SEE 2081 Karnali Province Compulsory Mathematics Solution | अनिवार्य गणित

Solution / समाधान:

a) The set of students who did not like any of these two subjects in cardinality notation:
$$n(\overline{E \cup M}) = 20$$

b) Venn diagram representation:

c) Here, $n(U) = 100$, $n(E) = 60$, $n(M) = 50$, $n(\overline{E \cup M}) = 20$
Let $n(E \cap M) = x$
From Venn diagram:
$$n(U) = n(E) + n(M) – n(E \cap M) + n(\overline{E \cup M})$$$$100 = 60 + 50 – x + 20$$$$100 = 130 – x$$$$x = 30$$ Number of students who liked both subjects $n(E \cap M) = 30$.

Number of students who liked exactly one subject:
$$n_0(E) + n_0(M) = (60 – x) + (50 – x)$$$$= (60 – 30) + (50 – 30)$$$$= 30 + 20$$$$= 50$$

d) New condition: $n(\overline{E \cup M}) = 30$
$$100 = 60 + 50 – x_{new} + 30$$$$100 = 140 – x_{new}$$$$x_{new} = 40$$ Previously $x = 30$, now $x = 40$.
Conclusion: $n(E \cap M)$ is increased by 10.
(पहिले $x = 30$ थियो, अहिले $x = 40$ भयो । निष्कर्ष: दुवै विषय मन पराउने विद्यार्थी सङ्ख्या १० ले बढ्छ ।)

Solution / समाधान:

a) Annual Compound Interest formula:
$$C.I. = P\left[\left(1 + \frac{R}{100}\right)^T – 1\right]$$

b) Principal $(P) = Rs. 1,00,000$
Time $(T) = 2$ years
Rate $(R) = 8\%$ p.a.
Semi-annual Compound Interest:
$$C.I. = P\left[\left(1 + \frac{R}{200}\right)^{2T} – 1\right]$$$$= 1,00,000 \left[\left(1 + \frac{8}{200}\right)^{2 \times 2} – 1\right]$$$$= 1,00,000 \left[(1.04)^4 – 1\right]$$$$= 1,00,000 \times (1.16985856 – 1)$$$$= 1,00,000 \times 0.16985856$$$$= Rs. 16,985.86$$ (Answer: Rs. 16,986)

c) Yearly Compound Interest:
$$C.I. = P\left[\left(1 + \frac{R}{100}\right)^T – 1\right]$$$$= 1,00,000 \left[\left(1 + \frac{8}{100}\right)^2 – 1\right]$$$$= 1,00,000 \times (1.1664 – 1)$$$$= Rs. 16,640$$ Loss = Semi-annual C.I. – Annual C.I.
$$= 16,985.86 – 16,640$$$$= Rs. 345.86$$ (Answer: Rs. 346)

Solution / समाधान:

a) Initial population $(P) = 10,000$
Growth rate $(R) = 4\%$
Out-migration = 100
Population after 1 year:
$$P_1 = P\left(1 + \frac{R}{100}\right)^1 – 100$$$$= 10,000(1 + 0.04) – 100$$$$= 10,400 – 100$$$$= 10,300$$

b) Initial population for 2nd year = 10,300
Population after 2 years (with no migration in 2nd year):
$$P_2 = 10,300 \left(1 + \frac{4}{100}\right)$$$$= 10,300 \times 1.04$$$$= 10,712$$

c) If nobody migrated in first year:
Population after 2 years:
$$P = 10,000(1.04)^2 = 10,816$$ Difference = $10,816 – 10,712 = 104$

Solution / समाधान:

a) The act of exchanging the currency of one country with the currency of another country is called currency exchange.
(एउटा देशको मुद्रासँग अर्को देशको मुद्रा साट्ने कार्यलाई मुद्रा विनिमय भनिन्छ ।)

b) Rate: $1 = NRs. 138.83$
Amount = $1500 \times 138.83$
$$= NRs. 2,08,245$$

c) New Rate after devaluation:
$$= 138.83 + 2\% \text{ of } 138.83$$$$= 138.83 + 2.7766$$$$= NRs. 141.6066$$ Exchanged Dollars = $\frac{7,08,033}{141.6066} \approx \$ 5,000$

