Question

"A technician has only two resistance coils.By using them either in series or in parallel he is able to obtain the resistance of 3,4,12 and 16 ohms.The resistance of two coils is"

Answer 1

With two resistances R1 and R2 we have the 4 following combinations possible:

R1  or  R2 if they are used individually. 
   let R2  be the larger of the two.

R1 R2 / (R1 + R2)  if used in parallel.      
     -- This is the smallest value.  It is  < R1 and < R2.

R1 + R2  if used in series.    This is the largest value.  SO it is  > R1 and > R2.

Hence, from the given values,
      R1 + R2 = 16 Ω
       R1 R2 / (R1 + R2) =  3 Ω

So, R1 and R2 are 4 Ω and 12 Ω respectively.

Answer 2

"The resistance of two coils R1 and R2 are 4ῼ and 12 ῼ respectively.

Given, A technician has only two resistance coils. By connecting them either in series or in parallel he is able to obtain the resistance of 3, 4, 12 and 16 ohms. The resistance of two coils is

The "two resistances" $R_1 and R_2$have the following combinations

1. $R_1 and R_2$ are used individually

If $R_2$ be the ""larger of the two""

2. $\frac { { { R }_{ 1 }{ R }_{ 2 } } }{ ({ R }_{ 1 }+{ R }_{ 2 }) }$ if used in parallel. (Then $<R_2 and <R_2$. Both have smallest values)

3. $R_1+R_2$used in series .(Here $>R_1 and >R_2.$Both have largest values)

Finally $R_1+R_2 = 16 \Omega$

$\frac { { { R }_{ 1 }{ R }_{ 2 } } }{ ({ R }_{ 1 }+{ R }_{ 2 }) } = 3 \Omega$

Hence, R1 and R2 are $4 \Omega and 12 \Omega$ respectively."