NEB Grade 12 Math Model Paper Solution

NEB Grade 12 Math Model Paper Solution

Check out the NEB Grade 12 Math Model Paper Solution of the New syllabus(revised curriculum) of 2078. Download NEB Grade 12 Math Model Paper here. Check out our Grade 12 chemistry notes here.
Total Time- 3hrs
Full marks- 75

Group ‘A’ (1×11=11)

Rewrite the correct option in your answer sheet.

1. If 1, ω, ω² are the cube roots of the unit then
(a) ω = ω²
(b) ω² = ω³
(c) 1 + ω + ω² = 0
(d) 1 + ω = ω²
Ans: C is the correct ans.

2. The number of ways that 7 beads of different colours can be strung together so as to form a necklace is
(a) 5040
(b) 2520
(c) 720
(d) 360
Ans: D is the correct ans.
Hint: The number of ways arranging n distinct beads in the necklace is (1/2)(n-1)!.
So, the required number of ways = (1/2)(7-1)!= (1/2) x 6! = 360

3. tan^-1(5/12) is equal to
a. sin^-1(12/13)
b. cos^-1(12/13)
c. sec^-1(12/13)
d. cosec^-1(12/13)
Ans: B is the correct ans.
Hint:

Let\ x=\tan ^{-1} \frac{5}{12} \Rightarrow \tan x=\frac{5}{12}=\frac{p}{b}\\
\therefore h=\sqrt{p^{2}+b^{2}}=\sqrt{5^{2}+12^{2}}=13\\ \sin x=\frac{p}{h}=\frac{5}{13}\\ \Rightarrow x=\sin ^{-1} \frac{5}{13}(Option\ not\ available)\\ \cos x=\frac{b}{h}=\frac{12}{13}\\ \Rightarrow x=\cos ^{-1} \frac{12}{13} (Option\ available)

4. If 2 cosθ + 1 = 0 is the trigonometric equation of the locus related to the string attached in the wall of a hall then the general value for θ is
a. nπ + (– 1)ⁿ (2π/3) for n ∈ Z
b. nπ + (2π/3) for n ∈ Z
c. 2nπ ± (2π/3) for n ∈ Z
d. 2nπ + (π/3) for n ∈ Z.
Ans: C is the correct ans.
Hint:

Here, \quad 2 \cos \theta+1=0\\
or, \quad \cos \theta=-\frac{1}{2}=\cos \frac{2 \pi}{3}\\
\text{The general solution of} \cos \theta=k\ is\\ \theta=2 n \pi \pm \alpha=2 n \pi \pm \frac{2 \pi}{3}

5. If vector a=2i and b = 3j where, vectors i, j and k unit vectors along X, Y, Z- axes respectively, then the value of vector b × a is equal to
a. -6k
b. 6k
c. 6i
d. 6j
Ans: A is the correct ans.

6. There is a large grassy area near the president’s house of Nepal. The area is the set of all points in a plane. The sum of whose distances from two fixed places (points) is constant. The conic section represented by the grassy area is…
(a) Circle
(b) Parabola
(c) Hyperbola
(d) Ellipse
Ans: D is the ellipse.

7. Four unbiased coins are tossed successively. The mean and variance of the distribution differed by
(a) 1
(b) 2
(c) 3
(d) 4
Ans: A is the correct ans.

8. The degree of the differential equation

\frac{\partial^3y }{\partial x^3}+5\left ( \frac{\partial^2y}{\partial x^2} \right)^{2}+4\left ( \frac{\mathrm{d}y}{\mathrm{d} x} \right )^{4}+6=0\ is

a. 1
b. 2
c. 3
d. 4
Ans: A is the correct ans.

9. According to L Hospital’s rule the value of

\displaystyle \lim_{x \to 0}\frac{x^{3}}{sinx}\ is\ equal\ to

a. 3/4
b. 0
c. 1/4
d. ∞
Ans: B is the correct ans.

10. When Gauss forward elimination method is used for solving the equations 3x +4y = 18…. (i) and 3y – x = 7 ….(ii), we apply the operation like….
a. eqn(i) + 4 eqn(ii)
b. eqn(i) + 3 eqn(ii)
c. eqn(i) + eqn(ii)
d. eqn(ii) + 3 eqn(i)
Ans: B is the correct ans.

