SEE 2081 (2025) Bagmati Province — अनिवार्य गणित

1

कुनै गाउँको १२० घरधुरीमा गरिएको सर्वेक्षणमा ७० घरधुरीले होमस्टेको व्यवसाय र ५० घरधुरीले कृषि व्यवसाय गर्छन्। ३० घरधुरीले अन्य काम गर्छन्। होमस्टे र कृषि व्यवसाय गर्ने घरधुरीहरूको समूहलाई क्रमशः H र A ले जनाइएका छन्।

In a survey of 120 households: 70 do homestay (H), 50 do agriculture (A), 30 do other work.

6 marks

a

होमस्टे र कृषिमध्ये कुनै पनि व्यवसाय नगर्नेको सङ्ख्यालाई गणनात्मकता सङ्केतमा लेख्नुहोस्।

Write the cardinality notation of the number of households who do not do any of the homestay and agriculture business.

[1 mark]

✓ Answer

Households doing other work = 30. These do neither homestay nor agriculture.

In set notation, this is the complement of H∪A:

n(H∪A)' = 30

b

माथिको जानकारीलाई भेनचित्रमा प्रस्तुत गर्नुहोस्।

Present the above information in a Venn-diagram.

[1 mark]

✓ Venn Diagram

n(U) = 120, n(H) = 70, n(A) = 50, n(H∪A)' = 30

n(H∪A) = 120 − 30 = 90

n(H∩A) = n(H) + n(A) − n(H∪A) = 70 + 50 − 90 = 30

Only H = 70 − 30 = 40 | Only A = 50 − 30 = 20

Venn diagram: H = Homestay, A = Agriculture, U = 120 households

c

एउटा मात्र व्यवसाय गर्ने घरधुरीको सङ्ख्या निकाल्नुहोस्।

Find the number of households doing only one business.

[3 marks]

✓ Solution

Given: n(U) = 120, n(H) = 70, n(A) = 50, n(H∪A)' = 30

n(H∪A) = 120 − 30 = 90

n(H∩A) = n(H) + n(A) − n(H∪A) = 70 + 50 − 90 = 30

Only H = 70 − 30 = 40

Only A = 50 − 30 = 20

Only one business = 40 + 20 = 60

∴ Number of households doing only one business = 60

d

अन्य काम गर्ने १० घरधुरीले होमस्टे व्यवसाय सुरु गर्न थाले भने होमस्टे र कृषि व्यवसाय गर्ने घरधुरीको अनुपात निकाल्नुहोस्।

If 10 households doing other work start homestay business, find the ratio of homestay and agriculture business households.

[1 mark]

✓ Solution

New homestay households = 70 + 10 = 80

Agriculture households remain = 50

Ratio = 80 : 50 = 8 : 5

∴ Ratio of homestay to agriculture = 8 : 5

2

रमेशले बैंकबाट रु.२,००,००० प्रति वर्ष ७% का दरले वार्षिक चक्रीय व्याजमा ऋण लिएछन्। केही समयपछि साँवा र व्याज सहित रु.२,२८,९८० बैंकमा बुझाएछन्।

Ramesh borrowed Rs.2,00,000 from a bank at annual compound interest rate of 7% per annum. After some time, he repaid Rs.2,28,980 including principal and interest.

4 marks

a

रमेशले व्याजमात्र कति बुझाएछन्? पत्ता लगाउनुहोस्।

How much interest has Ramesh paid? Find it.

[1 mark]

✓ Solution

Total paid = Rs.2,28,980 | Principal = Rs.2,00,000

Interest = 2,28,980 − 2,00,000 = Rs.28,980

∴ Interest paid by Ramesh = Rs.28,980

b

रमेशले उक्त ऋण कति वर्षको लागि प्रयोग गरेछन्? पत्ता लगाउनुहोस्।

For how many years has Ramesh used the loan? Find it.

[2 marks]

✓ Solution

A = P(1 + R/100)ᵀ → 2,28,980 = 2,00,000 × (1.07)ᵀ

(1.07)ᵀ = 2,28,980 / 2,00,000 = 1.1449

Check: (1.07)² = 1.1449 ✓

∴ Ramesh used the loan for 2 years

c

यदि व्याजदर १% ले घटायो भने रमेशले तिर्नुपर्ने मिश्रधन कति हुन्छ? पत्ता लगाउनुहोस्।

If the interest rate is reduced by 1%, how much amount would be paid by Ramesh? Find it.

