NEB Class 12 Science Applied Mathematics Question Paper 2082 Nepal

Karl Pearson's coefficient of skewness based on mean, median and standard deviation is $S_k = \dfrac{3(\text{mean} - \text{median})}{\text{standard deviation}}$.

Volume of a hemisphere $= \dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\pi y^3\ \text{cm}^3$.

VAT is levied on the selling price, so $\text{VAT} = \text{rate of VAT} \times \text{Selling price}$.

Cost of 20 pens $= 20x$ and cost of 30 books $= 30y$. The maximum (at most) total cost is Rs. 7000, so $20x + 30y \le 7000$.

EMV = (probability of finding oil × its contribution) + (probability of not finding oil × its contribution) $= 0.34 \times 3900 + 0.66 \times 3500$.

The word "RAPATI" has 6 letters: R, A, P, A, T, I. The letter A appears 2 times. So $P(\text{A}) = \dfrac{2}{6} = \dfrac{1}{3}$.

$\dfrac{(n-1)!}{(n+1)!} = \dfrac{(n-1)!}{(n+1)\,n\,(n-1)!} = \dfrac{1}{n(n+1)}$.

Given $P(A) = \dfrac{1}{4}$ and $2P(B) = \dfrac{1}{4}$ so $P(B) = \dfrac{1}{8}$. For independent events, $P(A \cap B) = P(A) \cdot P(B) = \dfrac{1}{4} \times \dfrac{1}{8} = \dfrac{1}{32}$.

When demand exceeds (or equals) the stock, all units in stock are sold, so the pay-off (profit per period) is the margin/price times the number of units in stock: pay off = MP × units in stock.

Putting $y=0$: $4x = 24 \Rightarrow x = 6$, so $A=(6,0)$. Putting $x=0$: $3y = 24 \Rightarrow y = 8$, so $B=(0,8)$. Area of right triangle AOB $= \dfrac{1}{2} \times 6 \times 8 = 24$ sq. unit.

A path equally inclined to both axes makes an angle of $45^\circ$ with the axes, giving slope $\pm 1$ and passing through the origin. Among the options, $x - y = 0$ (i.e. $y = x$) passes through the origin with slope $1$, equally inclined to both axes.

Let $x$ = marks in science, $y$ = marks in math, $n = 6$.

$\sum x = 30+40+50+60+70+80 = 330$, $\bar{x} = 55$. $\sum y = 30+40+60+40+80+90 = 340$, $\bar{y} = 56.67$ (approx).

Using deviations $u = x - 55$ and $v = y - 56.\overline{6}$, or directly with sums: $\sum x^2 = 900+1600+2500+3600+4900+6400 = 19900$. $\sum y^2 = 900+1600+3600+1600+6400+8100 = 22200$. $\sum xy = 900+1600+3000+2400+5600+7200 = 20700$.

$r = \dfrac{n\sum xy - \sum x \sum y}{\sqrt{n\sum x^2 - (\sum x)^2}\,\sqrt{n\sum y^2 - (\sum y)^2}}$

$= \dfrac{6(20700) - (330)(340)}{\sqrt{6(19900) - 330^2}\,\sqrt{6(22200) - 340^2}}$

$= \dfrac{124200 - 112200}{\sqrt{119400 - 108900}\,\sqrt{133200 - 115600}}$

$= \dfrac{12000}{\sqrt{10500}\,\sqrt{17600}} = \dfrac{12000}{102.47 \times 132.66} = \dfrac{12000}{13593.6} \approx 0.883$.

So the coefficient of correlation $r \approx 0.88$ (strong positive correlation).

Total expenditure $= 14000 + 7000 + 6000 + 5000 + 4000 = 36000$.

Central angle for each item $= \dfrac{\text{expenditure}}{36000} \times 360^\circ$:

Draw a circle and mark sectors of $140^\circ$, $70^\circ$, $60^\circ$, $50^\circ$ and $40^\circ$ for food, education, clothing, medicine and fuel respectively (total $360^\circ$).

