Is this simple ratio and proportion?

Let the resistances of $X$ , $Y$ and the variable resistor be $R_x$ , $R_y$ and $R$ respectively. By voltage division, we have: $\dfrac{R_y+R}{R_x}= \dfrac{V_A}{72-V_A}$ $\quad \Rightarrow (72-V_A) (R_y+R) = V_AR_x$

$\Rightarrow \begin{cases} (72-27)(R_y+0) = 27R_x & \Rightarrow 45R_y = 27R_x \\ & \Rightarrow \color{#3D99F6}{R_y = 0.6R_x} \\ (72-52)(R_y+1000) = 52R_x & \Rightarrow 20R_y + 20000 = 52R_x \\ & \Rightarrow 13R_x - \color{#3D99F6}{5R_y} = 5000 \\ & \Rightarrow 13R_x - \color{#3D99F6}{3R_x} = 5000 \\ & \Rightarrow \color{#D61F06}{R_x = 500} \Omega \\ (72-42)(R_y+R) = 42R_x & \Rightarrow \color{#3D99F6}{30R_y} + 30R = 42R_x \\ & \Rightarrow 30R = 42\color{#D61F06}{R_x} - \color{#3D99F6}{18}\color{#D61F06}{R_x} = 24(\color{#D61F06}{500}) \\ & \Rightarrow R = \boxed{400}\Omega \end{cases}$

By subtracting the given voltages from 72 V, the algebra becomes a tad easier than in Efren's solution.

Basic equation: $72 - V_A = 72\frac{R_X}{R_X+R_Y+R}.$ Given: $45 = 72\frac{R_X}{R_X+R_Y}\ \ \therefore\ \ \ 72R_X = 45(R_X+R_Y);$ $20 = 72\frac{R_X}{R_X+R_Y+1000}\ \ \therefore\ \ \ 72R_X = 20(R_X+R_Y) + 20\:000.$ Subtraction shows $25(R_X+R_Y) = 20\:000\ \ \ \therefore\ \ \ R_X+R_Y = 800.$ It is now easy to find $R_X = 500, R_Y = 300$ . For the final answer, solve $30 = 72\frac{R_X}{R_X+R_Y+R} = 72\frac{500}{800+R}\ \ \ \therefore\ \ \ 72\cdot 500 = 30\cdot (800+R).$ This results in $R = \boxed{400}$ .

Alternative solution:

Since $72-V_A$ is proportional to the current, it is inversely proportional to the total resistance of the circuit. Therefore $180/(72-V_A)$ is a linear function of $R$ . (I have chosen 180 to make nicer numbers; replace it by any constant you like.)

• If $R = 0$ , then $180/(72-V_A) = 180/45 = 4$ .

• If $R = 1000$ , then $180/(72-V_A) = 180/20 = 9$ .

We see that $180/(72-V_A) = 4 + R/200.$

Solve $4 + R/200 = 6$ , to find $R = 400$ .

Initially, it was 45 V : 27 V = 5 R: 3 R;

Later, it was 20 V : 52 V = 5 R : (3 R + 1000) = 5 : 13;

$\frac{5 R}{3 R + 1000} = \frac{5}{13}$

R is obviously 100 and therefore the two resistors are of 500 $\Omega$ and 300 $\Omega$ respectively;

Then, 30 V : 42 V = 500 : (300 + R') = 5 : 7;

Obviously, R' = 400 $\Omega$

Answer: 400 $\Omega$

This problem solely revolves on the principle of voltage divider theorem, from which we can derive equations for determining first the values of $X$ and $Y$ . Let $K$ be the resistance of the variable resistor.

When $K=0$ ,

$\large \frac {Y}{X+Y} = \frac {27}{72} = \frac {3}{8}$

when $K=1000$

$\large \frac {Y+1000}{X+Y+1000} = \frac {52}{72} = \frac {13}{18}$

from these equations, we get

$\large \begin{cases} Y = \frac {3}{5}X \\ 13X - 5Y = 5000 \end{cases}$

from which we get

$X = 500; \: Y= 300$

so from here, we now use these values to determine the desired value of $K$ .

$\frac { 300+K}{ 300 + 500 + K} = \frac {42}{72}$

from which we get $K = \boxed { 400 \: \Omega }$ .