G8_Pokhara_8_2081

1. If the sets \(A = \{3, 4, 5, 7\}\) and \(B = \{2, 3, 4, 9\}\),
1. Which of the sets \(A\) and \(B\) are overlapping or disjoint? Write it.[1]
2. Form any two proper subsets of set \(A\).[1]
3. If set \(C = \{2, 3, 4, 9\}\), then the set \(C\) is an improper subset of set \(B\). Give a reason.[1]
2. The marks obtained by \(12\) students of Grade Eight in the Annual Examination in Mathematics are given below: \(22, 30, 25, 26, 24, 29, 27, 27, 28, 31, 33, 23\).
1. Sita said, “Mean divides the given data into two equal parts.” But Gita said, “Median divides the given data into two equal parts.” Whose statement is true?[1]
2. Find the mean from the above data.[2]
3. The monthly salary of Hari Rana is \(\text{Rs.}\,36{,}000\).
1. Write the amount in scientific notation.[1]
2. Express the number in the quinary number system.[2]
3. If Hari’s salary is twice as much as Suraj’s salary and Suraj’s salary is thrice as much as Shyam’s salary, find the annual salary of Shyam.[2]
4. Ram has a stationery shop at his house. He sells a calculator with a marked price \(\text{Rs.}\,450\) at a discount of \(12\%\), and he sells a ball at \(\text{Rs.}\,850\) after giving \(15\%\) discount.
1. If marked price (\(M\)) and discount (\(D\)) are given, then write the formula to find the price after discount.[1]
2. How much is the discount amount of the calculator? Find it.[1]
3. Find the marked price of the ball.[1]
4. By how much is the discount amount on the calculator more or less than the discount amount on the ball?[1]
5. Rina took a loan of \(\text{Rs.}\,50{,}000\) from Pawan at \(18\%\) per annum simple interest for \(3\) years.
1. How much simple interest is paid by Rina in \(3\) years? Find it.[1]
2. Find the total amount of principal and interest to be paid by Rina.[1]
3. If Rina took the loan at \(12\%\) per annum, then how much less interest would she have to pay?[2]
6. Bindu decided to exchange her square-shaped land of \(42\,\text{m}\) side with a rectangular land of equal area.
1. What was the area of her square-shaped land?[1]
2. If the length of the rectangular land to be exchanged is \(72\,\text{m}\), find the breadth of the land.[1]
3. What will be the length of wire required to fence the rectangular land three times?[2]
4. To fence the above square and rectangular land one round each, which land needs more wire and by how much?[1]
1. Express \(x^m \times x^n\) as a power of \(x\).[1]
2. Simplify: \(\dfrac{a}{(a - b)(a - c)} + \dfrac{b}{(a - c)(b - a)}\).[2]
8. Two equations are given below: \(2x - y = 5\) and \(x - y = 1\).
1. Write the degree of the given equations.[1]
2. Solve the above equations by using graphical method.[2]
1. Find the H.C.F. of: \(3x^3 - 15x^2\) and \(2x^3 - 50x\).[2]
2. Write a quadratic equation having roots \(4\) and \(5\) of \(x\).[2]
10. In the adjoining figure, \(TU\) intersects straight lines \(PQ\) and \(RS\) at points \(V\) and \(W\) respectively.
1. Write a pair of alternate angles from the figure.[1]
2. What type of triangle is \(\triangle VWX\) according to the angles of the triangle?[2]
3. At what value of \(\angle QVX\) will the given line segments \(PQ\) and \(RS\) be parallel?[1]
1. Construct a parallelogram \(ABCD\) with \(AB = 6\,\text{cm}\), \(BC = 4.5\,\text{cm}\), and \(\angle ABC = 45^\circ\).[3]
2. In the given figure, if \(\triangle ABC \sim \triangle DEF\), find the measurement of \(DF\).[2]
1. Write down the bearing of point \(A\) from point \(O\).[1]
2. \(L(0,-2)\), \(M(5,-4)\), and \(N(2,5)\) are the vertices of \(\triangle LMN\). Find the coordinates of the vertices of its image when it is reflected about the \(x\)-axis. Also, draw the graph of the reflection.[3]
3. If the centre of a circle is \(A(4,4)\) and \(B(7,4)\) is any point on its circumference, find the area of the circle.[2]

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G_8_Bharatpur_8_2081

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]
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 \(\text{Rs.}\,3{,}750\); the marked price of ball \(B\) is \(\text{Rs.}\,2{,}500\), and a \(5\%\) discount is offered.
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 wanted to earn a \(10\%\) profit by selling ball \(B\) after the discount, at what price did he buy ball \(B\)?[2]
4. Sarita invested \(\text{Rs.}\,5{,}000\) 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 \(\text{Rs.}\,12{,}000\) at the same rate and for the same time?[1]
5. In the figure, \(ABCD\) is a kite-shaped plot where diagonals \(AC = 6\,\text{m}\) and \(BD = 5\,\text{m}\). Inside it, there is a circular well of radius \(50\,\text{cm}\).
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. 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 \(AC = 5.5\,\text{cm}\) and \(AF = 3.6\,\text{cm}\). 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:

Animal	Buffalo	Goat	Cow	Chicken
Number	50	200	450	200

1. Write the names of animals that represent the mode value.[1]
2. Find the average number of animals.[2]

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G8_Biratnagar_8_2081

1. Study the given Venn diagram and answer the following questions.
1. Define disjoint sets.[1]
2. Write any one proper subset of set \(P\).[1]
3. If the common element \(5\) is removed from the Venn diagram, then what will be the relation between sets \(P\) and \(Q\)? Write it.[1]
2. The marked price of a television is \(\text{Rs.}\,24{,}000\). If the shopkeeper got \(\text{Rs.}\,2{,}400\) profit after selling it with \(15\%\) discount, then
1. If marked price and discount amount are represented by \(MP\) and \(D\) respectively, write the formula to find the selling price after discount.[1]
2. How much discount had been given by the shopkeeper to sell the television? Find it.[2]
3. Find the cost price of the television.[2]
3. Sunil has deposited \(\text{Rs.}\,3{,}00{,}000\) in Rastriya Banijya Bank for \(3\) years at the rate of \(\text{Rs.}\,12\) simple interest per annum for every \(\text{Rs.}\,100\).
1. At what percent of interest rate per annum had Sunil deposited the amount?[1]
2. After \(3\) years, how much total money does Sunil get with principal and interest? Calculate it.[1]
3. If Sunil decides to distribute \(\text{Rs.}\,3{,}00{,}000\) to his brothers Chandan and Ram in the ratio \(2:3\), then compare the amount received by Chandan and Ram.[2]
4. Raju takes a bus to Dharan from Biratnagar. The wheel of the bus rotates \(35{,}750\) times in an hour.
1. Write \(35{,}750\) in scientific notation.[1]
2. How many times will the wheel rotate in \(90\) minutes?[1]
3. Find the value of \(\sqrt{48} + \sqrt{75} - \sqrt{3}\).[2]
4. Convert \(0.\overline{24}\) into a fraction.[1]
5. In the given figure, \(ABCD\) is a square and a circle is drawn inside it.
1. Write the formula to find the area of a circle.[1]
2. How much is the radius of the circle?[1]
3. Find the area of the shaded region.[2]
4. Compare the circumference of the circle and the perimeter of the square.[1]
1. Express \(x^m \times x^{-1}\) as a power of \(x\).[1]
2. Simplify: \(\dfrac{a}{(a - b)^2} + \dfrac{b}{(a - b)^2}\).[2]
7. Two equations are given: \(x + y = 6\) and \(x - y = 2\).
1. What is meant by simultaneous equations?[1]
2. Solve the given equations by using a graph.[2]
1. Find the L.C.M. of the algebraic expressions: \(x^2 - 7x + 12\) and \(3x^2 - 27\).[2]
2. Find the quadratic equation in which the values of \(x\) are \(2\) and \(3\).[2]
9. In the adjoining figure, when \(XY\) and \(XZ\) meet the line segments \(PQ\) and \(RS\), a \(\triangle XYZ\) is formed.
1. Write the relation between \(\angle XYZ\) and \(\angle XZY\).[1]
2. Find the value of \(x\).[2]
3. At which value of \(\angle PXY\) will the line segments \(PQ\) and \(RS\) be parallel?[1]
1. Construct a rectangle \(ABCD\) in which \(AB = 7\,\text{cm}\) and \(BC = 5\,\text{cm}\).[3]
2. In rectangle \(ABCD\), prove that \(\triangle ABC \cong \triangle ACD\) by drawing diagonal \(AC\).[2]
1. What is meant by regular tessellation?[1]
2. In the adjoining figure, if the bearing of point \(S\) from point \(R\) is \(060^\circ\), find the bearing of point \(R\) from point \(S\).[2]
3. Find the coordinates of the images \(M'\), \(N'\), and \(O'\) of \(\triangle MNO\) with vertices \(M(2,1)\), \(N(4,3)\), and \(O(-1,2)\) after reflection on the \(x\)-axis.[3]
12. The monthly expenses of Shital's meals are given in the table below:

