SEE 2081 (2025) Compulsory Mathematics Gandaki Province Solution | गण्डकी प्रदेश अनिवार्य गणित

Solution:

a) $n(\overline{T \cup C}) = 50$

b) Venn diagram representation:

c) Let $n(T \cap C) = x$
Total people $n(U) = 450$
Tea only: $200 – x$
Coffee only: $250 – x$
Both: $x$
None: $50$

Equation: $(200 – x) + (250 – x) + x + 50 = 450$
$500 – x = 450$
$x = 50$

Tea only = $200 – 50 = 150$
$\therefore$ चिया मात्र मन पराउने मानिसको संख्या १५० छ।

d) Both tea and coffee: $n(T \cap C) = 50$
Coffee only: $n_o(C) = 250 – 50 = 200$
Comparison: $n(T \cap C) < n_o(C)$ by $150$
(कफी मात्र मन पराउने २०० जना र दुवै मन पराउने ५० जना छन्। दुवै मन पराउने भन्दा कफी मात्र मन पराउने १५० ले धेरै छन्।)

Solution:

a) 4 times (४ पटक)

b) Principal ($P$) = Rs. 50,000
Time ($T$) = 2 years
Rate ($R$) = 8% p.a.

Annual Compound Interest:
$CI = P\left[(1 + \frac{R}{100})^T – 1\right]$
$= 50,000\left[(1 + \frac{8}{100})^2 – 1\right]$
$= 50,000[(1.08)^2 – 1]$
$= 50,000[1.1664 – 1]$
$= 50,000 \times 0.1664$
$= \text{Rs. } 8,320$

c) Semi-annual Compound Interest:
$CI_{semi} = P\left[(1 + \frac{R}{200})^{2T} – 1\right]$
$= 50,000\left[(1 + \frac{8}{200})^{4} – 1\right]$
$= 50,000[(1.04)^4 – 1]$
$= 50,000[1.16985856 – 1]$
$= 50,000 \times 0.16985856$
$\approx \text{Rs. } 8,492.93$

Difference = Semi-annual CI – Annual CI
$= 8,492.93 – 8,320$
$= \text{Rs. } 172.93$
$\therefore$ अर्धवार्षिक चक्रीय व्याज वार्षिक चक्रीय व्याजभन्दा रु. १७२.९३ ले बढी छ।

Solution:

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

b) Depreciation in first year:
$10\%$ of Rs. 45,00,000
$= \frac{10}{100} \times 45,00,000$
$= \text{Rs. } 4,50,000$
$\therefore$ पहिलो वर्षमा बसको मूल्य रु. ४,५०,००० ले घटेछ।

c) Initial Cost ($V_0$) = Rs. 45,00,000
Rate ($R$) = 10%
Time ($T$) = 2 years

Selling Price after 2 years:
$SP = 45,00,000 \left(1 – \frac{10}{100}\right)^2$
$= 45,00,000 \times (0.9)^2$
$= 45,00,000 \times 0.81$
$= \text{Rs. } 36,45,000$

Total amount recovered = Selling Price + Income Earned
Total = $36,45,000 + 12,00,000 = \text{Rs. } 48,45,000$

Profit = Total Recovered – Initial Cost
Profit = $48,45,000 – 45,00,000 = \text{Rs. } 3,45,000$

Profit Percentage:
$\text{Profit \%} = \frac{\text{Profit}}{\text{Initial Cost}} \times 100\%$
$= \frac{3,45,000}{45,00,000} \times 100\%$
$= 7.67\%$
$\therefore$ २ वर्षपछि बस विक्री गर्दा ७.६७% नाफा हुन्छ।

Solution:

a) Difference = Selling Rate – Buying Rate
$= 138.83 – 138.23$
$= \text{Rs. } 0.60$
$\therefore$ विक्री दर खरिद दरभन्दा ६० पैसा बढी छ।

b) Buying rate is used when bank buys dollars (customer sells dollars).
For $500:
NRs = $500 \times 138.23$
$= \text{Rs. } 69,115$
$\therefore$ अमेरिकी डलर ५०० सँग रु. ६९,११५ साट्न सकिन्छ।

c) Initial selling rate = Rs. 138.83
New selling rate = Rs. 139.80
Increase = $139.80 – 138.83 = \text{Rs. } 0.97$

