Show the above information in a Venn-diagram. [Reference: n(U) = 80; n(I) : n(A) = 3 : 2; both = 10; other kinds = 10]

n(I \cup A) = 80 - 10 = 70 n(I) + n(A) - n(I \cap A) = 70 n(I) + n(A) = 70 + 10 = 80 ┌─────────────────────────────────────────────┐ │ U = 80 │ │ ╭───────────╮ ╭───────────╮ │ │ │ I │ │ A │ │ │ │ only │\cap │ only │ │ │ │ 38 │10│ 22 │ │ │ ╰───────────╯ ╰───────────╯ │ │ 10 │ └─────────────────────────────────────────────┘

A farmer has taken Rs.5,00,000 loan from Agriculture development bank for his poultry farm. He aimed to pay back the loan in 2 years at the rate of 8% per annum. Bank calculated the interest compounded semiannually. How many times the compound interest semiannually is calculated in 2 years? Write it.

4 times. Semiannual means twice a year, so in 2 years interest is calculated 2 \times 2 = 4 times. (Each time the rate is 8/2 = 4%.)

What does P_T represent in the formula P_T = P_{0}(1 + R/100)^T? Write it.

P_T denotes the final value after time T — that is, the value or population reached after growth. (Other symbols in the formula: P_{0} = initial value, R = annual growth rate in percent, T = time.)

According to Nepal Rastra Bank's exchange rate, the buying and selling rate of 1 Australian dollar were NRs.93.20 and NRs.93.59 respectively in a certain day. How much Nepali currency can be exchanged with Australian dollars 5,000? Find it.

Here the customer is handing over dollars and taking Nepali rupees, that is, the bank is buying dollars. So the buying rate NRs. 93.20 is used. Amount = 5000 \times 93.20 = NRs. 4,66,000 Answer: 5,000 Australian dollars can be exchanged for NRs. 4,66,000.

The diagram along side is a square based pyramid. Its slant height is 20 cm. The area of the triangular surfaces of the pyramid is 1280 square cm. Write the relation among triangular surfaces of square based pyramid.

The four triangular faces of a pyramid with a square base are congruent (equal). That is: \triangle_{1} = \triangle_{2} = \triangle_{3} = \triangle_{4} and the total lateral surface area = 4 \times area of one triangular face Reason: The four sides of the base are equal and the slant heights of the four triangles are also the same, therefore all four areas are equal.

Find the ratio of total surface area and area of triangular surfaces of that pyramid. [side of the base 32 cm, area of the triangular faces 1280 sq. cm]

Area of the base = a^{2} = 32^{2} = 1024 sq. cm Area of the triangular (slant) faces = 1280 sq. cm Total surface area = 1024 + 1280 = 2304 sq. cm Ratio = 2304 : 1280 = 2304/64 : 1280/64 (dividing both by 64) = 36 : 20 = 9 : 5 Answer: the ratio of the total surface area to the area of the triangular faces is 9 : 5.

The given figure is a combined solid made up of a cylinder and cone. Diameter of base is 14 cm, height of cone is 24 cm and total height of solid object is 54 cm. Write down the formula to find the volume of the cone.

V = ⅓ \pi r^{2} h where r = radius of the cone's base, h = the cone's vertical height. (That is, the volume of a cone is one-third of the volume of a cylinder with the same base and the same height.)

If 'a', 'm' and 'b' are in arithmetic sequence, what is the value of arithmetic mean 'm'? Write it.

m = (a + b)/2 Reason: In an arithmetic progression the difference between consecutive terms 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 PT \parallel QR and PQ \parallel SR. Write the relation between the areas of a parallelogram and a triangle on the same base and between the same parallel lines. Write the relation between the area of parallelogram and triangle standing on the same base and between the same parallel lines.

Area of parallelogram = 2 \times area of triangle That is, the area of the triangle is half of the area of the parallelogram. (Theorem: A triangle on the same base and between the same parallel lines is half the area of the parallelogram — because both have the same base and the same height.)

In the given figure, O is the centre of the circle. Where \angleWXY = x^\circ and \angleWZY = 3x^\circ are given. What is the sum of the opposite angles of a cyclic quadrilateral? Write it. What is the sum of the opposite angles of a cyclic quadrilateral? Write it.

The sum of the opposite angles of a cyclic quadrilateral is 180^\circ — that is, they are supplementary. In this figure: \angleWXY + \angleWZY = 180^\circ and \angleXWZ + \angleXYZ = 180^\circ

Find the value of x from the given figure. [In the cyclic quadrilateral XWZY, \angleWXY = x^\circ and \angleWZY = 3x^\circ ]

\angleWXY + \angleWZY = 180^\circ x + 3x = 180^\circ 4x = 180^\circ x = 45^\circ

Why the area of parallelogram PQRS and triangle QTU are equal? Give reason.

Because the triangle's base is exactly twice the parallelogram's base, and both lie between the same parallels and so have the same height. Let the parallelogram have base QR = b and height h. Area (PQRS) = base \times height = b \times h The triangle's base QU = 2b and its height = h, since they lie between the same parallels. Area (\triangleQTU) = ½ \times base \times height = ½ \times 2b \times h = b \times h ∴ the two areas are equal. Here b = 6 cm and h = 3.46 cm, so both areas are about 20.78 sq. cm.