Month	Ashoj	Kartik	Mangsir	Poush	Magh
Expenditure (Rs.)	4000	2500	2000	1700	1800

1. What is the monthly average expenditure of Shital on her meals?[1]
2. Present Shital’s expenditure in a pie chart.[2]

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G8_Lalitpur_8_2081

1. If set \(A = \{\text{even numbers up to }15\}\) and set \(B = \{\text{prime numbers up to }15\}\),
1. Define overlapping sets.[1]
2. Make any two proper subsets from set \(B\).[1]
3. What change in the outcome of set \(B\) makes the two sets \(A\) and \(B\) disjoint?[1]
2. The ratio of boys and girls of class eight of a school is \(5:7\) respectively. If the total number of students is \(60\), then
1. How many girls are there? Find it.[1]
2. Change the total number of students into the binary number system.[1]
3. Write \(350{,}000\) in scientific notation.[1]
4. Convert \(1.\overline{24}\) into a fraction.[1]
3. Raju Lama visited a computer store to get \(5\) printers and a laptop. A set of \(5\) printers and a laptop is available for \(\text{Rs.}\,4{,}65{,}000\).
1. If \(15\%\) discount is allowed on those machineries, find the discount amount.[2]
2. If the shopkeeper earned \(20\%\) profit even after allowing \(15\%\) discount, at what price did the shopkeeper purchase such machineries?[2]
3. By how much is the marked price of a laptop less or more than \(\text{Rs.}\,95{,}000\), if the price of a printer is \(\text{Rs.}\,75{,}000\)?[1]
4. Dilliram deposited \(\text{Rs.}\,1{,}20{,}000\) in a bank at the rate of \(12\%\) per annum.
1. Write a formula to calculate simple interest.[1]
2. How much interest does Dilliram get after \(6\) years?[1]
3. If he has to pay \(5\%\) of interest as income tax, how much amount will he receive after \(6\) years?[2]
5. Mohan constructed a rectangular garden and a circular fish pond in the garden of his house with equal areas.
1. Find the area of a triangle having base \(b\,\text{cm}\) and height \(h\,\text{cm}\).[1]
2. Find the area of the circular fish pond.[1]
3. Find the perimeter of the rectangular garden.[2]
4. Which of the garden or fish pond needs more cost to fence at the same rate of cost?[1]
1. What exponent of \(y\) will be equal to \(1\)?[1]
2. Prove that: \((x^{a-b})^{a+b} \cdot (x^{b-c})^{b+c} \cdot (x^{c-a})^{c+a} = 1\).[2]
7. An algebraic fraction is given: \(\dfrac{x}{x^2 + 3x + 2}\) \(\div\) \(\dfrac{2}{x^2 - 1}\).
1. Find the L.C.M. of the denominators.[1]
2. Simplify the given fraction and reduce it to the lowest term.[2]
8. There are two numbers \(x\) and \(y\) such that their sum is \(8\) and difference is \(4\).
1. Construct the simultaneous equations based on the given statements.[1]
2. What are the numbers? Calculate by using graphical method.[2]
9. Study the given figure and answer the following questions.
1. Find the value of \(x\) and \(y\).[2]
2. Compare the angles \(x\) and \(y\).[1]
3. Draw a bearing angle of \(030^\circ\).[1]
10. Study the given figure and answer the following questions.
1. Define congruent figures.[1]
2. If \(\triangle ADE \sim \triangle ABC\), find the length of \(DE\).[2]
3. In a regular polyhedron, the number of vertices is \(8\) and the number of edges is \(12\). Calculate the number of faces by using Euler’s formula.[2]
1. Construct a parallelogram having adjacent sides \(7\,\text{cm}\) and \(4\,\text{cm}\) and the angle between them is \(60^\circ\).[3]
2. The vertices of \(\triangle ABC\) are \(A(-3,2)\), \(B(5,3)\), and \(C(1,6)\). Sketch it on graph paper and reflect it in the \(x\)-axis. Write down the coordinates of the image.[3]
12. The total expenditure of a family is \(\text{Rs.}\,36{,}000\) for four months. Expenditure of each month is shown in the adjoining pie chart.
1. Find the expenditure in each month.[2]
2. How much is the average expenditure of such a family in one month? Calculate it.[1]

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G8_Kathmandu_8_2081

1. Set \(A = \{2, 5\}\) and set \(B = \{5, 7\}\) are given.
1. Are sets \(A\) and \(B\) overlapping or disjoint? Write it.[1]
2. Write any two proper subsets that can be made from set \(B\).[2]
2. As announced on December 8, 2020, the height of Mount Everest, the highest peak in the world, was \(8848.86\) meter.
1. Write whether the number \(8848.86\) is a rational or irrational number.[1]
2. Convert the height of Mt. Everest in centimeter and write in scientific notation.[2]
3. Prove that: \(8848 = 240343_5\).[2]
3. Two friends, Ramnaresh and Mahesh invested \(\text{Rs.}\,50{,}00{,}000\) in a factory in the ratio of \(3:2\).
1. What is the difference in direct and indirect variation? Write one difference.[1]
2. How much amount has Ramnaresh invested in the factory? Find it.[1]
3. If Mahesh had deposited the amount invested in the industry in a bank at an annual interest rate of \(10\%\), how much simple interest would he have received after \(2\) years? Calculate.[2]
4. If \(3:2 = x:500\), find the value of \(x\).[1]
4. Ashim bought a machine for \(\text{Rs.}\,25{,}000\) and fixed the market price by increasing \(20\%\) on its price. He made a loss of \(\text{Rs.}\,1{,}000\) after selling the machine with some discount amount.
1. Find the marked price of the machine.[1]
2. At how much discount percentage was the machine sold? Find out.[2]
3. If Ashim wants to earn a profit of \(\text{Rs.}\,2{,}000\) by selling the machine, what discount rate should be maintained?[1]
5. In the figure, \(ABCD\) is a parallelogram where a right-angled triangle \(ADB\) with height \(AD = 5\,\text{cm}\) is formed on the semicircle having diameter \(13\,\text{cm}\).
1. Write the formula to find the area of parallelogram.[1]
2. Find the area of the semicircle.[1]
3. By how much is area of right-angled triangle \(ADB\) less than the area of semicircle? Calculate it.[2]
4. Is the area of parallelogram \(ABCD\) double the area of triangle \(ADB\)? Give logical answer.[1]
1. Write the expanded form of \((a - b)^2\).[1]
2. Simplify: \(x^{p(a-b)} \times x^{p(b-c)} \times x^{p(c-a)}\).[2]
7. A rectangular garden has length \(4\) meter more than its breadth. The area of the garden is \(96\) square meter.
1. If the breadth of the garden is \(x\) meters, what is the length? Write in terms of \(x\).[1]
2. What are the length and breadth of the garden? Find by making quadratic equation.[2]
1. Factorise: \(2x + xy - 2x - y\).[2]
2. Simplify: \(\dfrac{1}{x - 3} - \dfrac{x-3}{x^2 - 9}\).[2]
9. In the given graph, triangle \(ABC\) is shown.
1. Which formula do you use to find the distance between two points? Write it.[1]
2. Find the length of side \(BC\).[2]
3. Rotate the triangle \(ABC\) a quarter turn in the positive direction and write the coordinates of the vertices of image triangle.[2]
4. Experimentally verify that the sum of interior angles of a triangle is \(180^\circ\). (Two triangles of different measurements are required.)[3]
10. In the figure \(PQRS\) is a rectangle in which length \(QR = 6\,\text{cm}\) and breadth \(RS = 4\,\text{cm}\).
1. Construct the rectangle \(PQRS\) using compass according to the given dimensions.[3]
2. Are \(\triangle QPS\) and \(\triangle QRS\) congruent to each other? Write with reason.[1]
11. A point \(Q\), which is \(8\,\text{cm}\) away from a point \(P\), has a bearing of \(110^\circ\) in the scale \(1\,\text{cm} = 5\,\text{km}\).
1. Draw the bearing according to the above context.[1]
2. Compare the bearing of \(P\) from point \(Q\) and bearing of \(Q\) from \(P\).[2]
12. The following are the marks obtained by \(7\) students of grade eight in first terminal examination in mathematics: \(23, 30, 25, 26, 23, 28, 27\).
1. Find the mode from the above data.[1]
2. What is the average marks obtained by the \(7\) students in first terminal examination? Find it.[1]
3. Which among mean, median and mode divides the given data in two equal parts? Write with reason.[1]

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G8_Statistics_Questions

Q 12(a): BLE Exam

1. In a given data, if \( \sum x = 150 \) and number of items \( n = 15 \), find the value of \( \bar{X} \).[1U]
2. How many terms have sum 228 and mean 19?[1U]
3. In the given data, if \( \sum x = 50 \) and average \( (\bar{X}) = 10 \), find the number of items.[1U]
4. If the sum of the terms is \( \sum fx \), and sum of the frequencies is \( \sum f \), write down the formula to calculate mean \( (\bar{X}) \).[1K]
5. Write down the formula to find the mean of individual series where 'x' is variate value, 'n' the number of items and '\(\bar{X}\)' the mean.[1K]
6. In a data, if the sum of the terms is \( (\sum x) \) and number of terms is (n), then write down the formula to find arithmetic mean \( (\bar{X}) \).[1K]
7. If \( \sum fx = 100 \) and \( N = 20 \), find the mean \( (\bar{X}) \).[1U]
8. Which term will be the median of an ascending series containing 7 terms?[1K]
9. If seven workers are receiving daily wages of Rs. 350, 325, 350, 375, 350, 325, 350, find the mode wage.[1U]
10. Which is the modal letter in the word "curriculum"?[1K]
11. Find out the mode value of the given data: 2, 4, 6, 4, 3, 4, 5, 6, 4.[1K]
12. Find out the mode of the given data: d, f, e, f, e, d, f, d, e, f.[1K]
13. If the mode of the following data is 7, find the value of x: 3, 4, 1, 3, 7, 5, 6, 3, 7, x, 7, 2.[1K]
14. Find the range from the given data: 32 cm, 45 cm, 13 cm, 18 cm, 23 cm, 20 cm, 19 cm.[1K]
15. Find the range from the following data: 16 cm, 12 cm, 18 cm, 9 cm, 6 cm, 10 cm.[1K]
16. Find the range in given data: 28, 35, 21, 17, 10, 12, 3, 9, 34.[1U]
17. If \( \sum x = 42 \) and \( n = 6 \), find mean \( (\bar{X}) \).[1U]
18. The ages of Hari, Mohan, Sudip, Jiban and Niranjan are 12, 18, 13, 16 and 6 years respectively. Find their average age.[1U]
19. Find the average value \( (\bar{X}) \) from the given data: 2, 7, 9, 12 and 15.[1U]
20. Find the arithmetic mean of the given data: 10, 16, 14, 20, 22, 26.[1U]
21. Find the arithmetic mean of given data: 7, 5, 3, 4, 7, 10, 8, 13, 7, 6, 7, 5.[1U]
22. Find the median of the given data: 5, 10, 25, 20, 15.[1U]
23. Find the median from the given data: 11, 17, 9, 10, 14, 20, 16.[1U]
24. Find the range from the following given data: 62, 64, 60, 62, 64, 61, 65, 67, 63.[1U]
25. If the mean of 4, 10, 3, 6, a, 9 and 10 is 7, find the value of a.[1A]
26. If 7 is the mean of 3, 6, a, 9 and 10, find their median.[1A]
27. If the following numbers are in ascending order and median is 7, then find the value of x: 4, 6, x - 2, 9, 13.[1A]
28. The average age of 5 students is 9 years. Out of them the ages of 4 students are 5, 7, 8 and 15 years. What was the age of the remaining student?[1HA]
29. In the adjoining pie chart, the total number of students of class 6 to 9 is 600. How many students are there in class 8?[1HA]
30. Following numbers are in ascending order of magnitude. If their median is 36, then find the value of a: a + 2, a + 4, a + 6, a + 8, a + 10.[1HA]