[1 mark]

✓ Solution

New rate = 7 − 1 = 6%, T = 2 years, P = Rs.2,00,000

A = 2,00,000 × (1 + 6/100)² = 2,00,000 × (1.06)² = 2,00,000 × 1.1236

∴ New amount = Rs.2,24,720

3

गाउँ A को हालको जनसङ्ख्या ४५०० र गाउँ B को ५००० छ। गाउँ A को वार्षिक जनसङ्ख्या वृद्धिदर २% छ।

Present population of village A = 4500, village B = 5000. Annual growth rate of village A = 2%.

4 marks

a

T वर्षपछिको जनसङ्ख्या Pᵀ = P(1 + R/100)ᵀ मा P ले के जनाउँछ? लेख्नुहोस्।

What does P denote in Pᵀ = P(1 + R/100)ᵀ? Write it.

[1 mark]

✓ Answer

P denotes the present (current) population of the village — the initial population before growth is applied.

b

यदि गाउँ A मा १ वर्ष पछि २०० जना बसाईँसराइ गरी थपिन आए भने, १ वर्षपछि जनसङ्ख्या कति होला? पत्ता लगाउनुहोस्।

If 200 people are added by migration to village A after 1 year, what will the population be after 1 year?

[1 mark]

✓ Solution

Population after 1 year (natural growth) = 4500 × (1 + 2/100)¹ = 4500 × 1.02 = 4590

Adding migrants: 4590 + 200 = 4790

∴ Population of village A after 1 year = 4790

c

यदि गाउँ A कै वृद्धिदर बराबर गाउँ B को जनसङ्ख्या घटेमा २ वर्षपछि गाउँ B को जनसङ्ख्या कति हुन्छ? पत्ता लगाउनुहोस्।

If population of village B decreases by the same growth rate of village A, what will be village B's population after 2 years?

[2 marks]

✓ Solution

Decrease rate = 2%, P = 5000, T = 2 years

P₂ = P × (1 − R/100)² = 5000 × (1 − 0.02)² = 5000 × (0.98)²

= 5000 × 0.9604 = 4802

∴ Population of village B after 2 years = 4802

4

रीता अमेरिकाबाट नेपाल आउँदा साथमा $२५०० लिएर आएकी रहिछन्। नेपाल आउँदा खरिद दर $१ = ने.रु.१२७.३५ र विक्री दर $१ = ने.रु.१२७.९५ थियो।

Rita brought $2500 from America to Nepal. Buying rate: $1 = NRs.127.35; Selling rate: $1 = NRs.127.95.

5 marks

a

रीताले बैंकबाट नेपालमा रुपियाँ साट्दा कुन विनिमय दर प्रयोग हुन्छ? लेख्नुहोस्।

Which exchange rate is used to exchange Nepali currency by Rita from the bank? Write it.

[1 mark]

✓ Answer

Rita is selling USD to the bank (bank buys her dollars). So the buying rate ($1 = NRs.127.35) is used by the bank.

Rita receives: 2500 × 127.35 = NRs.3,18,375

b

रीताले उनीसँग भएको अमेरिकी डलरलाई नेपाली रुपैयाँमा साटी पुनः दुई दिन पछि नेपाली रुपैयाँलाई अमेरिकी डलरमा साट्दा उनलाई कति फाइदा वा घाटा हुन्छ? पत्ता लगाउनुहोस्।

What is the profit or loss for Rita if she exchanges USD to NRS and then exchanges NRS back to USD after two days?

[2 marks]

✓ Solution

Step 1: Rita sells $2500 to bank at buying rate (127.35)

NRs received = 2500 × 127.35 = NRs.3,18,375

Step 2: Rita buys back USD with NRs at selling rate (127.95)

USD obtained = 3,18,375 ÷ 127.95 = $2,488.27 (approx.)