For the cone, slant height $l = 10$ ft and height $h = 6$ ft. The base radius $r = \sqrt{l^2 - h^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8$ ft.

Area of the base (circle) $= \pi r^2 = \dfrac{22}{7} \times 8^2 = \dfrac{22}{7} \times 64 = \dfrac{1408}{7} \approx 201.14$ sq. ft.

Base radius $r = 8$ ft (from part a). Total height $= 21$ ft, cone height $= 6$ ft, so cylinder height $H = 21 - 6 = 15$ ft. Slant height of cone $l = 10$ ft.

Canvas required = curved surface of cone + curved surface of cylinder (the base is the ground, not covered):

CSA of cone $= \pi r l = \dfrac{22}{7} \times 8 \times 10 = \dfrac{1760}{7} \approx 251.43$ sq. ft.

CSA of cylinder $= 2\pi r H = 2 \times \dfrac{22}{7} \times 8 \times 15 = \dfrac{5280}{7} \approx 754.29$ sq. ft.

Total canvas $= \dfrac{1760}{7} + \dfrac{5280}{7} = \dfrac{7040}{7} \approx 1005.71$ sq. ft.

Original price $P = 80000$.

After first 2 years at 10% depreciation: $P_2 = 80000(1 - 0.10)^2 = 80000 \times (0.9)^2 = 80000 \times 0.81 = 64800$.

After next 3 years at 15% depreciation: $P_5 = 64800(1 - 0.15)^3 = 64800 \times (0.85)^3 = 64800 \times 0.614125 = 39795.3$ (approx).

So the price at the end of 5 years $\approx$ Rs. 39795.30.

Comparison: Total depreciation $= 80000 - 39795.30 = 40204.70$, i.e. the final price is about $\dfrac{39795.30}{80000} \times 100\% \approx 49.74\%$ of the original price (machine has lost roughly half its value).

(a) Efficiency:

Ram does $\dfrac{1}{3}$ of the work in 6 days, so the whole work takes Ram $6 \times 3 = 18$ days. Ram's one-day work $= \dfrac{1}{18}$.

Geeta does $\dfrac{1}{4}$ of the work in 5 days, so the whole work takes Geeta $5 \times 4 = 20$ days. Geeta's one-day work $= \dfrac{1}{20}$.

Ratio of efficiencies (Ram : Geeta) $= \dfrac{1}{18} : \dfrac{1}{20} = 20 : 18 = 10 : 9$.

(b) Division of Rs. 18000:

Wages are divided in the ratio of efficiency (work done) $= 10 : 9$. Total parts $= 19$.

Ram's share $= \dfrac{10}{19} \times 18000 = \dfrac{180000}{19} \approx \text{Rs. } 9473.68$.

Geeta's share $= \dfrac{9}{19} \times 18000 = \dfrac{162000}{19} \approx \text{Rs. } 8526.32$.

(a) Equal number of male and female: committee of 8 with 4 male and 4 female.

Number of ways $= \binom{12}{4} \times \binom{10}{4} = 495 \times 210 = 103950$.

($\binom{12}{4} = \dfrac{12!}{4!\,8!} = 495$, $\binom{10}{4} = \dfrac{10!}{4!\,6!} = 210$.)

(b) 75% male and 25% female: of 8 people, 75% = 6 male and 25% = 2 female.

Number of ways $= \binom{12}{6} \times \binom{10}{2} = 924 \times 45 = 41580$.

($\binom{12}{6} = 924$, $\binom{10}{2} = 45$.)

The question asks the student to frame a linear programming problem (objective function + constraints) for maximum profit and to solve it graphically. As stated, no specific data (resources, profit per unit) is printed, so a representative LPP is set up.

Example formulation: Let a firm produce $x$ units of product I and $y$ units of product II with profits Rs. 30 and Rs. 20 per unit.

Objective function: Maximize $Z = 30x + 20y$.

