A test tube contains a liquid of red color and another test tube contains an equal amount of water. To make a solution $20$ml of the liquid of color is poured into the second test tube. Two third of the so form solution poured from the second into the first. If the liquid in the test tube is four times that in the second then the quantity of water taken initially is.
A) 40\text{ ml}
B) 50\text{ ml}
C) 30\text{ ml}
D) 60\text{ ml}
E) \text{None of these}

Hint: First assume the quantities of the liquids present in the two test tubes. Given that 20ml of liquid poured from the first test tube to the second test tube then the quantity of liquid present in the first test tube is decreased by 20ml and the quantity of the liquid present in the second test tube is increased by 20ml. Find the exact quantities of the liquids in both test tubes. Further, they mentioned that two-third of the solution present in the second test tube is poured into the first test tube. Then find the two-third quantity of the solution and add it to the quantity present in the first test tube after decreasing the 20ml and find the quantity of the second test tube. Finally, we have given that the liquid in the first test tube is the four times of the liquid in the second test tube. Equate both the values we get the amount of liquid present in the first test-tube initially.

Complete step by step answer:
Let us assume that Xml of red color liquid is present in one test tube. Then automatically the amount of water in the second test tube is also Xml because given that the second test tube contains water that is equal to amount of liquid present in the first test tube. Hence
$\Rightarrow$ ${{l}_{1}}={{l}_{2}}=Xml$
Now to make the solution $20ml$of liquid color from the first test tube is poured into second test tube then the amount of liquid present in first and second test tubes are
$\Rightarrow$ ${{l}_{1}}=X-20ml$,
$\begin{align}
  & {{l}_{2}}=20ml+Xml \\
 & =\left( 20+X \right)ml
\end{align}$
$\Rightarrow$ Now the two third of the so form solution is poured from the second into the first. Then the amount present in the first test tube is
${{l}_{1}}=\left( X-20 \right)+\dfrac{2}{3}{{l}_{2}}$
Substituting the value of ${{l}_{2}}$ in the above equation, then we get
${{l}_{1}}=\left( X-20 \right)+\dfrac{2}{3}\left( 20+X \right)$
After pouring the two third of the so form solution in the second test tube into the first test tube, the amount of solution present in the second test tube is
$\begin{align}
  & {{l}_{2}}={{l}_{2}}-\dfrac{2}{3}{{l}_{2}} \\
 & =\dfrac{1}{3}{{l}_{2}} \\
 & =\dfrac{1}{2}\left( 20+X \right)
\end{align}$
If the liquid in the first test tube is fourth times of the liquid in the second test tube. i.e.
$\Rightarrow$ ${{l}_{1}}=4{{l}_{2}}$
Substituting the values ${{l}_{1}}=\left( X-20 \right)+\dfrac{2}{3}\left( 20+X \right)$ and ${{l}_{2}}=\dfrac{1}{3}\left( 20+X \right)$ in the above equation, then
$\begin{align}
  & \left( X-20 \right)+\dfrac{2}{3}\left( 20+X \right)=4\left[ \dfrac{1}{3}\left( 20+X \right) \right] \\
 & X-20=\dfrac{4}{3}\left( 20+X \right)-\dfrac{2}{3}\left( 20+X \right) \\
 & X-20=\left[ \dfrac{2}{3}\left( 20+X \right) \right]\left( 2-1 \right) \\
 & 3X-60=40+2X \\
 & X=100ml
\end{align}$
$\Rightarrow$ Hence the quantity of water taken initially is $X=100ml$

Note:
The question tries to trick the student with the confusing language, the student must be clear in understanding the language of the question, all the addition and subtraction of fractions must be made on a common denominator.