Rupali Bank Ltd. || Officer (22-11-2019) || 2019

Mr. Hossain invested n taka at 9% interest

$\therefore$He invested (18,000-n) taka at 10% interest

According to question, $\frac{(18000-n)\times 10}{100}+\frac{n\times 9}{100}=1700$

= (18000 - n) $\times$ 10 + 9n = 170000

=1,80,000-10n+9n = 1,70,000

$$\begin{gathered} \therefore n = 180000 - 170000 = 10,000 \\ \therefore He invested 10,000 at 9\% interest and (18000 - 10000) = 8000 taka at 10\% interest. \left(answer\right) \end{gathered}$$

Daily workers complete in 40 days 30 x 40 =1,200

$\therefore$Alternative workers complete in 40 days = 20 $\times \frac{40}{2}$=400

Now assume, they work x days after 40 days.

That means, alternate workers will work for $\frac{x}{2}$ days.

$\therefore$Total task can be written as

= 1, 200 + 400 + 30x + 20 $\times \frac{x}{2}$ = 2000

= 1, 600 + 30x + 10x = 2

=40x = 400

$\therefore$$x=\frac{400}{40}=10 days.$

Remaining papers = 35-5=30

$\therefore$Remaining time = 6-1=5 hours

$\therefore$Required new efficiency should be $\frac{30}{5}$ = 6 papers per hour

$\therefore$He must work = $\frac{(6-5)\times 100}{5}\%=\frac{1\times 100}{5}\%$ = 20%  faster. (answer)

Let, unit digit is y and tenth digit is x

$\therefore$The number is (10x + y)

From 1st condition, 10x+y= 6(x + y).......(i)

And from 2nd condition, (10x + y) - (10y + x) = 9 ............... (ii)

Now, from equation (i)

10x + y = 6x + 6y

=4x = 5y

$\therefore$x = $\frac{5y}{4}$…………………(iii)

Now, putting the value of x in equation (ii)

10x + y - (10y + x) = 9

= 10x + y - 10y - x = 9

=9x - 9y = 9

=(x - y) $\times$ 9 = 9

=x - y = 1

=$\frac{5y}{4}$ - y = 1

= $\frac{5y-4y}{4}$ = 1

$\therefore$y = 4

Putting y = 4 in equation

(iii), we get

$\therefore$x$\frac{5\times 4}{4}=5$

$\therefore$The original number is 10x + y = (10 $\times$ 5) + 4 = 50 + 4 = 54 (answer)

$$\begin{gathered} Given, x^3-21x+20 \\ let, f\left(x\right)=x^3-21x+20 \\ Here, if x=1, then f\left(x\right) = 0 \\ \therefore (x-1) will be the factor of f\left(x\right) \\ \therefore x^3-21x+2 \\ =x^3-x^2+x^2-x-20x+20 \\ =x^2(x-1)+x(x-1)-20(x-1) \\ =(x-1)(x^2+x-20) \\ =(x-1)(x^2+5x-4x-20) \\ =(x-1)\left\{\right(x(x+5)-4(x+5)\} \\ =(x-1) (x+5) (x-4) (answer) \end{gathered}$$

Here given, radius = 2$\sqrt{41}$ m

$\therefore$Diameter of the circle will be = 2 $\times$ 2$\sqrt{41}$ m = 4$\sqrt{41}$

$\therefore$4$\sqrt{41}$ will be the diagnal of the rectangular playground.

Now let, length be 4x m and width be 5x m.

According to Pythagorean theorem,

(4x)² + (5x)² = (4$\sqrt{41}$)²

⁼ 16x² + 25x² = 16 $\times$ 41

=41x² = 16 $\times$ 41

$\therefore$x² = $\frac{16\times 41}{31}$= 16

$\therefore$x = $\sqrt{16}$ = ${\sqrt{4}}^2$ = 4

$\therefore$Area will e = 4x $\times$ 5x = 20x² = 20 $\times$ 4² = 20 $\times$ 16 = 320m²

Let, one number is x and another one is y

According to question

x + y = 30 .......... (i)

And x² + y² = 468 ............(ii)

From equation (i), x + y = 30

⇒x=30-y........ (iii)

putting x = 30 - y in equation (ii)

x² + y² = 468

=(30 - y)² + y² = 468

= 900 - 60y + y² + y² = 468

=2y² - 60y + 432 = 0

=y² - 30y + 216 = 0

=y² - 18y - 12y + 216 = 0

=y(y - 18) - 12(y - 18) = 0

=(y - 18)(y - 12) = 0

Here, y - 18 = 0

$\therefore$y = 18

Or, y - 12 = 0

$\therefore$ y = 12

Putting y = 18 in equation (iii), we get

x = 30 - 18 = 12

$\therefore$x = 12

Putting y = 12 , in equation (iii) we get

x = 30 - 12 = 18

$\therefore$x = 18

(answer) (x, y) = (12, 18) Or, (18, 12).