(i) (b) : In 2019,dealer A sold 120 Hatchback, 50 Sedan and 10 SUV; dealer B sold 100 Hatchback, 30 Sedan and 5 SUV and dealer C sold 90 Hatchback, 40 Sedan and 2 SUV
\(\therefore\)  Required matrix, say P, is given by
 
(ii) (a) : In 2020,
dealer A sold 300 Hatchback, 150 Sedan, 20 SUV
dealer B sold 200 Hatchback, 50 sedan, 6 SUV
dealer C sold 100 Hatchback, 60 sedan,S SUV
\(\therefore\) Required matrix, say Q, is given by
 
(iii) (c) : Total number of cars sold in two given years, by each dealer, is given by
 
 
(iv) (c): The increase in sales from 2019 to 2020 is given by
 
 
(v) (c) : The amount of profit in 2020 received by each dealer is given by the matrix
 
\(\begin{array}{c} A \\ B \\ C \end{array}\left[\begin{array}{c} 15000000+15000000+4000000 \\ 10000000+5000000+1200000 \\ 5000000+6000000+1000000 \end{array}\right]\) 
\(\begin{array}{r} A \\ =B \\ C \end{array}\left[\begin{array}{l} 34000000 \\ 16200000 \\ 12000000 \end{array}\right]\)

(i) (a) : Clearly,

(ii) (d) : Since Q is a 3 x 1 matrix, therefore
 
(iii) (a) : Clearly, total funds collected by each school is given by the matrix.
\(P Q=\left[\begin{array}{ccc} 40 & 50 & 20 \\ 25 & 40 & 30 \\ 35 & 50 & 40 \end{array}\right]\left[\begin{array}{c} 25 \\ 100 \\ 50 \end{array}\right]\) 
\(=\left[\begin{array}{c} 1000+5000+1000 \\ 625+4000+1500 \\ 875+5000+2000 \end{array}\right]=\left[\begin{array}{l} 7000 \\ 6125 \\ 7875 \end{array}\right]\) 
\(\therefore\) Funds collected by school A is Rs. 7000
Funds collected by school B is Rs. 6125
Pimds collected by school C is Rs. 7875
(iv) (b) 
(v) (b) : Total funds collected for the required purpose
= Rs. (7000 + 6125 + 7875) = Rs. 21000

Since v(3) = 64, v(6) = 133 and v(9) = 208, we get the following system of linear equations
9a + 3b + c = 64
36a + 6b + c = 133
81a + 9b + c = 208
This can be written in matrix form as
\(\begin{equation} \left[\begin{array}{ccc} 9 & 3 & 1 \\ 36 & 6 & 1 \\ 81 & 9 & 1 \end{array}\right]\left[\begin{array}{l} a \\ b \\ c \end{array}\right]=\left[\begin{array}{c} 64 \\ 133 \\ 208 \end{array}\right] \end{equation}\) 
or AX = B
Since,\(\begin{equation} A^{-1}=\frac{1}{18}\left(\begin{array}{ccc} 1 & -2 & 1 \\ -15 & 24 & -9 \\ 54 & -54 & 18 \end{array}\right) \end{equation}\) 
 \(\begin{equation} \therefore \quad X=\left[\begin{array}{l} a \\ b \\ c \end{array}\right]=A^{-1} B=\frac{1}{18}\left(\begin{array}{ccc} 1 & -2 & 1 \\ -15 & 24 & -9 \\ 54 & -54 & 18 \end{array}\right)\left(\begin{array}{c} 64 \\ 133 \\ 208 \end{array}\right) \end{equation}\) 
\(\begin{equation} =\frac{1}{18}\left(\begin{array}{c} 64-266+208 \\ -960+3192-1872 \\ 3456-7182+3744 \end{array}\right)=\frac{1}{18}\left(\begin{array}{c} 6 \\ 360 \\ 18 \end{array}\right)=\left(\begin{array}{c} 1 / 3 \\ 20 \\ 1 \end{array}\right) \end{equation}\) 
Thus,\(\begin{equation} a=\frac{1}{3} \end{equation}\) ,b = 20 and c = 1
(i) (b)
(ii) (c)
(iii) (b) : \(\begin{equation} v(t)=\frac{1}{3} t^{2}+20 t+1 \end{equation}\) 
(iv) (d) : Clearly, required speed = v(15)
\(\begin{equation} =\frac{1}{3} \times 225+20 \times 15+1 \end{equation}\) 
= 75 + 300 + 1 = 376 miles per second
(v) (d) : Consider, v(t) = 784
\(\begin{equation} \Rightarrow \frac{1}{3} t^{2}+20 t+1=784 \end{equation}\) 
\(\begin{equation} \Rightarrow \end{equation}\) t² + 60t = 2349
\(\begin{equation} \Rightarrow \end{equation}\) t²+ (87 - 27)t - 2349 = 0
\(\begin{equation} \Rightarrow \end{equation}\) t(t + 87) - 27(t + 87) = 0
\(\begin{equation} \Rightarrow \end{equation}\) (t - 27)(t + 87) = 0
\(\begin{equation} \Rightarrow \end{equation}\) = 27 seconds [\(\therefore\) Time can't be negative]

