Rupali Bank Ltd. || Officer (Cash) (28-04-2018) || 2018

প্রশ্নে বলা হচ্ছে, X সপ্তাহের জন্য একটি পরিবারের নিকট 480kg চাল রয়েছে। তারা যদি ঐ চাল আরো  4 সপ্তাহ চালাতে চায়, তবে তাদের সাপ্তাহিক চালের ভোগের পরিমাণ 4kg কমাতে হবে। X এর মান কত?

Here, X is the number of weeks.

The amount of rice is 480kg.

∴ If they need to run 4 more weeks, they need to imply following equation.

$$\begin{gathered} \frac{480}{x} - \frac{480}{\left(x+4\right)} = 4 \\ \Rightarrow 480 \left(\frac{1}{x} - \frac{1}{\left(x+4\right)}\right) = 4 \\ \Rightarrow \frac{1}{x} - \frac{1}{\left(x+4\right)} = \frac{4}{480} = \frac{1}{120} \\ \Rightarrow \frac{x+4-x}{x\left(x+4\right)} = \frac{1}{120} \\ \Rightarrow \frac{4}{x^2 + 4x} = \frac{1}{120} \\ \Rightarrow x^2+4x = 480 \\ \Rightarrow x^2+4x-480 = 0 \\ \Rightarrow x^2+24x-20x-480 = 0 \\ \Rightarrow x(x+24)-20(x +24) = 0 \\ \Rightarrow (x+24) (x-20) = 0 \\ Here, x-20=0 \\ \therefore x = 20 \\ But, x=-24 is not acceptable. \\ \therefore The value of x is 20. \end{gathered}$$

$$\begin{gathered} Given that, \cos (A + B) = \frac{1}{\sqrt{2}} \\ \Rightarrow \cos (A+B) = \cos 45° \left[As \cos 45° = \frac{1}{\sqrt{2}}\right] \\ \Rightarrow A+B=45° . . . . . . . . . . . \left(i\right) \\ And, \cos (A-B) = \frac{\sqrt{3}}{2} \\ \Rightarrow \cos (A-B) = \cos 30° \left[As \cos 30° = \frac{\sqrt{3}}{2}\right] \\ \Rightarrow A-B=30° . . . . . . . .\left(ii\right) \\ Adding, equation \left(i\right) and \left(ii\right) we get \\ \frac{A + B = 45° \\ A - B=30°}{2A = 75°} \\ \Rightarrow A = \frac{75°}{2} = 37.5° \\ Now, we put A = 37.5° in equation \left(i\right) \\ A+B=45° \\ \Rightarrow 37.5° + B = 45° \\ \therefore B=45° -37.5° = 7.5° \\ \therefore A=37.5° and B = 7.5° \left(answer\right) \end{gathered}$$

$$\begin{gathered} \frac{1}{2x} + \frac{6}{y} = 3 . . . . . . . . \left(i\right) \\ \frac{5}{x} + \frac{3}{y} = 11 . . . . . . . . \left(ii\right) \\ Multiplying equation \left(ii\right) by 2 and subtracting it from equation \left(i\right) we get \\ \frac{{\displaystyle \frac{1}{2x}}+ {\displaystyle \frac{6}{y}} = 3 \\ {\displaystyle \frac{10}{x}} + {\displaystyle \frac{6}{y}} = 22 }{{\displaystyle \frac{1}{2x}} - {\displaystyle \frac{10}{x}} = -19} \\ \Rightarrow \frac{1-20}{2x} = -19 \\ \Rightarrow \frac{-19}{2x } = -19 \\ \Rightarrow \frac{1}{2x} = 1 [Dividing both sides by - 19] \\ \Rightarrow 2x = 1 \\ \therefore x = \frac{1}{2} \\ Putting the value of x in equation \left(i\right), we get \\ \frac{1}{2x} + \frac{6}{y} = 3 \\ \Rightarrow \frac{1}{2 \times {\displaystyle \frac{1}{2}}}+ \frac{6}{y} = 3 \\ \Rightarrow 1 + \frac{6}{y}= 3 \\ \Rightarrow \frac{6}{y} = 3-1 = 2 \\ \Rightarrow \frac{6}{y} = 2 \\ \Rightarrow \frac{3}{y} = 1 [Dividing both sides by 2] \\ \therefore y = 3 \\ \therefore \left(x,y\right)= \left(\frac{1}{2}, 3\right) \left(Answer\right) \end{gathered}$$