Loss in USD = 2500 − 2488.27 = $11.73

Or: Loss in NRs = 2500 × (127.95 − 127.35) = 2500 × 0.60 = NRs.1,500

∴ Rita incurs a loss of NRs.1,500 (due to bank's buy-sell spread)

c

सोही दिन $१ = ने.रु.१२७.३५ र ने.रु.१६० = भा.रु.१०० थियो भने भा.रु.१,२०,००० सँग कति डलर साट्न सकिन्छ? निकाल्नुहोस्।

On the same day, $1 = NRs.127.35 and NRs.160 = IRs.100. How many dollars can be exchanged with IRs.1,20,000?

[2 marks]

✓ Solution

NRs.160 = IRs.100 → IRs.1,20,000 = ? NRs

NRs = (1,20,000 × 160) / 100 = 1,92,000 NRs

USD = 1,92,000 ÷ 127.35 = $1507.26 (approx.)

∴ IRs.1,20,000 can be exchanged for approximately $1507.26

5

संगैको चित्र, वर्ग आधार भएको एउटा पिरामिडको हो। त्यसको आधारको प्रत्येक भुजा ६ मिटर र ठाडो उचाइ ४ मिटर छ।

A square-based pyramid with base side = 6 m and vertical height = 4 m.

4 marks

Square-based pyramid: base = 6 m × 6 m, vertical height h = 4 m

a

पिरामिडमा भएको सबै त्रिभुजाकार सतहहरूको क्षेत्रफल पत्ता लगाउने सूत्र लेख्नुहोस्।

Write the formula for finding the area of all triangular surfaces of pyramid.

[1 mark]

✓ Formula

Lateral Surface Area = 4 × (1/2) × base × slant height = 2 × l × l'

Where l = base side length, l' = slant height = √(h² + (l/2)²)

b

उक्त पिरामिडको आयतन पत्ता लगाउनुहोस्।

Find the volume of the pyramid.

[1 mark]

✓ Solution

l = 6 m, h = 4 m

V = (1/3) × l² × h = (1/3) × 36 × 4 = 48 m³

∴ Volume = 48 m³

c

पिरामिडको पूरा सतहको क्षेत्रफल निकाल्नुहोस्।

Find the total surface area of the pyramid.

[2 marks]

✓ Solution

Slant height l' = √(h² + (l/2)²) = √(4² + 3²) = √(16 + 9) = √25 = 5 m

Lateral Surface Area = 4 × (1/2) × 6 × 5 = 4 × 15 = 60 m²

Base area = l² = 6² = 36 m²

Total Surface Area = 60 + 36 = 96 m²

∴ Total Surface Area = 96 m²

6

दिइएको ठोस वस्तु सोली र अर्धगोला मिली बनेको छ। सोली र अर्धगोलाका व्यासहरू बराबर छन् र ठोस वस्तुको जम्मा उचाइ १७ से.मी. र आधारको व्यास १० से.मी. छन्।

A solid object made of cone + hemisphere. Diameters are equal, total height = 17 cm, diameter = 10 cm.

4 marks

Cone + hemisphere: total height = 17 cm, diameter = 10 cm (r = 5 cm)

a

अर्धगोलाको उचाइ र अर्धव्यासको सम्बन्ध लेख्नुहोस्।

Write the relation between height and radius of hemisphere.

[1 mark]

✓ Answer

Height of hemisphere = radius of hemisphere → h = r

A hemisphere's height equals its radius because it is exactly half of a complete sphere.

b

उक्त ठोस वस्तुको पूरा सतहको क्षेत्रफल पत्ता लगाउनुहोस्।

Find the total surface area of the solid object.

[3 marks]

✓ Solution

Diameter = 10 cm → r = 5 cm

Height of hemisphere = r = 5 cm

Height of cone = total height − r = 17 − 5 = 12 cm

Slant height of cone: l = √(r² + h²) = √(25 + 144) = √169 = 13 cm

Curved Surface of cone = πrl = π × 5 × 13 = 65π

Curved Surface of hemisphere = 2πr² = 2π × 25 = 50π

Total Surface Area = 65π + 50π = 115π = 361.28 cm²

∴ Total Surface Area = 115π ≈ 361.28 cm²

c

उक्त ठोस वस्तुको सतहमा प्रति वर्ग सेन्टिमिटर ४० पैसाका दरले रङ लगाउँदा रु.१५० पर्याप्त हुन्छ? हिसाब गरी कारण दिनुहोस्।

Will Rs.150 be sufficient to color the surface at 40 paisa per sq.cm? Give reason with calculation.