11. The amount of gravity exerted by the earth on the mass 10 kg (g = 9.8 ms^–2) is …
(a) 9.8 Joule
(b) 9.8 Newton
(c) 98 Joule
(d) 98 Newton
Ans: D is the correct ans.

OR

For the quadratic function f(Q) = aQ² + bQ + C for real numbers a, b, c and ≠ 0, the maximum value attained at

\begin{align*}a.\ \left ( \frac{b}{2a},\frac{4ac-b^{2}}{4a} \right )\ b.\ \left ( -\frac{b}{2a},\frac{4ac-b^{2}}{4a} \right )\\c.\ \left ( -\frac{b}{2a},\frac{b^{2}-4ac}{4a} \right )\ d.\ \left ( \frac{b}{2a},\frac{b^{2}-4ac}{4a} \right ) \end{align*}

Ans: B is the correct ans.

Group ‘B’ [5 × 8 = 40]

12. The binomial expression for two algebraic terms a and x is given as (a + x)ⁿ.

a) Write the binomial theorem for any positive integer n in expansion form.
Ans: The binomial theorem for any positive integer n in expansion form is:

(a+x)ⁿ = C(n,0)aⁿ + C(n,1)a^n-1x + C(n,2)a^n-2x² + …. + C(n,r)a^n-rx^r + … + C(n,n)xⁿ

b) Write the general term of the expansion.
Ans: The general term of the expansion is:

t_{r+1}=C(n, r)\ a^{n-r}{x}^{r}

c) Write any one property of binomial coefficients.
Ans: A property of the binomial coefficients is:

C_{0}+C_{1}+C_{2}+ \cdots +C_{n}=2^{n}

d) Write the single term for C(n, r) + C(n, r – 1).
Ans: the single term for C(n, r) + C(n, r – 1) is C(n+1,r).

e) How many terms are there in the expression?
Ans: The number of terms is n+1.

13. Given n⁴ < 10ⁿ for a fixed positive integer n ≥ 2, prove that (n + 1)⁴ < 10ⁿ + 1 using principle of mathematical induction.
Solution:
Let P(n) be given statement. Then,

P(n): n^{4}<10^{n} \quad \ldots (i)\\
When\ n=2 :\\
L.H.S. =2^{4}=16\\
R.H.S. =10^{2}=100\\
\therefore L.H.S. < R.H.S.\\ i.e. \text{P(2) is true.}

Now, we shall show that P(n + 1) : (n + 1) ⁴ < 10ⁿ+1 is true when P(n) is true. For this, we multiply by 10 on both sides of (i), we get

10n⁴ < 10.10ⁿ
or, 10n⁴ < 10ⁿ⁺¹ —-(ii)

Given\ n \geq 2,\\
So,\ \frac{1}{n} \leq \frac{1}{2}\\
or, \ 1+\frac{1}{n} \leq 1+\frac{1}{2}\\
or, \ 1+\frac{1}{n} \leq \frac{3}{2}\\
or, \ \left(1+\frac{1}{n}\right)^{4} \leq\left(\frac{3}{2}\right)^{4}\\
or, \ \quad\left(\frac{n+1}{n}\right)^{4} \leq 5.06\\
or, \ \quad \frac{(n+1)^{4}}{n^{4}}<10\\
\therefore \quad(n+1)^{4}<10 n^{4}\\
\text{From eq. (ii) and (iii), we get}\\
(n+1)^{4}<10 n^{4}<10^{n+1}\\
\therefore (n+1)^{4}<10^{n+1}

Thus, the above equation shows that P(n+1):(n+1)⁴ < 10ⁿ⁺¹ is true whenever P(n) is true.

