A shopkeeper sells two shirts at the same price. He makes 10% profit on one and loses 10 the other. How much in percentage does he gain or lose?

$$\begin{gathered} \frac{5x+2}{x^2-x-20}+ \frac{2x-1}{x^2-4x-5} \\ \Rightarrow \frac{5x+2}{x^2-5x+4x-20}+\frac{2x-1}{x^2-5x+x-5} \\ \Rightarrow \frac{5x+2}{x\left(x-5\right)+4\left(x-5\right)}+\frac{2x-1}{x\left(x-5\right)+1\left(x-5\right)} \\ \Rightarrow \frac{5x+2}{\left(x-5\right)\left(x+4\right)}+\frac{2x-1}{ \left(x-5\right)\left(x+1\right)} \\ \Rightarrow \frac{\left(5x+2\right)\left(x+1\right)+\left(2x-1\right)\left(x+4\right)}{\left(x-5\right)\left(x+4\right)\left(x+1\right)} \\ \Rightarrow \frac{5x^2+5x+2x+2+2x^2+8x-x-4}{\left(x-5\right)\left(x+4\right)\left(x+1\right)} \\ \Rightarrow \frac{7x^2+14x-2}{\left(x-5\right)\left(x+4\right)\left(x+1\right)} \end{gathered}$$

Ler, total number of travelers $=n\left(\cup \right)$

Number of traveler who had been to Spain, n(A) = 55

 Number of traveler who had been to France, n(B) = 53

 Number of traveler who had been to Germany, n(C) =79

Number of traveler who had been to Spain and France, (A∩B) = 18 

Number of traveler who had been to Spain and Germany, n (A ∩ C)=17 

Number of traveler who had been to France and Germany, n( B ∩ C )=25 

Number of traveler who had been to all three countries, n (A ∩ B ∩ C)=10 

We know, n(∪)= n(A)+n(B)+n(C)-n(A ∩ B)-n(B ∩ C)-n(A ∩ C)+n(A ∩ B ∩ C) = 55 + 53 + 79 - 18 - 17 - 25 + 10 = 197-60 = 137 ans. 

Let, the tangent AB = 5cm

 Distance of the point A from the centre the circle OA = 4cm 

Now drawing a perpendicular from the tangent on the center of the circle OB, we get a right angled triangle $∆AOB$

apply Pythagorean thorium, 

$$\begin{gathered} O{A}^2+O{B}^2=A{B}^2 \\ \Rightarrow 4^2+O{B}^2=5^2 \\ \Rightarrow O{B}^2=25-16 \\ \Rightarrow O{B}^2=9 \\ \therefore OB=3 \end{gathered}$$

The radius of the circle is 3cm.

1^st part,

$$\begin{gathered} x^3-16x \\ =x\left(x^2-16\right) \\ =x\left\{{\left(x\right)}^2-{\left(4\right)}^2\right\} \\ =x\left(x+4\right)\left(x-4\right) \end{gathered}$$

2^nd part 

$$\begin{gathered} 2x^3+9x^2+4x \\ =x\left(2x^2+9x+4\right) \\ =x\left(2x^2+8x+x+4\right) \\ =x\left\{2x\left(x+4\right)+1\left(x+4\right)\right\} \\ =x\left(x+4\right)\left(2x+1\right) \end{gathered}$$

3^rd part 

$$\begin{gathered} 2x^3+x^2-28x \\ =x\left(2x^2+x-28\right) \\ =x\left(2x^2+8x-7x-28\right) \\ =x\left\{2x\left(x+4\right)\left(2x-7\right)\right\} \\ =x\left(x+4\right)\left(2x-7\right) \end{gathered}$$

H.C.F of these three part will be x(x + 4).