(i) (a) : Let \(\begin{equation} \Delta=\left|\begin{array}{cc} 1 & -2 \\ 4 & 3 \end{array}\right| \end{equation}\) 
Cofactor of 1 = 3, cofactor of -2 =-4
Cofactor of 4 = 2, cofactor of 3 = 1
\(\therefore\) Required sum = 3 - 4 + 2 + 1 = 2
(ii) (b) : Let \(\begin{equation} \Delta=\left|\begin{array}{ccc} 5 & 6 & -3 \\ -4 & 3 & 2 \\ -4 & -7 & 3 \end{array}\right| \end{equation}\) 
Minor of \(\begin{equation} a_{21}=\left|\begin{array}{cc} 6 & -3 \\ -7 & 3 \end{array}\right|=18-21=-3 \end{equation}\) 
(iii) (c) : Let \(\begin{equation} \Delta=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right| \end{equation}\) 
Clearly, a₃₂ = 5
and A₃₂ = cofactor of a₃₂ in \(\begin{equation} \Delta=(-1)^{3+2}\left|\begin{array}{ll} 2 & 5 \\ 6 & 4 \end{array}\right| \end{equation}\) 
= (-1)(8-30) = 22
\(\begin{equation} \therefore \end{equation}\) a₃₂·A₃₂ = 5 x 22 = 110
(iv) (d) : Here, \(\begin{equation} \Delta=\left|\begin{array}{lll} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right| \end{equation}\) 
\(\therefore\) Minor of  \(\begin{equation} a_{23}=\left|\begin{array}{ll} 5 & 3 \\ 1 & 2 \end{array}\right|=10-3=7 \end{equation}\) 
(v) (b) : Here,\(\begin{equation} \Delta=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right| \end{equation}\) 
 \(\begin{equation} A_{11}=(-1)^{1+1}\left|\begin{array}{cc} 0 & 4 \\ 5 & -7 \end{array}\right|=1(0-20)=-20 \end{equation}\) 
\(\begin{equation} A_{12}=(-1)^{1+2}\left|\begin{array}{cc} 6 & 4 \\ 1 & -7 \end{array}\right|=-1(-42-4)=46 \end{equation}\) 
\(\begin{equation} A_{13}=(-1)^{1+3}\left|\begin{array}{ll} 6 & 0 \\ 1 & 5 \end{array}\right|=1(30-0)=30 \end{equation}\) 
\(\begin{equation} \therefore \quad \Delta=a_{11} A_{11}+a_{12} a_{12}+a_{13} A_{13} \end{equation}\) 
= 2(-20) -3(46) + 5(30) = -28
\(\begin{equation} \Rightarrow|\Delta|=28 \end{equation}\)

(i) (c)
(ii) (b)
(iii) (b)
(iv) (b)
(v) (a)

