Solve the following mathematical problems:

প্রশ্নে বলা হচ্ছে, একটি  নির্দিষ্ট দূরত্ব অতিক্রম করতে স্রোতের প্রতিকূলে একটি নৌকা সময় নেয় 8 ঘণ্টা 48 মিনিট এবং অনুকূলে ফিরে আসতে সময় নেয় 4 ঘণ্টা। নৌকার বেগ ও স্রোতের বেগের অনুপাত কত?

ধরি, স্রোতের প্রতিকূলে লোকটির গতিবেগ x kmph এবং অনুকূলে গতিবেগ y kmph.

 স্রোতের প্রতিকূলে 8 ঘণ্টা 48 মিনিটে যত দূরত্ব অতিক্রম করে স্রোতের অনুকূলে 4 ঘন্টায় সেই দূরত্ব অতিক্রম করে।

$$\begin{gathered} \therefore x\times 8\frac{4}{5}=y\times 4 \\ \Rightarrow \frac{44x}{5}=4y \\ \therefore y= \frac{44x}{5\times 4}=\frac{11x}{5} \end{gathered}$$

নৌকার বেগ : স্রোতের বেগ = $$\begin{gathered} \frac{y+x}{2} : \frac{y-x}{2} \\ =\left(y+x\right)\times \frac{1}{2} : \left(y-x\right)\times \frac{1}{2} \\ =\left(\frac{11}{5}+x\right)\frac{1}{2} : \left(\frac{11x}{5}-x\right)\frac{1}{2} \\ =\frac{16x}{5}\times \frac{1}{2} : \frac{6x}{5}\times \frac{1}{2} \\ =\frac{8x}{5} : \frac{3x}{5} \\ =8 : 3 \end{gathered}$$

নৌকার বেগ ও স্রোতের বেগ = 8 : 3 ans.

Total sample space after tossing 3 coins randomly will be = (HHH, HHT, HTH, THH, HTT, THI, TTH, TTT),

Now, we will find the probability of getting one head and two tails:

In the sample space we can see, a total of 8 types of outcome is possible. 

Among these 8 types of outcomes, the combinations with one head and two tails are (HTT, THT TTH). That means, 3 outcomes.

So, the required probability is 3/8 ans.

(ii) Probability of getting one tail: 

Above the sample space we can see that, a total of 8 types of outcome is possible. 

Among these 1 types of outcomes, the combinations with one tail are (HHT, HTH, THH). That means, 3 outcomes

So, the required probability is 3/8 ans.

(iii) Probability of getting one tail and two heads: 

Above the sample space we can see that, a total of 8 types of outcome is possible.

Among these 8 types of outcomes, the combinations with one tail and two heads are: (HHT, HTH, THH)

i.e: 3 outcomes: So, the required probability is 3/8 ans.

In first 6 days, C will work just two days but A and B will work 6 days.

So total work done by them in 6 days will

= $$\begin{gathered} \left(6\times \frac{1}{12}\right)+\left(6\times \frac{1}{15}\right)+\left(2\times \frac{1}{25}\right) \\ = \frac{1}{2}+\frac{2}{5}+\frac{2}{25} \\ = \frac{25+20+4}{50} \\ =\frac{49}{50 } \end{gathered}$$

Remaining work =$1-\frac{49}{50}=\frac{1}{50}$ part of the work.

So. (A and B) 1 day's work= $\frac{1}{12}+\frac{1}{15}=\frac{4+5}{60}=\frac{9}{60}=\frac{3}{20}$

In 7^th day. C will not help them. Only A and B will do the $\frac{1}{50}$  part of the work because $\frac{3}{20}>\frac{1}{50}$ 

Now, $\frac{3}{20}$ part of the work is done by A and B in 1 day 

$\frac{1}{50}$ part of the work is done by A and B in = $\frac{20}{3}\times \frac{1}{50}=\frac{2}{15}$ day 

Total work will be completed in = $\left(6+\frac{2}{15}\right)=6\frac{2}{15}$ days 

ans. The number of days required to complete the work is $6\frac{2}{15}$ days 

Suppose, the price of a shirt Tk. x

 and the price of a pant is Tk. y

According to the question:

$$\begin{gathered} x+y=1300 ..............\left(i\right) \\ 1.05x+1.1y=1405.............\left(ii\right) \end{gathered}$$

Now, we multiply equation (i) by 1.1 and subtract equation (ii) from equation (i)

$1.1x+1.1y=14301.05x+1.1y=14050.5=25$

$$\begin{gathered} x=\frac{25}{0.05} \\ x=500 \end{gathered}$$

Putting the value of x in equation (i) we get.

500+y=1300 

y=1300-500 = 800 

ans. Shirt Tk. 500, Pant Tk. 800.

Let, the length of the rectangle is x meter 

And the width of the rectangle is y meter

According to question,

$$\begin{gathered} 2y=x+10 ............ \left(i\right) \\ And xy=600 ....... \left(ii\right) \\ From equation \left(ii\right) \\ xy=600 \\ x=\frac{600}{y}...........\left(iii\right) \\ Putting the value of x in equation \left(i\right) \\ 2y=x+10 \\ \Rightarrow 2y=\frac{600}{y}+10 \\ \Rightarrow 2y^2-10y-600=0 \\ \Rightarrow y^2-5y-300=0 \left[Dividing both sides by 2\right] \\ \Rightarrow y^2-20y+15y-300=0 \\ \Rightarrow y\left(y-20\right)+15\left(y-20\right)=0 \\ \Rightarrow \left(y-20\right)\left(y+15\right)=0 \\ \therefore \left(y-20\right)\left(y+15\right)=0 \end{gathered}$$

Since width cannot be (-)ve. So we find y=20

That means the width of the rectangle is 20 meters

The length of the rectangle is $x=\frac{600}{y}$ 

$\therefore x=\frac{600}{20}=30$

The length of the rectangle is 30 meters.

 The perimeter of the rectangle = $2\times (30+20) = 2\times 50=100$ meters. 

ans. The perimeter is 100m. 

Let, the initial price of each pencil is Tk. x

And the initial price of each eraser is Tk y

 According the questions, 5x+6y= 80 …………..(i) 

After price increase of pencil, the new price I pencil is (x+20% of x) Tk.

$x+\frac{20x}{100}=\frac{6x}{5}$ tk. 

According to the question, 

$$\begin{gathered} \left(2\times \frac{6x}{5}\right)+\left(3\times y\right)=39 \\ \Rightarrow \frac{12x}{5}+3y=39 \\ \Rightarrow \frac{12x+15y}{5}=39 \\ \Rightarrow 12x+15y=195 ...........\left(ii\right) \end{gathered}$$ 

Now, multiplying equation (i) by 5 and equation (ii) by 2 and subtracting equation (ii) from equation (i) we get , 

$25x+30y=40024x+30y=390x=10$

New price of a pencil will be = $\frac{6x}{5}=\frac{6\times 10}{5}=12$ tk. 

ans. The new price of each pencil is 12 Tk.