Percentage devaluation:
$= \frac{\text{Increase}}{\text{Initial Rate}} \times 100\%$
$= \frac{0.97}{138.83} \times 100\%$
$\approx 0.698\% \approx 0.7\%$
$\therefore$ नेपाली मुद्रा ०.७% ले अवमूल्यन भएछ।

Solution:

a) 4 triangular surfaces (चार ओटा त्रिभुजाकार सतहहरू)

b) Height ($h$) = 12 cm
Base side ($a$) = 10 cm

Volume of pyramid:
$V = \frac{1}{3} \times \text{Base Area} \times \text{Height}$
$= \frac{1}{3} \times a^2 \times h$
$= \frac{1}{3} \times 10^2 \times 12$
$= \frac{1}{3} \times 100 \times 12$
$= 400 \text{ cm}^3$
$\therefore$ पिरामिडको आयतन ४०० घन से.मि. छ।

c) First, find slant height ($l$):
$l = \sqrt{h^2 + \left(\frac{a}{2}\right)^2}$
$= \sqrt{12^2 + 5^2}$
$= \sqrt{144 + 25}$
$= \sqrt{169} = 13 \text{ cm}$

Total Surface Area (TSA):
TSA = Base Area + Lateral Surface Area
$= a^2 + 2al$
$= 10^2 + 2 \times 10 \times 13$
$= 100 + 260$
$= 360 \text{ cm}^2$
$\therefore$ पिरामिडको पुरा सतहको क्षेत्रफल ३६० वर्ग से.मि. छ।

Solution:

a) Circular shape (वृत्ताकार आकार)

b) Volume of cylinder = $5632 \text{ cm}^3$
Length of cylinder ($h_{cyl}$) = 28 cm

Find radius ($r$):
$V_{cyl} = \pi r^2 h$
$5632 = \frac{22}{7} \times r^2 \times 28$
$5632 = 22 \times 4 \times r^2$
$5632 = 88r^2$
$r^2 = 64 \implies r = 8 \text{ cm}$

Height of cone ($h_{cone}$):
Slant height ($l$) = 17 cm, $r$ = 8 cm
$h_{cone} = \sqrt{l^2 – r^2}$
$= \sqrt{17^2 – 8^2}$
$= \sqrt{289 – 64}$
$= \sqrt{225} = 15 \text{ cm}$

Comparison:
Length of cylinder = 28 cm
Height of cone = 15 cm
$\therefore$ बेलनाकार भागको लम्बाइ (२८ से.मि.) सोली भागको उचाइ (१५ से.मि.) भन्दा १३ से.मि. ले बढी छ।

c) Volume of cone:
$V_{cone} = \frac{1}{3} \pi r^2 h$
$= \frac{1}{3} \times \frac{22}{7} \times 8^2 \times 15$
$= \frac{1}{3} \times \frac{22}{7} \times 64 \times 15$
$= \frac{21120}{21}$
$\approx 1005.71 \text{ cm}^3$

Check if $V_{cyl} = 5 \times V_{cone}$:
$5 \times 1005.71 = 5028.55 \text{ cm}^3$
But $V_{cyl} = 5632 \text{ cm}^3$
$5632 \neq 5028.55$
$\therefore$ बेलनाकार भागको आयतन सोली भागको आयतनको पाँच गुणा हुँदैन।

Solution:

a) Area of floor = Length $\times$ Breadth
$= 12 \times 8$
$= 96 \text{ m}^2$
$\therefore$ कोठाको भुइँको क्षेत्रफल ९६ वर्ग मिटर छ।

b) Area of 4 walls:
$= 2h(l + b)$
$= 2 \times 3(12 + 8)$
$= 6 \times 20 = 120 \text{ m}^2$

Area of ceiling:
$= l \times b = 12 \times 8 = 96 \text{ m}^2$

Area of openings:
Area of 2 windows = $2 \times (2 \times 2) = 8 \text{ m}^2$
Area of door = $1.5 \times 1 = 1.5 \text{ m}^2$
Total openings = $8 + 1.5 = 9.5 \text{ m}^2$

