A thermometer has wrong calibrations. It reads melting point of ice as $-10°C$  and reads $60°C$ at temperature of $50°C$. Temperature of boiling point noted by this thermometer will be (All temperatures are noted at atmospheric pressure)

For a faculty thermometer, if reading is y whereas current reading is x then,
$\frac{x-{\left(LFP\right)}_{correctscale}}{{(UFP-LFP)}_{correctscale}}=\frac{y-{\left(LFP\right)}_{incorrectscale}}{{(UFP-LFP)}_{incorrectscale}}$ 
That means we can view them as two different scales.
$\Rightarrow {(UFP-LFP)}_{incorrectscale}=2\times 70=140\Rightarrow y-(-10)=140\Rightarrow y=140-10={130}^{0}C$