NEB Class 11 Science Applied Mathematics Question Paper 2078 Nepal

The 'most demanding' (most frequent) shoe size is given by the mode, which identifies the most frequently occurring value. Answer: (c) Mode.

1 Kattha = 20 Dhur, and in the Ropani–Aana system 1 Kattha ≈ 5.0 Aana... however the standard conversion used here gives 1 Kattha ≈ 10.65 Aana. Answer: (a) 10.65.

Length $=\dfrac{\text{area}}{\text{breadth}}=\dfrac{24x^2y-18xy^2}{4x-3y}=\dfrac{6xy(4x-3y)}{4x-3y}=6xy$. Answer: (a) 6xy.

Cost: 2 kg tea = 3.3 kg sugar; 2.2 kg sugar = 0.5 kg coffee. Convert 2 kg coffee → sugar: $2\,\text{kg coffee}=\dfrac{2.2}{0.5}\times2=8.8\,\text{kg sugar}$. Convert sugar → tea: $8.8\,\text{kg sugar}=\dfrac{2}{3.3}\times8.8=5.33\,\text{kg tea}$. Answer: (a) 5.3 kg.

First child: 5 choices; second child: 4 remaining choices (different college). Total $=5\times4=20$ ways. Answer: (b) 20.

The total cost cannot exceed the budget: $50x+200y\le 20000$. Answer: (b).

Getting grade A and grade B on the same test are mutually exclusive events, so they cannot both occur: $P(\text{both})=0$. Answer: (c) 0.

A 30-day month has 4 complete weeks (28 days) + 2 extra days. There are 7 equally likely pairs of consecutive extra days; Saturday appears in 2 of these pairs (Fri-Sat and Sat-Sun). So $P(5\text{ Saturdays})=\dfrac{2}{7}$. Answer: (a) 2/7.

Numbers < 500 with three different digits (digits 0–9, no repetition). Hundreds digit can be 1,2,3,4 (4 choices, can't be 0). Tens digit: any of remaining 9 digits. Units digit: any of remaining 8. Total $=4\times9\times8=288$. Answer: (b) 288.

The relation is $Y=4X+2$: $4(2)+2=10$, $4(5)+2=22$, $4(7)+2=30$. So for $X=10$: $Y=4(10)+2=42$. Answer: (c) 42.

The line $y=x$ passes through the origin with slope 1, bisecting the first and third quadrants into two equal halves. Answer: (c).

Consistency is measured by the coefficient of variation (CV); the smaller CV is more consistent.

X: mean $\bar{X}=\dfrac{56+72+48+69+64+81}{6}=\dfrac{390}{6}=65$. Deviations: −9,7,−17,4,−1,16; squares: 81,49,289,16,1,256; sum=692; $\sigma_X=\sqrt{692/6}=\sqrt{115.33}=10.74$. $CV_X=\dfrac{10.74}{65}\times100=16.5\%$.

Y: mean $\bar{Y}=\dfrac{63+74+45+57+82+63}{6}=\dfrac{384}{6}=64$. Deviations: −1,10,−19,−7,18,−1; squares: 1,100,361,49,324,1; sum=836; $\sigma_Y=\sqrt{836/6}=\sqrt{139.33}=11.80$. $CV_Y=\dfrac{11.80}{64}\times100=18.4\%$.

Since $CV_X(16.5\%)<CV_Y(18.4\%)$, student X is more consistent and should get the award.

Read the applicable slab rate and minimum charge for each meter capacity from the NEA tariff table, then compute each bill:

• A (15 A, 188 units): bill = (energy charge per unit for the 15 A slab covering 188 units × 188) + the 15 A minimum/service charge.
• B (30 A, 175 units): bill = (energy charge per unit for the 30 A slab covering 175 units × 175) + the 30 A minimum/service charge.

Compare the two totals; the one with the larger total pays more. (Exact figures require the printed per-unit rates and minimum charges in the tariff table image.)

Independent events: Two events are independent if the occurrence of one does not affect the probability of the other, i.e. $P(A\cap B)=P(A)\times P(B)$. Example: tossing a coin and rolling a die — the coin outcome does not affect the die outcome.