Q 12(b): BLE Exam

1. 50 students in a school pay Rs. 1500 per month on an average, what is total yearly income of the school?[1U]
2. Find the mean from the given data:
   

   प्राप्तांक (Marks obtained):	10	15	20	25	30	35
   विद्यार्थी सङ्ख्या (No. of students):	2	4	7	5	4	3

   [2A]
3. Find the mean:
   

   प्राप्तांक (Marks obtained):	30	31	32	33	34
   विद्यार्थी सङ्ख्या (No. of students):	8	10	15	8	9

   [2A]
4. If the mean of the following data is 17, find the value of \( f_1 \):
   

   X	5	10	15	20	25	30
   f	2	5	10	\( f_1 \)	4	2

   [2A]
5. If the arithmetic mean of the data is 41, find the value of 'x':
   

   x	20	30	40	50	60	70
   f	3	4	x	4	1	2

   [2A]
6. Find the value of y when the average mark is 18 in the table given below:
   

   x	5	10	15	20	25	30
   f	7	y	8	4	5	10

   [2A]
7. Find the median:
   

   प्राप्तांक (Marks obtained):	26	30	35	40	45	50	55	65
   विद्यार्थी सङ्ख्या (No. of students):	2	3	6	10	12	13	3	4

   [2A]
8. Find the median:
   

   X	19	20	22	25	30	31	33
   f	7	9	8	11	5	6	7

   [2A]
9. The total monthly income of a family is Rs. 3600. The sources of income are shown in the given pie chart. Then find:
   (क) What is the income of the family from business?
   (ख) What is the income from agriculture?[1A]
10. The monthly expenditure of a family is given in the table. Represent it in a pie chart.
    

    शीर्षक (Headings):	खाना (Food)	कपडा (Clothing)	शिक्षा (Education)	स्वास्थ्य (Health)	बचत (Saving)
    खर्च रु. (Expenditure Rs.):	2000	3000	4000	2500	6500

    [2A]
11. Show the annual budget of a family as given below in a pie chart.
    

    शीर्षक (Items):	खाना (Food)	कपडा (Clothing)	शिक्षा (Education)	औषधी (Medicines)
    खर्चरु. (Expenditures):	Rs. 150000	Rs. 50000	Rs. 100000	Rs. 60000

    [2A]
12. The monthly expenditure of Ram's family is Rs. 3600. Represent the given information in pie chart.
    

    खर्च (Expenditure):	खाना (Food)	कपडा (Clothing)	शिक्षा (Education)	अन्य (Other)
    रकम रु. (Amount in Rs.):	1200	1500	x	300

    [2A]
13. Represent each of the following data in a pie chart.
    

    ग्यासहरू (Gases):	नाइट्रोजन (Nitrogen)	अक्सिजन (Oxygen)	अन्य (Others)
    प्रतिशत (Percentage):	78%	20%	2%

    [2A]
14. The monthly budget of a family is given below. Represent it in pie chart.
    

    खाना (Food)	शिक्षा (Education)	स्वास्थ्य (Health)	कपडा (Clothing)	अन्य (Others)
    Rs. 5000	Rs. 2000	Rs. 3000	Rs. 4000	Rs. 4000

    [2A]
15. The number of students studying from class 5 to class 8 in the Darbar High School is given in the table below.
    

    कक्षा (Class)	5	6	7	8
    विद्यार्थी सङ्ख्या (No. of students)	50	60	55	35

    
    (a) Represent the above data in a pie chart.
    (b) Find the mean of the data.[1A]
16. Annual expenditure of Arun's family is given below. Represent this information in a pie chart.
    

    शीर्षक (Title)	शिक्षा (Education)	स्वास्थ्य (Health)	लताकपडा (Clothing)	खाना (Food)
    खर्च रु. (Expenditure (Rs.))	96,000	40,000	64,000	1,20,000

    [2A]
17. In a data, if \( \sum x = 77 + m \), \( n = 10 \) and mean \( (\bar{X}) = 8 \), find the value of \( m \).[1A]
18. The monthly expenditure of a family is given in the table below.
    

    शीर्षक (Title)	खाना (Food)	शिक्षा (Education)	लताकपडा (Clothing)	अन्य खर्च (Other expenses)
    खर्च रकम (Expenditure amount)	Rs. 18,000	Rs. 10,000	Rs. 5,000	Rs. 3,000

    
    (a) Show the above data in a pie chart.
    (b) Find the median of the data given below: 10, 12, 8, 9, 17, 16, 18, 20, 22, 15, 13.[2A]
19. Marks obtained by 30 students of a class in the first term examination in Mathematics are given in the table below.
    

    प्राप्तांक (Marks)	20	25	30	35	40	45	50
    विद्यार्थी सङ्ख्या (No. of students)	3	4	5	8	5	4	1

    
    (a) Find the average marks in Mathematics.
    (b) Find the median.[2A]

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G8_Construction_Geometry

Q.No.10(a): Construction, in BLE Exam

1. Construct a rectangle \(ABCD\) in which the adjacent sides \(AB\) and \(BC\) are \(6\) cm and \(5\) cm respectively. [3A]
2. Construct a rectangle \(ABCD\) having adjacent sides \(AB = 8\) cm, \(BC = 6\) cm and diagonal \(AC = 10\) cm. [3A]
3. Construct parallelogram \(ABCD\) having \(AB = 5\) cm, \(BC = 4\) cm and \(\angle ABC = 60^\circ\). [3A]
4. Construct a square \(ABCD\) having one side \(AB = 5\) cm. [3A]
5. Construct a rectangle \(ABCD\) in which \(AB = 8\) cm and \(BC = 6\) cm. [3A]
6. Construct a rectangle where diagonals \(AC = BD = 6\) cm, \(\angle BOC = 30^\circ\) and \(O\) is the point of intersection of the diagonals. [3A]
7. Construct a rectangle \(ABCD\) where \(BC = 5\) cm, \(BD = 10\) cm and \(\angle CBD = 60^\circ\). [3A]
8. Construct a rectangle \(PQRS\) where \(PQ = 4.8\) cm and \(PR = 6.2\) cm. [3A]
9. Construct a rectangle \(ABCD\) having \(BC = 7.1\) cm, \(BD = 10\) cm and \(\angle DBC = 45^\circ\). [3A]
10. Construct a rectangle \(ABCD\), where \(AC = BD = 7\) cm, and \(\angle COD = 45^\circ\), \(AC\) and \(BD\) intersect at the point \(O\). [3A]
11. Construct a parallelogram \(PQRS\), in which \(OR = 4\) cm, diagonal \(QS = 5\) cm and \(\angle SOR = 30^\circ\). [3A]
12. Construct a rectangle with one side \(3\) cm and diagonal \(5\) cm. [3A]
13. Construct a rectangle of a diagonal \(6\) cm and the angle between the diagonals is \(60^\circ\). [3A]
14. Construct a rectangle so that the diagonal is \(8\) cm and angle between diagonals is \(30^\circ\). [3A]
15. Construct a square having the length of a side \(4\) cm. [3A]

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G8_Experimental Verification_Geometry

Q.No.10(b): Experimental, in BLE Exam

1. Verify experimentally that the sum of three angles of a triangle is equal to that of two right angles. (Two figures of different measurements are required.) [2A]
2. Verify experimentally that the base angles of an isosceles right angled triangle are each equal to \(45^\circ\). (Two figures of different measurements are required.) [2A]
3. Verify experimentally that if two sides of a triangle are equal, then the opposite angles to them are also equal. (Two figures with different measures are necessary.) [2A]
4. Verify experimentally that all the interior angles of an equilateral triangle are equal and the value of each angle is \(60^\circ\). (Two figures of different measurements are required.) [2A]
5. Verify experimentally that the base angles of an isosceles triangle are equal. (Two figures of different sizes are necessary.) [2A]
6. Verify experimentally that the line joining the vertex and mid-point of the base of an isosceles triangle is perpendicular to the base. (Two figures of different measurements are required.) [2A]
7. Verify experimentally that the diagonals of a rectangle are equal. (Two figures of different measurements are required.) [2A]
8. Verify experimentally that all the angles of a rectangle are equal and each angle is right angle. (Two figures of different measurements are required.) [2A]
9. Verify experimentally that the opposite angles of a parallelogram are equal. (Two figures with different measurements are necessary.) [2A]
10. Verify experimentally that the opposite sides of a parallelogram are equal. (Two figures with different measurements are necessary.) [2A]
11. Verify experimentally that the diagonals of a parallelogram bisect each other. (Two figures of different measurements are required.) [2A]
12. Verify experimentally that the diagonals of a square are equal. (Two figures of different measurements are required.) [2A]
13. Verify experimentally that the diagonals of a square bisect the vertical angle. (Two figures of different measurements are required.) [2A]
14. Verify experimentally that the diagonals of a rhombus bisect at right angle. [2A]

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G8_Algebra_8_Questions

आनुपातिक बीजीय भिन्न (Rational Algebraic Fraction)

For Q.No.6 (b) in BLE Examination

1. सरल गर्नुहोस् (Simplify): \( \dfrac{1}{a - 2} - \dfrac{4}{a^2 - 4} \) [2U]
 
   The simplification is
\( \dfrac{1}{a - 2} - \dfrac{4}{a^2 - 4} \)
or \( \dfrac{1}{a - 2} - \dfrac{4}{(a - 2)(a + 2)} \)
or \( \dfrac{1 \cdot (a + 2) - 4}{(a - 2)(a + 2)} \)
or \( \dfrac{a + 2 - 4}{(a - 2)(a + 2)} \)
or \( \dfrac{(a - 2)}{(a - 2)(a + 2)} \)
or \( \dfrac{1}{a + 2} \)
2. सरल गर्नुहोस् (Simplify): \( \dfrac{a^2 + b^2}{a^2 - b^2} - \dfrac{a - b}{a + b} \) [2U]
 
   The simplification is
\( \dfrac{a^2 + b^2}{a^2 - b^2} - \dfrac{a - b}{a + b} \)
or \( \dfrac{a^2 + b^2}{(a - b)(a + b)} - \dfrac{a - b}{a + b} \)
or \( \dfrac{(a^2 + b^2) - (a - b)(a - b)}{(a - b)(a + b)} \)
or \( \dfrac{a^2 + b^2 - (a - b)^2}{a^2 - b^2} \)
or \( \dfrac{a^2 + b^2 - (a^2 - 2ab + b^2)}{a^2 - b^2} \)
or \( \dfrac{a^2 + b^2 - a^2 + 2ab - b^2}{a^2 - b^2} \)
or \( \dfrac{2ab}{a^2 - b^2} \)
3. सरल गर्नुहोस् (Simplify): \( \dfrac{x}{x^2 + 3x + 2} - \dfrac{2}{x^2 - 1} \) [2U]
 
