Illustrate the above given information in a Venn-diagram. [Reference: n(U) = 80; M only : S only = 4 : 3; both = 5; neither = 12]

┌─────────────────────────────────────────────┐ │ U = 80 │ │ ╭───────────╮ ╭───────────╮ │ │ │ M │ │ S │ │ │ │ only │\cap │ only │ │ │ │ 36 │ 5│ 27 │ │ │ ╰───────────╯ ╰───────────╯ │ │ 12 │ └─────────────────────────────────────────────┘

Compare the number of students who take extra class of at least one subject and who take extra class of at most one subject in ratio. [n(U) = 80; M only = 36; S only = 27; both = 5; neither = 12]

M \cup S = 80 - 12 = 68= 63 + 12 = 75

Find the principal. [compound amount after 2 years Rs. 14,580; compound rate 8% per annum]

A_{2} = P(1 + R/100)^{2} 14580 = P(1 + 8/100)^{2} 14580 = P \times (1.08)^{2} 14580 = P \times 1.1664 P = 14580 / 1.1664

Write the formula to find compound depreciation.

D = P[1 - (1 - R/100)^T] where P = original price of the item, R = annual rate of depreciation (%), T = time (in years), D = compound depreciation. (After-depreciation value V_T = P(1 - R/100)^T, and D = P - V_T can also be written.)

A businessman bought a truck for Rs.40,00,000. Its value was depreciated at the rate of 15% p.a. Find the selling price of the truck after 3 years.

V_T = P(1 - R/100)^T = 4000000 \times (1 - 15/100)^{3} = 4000000 \times (0.85)^{3} = 4000000 \times 0.614125

Define currency exchange rate.

The rate used when exchanging one country’s currency for another country’s currency is the exchange rate — that is, the ratio showing how much domestic currency is required to buy or sell one unit of foreign currency. In Nepal this rate is published daily by Nepal Rastra Bank, and it consists of two rates: the buying rate (the rate at which the bank buys foreign currency) and the selling rate (the rate at which the bank sells foreign currency).

Pemba went to the bank to exchange American dollars in order to go abroad. According to currency exchange rate on that day, the buying rate and the selling rate of 1 American dollar are NRs.142.84 and NRs.143.44 respectively. How many American dollar will Pemba get to exchange NRs.2,15,160? Find it.

Here Pemba is handing over rupees and buying dollars, that is, the bank is selling dollars. So the selling rate NRs. 143.44 is used. Dollars = total amount ÷ selling rate = 215160 / 143.44 = 1,500 Answer: Pemba receives 1,500 — one thousand five hundred American dollars.

Due to different circumstances, Pemba can't go abroad. Then how many Nepali rupees will he get back after exchanging American dollars on the same day? Find it. [amount exchanged 1,500; buying rate NPR 142.84, selling rate NPR 143.44]

Now Pemba is returning the dollars and receiving rupees, that is, the bank is buying the dollars. Therefore the buy rate Ne. ru. 142.84 is used. Rupees = 1500 \times 142.84 = Ne. ru. 2,14,260 Answer: Pemba receives Ne. ru. 2,14,260 back. (She had given 2,15,160 and received 2,14,260 — by exchanging on the same day she lost Ne. ru. 900. This is the difference between the buy and sell rates, which is the bank's income.)

How many triangular surfaces are there in a square based pyramid? Write it.

4. A square base has four edges, and from each edge to the apex a triangular face is formed. Therefore the lateral triangular faces are four (the base is square, separate).

A square based pyramid has slant height 10 cm and the length of a side of base is 12 cm. What is the total surface area of the pyramid? Find it.

= ½ \times (4 \times 12) \times 10 = ½ \times 48 \times 10

If it is melted to form a cylindrical wire of radius 3.5 mm, what is the length of the wire? Find it. [volume of the solid 5852 cu. cm]

\pi r^{2}L = 5852 (22/7) \times (0.35)^{2} \times L = 5852 (22/7) \times 0.1225 \times L = 5852 0.385 \times L = 5852 L = 5852 / 0.385

What is the arithmetic mean between two numbers a and b? Write it.