(i) (b) : x³ + x²y+ xy²+y³= 81
\(\begin{equation} \Rightarrow 3 x^{2}+x^{2} \frac{d y}{d x}+2 x y+2 x y \frac{d y}{d x}+y^{2}+3 y^{2} \frac{d y}{d x}=0 \end{equation}\) 
\(\begin{equation} \Rightarrow \left(x^{2}+2 x y+3 y^{2}\right) \frac{d y}{d x}=-3 x^{2}-2 x y-y^{2} \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{d y}{d x}=\frac{-\left(3 x^{2}+2 x y+y^{2}\right)}{x^{2}+2 x y+3 y^{2}} \end{equation}\) 
(ii) (c) : x^y = e^x-y⇒y log x = x - y
\(\begin{equation} \Rightarrow y \times \frac{1}{x}+\log x \cdot \frac{d y}{d x}=1-\frac{d y}{d x} \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{d y}{d x}[\log x+1]=1-\frac{y}{x} \Rightarrow \frac{d y}{d x}=\frac{x-y}{x[1+\log x]} \end{equation}\) 
(iii) (d) : \(\begin{equation} e^{\sin y}=x y \Rightarrow \sin y=\log x+\log y \end{equation}\) 
\(\begin{equation} \Rightarrow \cos y \frac{d y}{d x}=\frac{1}{x}+\frac{1}{y} \frac{d y}{d x} \Rightarrow \frac{d y}{d x}\left[\cos y-\frac{1}{y}\right]=\frac{1}{x} \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{d y}{d x}=\frac{y}{x(y \cos y-1)} \end{equation}\) 
(iv) (d) : sin²x + cos²y = 1
\(\begin{equation} \Rightarrow \quad 2 \sin x \cos x+2 \cos y\left(-\sin y \frac{d y}{d x}\right)=0 \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{d y}{d x}=\frac{-\sin 2 x}{-\sin 2 y}=\frac{\sin 2 x}{\sin 2 y} \end{equation}\) 
(v) (d) : \(\begin{equation} y=(\sqrt{x})^{\sqrt{x}} \quad \Rightarrow y=(\sqrt{x})^{y} \end{equation}\) 
\(\begin{equation} \Rightarrow \log y=y(\log \sqrt{x}) \Rightarrow \log y=\frac{1}{2}(y \log x) \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{2}\left[y \times \frac{1}{x}+\log x\left(\frac{d y}{d x}\right)\right] \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{d y}{d x}\left\{\frac{1}{y}-\frac{1}{2} \log x\right\}=\frac{1}{2} \frac{y}{x} \end{equation}\) 
\(\begin{equation} \Rightarrow \frac{d y}{d x}=\frac{y}{2 x} \times \frac{2 y}{(2-y \log x)}=\frac{y^{2}}{x(2-y \log x)} \end{equation}\)

(i) (a) : Let x be the number of extra days after 1^st July.
\(\therefore\) Price = Rs.(300 - 3Xx) = Rs.(300 - 3x)
Quantity = 80 quintals + x(1 quintal per day)
= (80 + x) quintals
(ii) (b) : R(x) = Quantity x Price
= (80 + x) (300 - 3x) = 24000 - 240x + 300x -3x²
= 24000 + 60x - 3x²
(iii) (a) : We have, R(x) = 24000 + 60x - 3x²
\(\begin{equation} \Rightarrow R^{\prime}(x)=60-6 x \Rightarrow R^{\prime \prime}(x)=-6 \end{equation}\)
For R(x) to be maximum, R'(x) = 0 and R"(x) < 0
\(\begin{equation} \Rightarrow 60-6 x=0 \Rightarrow x=10 \end{equation}\)
(iv) (a) : Shyams father will attain maximum revenue after 10 days.
So, he should harvest the onions after 10 days of 1^st July i.e., on 11^th July.
(v) (c) : Maximum revenue is collected by Shyams father when x = 10
\(\therefore\) Maximum revenue = R(10)
= 24000 + 60(10) - 3(10)2 = 24000 + 600 - 300 = 24300

(i) (c): When we solve an L.P.P. graphically, the optimal (or optimum) value of the objective function is attained at corner points of the feasible region.
(ii) (d): From the graph of 3x + 4y < 12 it is clear that it contains the origin but not the points on the line 3x + 4y = 12.