   The simplification is
\( \dfrac{x}{x^2 + 3x + 2} - \dfrac{2}{x^2 - 1} \)
or\( \dfrac{x}{(x + 2)(x + 1)} - \dfrac{2}{(x - 1)(x + 1)} \)
or \( \dfrac{x(x - 1) - 2(x + 2)}{(x + 2)(x + 1)(x - 1)} \)
or \( \dfrac{x^2 - x - 2x - 4}{(x + 2)(x^2 - 1)} \)
or \( \dfrac{x^2 - 3x - 4}{(x + 2)(x^2 - 1)} \)
or \( \dfrac{(x - 4)(x + 1)}{(x + 2)(x + 1)(x - 1)} \)
or \( \dfrac{x - 4}{(x + 2)(x - 1)} \)

4. सरल गर्नुहोस् (Simplify): \( \dfrac{2x}{x - 2} - \dfrac{4}{x - 2} \) [2U]
 
   The simplification is
\( \dfrac{2x}{x - 2} - \dfrac{4}{x - 2} \)
or \( \dfrac{2x - 4}{x - 2} \)
or \( \dfrac{2(x - 2)}{(x - 2)} \)
or \( 2 \)
5. सरल गर्नुहोस् (Simplify): \( \dfrac{3}{a^2 - 4} + \dfrac{1}{(a - 2)^2} \) [2U]
 
   The simplification is
\( \dfrac{3}{a^2 - 4} + \dfrac{1}{(a - 2)^2} \)
or \( \dfrac{3}{(a - 2)(a + 2)} + \dfrac{1}{(a - 2)(a - 2)} \)
or \( \dfrac{3(a - 2) + 1(a + 2)}{(a - 2)^2 (a + 2)} \)
or \( \dfrac{3a - 6 + a + 2}{(a - 2)^2 (a + 2)} \)
or \( \dfrac{4a - 4}{(a - 2)^2 (a + 2)} \)
or \( \dfrac{4(a - 1)}{(a - 2)^2 (a + 2)} \)
6. सरल गर्नुहोस् (Simplify): \( \dfrac{3a^2 - 6a}{a^2 - 1} + \dfrac{3}{a^2 - 1} \) [2U]
 
   The simplification is
\( \dfrac{3a^2 - 6a}{a^2 - 1} + \dfrac{3}{a^2 - 1} \)
or \( \dfrac{3a^2 - 6a + 3}{a^2 - 1} \)
or \( \dfrac{3(a^2 - 2a + 1)}{a^2 - 1} \)
or \( \dfrac{3(a - 1)^2}{(a - 1)(a + 1)} \)
or \( \dfrac{3 \cdot (a - 1) \cdot (a - 1)}{(a - 1) \cdot (a + 1)} \)
or \( \dfrac{3(a - 1)}{a + 1} \)
7. सरल गर्नुहोस् (Simplify): \( \dfrac{2x^2 + x}{2x + 1} - \dfrac{2xy + y}{2x + 1} \) [2U]
 
   The simplification is
\( \dfrac{2x^2 + x}{2x + 1} - \dfrac{2xy + y}{2x + 1} \)
or \( \dfrac{(2x^2 + x) - (2xy + y)}{2x + 1} \)
or \( \dfrac{x(2x + 1) - y(2x + 1)}{2x + 1} \)
or \( \dfrac{(2x + 1)(x - y)}{(2x + 1)} \)
or \( x - y \)
8. सरल गर्नुहोस् (Simplify): \( \dfrac{x^2 - 2xy + y^2}{x^2 - y^2} \times \dfrac{x + y}{x - y} \) [2U]
 
   The simplification is
\( \dfrac{x^2 - 2xy + y^2}{x^2 - y^2} \times \dfrac{x + y}{x - y} \)
or \( \dfrac{(x - y)^2}{(x - y)(x + y)} \times \dfrac{x + y}{x - y} \)
or \( 1 \)
9. सरल गर्नुहोस् (Simplify): \( \dfrac{a^2 - 4a}{a^2 - 4} - \dfrac{4}{4 - a^2} \) [2U]
 
   The simplification is
\( \dfrac{a^2 - 4a}{a^2 - 4} - \dfrac{4}{4 - a^2} \)
or \( \dfrac{a^2 - 4a}{a^2 - 4} - \dfrac{4}{-(a^2 - 4)} \)
or \( \dfrac{a^2 - 4a}{a^2 - 4} + \dfrac{4}{a^2 - 4} \)
or \( \dfrac{a^2 - 4a + 4}{a^2 - 4} \)
or \( \dfrac{(a - 2)^2}{(a - 2)(a + 2)} \)
or \( \dfrac{a - 2}{a + 2} \)
10. सरल गर्नुहोस् (Simplify): \( \dfrac{x^2 + y^2}{x - y} + \dfrac{2xy}{y - x} \) [2U]
 
   The simplification is
\( \dfrac{x^2 + y^2}{x - y} + \dfrac{2xy}{y - x} \)
or \( \dfrac{x^2 + y^2}{x - y} + \dfrac{2xy}{-(x - y)} \)
or \( \dfrac{x^2 + y^2}{x - y} - \dfrac{2xy}{x - y} \)
or \( \dfrac{x^2 + y^2 - 2xy}{x - y} \)
Factorizing the numerator:
or \( \dfrac{(x - y)^2}{x - y} \)
or \( x - y \)
11. सरल गर्नुहोस् (Simplify): \( \dfrac{p^2}{p - q} + \dfrac{q^2}{q - p} \) [2U]
 
   The simplification is
\( \dfrac{p^2}{p - q} + \dfrac{q^2}{q - p} \)
or \( \dfrac{p^2}{p - q} + \dfrac{q^2}{-(p - q)} \)
or \( \dfrac{p^2}{p - q} - \dfrac{q^2}{p - q} \)
The denominators are the same, so we subtract the numerators.
or \( \dfrac{p^2 - q^2}{p - q} \)
or \( \dfrac{(p + q) \cdot (p - q)}{(p - q)} \)
or \( p + q \)
12. सरल गर्नुहोस् (Simplify): \( \dfrac{x - 15}{x^2 - 9} + \dfrac{18}{x^2 - 9} \) [2U]
 
   The simplification is
\( \dfrac{x - 15}{x^2 - 9} + \dfrac{18}{x^2 - 9} \)
or \( \dfrac{x - 15 + 18}{x^2 - 9} \)
or \( \dfrac{x + 3}{x^2 - 3^2} \)
or \( \dfrac{x + 3}{(x - 3) \cdot (x + 3)} \)
or \( \dfrac{1}{x - 3} \)
13. सरल गर्नुहोस् (Simplify): \( \dfrac{1}{x - 1} + \dfrac{2x}{1 - x^2} \) [2U]
 
   The simplification is
\( \dfrac{1}{x - 1} + \dfrac{2x}{1 - x^2} \)
or \( \dfrac{1}{x - 1} + \dfrac{2x}{-(x^2 - 1)} \)
or \( \dfrac{1}{x - 1} - \dfrac{2x}{(x - 1)(x + 1)} \)
or \( \dfrac{1 \cdot (x + 1) - 2x}{(x - 1)(x + 1)} \)
or \( \dfrac{x + 1 - 2x}{x^2 - 1} \)
or \( \dfrac{1 - x}{x^2 - 1} \)
or \( \dfrac{-(x - 1)}{(x - 1)(x + 1)} \)
or \( \dfrac{-1}{x + 1} \)
14. सरल गर्नुहोस् (Simplify): \( \dfrac{2}{a + 1} + \dfrac{2a}{a - 1} - \dfrac{a^2 + 3}{a^2 - 1} \) [2U]
 
   The simplification is
\( \dfrac{2}{a + 1} + \dfrac{2a}{a - 1} - \dfrac{a^2 + 3}{a^2 - 1} \)
or \( \dfrac{2}{a + 1} + \dfrac{2a}{a - 1} - \dfrac{a^2 + 3}{(a - 1)(a + 1)} \)
or \( \dfrac{2(a - 1) + 2a(a + 1) - (a^2 + 3)}{(a - 1)(a + 1)} \)
or \( \dfrac{2a - 2 + 2a^2 + 2a - a^2 - 3}{a^2 - 1} \)
or \( \dfrac{(2a^2 - a^2) + (2a + 2a) + (- 2 - 3)}{a^2 - 1} \)
or \( \dfrac{a^2 + 4a - 5}{a^2 - 1} \)
or \( \dfrac{(a + 5) \cdot (a - 1)}{(a + 1) \cdot (a - 1)} \)
or \( \dfrac{a + 5}{a + 1} \)
15. सरल गर्नुहोस् (Simplify): \( \dfrac{1}{x + y} + \dfrac{1}{x - y} + \dfrac{2y}{x^2 - y^2} \) [2U]
 
   The simplification is
\( \dfrac{1}{x + y} + \dfrac{1}{x - y} + \dfrac{2y}{x^2 - y^2} \)
or \( \dfrac{1}{x + y} + \dfrac{1}{x - y} + \dfrac{2y}{(x + y)(x - y)} \)
or \( \dfrac{1 \cdot (x - y) + 1 \cdot (x + y) + 2y}{(x + y)(x - y)} \)
or \( \dfrac{x - y + x + y + 2y}{x^2 - y^2} \)
or \( \dfrac{(x + x) + (-y + y + 2y)}{x^2 - y^2} \)
or \( \dfrac{2x + 2y}{x^2 - y^2} \)
or \( \dfrac{2(x + y)}{(x - y)(x + y)} \)
or \( \dfrac{2}{x - y} \)
16. सरल गर्नुहोस् (Simplify):
    \( \dfrac{x + y}{(y - z)(z - x)} - \dfrac{y + z}{(x - z)(x - y)} + \dfrac{z + x}{(z - y)(y - x)} \) [3U]
 