14. a. Evaluate:

cos\left ( sin^{-1}\frac{3}{5}\ +\ sin^{-1}\frac{5}{13} \right )

Solution:

\text{Let,}\ A=\sin ^{-1} \frac{3}{5}\ \& \ B=\sin ^{-1} \frac{5}{13}.\\
\text{Then,}\ sin A=\frac{3}{5}\ \&\ sin B=\frac{5}{13}\\
\cos A=\sqrt{1-\sin ^{2} A}=\sqrt{1-\left(\frac{3}{5}\right)^{2}}\\ =\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5} \\
\cos B=\sqrt{1-\sin ^{2} B}=\sqrt{1-\left(\frac{5}{13}\right)^{2}}\\ =\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}\\
\text{Now,}\\
\begin{align*}
\cos &\left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right) =\cos (A+B) \\
&=\cos A \cos B-\sin A \sin B \\
&=\frac{4}{5} \cdot \frac{12}{13}-\frac{3}{5} \cdot \frac{5}{13} \\
&=\frac{48}{65}-\frac{15}{65} \\
&=\frac{33}{65}
 \end{align*}

b. Using the vector method, find the area of the triangle with vertices A(1, 4, 6), B(– 2, 5, 1) and C(1, – 1, 1).
Solution:

\text{Let \(O\) be the origin. Then,}\\
\overrightarrow{O A}=(1,4,6) \\
\overrightarrow{O B}=(-2,5,1) \\
\overrightarrow{O C}=(1,-1,1)

Now,\\
\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}\\=(-2,5,1)-(1,4,6)=(-3,1,-5) \\
\overrightarrow{A C}=\overrightarrow{O C}-\overrightarrow{O A}\\=(1,-1,1)-(1,4,6)=(0,-5,-5) \\
\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
-3 & 1 & -5 \\
0 & -5 & -5
\end{array}\right|\\=(-5-25) \vec{i}-(15-0) \vec{k}+(15-0) \vec{k}\\=-30 \vec{i}-15 \vec{j}+15 \vec{k} \\
|\overrightarrow{A B} \times \overrightarrow{A C}|=|-30 \vec{i}-15 \vec{j}+15 \vec{k}|\\=\sqrt{(-30)^{2}+(-15)^{2}+15^{2}}=15 \sqrt{6}  \\
\therefore \quad \text { Area of triangle }=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|\\=\frac{1}{2} \times 15 \sqrt{6}=18.37 \text { sq. units }

15. The information given below relates to the advertisement and sales of a departmental store in lakhs of Nepalese rupees.

a) Find the two regression equations related to the above data.
Solution:
Here, x̄ = 20, ȳ = 100, σ_x=3, σ_y= 12, and r = 0.8.
Now,

b_{y X}=r \frac{\sigma_{Y}}{\sigma_{X}}=0.8 \times \frac{12}{3}=3.2\\
b_{X Y}=r \frac{\sigma_{X}}{\sigma_{Y}}=0.8 \times \frac{3}{12}=0.2\\
\text{The regression equation of Y on X is}\\
\begin{align*} 
& Y-\bar{Y}=b_{r X}(X-\bar{X}) \\
\text { or, } & Y-100=3.2(X-20) \\
\text { or, } & Y=3.2 X-64+100 \\
\therefore & Y=3.2 X+36
\end{align*}\\
\text{Again, the regression equation of X on Y is} \\
\begin{align*}
& X-\bar{X}=b_{X Y}(Y-\bar{Y}) \\
\text { or, } & X-20=0.2(Y-100) \\
\text { or, } & X=0.2 Y-20+20 \\
\therefore & X=0.2 Y
\end{align*}

b) What should be the advertisement expenditure if the department store wants to attain a sales target of Rs. 200 lakhs.
Ans: Here, Sales (Y) = Rs 200 lakhs
Then, advertisement expenditure(X) = 0.2Y = 0.2 x Rs. 200 = Rs. 40 laksh.

16. Suman and Nikita are studying the application of derivatives and integration in a class. They ask each other the quiz questions as given below. On the basis of these questions answer the following.

a. f ‘(x) and g'(x) are derivatives of the functions f(x) and g(x). What is the expression equal to according to L’Hospital’s rule for form ∞/∞?
Ans: According to L-Hospital’s rule,

\begin{align*} \lim_{x\rightarrow a}\frac{f(x)}{g(x)}&=\frac{\lim_{x\rightarrow a}f'(x)}{\lim_{x\rightarrow a}g'(x)}\\
&=\frac{f'(a)}{g'(a)},\\
\text{provided that}&\ g'(a)\neq 0
\end{align*}

b. State Rolle’s Theorem.
Ans: Rolle’s theorem states that if the function f(x) is:

• continuous in the closed interval [a,b]
• differentiable in the open interval (a,b)
• f(a) = f(b)

then there exists at least one point c ∈ (a,b) such that f'(c) = 0.

c. What is the expression equal to ∫(1/(x²+a²))dx
Ans: The given expression is equal to:

\frac{1}{a}\ tan^{-1}\frac{x}{a}+c

d. What does ‘C’ represent in the expression

\int\frac{dx}{3sinx + 4cosx}=\frac{1}{5}\ ln\\\left|tan\left ( \frac{x}{2}+\frac{1}{2}tan^{-1}\frac{4}{3} \right ) \right|+C 

Ans: In the above expression, the C represents the constant of integratino.

e. Write a difference between derivative and antiderivative?
Ans: A difference between derivative and antiderivative is:

The derivative of a function f(x) at a point x is defined as

\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\\
\text{provided the limit exists.}

The antiderivative is the opposite of a derivative. That is, if F'(x) = f(x) then F(x) is antiderivative of f(x).

17. Integrate ∫1/(x⁴ – 1) dx using the concept of partial fraction. Also, give an example of proper rational fraction and improper rational fraction.
solution:

Let x^{2}=y. Then,\\
\frac{1}{x^{4}-1}=\frac{1}{y^{2}-1}=\frac{1}{(y-1)(y+1)}\\
\text{Using the concept of partial fraction,}\\
\begin{align*}
\frac{1}{(y-1)(y+1)}&=\frac{A}{y-1}+\frac{B}{y+1}\\
\text{ or, } \frac{1}{(y-1)(y+1)}&=\frac{A(y+1)+B(y-1)}{(y-1)(y+1)}\\
\text{ or, } \quad 1=A(y+1)+&B(y-1)---(i)
\end{align*}

Now putting y = 1 in eqn (i), we get,
1 = 2A
∴ A = 1/2

Again, putting y =-1 in eqn (i), we get
1 = -2B
∴ B = -(1/2)
Now,

\begin{aligned}
\frac{1}{x^{4}-1}=\frac{1}{(y-1)(y+1)} &=\frac{A}{y-1}+\frac{B}{y+1} \\
&=\frac{\frac{1}{2}}{y-1}+\frac{-\frac{1}{2}}{y+1} \\
&=\frac{1}{2} \cdot \frac{1}{y-1}-\frac{1}{2} \cdot \frac{1}{y+1} \\
&=\frac{1}{2} \cdot \frac{1}{x^{2}-1}-\frac{1}{2} \cdot \frac{1}{x^{2}+1}
\end{aligned}\\
Again,\\
\begin{aligned}
\int \frac{1}{x^{4}-1} d x &=\int\left(\frac{1}{2} \cdot \frac{1}{x^{2}-1}-\frac{1}{2} \cdot \frac{1}{x^{2}+1}\right) d x \\
&=\frac{1}{2} \int \frac{1}{x^{2}-1^{2}} d x-\frac{1}{2} \int \frac{1}{x^{2}+1^{2}} d x \\
&=\frac{1}{2}\left(\frac{1}{2 \cdot 1} \ln \left(\frac{x-1}{x+1}\right)-\frac{1}{2}\left(\frac{1}{1} \tan ^{-1} \frac{x}{1}\right)+c\right.\\
&=\frac{1}{4} \ln \left(\frac{x-1}{x+1}\right)-\frac{1}{2} \tan ^{-1} x+c
\end{aligned}

Example of proper rational fraction:

\frac{x+1}{x^{2}-2x-8}

Example of improper rational fraction:

\frac{x^{3}-1}{x^{2}-3x+2}

19. Write any one difference between like parallel forces and unlike parallel forces. A heavy uniform beam whose mass is 60 kg is suspended in a horizontal position by two vertical strings each of which can sustain a tension of 52.5 kg wt. How far from the centre of the beam must a body of mass 30 kg placed so that one of the strings may just break?
Ans: One difference between like parallel and unlike parallel forces is:

Now,
Let AB be the uniform beam and C be its centre where its weight 60kg acts. Let a body of mass 30kg placed at D where CD=x. Let AB=l be the length of the beam.
Then, AC=BC= l/2,
AD= (l/2)+x and BD = (l/2)-x.

