QUESTION 2
Easy
A gym charges a fixed monthly fee and an additional cost per hour for using the badminton court. A student using the gym observed that when she used the badminton court for 10 hours, her bill was ₹800. When she used it for 15 hours, her bill was ₹1100. If the monthly bill y depends on the hours of the use of the badminton court, x, according to the relation y = ax + b. Find the values of a and b.
SOLUTION
1
Consider the given data
We have: \(y = ax + b\)
\(x = \) hours using badminton court
\(y = \) monthly bill(₹)
Given:
When \(x = 10, y = 800\)
When \(x = 15, y = 1100\)
Substitute these in \(y = ax + b\) to arrive at the following equations:
\(800 = 10a + b\)
and
\(1100 = 15a + b\)
\(x = \) hours using badminton court
\(y = \) monthly bill(₹)
Given:
When \(x = 10, y = 800\)
When \(x = 15, y = 1100\)
Substitute these in \(y = ax + b\) to arrive at the following equations:
\(800 = 10a + b\)
and
\(1100 = 15a + b\)
2
Solve the equations to find \(a\) and \(b\)
Let \(b = 800 − 10a\) [from first equation] and substitute it in the second equation to get
\(1100 = 15a + (800 − 10a)\)
⇒ \(1100 = 5a + 800\)
⇒ \(5a = 300\)
⇒ \(a = 60\)
So,
\(\begin{aligned}b &= 800 − 10(60) \\ &= 800 − 600 \\ &= 200 \end{aligned}\)
\(1100 = 15a + (800 − 10a)\)
⇒ \(1100 = 5a + 800\)
⇒ \(5a = 300\)
⇒ \(a = 60\)
So,
\(\begin{aligned}b &= 800 − 10(60) \\ &= 800 − 600 \\ &= 200 \end{aligned}\)
3
Write the final equation
\(y = 60x + 200\)
🏆
Final Answer : \(a = 60, b = 200\)
Concept Note
The bill has two parts: the fixed component(same every month) and the variable component(depends on usage). This gives a linear relationship of the form \(y = ax + b\), where
\(a = \)rate per unit of usage(slope)
\(b = \)fixed charge(y-intercept)
\(x = \)units of usage
\(y = \)total cost