   The simplification is
\( \dfrac{x + y}{(y - z)(z - x)} + \dfrac{y + z}{(z - x)(x - y)} + \dfrac{z + x}{(y - z)(x - y)} \)
or \( \dfrac{(x + y)(x - y) + (y + z)(y - z) + (z + x)(z - x)}{(x - y)(y - z)(z - x)} \)
\( \dfrac{(x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2)}{(x - y)(y - z)(z - x)} \)
\( \dfrac{\cancel{x^2} - \cancel{y^2} + \cancel{y^2} - \cancel{z^2} + \cancel{z^2} - \cancel{x^2}}{(x - y)(y - z)(z - x)} \)
or \( \dfrac{0}{(x - y)(y - z)(z - x)} \)
or \( 0 \)
17. सरल गर्नुहोस् (Simplify): \( \dfrac{2}{a + 1} - \dfrac{2}{a - 1} + \dfrac{4}{a^2 + 1} \) [2U]
 
   The simplification is
\( \dfrac{2}{a + 1} - \dfrac{2}{a - 1} + \dfrac{4}{a^2 + 1} \)
or \( \dfrac{2(a - 1) - 2(a + 1)}{(a + 1)(a - 1)} + \dfrac{4}{a^2 + 1} \)
or \( \dfrac{2a - 2 - 2a - 2}{a^2 - 1} + \dfrac{4}{a^2 + 1} \)
or \( \dfrac{- 4}{a^2 - 1} + \dfrac{4}{a^2 + 1} \)
or \( \dfrac{- 4(a^2 + 1) + 4(a^2 - 1)}{(a^2 - 1)(a^2 + 1)} \)
or \( \dfrac{- 4a^2 - 4 + 4a^2 - 4}{(a^2)^2 - 1^2} \)
or \( \dfrac{\cancel{- 4a^2} - 4 + \cancel{4a^2} - 4}{a^4 - 1} \)
or \( \dfrac{- 8}{a^4 - 1} \)
or \( - \dfrac{8}{a^4 - 1} \)
18. सरल गर्नुहोस् (Simplify):
    \( \dfrac{1}{(x - 3)(x - 4)} + \dfrac{1}{(x - 4)(x - 5)} + \dfrac{1}{(x - 5)(x - 3)} \) [3U]
 
   The simplification is
\( \dfrac{1}{(x - 3)(x - 4)} + \dfrac{1}{(x - 4)(x - 5)} + \dfrac{1}{(x - 5)(x - 3)} \)
or \( \dfrac{1 \cdot (x - 5) + 1 \cdot (x - 3) + 1 \cdot (x - 4)}{(x - 3)(x - 4)(x - 5)} \)
or \( \dfrac{x - 5 + x - 3 + x - 4}{(x - 3)(x - 4)(x - 5)} \)
or \( \dfrac{(x + x + x) + (- 5 - 3 - 4)}{(x - 3)(x - 4)(x - 5)} \)
or \( \dfrac{3x - 12}{(x - 3)(x - 4)(x - 5)} \)
or \( \dfrac{3 \cdot \boxed{(x - 4)}}{(x - 3) \cdot \boxed{(x - 4)} \cdot (x - 5)} \)
or \( \dfrac{3}{(x - 3)(x - 5)} \)
19. सरल गर्नुहोस् (Simplify):
    \( \dfrac{a}{(a - b)(a - c)} + \dfrac{b}{(b - a)(b - c)} + \dfrac{c}{(c - a)(c - b)} \) [3U]
 
   The simplification is
\( \dfrac{a}{(a - b)(a - c)} + \dfrac{b}{-(a - b)(b - c)} + \dfrac{c}{-(a - c) \cdot [-(b - c)]} \)
or\( \dfrac{a}{(a - b)(a - c)} - \dfrac{b}{(a - b)(b - c)} + \dfrac{c}{(a - c)(b - c)} \)
or\( \dfrac{a \cdot (b - c) - b \cdot (a - c) + c \cdot (a - b)}{(a - b)(b - c)(a - c)} \)
or\( \dfrac{ab - ac - ab + bc + ac - bc}{(a - b)(b - c)(a - c)} \)
or\( \dfrac{\cancel{ab} - \cancel{ac} - \cancel{ab} + \cancel{bc} + \cancel{ac} - \cancel{bc}}{(a - b)(b - c)(a - c)} \)
or \( \dfrac{0}{(a - b)(b - c)(a - c)} \)
or \( 0 \)
20. सरल गर्नुहोस् (Simplify): \( \dfrac{1}{1 + x} - \dfrac{1}{x - 1} + \dfrac{2}{1 + x^2} \) [2U]
 
   The simplification is
\( \dfrac{1}{1 + x} - \dfrac{1}{x - 1} + \dfrac{2}{1 + x^2} \)
or \( \dfrac{1}{1 + x} + \dfrac{1}{1 - x} + \dfrac{2}{1 + x^2} \)
or \( \dfrac{1 \cdot (1 - x) + 1 \cdot (1 + x)}{(1 + x)(1 - x)} + \dfrac{2}{1 + x^2} \)
or \( \dfrac{1 - \cancel{x} + 1 + \cancel{x}}{1 - x^2} + \dfrac{2}{1 + x^2} \)
or \( \dfrac{2}{1 - x^2} + \dfrac{2}{1 + x^2} \)
or \( \dfrac{2(1 + x^2) + 2(1 - x^2)}{(1 - x^2)(1 + x^2)} \)
or \( \dfrac{2 + \cancel{2x^2} + 2 - \cancel{2x^2}}{1 - x^4} \)
or \( \dfrac{4}{1 - x^4} \)
21. सरल गर्नुहोस् (Simplify):
    \( \dfrac{x}{x - 2y} + \dfrac{x}{x + 2y} + \dfrac{2x^2}{x^2 + 4y^2} \) [2U]
 
   The simplification is
\( \dfrac{x}{x - 2y} + \dfrac{x}{x + 2y} + \dfrac{2x^2}{x^2 + 4y^2} \)
or \( \dfrac{x(x + 2y) + x(x - 2y)}{(x - 2y)(x + 2y)} + \dfrac{2x^2}{x^2 + 4y^2} \)
or \( \dfrac{x^2 + \cancel{2xy} + x^2 - \cancel{2xy}}{x^2 - 4y^2} + \dfrac{2x^2}{x^2 + 4y^2} \)
or \( \dfrac{2x^2}{x^2 - 4y^2} + \dfrac{2x^2}{x^2 + 4y^2} \)
or \( \dfrac{2x^2(x^2 + 4y^2) + 2x^2(x^2 - 4y^2)}{(x^2 - 4y^2)(x^2 + 4y^2)} \)
or \( \dfrac{2x^4 + \cancel{8x^2y^2} + 2x^4 - \cancel{8x^2y^2}}{(x^2)^2 - (4y^2)^2} \)
or \( \dfrac{4x^4}{x^4 - 16y^4} \)
22. सरल गर्नुहोस् (Simplify): \( \dfrac{x}{x + y} + \dfrac{y}{x - y} - \dfrac{2xy}{x^2 - y^2} \) [2U]
 
   The simplification is
\( \dfrac{x}{x + y} + \dfrac{y}{x - y} - \dfrac{2xy}{x^2 - y^2} \)
or\( \dfrac{x}{x + y} + \dfrac{y}{x - y} - \dfrac{2xy}{(x + y)(x - y)} \)
or \( \dfrac{x(x - y) + y(x + y) - 2xy}{(x + y)(x - y)} \)
or \( \dfrac{x^2 - xy + xy + y^2 - 2xy}{x^2 - y^2} \)
or \( \dfrac{x^2 + y^2 + \cancel{xy} - \cancel{xy} - 2xy}{x^2 - y^2} \)
or \( \dfrac{x^2 - 2xy + y^2}{x^2 - y^2} \)
or \( \dfrac{(x - y)^2}{(x + y)(x - y)} \)
or \( \dfrac{x - y}{x + y} \)
23. सरल गर्नुहोस् (Simplify): \( \dfrac{1}{a^2 - 3a + 2} + \dfrac{1}{a^2 - 5a + 6} \) [2U]
 
   The simplification is
\( \dfrac{1}{a^2 - 3a + 2} + \dfrac{1}{a^2 - 5a + 6} \)
or \( \dfrac{1}{(a - 1)(a - 2)} + \dfrac{1}{(a - 2)(a - 3)} \)
or \( \dfrac{1 \cdot (a - 3) + 1 \cdot (a - 1)}{(a - 1)(a - 2)(a - 3)} \)
or \( \dfrac{a - 3 + a - 1}{(a - 1)(a - 2)(a - 3)} \)
or \( \dfrac{2a - 4}{(a - 1)(a - 2)(a - 3)} \)
or \( \dfrac{2 \cdot (a - 2)}{(a - 1) \cdot (a - 2) \cdot (a - 3)} \)
or \( \dfrac{2}{(a - 1)(a - 3)} \)
24. सरल गर्नुहोस् (Simplify):
    \( \dfrac{2}{x^2 + 3x + 2} + \dfrac{5x}{x^2 - x - 6} - \dfrac{x + 2}{x^2 - 2x - 3} \) [3U]
 
   The simplification is
\( \dfrac{2}{x^2 + 3x + 2} + \dfrac{5x}{x^2 - x - 6} - \dfrac{x + 2}{x^2 - 2x - 3} \)
or\( \dfrac{2}{(x + 2)(x + 1)} + \dfrac{5x}{(x - 3)(x + 2)} - \dfrac{x + 2}{(x - 3)(x + 1)} \)
or \( \dfrac{2(x - 3) + 5x(x + 1) - (x + 2)(x + 2)}{(x + 2)(x + 1)(x - 3)} \)
or \( \dfrac{2x - 6 + 5x^2 + 5x - (x^2 + 4x + 4)}{(x + 2)(x + 1)(x - 3)} \)
or \( \dfrac{5x^2 + 7x - 6 - x^2 - 4x - 4}{(x + 2)(x + 1)(x - 3)} \)
or \( \dfrac{(5x^2 - x^2) + (7x - 4x) + (- 6 - 4)}{(x + 2)(x + 1)(x - 3)} \)
or \( \dfrac{4x^2 + 3x - 10}{(x + 2)(x + 1)(x - 3)} \)
or \( \dfrac{(4x - 5) \cdot (x + 2)}{(x + 2) \cdot (x + 1) \cdot (x - 3)} \)
or \( \dfrac{4x - 5}{(x + 1)(x - 3)} \)
25. सरल गर्नुहोस् (Simplify):
    \( \dfrac{2(a + 4)}{a + 3} \cdot \dfrac{a^2 + 2a - 24}{a^2 - a - 12} \) [2U]
 
