Introduction to Linear Polynomials | Exercise 2.5

Question 2

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 yy depends on the hours of the use of the badminton court, xx, according to the relation y=ax+by = ax + b, find the values of aa and bb.

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Solution
Understand the Question
  • The monthly bill yy and hours of court use xx follow the linear relation y=ax+by = ax + b, where aa is the hourly rate and bb is the fixed monthly fee.
  • We are given two pairs of values: (x1,y1)=(10,800)(x_1, y_1) = (10, 800) and (x2,y2)=(15,1100)(x_2, y_2) = (15, 1100).
  • Substituting these pairs into the relation gives a system of two linear equations in aa and bb, which can be solved using elimination.

Step 1 · Formulate Linear Equations

Given the relation y=ax+by = ax + b

For x=10x = 10 and y=800y = 800

800=a(10)+b10a+b=800(1)\begin{aligned} 800 &= a(10) + b \\[0.6em] 10a + b &= 800 \quad \dots (1) \end{aligned}

For x=15x = 15 and y=1100y = 1100

1100=a(15)+b15a+b=1100(2)\begin{aligned} 1100 &= a(15) + b \\[0.6em] 15a + b &= 1100 \quad \dots (2) \end{aligned}

Step 2 · Solve for aa

Subtract equation (1)(1) from equation (2)(2)

(15a+b)(10a+b)=110080015a10a+bb=3005a=300a=3005a=60\begin{aligned} (15a + b) - (10a + b) &= 1100 - 800 \\[0.6em] 15a - 10a + b - b &= 300 \\[0.6em] 5a &= 300 \\[0.6em] a &= \dfrac{300}{5} \\[0.6em] a &= 60 \end{aligned}

Step 3 · Solve for bb

Substitute a=60a = 60 into equation (1)(1)

10(60)+b=800600+b=800b=800600b=200\begin{aligned} 10(60) + b &= 800 \\[0.6em] 600 + b &= 800 \\[0.6em] b &= 800 - 600 \\[0.6em] b &= 200 \end{aligned}
Answer

a=60a = 60 and b=200b = 200

Common Mistakes
  • Swapping Variables: Confusing the independent variable xx (hours) with the dependent variable yy (total bill).
  • Sign Errors in Subtraction: Forgetting to distribute the negative sign to both terms when calculating (15a+b)(10a+b)(15a + b) - (10a + b).

More questions in Exercise 2.5

Q1

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 yy depends on the number of modules accessed, xx, according to the relation y=ax+by = ax + b, find the values of aa and bb.

Q2

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 yy depends on the hours of the use of the badminton court, xx, according to the relation y=ax+by = ax + b, find the values of aa and bb.

Q3

Consider the relationship between temperature measured in degrees Celsius (C^\circ\text{C}) and degrees Fahrenheit (F^\circ\text{F}), which is given by C=aF+b^\circ\text{C} = a^\circ\text{F} + b. Find aa and bb, given that ice melts at 00 degrees Celsius and 3232 degrees Fahrenheit, and water boils at 100100 degrees Celsius and 212212 degrees Fahrenheit. (Hint: When C=0^\circ\text{C} = 0, F=32^\circ\text{F} = 32 and when C=100^\circ\text{C} = 100, F=212^\circ\text{F} = 212. Use this information to find aa and bb, and thus, the linear relationship between C^\circ\text{C} and F^\circ\text{F}.)

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