Subject to constraints (resource/time limits), e.g.:

Graphical method: plot the lines $x + y = 8$ and $3x + y = 18$, shade the feasible region (intersection of half-planes with $x,y \ge 0$), find the corner points (e.g. $(0,0)$, $(6,0)$, $(5,3)$, $(0,8)$), evaluate $Z$ at each corner, and the maximum value of $Z$ gives the optimal solution.

We select 5 balls (2 red and 3 white) from one of the boxes.

Box A (2 red, 8 white, total 10): number of ways to choose 5 from 10 $= \binom{10}{5} = 252$. Ways to get exactly 2 red and 3 white $= \binom{2}{2}\binom{8}{3} = 1 \times 56 = 56$.

Box B (6 red, 6 white, total 12): number of ways to choose 5 from 12 $= \binom{12}{5} = 792$. Ways to get exactly 2 red and 3 white $= \binom{6}{2}\binom{6}{3} = 15 \times 20 = 300$.

Thus the two possible cases (drawing from box A or from box B) have probabilities $\dfrac{2}{9} \approx 0.222$ and $\dfrac{25}{66} \approx 0.379$ respectively.

Let $x$ = Maths, $y$ = Nepali, $n = 6$. We find the regression line of $y$ on $x$.

$\sum x = 50+45+65+30+60+80 = 330$, $\bar{x} = 55$. $\sum y = 40+60+55+45+70+75 = 345$, $\bar{y} = 57.5$.

$\sum x^2 = 2500+2025+4225+900+3600+6400 = 19650$. $\sum xy = 2000+2700+3575+1350+4200+6000 = 19825$.

Regression coefficient of $y$ on $x$:

Regression line of $y$ on $x$: $y - \bar{y} = b_{yx}(x - \bar{x})$

When $x = 55$: $y - 57.5 = 0.5667(0) = 0$, so $y = 57.5$.

The marks in Nepali when Maths is 55 $\approx 57.5$ (since 55 is the mean of Maths, the predicted value equals $\bar{y}$).

Multiply the chained ratios:

Therefore $a : d = 15 : 14$.

Cost of one wheel (cement ring) $=$ Rs. 1200, and 15 wheels are required.

Total cost $= 15 \times 1200 = \text{Rs. } 18000$.

Diameter $= 44$ cm, so radius $r = 22$ cm. Height of each wheel $= 21$ cm. For 12 wheels stacked, total height $H = 12 \times 21 = 252$ cm.

Volume of water (cylinder) $= \pi r^2 H = \dfrac{22}{7} \times 22^2 \times 252 = \dfrac{22}{7} \times 484 \times 252$.

$= \dfrac{22}{7} \times 121968 = 22 \times 17424 = 383328\ \text{cm}^3$.

Since $1$ litre $= 1000\ \text{cm}^3$, volume $= \dfrac{383328}{1000} = 383.328$ litres.

Rate $= 100$ litres per 5 minutes $= 20$ litres per minute.

Time to fill 900 litres $= \dfrac{900}{20} = 45$ minutes.

(Equivalently, $\dfrac{900}{100} \times 5 = 9 \times 5 = 45$ minutes.)

For 2 years, the difference between compound interest and simple interest is given by

Here difference $= 250$, $R = 8\%$:

The sum of money is Rs. 39062.50.

The bag has 20 bulbs, of which 8 are defective. The number of defective bulbs in the bag is fixed at 8, not more than 6 — however, as stated the question is ambiguous (likely intends drawing a sample of bulbs and finding the probability that more than 6 are defective).

Interpreting it as a single draw, the probability of selecting a defective bulb is

If instead the intended question is the probability that a selected single bulb is defective, the answer is $\dfrac{2}{5}$. The phrase 'more than 6 bulbs are defective' is unclear for a single selection, so this part needs review against the intended wording.

Profit is shared in the ratio of investments (for equal time periods):

Total parts $= 4 + 5 = 9$.

A's share $= \dfrac{4}{9} \times 10000 = \dfrac{40000}{9} \approx \text{Rs. } 4444.44$.

B's share $= \dfrac{5}{9} \times 10000 = \dfrac{50000}{9} \approx \text{Rs. } 5555.56$.