Solution / समाधान:

a) Here, EF is the base side $(a)$ and HD is half of the base side.
Relation: $HD = \frac{1}{2} EF$ (or $EF = 2 \times HD$)

b) In right-angled triangle ADH:
$$HD = \sqrt{AH^2 – AD^2} = \sqrt{26^2 – 24^2}$$$$= \sqrt{676 – 576} = \sqrt{100} = 10 \text{ cm}$$$$EF = 2 \times HD = 2 \times 10 = 20 \text{ cm}$$

c) Base side $(a) = 20$ cm
Slant height $(l) = 26$ cm
Total Surface Area (TSA) $= a^2 + 2al$
$$= (20)^2 + 2 \times 20 \times 26$$$$= 400 + 1040$$$$= 1440 \text{ cm}^2$$

Solution / समाधान:

a) Volume of solid object:
$$V = \pi r^2 h + \frac{2}{3}\pi r^3$$

b) Circumference $2\pi r = 44$
$$2 \times \frac{22}{7} \times r = 44 \implies r = 7 \text{ cm}$$ Volume of Hemisphere = $\frac{2}{3}\pi r^3$
$$= \frac{2}{3} \times \frac{22}{7} \times 7^3$$$$= 718.67 \text{ cm}^3$$

c) Height of cylinder $h = 17 – 7 = 10 \text{ cm}$
Volume of Cylinder $= \pi r^2 h$
$$= \frac{22}{7} \times 49 \times 10 = 1540 \text{ cm}^3$$ Difference $= 1540 – 718.67 = 821.33 \text{ cm}^3$
The volume of cylinder is $821.33 \text{ cm}^3$ more than the volume of hemisphere.
(बेलनाको आयतन अर्धगोलाको आयतन भन्दा $821.33 \text{ cm}^3$ ले बढी छ ।)

Solution / समाधान:

a) Volume of tank $= 5 \times 1 \times 4 = 20 \text{ m}^3$
In Liters $= 20 \times 1000 = 20,000 \text{ L}$
Total Cost $= 20,000 \times \text{Rs. } 0.50 = \text{Rs. } 10,000$
Cost for 1 family/month $= \frac{10,000}{20} = \text{Rs. } 500$
Cost for 1 family/year $= 500 \times 12 = \text{Rs. } 6,000$

b) New dimensions: $6 \text{ m} \times 2 \text{ m} \times 5 \text{ m}$
New Volume $= 6 \times 2 \times 5 = 60 \text{ m}^3$
Ratio $= \frac{60}{20} = 3$ times

Solution / समाधान:

a) Geometric Sequence (गुणोत्तर अनुक्रम)

b) Here $a=3, r=2, n=8$
Sum $S_8 = \frac{a(r^n – 1)}{r – 1}$
$$= \frac{3(2^8 – 1)}{2 – 1}$$$$= 3(256 – 1) = 3 \times 255 = 765$$ The child learns 765 words up to 8 days.

c) $$S_n = 6141$$$$3(2^n – 1) = 6141$$$$2^n – 1 = 2047$$$$2^n = 2048$$$$2^n = 2^{11}$$ $\therefore n = 11$ days

Solution / समाधान:

a) $$10x + y$$

b) Condition 1: $10x + y = 4(x + y) \Rightarrow 6x = 3y \Rightarrow y = 2x$
Condition 2: $10x + y = 3xy$
Substitute $y=2x$: $10x + 2x = 3x(2x)$
$$12x = 6x^2$$$$6x^2 – 12x = 0 \text{ or } x^2 – 2x = 0$$

c) $$x(x – 2) = 0$$ Since $x \neq 0$, $x = 2$
$y = 2x = 4$
Number $= 24$

Solution / समाधान:

a) $$\frac{1}{b-1} – \frac{1}{b+1}$$$$= \frac{(b+1) – (b-1)}{(b-1)(b+1)}$$$$= \frac{b + 1 – b + 1}{b^2 – 1}$$$$= \frac{2}{b^2 – 1}$$

b) $$7^x + \frac{1}{7^x} = \frac{50}{7}$$ Let $7^x = a$, then $a + \frac{1}{a} = \frac{50}{7}$
$$7a^2 – 50a + 7 = 0$$$$(7a – 1)(a – 7) = 0$$ $a = 7 \Rightarrow 7^x = 7^1 \Rightarrow x = 1$
$a = \frac{1}{7} \Rightarrow 7^x = 7^{-1} \Rightarrow x = -1$
Answer: $x = \pm 1$