(iii) (d): Maximum of objective function occurs at corner points.

(iv) (d): Value of Z = px + qy at (15, 15) = 15p + 15q and that at (0, 20) = 20 q. According to given condition, we have
15p + 15q = 20q \(\Rightarrow\) 15p = 5q \(\Rightarrow\) q = 3p
(v) (b): Construct the following table of values of the objective function:

(i) (d) : Given, r cm is the radius and h cm is the height of required cylindrical can
Given that, volume = 3 l= 3000 cm³ \(\left(\because 1 l=1000 \mathrm{~cm}^{3}\right)\) 
\(\Rightarrow \pi r^{2} h=3000 \Rightarrow h=\frac{3000}{\pi r^{2}}\) 
Now, the surface area, as a function of r is given by
\(S(r)=2 \pi r^{2}+2 \pi r h=2 \pi r^{2}+2 \pi r\left(\frac{3000}{\pi r^{2}}\right)\) 
\(=2 \pi r^{2}+\frac{6000}{r}\) 
(ii) (c) : Now, \(S(r)=2 \pi r^{2}+\frac{6000}{r}\) 
\(\Rightarrow S^{\prime}(r)=4 \pi r-\frac{6000}{r^{2}}\) 
To find criti£al points, put S'(r) = 0
\(\Rightarrow \frac{4 \pi r^{3}-6000}{r^{2}}=0\) 
\(\Rightarrow r^{3}=\frac{6000}{4 \pi} \Rightarrow r=\left(\frac{1500}{\pi}\right)^{1 / 3}\) 
Also, \(\left.S^{\prime \prime}(r)\right|_{r=} \sqrt[3]{\frac{1500}{\pi}}=4 \pi+\frac{12000 \times \pi}{1500}\) 
\(=4 \pi+8 \pi=12 \pi>0\) 
Thus, the critical point is the point of minima.
(iii) (b) : The cost of material for the tin can is minimized when \(r=\sqrt[3]{\frac{1500}{\pi}} \mathrm{cm}\) and the height is  \(\frac{3000}{\pi\left(\sqrt[3]{\frac{1500}{\pi}}\right)^{2}}=2 \sqrt[3]{\frac{1500}{\pi}} \mathrm{cm} .\) .
(iv) (a) : We have,minimum surface area = \(\frac{2 \pi r^{3}+6000}{r}\) .
\(=\frac{2 \pi \cdot \frac{1500}{\pi}+6000}{\sqrt[3]{\frac{1500}{\pi}}}=\frac{9000}{7.8}=1153.84 \mathrm{~cm}^{2}\) 
Cost of 1 m² material = Rs.100
\(\therefore \ \text { Cost of } 1 \mathrm{~cm}^{2} \text { material }=Rs. \frac{1}{100}\) 
\(\therefore \ \text { Minimum cost }=Rs. \frac{1153.84}{100}=Rs. 11.538\) 
(v) (c) : To minimize the cost we need to minimize the total surface area.

(i) (c)
(ii) (b): Since (8, 12) satisfy all the inequalities therefore (8, .1.2) is the point in its feasible region.
(iii) (c) : At (0, 0), z = 0
At (16, 0), z = 352
At (8, 12), z = 392
At (0, 20), z = 360
It can be observed that max z occur at (8, 12). Thus, z will attain its optimal value at (8, 12).
(iv) (c) : We have, x + y = 20 ... (i)
and 3x + 2y = 48.. (ii)
On solving (i) and (ii), we get
X = 8, y = 12.
Thus, the coordinates of P are (8,12) and hence (8, 12) is one of its corner points.
(v) (b): The optimal solution occurs at every point on the line joining these two points.

