Rs. 30,380
   Interest (I) = A - P = 30,380 - 24,500 = Rs. 5,880
   Time (T) = 2 years
   Thus,
   Rate (R) = \( \dfrac{I \times 100}{P \times T} = \dfrac{5,880 \times 100}{24,500 \times 2} = 12\% \)
   So, the interest rate is 12% per annum.
3. Total amount after 3 years:
   Time (T) = 3 years
   Interest (I) = \( \dfrac{24,500 \times 3 \times 12}{100} = 8,820 \)
   Thus,
   Amount (A) = P + I = 24,500 + 8,820 = 33,320
   So, the amount returned after 3 years would be Rs. 33,320.
4. Distribution between Ganesh and Mahesh:
   Total Amount = Rs. 30,380
   Ratio = 3:7 (Ganesh : Mahesh)
   Sum of ratios = 3 + 7 = 10
   Ganesh's share = \( \dfrac{3}{10} \times 30,380 = 9,114 \)
   Mahesh's share = \( \dfrac{7}{10} \times 30,380 = 21,266 \)
   Thus,
   Difference = 21,266 - 9,114 = 12,152
   So, Mahesh will receive Rs. 12,152 more than Ganesh.
5. The length and width of the floor of a rectangular room are 5 m and 4 m respectively. 500 circular tiles with a diameter of 20 cm have been installed on it.
   1. Write the formula to find the area of a rectangle. [1]
   2. How much square cm space of floor does a tile cover? (\(\pi = 3.14\)) [1]
   3. Calculate the area of the room's floor excluding the tiles. [2]
   4. If a circular tile is being replaced with a square tile that has a side length of 25 cm, calculate the total cost of tiling the room where each tile costs Rs 20. [1]
1. Area of rectangle formula
   We know that,
   Area (\( A \)) = Length (\( l \)) \(\times\) Breadth (\( b \))
2. Space covered by one tile
   Given: Diameter (\( d \)) = 20 cm
   Radius (\( r \)) = 10 cm
   Area (\( a \)) = \( \pi r^2 = 3.14 \times 10^2 = 314 \) cm²
3. Floor area excluding tiles
   Total area of the floor = \( 5 \text{ m} \times 4 \text{ m} = 20 \text{ m}^2 = 2,00,000 \) cm²
   Total area covered by 500 tiles = \( 500 \times 314 = 1,57,000 \) cm²
   Remaining Area = \( 2,00,000 - 1,57,000 = 43,000 \) cm²
4. Cost with square tiles
   Area of one square tile = \( 25 \text{ cm} \times 25 \text{ cm} = 625 \) cm²
   Number of tiles required = \( \frac{2,00,000}{625} = 320 \) tiles
   Total Cost = \( 320 \times 20 = \) Rs 6,400
1. What is the value of \((30x^3)^0\)?[1]
2. Simplify: \(\dfrac{1}{a + b} + \dfrac{2b}{a^2 - b^2}\).[2]
1. What value should be written in the blank space of \(x + \ldots + y^2\) to make a perfect square?[1]
2. Find the H.C.F. of the given algebraic expressions: \(4x^2 - 9\) and \(2x^2 + x - 3\).[2]
1. Solve: \(\dfrac{x + 15}{15x + 25} = \dfrac{1}{x}\).[2]
2. Solve the given equations by using graphical method: \(2x + y = 7\) and \(x - 2y = -4\).[2]
9. In the adjacent figure, the line \(CR\) intersects the straight lines \(EF\) and \(GH\) at points \(A\) and \(C\), while the line \(TS\) intersects the straight lines \(EF\) and \(GH\) at points \(B\) and \(C\).
1. Write a pair of co-interior angles from the figure.[1]
2. Find the value of \(x\).[2]
1. Construct a parallelogram \(ABCD\) with \(AB = 6.7cm\), \(\angle ABC = 60^\circ\), and \(BC = 5.6cm\).[3]
2. In the adjoining figure, if \(\triangle PMN \sim \triangle PQR\), find the length of \(MN\).[2]
3. Construct two triangles \(\triangle ABC\) and \(\triangle ADC\) from parallelogram \(ABCD\) and prove that \(\triangle ABC \cong \triangle ADC\).[2]
1. Which type of triangle is used to make a regular tessellation?[1]
2. In the adjoining figure, the bearing of \(B\) from \(A\) is \(106^\circ\). Find the bearing of \(A\) from \(B\).[2]
3. The triangle \(\triangle ABC\) with vertices \(A(2,3)\), \(B(5,6)\), and \(C(6,1)\) is reflected on the \(y\)-axis. Find the coordinates of the image triangle \(\triangle A'B'C'\).[3]