Compulsory Mathematics SEE 2081 Madhesh Province Solution | अनिवार्य गणित

Solution:

(a) Here / यहाँ:
$n(U) = 160$
Only Apple / स्याउ मात्र: $n_o(A) = 75$
Only Orange / सुन्तला मात्र: $n_o(O) = 45$
Neither / दुवै मन नपराउने: $23$

Cardinality Notation / गणनात्मकता संकेत:
$$n(\overline{A \cup O}) \text{ or } n(A \cup O)’ = 23$$

(b) Venn diagram representation:

(c) Let $n(A \cap O) = x$ / मानौँ दुवै मन पराउनेको सङ्ख्या $x$ छ ।

From Venn diagram / भेन चित्रबाट:
$$n(U) = n_o(A) + n_o(O) + n(A \cap O) + n(\overline{A \cup O})$$$$160 = 75 + 45 + x + 23$$$$160 = 143 + x$$$$x = 160 – 143$$$$\therefore x = 17$$
Number of people liking apple / स्याउ मन पराउने मानिसको सङ्ख्या:
$$n(A) = n_o(A) + n(A \cap O)$$$$n(A) = 75 + 17$$$$n(A) = 92$$

(d) Number of people liking orange / सुन्तला मन पराउने सङ्ख्या:
$$n(O) = 45 + 17 = 62$$
Ratio / अनुपात of $n(A)$ and $n(O)$:
$$n(A) : n(O) = 92 : 62 = 46 : 31$$

Solution:

(a) Relation / सम्बन्ध:
$$CI = P\left[\left(1+\frac{R}{100}\right)^{T}-1\right]$$

(b) For 1st year / पहिलो वर्षको लागि:
$P = Rs. 4,00,000$, $R = 10\%$, $T = 1$

$$CI_1 = \frac{P \times T \times R}{100}$$$$= \frac{400000 \times 1 \times 10}{100}$$$$= Rs. 40,000$$

(c) Amount after 1st year / पहिलो वर्षको अन्त्यमा मिश्रधन:
$$= 4,00,000 + 40,000 = Rs. 4,40,000$$
Paid amount / तिरेको रकम: $Rs. 2,40,000$
Principal for 2nd year / दोस्रो वर्षको लागि बाँकी सावाँ:
$$= 4,40,000 – 2,40,000 = Rs. 2,00,000$$
Interest for 2nd year / दोस्रो वर्षको ब्याज:
$$CI_2 = \frac{200000 \times 1 \times 10}{100} = Rs. 20,000$$
Total Interest Paid / जम्मा तिरेको ब्याज:
$$= CI_1 + CI_2 = 40,000 + 20,000 = Rs. 60,000$$

Solution:

(a) Depreciation amount (D) / ह्रास रकम:
$$D = P \times R\% = 40,000 \times \frac{5}{100} = Rs. 2,000$$

(b) Using depreciation formula / ह्रास सूत्र प्रयोग गर्दा:
$$SP = P(1 – \frac{R}{100})^T$$$$36,100 = 40,000(1 – \frac{5}{100})^T$$$$\frac{36100}{40000} = (0.95)^T$$$$0.9025 = (0.95)^T$$$$(0.95)^2 = (0.95)^T$$$$\therefore T = 2 \text{ years (वर्ष)}$$

(c) Cost Price (C.P.) / क्रय मूल्य: $Rs. 40,000$
Total amount received / जम्मा प्राप्त रकम:
$$= Selling Price + Rent = 36,100 + 4,900 = Rs. 41,000$$
Profit / नाफा:
$$= 41,000 – 40,000 = Rs. 1,000$$
Profit % / नाफा प्रतिशत:
$$= \frac{\text{Profit}}{C.P.} \times 100\% = \frac{1000}{40000} \times 100\% = 2.5\%$$

Solution:

(a) Selling rate / बिक्री दर प्रयोग गरिन्छ (because bank sells dollars / किनकि बैंकले डलर बेच्छ)।

(b) Amount ($) / डलर रकम:
$$= \frac{\text{NRs. } 2,07,345}{\text{Selling Rate}} = \frac{2,07,345}{138.83}$$$$= \$ 1,493.52$$

(c) Difference / फरक:
$$= 140.2183 – 138.83 = Rs. 1.3883$$
Devaluation % / अवमूल्यन प्रतिशत:
$$= \frac{\text{Diff}}{\text{Original Rate}} \times 100\% = \frac{1.3883}{138.83} \times 100\% = 1\%$$

