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NEB Class 12 Management Business Mathematics Question Paper 2077 Nepal

NEB Class 12ManagementModel questions
40marks
90min
12questions

This is the official NEB Class 12 (Management stream) Business Mathematics question paper for 2077, as set in the Model questions examination. It carries 40 full marks and a time allowance of 90 minutes, across 12 questions. On Kekkei you can attempt this Business Mathematics past paper online with a timer, get instant AI feedback and step-by-step solutions, and track the topics where you lose marks — completely free. Whether you are revising for your NEB Class 12 Business Mathematics exam or solving previous years' question papers, this 2077 paper is a great way to practise under real exam conditions.

LevelNEB Class 12
StreamManagement
SubjectBusiness Mathematics
Year2077 BS
Exam sessionModel questions
Full marks40
Time allowed90 minutes
Questions12, all with step-by-step solutions
A

Group 'A'

Attempt all questions.

8 questions·2 marks each
1aShort answer2 marks

Find the domain and range in relation R={(4,8),(4,12),(5,10),(6,12),(7,7)}R=\{(4,8),(4,12),(5,10),(6,12),(7,7)\}.

Domain = set of first elements ={4,5,6,7}=\{4,5,6,7\}.

Range = set of second elements ={8,12,10,7}={7,8,10,12}=\{8,12,10,7\}=\{7,8,10,12\}.

relationsdomain-and-range
1bNumeric answer2 marks

Find the first term of a G.P. whose fifth term is 243 and common ratio is 3.

Numeric answer

sequence-and-seriesgeometric-progression
2aShort answer2 marks

Find the slope and intercepts of the line 2x−3y+10=02x-3y+10=0.

Rewrite as y=23x+103y=\dfrac{2}{3}x+\dfrac{10}{3}.

Slope m=23m=\dfrac{2}{3}.

yy-intercept: set x=0⇒y=103x=0\Rightarrow y=\dfrac{10}{3}.

xx-intercept: set y=0⇒2x+10=0⇒x=−5y=0\Rightarrow 2x+10=0\Rightarrow x=-5.

coordinate-geometrystraight-line
2bNumeric answer2 marks

Evaluate: lim⁡x→3x−3x−2−4−x\displaystyle\lim_{x\to 3}\dfrac{x-3}{\sqrt{x-2}-\sqrt{4-x}}.

Numeric answer

limits
3aShort answer2 marks

Find dydx\dfrac{dy}{dx} when x=2atx=2at, y=at2y=at^2.

dxdt=2a\dfrac{dx}{dt}=2a, dydt=2at\dfrac{dy}{dt}=2at.

dydx=dy/dtdx/dt=2at2a=t\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{2at}{2a}=t.

differentiationparametric
3bShort answer2 marks

If the marginal revenue function for output xx is MR=x3+4x−2MR=x^3+4x-2, find the total revenue function.

Total revenue R=∫MR dx=∫(x3+4x−2) dx=x44+2x2−2x+CR=\displaystyle\int MR\,dx=\int (x^3+4x-2)\,dx=\dfrac{x^4}{4}+2x^2-2x+C.

Since R=0R=0 when x=0x=0, the constant C=0C=0. Hence R=x44+2x2−2xR=\dfrac{x^4}{4}+2x^2-2x.

business-mathrevenue-functionintegration
4aNumeric answer2 marks

Find the mean deviation from mean of the following observations.

x68101214161820
f114252718942

Numeric answer

statisticsmean-deviation
4bNumeric answer2 marks

If 200 men can make an embankment 3 km long in 25 days, how much overtime per day must 60 men work to complete an embankment 2 km long in 32 days; 12 hours being a day's work?

Numeric answer (hours/day)

work-timevariation
B

Group 'B'

Attempt all questions.

4 questions·4 marks each
5Long answer4 marks

Solve by Cramer's rule: x+y+z=6x+y+z=6, 2x+3y−z=52x+3y-z=5, 3x+4y+5z=263x+4y+5z=26.

Coefficient determinant D=∣11123−1345∣=1(15+4)−1(10+3)+1(8−9)=19−13−1=5D=\begin{vmatrix}1&1&1\\2&3&-1\\3&4&5\end{vmatrix}=1(15+4)-1(10+3)+1(8-9)=19-13-1=5.

Dx=∣61153−12645∣=6(15+4)−1(25+26)+1(20−78)=114−51−58=5D_x=\begin{vmatrix}6&1&1\\5&3&-1\\26&4&5\end{vmatrix}=6(15+4)-1(25+26)+1(20-78)=114-51-58=5, so x=1x=1.