Probability of winning: $P(\text{head})=\dfrac{1}{2}$; $P(\text{even number on die})=\dfrac{3}{6}=\dfrac{1}{2}$. Since independent, $P(\text{win})=\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{1}{4}$.

Let $x$ = bags of M, $y$ = bags of N.

Model: Minimize $C=50x+40y$ subject to $6x+3y\ge 30$ (nitrogen), $3x+3y\ge 18$ (potassium), $x,y\ge 0$.

Simplify: $2x+y\ge 10$, $x+y\ge 6$.

Corner points: intersection of $2x+y=10$ and $x+y=6$ → $x=4,y=2$. Also $(0,10)$ from $2x+y=10$ at $x=0$, and $(6,0)$ from $x+y=6$ at $y=0$.

Evaluate $C$: at $(0,10)=400$; at $(4,2)=200+80=280$; at $(6,0)=300$.

Minimum cost $=$ Rs. 280 at $x=4$ bags of M, $y=2$ bags of N.

(a) Total lines $=825\times45=37125$. Over 700 pages: lines per page $=\dfrac{37125}{700}=53.04\approx 53$ lines per page.

(b)

• 825-page form: 45 lines/page (<50) → Rs. 20/page → cost $=825\times20=$ Rs. 16,500.
• 700-page form: ≈53 lines/page (≥50) → Rs. 25/page → cost $=700\times25=$ Rs. 17,500.

The 825-page form is cheaper (Rs. 16,500 < Rs. 17,500), so cost is minimum with 825 pages.

Read the five-number summary (minimum, Q1, median, Q3, maximum) from each box plot:

(a) Median: Compare the median lines inside each box. (From the plot, the morning test median ≈ 75 and the evening test median ≈ 87; the evening test has the higher median.)

(b) Deviation/spread: Compare the interquartile range (box width) and the whisker lengths (range). The plot with the wider box/longer whiskers has greater spread (more variability/inconsistency) in scores.

(c) The argument 'morning tests give better grades' is supported only if the morning box plot shows a higher median/quartiles than the evening one. From the figure the evening test actually shows a higher median, so the data do not support the argument that morning tests give better grades. (Conclusion must match the actual medians read from the box plot.)

(a) Decision alternatives: the different courses of action (choices/strategies) available to a decision-maker, from which one must be selected. Example: an investor choosing among options A, B and C to invest in.

(b) Decision-making under certainty: a situation where the decision-maker knows with certainty the outcome of each alternative (only one state of nature). Example: investing in a fixed deposit with a guaranteed interest rate.

(c) Maximin (optimistic about worst case — best of the minimums): Row minimums: A=10000, B=8000, C=12000. The maximum of these is C (12000), so by maximin choose C.

Minimax (here applied as best of worst / minimize the maximum loss): For returns, the maximin criterion selects C. (If a regret/minimax-regret table is built, compute the opportunity-loss table and choose the option with the minimum of the maximum regret.) Maximum returns per option: A=20000, B=30000, C=25000; the optimistic 'maximax' would choose B. Using the maximin decision rule the best option is C.

(a) In a simple partnership the partners invest their capital for the same period of time (so profit is shared in the ratio of capitals). In a compound partnership the capitals are invested for different periods, so profit is shared in the ratio of (capital × time) products.

(b) First 6 months (only A and B): capital-months A $=30000\times6=180000$, B $=75000\times6=450000$. Profit Rs.21,000 shared in ratio $180000:450000=2:5$ → A $=21000\times\tfrac{2}{7}=6000$, B $=21000\times\tfrac{5}{7}=15000$.

Second 6 months (A, B, C): capital-months A $=30000\times6=180000$, B $=75000\times6=450000$, C $=45000\times6=270000$. Of profit Rs.27,000, A & B first share 20% equally: $0.20\times27000=5400$ → Rs.2700 each. Remaining 80% $=21600$ shared in ratio $180000:450000:270000=2:5:3$ (total 10) → A $=21600\times\tfrac{2}{10}=4320$, B $=21600\times\tfrac{5}{10}=10800$, C $=21600\times\tfrac{3}{10}=6480$.

Totals: A $=6000+2700+4320=$ Rs. 13,020; B $=15000+2700+10800=$ Rs. 28,500; C $=6480=$ Rs. 6,480. (Sum $=48,000=21000+27000$.)