QUESTION 1
Easy
A learning platform charges a fixed monthly fee and an additional cost per digital learning module accessed.A student observes that when she accessed 10 modules, her bill was ₹400. When she accessed 14 modules, her bill was ₹500. If the monthly bill y depends on the number of modules accessed, x, according to the relation y = ax + b, find the values of a and b.
SOLUTION
1
Consider the given data
We have: \(y = ax + b\)
\(x = \) number of modules accessed
\(y = \) monthly bill(₹)
Given:
When \(x = 10, y = 400\)
When \(x = 14, y = 500\)
Substitute these in \(y = ax + b\) to arrive at the following equations:
\(400 = 10a + b\)
and
\(500 = 14a + b\)
\(x = \) number of modules accessed
\(y = \) monthly bill(₹)
Given:
When \(x = 10, y = 400\)
When \(x = 14, y = 500\)
Substitute these in \(y = ax + b\) to arrive at the following equations:
\(400 = 10a + b\)
and
\(500 = 14a + b\)
2
Solve the equations to find \(a\) and \(b\)
Let \(b = 400 − 10a\) [from first equation] and substitute it in the second equation to get
\(500 = 14a + (400 − 10a)\)
⇒ \(500 = 4a + 400\)
⇒ \(4a = 100\)
⇒ \(a = 25\)
So,
\(\begin{aligned}b &= 400 − 10(25) \\ &= 400 − 250 \\ &= 150 \end{aligned}\)
\(500 = 14a + (400 − 10a)\)
⇒ \(500 = 4a + 400\)
⇒ \(4a = 100\)
⇒ \(a = 25\)
So,
\(\begin{aligned}b &= 400 − 10(25) \\ &= 400 − 250 \\ &= 150 \end{aligned}\)
3
Write the final equation
\(y = 25x + 150\)
🏆
Final Answer : \(a = 25, b = 150\)
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