A solution is be kept between `40^(@)C and 45^(@)C`. What is the range of temperature in degree fahrenheit, if the conversion formula is `F = 9/5 C +32`?

Explanation

Let the required temperature be `x^(@)F`.
Given that, `F=9/5 C+32`
`rArr 5F = 9C+32 xx 5`
`rArr 9C=5F-32 xx 5`
`:. C=(5F-160)/9`
Since, temperature in degree calcius lies between `40^(@)C to 45^(@)C`.
Therefore, ` 40 lt (5F-160)/9 lt 45`
`rArr 40 lt (5x-160)/9 lt 45`
`rArr 40 xx 9 lt 5x - 160 lt 45 xx 9 ` [ multiplying throughout by 9]
`rArr 360 lt 5x - 160 lt 405` [ adding 160 throughout]
`rArr 360 + 160 lt 5x lt 405 + 160`
`rArr 520 lt 5x lt 565`
` rArr (520)/5 lt x lt (565)/5 ` [ divide throughout by 5}
`rArr 104 lt x lt 113`
Hence, the range of temperature in degree fahrenheit is ` 104^(@)F to 113^(@)F`.