[1 mark]

✓ Solution

Total surface area ≈ 361.28 cm²

Cost = 361.28 × 0.40 = Rs.144.51

144.51 < 150 ✓

✅ Yes, Rs.150 is sufficient. The cost is Rs.144.51 which is less than Rs.150.

7

एउटा आयताकार कोठाको लम्बाइ १२ फिट, चौडाइ ८ फिट र उचाइ ९ फिट छन्। उक्त कोठामा २.५ फिट × ३ फिटका दुईओटा भ्यालहरू र ६ फिट × २ फिटका दुईओटा ढोकाहरू छन्।

Rectangular room: length = 12 ft, breadth = 8 ft, height = 9 ft. Two windows (2.5×3 ft) and two doors (6×2 ft).

4 marks

a

चार भित्ता, भुईँ र सिलिङको जम्मा क्षेत्रफल निकाल्नुहोस्।

Find the total area of four walls, floor and ceiling.

[2 marks]

✓ Solution

l = 12 ft, b = 8 ft, h = 9 ft

Area of 4 walls = 2(l + b) × h = 2(12 + 8) × 9 = 2 × 20 × 9 = 360 sq.ft

Floor area = l × b = 12 × 8 = 96 sq.ft

Ceiling area = l × b = 12 × 8 = 96 sq.ft

Total = 360 + 96 + 96 = 552 sq.ft

∴ Total area of 4 walls + floor + ceiling = 552 sq.ft

b

भ्याल ढोका बाहेक चार भित्तामा प्रति वर्ग फिट रु.१७५ का दरले रङ लगाउँदा रु.५०,००० भन्दा कति कम वा बढी खर्च लाग्ला? पत्ता लगाउनुहोस्।

How much less or more is the cost of coloring the four walls (excluding doors/windows) at Rs.175/sq.ft compared to Rs.50,000?

[2 marks]

✓ Solution

Wall area = 360 sq.ft

Area of 2 windows = 2 × (2.5 × 3) = 15 sq.ft

Area of 2 doors = 2 × (6 × 2) = 24 sq.ft

Net paintable area = 360 − 15 − 24 = 321 sq.ft

Cost = 321 × 175 = Rs.56,175

Difference = 56,175 − 50,000 = Rs.6,175 more

∴ Cost is Rs.6,175 more than Rs.50,000

8

एउटा पसलले जाडोको बेलामा जुत्ता र चप्पल (जोडीमा) निम्नलिखित तालिका अनुसार बिक्री गरेको रहेछ। (जुत्ता: 2, 4, 8, 16, … | चप्पल: 3, 6, 9, 12, …)

A shop sells shoes (2,4,8,16,…) and slippers (3,6,9,12,…) in winter season.

5 marks

Day	1st	2nd	3rd	4th	5th
Shoes	2	4	8	16	…
Slippers	3	6	9	12	…

a

समानान्तरीय मध्यमा र गुणोत्तर मध्यमाको सम्बन्ध लेख्नुहोस्।

Write the relationship between arithmetic mean and geometric mean.

[1 mark]

✓ Answer

AM ≥ GM (Arithmetic Mean ≥ Geometric Mean)

For two positive numbers a and b: (a+b)/2 ≥ √(ab)

b

८ औँ दिनसम्म जम्मा चप्पलहरू कतिवटा बिक्री भएछन्? पत्ता लगाउनुहोस्।

How many slippers are sold up to the 8th day? Find it.

[2 marks]

✓ Solution

Slippers: 3, 6, 9, 12, … → AP with a = 3, d = 3, n = 8

Sₙ = (n/2)[2a + (n−1)d] = (8/2)[2×3 + 7×3] = 4[6 + 21] = 4 × 27 = 108

∴ Total slippers sold up to 8th day = 108 pairs

c

८ औँ दिनसम्म बिक्री भएका जम्मा जुत्ता र चप्पलको सङ्ख्या तुलना गर्नुहोस्।

Compare the total number of shoes and slippers sold up to the 8th day.

