Transformers and resistors! Help!

In a transformer, the load resistance is 50 ohms. The turns ratio N1/N2 is 2.5-, and the rms source voltage is 80 V. IF a voltmeter across the load resistance measures an rms voltage of 25.0 V, what is the source resistance? My professor says the answer is 250 ohms but I can't come close to that.

If losses are not given, assume that the transformer is ideal.

The load current is:

I = V/R = 25/50 = 0.5 A

The primary voltage is:

Vp = Vs*N1/N2 = 25*2.5 = 62.5 Vrms

The voltage drop across the source resistance is:

Vd = 80 - 62.5 = 17.5 Vrms

The primary current is:

I1 = I2 / ( N1/N2 ) = .5 / 2.5 = .20 Arms

This same current flows in the source resistance, so the source resistance is:

Rs = Vd/I = 17.5 / .20 = 87.5 ohms

I don't agree with your prof on this one.

Hello Sam some data is missing. Hope efficiency is to be given. As you said that the output and input volts as 25 and 80, then the turn ratio ie 25/80 has to be 5/16. Please note that the turn ratio is always the ratio of secondary turns to that of primary. ie N2/N1.
Move over if transformer is not 100% efficient, then some power loss would be there. Hence efficiency is required to solve the given problem. If some one knows the way to tackle the same then let me also learn the same and let us both be thankful to him/her.
Current is got by V/R. So secondary current i2 could be got. So V2i2 would give the output power
Hence input power = efficiency x out put power.
Right from input power i1 can be calculated from input power/80. Hence the resistance R1 is to be got from V1/i1. THis is how I think. Ok. Let us wait.