logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Linear Inequalities
0 votes

A solution is to be kept between $68^{\circ} F$ and $ 77^{\circ} F$ . What is the range in to temperature in degree ceisius (C) if the ceisius (C) if the ceisius / Fahrenheit (F) conversion. formula is given by $F= \large\frac{9}{8} $$c +32?$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Same Quantity can be added (a subtracted ) to (from ) both sides of the inequality with out changing the sign of the in equality.
  • Same positive quantities can be multiplied or divided to both side of the in equality with out changing the sign of the inequality.
  • If same negative quantity is multiplied or divided to both sides of the inequality is reversed i.e $ '>'$ sign changes to $'<' $ and $'<'$ changes $'>'$ .
Step 1:
Since the solution is to be kept between $68^{\circ}$ and $77^{\circ}$ F
We have that, $68 < F <77$
Substituting for $F= \large\frac{9}{5} $$C +32$
We get $68 < \large\frac{9}{5}$$c +32 < 77$
Step 2:
Subtracting 32 from both sides of inequality
$=> 68-32 < \large\frac{9}{5} $$ C < 77 -32$
$=> 36 < \large\frac{9}{5} $$C<45$
Multiplying by positive number $\large\frac{5}{9}$ on both sides, of inequality
$=> 36 \times \large\frac{5}{9} < \frac{9}{5} \times \frac{5}{9} $$C < 45 \times \large\frac{5}{9}$
$=> 20 < C <25$
Step 3:
The required range of temperature in degree ceisius is between $20^{\circ}$ and $25^{\circ}C$
answered Aug 5, 2014 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...