(i) (d) : In order to make least expensive water tank, Nitin need to minimize its cost.
(ii) (d) : Let 1ft be the length and h ft be the height of the tank. Since breadth is equal to 5 ft. (Given)
\(\therefore\) Two sides will be 5h sq. feet and two sides will be
1h sq. feet. So, the total area of the sides is (10 h + 2 1h)ft²
Cost of the sides is Rs.10 per sq. foot. So, the cost to build the sides is (10h + 21h) x 10 = Rs.(100h + 20lh)
Also, cost of base = (5l) x 20 = Rs. 100 l
\(\therefore\) Total cost of the tank in Rs. is given by
c = 100 h + 20 I h + 100 l
Since, volume of tank = 80ft³
\(\therefore \quad 5 l h=80 \mathrm{ft}^{3} \quad \therefore l=\frac{80}{5 h}=\frac{16}{h}\) 
\(\therefore \quad c(h)=100 h+20\left(\frac{16}{h}\right) h+100\left(\frac{16}{h}\right)\) 
\(=100 h+320+\frac{1600}{h}\) 
(iii) (b) : Since, all side lengths must be positive
\(\therefore \quad h>0\) and \(\frac{16}{h}>0\) 
Since, \(\frac{16}{h}>0, \text { whenever } h>0\) 
\(\therefore \text { Range of } h \text { is }(0, \infty)\) 
(iv) (a) : To minimize cost, \(\frac{d c}{d h}=0\) 
\(\Rightarrow \quad 100-\frac{1600}{h^{2}}=0\) 
\(\Rightarrow 100 h^{2}=1600 \Rightarrow h^{2}=16 \Rightarrow h=\pm 4\) 
\(\Rightarrow h=4\)  [\(\therefore\) height can not be negative]
(v) (c) : Cost of least expensive tank is given by
\(c(4)=400+320+\frac{1600}{4}\) 
= 720 + 400 = Rs. 1120

Let E be the event that the doctor visit the patient late and let A₁, A₂, A₃, A₄ be the events that the doctor comes by cab, metro, bike and other means of transport respectively.
\(P\left(A_{1}\right)=0.3, P\left(A_{2}\right)=0.2, P\left(A_{3}\right)=0.1, P\left(A_{4}\right)=0.4\)
P(E I A₁) = Probability that the doctor arriving late when he comes by cab = 0.25
Similarly, P ( E I A₂) = 0.3, P (E I A₃) = 0.35 and P ( E I A₃) = 0.1
(i) (b): P(A₂ | E) = Probability that the doctor arriving late and he comes by metro
\(=\frac{P\left(A_{2}\right) P\left(E \mid A_{2}\right)}{\sum P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\)
\(=\frac{(0.2)(0.3)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)
\(=\frac{0.06}{0.21}=\frac{2}{7}\)
(ii) (c): P(A₁ | E) = Probability that the doctor arriving late and he comes by cab
\(=\frac{P\left(A_{1}\right) P\left(E \mid A_{1}\right)}{\Sigma P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\)
\(=\frac{(0.3)(0,25)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)
\(=\frac{0.075}{0.21}=\frac{5}{14}\)
(iii) (d): P(A₃| E) = Probability that the doctor arriving late and he comes by bike
\(=\frac{P\left(A_{3}\right) P\left(E \mid A_{3}\right)}{\sum P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\)
\(=\frac{(0.1)(0.35)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)
\(=\frac{0.035}{0.21}=\frac{1}{6}\)
(iv) (c): P(A₄ | E) = Probability that the doctor arriving late and he comes by other means of transport
\(=\frac{P\left(A_{4}\right) P\left(E \mid A_{4}\right)}{\Sigma P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\)
\(=\frac{(0.4)(0.1)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)
\(=\frac{0.04}{0.21}=\frac{4}{21}\)
(v) (a): Probability that the doctor is late by any means
\(=\frac{2}{7}+\frac{5}{14}+\frac{1}{6}+\frac{4}{21}=1\)

