Compulsory Mathematics SEE 2081 Sudurpaschim Province Solution | Sudurpashchim Province | अनिवार्य गणित

Solution:

a) Given: Total students $n(U) = 120$
Liked Cricket $n(C) = 60$
Liked Basketball $n(B) = 55$
Liked none $n(\overline{B \cup C}) = 20$
The cardinality of $n(\overline{B\cup C})$ is:
$$n(\overline{B\cup C}) = 20$$

b) Venn diagram representation:

c) We know: $n(U) = n(C \cup B) + n(\overline{C \cup B})$
$$120 = n(C \cup B) + 20$$$$\therefore n(C \cup B) = 100$$ Using formula: $n(C \cup B) = n(C) + n(B) – n(C \cap B)$
$$100 = 60 + 55 – n(C \cap B)$$$$100 = 115 – n(C \cap B)$$$$n(C \cap B) = 15$$ Number of students who liked cricket only:
$$n_0(C) = n(C) – n(C \cap B)$$$$n_0(C) = 60 – 15 = 45$$

d) Basketball only: $n_0(B) = n(B) – n(C \cap B) = 55 – 15 = 40$
Cricket only: $n_0(C) = 45$
Comparison: $n_0(C) > n_0(B)$ by 5
The number of students who like cricket only is 5 more than those who like basketball only.
(क्रिकेट मात्र मन पराउने विद्यार्थी सङ्ख्या, बास्केटबल मात्र मन पराउने भन्दा ५ ले बढी छ ।)

Solution:

a) Quarterly Compound Interest yields more interest. (त्रैमासिक चक्रीय व्याज अनुसार बढी व्याज प्राप्त हुन्छ ।)

b) Principal ($P$) = Rs. 4,00,000
Time ($T$) = 2 years
Rate ($R$) = 10% p.a.
For Semi-annual Compound Interest:
$$C.I = P \left[ \left(1 + \frac{R}{200}\right)^{2T} – 1 \right]$$$$C.I = 4,00,000 \left[ \left(1 + \frac{10}{200}\right)^{4} – 1 \right]$$$$C.I = 4,00,000 \left[ (1.05)^4 – 1 \right]$$$$C.I = 4,00,000 \times 0.21550625$$$$C.I = \text{Rs. } 86,202.50$$

c) Calculating Quarterly C.I. for 1 year:
$$C.I_{qt} = P \left[ \left(1 + \frac{R}{400}\right)^{4T} – 1 \right]$$$$C.I_{qt} = 4,00,000 \left[ \left(1 + \frac{10}{400}\right)^{4 \times 1} – 1 \right]$$$$C.I_{qt} = 4,00,000 \left[ (1.025)^4 – 1 \right]$$$$C.I_{qt} \approx \text{Rs. } 41,525.16$$ Comparison: Double of Quarterly C.I. = $2 \times 41,525.16 = \text{Rs. } 83,050.32$
Since Rs. 86,202.50 $\neq$ Rs. 83,050.32
Conclusion: No, the semi-annual C.I. in 2 years is not exactly double the quarterly C.I. in 1 year.
(होइन, २ वर्षको अर्धवार्षिक चक्रीय व्याज १ वर्षको त्रैमासिक चक्रीय व्याजको दोब्बर हुँदैन ।)

Solution:

a) $$V_T = V_0 \left(1 – \frac{R}{100}\right)^T$$

b) Initial Cost ($V_0$) = Rs. 80,000
Rate ($R$) = 20%
Time ($T$) = 2 years
Income earned = Rs. 30,000
Selling Price after 2 years ($V_T$):
$$SP = 80,000 \left(1 – \frac{20}{100}\right)^2$$$$SP = 80,000 \times (0.8)^2 = 80,000 \times 0.64 = \text{Rs. } 51,200$$ Total amount recovered = Selling Price + Income Earned
Total = $51,200 + 30,000 = \text{Rs. } 81,200$
Profit ($Gain$) = Total Recovered – Initial Cost
Profit = $81,200 – 80,000 = \text{Rs. } 1,200$

