Example 19 - Chapter 6 Class 11 Linear Inequalities (Term 2)
Last updated at July 5, 2019 by Teachoo
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Example 19 You are here
Example 20 Important
Chapter 6 Class 11 Linear Inequalities (Term 2)
Serial order wise
- Ex 6.1
- Ex 6.2
- Ex 6.3
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Examples
- Miscellaneous
Transcript
Example 19 (Method 1) In an experiment, a solution of hydrochloric acid is to be kept between 30° and 35° Celsius. What is the range of temperature in degree Fahrenheit if conversion formula is given by C = 59 (F – 32), where C and F represent temperature in degree F = 95 C + 32 Given that the solution is to be kept between 30°C and 35°C, Hence, 86 < F < 95 Hence when value of C is between 30°C and 35°C, value of F is between 86 °F & 95 °F Example 19 (Method 2) In an experiment, a solution of hydrochloric acid is to be kept between 30° and 35° Celsius. What is the range of temperature in degree Fahrenheit if conversion formula is given by C = 59 (F – 32), where C and F represent temperature in degree Given F = 95 C + 32 F – 32 = 95 C 95 C = F – 32 C = 59 (F – 32) Given that the solution is to be kept between 30°C and 35°C, 30 < C < 35 30 < 59 (F – 32) < 35, Multiplying 95 both sides (Eliminating 59) 95 × (30) < 59 (F – 32) × 95 < 95 × (35) 9 × 6 < F – 32 < 9 × 7 54 < F – 32 < 63 Adding 32 both sides (Eliminating 32) 54 + 32 < F – 32 + 32 < 63 + 32 86 < F < 95 Hence when value of C is between 30°C and 35°C, value of F is between 86 °F & 95 °F