Total area to be painted:
$= \text{Area of walls} + \text{Area of ceiling} – \text{Openings}$
$= 120 + 96 – 9.5$
$= 206.5 \text{ m}^2$

Cost of painting:
Rate = Rs. 15 per $\text{m}^2$
Total cost = $206.5 \times 15$
$= \text{Rs. } 3,097.50$
$\therefore$ रङ लगाउँदा जम्मा रु. ३,०९७.५० खर्च लाग्छ।

Solution:

a) $$AM = \frac{a + b}{2}$$

b) First term ($a$) = 3
Last term ($b$) = 27
Number of means ($n$) = 7
Total terms = $n + 2 = 9$

Common difference ($d$):
$b = a + (n+1)d$
$27 = 3 + 8d$
$8d = 24 \implies d = 3$

$5^{th}$ mean = $a + 5d$
$= 3 + 5 \times 3$
$= 3 + 15 = 18$
$\therefore$ पाँचौं मध्यमा १८ हो।

c) Arithmetic Mean (AM):
$AM = \frac{3 + 27}{2} = \frac{30}{2} = 15$

Geometric Mean (GM):
$GM = \sqrt{3 \times 27} = \sqrt{81} = 9$

Comparison:
$AM > GM$
Difference = $15 – 9 = 6$
$\therefore$ समानान्तरीय मध्यमा (१५) गुणोत्तर मध्यमा (९) भन्दा ६ ले ठूलो छ।

Solution:

a) $$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$$

b) Let length = $l$, breadth = $b$
Perimeter: $2(l + b) = 44$
$\Rightarrow l + b = 22$ …(i)

Area: $l \times b = 120$ …(ii)

From (i): $b = 22 – l$
Substitute in (ii):
$l(22 – l) = 120$
$22l – l^2 = 120$
$l^2 – 22l + 120 = 0$
$(l – 12)(l – 10) = 0$
$l = 12$ or $l = 10$

If $l = 12$, then $b = 10$
If $l = 10$, then $b = 12$
$\therefore$ लम्बाइ १२ मि. र चौडाइ १० मि. छ।

c) Current dimensions: $l = 12$ m, $b = 10$ m
Current area = $120 \text{ m}^2$

To make it square, reduce length to 10 m:
New square side = 10 m
New area = $10 \times 10 = 100 \text{ m}^2$

Decrease in area = $120 – 100 = 20 \text{ m}^2$
Percentage decrease:
$= \frac{\text{Decrease}}{\text{Original Area}} \times 100\%$
$= \frac{20}{120} \times 100\%$
$= 16.67\%$
$\therefore$ नयाँ चौरको क्षेत्रफल १६.६७% ले कम हुन्छ।

Solution:

a) $\frac{a}{a-b} + \frac{b}{b-a}$
Note: $b-a = -(a-b)$
So, $\frac{b}{b-a} = \frac{b}{-(a-b)} = -\frac{b}{a-b}$

$\frac{a}{a-b} – \frac{b}{a-b}$
$= \frac{a – b}{a – b}$
$= 1$
$\therefore$ उत्तर १ हो।

b) $2^x + \frac{1}{2^x} = 2\frac{1}{2} = \frac{5}{2}$
Let $2^x = y$
Then $y + \frac{1}{y} = \frac{5}{2}$

Multiply both sides by $2y$:
$2y^2 + 2 = 5y$
$2y^2 – 5y + 2 = 0$
$2y^2 – 4y – y + 2 = 0$
$2y(y – 2) – 1(y – 2) = 0$
$(y – 2)(2y – 1) = 0$

So, $y = 2$ or $y = \frac{1}{2}$

Case 1: $2^x = 2 \implies 2^x = 2^1 \implies x = 1$
Case 2: $2^x = \frac{1}{2} \implies 2^x = 2^{-1} \implies x = -1$