[2 marks]

✓ Solution

Shoes: 2, 4, 8, 16, … → GP with a = 2, r = 2, n = 8

Sₙ = a(rⁿ − 1)/(r−1) = 2(2⁸ − 1)/(2−1) = 2(256−1)/1 = 2 × 255 = 510

Slippers (from b) = 108

Difference = 510 − 108 = 402

∴ Total shoes (510) > Total slippers (108) by 402 pairs

9

एउटा आयताकार जग्गाको क्षेत्रफल र परिमिति क्रमशः १५० वर्ग.मि. र ५० मि. छन्।

A rectangular field has area = 150 sq.m and perimeter = 50 m.

5 marks

a

उक्त जग्गाको लम्बाइ र चौडाइ कति हुन्छ? वर्ग समीकरण बनाई पत्ता लगाउनुहोस्।

What is the length and breadth of the field? Find it by making quadratic equation.

[3 marks]

✓ Solution

Let length = l, breadth = b

Perimeter: 2(l + b) = 50 → l + b = 25 → b = 25 − l

Area: l × b = 150 → l(25 − l) = 150

25l − l² = 150 → l² − 25l + 150 = 0

Factoring: (l − 15)(l − 10) = 0 → l = 15 or l = 10

Taking l = 15 m (larger): b = 25 − 15 = 10 m

∴ Length = 15 m, Breadth = 10 m

b

उक्त जग्गाको लम्बाइ र चौडाइबाट कति बराबर भाग घटाउँदा क्षेत्रफल ८४ वर्ग मिटर हुन्छ? हल गर्नुहोस्।

How much equal part should be subtracted from the length and breadth to get area 84 sq.m? Solve it.

[2 marks]

✓ Solution

Let x be subtracted from each: new dimensions = (15−x) and (10−x)

(15−x)(10−x) = 84

150 − 15x − 10x + x² = 84

x² − 25x + 66 = 0

(x − 22)(x − 3) = 0 → x = 3 (since x = 22 is invalid)

∴ 3 m should be subtracted from each side (gives 12 m × 7 m = 84 m² ✓)

10

Algebra and Indices

5 marks

a

1/x⁻⁶ लाई x को धनात्मक घाताङ्कमा लेख्नुहोस्।

Write 1/x⁻⁶ in the positive index of x.

[1 mark]

✓ Answer

1/x⁻⁶ = x⁶ (since 1/xⁿ = x⁻ⁿ, so 1/x⁻⁶ = x⁶)

∴ 1/x⁻⁶ = x⁶

b

सरल गर्नुहोस् (Simplify): a²/(a−2b) + 4b²/(2b−a)

Simplify: a²/(a−2b) + 4b²/(2b−a)

[2 marks]

✓ Solution

Note: (2b − a) = −(a − 2b)

= a²/(a−2b) + 4b²/(−(a−2b))

= a²/(a−2b) − 4b²/(a−2b)

= (a² − 4b²)/(a−2b)

= (a+2b)(a−2b)/(a−2b)

∴ = a + 2b

c

हल गर्नुहोस् (Solve): 4^(x−2) = 0.25

Solve: 4^(x−2) = 0.25

[2 marks]

✓ Solution

0.25 = 1/4 = 4⁻¹

4^(x−2) = 4⁻¹

Bases equal → x − 2 = −1

∴ x = 1

11

दिइएको चित्रमा AE ∥ BD, ED ∥ AC र BE ∥ CD छन्।

In the given figure, AE ∥ BD, ED ∥ AC and BE ∥ CD.

4 marks

AE ∥ BD, ED ∥ AC, BE ∥ CD

a

एउटै आधार र उही समानान्तर रेखाहरू बीच रहेका समानान्तर चतुर्भुज र त्रिभुजको क्षेत्रफल बिचको सम्बन्ध लेख्नुहोस्।

Write the relationship between the area of a parallelogram and triangle on the same base and between the same parallel lines.

[1 mark]

✓ Answer

Area of triangle = (1/2) × Area of parallelogram

When a triangle and a parallelogram are on the same base and between the same parallel lines, the area of the triangle is half the area of the parallelogram.

b

प्रमाणित गर्नुहोस् (Prove that): △ABE ≅ △BCD

Prove that: △ABE = △BCD (in area)

[2 marks]

✓ Proof

Given: AE ∥ BD, ED ∥ AC, BE ∥ CD

Since AE ∥ BD and ED ∥ AB, ABDE is a parallelogram.

Since BE ∥ CD and BD ∥ EC, BCDE is a parallelogram.