Solution:

(a) 5 surfaces (५ वटा सतहहरू)

(b) Volume formula / आयतन सूत्र:
$$V = \frac{1}{3} a^2 h$$$$512 = \frac{1}{3} \times (16)^2 \times h$$$$512 \times 3 = 256 \times h$$$$1536 = 256h$$$$\therefore h = 6 \text{ cm}$$

(c) Slant height / छड्के उचाइ:
$$l = \sqrt{h^2 + (a/2)^2} = \sqrt{6^2 + (16/2)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm}$$
Total Surface Area (T.S.A.) / पूरा सतहको क्षेत्रफल:
$$= a^2 + 2al = (16)^2 + 2 \times 16 \times 10$$$$= 256 + 320 = 576 \text{ cm}^2$$

Solution:

(a) 2 curved surfaces (२ वटा बक्र सतहहरू)

(b) Height of cylinder / बेलनाको उचाइ:
$$h_{cyl} = \text{Total Height} – \text{Radius} = 3.5 – 1.05 = 2.45 \text{ m}$$
Volume / आयतन:
$$V = \text{Vol of Cylinder} + \text{Vol of Hemisphere}$$$$= \pi r^2 h_{cyl} + \frac{2}{3} \pi r^3$$$$= \pi r^2 (h_{cyl} + \frac{2r}{3})$$$$= \frac{22}{7} \times 1.05 \times 1.05 \left(2.45 + \frac{2 \times 1.05}{3}\right)$$$$= 3.465 \times (2.45 + 0.7)$$$$= 3.465 \times 3.15$$$$= 10.91475 \text{ m}^3$$

(c) Conversion / रूपान्तरण: $1 \text{ m}^3 = 1000 \text{ Liters}$
Capacity / क्षमता:
$$= 10.91475 \times 1000 = 10,914.75 \text{ Liters}$$
Or / अथवा: $10,914 \text{ Liters and } 750 \text{ ml}$

Solution:

(a) Area of floor / भुइँको क्षेत्रफल:
$$A = l \times b = 16 \times 12 = 192 \text{ sq. ft.}$$
Cost / खर्च:
$$= \text{Area} \times \text{Rate} = 192 \times 300 = Rs. 57,600$$

(b) Gross Area (Walls + Ceiling) / (भित्ता + छाना) को जम्मा क्षेत्रफल:
$$= 2h(l+b) + lb = 2 \times 9(16+12) + 192$$$$= 18(28) + 192 = 504 + 192 = 696 \text{ sq. ft.}$$
Area of Openings (Windows + Door) / (झ्याल + ढोका) को क्षेत्रफल:
$$= 2 \times (4^2) + (6 \times 2) = 2 \times 16 + 12 = 32 + 12 = 44 \text{ sq. ft.}$$
Net Area for painting / रङ लगाउने शुद्ध क्षेत्रफल:
$$= 696 – 44 = 652 \text{ sq. ft.}$$
Rate / दर:
$$= \frac{\text{Total Cost}}{\text{Area}} = \frac{19,560}{652} = Rs. 30 \text{ per sq. ft.}$$

Solution:

(a) दुई राशिहरू बिच पर्ने राशि वा पदलाई समानान्तरीय मध्यमा भनिन्छ ।
(The term lying between two terms in an arithmetic progression is called arithmetic mean.)

(b) This is an Arithmetic Progression with:
First term / पहिलो पद ($a$) = $1000$
Common difference / साधारण अन्तर ($d$) = $1000$
Number of terms / पद सङ्ख्या ($n$) = $10$

Sum of first 10 terms / पहिलो 10 पदहरूको योगफल:
$$S_{10} = \frac{n}{2}[2a + (n-1)d]$$$$= \frac{10}{2}[2(1000) + (10-1)1000]$$$$= 5[2000 + 9000] = 5 \times 11000 = Rs. 55,000$$

(c) Sum up to 11th birthday / 11 औं जन्मदिनसम्मको योगफल:
$$S_{11} = \frac{11}{2}[2(1000) + 10(1000)] = 5.5[12000] = 66,000$$
Yes, it is deposited / हो, जम्मा हुन्छ ।

