PABSON Kathmandu Compulsory Mathematics 2082 Pre-Qualifying Examination Solution Class 10 | अनिवार्य गणित

Solution:

a) $n(L\cup T)=200$
Explanation: Total people = 300, 1/3 dislike both = 100. So, people who like at least one item = 300 – 100 = 200.

b) Venn diagram representation:

c) Let $x$ be the number of people who like both Tablet and Laptop.
$120 + 100 – x + 100 = 300$
$320 – x = 300$
$x = 20$
Only Tablet = $120 – 20 = 100$
Only Laptop = $100 – 20 = 80$
Number of people who like only one item = $100 + 80 = \mathbf{180}$.

d) Number of people who like at most one item = Only one item + None = $180 + 100 = 280$
Percentage = $\frac{280}{300} \times 100\% = \mathbf{93.33\%}$.

Solution:

a) Quarterly compound interest is more profitable because interest is compounded more frequently.

b) For half-yearly compounding:
Principal (P) = Rs 2,00,000
Annual rate (R) = 8%
Half-yearly rate = 8/2 = 4%
Time (T) = 2 years = 4 half-years
Amount (A) = $P \left(1 + \frac{R}{100}\right)^n = 200000 \left(1 + \frac{4}{100}\right)^4$
= $200000 \times (1.04)^4$
= $200000 \times 1.16985856$
= Rs 2,33,971.71
Compound Amount = Rs 2,33,971.71

c) First year (half-yearly):
Amount after 1 year = $200000 \left(1 + \frac{4}{100}\right)^2 = 200000 \times 1.0816 = Rs 2,16,320$
Second year (quarterly):
Principal = Rs 2,16,320
Quarterly rate = 8/4 = 2%
Number of quarters = 4
Amount after 2nd year = $216320 \left(1 + \frac{2}{100}\right)^4 = 216320 \times 1.08243216 = Rs 2,34,151.72$
Total interest in two years = Final Amount – Principal = $2,34,151.72 – 2,00,000 = \mathbf{Rs~34,151.72}$.

Solution:

a) $P_T = P \left(1 + \frac{R}{100}\right)^T$, where $P_T$ = population after T years, $P$ = initial population, $R$ = growth rate.

b) Let population 3 years ago be $P$.
$59,895 = P \left(1 + \frac{10}{100}\right)^3$
$59,895 = P \times (1.1)^3$
$59,895 = P \times 1.331$
$P = \frac{59895}{1.331} = \mathbf{45,000}$
Population 3 years ago = 45,000.

c) Population 2 years ago = 45,000
Population 1 year ago = $45,000 \times 1.1 = 49,500$
If growth rate was 8% last year:
Present population = $49,500 \times 1.08 = \mathbf{53,460}$.

Solution:

a) Amount of Chinese Yuan = $\frac{9,56,000}{19.12} = \mathbf{50,000}$ Yuan.

b) After devaluation, new rate: 1 Yuan = $19.12 \times 1.01 = Rs 19.3112$
Nepali rupees received = $50,000 \times 19.3112 = \mathbf{Rs~9,65,560}$.

c) Profit = Final amount – Initial amount = $9,65,560 – 9,56,000 = Rs 9,560$
Profit percent = $\frac{9,560}{9,56,000} \times 100\% = \mathbf{1\%}$.

Solution:

a) Lateral surface area = $2al$, where $a$ = base side, $l$ = slant height.

b) Volume = $\frac{1}{3} \times \text{Base Area} \times \text{Height}$
$384 = \frac{1}{3} \times (12)^2 \times h$
$384 = \frac{1}{3} \times 144 \times h$
$h = \frac{384 \times 3}{144} = 8$ cm
Slant height $(l) = \sqrt{h^2 + \left(\frac{a}{2}\right)^2} = \sqrt{8^2 + 6^2} = \sqrt{64+36} = \sqrt{100} = 10$ cm
Lateral surface area = $2 \times 12 \times 10 = 240$ cm²
Base area = $12^2 = 144$ cm²
Total surface area = $240 + 144 = 384$ cm²
Cost = $384 \times 2 = \mathbf{Rs~768}$.

c) Volume remains same = 384 cm³
For new pyramid: $\frac{1}{3} \times a^2 \times 18 = 384$
$6a^2 = 384$
$a^2 = 64$
$a = \mathbf{8}$ cm.

