On a new temperature scale, the melting point | vedclass

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(B) The relationship between two temperature scales is given by the formula:$\frac{	ext{Reading on scale} - 	ext{LFP}}{	ext{UFP} - 	ext{LFP}} =

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$56$

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(B) The relationship between two temperature scales is given by the formula:$\frac{	ext{Reading on scale} - 	ext{LFP}}{	ext{UFP} - 	ext{LFP}} = 	ext{constant}$For the Celsius scale, the Lower Fixed Point $(LFP)$ is $0^{\circ} C$ and the Upper Fixed Point $(UFP)$ is $100^{\circ} C$.For the new scale $X$, the $LFP$ is $20^{\circ} X$ and the $UFP$ is $110^{\circ} X$.Let the temperature on the new scale be $z^{\circ} X$ corresponding to $40^{\circ} C$.$\frac{z - 20}{110 - 20} = \frac{40 - 0}{100 - 0}$$\frac{z - 20}{90} = \frac{40}{100}$$\frac{z - 20}{90} = 0.4$$z - 20 = 0.4 	imes 90$$z - 20 = 36$$z = 36 + 20 = 56^{\circ} X$Therefore, $40^{\circ} C$ corresponds to $56^{\circ} X$.

On a new temperature scale, the melting point of ice is $20^{\circ} X$ and the boiling point of water is $110^{\circ} X$. $A$ temperature of $40^{\circ} C$ would be indicated on this new temperature scale as (in $^{\circ} X$)

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