Mathematices

$$\begin{gathered} Let, Sana earns 2x Tk. monthly \\ \therefore Ripa earns x Tk. monthly \\ And again, Sana spends 5y Tk \\ \therefore Ripa spends 3y Tk \\ According to question \\ (2x - 5y) : (x - 3y) = 4:1 \end{gathered}$$

$$\begin{gathered} =\frac{2x-5y}{x-3y}=\frac{4}{1} \\ =4x-12y=2x-5y \\ =2x=7y \\ \therefore y=\frac{2x}{7} \end{gathered}$$

$$\begin{gathered} Again \left(\right(2x - 5y\left)\right) + (x - 3y) = 5000 \\ = 3x - 8y = 5000 \\ =3x-8\times \frac{2x}{7}=5000 \\ =3x-\frac{16x}{7}=5000 \\ =\frac{21x-16x}{7}=5000 \\ =5x=5000\times 7 \\ \therefore x=7000 \\ \therefore Ripa earns 7,000 Tk. monthly. \end{gathered}$$

$$\begin{gathered} Let, the principal is 100 Tk. \\ After 6 years, the amount will be = 100 of 160\% = 160Tk. \end{gathered}$$

$$\begin{gathered} \therefore Simple interest = 160 - 100 = 60Tk . \\ \therefore 1 year interest: 60 =\frac{60}{6}= 10 Tk. \\ \therefore Rate of interest = 10\% \\ \therefore The amount of 12,000 Tk. for 3 years at 10\% compounded annually is \\ =12000{(1+\frac{1}{10})}^3=12000{\left(\frac{11}{10}\right)}^3=12000\times \frac{11}{10}\times \frac{11}{10}\times \frac{11}{10}=15972 Tk. \end{gathered}$$

$\therefore Required interest = 15 ,972-12,000=3,972 Tk.$

$$\begin{gathered} Given, One side of the square = 30 m \\ \therefore Area of the square {\left(30\right)}^2 m^2 = 900 m^2 \\ \therefore Area covered by the 4 cows in the over field=4\times \frac{1}{4}{\mathrm{\pi r}}^2{\mathrm{m}}^2 \\ =\frac{22}{7}\times {\left(14\right)}^2 {\mathrm{m}}^2 \\ =(\frac{22}{7}\times 14\times 14) {\mathrm{m}}^2 \\ =616 {\mathrm{m}}^2 \\ \therefore \mathrm{Ungrazed} \mathrm{area}=900-616=284{\mathrm{m}}^2 \end{gathered}$$

$$\begin{gathered} Let, the number of students in rooms A and B be x and y respectively \\ Then, x - 10 = y + 10 \\ =x - y = 20 ..............\left(1\right) \\ and x+20=2(y-20) \\ =x - 2y =-60.......\left(ii\right) \\ Solving \left(i\right) and \left(ii\right) we get: x = 100 v = 80. \\ ans: The number of students in room A = 100 \end{gathered}$$

$m^2=121$ বলে চেষ্টা করুন 121 কে বর্গ আকারে প্রকাশ করতে,

সেক্ষেত্রে $m^n={11}^2$ তাই m=11 এবং n=2

$\therefore {(m-1)}^{n+1}={(11-1)}^{2+1}={10}^3=1000$.