In the figure along side, from the top of the 45 meter high view tower (AB) an angle of 60^\circ is formed when looking at the roof of a house at a distance of 15 meter to the east of the tower. What angle is formed when the top of the view tower is seen from the roof of the house? Write it. What is the angle formed when observed from the roof of the house to the top of the view tower? Write it.

60^\circ . Reason: AF (the horizontal line from the top of the tower) and EC (the horizontal line from the roof of the house) are both horizontal, therefore they are parallel to each other, and AC is a transversal line. \angleFAC (an alternate interior angle) and \angleACE (an alternate interior angle) are alternate interior angles, therefore equal. ∴ The angle of elevation from the house roof to the top of the tower = 60^\circ

What is the height of the part AE of the tower? Find it. [tower AB = 45 metres, BD = 15 metres, \angleFAC = 60^\circ (angle of depression)]

\angleAEC = 90^\circ \tan \angleACE = AE / EC \tan 60^\circ = AE / 15 \sqrt{3} = AE / 15 AE = 15\sqrt{3} = 15 \times 1.732

What is the height of the house? Find it. [tower AB = 45 metres; AE = 15\sqrt{3} \approx 25.98 metres]

EB = AB - AE = 45 - 15\sqrt{3} = 45 - 25.98

What distance does the observer have to come down from the top of the view tower to observe the roof of the house such that the angle of depression is 45^\circ? Find it. [horizontal distance 15 metres; the present vertical difference from the top to the roof AE = 15\sqrt{3} \approx 25.98 metres]

\tan 45^\circ = A′E′ / 15 1 = A′E′ / 15 = 25.98 - 15

Median of the data given below is 24. Class Interval │ 0-10 │ 10-20 │ 20-30 │ 30-40 │ 40-50 ─────────────────────────┼──────┼───────┼───────┼───────┼─────── Frequency │ 4 │ m+2 │ 10 │ 9 │ 5 What is the class interval of median? Write it.

Median class = 20 - 30 Reason: The value of the median is given as 24, and 24 lies exactly in the interval 20 to 30. The class interval in which the median lies is the median class.

Is the model class and the first quartile class in the given data same? If it is, then justify it. [frequencies: 4, 12, 10, 9, 5; total 40]

Yes, both are the same — 10 – 20 class interval. Verification: a. Mode class: Among the frequencies 4, 12, 10, 9, 5 the greatest is 12, which is of the 10 – 20 class interval. Therefore mode class = 10 - 20. b. First quartile (Q_{1}) class interval: Cumulative frequency: 0 – 10 : 4 10 – 20 : 16 20 – 30 : 26 30 – 40 : 35 40 – 50 : 40 Q_{1} position = N/4 = 40/4 = 10th term The 10th term falls between cumulative frequencies 4 and 16, that is in the 10 – 20 class interval. Therefore Q_{1} class = 10 - 20. Conclusion: Since the mode class and the Q_{1} class are both 10 – 20, they are the same.

Find the ratio of the probability of getting balls of the same colour to the different colour. [5 green and 7 blue balls; with replacement]

Probability of same color: P(same color) = P(GG) + P(BB) = 25/144 + 49/144 = 74/144 Probability of different color: P(different color) = P(GB) + P(BG) = 35/144 + 35/144 = 70/144 Ratio: P(same color) : P(different color) = 74/144 : 70/144 = 74 : 70 = 37 : 35 (dividing both by 2) Answer: The ratio is 37 : 35. (Check: 74/144 + 70/144 = 144/144 = 1 ✓)

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

49 questions worth 75 marks. There are 27 of 1 mark, 18 of 2 marks and 4 of 3 marks.

Secondary Education Examination · Class 10

SEE 2082 Mathematics question paper, Sudurpashchim Province

SEE 2082 Mathematics (Sudurpashchim Province, RE-1031'SP') has 49 questions worth 75 marks, with 3 hours to answer them. It has 27 of 1 mark, 18 of 2 marks and 4 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'SP'

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Questions

Answers are shown for the 27 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 among 80 youths of a community, it is found that the ratio of the number of youths who liked iPhone and Android phone was 3:2. Among them 10 liked both types and 10 liked other types of phone. If I and A denote the sets of youths who liked iPhone and Android Phone respectively, write the value of $n(I \cap A)$ .
एउटा समुदायमा रहेका 80 जना युवाहरूको समूहमा गरिएको सर्वेक्षणमा आइफोन र एन्ड्रोइड फोन मन पराउने युवाहरूको अनुपात 3:2 पाइयो । तीमध्ये 10 जनाले दुवै प्रकारको र 10 जनाले अन्य प्रकारको फोन मन पराएको पाइयो । (a) यदि I र A ले क्रमशः आइफोन र एन्ड्रोइड फोन मन पराउने युवाहरूको समूह जनाएको भए $n(I \cap A)$

Questions learners ask

How many questions and marks are in the SEE 2082 Mathematics (Sudurpashchim Province, RE-1031'SP') paper?: 49 questions worth 75 marks. There are 27 of 1 mark, 18 of 2 marks and 4 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 Sudurpashchim Province, code RE-1031'SP', typed up from the original and checked by the GyanMul team.
Are the answers included?: Answers to the 27 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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