(d) We need to find $n$ when $S_n = 1,05,000$:
$$105000 = \frac{n}{2}[2000 + (n-1)1000]$$$$210000 = n[2000 + 1000n – 1000]$$$$210000 = 1000n^2 + 1000n$$$$n^2 + n – 210 = 0$$$$n^2 + 15n – 14n – 210 = 0$$$$(n+15)(n-14) = 0$$
Since $n$ cannot be negative / $n$ ऋणात्मक हुन सक्दैन:
$$\therefore n = 14 \text{ years (वर्ष)}$$

Solution:

(a) Standard form / स्तरीय स्वरूप:
$$ax^2 + bx + c = 0, \text{ where } a \neq 0$$

(b) Let breadth / चौडाइ ($b$) = $x$ ft
Then length / लम्बाइ ($l$) = $2x$ ft

Area / क्षेत्रफल:
$$l \times b = 800$$$$2x \times x = 800$$$$2x^2 = 800$$$$x^2 = 400$$$$\therefore x = 20 \text{ ft}$$
Length / लम्बाइ: $l = 2 \times 20 = 40 \text{ ft}$

Perimeter (Wire length) / परिमिति (तारको लम्बाइ):
$$= 2(l+b) = 2(40+20) = 120 \text{ ft}$$

Solution:

(a) $$\frac{x}{xy-y^{2}} + \frac{y}{xy-x^{2}}$$$$= \frac{x}{y(x-y)} + \frac{y}{x(y-x)}$$$$= \frac{x}{y(x-y)} – \frac{y}{x(x-y)}$$$$= \frac{x^2 – y^2}{xy(x-y)}$$$$= \frac{(x-y)(x+y)}{xy(x-y)} = \frac{x+y}{xy}$$

(b) Let $2^x = a$
Then / त्यसपछि:
$$a + \frac{16}{a} = 10$$$$a^2 + 16 = 10a$$$$a^2 – 10a + 16 = 0$$$$(a-8)(a-2) = 0$$
Either / या त:
$a=8 \implies 2^x = 2^3 \implies x=3$
Or / अथवा:
$a=2 \implies 2^x = 2^1 \implies x=1$

$$\therefore x = 1, 3$$

Solution:

(a) Area of $\Box PQRS$ = Area of $\Box PQUT$
(Standing on same base PQ and between same parallel lines / एउटै आधार PQ र उही समानान्तर रेखाहरूबिच बनेका)

(b) Area of $\triangle PQT = \frac{1}{2} \times \text{base} \times \text{height}$
Area of $\Box PQRS = \text{base} \times \text{height}$
$\therefore$ Area of $\triangle PQT = \frac{1}{2}$ Area of $\Box PQRS$

(c) Yes / हो, Area($\triangle APD$) = Area($\triangle BPQ$)

Reason / कारण:
$\Delta ABD$ and $\Delta ABQ$ stand on same base AB and between same parallels.
So / त्यसैले, Area($\Delta ABD$) = Area($\Delta ABQ$).
Subtracting common Area($\Delta ABP$) from both sides / दुबै तिरबाट समान क्षेत्रफल $\Delta ABP$ घटाउँदा:
Area($\Delta APD$) = Area($\Delta BPQ$)

Solution:

(a) वृत्तको दुई जीवाहरू परिधिमा मिलि बन्ने कोणलाई परिधि कोण भनिन्छ ।
(The angle formed by the intersection of two chords at the circumference of a circle is called an inscribed angle.)

(b) Central Angle = $2 \times$ Inscribed Angle / केन्द्रीय कोण = $2 \times$ परिधि कोण
$$9x + 2 = 2(4x + 5)$$$$9x + 2 = 8x + 10$$$$9x – 8x = 10 – 2$$$$\therefore x = 8$$

(c) Experimental Verification / प्रयोगात्मक प्रमाण:
1. Construct two circles with radii $\ge 3$ cm / 3 से.मि. भन्दा बढी अर्धव्यास भएका दुई वृत्तहरू बनाउनुहोस् ।
2. Draw points L, M, N, P on circumference / परिधिमा L, M, N, P बिन्दुहरू खिच्नुहोस् ।
3. Measure $\angle LMP$ and $\angle LNP$ / $\angle LMP$ र $\angle LNP$ को मापन गर्नुहोस् ।
4. Both angles will be equal / दुबै कोण बराबर हुनेछन् ।

Solution:

(a) Construction steps / रचना चरणहरू:

1. Draw $PQ = 5.4$ cm
2. Construct $\angle PQR = 75^{\circ}$ at Q
3. Cut off $QR = 5.6$ cm
4. With center R, draw an arc of radius 5.4 cm
5. With center P, draw an arc of radius 6.8 cm
6. Intersection gives point S
7. Join PS and RS to complete quadrilateral PQRS
8. For equal area triangle, draw diagonal PR
9. Construct triangle PSM with same base and between same parallels

(b) Since $AE = BE$, E is the midpoint of AB / $AE = BE$ भएकोले E ले AB लाई दुई भागमा बराबर बाँड्छ ।

Area of $\triangle BEC$ / $\triangle BEC$ को क्षेत्रफल:
$$= \frac{1}{2} \times EB \times h = \frac{1}{2} \times (\frac{1}{2} AB) \times h = \frac{1}{4} (AB \times h)$$
Area of Parallelogram ABCD / समानान्तर चतुर्भुज ABCD को क्षेत्रफल:
$$= AB \times h$$
Percentage / प्रतिशत:
$$= \frac{\frac{1}{4} (AB \times h)}{AB \times h} \times 100\% = 25\%$$

Solution:

(a) $\angle RPT$

(b) $$TR = RS – TS = RS – PQ = 72 – 30 = 42 \text{ m}$$

(c) In right triangle $\Delta PTR$:
$$\tan 60^{\circ} = \frac{TR}{PT}$$$$\sqrt{3} = \frac{42}{PT}$$$$PT = \frac{42}{\sqrt{3}} = 14\sqrt{3} \text{ m}$$$$\approx 24.24 \text{ m}$$

(d) Height of observer from top / टुप्पोबाट दर्शकको उचाइ: $28m$
Height of observer from ground / जमिनबाट दर्शकको उचाइ: $72-28 = 44m$
Height relative to house roof / घरको छतको सापेक्ष उचाइ: $44 – 30 = 14m$

$$\tan \theta = \frac{\text{Perpendicular}}{\text{Base}} = \frac{14}{14\sqrt{3}} = \frac{1}{\sqrt{3}}$$$$\therefore \theta = 30^{\circ}$$
Yes, the statement is correct / हो, भनिएको कुरा सही छ ।

Solution:

(a) Since Median = 29, which lies in 20-30 class / मध्यिका 29 भएकोले 20-30 श्रेणीमा पर्छ ।
Median class / मध्यिका श्रेणी: $20-30$

(b) Total frequency / जम्मा बारम्बारता:
$$N = 3 + 7 + 10 + x + 10 = 30 + x$$
Using median formula / मध्यिका सूत्र प्रयोग गर्दा:
$$Md = L + \frac{\frac{N}{2} – c.f.}{f} \times i$$$$29 = 20 + \frac{\frac{30+x}{2} – 10}{10} \times 10$$$$9 = \frac{30+x-20}{2}$$$$18 = 10 + x$$$$\therefore x = 8$$

(c) With $x=8$:
$N = 30 + 8 = 38$

Calculation table / गणना तालिका:

Mean / मध्यक:
$$\overline{X} = \frac{\sum fx}{N} = \frac{1100}{38} = 28.94$$

(d) Marks $<20$: $3+7=10$
Marks $\ge 20$: $10+8+10=28$

Ratio / अनुपात: $10:28 = 5:14$

Solution:

(a) For independent events / अनाश्रित घटनाहरूको लागि:
$$P(B \cap R) = P(B) \times P(R)$$

(b) Tree diagram:

(c) Probability of both black / दुवै बल कालो नै पर्ने सम्भाव्यता:
$$P(BB) = \frac{7}{11} \times \frac{6}{10} = \frac{42}{110} = \frac{21}{55}$$

(d) Probability of both red / दुवै बल रातो नै पर्ने सम्भाव्यता:
$$P(RR) = \frac{4}{11} \times \frac{3}{10} = \frac{12}{110} = \frac{6}{55}$$
Difference / फरक:
$$\frac{21}{55} – \frac{6}{55} = \frac{15}{55} = \frac{3}{11}$$
$P(RR)$ is less than $P(BB)$ by $\frac{3}{11}$ / $P(RR)$ ले $P(BB)$ भन्दा $\frac{3}{11}$ ले कम छ ।