Arithmetic mean (A.M.) = (a + b)/2 Reason: When a, m, b are in arithmetic progression the common difference is equal — m - a = b - m ⟹ 2m = a + b ⟹ m = (a + b)/2.

Write the roots of quadratic equation ax^{2} + bx + c = 0; a \ne 0.

x = [-b \pm \sqrt{b^{2} - 4ac}] / 2a

In the given figure AD \parallel BC and BE \parallel CD. Write the relation between the areas of a triangle and a parallelogram on the same base and between the same parallel lines. Write the relation between the area of triangle and parallelogram standing on the same base and between the same parallel lines.

Area of the triangle = ½ \times area of the parallelogram That is, the area of the parallelogram is twice the area of the triangle. (Theorem: a triangle on the same base as a parallelogram and between the same parallels has half its area, because the two share a base and a height.)

In the given figure, O is the centre of the circle. Where \anglePSR = x^\circ and \anglePQR = 2x^\circ are given. Write the relation between the opposite angles of a cyclic quadrilateral. Write the relation between the opposite angles of a cyclic quadrilateral.

Opposite angles of a cyclic quadrilateral are supplementary — that is, their sum is 180^\circ. In this figure: \anglePSR + \anglePQR = 180^\circ and \angleSPQ + \angleSRQ = 180^\circ.

Find the value of x from the given figure. [In the cyclic quadrilateral PSRQ, \anglePSR = x^\circ and \anglePQR = 2x^\circ ]

\anglePSR + \anglePQR = 180^\circ x + 2x = 180^\circ 3x = 180^\circ x = 60^\circ

Why are the area of \triangleNQR and rectangle MNOP equal? Give reason.

Because the triangle's base is twice the rectangle's base, while both lie between the same parallels and so have the same height. Let the rectangle have base b and height h. Area (rectangle) = b \times h The triangle's base = 2b and its height = h, since they lie between the same parallels. Area (triangle) = ½ \times 2b \times h = b \times h ∴ the two areas are equal. Here b = 6 cm and h = 4 cm, so both areas are 24 sq. cm.

Give an example of angle of depression.

An angle of depression is the angle formed between the line of sight and the horizontal line when looking at an object that lies below the horizontal line. Example: A person standing on the roof of a house looks down at a car standing on the road; the angle formed between his line of sight and the horizontal line is the angle of depression. (Other examples: the angle formed when a passenger seated in an airplane looks down at the runway on the ground, or when looking down at a boat on a river from a bridge.)

A man of height 1.75 m is flying a kite. The length of string on the air is 105\sqrt{3} m and string formed an angle 60^\circ with horizon. Draw a figure according to given context.

╱│ ╱ │ ╱ │ ╱ 60^\circ │ C ──────────╱──────────────┤ B │ │ \cdot \angleABC = 90^\circ

What is the height of the kite from the ground level? Find it. [length of the thread 105\sqrt{3} m, angle with the horizontal 60^\circ, height of the man 1.75 m]

\sin 60^\circ = AB / CA \sqrt{3}/2 = AB / (105\sqrt{3}) AB = 105\sqrt{3} \times \sqrt{3}/2 = 105 \times 3 / 2 = 157.5 + 1.75

If the angle formed by string with horizon decreases from 60^\circ to 45^\circ then what is the height of the kite from the ground level now? Find it. [the length of the thread stays 105\sqrt{3} m; height of the man 1.75 m]

\sin 45^\circ = AB′ / CA 1/\sqrt{2} = AB′ / (105\sqrt{3}) AB′ = 105\sqrt{3} / \sqrt{2} = 105\sqrt{3/2} = 105\sqrt{6} / 2 = 105 \times 2.449489 / 2 = 128.598 + 1.75

Write the formula to calculate mode of the continuous series.