Solution:

a) Volume of hemisphere = $\frac{2}{3}\pi r^3$.

b) Height of hemisphere = Total height – Cylinder height = $94 – 80 = 14$ cm
Radius $(r) = 14$ cm
Cylinder surface area = $2\pi r h + \pi r^2$ (only one circle included as top is covered by hemisphere)
= $2\pi \times 14 \times 80 + \pi \times 14^2$
= $2240\pi + 196\pi = 2436\pi$ cm²
Hemisphere curved surface area = $2\pi r^2 = 2\pi \times 14^2 = 392\pi$ cm²
Total surface area = $2436\pi + 392\pi = 2828\pi$ cm²
Using $\pi = \frac{22}{7}$: $2828 \times \frac{22}{7} = 2828 \times 3.142857 \approx \mathbf{8888}$ cm².

Solution:

a) Volume = Length × Width × Height = $20 \times 0.5 \times 3 = \mathbf{30}$ m³.

b) Volume of one brick = $0.2 \times 0.1 \times 0.08 = 0.0016$ m³
Number of bricks = $\frac{30}{0.0016} = \mathbf{18,750}$ bricks.

c) Cost per brick = $\frac{13,500}{1000} = Rs 13.5$
Total cost = $18,750 \times 13.5 = \mathbf{Rs~2,53,125}$.

Solution:

a) Arithmetic mean between a and b = $\frac{a+b}{2}$.

b) $t_5 = a + 4d = 34$ … (i)
$t_{12} = a + 11d = 83$ … (ii)
Subtracting (i) from (ii): $(a+11d) – (a+4d) = 83 – 34$
$7d = 49 \Rightarrow d = \mathbf{7}$
Substitute in (i): $a + 4(7) = 34 \Rightarrow a = 34 – 28 = \mathbf{6}$
Common difference = 7, First term = 6.

c) Sum of first 5 terms = $\frac{n}{2}[2a + (n-1)d] = \frac{5}{2}[2\times6 + (5-1)\times7]$
= $\frac{5}{2}[12 + 28] = \frac{5}{2} \times 40 = \mathbf{100}$.

Solution:

a) Roots are given by: $x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$.

b) Let Ramu’s age = $x$, Jinu’s age = $y$
$x + y = 19$ … (i)
After 1 year: $(x+1)(y+1) = 104$
$xy + x + y + 1 = 104$
$xy + 19 + 1 = 104$ (from i)
$xy = 84$ … (ii)
From (i) and (ii): $x(19-x) = 84$
$19x – x^2 = 84$
$x^2 – 19x + 84 = 0$
$(x-12)(x-7) = 0$
$x = 12$ or $x = 7$
If $x=12$, $y=7$; if $x=7$, $y=12$
Present ages: Ramu = 12 years, Jinu = 7 years.

c) Current sum = 19
Required increase = 25 – 19 = 6 years
This increase will be shared by both, so time = $\frac{6}{2} = \mathbf{3}$ years.

Solution:

a) $\frac{1}{(x+1)^2} + \frac{1}{(x+2)^2} – \frac{1}{(x+1)^2(x+2)^2}$
= $\frac{(x+2)^2 + (x+1)^2}{(x+1)^2(x+2)^2} – \frac{1}{(x+1)^2(x+2)^2}$
= $\frac{x^2+4x+4 + x^2+2x+1 – 1}{(x+1)^2(x+2)^2}$
= $\frac{2x^2+6x+4}{(x+1)^2(x+2)^2}$
= $\frac{2(x^2+3x+2)}{(x+1)^2(x+2)^2}$
= $\frac{2(x+1)(x+2)}{(x+1)^2(x+2)^2}$
= $\mathbf{\frac{2}{(x+1)(x+2)}}$.

b) $9^{x-2} + 2\times3^{2x-3} = 63$
$3^{2(x-2)} + 2\times3^{2x-3} = 63$
$3^{2x-4} + 2\times3^{2x-3} = 63$
Let $y = 3^{2x}$, then:
$\frac{y}{3^4} + 2\times\frac{y}{3^3} = 63$
$\frac{y}{81} + \frac{2y}{27} = 63$
$\frac{y + 6y}{81} = 63$
$\frac{7y}{81} = 63$
$7y = 63 \times 81$
$y = 9 \times 81 = 729$
$3^{2x} = 729 = 3^6$
$2x = 6$
$\mathbf{x = 3}$.

Solution:

a) $x = 2y$ or $y = \frac{x}{2}$.

b) Proof: Parallelogram PQRS and triangle PQM stand on same base PQ and between same parallels PQ and SM.
Area of parallelogram = base × height = PQ × h
Area of triangle = $\frac{1}{2}$ × base × height = $\frac{1}{2}$ × PQ × h
Thus, area of parallelogram = 2 × area of triangle.
Hence, $x = 2y$.

c) Construction steps:
1. Draw AB = 4.2 cm
2. With center A and radius 4.8 cm, draw an arc
3. With center B and radius 6.5 cm, draw another arc intersecting first arc at D
4. Join AD and BD
5. With center B and radius 5.6 cm, draw an arc
6. With center D and radius 5 cm, draw another arc intersecting at C
7. Join BC, CD to complete quadrilateral ABCD
8. To construct triangle ADP with equal area: Draw line through C parallel to BD meeting AD extended at P
9. Join AP to get triangle ADP.