Both parallelograms ABDE and BCDE share base BD and lie between same parallels AE ∥ BD (and the line through C ∥ AE).

Area(ABDE) = Area(BCDE) [same base BD, same parallels]

△ABE = (1/2) × □ABDE [diagonal BE divides ABDE]

△BCD = (1/2) × □BCDE [diagonal BD divides BCDE]

∴ △ABE = △BCD (Proved)

c

त्रिभुज ABE र समलम्ब चतुर्भुज ACDE को क्षेत्रफलबिच तुलना गर्नुहोस्।

Compare between the area of triangle ABE and trapezium ACDE.

[1 mark]

✓ Solution

Area(ACDE) = Area(△ABE) + Area(△BCD) + Area(△BDE) − ... let's use direct approach.

Trapezium ACDE = △ABE + △BCD + Area between (noting from the figure ACDE has vertices A, C, D, E)

Area(ACDE) = Area(□ABDE) + Area(△BCD) = 2△ABE + △ABE [since △BCD = △ABE]

∴ Area(ACDE) = 3 × Area(△ABE)

∴ Area of trapezium ACDE = 3 × Area of △ABE

12

दिइएको चित्रमा, एउटा चक्रिय चतुर्भुज PQRS छ। भुजा PQ लाई बिन्दु T सम्म लम्बाइएको छ।

In the given figure, PQRS is a cyclic quadrilateral. The side PQ is produced to the point T.

5 marks

Cyclic quadrilateral PQRS with PQ produced to T

a

∠PSQ र ∠PRQ को सम्बन्ध लेख्नुहोस्।

Write the relation of ∠PSQ and ∠PRQ.

[1 mark]

✓ Answer

∠PSQ and ∠PRQ both subtend arc PQ from the same side of the chord PQ.

∠PSQ = ∠PRQ (Angles in the same segment are equal)

b

∠SPR र ∠SQR बराबर हुन्छन् भनी प्रयोगद्वारा प्रमाणित गर्नुहोस्। (कम्तीमा ३ से.मि. अर्धव्यास भएका दुईओटा वृत्तहरू अनिवार्य छन्।)

Verify experimentally that ∠SPR and ∠SQR are equal. (Two circles with radii at least 3 cm are necessary.)

[2 marks]

✓ Experimental Verification

Experimental verification using two circles: ∠SPR = ∠SQR (angles in the same segment)

Steps:
1. Draw a circle of radius ≥ 3 cm. Mark four points S, P, Q, R on the circumference forming cyclic quadrilateral PQRS.
2. Draw chords SP, SQ, RP, RQ (the diagonals).
3. Measure ∠SPR and ∠SQR using a protractor.
4. Repeat with a second circle of different radius with different positions of points.
Observation: In both circles, ∠SPR = ∠SQR.
Conclusion: Angles in the same segment of a circle are equal. ✓

c

प्रमाणित गर्नुहोस् (Prove that): ∠RQT = ∠PSR

Prove that: ∠RQT = ∠PSR

[2 marks]

✓ Proof

Given: PQRS is a cyclic quadrilateral; PQ produced to T.

To prove: ∠RQT = ∠PSR

PQRS is a cyclic quadrilateral. Opposite angles are supplementary:

∠PQR + ∠PSR = 180° ...(i)

∠PQR and ∠RQT are supplementary (angles on a straight line PQT):

∠PQR + ∠RQT = 180° ...(ii)

From (i) and (ii): ∠PQR + ∠PSR = ∠PQR + ∠RQT

∴ ∠RQT = ∠PSR (Proved) — Exterior angle of cyclic quad = opposite interior angle

13

चतुर्भुज PQRS मा PQ = ५ से.मी., QR = ४.५ से.मी., RS = SP = ६ से.मी. र QS = ६.५ से.मी. छन्।

Quadrilateral PQRS: PQ = 5 cm, QR = 4.5 cm, RS = SP = 6 cm, QS = 6.5 cm.

4 marks

a

माथिको नाप अनुसारको चतुर्भुज PQRS रचना गर्नुहोस् र उक्त चतुर्भुजको क्षेत्रफलसँग बराबर हुने गरी एउटा त्रिभुजको पनि रचना गर्नुहोस्।

Construct a quadrilateral PQRS according to the above measurements and then construct a triangle equal in area to the quadrilateral.