Here, BD being less than AC, the tension of the string at B will be greater than that at A.
∴ Tension at B = maximum tension = 52.5 kg
Now, taking moments about A, we get,

\begin{array}{ll} & 60 \times A C+30 \times A D=52.5 \times A B \\ \text { or, } & 60 \times \frac{l}{2}+30 \times\left(\frac{l}{2}+x\right)=52.5 \times l \\ \text { or, } & 30 l+15 l+30 x=52.5 l \\ \text { or, } \quad & 45 l+30 x=52.5 l \\ \text { or, } & 30 x=52.5 l-45 l \\ \text { or, } & 30 x=7.5 l \\ \text { or, } & x=\frac{7.5 l}{30} \\ \therefore & x=\frac{l}{4}\end{array}

Hence, a body of mass 30kg should be placed at 1/4th of the length of the beam from the centre.

OR

If the demand function P = 85 – 4Q – Q², find the consumer’s surplus at demand 4 units and price 64 units. Also make a revenue function for demand equation P = 20 + 5Q – Q². Obtain the standard quadratic equation for marginal revenue. Q represents the number of units demands and P represent the price.
Solution:

Here, P =85-4Q-Q²
a) When Q = 4
P = 85-4 x 4-4² = 53
∴ (P₀, Q₀)= (53,4)
Now,
Consumer’s surplus is:

\begin{align*} &=\int_{0}^{Q} P d Q-P_{o} Q_{o}\\
&=\int_{0}^{4}\left(85-4 Q-Q^{2}\right) d Q-53 \times 4\\
&=\left[85 Q-\frac{4 Q^{2}}{2}-\frac{Q^{3}}{3}\right]_{0}^{4}-212\\
&=85 \times 4-2 \times 4^{2}-\frac{4^{3}}{3}-212\\
&=\frac{224}{3}
\end{align*}

b) When P = 64

\begin{array}{ll} 
& 64=85-4 Q-Q^{2} \\
\text { or, } & Q^{2}+4 Q-21=0 \\
\text { or, } & Q^{2}+7 Q-3 Q-21=0 \\
\text { or, } & Q(Q+7)-3(Q+7)=0 \\
\text { or, } & (Q-3)(Q+7)=0 \\
\text { or, } & Q=3,-7 \text { (not possible) } \\
\therefore & Q=3 \\
\therefore & \left(P_{o}, Q_{o}\right)=(64,3)
\end{array}\\

Now,
Consumer’s surplus is:

\begin{aligned}
&=\int_{0}^{Q} P d Q-P_{o} Q_{o} \\
&=\int_{0}^{3}\left(85-4 Q-Q^{2}\right) d Q-64 \times 3 \\
&=\left[85 Q-\frac{4 Q^{2}}{2}-\frac{Q^{3}}{3}\right]_{0}^{3}-192 \\
&=85 \times 3-2 \times 3^{2}-\frac{3^{3}}{3}-192 \\
&=36
\end{aligned}

Now, revenue function for demand equation P = 20 + 5Q – Q² is given by:
R = PQ = (20 + 5Q – Q² )Q = 20Q + 5Q² – Q³

Again, material revenue is given by:

\begin{aligned} MR&=\frac{dR}{dQ}\\
&= \frac{d(20Q + 5Q^{2} - Q^{3})}{dQ}\\
&=20 + 10Q - 3Q^{2}
\end{aligned}

Group ‘C’ [8 × 3 = 24]

20. A mixture is to be made of three foods, A, B and C which contain nutrients P, Q, R as shown in the table below. The quantity of P, Q, R is 45 units, 54 units and 45 units respectively.

(a) Express the information in equation form.
Ans: Let the food needed be x kg of A, y kg of B, and z kg of C.
Now, writing the given information in equation form:

P: 2x + 3y + 4z = 45 —-(i)
Q: 2x + 5y + 3z = 54 —-(ii)
R: 4x + 5z = 45 —-(iii)

(c) If the cost per kg of the foods A, B, C are Rs. 300, Rs. 240 and Rs. 180 respectively, find the total cost of the mixture by matrix method.
Solution:

Let C be per unit cost matrix. Then, C = (300 240 180)
Total cost of mixture = CX

\begin{aligned} &=\begin{pmatrix}
300 & 240 & 180
\end{pmatrix}\begin{pmatrix}
7.5\\ 6
\\ 3

\end{pmatrix}\\
&=300\times 7.5+240\times 6+180\times 3\\
&=4230
\end{aligned}

Hence, the total cost of the mixture is Rs. 4,230.