Dy=∣16125−13265∣=1(25+26)−6(10+3)+1(52−15)=51−78+37=10D_y=\begin{vmatrix}1&6&1\\2&5&-1\\3&26&5\end{vmatrix}=1(25+26)-6(10+3)+1(52-15)=51-78+37=10, so y=2y=2.

Dz=∣1162353426∣=1(78−20)−1(52−15)+6(8−9)=58−37−6=15D_z=\begin{vmatrix}1&1&6\\2&3&5\\3&4&26\end{vmatrix}=1(78-20)-1(52-15)+6(8-9)=58-37-6=15, so z=3z=3.

Solution: x=1, y=2, z=3x=1,\ y=2,\ z=3.

matricescramers-rulesystem-of-equations
6Long answer4 marks

Find the derivative from first principles: y=1xy=\dfrac{1}{x}.

dydx=lim⁡h→0f(x+h)−f(x)h=lim⁡h→01x+h−1xh=lim⁡h→0x−(x+h)h x(x+h)=lim⁡h→0−hh x(x+h)=lim⁡h→0−1x(x+h)=−1x2\dfrac{dy}{dx}=\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\lim_{h\to0}\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}=\lim_{h\to0}\dfrac{x-(x+h)}{h\,x(x+h)}=\lim_{h\to0}\dfrac{-h}{h\,x(x+h)}=\lim_{h\to0}\dfrac{-1}{x(x+h)}=-\dfrac{1}{x^2}.

differentiationfirst-principles
7Long answer4 marks

Find the extreme values of the function FF defined by F=10x+15yF=10x+15y subject to constraints x+2y≤20x+2y\le 20, x+y≤16x+y\le 16, x≥0x\ge 0, y≥0y\ge 0.

Corner points of the feasible region: intersection of x+2y=20x+2y=20 and x+y=16x+y=16: subtracting gives y=4y=4, x=12x=12. Vertices: (0,0)(0,0), (16,0)(16,0), (12,4)(12,4), (0,10)(0,10).

Evaluate F=10x+15yF=10x+15y: at (0,0)=0(0,0)=0; at (16,0)=160(16,0)=160; at (12,4)=120+60=180(12,4)=120+60=180; at (0,10)=150(0,10)=150.

Maximum F=180F=180 at (12,4)(12,4); minimum F=0F=0 at (0,0)(0,0).

linear-programming
8Long answer4 marks

Ram, Shyam and Hari invest in a business with capitals Rs. 5000, Rs. 4500 and Rs. 6500 respectively. After six months, Ram doubles his capital and after next three months Shyam trebles his capital. If the profit at the end of the year amounted to Rs. 8300, find the profit obtained by each.

Compute capital × months (in months over the year):

Ram: Rs.5000 for 6 months ++ Rs.10000 (doubled) for 6 months =5000×6+10000×6=30000+60000=90000=5000\times6+10000\times6=30000+60000=90000.

Shyam: Rs.4500 for 9 months ++ Rs.13500 (trebled) for 3 months =4500×9+13500×3=40500+40500=81000=4500\times9+13500\times3=40500+40500=81000.

Hari: Rs.6500 for 12 months =78000=78000.

Ratio =90000:81000:78000=90:81:78=30:27:26=90000:81000:78000=90:81:78=30:27:26. Total parts =83=83.

Profit Rs.8300 ÷ 83 = Rs.100 per part.

Ram =30×100==30\times100= Rs.3000; Shyam =27×100==27\times100= Rs.2700; Hari =26×100==26\times100= Rs.2600.

partnershipprofit-sharing

Frequently asked questions

Where can I find the NEB Class 12 Business Mathematics question paper 2077?
The full NEB Class 12 Business Mathematics 2077 (Model questions) question paper is available free on Kekkei. You can read every question online and attempt the paper under timed exam conditions.
Does the Business Mathematics 2077 paper come with solutions?
Yes. Every question on this Business Mathematics past paper includes a step-by-step solution, plus instant AI feedback when you attempt it on Kekkei.
How many marks is the NEB Class 12 Business Mathematics 2077 paper?
The NEB Class 12 Business Mathematics 2077 paper carries 40 full marks and is meant to be completed in 90 minutes, across 12 questions.
Is practising this Business Mathematics past paper free?
Yes — reading and attempting this Business Mathematics past paper on Kekkei is completely free.