[3 marks]

✓ Construction

Quadrilateral PQRS and equal-area triangle QST constructed (schematic)

Construction Steps:
1. Draw diagonal QS = 6.5 cm as base line.
2. Construct △PQS: Arc from Q (radius 5 cm) and arc from S (radius 6 cm) intersect at P above QS.
3. Construct △QRS: Arc from Q (radius 4.5 cm) and arc from S (radius 6 cm) intersect at R below QS.
4. Join all vertices to complete quadrilateral PQRS.
5. Equal-area triangle: Through R, draw a line parallel to QS.
6. Extend SP to meet this parallel line at point T.
7. Triangle QST has the same area as quadrilateral PQRS.

b

यसरी बनेको चतुर्भुज र त्रिभुजको क्षेत्रफल किन बराबर हुन्छन्? कारण दिनुहोस्।

Why are the areas of the quadrilateral and triangle equal? Give reason.

[1 mark]

✓ Reason

When line RT is drawn through R parallel to QS and SP is extended to T:

△RSQ and △TSQ lie on same base QS and between same parallels QS and RT.

∴ Area(△RSQ) = Area(△TSQ) [same base, same parallels]

Adding △PQS to both sides:

△PQS + △RSQ = △PQS + △TSQ

∴ Area(Quad PQRS) = Area(△QST) — the triangles on same base QS between same parallels have equal area, so replacing △RSQ with equal △TSQ gives the equal-area triangle.

14

चित्रमा टावरको उचाइ (CD) 15√3 मिटर र घरको उचाइ (AB) 10√3 मिटर छन्। घरको छतबाट टावरको टुप्पामा हेर्दा ३०° को कोण बन्दछ।

Height of tower CD = 15√3 m, height of house AB = 10√3 m. Angle from roof of house to top of tower = 30°.

4 marks

Tower CD = 15√3 m, House AB = 10√3 m, Angle of elevation from B to D = 30°; E is on CD at height of B

a

घरको छतबाट टावरको टुप्पामा हेर्दा बन्ने कोण कुन प्रकारको कोण हो? लेख्नुहोस्।

What type of angle is formed when the top of the tower is observed from the roof of the house? Write it.

[1 mark]

✓ Answer

The tower top (D) is higher than the house roof (B), so looking upward from B to D forms an angle of elevation (उन्नतांश कोण).

b

टावरको भाग DE को उचाइ पत्ता लगाउनुहोस्।

Find the height of part DE of the tower.

[1 mark]

✓ Solution

E is at the same level as B (horizontal line BE).

DE = CD − CE = CD − AB = 15√3 − 10√3 = 5√3 m

∴ Height of DE = 5√3 m

c

घर र टावरबीचको दूरी निकाल्नुहोस्।

Calculate the distance between house and tower.

[1 mark]

✓ Solution

From roof B, angle of elevation to D = 30°, DE = 5√3 m

In right triangle BDE: tan(30°) = DE / BE

1/√3 = 5√3 / BE → BE = 5√3 × √3 = 5 × 3 = 15 m

Distance between house and tower = BC = BE = 15 m

∴ Distance between house and tower = 15 m

d

घरको आधारबाट टावरको टुप्पामा हेर्दा बन्ने कोण पत्ता लगाउनुहोस्।

Find the angle of the top of the tower from the basement of the house.

[1 mark]

✓ Solution

From basement A, looking to top of tower D: AD/AC = ?

AC = BC = 15 m (horizontal), CD = 15√3 m

tan(θ) = CD / AC = 15√3 / 15 = √3

θ = tan⁻¹(√3) = 60°

∴ Angle from basement of house to top of tower = 60°

15

एउटा कक्षाका ५० जना विद्यार्थीहरूले गणित विषयको ६० पूर्णाङ्कको परीक्षामा प्राप्त गरेको प्राप्ताङ्कलाई तालिकामा दिइएको छ।

Marks obtained by 50 students of a class in Maths exam (full marks 60).

6 marks

Marks (Obtained)	10–20	20–30	30–40	40–50	50–60
No. of students	7	13	15	10	5

a

रीत पर्ने श्रेणी लेख्नुहोस्।

Write the class of mode.