   The simplification is
\( \dfrac{2(a + 4)}{a + 3} \cdot \dfrac{a^2 + 2a - 24}{a^2 - a - 12} \)
or \( \dfrac{2(a + 4)}{a + 3} \cdot \dfrac{(a + 6) \cdot (a - 4)}{(a - 4) \cdot (a + 3)} \)
or \( \dfrac{2(a + 4)(a + 6)}{(a + 3)^2} \)
26. सरल गर्नुहोस् (Simplify):
    \( \dfrac{1}{x + 2} + \dfrac{2}{2(x - 2)} - \dfrac{4}{x^2 - 4} \) [2U]
 
   The simplification is
\( \dfrac{1}{x + 2} + \dfrac{2}{2(x - 2)} - \dfrac{4}{x^2 - 4} \)
or\( \dfrac{1}{x + 2} + \dfrac{2}{2(x - 2)} - \dfrac{4}{(x + 2)(x - 2)} \)
or\( \dfrac{1 \cdot (x - 2) + 1 \cdot (x + 2) - 4}{(x + 2)(x - 2)} \)
or \( \dfrac{x - \cancel{2} + x + \cancel{2} - 4}{x^2 - 4} \)
or \( \dfrac{2x - 4}{x^2 - 4} \)
or \( \dfrac{2(x - 2)}{(x + 2)(x - 2)} \)
or \( \dfrac{2 \cdot (x - 2)}{(x + 2) \cdot (x - 2)} \)
or \( \dfrac{2}{x + 2} \)
27. सरल गर्नुहोस् (Simplify):
    \( \dfrac{1}{a - 3} - \dfrac{1}{2(a + 3)} + \dfrac{a}{a^2 - 9} \) [2U]
 
   The simplification is
\( \dfrac{1}{a - 3} - \dfrac{1}{2(a + 3)} + \dfrac{a}{a^2 - 9} \)
or\( \dfrac{1}{a - 3} - \dfrac{1}{2(a + 3)} + \dfrac{a}{(a - 3)(a + 3)} \)
or \( \dfrac{1 \cdot 2(a + 3) - 1 \cdot (a - 3) + a \cdot 2}{2(a - 3)(a + 3)} \)
or \( \dfrac{2a + 6 - a + 3 + 2a}{2(a^2 - 9)} \)
or \( \dfrac{(2a - a + 2a) + (6 + 3)}{2(a^2 - 9)} \)
or \( \dfrac{3a + 9}{2(a^2 - 9)} \)
or \( \dfrac{3(a + 3)}{2(a - 3)(a + 3)} \)
or \( \dfrac{3 \cdot (a + 3)}{2(a - 3) \cdot (a + 3)} \)
or \( \dfrac{3}{2(a - 3)} \)
28. सरल गर्नुहोस् (Simplify): \( \dfrac{2x + 6}{x^2 - 9} - \dfrac{4x}{2x^2 - 6x} \) [2U]
 
   The simplification is
\( \dfrac{2x + 6}{x^2 - 9} - \dfrac{4x}{2x^2 - 6x} \)
or \( \dfrac{2(x + 3)}{(x+3)(x-3)} - \dfrac{4x}{2x(x - 3)} \)
or \( \dfrac{2}{(x-3)} - \dfrac{2}{(x - 3)} \)
or \( 0 \)

Unit 10: समीकरण र लेखाचित्र (Equation and Graph)

For Q.No.7 (a) in BLE Examination

1. कस्ता समीकरणहरूलाई युगपतरैखीय समीकरण भनिन्छ ?
   What type of equations are called simultaneous equations? [1K]
 
Two or more linear equations involving the same variables and that are solved together to find a common solution, is called Simultaneous linear equations
2. कस्ता समीकरणहरूलाई वर्ग समीकरण भनिन्छ ?
   What type of equations are called quadratic equations? [1K]
 
An equation in which the highest power (degree) of the variable is two is called a quadratic equation. The standard form is \( ax^2 + bx + c = 0 \), where \( a \neq 0 \).
3. हल गर्नुहोस् (Solve): \( 2 - x = 17 - 4x \) [1A]
 
The solution is
\( 2 - x = 17 - 4x \)
or \( -x + 4x = 17 - 2 \)
or \( 3x = 15 \)
or \( x = \dfrac{15}{3} \)
or \( x = 5 \)
4. हल गर्नुहोस् (Solve): \( \dfrac{7x + 3}{4} = 6 \) [1A]
 
The solution is
\( \dfrac{7x + 3}{4} = 6 \)
or \( 7x + 3 = 6 \cdot 4 \)
or \( 7x + 3 = 24 \)
or \( 7x = 24 - 3 \)
or \( 7x = 21 \)
or \( x = \dfrac{21}{7} \)
or \( x = 3 \)
5. x को मान कति हुँदा समीकरण \( 8x + 1 = 57 \) सत्य हुन्छ ?
   For what value of x the equation \( 8x + 1 = 57 \) becomes true? [1A]
 
The solution is
\( 8x + 1 = 57 \)
or \( 8x = 57 - 1 \)
or \( 8x = 56 \)
or \( x = \dfrac{56}{8} \)
or \( x = 7 \) For what value of \(x=7\) the equation \( 8x + 1 = 57 \) becomes true.
6. यदि \( 5x + 8 = 13 \), x को मान पत्ता लगाउनुहोस्।
   If \( 5x + 8 = 13 \), find the value of x. [1A]
 
The solution is
\( 5x + 8 = 13 \)
or \( 5x = 13 - 8 \)
or \( 5x = 5 \)
or \( x = \dfrac{5}{5} \)
or \( x = 1 \)
7. वर्ग समीकरण \( x^2 = 16 \) मा x का मानहरू के के हुन् ?
   What are the values of x in the quadratic equation \( x^2 = 16 \)? [1A]
 
The solution is
\( x^2 = 16 \)
or \( x = \pm \sqrt{16} \)
or \( x = \pm 4 \)

8. x को मान पत्ता लगाउनुहोस् (Find the value of x): \( \dfrac{x}{3} = \dfrac{3}{x} \) [1A]
 
The solution is
\( \dfrac{x}{3} = \dfrac{3}{x} \)
or \( x \cdot x = 3 \cdot 3 \)
or \( x^2 = 9 \)
or \( x = \pm \sqrt{9} \)
or \( x = \pm 3 \)

9. हल गर्नुहोस् (Solve): \( x^2 - 4 = 21 \) [1A]
 
The solution is
\( x^2 - 4 = 21 \)
or \( x^2 = 21 + 4 \)
or \( x^2 = 25 \)
or \( x = \pm \sqrt{25} \)
or \( x = \pm 5 \)

10. दिइएको समीकरण हल गर्नुहोस् (Solve the given equation): \( x^2 + 2x = 0 \) [1A]
 
The solution is
\( x^2 + 2x = 0 \)
or \( x(x + 2) = 0 \)
Either \( x = 0 \) or \( x + 2 = 0 \).
Therefore, the values of \( x \) are
\( x=0 \) and \( x=-2 \)
11. हल गर्नुहोस् (Solve): \( \dfrac{1}{x} = \dfrac{x}{16} \) [1A]
 
The solution is
\( \dfrac{1}{x} = \dfrac{x}{16} \)
or \( x \cdot x = 1 \cdot 16 \)
or \( x^2 = 16 \)
or \( x = \pm \sqrt{16} \)
or \( x = \pm 4 \)
   
12. हल गर्नुहोस् (Solve): \( 3x^2 - 4x = 0 \) [1A]
 
The solution is
\( 3x^2 - 4x = 0 \)
or \( x(3x - 4) = 0 \)
Either \( x = 0 \) or \( 3x - 4 = 0 \).
Therefore, the values of \( x \) are
\(x= 0 \) and \( x=\dfrac{4}{3} \).
   
13. x को मान २ र ३ हुने वर्ग समीकरण पत्ता लगाउनुहोस्।
    Find the quadratic equation in which values of x are 2 and 3. [1U]
 
The quadratic equation is
\( x^2 - (sum)x + (product) = 0 \)
or \( x^2 - (5)x + 6 = 0 \)
or \( x^2 - 5x + 6 = 0 \)
   
14. मूलहरू १ र २ हुने वर्ग समीकरण लेख्नुहोस्।
    Write the quadratic equations whose roots are 1 and 2. [1A]
 
The quadratic equation is
\( x^2 - (sum)x + (product) = 0 \)
or \( x^2 - (3)x + 2 = 0 \)
or \( x^2 - 3x + 2 = 0 \)
   
15. एउटा वर्ग समीकरणका जम्मा कतिओटा मूलहरू हुन्छन् ?
    How many roots are there in a quadratic equation? [1A]
 
There are two roots in a quadratic equation.

For Q.No.7(b) in BLE Examination

1. हल गर्नुहोस् (Solve): \( \dfrac{x + 1}{2} = \dfrac{7x - 3}{3x} \) [2A]
 
The solution is
\( \dfrac{x + 1}{2} = \dfrac{7x - 3}{3x} \)
or \( 3x(x + 1) = 2(7x - 3) \)
or \( 3x^2 + 3x = 14x - 6 \)
or \( 3x^2 + 3x - 14x + 6 = 0 \)
or \( 3x^2 - 11x + 6 = 0 \)
or \( 3x^2 - (9x + 2x) + 6 = 0 \)
or \( 3x(x - 3) - 2(x - 3) = 0 \)
or \( (x - 3)(3x - 2) = 0 \)
Therefore, the values of \( x \) are
or\(x= 3 \) and \(x= \dfrac{2}{3} \).
2. हल गर्नुहोस् (Solve): \( 8x^2 - 32 = 0 \) [2A]
 
The solution is
\( 8x^2 - 32 = 0 \)
or \( 8x^2 = 32 \)
or \( x^2 = \dfrac{32}{8} \)
or \( x^2 = 4 \)
or \( x = \pm \sqrt{4} \)
Therefore, the values of \( x \) are
or \( x = 2 \) and \( x = -2 \).
   
3. हल गर्नुहोस् (Solve): \( 3x - 9x^2 = 0 \) [2A]
 
The solution is
\( 3x - 9x^2 = 0 \)
or \( 3x(1 - 3x) = 0 \)
Therefore, the values of \( x \) are
or \( x = 0 \) and \( x = \dfrac{1}{3} \).
   
4. हल गर्नुहोस् (Solve): \( \dfrac{x^2}{4} - 25 = 0 \) [2A]
 
The solution is
\( \dfrac{x^2}{4} - 25 = 0 \)
or \( \dfrac{x^2}{4} = 25 \)
or \( x^2 = 25 \times 4 \)
or \( x^2 = 100 \)
or \( x = \pm \sqrt{100} \)
Therefore, the values of \( x \) are
or \( x = 10 \) and \( x = -10 \).
   