22. A college hostel accommodating 1000 students; one of them came from abroad with infection of coronavirus, then the hostel was isolated. If the rate at which the virus spreads is assumed to be proportional to the product of the number ‘N’ of infected students and number of non-infected students and the number of infected students is 50 after 4 days.

a. Express the above information in the form of the differential equation.
Solution:

Here,
Total number of students = 1000
Number of students infected = N
Number of non-infected students = 1000-N
By question,

\begin{aligned} \frac{dN}{dt}\ &\alpha\ N(1000-N)\\
or,\ \frac{dN}{dt}\ &= kN(1000-N)--(i)\\
\end{aligned}\\
\text{Eq. i is the required differential equation.}

b. Solve the differential equation.
Solution:
Here,

\frac{dN}{dt}\ = kN(1000-N)\\
\text{separating the variable, we get}\\
\frac{dN}{N(1000-N)}\ = kdt

Integrating on both sides, we get:

\begin{array}{ll} \int \frac{d N}{N(1000-N)}=\int k d t \\
\text { or, } \frac{1}{1000} \int \frac{(1000-N)+N}{N(1000-N)} d N=k \int d t \\
\text { or, }  \int\left[\frac{(1000-N)}{N(1000-N)}+\frac{N}{N(1000-N)}\right] d N=1000 k \int d t \\ \text { or, } \int\left[\frac{1}{N}+\frac{1}{1000-N}\right] d N=1000 k \int d t \\
\text { or, } \ln |N|-\ln |100-N|=1000 k t+c \\
\text { or, } \ln \left | \frac{N}{1000-N } \right |=1000 k t+c \ldots(ii)\end{array}

At initial, one student was infected i.e., when t=0, N=1.
Using this initial condition in eq (ii), we get,

\begin{aligned} ln\left | \frac{1}{1000-1} \right |&=1000k\times 0+c\\
or,\ ln\left | \frac{1}{999} \right |&= c\\
\therefore c&=ln\left | \frac{1}{999} \right |
\end{aligned}

Substituting the value of c in eqn (ii), we get,

\begin{array}{ll} 
\ln \left|\frac{N}{1000-N}\right|=1000 k t+\ln \left|\frac{1}{999}\right| \\
\text { or, } \quad \ln \left|\frac{N}{1000-N}\right|-\ln \left|\frac{1}{999}\right|=1000 \mathrm{kt} \\
\text { or, } \quad \ln \left|\frac{999 N}{1000-N}\right|=1000 k t \ldots\text { (iii) }
\end{array}

Again, when t = 4, N = 50. Using this condition in eq. (iii), we get,

\ln \left| \frac{999 \times 50}{1000-50}\right|=1000 k \times 4\\
\begin{array}{l}
or,\ \ln \left|\frac{99}{19}\right|=4000 k \\
\therefore \quad k \approx 0.0009906
\end{array}

Substituting the value of k in eqn (iii), we get,

\ln \left|\frac{999 N}{1000-N}\right|=0.9906 t\\
or, \quad \frac{999 N}{1000-N}=e^{0.9906 t}\\
or, \quad 999 N=1000 e^{0.9906 t}-N e^{0.9906 t}\\
or, N e^{0.9906 t}+999 N=1000 e^{0.9906 t}\\
or, \quad N\left(e^{0.9906 t}+999\right)=1000 e^{0.9906 t}\\
or, N=\frac{1000 e^{0.9906 t}}{e^{0.9906 t}+999}\\
\therefore \quad N=\frac{1000}{1+999 e^{-0.9906 t}}---(iv)

c. Show that more than 95% of students will be infected after 10 days.
Solution:

When t = 10 in eqn (iv), we get,

N=\frac{1000}{1+999 e^{-0.9906 \times 10}} \approx 953\\
\therefore Infected\ \%=\frac{N}{1000} \times 100 \%\\
=\frac{953}{1000} \times 100 \%=95.3 \%

Hence, more than 95% will be infected after 10 days.