Solution / समाधान:

a) Both triangles stand on the same base PQ and between the same parallel lines, so their areas are equal.
Area of $\triangle APQ$ = Area of $\triangle BPQ$

b) Area $= \frac{1}{2} \times \text{base} \times \text{height}$
$$= \frac{1}{2} \times AB \times h$$$$= \frac{1}{2} \times 10 \times 8$$$$= 40 \text{ cm}^2$$

Solution / समाधान:

a) Construction diagram:
(Note: This is a geometrical construction, students should do it themselves based on measurements: PQ=5.1, QR=7, RS=4.6, SP=5.4, QS=6.6)

b) Because triangles standing on the same base and between the same parallel lines are equal in area, this theorem is used in the construction.
(एउटै आधार र उही समानान्तर रेखाहरुबिच रहेका त्रिभुजहरुको क्षेत्रफल बराबर हुने साध्य प्रयोग गरिएकोले ।)

Solution / समाधान:

a) They are equal. (तिनीहरू बराबर हुन्छन् ।)

b) Central angle is double of inscribed angle standing on the same arc.
$$\angle BOC = 2 \times \angle BAC = 2 \times 35^{\circ} = 70^{\circ}$$

c) Arc BDC = Arc ACD
Subtracting Arc CD from both: Arc BD = Arc AC
If arcs cut by two chords between parallel lines are equal, the lines are parallel.
Or, corresponding chords BD = AC, implies diagonals equal in cyclic quadrilateral, implies isosceles trapezoid, so $AB \parallel CD$.

d) (Experimental verification with two circles of radius ≥ 3 cm).
Note: In a cyclic quadrilateral, opposite angles are supplementary. $\angle BAC$ and $\angle BDC$ are opposite angles in cyclic quadrilateral ABDC, so they are supplementary (sum = 180°).

Solution / समाधान:

a) Angle of elevation (उन्नतांश कोण)

b) Figure:

c) Let building height above eye level be $p$
Distance $b = 36$ m
$$\tan 30^{\circ} = \frac{p}{36}$$$$p = 36 \times \frac{1}{\sqrt{3}} = 12\sqrt{3} = 20.78 \text{ m}$$ Total Height = $20.78 + 1.22 = 22$ m

d) For $60^{\circ}$ angle:
$$\tan 60^{\circ} = \frac{20.78}{x} \Rightarrow \sqrt{3} = \frac{20.78}{x} \Rightarrow x = 12 \text{ m}$$ Distance to walk = $36 – 12 = 24$ m (Towards the building)
Himali should walk 24 m nearer to the building.

Solution / समाधान:

a) $c.f.$ denotes cumulative frequency of the class preceding the first quartile class.
(पहिलो चतुर्थांश पर्ने श्रेणी भन्दा अघिल्लो श्रेणीको सञ्चित बारम्बारता ।)

b) Calculation table for mean:
Mid-values: 2, 6, 10, 14, 18
$\Sigma fm = (50×2) + (65×6) + (75×10) + (60×14) + (50×18) = 100 + 390 + 750 + 840 + 900 = 2980$
$N = 300$
Mean $(\overline{X}) = \frac{\Sigma fm}{N} = \frac{2980}{300} = 9.93$

c) Position = $N/2 = 300/2 = 150$
Median class = 8-12 (cumulative frequency reaches 150 in this class)
$L=8, N/2=150, cf=115 (50+65), f=75, i=4$
$$Md = L + \frac{\frac{N}{2} – cf}{f} \times i$$$$= 8 + \frac{150 – 115}{75} \times 4$$$$= 8 + \frac{35}{75} \times 4$$$$= 8 + 1.87 = 9.87$$

d) Students above median class (8-12):
Classes 12-16 (60) and 16-20 (50) = Total 110
Percentage = $\frac{110}{300} \times 100\% = 36.67\%$

Solution / समाधान:

a) If A and B are independent events: $P(A \cap B) = P(A) \times P(B)$
For dependent events: $P(A \cap B) = P(A) \times P(B/A)$

b) Tree diagram:

Total face cards = 12, Non-face cards = 40

c) $$P(F \cap F) = \frac{12}{52} \times \frac{11}{51} = \frac{11}{221}$$

d) Probability of both non-face cards:
$$P(\overline{F} \cap \overline{F}) = \frac{40}{52} \times \frac{39}{51} = \frac{10}{17} = \frac{130}{221}$$ Ratio (Both Face : Both Non-Face) = $11 : 130$
The probability of getting both face cards is much smaller than getting both non-face cards.