Sonali Bank Ltd. || Senior Officer (29-06-2018) || 2018

প্রশ্নে বলা হচ্ছে যে, একজন শিক্ষককে 35 খাতার দেখার জন্য 3 ঘণ্টা সময় দেয়া হলো। প্রথম 30 মিনিটে সে 5 টি খাতা দেখল ৷ প্রদত্ত সময়ে শেষ করতে হলে বাকি খাতাগুলো কি হারে দেখতে হবে?

Given, total allotted time = 3 hours

$= 3 \times 60 = 180 minutes$

∴ Remaining time = 180-30 = 150 minutes

∴ Remaining papers = 35 - 5 = 30 minutes

Per hour speed of grading first 5 papers $= \frac{60 \times 5}{30}$ = 10 papers

& to complete the work in time, required per hour speed of grading 30 papers $= \frac{30 \times 60}{150}$ = 12 papers

So, she should enhance the work speed by $= \frac{12-10}{10} \times 100 = 20\%$

প্রশ্নে বলা হচ্ছে, দুই অঙ্ক বিশিষ্ট কোন সংখ্যা তার অঙ্ক দুটির যোগফলের 4 গুণ। অঙ্ক দুটিকে স্থান পরিবর্তন করলে যে সংখ্যা পাওয়া যায় তা মূল সংখ্যার চেয়ে 18 বেশি। মূল সংখ্যাটি কত?

Let, Tenth digit is x and unit digit is y

∴ The number = (10x + y)

According to first condition, $4(x + y) = 10x + y . . . . . . . . . . . . \left(i\right)$

And second condition, $(10x + y) + 18 = 10y + x . . . . . . . . . . . . . \left(ii\right)$

From equation (ii), $10x + y + 18 = 10y + x$

$$\begin{gathered} \Rightarrow 9x-9y=-18 \\ \Rightarrow 9y-9x=18 [Multiplying by - 1] \\ \Rightarrow 9(y-x) = 18 \\ \Rightarrow (y-x) = \frac{18}{9} = 2 \\ \Rightarrow y-x = 2 \\ \therefore y = x+2 . . . . . . . . . \left(iii\right) \end{gathered}$$

From equation (i), we get $4(x + y) = 10x + y$

$$\begin{gathered} \Rightarrow 4x+4y = 10x + y \\ \Rightarrow 3y = 6x \\ \Rightarrow y = 2x \\ \Rightarrow x+2 = 2x [Putting the value of y] \\ \therefore x = 2 \end{gathered}$$

Now, putting the value of x in equation (iii) we get

$$\begin{gathered} y= x+2 \\ \Rightarrow y = 2+2 = 4 \\ \therefore y = 4 \\ The number is (10x + y) = (10 \times 2)+4=20+4=24. ans. \end{gathered}$$

$$\begin{gathered} Given, x^2-yx = 7.............. \left(i\right) \\ \Rightarrow xy=x^2-7 \\ \Rightarrow y = \frac{x^2-7}{x} \\ \Rightarrow y = x- \frac{7}{x} . . . . . . . . . \left(ii\right) \end{gathered}$$

Another equation is $y^2+xy = 30 . . . . . . . . . . \left(iii\right)$

Adding equation (i) and (iii) we get

$$\begin{gathered} \frac{x^2 - xy = 7 \\ {y}^2 + xy = 30}{ x^2 + y^2 = 37. . . . .\left(iv\right) } \end{gathered}$$

Now, putting the value of y in equation (iv) we get

$$\begin{gathered} x^2 + {\left(x - \frac{7}{x}\right)}^2 = 37 \\ \Rightarrow x^2 + x^2 - 2 \times x \times \frac{7}{x} + {\left(\frac{7}{x}\right)}^2 = 37 \\ \Rightarrow 2x^2 -14 + \frac{49}{x^2} = 37 \\ \Rightarrow 2x^2 + \frac{49}{x^2} = 51 \\ \Rightarrow \frac{2x^4 + 49}{x^2} = 51 \\ \Rightarrow 2x^4 - 51x^2 + 49 = 0 \\ \Rightarrow 2x^4 - 2x^2 - 49x^2 + 49 = 0 \\ \Rightarrow 2x^2 \left(x^2 - 1\right) - 49 \left(x^2 - 1\right) = 0 \\ \Rightarrow \left(x^2 - 1\right) \left(2x^2 - 49\right) = 0 \\ Here, x^2 - 1 = 0 \\ \Rightarrow x^2 = 1 \\ \Rightarrow x = \sqrt{1} \\ \therefore x =1 \\ And, 2x^2 - 49 = 0 \\ \Rightarrow x^2 = \frac{49}{2} \\ \therefore x = \sqrt{\frac{49}{2}} = \frac{7}{\sqrt{2}} \\ Thus we get z = 1, \frac{7}{\sqrt{2}} \\ When x = 1, then y = 1 - \frac{7}{1} = 1 - 7 = -6 \\ And when x = \frac{7}{\sqrt{2}}, then y = \frac{7}{\sqrt{2}} - \frac{7}{7} = \frac{7}{\sqrt{2}} - \frac{7}{1} \times \frac{\sqrt{2}}{7} =\frac{7}{\sqrt{2}} - \sqrt{2} = \frac{7-{\left(\sqrt{2}\right)}^2}{\sqrt{2}} = \frac{7-2}{\sqrt{2}}= \frac{5}{\sqrt{2}} \\ \therefore \left(x,y\right) = \left(1, -6\right) or \left(\frac{7}{\sqrt{2}}, \frac{5}{\sqrt{2}}\right)ans. \end{gathered}$$