5. हल गर्नुहोस् (Solve): \( x^2 - x - 2 = 0 \) [2A]
 
The solution is
\( x^2 - x - 2 = 0 \)
or \( x^2 - (2x - x) - 2 = 0 \)
or \( x^2 - 2x + x - 2 = 0 \)
or \( x(x - 2) + 1(x - 2) = 0 \)
or \( (x - 2)(x + 1) = 0 \)
Therefore, the values of \( x \) are
or \( x = 2 \) and \( x = -1 \).
   
6. हल गर्नुहोस् (Solve): \( x^2 - x - 6 = 0 \) [2A]
 
The solution is
\( x^2 - x - 6 = 0 \)
or \( x^2 - (3x - 2x) - 6 = 0 \)
or \( x^2 - 3x + 2x - 6 = 0 \)
or \( x(x - 3) + 2(x - 3) = 0 \)
or \( (x - 3)(x + 2) = 0 \)
Therefore, the values of \( x \) are
or \( x = 3 \) and \( x = -2 \).
7. समीकरण \( x + y = 4 \) लाई लेखाचित्रमा देखाउनुहोस्।
   Show the equation \( x + y = 4 \) in a graph. [2A]
 
Given lines are
\( x + y = 4 \)
The table value of the line is
\( x \)	0	4	2
\( y \)	4	0	2
The line is shown in the graph.
8. दिइएको समीकरणलाई लेखाचित्रमा देखाउनुहोस्।
   (Show the given equation in a graph): \( x = y + 4 \) [2A]
 
  
    
  Given line is
  \( x = y + 4 \) वा \( y = x - 4 \)
  The table value of the line is                                        
\( x \)	0	4	2
\( y \)	-4	0	-2
The line is shown in the graph.   
  
   
 
 
9. दिइएको समीकरणलाई लेखाचित्रमा देखाउनुहोस्।
   (Show the given equation in a graph): \( y = 4x + 8 \) [2A]
 
  
    
  Given line is
  \( y = 4x + 8 \)
  The table value of the line is                                        
\( x \)	0	-2	-1
\( y \)	8	0	4
The line is shown in the graph.   
  
 
 
 
10. दिइएको समीकरणलाई लेखाचित्रमा देखाउनुहोस्।
    (Show the given equation in a graph): \( 3x + 4y = 12 \) [2A]
 
  
    
  Given line is
\( 3x + 4y = 12 \)
\( y = \dfrac{12 - 3x}{4} \)
  The table value of the line is                                        
\( x \)	0	4	-4
\( y \)	3	0	6
The line is shown in the graph.   
 
 
 
11. हल गर्नुहोस् (Solve): \( x(x + 1) = 4 + x \) [2A]
 
The solution is
\( x(x + 1) = 4 + x \)
or \( x^2 + x = 4 + x \)
or \( x^2 + x - x - 4 = 0 \)
or \( x^2 - 4 = 0 \)
or \( x^2 = 4 \)
or \( x = \pm \sqrt{4} \)
Therefore, the values of \( x \) are
or \( x = 2 \) and \( x = -2 \).
   
12. हल गर्नुहोस् (Solve): \( (x - 7)^2 - 64 = 0 \) [2HA]
 
The solution is
\( (x - 7)^2 - 64 = 0 \)
or \( (x - 7)^2 = 64 \)
or \( x - 7 = \pm \sqrt{64} \)
or \( x - 7 = \pm 8 \)
Taking the positive sign (+)
\( x - 7 = 8 \Rightarrow x = 15 \)
Taking the negative sign (-)
\( x - 7 = -8 \Rightarrow x = -1 \)
Therefore, the values of \( x \) are
or \( x = 15 \) and \( x = -1 \).
   
13. हल गर्नुहोस् (Solve): \( 7x^2 + 13x - 2 = 0 \) [2HA]
 
The solution is
\( 7x^2 + 13x - 2 = 0 \)
or \( 7x^2 + (14x - x) - 2 = 0 \)
or \( 7x^2 + 14x - x - 2 = 0 \)
or \( 7x(x + 2) - 1(x + 2) = 0 \)
or \( (x + 2)(7x - 1) = 0 \)
Either \( x + 2 = 0 \) or \( 7x - 1 = 0 \)
Therefore, the values of \( x \) are
or \( x = -2 \) and \( x = \dfrac{1}{7} \).
   
14. x को मान कति हुँदा, \( x^2 - 5x + 6 \) को मान शून्य हुन्छ ?
    For what value of x, the value of \( x^2 - 5x + 6 \) is zero? [2HA]
 
The solution is
\( x^2 - 5x + 6 = 0 \)
or \( x^2 - (3x + 2x) + 6 = 0 \)
or \( x^2 - 3x - 2x + 6 = 0 \)
or \( x(x - 3) - 2(x - 3) = 0 \)
or \( (x - 3)(x - 2) = 0 \)
Either \( x - 3 = 0 \) or \( x - 2 = 0 \)
Therefore, the values of \( x \) are
or \( x = 3 \) and \( x = 2 \).
   
15. x को मान कति हुँदा, \( 2x^2 - x - 6 \) को मान शून्य हुन्छ ?
    For what value of x, the value of \( 2x^2 - x - 6 \) is zero? [2HA]
 
The solution is
\( 2x^2 - x - 6 = 0 \)
or \( 2x^2 - (4x - 3x) - 6 = 0 \)
or \( 2x^2 - 4x + 3x - 6 = 0 \)
or \( 2x(x - 2) + 3(x - 2) = 0 \)
or \( (x - 2)(2x + 3) = 0 \)
Either \( x - 2 = 0 \) or \( 2x + 3 = 0 \)
Therefore, the values of \( x \) are
or \( x = 2 \) and \( x = -\dfrac{3}{2} \).

For Q.No.8 (b) in BLE Examination

1. हल गर्नुहोस् (Solve): \( x^2 + \dfrac{1}{16} = \dfrac{x}{2} \) [2A]
 
The solution is
\( x^2 + \dfrac{1}{16} = \dfrac{x}{2} \)
or \( 16x^2 + 16 \cdot \dfrac{1}{16} = 16 \cdot \dfrac{x}{2} \)
or \( 16x^2 + 1 = 8x \)
or \( 16x^2 - 8x + 1 = 0 \)
or\( (4x)^2 - 2(4x)(1) + (1)^2 = 0 \)
or \( (4x - 1)^2 = 0 \)
or \( 4x - 1 = 0 \)
or \( 4x = 1 \)
or \( x = \dfrac{1}{4} \)
2. हल गर्नुहोस् (Solve): \( \dfrac{3 - 4x}{5} - \dfrac{4 + 5x}{9} + \dfrac{7x + 11}{15} = 0 \) [2A]
 
The solution is
\( \dfrac{3 - 4x}{5} - \dfrac{4 + 5x}{9} + \dfrac{7x + 11}{15} = 0 \)
or \( 9(3 - 4x) - 5(4 + 5x) + 3(7x + 11) = 0 \)
or \( 27 - 36x - 20 - 25x + 21x + 33 = 0 \)
or \( (-36x - 25x + 21x) + (27 - 20 + 33) = 0 \)
or \( (-61x + 21x) + (40) = 0 \)
or \( -40x + 40 = 0 \)
or \( 40 = 40x \)
or \( x = 1 \).
   
3. हल गर्नुहोस् (Solve): \( (x + 1)(x + 2) = x(x + 7) - 6 \) [2A]
 
The solution is
\( (x + 1)(x + 2) = x(x + 7) - 6 \)
or \( x^2 + 2x + x + 2 = x^2 + 7x - 6 \)
or \( x^2 + 3x + 2 = x^2 + 7x - 6 \)
or \( 3x + 2 = 7x - 6 \)
or \( 2 + 6 = 7x - 3x \)
or \( 8 = 4x \)
or \( x = 2 \)
4. हल गर्नुहोस् (Solve): \( \dfrac{6x - 3}{2x + 7} = \dfrac{3x - 2}{x + 5} \) [2A]
 
The solution is
\( \dfrac{6x - 3}{2x + 7} = \dfrac{3x - 2}{x + 5} \)
or \( (6x - 3)(x + 5) = (3x - 2)(2x + 7) \)
or \( 6x^2 + 30x - 3x - 15 = 6x^2 + 21x - 4x - 14 \)
or \( 6x^2 + 27x - 15 = 6x^2 + 17x - 14 \)
or \( 27x - 17x = -14 + 15 \)
or \( 10x = 1 \)
or \( x = \dfrac{1}{10} \).
      
5. हल गर्नुहोस् (Solve): \( 2x + y = 6 \), \( 3x + y = 7 \) [2A]
 
The solution is
Given equations are
\( 2x + y = 6 \) (i)
\( 3x + y = 7 \) (ii)
Subtracting equation (i) from (ii), we get
\( 3x + y=7 \)
\( _{(-)}2x +_{(-)} y = _{(-)}6 \)

---
\( x = 1 \)
Now, substituting the value of \( x \) in equation (i), we get
\( 2x + y = 6 \)
or \( 2(1) + y = 6 \)
or \( 2 + y = 6 \)
or \( y = 6 - 2 \)
or \( y = 4 \)
Therefore, the values of \( x \) and \( y \) are
\( x = 1 \) and \( y = 4 \)
   
6. हल गर्नुहोस् (Solve): \( \dfrac{x}{2} + \dfrac{2}{x} = 2 \) [2A]
 
The solution is
\( \dfrac{x}{2} + \dfrac{2}{x} = 2 \)
or \( \dfrac{x \cdot x + 2 \cdot 2}{2x} = 2 \)
or \( \dfrac{x^2 + 4}{2x} = 2 \)
or \( x^2 + 4 = 2 \cdot 2x \)
or \( x^2 + 4 = 4x \)
or \( x^2 - 4x + 4 = 0 \)
or \( (x - 2)^2 = 0 \)
or \( x - 2 = 0 \)
Therefore, the value of \( x \) is
or \( x = 2 \).
7. हल गर्नुहोस् (Solve): \( \dfrac{x - 4}{4} + \dfrac{6}{x + 1} = \dfrac{5}{4} \) [2A]
 