[1 mark]

✓ Answer

The class with the highest frequency is 30–40 (frequency = 15).

∴ Modal class = 30–40

b

दिइएको तथ्याङ्कबाट पहिलो चतुर्थांश (Q₁) को मान निकाल्नुहोस्।

Find the value of first quartile (Q₁) from the given data.

[2 marks]

✓ Solution

N = 50, Q₁ position = N/4 = 50/4 = 12.5th value

Cumulative frequencies: 0–10: 0, 10–20: 7, 10–30: 20 → Q₁ lies in class 20–30

L = 20, c.f. (preceding) = 7, f = 13, i = 10

Q₁ = L + [(N/4 − c.f.) / f] × i = 20 + [(12.5 − 7) / 13] × 10

= 20 + [5.5/13] × 10 = 20 + 4.23 ≈ 24.23

∴ First Quartile Q₁ = 24.23 marks

c

दिइएको तथ्याङ्कबाट औसत प्राप्ताङ्क गणना गर्नुहोस्।

Calculate the average mark from the given data.

[2 marks]

✓ Solution

Class	f	Mid-pt (m)	f × m
10–20	7	15	105
20–30	13	25	325
30–40	15	35	525
40–50	10	45	450
50–60	5	55	275
Total	50		1680

Mean = Σ(f×m) / Σf = 1680 / 50 = 33.6

∴ Average (Mean) marks = 33.6

d

४० वा ४० भन्दा बढी प्राप्ताङ्क ल्याउने विद्यार्थीहरूको औसत प्राप्ताङ्क कति छ? पत्ता लगाउनुहोस्।

What is the average mark of students who obtained 40 or more than 40 marks? Find it.

[1 mark]

✓ Solution

Students with marks ≥ 40: class 40–50 (f=10, m=45) and 50–60 (f=5, m=55)

Σ(f×m) = 10×45 + 5×55 = 450 + 275 = 725

Σf = 10 + 5 = 15

Mean = 725 / 15 = 48.33

∴ Average mark of students scoring ≥ 40 = 48.33 marks

16

एउटा बाकसमा उनै र उस्तै आकारका ५ ओटा राता र ३ ओटा सेता बलहरू राखिएका छन्।

In a box, 5 red and 3 white balls of same size and shape are kept.

5 marks

a

पारस्परिक निषेधक घटनाहरूको सम्भाव्यताको जोड नियम लेख्नुहोस्।

Write the addition law of probability of mutually exclusive events.

[1 mark]

✓ Answer

For mutually exclusive events A and B (P(A∩B) = 0):

P(A∪B) = P(A) + P(B)

i.e., if two events cannot occur simultaneously, the probability of either occurring equals the sum of their individual probabilities.

b

एउटा बल अनियमित रूपमा उठाउँदा रातो बल नपर्ने सम्भाव्यता कति हो? पत्ता लगाउनुहोस्।

If one ball is drawn at random, what is the probability of NOT getting a red ball? Find it.

[1 mark]

✓ Solution

Total balls = 5 + 3 = 8

P(red) = 5/8

P(not red) = 1 − 5/8 = 3/8

∴ P(not red) = 3/8

c

दुईओटा बलहरू एकपछि अर्को नहेरीकन र नराखीकन (without replacement) उठाउँदा दुवैओटा रातो हुने सम्भाव्यता पत्ता लगाउनुहोस्।

Two balls are drawn one after another without replacement. Find the probability that both are red.

[2 marks]

✓ Solution

P(1st ball red) = 5/8

P(2nd ball red | 1st was red) = 4/7 (without replacement)

P(both red) = 5/8 × 4/7 = 20/56 = 5/14

∴ P(both red, without replacement) = 5/14

d

दुईओटा बलहरू एकपछि अर्को नहेरीकन र राखेर (with replacement) उठाउँदा एउटा रातो र अर्को सेतो हुने सम्भाव्यता पत्ता लगाउनुहोस्।

Two balls are drawn one after another with replacement. Find the probability of getting one red and one white.

[1 mark]

✓ Solution

P(Red then White) = 5/8 × 3/8 = 15/64

P(White then Red) = 3/8 × 5/8 = 15/64

P(one red, one white) = 15/64 + 15/64 = 30/64 = 15/32

∴ P(one red, one white) with replacement = 15/32