Solution:

a) In parallelogram ABCD: $\angle DAB + \angle ABC = 180^\circ$ (co-interior angles) … (i)
In cyclic quadrilateral APQD: $\angle DAP + \angle PQD = 180^\circ$ (opposite angles) … (ii)
But $\angle DAP = \angle DAB$ (same angle)
From (i) and (ii): $\angle DAB + \angle ABC = \angle DAB + \angle PQD$
Therefore, $\angle ABC = \angle PQD$.

b) Yes, PBCQ is a cyclic quadrilateral.
Reason: $\angle PQD = \angle ABC$ (proved above) and $\angle PQD$ is the exterior angle of quadrilateral PBCQ at vertex Q, which equals the opposite interior angle at C ($\angle ABC$). When exterior angle equals opposite interior angle, the quadrilateral is cyclic.

Solution:

a) $\angle PSQ = 90^\circ$ (Angle in a semicircle).

b) In cyclic quadrilateral PQRS: $\angle PQS + \angle PRS = 180^\circ$
$45^\circ + \angle PRS = 180^\circ \Rightarrow \angle PRS = 135^\circ$
$\angle QRS = \angle PRS – \angle PRQ$
But $\angle PRQ = \angle PSQ = 90^\circ$ (angles in same segment)
$\angle QRS = 135^\circ – 90^\circ = \mathbf{45^\circ}$.

c) Experimental verification:
1. Draw two circles with radii > 3 cm
2. In each circle, mark an arc AB
3. Measure central angle AOB and circumference angle ACB standing on same arc AB
4. Tabulate results:

Circle	Central angle	Circumference angle	Ratio
1	80°	40°	2:1
2	120°	60°	2:1

Conclusion: Central angle is twice the circumference angle.

Solution:

a) The angle of depression is the angle formed between the horizontal line from the observer’s eye and the line of sight to an object below the horizontal level.

b)

c) Let tree height = $h$ m
From diagram: $\tan 30^\circ = \frac{h-10}{12}$
$\frac{1}{\sqrt{3}} = \frac{h-10}{12}$
$h-10 = \frac{12}{\sqrt{3}} = 4\sqrt{3} \approx 6.928$
$h = 10 + 6.928 = \mathbf{16.93}$ m.

d) Shortest distance = $\sqrt{12^2 + (16.93)^2}$
= $\sqrt{144 + 286.62}$
= $\sqrt{430.62} \approx \mathbf{20.75}$ m.

Solution:

a) Arithmetic mean = $\frac{\sum fm}{N}$, where f = frequency, m = mid-value, N = total frequency.

b)

Marks	f	m	fm
0-10	12	5	60
10-20	18	15	270
20-30	27	25	675
30-40	x	35	35x
40-50	17	45	765
50-60	6	55	330
Total	80+x		2100+35x

Mean = $\frac{2100+35x}{80+x} = 28$
$2100+35x = 28(80+x)$
$2100+35x = 2240+28x$
$7x = 140$
$\mathbf{x = 20}$.

c) Total students = 80+20 = 100
Median position = $\frac{N}{2} = \frac{100}{2} = 50$
Cumulative frequencies: 12, 30, 57, 77, 94, 100
Median class = 20-30 (where cumulative frequency reaches 50)
$L = 20$, $c.f. = 30$, $f = 27$, $i = 10$
Median = $L + \frac{\frac{N}{2} – c.f.}{f} \times i$
= $20 + \frac{50-30}{27} \times 10$
= $20 + \frac{20}{27} \times 10$
= $20 + 7.41 = \mathbf{27.41}$.

d) Failed students = students with marks < 30 = 12+18+27 = 57
Percentage = $\frac{57}{100} \times 100\% = \mathbf{57\%}$.

Solution:

a) A and B are independent events.

b) Tree diagram:

c) Probability (both blue) = $\frac{5}{9} \times \frac{4}{8} = \frac{20}{72} = \mathbf{\frac{5}{18}}$.

d) Probability (same color) = P(RR) + P(BB)
P(RR) = $\frac{4}{9} \times \frac{3}{8} = \frac{12}{72} = \frac{1}{6}$
P(BB) = $\frac{5}{18}$
P(same) = $\frac{1}{6} + \frac{5}{18} = \frac{3+5}{18} = \frac{8}{18} = \frac{4}{9}$
Probability (different color) = 1 – P(same) = $1 – \frac{4}{9} = \frac{5}{9}$
Comparison: $\frac{5}{9} > \frac{4}{9}$, so probability of different colors is higher.