c) Price after 3 years (T=3):
$$SP = 80,000 \left(1 – \frac{20}{100}\right)^3$$$$SP = 80,000 \times (0.8)^3 = 80,000 \times 0.512 = \text{Rs. } 40,960$$ Comparison with Purchase Price ($CP$):
Difference = $CP – SP = 80,000 – 40,960 = \text{Rs. } 39,040$
The selling price would be Rs. 39,040 less than the purchased price.
(बिक्री मूल्य खरिद मूल्यभन्दा रु. ३९,०४० ले कम हुन्छ ।)

Solution:

a) Exchanged Amount = $\frac{\text{Total NRs}}{\text{Rate}}$
$$= \frac{1,29,090}{86.06}$$$$= 1500 \text{ Australian Dollars}$$

b) Revaluation of Nepali currency by 2% means the exchange rate decreases by 2%.
New Rate = $86.06 \times (1 – \frac{2}{100}) = 86.06 \times 0.98 = \text{Rs. } 84.3388$
Amount received for AUD 1500:
$$= 1500 \times 84.3388$$$$= \text{Rs. } 1,26,508.20$$

c) Initial NRs = 1,29,090
Final NRs = 1,26,508.20
Loss Amount = $1,29,090 – 1,26,508.20 = \text{Rs. } 2,581.80$
$$\text{Loss \%} = \frac{\text{Loss}}{\text{Initial Amount}} \times 100\%$$$$\text{Loss \%} = \frac{2,581.80}{1,29,090} \times 100\% = 2\%$$

Solution:

a) $$V = \frac{1}{3}a^2h$$ (Where $a$ is base side length and $h$ is vertical height)

b) Here, $h = 24 \text{ cm}$, $a = 20 \text{ cm}$.
First, find Slant Height ($l$):
$$l = \sqrt{h^2 + \left(\frac{a}{2}\right)^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26 \text{ cm}$$ Total Surface Area (TSA):
$$TSA = a^2 + 2al$$$$TSA = (20)^2 + 2 \times 20 \times 26$$$$TSA = 400 + 1040$$$$TSA = 1440 \text{ cm}^2$$

Solution:

a) $$l = \sqrt{h^2 + r^2}$$

b) Radius ($r$) = $14/2 = 7 \text{ cm}$.
Height of cone ($h$) = 24 cm.
Total Volume = Volume of Cone + Volume of Hemisphere
$$V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3$$$$V = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24 + \frac{2}{3} \times \frac{22}{7} \times 7^3$$$$V = 1232 + 718.67$$$$V \approx 1950.67 \text{ cm}^3$$ (Using alternative calculation: $\frac{1}{3}\pi r^2(h + 2r) = \frac{1}{3} \times \frac{22}{7} \times 49 (24 + 14) = 1950.67$)

c) Volume of Cylinder = Volume of Solid
$$\pi R^2 H = 1950.67$$$$\frac{22}{7} \times 7^2 \times H = 1950.67$$$$154 \times H = 1950.67$$$$H = 12.67 \text{ cm}$$ $\therefore$ Height of the cylinder is 12.67 cm.

Solution:

a) Volume ($V$) = 75 $m^3$, Height ($h$) = 3 m.
Base Area ($A$) = $V/h = 75/3 = 25 \text{ m}^2$.
Since base is square ($l^2 = 25$), length ($l$) = 5 m.
Area of 4 walls = $2h(l + l) = 4lh = 4 \times 5 \times 3 = 60 \text{ m}^2$.
Area of door/windows = 6 $m^2$.
Plastering Area = $60 – 6 = 54 \text{ m}^2$.
Total Cost = Rate $\times$ Area
$$= 200 \times 54 = \text{Rs. } 10,800$$

b) New Rate = $200 + \frac{1}{4} \times 200 = 200 + 50 = \text{Rs. } 250$.
New Cost = $250 \times 54 = \text{Rs. } 13,500$.
Increment = $13,500 – 10,800 = \text{Rs. } 2,700$.

c) Area of floor = $l \times l = 5 \times 5 = 25 \text{ m}^2$.
Area of Carpet = Length $\times$ Width
$$25 = L \times 2$$$$L = 12.5 \text{ m}$$ Required length is 12.5 meters.