প্রশ্নে বলা হচ্ছে, এক লোক ‘M’ কোম্পানিতে 10% সুদে টাকা এবং 'N' কোম্পানিতে 9% সুদে টাকা বিনিয়োগ y করে। সে মোট 9,000 টাকা দুই কোম্পানিতে বিনিয়োগ করে 850 টাকা সুদ হিসেবে নিতে চায়। কোন কোম্পানিতে কত টাকা বিনিয়োগ করবে?

Let, He invested Tk. x in company 'M' at 10%

∴ In 'N' Company, he invested Tk. (9,000-x) at 9%

We know, $I = \frac{P \times r \times n}{100}$

∴ The amount of interest at 10% of Tk. x $= \frac{x \times 10 \times 1}{100} = \frac{x}{10} Tk.$

And the amount of interest at 9% of Tk. (9,000 - x) $= \frac{\left(9000 - x\right) \times 9 \times 1}{100} = \frac{9 \left(9000 -x\right)}{100}Tk.$

According to question $\frac{x}{10} + \frac{9 \left(9000-x\right)}{100} =850$

$$\begin{gathered} \Rightarrow \frac{10x + 81000 -9x}{100}= 850 \\ \Rightarrow x+ 81,000 = = 85000 \\ \Rightarrow x = 85,000 - 81,000 \\ \therefore x = 4, 000 \end{gathered}$$

He will invest 4,000 Tk. in company 'M' and (9,000 - 4,000) = 5,000 Tk. in company 'N'. ans.

প্রশ্নে বলা হচ্ছে যে, একটি সমকোণী সমদ্বিবাহু ত্রিভুজে একটি বৃত্ত অন্তর্লিখিত আছে। ত্রিভুজটির একবাহুর দৈর্ঘ্য  (সমান সমান বাহুর একটি) 4cm হলে বৃত্তটির ক্ষেত্রফল কত?

Depict the following figure according to question

Area of ABC triangle $= \frac{1}{2} \times base \times height$

$= \frac{1}{2} \times 4 \times 4 = 8c{m}^2$

Now, $A{C}^2 = {\left(AB\right)}^2 + {\left(BC\right)}^2 = 4^2 +4^2 = 16 +16 = 32$

$\therefore AC = \sqrt{32}$

The radius of a circle inscribed in a triangle $= \frac{2 \times Area of triangle}{Perimeter of triangle } = \frac{2 \times 8}{\left(4+4\right) + \sqrt{32}} = \frac{16}{8 + \left(\sqrt{2\times 4^2}\right)}$

$$\begin{gathered} = \frac{16}{8 + 4\sqrt{2}} = \frac{16}{4 \left(2 + \sqrt{2}\right)} = \frac{4}{2 + \sqrt{2}} = \frac{2 \times \sqrt{2} \times \sqrt{2}}{\sqrt{2} \left(\sqrt{2} +1\right)} = \frac{2\sqrt{2}}{\sqrt{2}+1} \\ = \frac{2\sqrt{2} \left(\sqrt{2} -1\right) }{\left(\sqrt{2} +1\right) \left(\sqrt{2} -1\right)} =\frac{2\sqrt{2} \left(\sqrt{2} -1\right) }{2-1} = 2\sqrt{2} \left(\sqrt{2} -1\right) \\ \therefore Radius = 2\sqrt{2} \left(\sqrt{2} -1\right) cm \\ \therefore Area of circle = {\mathrm{\pi R}}^2 = \mathrm{\pi }{\left\{2\sqrt{2} \left(\sqrt{2} -1\right) \right\}}^2 = \mathrm{\pi }4 \times 2{\left(\sqrt{2} -1\right)}^2 {\mathrm{cm}}^2 = 8\mathrm{\pi } 2\sqrt{2} {\left(\sqrt{2} -1\right)}^2 {\mathrm{cm}}^2 (\mathrm{ans}.) \end{gathered}$$

প্রশ্নে বলা হচ্ছে যে, বৃত্তের বহিঃস্থ একটি বিন্দু A হতে বৃত্তের উপর অঙ্কিত স্পর্শকের দূরত্ব 12cm এবং কেন্দ্র থেকে A বিন্দুর দূরত্ব 13cm হলে বৃত্তটির ব্যাস কত?

We depict the following figure according to question.

Here, OB is perpendicular on base AB.

So, clearly AO is hypotenuse

Now, applying Pythagorean law, we get

$O{A}^2 = O{B}^2 + A{B}^2$

$$\begin{gathered} \Rightarrow O{B}^2 = O{A}^2-A{B}^2 = {13}^2-{12}^2 = 169-144 = 25 \\ \therefore OB = \sqrt{25} = \sqrt{5^2} = 5 \\ \therefore OB = 5cm \end{gathered}$$

Here, OB is radius of the circle and we know that the diameter of a circle is twice of the radius.

$\therefore Diameter = 5 \times 2 = 10cm. (ans.)$