G8_Parshuram Minicipality_8_2081

1. Let set \( M = \{1, 2, 3\} \) be given.
1. Write all possible subsets that can be formed from set \(M\).[2]
2. Distinguish the proper and improper subsets from these subsets.[1]
1. Given that \(M = \{1, 2, 3\}\), so M has \(2^3 = 8\) total subsets. They are listed below .

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

2. From the subsets given above
   Proper subsets are all subsets except \(M\) itself.
   \(\emptyset, \{1\}, \{2\}, \{3\}, \{1,2\}, \{1,3\}, \{2,3\}\)   (7 subsets)
   Improper subsetis the set \(M\) itself
   \(\{1, 2, 3\}\)   (1 subset)
2. Ramdev went to utensil shop to buy a rice cooker. The marked price of a rice cooker is Rs. 6000.
1. If marked price and discount are represented by (MP) and (D) respectively, write the formula to find the selling price.[1]
2. How much discount did Ramdev get while buying a rice cooker on discount of 15%?[1]
3. The shopkeeper got 15% profit after selling it at 15% discount. What was the cost price of the rice cooker?[2]
1. Formula to find the selling price (SP) using marked price (MP) and discount (D):
   SP = MP – D
   or, if discount is given as a percentage \(D\%\):
   SP = \( (100-D)\% \times MP \)
2. Discount amount received by Ramdev.
   Marked Price (MP) = Rs. 6,000
   Discount = \(15\%\)
   Thus
   Discount = \(15\% \times MP= \dfrac{15}{100} \times 6,000 = 900 \)
   So, Ramdev got a discount of Rs. \(900\).
3. Cost price of the rice cooker
   We know that
   Selling Price (SP) = MP – Discount = \(6,000 - 900 = 5,100\)
   Given that
   Profit = \(20\%\)
   Thus
   SP= 120% of CP
   or\(5100= \dfrac{120}{100} \times\) CP
   or \(\text{CP} = \dfrac{5100 \times 100}{120} = 4250\)
   So, the cost price of the rice cooker was Rs. 4250.
3. Hari took a loan of Rs. 1,00,000 from a bank to go for abroad study at the rate of 12% p.a. simple interest. If he cleared the loan after 10 years then:
   1. Define amount.[1]
   2. What is the interest paid by him in 10 years?[2]
   3. If the interest rate was 10% p.a., what amount of money had he to pay in total?[2]
1. Definition of Amount:
   The sum of the principal and the interest earned on it over a certain period of time is called the amount. (i.e., Amount = Principal + Interest)
2. Interest for 10 years:
   Principal (P) = Rs. 1,00,000
   Time (T) = 10 years
   Rate (R) = 12%
   Thus,
   Interest (I) = \( \dfrac{P \times T \times R}{100} = \dfrac{1,00,000 \times 10 \times 12}{100} = 1,20,000 \)
   So, the interest paid by him is Rs. 1,20,000.
3. Total amount to pay if the rate was 10%:
   New Rate (R) = 10%
   Interest (I) = \( \dfrac{1,00,000 \times 10 \times 10}{100} = 1,00,000 \)
   Total Amount (A) = P + I = 1,00,000 + 1,00,000 = 2,00,000
   So, he would have to pay a total of Rs. 2,00,000.
4. The capacity of a water tank to supply water to a town is 480,000 liters.
1. Write the capacity of the water tank in scientific notation.[1]
2. A family of that town has consumed 3848 liters of water in a month. If the cost of 1 liter water is 50 paisa, how much rupees have to be paid for the consumption of water in one month?[2]
3. Convert \(0.\overline{35}\) into fraction.[1]
5. A hotel has a rectangular garden and a circular swimming pool having equal areas as shown in figure.
   
   1. Write the formula to calculate perimeter of rectangle. [1]
   2. Find the area of circular swimming pool. [1]
   3. Calculate the perimeter of the rectangular garden. [2]
   4. Which would have to spend more for fencing, a garden or a swimming pool? [1]
1. Perimeter of rectangle formula
   We know that,
   Perimeter (\( P \)) = \( 2(l + b) \)
2. Area of swimming pool
   From the figure, Diameter (\( d \)) = 28 cm.
   Radius (\( r \)) = \( \frac{28}{2} = 14 \) cm.
   Using the formula,
   Area (\( A_c \)) = \( \pi r^2 = \frac{22}{7} \times 14^2 = 616 \) cm²
3. Perimeter of rectangular garden
   Given, Area of garden = Area of pool = 616 cm².
   Length (\( l \)) = 40 cm.
   Breadth (\( b \)) = \( \frac{616}{40} = 15.4 \) cm.
   Now calculating perimeter,
   Perimeter (\( P \)) = \( 2(40 + 15.4) = 2 \times 55.4 = 110.8 \) cm
4. Comparison for fencing cost
   Circumference of pool (\( C \)) = \( 2\pi r = 2 \times \frac{22}{7} \times 14 = 88 \) cm.
   Perimeter of garden = 110.8 cm.
   Since the perimeter of the garden is greater than the pool's circumference,
   the garden would spend more for fencing.
1. Express \(\frac{x^a}{x^b}\) as power of \(x\).[1]
2. Simplify: \(\frac{x}{x-y} + \frac{x}{x+y}\)[2]
7. If two algebraic expressions are \(x^2 + 8x + 12\) and \(2x^2 - 72\),
1. Find the highest common factor of given algebraic expressions.[2]
2. For what value of \(x\), \(x^2 + 8x + 12 = 0\) becomes true?[2]
1. What types of equations are the given equations called? \(2x + 12 + y = 7\) and \(x + 2y = 8\)[1]
2. Solve the above equations by using graph.[3]
1. Observe the given figure and find the value of \(x\) and \(y\).[2]
2. Construct a rectangle \(ABCD\) in which \(AB = 7\) cm and \(BC = 4\) cm.[3]
3. Given triangles are the pair of congruent triangles. Find the values of \(x\) and \(y\).[2]
1. Define regular tessellation.[1]
2. Present Lakhans family expenditure in a pie chart.[2]
1. Write down the bearing of a point \(P\) from a point \(O\).[1]
2. The vertices of \(\triangle ABC\) are \(A(2,2)\), \(B(4,6)\) and \(C(5,2)\). Sketch it on a graph paper and reflect in y-axis. Write down the co-ordinates of the image.[3]
3. If the distance between \(P(0,6)\) and \(Q(a,0)\) is 6 units, then find the value of \(a\).[1]
12. The following are the marks obtained by 7 students of grade 8 in first terminal examination in mathematics: 22, 23, 44, 55, 66, 77, 33.
1. Find the mean from above data.[1]
2. Find the median of the above data.[1]
3. Find the mode from above data.[1]