A thermometer has wroing calibration. It reads the melting point of ice as -10. It reads `60^@` in place of `50^@`C. What is the temperature of the boiling point of water on the scale.

Explanation

`("Reading of faulty thermometer"-L.F.P.)/(U.F.P-L.F.P)=(C-0)/(100-0)`
`(60-(-10))/(U.F.P-L.F.P)=50/100=1/2`
or, U.F.P - L.F.P.=`2xx70=140`
Again `[x-(-10)]//140=100//100`
`therefore x=130^@C`
Thus it will read the boiling point of water as `130^@`C