Solution:

a) Day 2 amount = 20. Day 4 amount = 80.
This is a Geometric Sequence ($a=10, r=2$).
The term between 20 and 80 (Day 3) is the Geometric Mean.
Mean Value = $\sqrt{20 \times 80} = \sqrt{1600} = 40$.
Answer: Rs. 40.

b) This is a Geometric Progression with $a=10, r=2$.
Sum of first 10 terms ($S_{10}$):
$$S_{10} = \frac{a(r^n – 1)}{r – 1}$$$$S_{10} = \frac{10(2^{10} – 1)}{2 – 1}$$$$S_{10} = 10(1024 – 1)$$$$S_{10} = 10 \times 1023 = \text{Rs. } 10,230$$

c) Here, $S_n = 1,63,830$.
$$1,63,830 = \frac{10(2^n – 1)}{2 – 1}$$$$16383 = 2^n – 1$$$$2^n = 16384$$$$2^n = 2^{14}$$$$n = 14$$ Answer: 14 days (14 औं दिनसम्म).

Solution:

a) $$ax^2 + bx + c = 0 \text{ where } a \neq 0$$

b) Let breadth ($b$) = $x$. Then length ($l$) = $2x$.
Area = $l \times b = 200$
$$2x \times x = 200$$$$2x^2 = 200$$$$x^2 = 100 \implies x = 10$$ Breadth = 10 m, Length = 20 m.

c) Area of one piece = $5 \times 4 = 20 \text{ m}^2$.
Total Area = $200 \text{ m}^2$.
Number of pieces = $200 / 20 = 10$.
Diagram:

Solution:

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

b) Given, $x^2 = (3^{1/3})^2 – 2(3^{1/3})(3^{-1/3}) + (3^{-1/3})^2$
$= (3^{1/3} – 3^{-1/3})^2$
Taking square root: $x = 3^{1/3} – 3^{-1/3}$
Cubing both sides:
$$x^3 = (3^{1/3})^3 – (3^{-1/3})^3 – 3(3^{1/3})(3^{-1/3})(3^{1/3} – 3^{-1/3})$$$$x^3 = 3 – \frac{1}{3} – 3(1)(x)$$$$x^3 = \frac{9-1}{3} – 3x$$$$3x^3 = 8 – 9x$$$$3x^3 + 9x = 8 \text{ Proved.}$$

Solution:

a) They are equal in area. (तिनीहरूको क्षेत्रफल बराबर हुन्छ ।)

b) We know that area of a parallelogram is Base $\times$ Height.
Area of EBCF = Base $BC \times$ Height $AB$ (since AB is perpendicular to BC in square ABCD).
Area of Square ABCD = Base $BC \times$ Height $AB$.
Therefore, Area of EBCF = Area of Square ABCD.

c) Proof steps:
1. $\Delta PQT + \Delta SRT = \Delta TQR$ (Area of triangles on same base QR and between parallels is half of parallelogram? No. Here T is a point. $\Delta TQR = \frac{1}{2} \Box PQRS$). Also $\Delta PQT + \Delta SRT$ makes up the rest of the parallelogram if T is on PS? (Assumption based on key: $\Delta PQT + \Delta SRT = \Delta TQR$).
2. In $\Delta TQR$, M is mid-point of TR. So QM is the median.
3. Median divides triangle into equal areas. $\Delta TQM = \frac{1}{2} \Delta TQR$.
4. From 1 and 3, $\Delta TQM = \frac{1}{2} (\Delta PQT + \Delta SRT)$.