The solution is
\( \dfrac{x - 4}{4} + \dfrac{6}{x + 1} = \dfrac{5}{4} \)
Isolating the fraction with x in the denominator:
or \( \dfrac{6}{x + 1} = \dfrac{5}{4} - \dfrac{x - 4}{4} \)
or \( \dfrac{6}{x + 1} = \dfrac{5 - (x - 4)}{4} \)
or \( \dfrac{6}{x + 1} = \dfrac{5 - x + 4}{4} \)
or \( \dfrac{6}{x + 1} = \dfrac{9 - x}{4} \)
or \( 6 \cdot 4 = (9 - x)(x + 1) \)
or \( 24 = 9x + 9 - x^2 - x \)
or \( 24 = -x^2 + 8x + 9 \)
or \( x^2 - 8x + 24 - 9 = 0 \)
or \( x^2 - 8x + 15 = 0 \)
or \( x^2 - (5x + 3x) + 15 = 0 \)
or \( x^2 - 5x - 3x + 15 = 0 \)
or \( x(x - 5) - 3(x - 5) = 0 \)
or \( (x - 5)(x - 3) = 0 \)
Therefore, the values of \( x \) are
or \( x = 5 \) and \( x = 3 \).
8. लेखाचित्रद्वारा हल गर्नुहोस् (Solve graphically):
   \( 2x - y = 5 \) and \( x - y = 1 \) [2A]
 
  
    
  Given lines are:
\( 2x - y = 5 \)
\( y = 2x - 5 \)) (i)
                               
\( x \)	4	2	3
\( y \)	3	-1	1
Next
\( x - y = 1 \)
\( y = x - 1 \) (ii)
                               
\( x \)	0	1	3
\( y \)	-1	0	2
The solution is \( x = 4 \) and \( y = 3 \)    
  
 
 
 
9. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by graphical method):
   \( x + y = 6 \), \( 2x - y = 9 \) [2A]
Given lines are:
\( x + y = 6 \)
\( y = -x + 6 \) (i)

\( x \)	0	3	6
\( y \)	6	3	0
Next
\( 2x - y = 9 \)
\( y = 2x - 9 \) (ii)

\( x \)	3	4	5
\( y \)	-3	-1	1
The solution is \( x = 5 \) and \( y = 1 \)
10. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by graphical method):
    \( 5x - 3y = 5 \), \( -3x + 2y = -2 \) [2A]
Given lines are:
\( 5x - 3y = 5 \)
\( y = \frac{5x-5}{3} \) (i)

\( x \)	1	4	-5
\( y \)	0	5	-10
Next
\( -3x + 2y = -2 \)
\( y = \frac{3x-2}{2}\) (ii)

\( x \)	0	2	4
\( y \)	-1	2	5
The solution is \( x = 4 \) and \( y = 5 \)
11. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by graphical method):
    \( x + y = 5 \), \( x - y = 3 \) [2A]
Given lines are:
\( x + y = 5 \)
\( y = -x + 5 \) (i)

\( x \)	0	2	5
\( y \)	5	3	0
Next
\( x - y = 3 \)
\( y = x - 3 \) (ii)

\( x \)	0	3	4
\( y \)	-3	0	1
The solution is \( x = 4 \) and \( y = 1 \)
12. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by graphical method):
    \( x + y = 6 \), \( y = x - 4 \) [2A]
Given lines are:
\( x + y = 6 \)
\( y = -x + 6 \) (i)

\( x \)	0	3	6
\( y \)	6	3	0
Next
\( y = x - 4 \)
\( y = x - 4 \) (ii)

\( x \)	0	4	5
\( y \)	-4	0	1
The solution is \( x = 5 \) and \( y = 1 \)
13. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by graphical method):
    \( 2x - 1 = y \), \( 3x - 2y = 0 \) [2A]
Given lines are:
\( 2x - 1 = y \)
\( y = 2x - 1 \) (i)

\( x \)	0	1	2
\( y \)	-1	1	3
Next
\( 3x - 2y = 0 \)
\( y = \frac{3}{2}x \) (ii)

\( x \)	0	2	-2
\( y \)	0	3	-3
The solution is \( x = 2 \) and \( y = 3 \)
14. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by using graph):
    \( 5x + 7y = 1 \) and \( x + 4y = -5 \) [2A]
Given lines are:
\( 5x + 7y = 1 \)
\( y = \frac{1-5x}{7} \) (i)

\( x \)	3	-4
\( y \)	-2	3
Next
\( x + 4y = -5 \)
\( x = -5-4y \) (ii)

\( x \)	-5	-1	3
\( y \)	0	-1	-2
The solution is \( x = 3 \) and \( y = -2 \)
15. लेखाचित्र विधिबाट हल गर्नुहोस् (Solve by graphical method):
    \( 2x + y = 2 \), \( x - y = -5 \) [2A]
Given lines are:
\( 2x + y = 2 \)
\( y = -2x + 2 \) (i)

\( x \)	0	1	-1
\( y \)	2	0	4
Next
\( x - y = -5 \)
\( y = x + 5 \) (ii)

\( x \)	-2	0	2
\( y \)	3	5	7
The solution is \( x = -1 \) and \( y = 4 \)
16. बाबुको उमेर छोराको उमेरको ३ गुणा छ। तिनीहरूको उमेरको योगफल ४० वर्ष भए तिनीहरूको उमेर पत्ता लगाउनुहोस्।
    Father's age is three times the son's age. If the sum of their ages is 40 years, find their ages. [2A]
 
The solution is
Let the son's age be \( x \) years.
The father's age is \( y \) years.
Then
First Condition
Father's age is three times the son's age.
\(y=3x\)(i)
Second Condition
Sum of their ages is 40 years
\(x+y=40\)(ii)
Solving (i) and (ii), we get
\(x+y=40\)
or\(x+3x=40\)
or \( 4x = 40 \)
or \( x = 10 \)
\(y=3x = 3 \times 10 = 30 \)
Therefore,
The son's age \( x =10\) years.
The father's age is \( y=30 \) years.
17. ६ ओटा कलम र ३ ओटा पेन्सिलको संयुक्त मूल्य रु. ६० छ। ५ ओटा कलम र २ ओटा पेन्सिलको संयुक्त मूल्य रु. ४८ छ। प्रत्येक कलम र प्रत्येक पेन्सिलको मूल्य कति पर्छ ?
    The combined price of 6 pens and 3 pencils is Rs. 60. The combined price of 5 pens and 2 pencils is Rs. 48. What is the price of each pen and a pencil? [2HA]
 
The solution is
Let the price of a pen be \( x \) (Rs.)
Let the price of a pencil be \( y \) (Rs.)
Then
First Condition (6 pens and 3 pencils cost Rs. 60):
\( 6x + 3y = 60 \)
or\( 2x + y = 20 \)(i)
Second Condition (5 pens and 2 pencils cost Rs. 48)
\( 5x + 2y = 48 \)(ii)
To solve (i) and (ii), we multiply (i) by (2), and subtract from (ii), then we get
\(5x + 2y = 48 \)
\( _{(-)}4x +_{(-)} 2y = _{(-)}40 \)

---
\( x = 8 \)
Substituting \( x=8 \) back into (i), we get
\( 2x + y = 20 \)
or\( 2(8) + y = 20 \)
or\(16 + y = 20 \)
or\(y = 4 \)
Therefore,
  The price of each pen \( x = \) Rs. 8.
The price of each pencil \( y = \) Rs. 4.
18. यदि दुईओटा सङ्ख्याहरूको योगफल २५ र फरक १५ छ भने ती सङ्ख्याहरू पत्ता लगाउनुहोस्।
    If the sum of two numbers is 25 and their difference is 15, find the numbers. [2HA]
 
The solution is
Let the two numbers be \( x \) and \( y \).
Then
First Condition (The sum of two numbers is 25):
\( x + y = 25 \)(i)
Second Condition (Their difference is 15):
\( x - y = 15 \)(ii)
To solve (i) and (ii), we add equation (i) and (ii):
\( x + y = 25 \)
\( x - y = 15 \)
  
---
\( 2x = 40 \)
or \( x = 20 \)
   Substituting \( x=20 \) back into (i), we get
\( x + y = 25 \)
or\( 20 + y = 25 \)
or\(y = 25 - 20 \)
  or\(y = 5 \)
Therefore,
  The first number \( x = 20 \).
The second number \( y = 5 \).
19. दुईओटा सङ्ख्याहरूको अन्तर २८ छ। यदि ठूलो सङ्ख्या सानो सङ्ख्याको ३ गुणा छ भने उक्त सङ्ख्याहरू पत्ता लगाउनुहोस्।
    The difference of two numbers is 28. If the larger number is 3 times the smaller, find the numbers. [2A]
 
The solution is
Let the larger number be \( x \).
Let the smaller number be \( y \).
Then
First Condition (The difference of two numbers is 28):
\( x - y = 28 \)(i)
Second Condition (The larger number is 3 times the smaller):
\( x = 3y \)(ii)
To solve (i) and (ii), we substitute (ii) into (i):
\( x - y = 28 \)
or \( 3y - y = 28 \)
or \( 2y = 28 \)
or \( y = 14 \)
   Substituting \( y=14 \) back into (ii), we get
\( x = 3y \)
or\( x = 3 \times 14 \)
or\( x = 42 \)
Therefore,
  The larger number \( x = 42 \).
The smaller number \( y = 14 \).
20. दुईओटा सङ्ख्याहरू ३:५ को अनुपातमा छन्। यदि तिनीहरूको योगफल ८० भए ती सङ्ख्याहरू पत्ता लगाउनुहोस्।
    Two numbers are in the ratio 3:5. If the sum is 80, find the numbers. [2A]
 
The solution is
Let the two numbers be \( 3x \) and \( 5x \).
Condition (The sum is 80):
\( 3x + 5x = 80 \)
or \( 8x = 80 \)
or \( x = 10 \)
Therefore,
  The first number \( 3x = 30 \).
The second number \( 5x = 50 \).
21. दुई सङ्ख्याहरूको योग १७ र अन्तर ३ भए ती सङ्ख्याहरू पत्ता लगाउनुहोस्।
    If the sum of two numbers is 17 and their difference is 3, find the numbers. [2A]
 
The solution is
Let the two numbers be \( x \) and \( y \).
Then
First Condition (The sum of two numbers is 17):
\( x + y = 17 \)(i)
Second Condition (Their difference is 3):
\( x - y = 3 \)(ii)
To solve (i) and (ii), we add equation (i) and (ii):
\( x + y = 17 \)
\( x - y = 3 \)
  
---
\( 2x = 20 \)
or \( x = 10 \)
   Substituting \( x=10 \) back into (i), we get
\( x + y = 17 \)
or\( 10 + y = 17 \)
or\(y = 17 - 10 \)
  or\(y = 7 \)
Therefore,
  The first number \( x = 10 \).
The second number \( y = 7 \).

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