Mode = L + [(f_{1} - f_{0}) / (2f_{1} - f_{0} - f_{2})] \times h where L = lower limit of the modal class f_{1} = frequency of the modal class (the highest) f_{0} = frequency of the class immediately before the modal class f_{2} = frequency of the class immediately after the modal class h = class interval

Find the average height of the students above modal class. [modal class 130-140; 12 in 140-150, 4 in 150-160]

───────────┼───────────────┼───────────────┼─────── 140 – 150 │ 145 │ 12 │ 1740 150 – 160 │ 155 │ 4 │ 620 ───────────┴───────────────┼───────────────┼─────── = 2360 / 16 = 147.5

Two events G and B are not mutually exclusive events, then write the formula of P(G \cup B).

P(G \cup B) = P(G) + P(B) - P(G \cap B)

Find the probability of selecting same gender student. [27 girls, 23 boys; two of them, without repetition]

P(GG) = 27/50 \times 26/49 = 702/2450 P(BB) = 23/50 \times 22/49 = 506/2450 = 1208/2450 = 604/1225 \approx 0.493

By how much is the probability of selecting both girls less than the total probability? Find it. [27 girls, 23 boys; two of them, without repetition]

= (1225 - 351) / 1225 = 874/1225 \approx 0.713

How many questions and marks are in the SEE 2082 Mathematics (Lumbini Province, RE-1031'LP') paper?

49 questions worth 75 marks. There are 28 of 1 mark, 16 of 2 marks and 5 of 3 marks.

Secondary Education Examination · Class 10

SEE 2082 Mathematics question paper, Lumbini Province

SEE 2082 Mathematics (Lumbini Province, RE-1031'LP') has 49 questions worth 75 marks, with 3 hours to answer them. It has 28 of 1 mark, 16 of 2 marks and 5 of 3 marks.

The SEE examination is conducted by the National Examinations Board (NEB), Nepal. This paper was typed up from the original and checked by the GyanMul team.

Every question is given in Nepali and English, as the paper prints it.

49 questions
75 marks
3 hours
Code RE-1031'LP'

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Questions

Answers are shown for the 28 one-mark questions. They are GyanMul's model answers: the NEB does not publish an answer key, so these are not the board's answers.

Question 11 mark
In a survey of 80 students of class ten, the following information was found. The ratio of students who take extra class of only Maths subject and only Science subject is 4:3. 5 students take extra class of both subjects. 12 students don't take extra class of any subject. If M and S denote the set of students who take extra class of Maths and Science respectively, write the set of students who did not take extra class in any of these two subjects in cardinality notation.
कक्षा दशका 80 जना विद्यार्थीहरूमा गरिएको सर्वेक्षणबाट निम्न जानकारी प्राप्त भयो : $\cdot$ गणित विषय मात्र र विज्ञान विषय मात्रको अतिरिक्त कक्षा लिने विद्यार्थीको अनुपात 4:3 छ । $\cdot$ दुवै विषयको अतिरिक्त कक्षा लिने 5 जना विद्यार्थीहरू छन् । $\cdot$ कुनै पनि विषयको अतिरिक्त कक्षा नलिने 12 जना विद्यार्थीहरू छन् । (a) यदि M र S ले गणित र विज्ञान विषयको अतिरिक्त कक्षा लिने विद्यार्थीहरूको समूहलाई जनाउँछ भने यी दुवै विषयमध्ये कुनै पनि विषयमा अतिरिक्त कक्षा नलिने विद्यार्थीको समूहलाई गणनात्मकता सङ्केतमा लेख्नुहोस् ।
Answer:

Questions learners ask

How many questions and marks are in the SEE 2082 Mathematics (Lumbini Province, RE-1031'LP') paper?: 49 questions worth 75 marks. There are 28 of 1 mark, 16 of 2 marks and 5 of 3 marks.
How much time is given for this paper?: 3 hours (180 minutes).
Is this the real board paper?: Yes. It is the SEE 2082 paper printed for Lumbini Province, code RE-1031'LP', typed up from the original and checked by the GyanMul team.
Are the answers included?: Answers to the 28 one-mark questions are on this page (GyanMul's model answers; the NEB does not publish an answer key). Worked answers and the marking for every other question are in GyanMul Prep, free with an account.
Can I download this paper as a PDF?: Yes. Press "Print / Save as PDF" at the top of the page and choose "Save as PDF" in your browser's print window.

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