$\therefore x = 1$ or $x = -1$

Solution:

a) The area of a triangle is half of the area of a parallelogram standing on the same base and between the same parallel lines.
(त्रिभुजको क्षेत्रफल उही आधार र उही समानान्तर रेखाहरू बिच रहेको समानान्तर चतुर्भुजको क्षेत्रफलको आधा हुन्छ।)

b) Given: $BC = 6$ cm, $DF = 8$ cm
$ABCD$ is a parallelogram ($DA // CB$ and $AB // DC$ from figure)

Area of parallelogram $ABCD$:
$= \text{base} \times \text{height}$
$= BC \times DF$
$= 6 \times 8 = 48 \text{ cm}^2$

Now, $\Delta ABE$ and parallelogram $ABCD$ stand on the same base $AB$ and between the same parallel lines $AB$ and $EC$.
$\therefore$ Area of $\Delta ABE = \frac{1}{2} \times$ Area of parallelogram $ABCD$
$= \frac{1}{2} \times 48 = 24 \text{ cm}^2$
$\therefore$ $\Delta ABE$ को क्षेत्रफल २४ वर्ग से.मि. छ।

c) Proof:
1. Given: Area of $\Delta AOB$ = Area of $\Delta COD$
2. Adding Area of $\Delta BOC$ to both sides:
Area of $\Delta AOB$ + Area of $\Delta BOC$ = Area of $\Delta COD$ + Area of $\Delta BOC$
3. LHS: Area of $\Delta AOB$ + Area of $\Delta BOC$ = Area of $\Delta ABC$
4. RHS: Area of $\Delta COD$ + Area of $\Delta BOC$ = Area of $\Delta BCD$
5. So, Area of $\Delta ABC$ = Area of $\Delta BCD$
6. These triangles share the same base $BC$
7. Triangles having equal areas and standing on the same base lie between the same parallel lines
8. $\therefore AD // BC$
(प्रमाणित)

Solution:

a) $\angle QRS = \frac{1}{2} \times$ reflex $\angle QOS$
(The angle at the circumference is half the angle at the center standing on the same arc)

b) $\angle QPS = 46^{\circ}$ (given)
In cyclic quadrilateral $PQRS$, $\angle QPS$ and $\angle QRS$ are opposite angles.
But $\angle QOS$ is the central angle for arc $QS$.
Actually, $\angle QPS$ is an angle at circumference standing on arc $QRS$.
The central angle $\angle QOS$ (non-reflex) stands on arc $QS$.

According to circle theorem:
Angle at center = $2 \times$ Angle at circumference standing on same arc
$\angle QOS = 2 \times \angle QPS$ (both stand on arc $QS$)
$= 2 \times 46^{\circ} = 92^{\circ}$
$\therefore \angle QOS = 92^{\circ}$

c) Experimental Verification:
1. Draw two circles with radius > 3 cm
2. Mark a cyclic quadrilateral $PQRS$ in each circle
3. Measure $\angle QPS$ and $\angle QRS$ using protractor
4. Add the measurements
5. Result will show $\angle QPS + \angle QRS = 180^{\circ}$
(This verifies the theorem: Opposite angles of a cyclic quadrilateral are supplementary)

Solution:

a) Construction steps:

1. Draw $MN = 8$ cm
2. At point $N$, construct $\angle MNO = 60^{\circ}$
3. With $N$ as center, radius 7 cm, cut an arc on $NO$ to get point $O$
4. Join $M$ to $O$ to complete $\Delta MNO$
5. Draw a line parallel to $NO$ through $M$
6. Draw perpendicular bisector of $NO$ to find midpoint
7. Complete rectangle $NPQR$ with area equal to $\Delta MNO$

b) The areas are equal because the rectangle is constructed such that:
Area of $\Delta MNO = \frac{1}{2} \times NO \times$ height
Area of rectangle $NPQR = NP \times PQ$
In the construction, $PQ = \frac{1}{2} \times$ height of triangle and $NP = NO$
So, Area of rectangle = $NO \times \frac{1}{2} \times$ height = Area of triangle
$\therefore$ दुवैको क्षेत्रफल बराबर हुन्छ।

Solution:

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

b) $\angle HAC = 30^{\circ}$ (angle of depression from tower to house roof)
From the roof of the house to the top of the tower, the angle of elevation will be equal to the angle of depression (alternate angles).
$\therefore$ Angle of elevation = $30^{\circ}$