(i) (b): Required probability = \(\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}}\)
\(=\frac{12 \times 11}{51 \times 50}=\frac{22}{425}\)
(ii) (a): Required probability = \(\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}}=\frac{13 \times 12}{51 \times 50}=\frac{26}{425}\)
(iii) (b): Clearly, P(A l E₁) = \(\frac{{ }^{12} C_{2}}{{ }^{51} C_{2}}=\frac{22}{425}\)
\(P\left(A \mid E_{2}\right)=\frac{{ }^{13} C_{2}}{{ }^{51} C_{2}}=\frac{26}{425} \)
\(P\left(A \mid E_{3}\right)=P\left(A \mid E_{4}\right)=\frac{26}{425}\)
\(\therefore \sum_{i=1}^{4} P\left(A \mid E_{i}\right)=\frac{22}{425}+\frac{26}{425}+\frac{26}{425}+\frac{26}{425}=\frac{100}{425}=0.24\)
(iv) (b): P(getting both king) = \(\frac{{ }^{4} C_{2}}{{ }^{52} C_{2}}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221}\)
(v) (a): P(drawing a king) = \(\frac{4}{52}=\frac{1}{13}\)
\( \therefore \quad P(\text { not drawing a king })=1-\frac{1}{13}=\frac{12}{13} \)
\(\therefore \quad \text { Required probability }=\frac{12}{13} \times \frac{12}{13}=\frac{144}{169}\)

(i) (b): Total number of tickets = 50
Let event A = First ticket shows even number and B = Second ticket shows even number
Now, P(Both tickets show even number) = P(A)・P(B | A)
\(=\frac{25}{50} \cdot \frac{24}{49}=\frac{12}{49}\)
(ii) (c): Let the event A = First ticket shows odd number and B = Second ticket shows odd number
P (Both tickets show odd number)
\(=\frac{25}{50} \times \frac{24}{49}=\frac{12}{49}\)
(iii) (c): Required probability = P(one number is a multiple of 4 and other is a multiple of 5)
= P(multiple of 5 on first ticket and multiple of 4 on second ticket) + P(multiple of 4 on first ticket and multiple of 5 on second ticket)
\(=\frac{10}{50} \cdot \frac{12}{49}+\frac{12}{50} \times \frac{10}{49}=\frac{12}{245}+\frac{12}{245}=\frac{24}{245}\)
(iv) (d): Required probability = P(one ticket with prime number and other ticket with a multiple of 4)
\(=2\left(\frac{15}{50} \times \frac{12}{49}\right)=\frac{36}{245}\)
(v) (b): Let the event A = First ticket shows even number and B = Second ticket shows odd number.
Now, P(First ticket shows an even number and second ticket shows an odd number) = P(A)・P(B | A)
\(=\frac{25}{50} \times \frac{25}{49}=\frac{25}{98}\)

(i) (d) : If x be the amount of increase in annual charges, then number of subscriber reduces to 5000 - x.
\(\therefore\) Revenue, R(x) = (3000 + x) (5000 - x)
\(=15000000+2000 x-x^{2}, 0 
(ii) (a) : Clearly, at x = 500
R(500) = 15000000 + 2000(500) - (500)²
= 15000000 + 1000000 - 250000 = Rs.15750000
(iii) (c) : Since, 15000000 + 2000x - x² = 15640000 (Given)
\(\Rightarrow x^{2}-2000 x+640000=0\) 
\(\Rightarrow \quad x^{2}-1600 x-400 x+640000=0\) 
\(\Rightarrow x(x-1600)-400(x-1600)=0 \Rightarrow x=400,1600\) 
(iv) (a) : \(\frac{d R}{d x}=2000-2 x\) and \(\frac{d^{2} R}{d x^{2}}=-2<0\) 
For maximum revenue, \(\frac{d R}{d x}=0 \Rightarrow x=1000\) 
\(\therefore\) Required amount = Rs. 1000
(v) (b) : Maximum revenue = R(1000)
= (3000 + 1000) (5000 - 1000)
= 4000 x 4000 = ~ 16000000