Solution:

a) Construction diagram:

b) Because they stand on the same base (AB) and lie between the same parallel lines.
(किनभने तिनीहरू एउटै आधार र उही समानान्तर रेखाहरू बीच रहेका छन् ।)

Solution:

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

b) (Experimental Verification: Draw two circles with radius $\ge 3cm$. Draw arc AB, center O, and point C on circumference. Measure $\angle AOB$ and $\angle ACB$. Result will show $\angle AOB = 2 \angle ACB$.)

c) We know, Central Angle = $2 \times$ Inscribed Angle
$$5x = 2(2x + 10)$$$$5x = 4x + 20$$$$x = 20$$ Answer: $x = 20$.

Solution:

a) Angle formed by the line of sight with the horizontal line when viewing an object above the observer’s eye level.
(दर्शकले आफूभन्दा माथि रहेको वस्तुलाई हेर्दा दृष्टि रेखाले क्षितिज रेखासँग बनाएको कोणलाई उन्नतांश कोण भनिन्छ ।)

b) Diagram:

c) In the right-angled triangle formed:
Perpendicular ($p$) = 6 m
Hypotenuse ($h$) = Broken part = $(x – 6)$ m
Angle = $30^{\circ}$
$$\sin 30^{\circ} = \frac{p}{h}$$$$\frac{1}{2} = \frac{6}{x – 6}$$$$x – 6 = 12$$$$x = 18$$ Height of tree = 18 m.

d) Total height = 18 m.
Let it break at height $h$.
Perpendicular = $h$.
Hypotenuse (Broken part) = $18 – h$.
Angle = $45^{\circ}$.
$$\sin 45^{\circ} = \frac{h}{18 – h}$$$$\frac{1}{\sqrt{2}} = \frac{h}{18 – h}$$$$18 – h = \sqrt{2}h$$$$18 = h(\sqrt{2} + 1)$$$$h = \frac{18}{2.414} \approx 7.46 \text{ m}$$

Solution:

a) Since $Q_1 = 35$, which lies between 20 and 40.
$Q_1$ Class is 20-40.

b) $N = 2 + X + 8 + 5 + 1 = 16 + X$
$cf$ of preceding class = 2.
$f$ of class = $X$.
$i$ = 20.
$L$ = 20.
$$Q_1 = L + \frac{\frac{N}{4} – cf}{f} \times i$$$$35 = 20 + \frac{\frac{16+X}{4} – 2}{X} \times 20$$$$15 = \frac{16+X-8}{4X} \times 20$$$$15 = \frac{8+X}{X} \times 5$$$$3X = 8 + X \implies 2X = 8 \implies X = 4$$

c) Max frequency is 8 (Class 40-60).
Mode Class = 40-60.
$L=40, f_1=8, f_0=X=4, f_2=5, i=20$.
$$Mode = L + \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \times i$$$$Mode = 40 + \frac{8 – 4}{16 – 4 – 5} \times 20$$$$Mode = 40 + \frac{4}{7} \times 20$$$$Mode = 40 + 11.43 = 51.43$$

d) Students below Q1 class (0-20) = 2.
Students above Q1 class (40-60, 60-80, 80-100) = $8 + 5 + 1 = 14$.
Ratio (Above : Below) = $14 : 2 = 7 : 1$.

Solution:

a) Two events are said to be independent if the occurrence of one event does not affect the probability of the occurrence of the other.
(यदि एउटा घटना घट्दा अर्को घटनालाई कुनै असर पर्दैन भने तिनीहरूलाई अनाश्रित घटना भनिन्छ ।)

b) Tree diagram:

c) $P(DD) = P(D) \times P(D) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

d) Max probability (Total Probability) = 1.
Probability of both sons $P(SS) = 1/4$.
Difference = $1 – 1/4 = 3/4$.
It is less by $3/4$.
(यो अधिकतम सम्भाव्यता भन्दा ३/४ ले कम छ ।)