Let the original speed of the train be s'  km / hr and the original time taken be 't' hours. 

We know that distance = speed $\times$ time. 

Therefore: 300 = s$\times$t ______ (i) 

If the speed had been 10 km/hr. faster, the new speed would be (s + 10) km / hr and the time taken would be (t - 1) hours. The distance remains the same: 

Therefore: 300 = (s + 10)$\times$(t - 1) 

$\Rightarrow 300 = st - s + 10t - 10$

$\Rightarrow 300 = 300 - s + 10t - 10$

$\Rightarrow 300 - 300 = - s + 10t - 10$

$\Rightarrow s = 10t - 10$

$\therefore t=\frac{\left(s+10\right)}{10}$

Now substitute this value of 't' back into equation (i) 

$300 = s \times \frac{(s + 10)}{10}$

$\Rightarrow 3000 = s(s + 10)$[Multiply both sides by 10] 

$\Rightarrow$ 3000 = s² + 10s 

$\Rightarrow$ s² + 10s - 3000 = 0 

$\Rightarrow$ s² + 60s - 10s - 3000 = 0 

∴ (s + 60)(s - 50) = 0 

This gives us two possible solutions for 's': s = - 60 or s = 50 

Since speed cannot be negative, the original speed of the train is 50 km/hr. 

Answer: The original speed of the train was 50 km/hr.

Here, $x^2-3x+1=0$

$\Rightarrow x\left(x+\frac{1}{x}\right) = 3x$

∴ $\left(x+\frac{1}{x}\right) = 3$

Now, ${\left(x+\frac{1}{x}\right)}^2={\left(x+\frac{1}{x}\right)}^2-4.\frac{1}{x}=3^2-4=5$

$\therefore \left(x+\frac{1}{x}\right) =\sqrt{5}$

So, $x^2-\frac{1}{x^2}=\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x}\right)=3\sqrt{5}$

Let 'x' be the quantity (in kg) of rice costing Tk. 40/kg. 

Let 'y' be the quantity (in kg) of rice costing Tk. 60/kg. 

Set up Equations: 

Equation (i) (Total weight): x + y = 50 (The total weight of the mixture is 50 kg) 

Equation (ii) (Total cost): 40x + 60y = 52 $\times$ 50 (The total cost of the mixture is the weighted average of the costs of the two types of rice multiplied by the total weight.) 

Simplify Equation (ii): 

40x + 60y = 2600 [Divide the entire equation by 20 to make it simpler] 

$\Rightarrow$ 2x + 3y = 130 ---------(iii) 

From Equation (i): x = 50 - y 

Substitute this value of x into the simplified Equation (iii): 

2(50 - y) + 3y = 130 

$\Rightarrow$ 100 - 2y + 3y = 130 

∴ y = 30 

Substitute the value of y back into the equation x = 50-y: 

x = 50 - 30 

∴ x = 20 

Answer: The shopkeeper used 20 kg of rice costing Tk. 40/kg and 30 kg of rice costing Tk. 60/kg. 

Let's break down the problem step by step. 

Initial Positions: 

A is 10th from the left. 

B is 9th from the right. 

After Interchange: 

A takes B's position and becomes 15th from the left. 

Determine Total Boys in the Row 

Since A moves to the position where B was, the 15^th position from the left must be B's original position. This means B's original position is the 15th position from the left. 

To find the total number of boys in the row, we need to determine the position of B from the left and right: 

• From the right, B was originally 9^th. 
• From the left, B's position is now known to be 15^th. 

The total number of boys in the row can be calculated using the formula: 

• Total number of boys = Position from the left + Position from the right-1 

Total number of boys = 15+9-1 = 23 

So, there are 23 boys in the row.

যখন ট্রেনটি কোথাও না থামে তখন ট্রেনটি গন্তব্যস্থানে পৌঁছাতে সময় লাগে 

$T=\frac{D}{D}=\frac{600}{100}$= 6 hours 

এবং প্রতি 75 কিমি যাওয়ার পর ট্রেনটি 3 মিনিট করে থামলে, মোট থামে (8-1) 7 বার 

এবং থামার সময় = (7 $\times$ 3) = 21 মিনিট 

অতএব, গন্তব্যে পৌঁছাতে মোট সময় লাগবে ট্রেনটি পৌছার সময় + থামার সময় 

= 6 hrs 21 minutes 

= 6 hours + 21 minutes 

Answer: The train took 6 hours and 21 minutes to reach its