Introduction to Linear Polynomials | Exercise 2.4

Question 1

Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.

(i) Find the height after 7 months.

(ii) Make a table of values for tt varying from 0 to 10 months and show how the height, hh, increases every month.

(iii) Find an expression that relates hh and tt, and explain why it represents linear growth.

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Solution
Understand the Question
  • Initial height: 1.75 feet1.75\text{ feet} (at month t=0t = 0).
  • Growth rate: 0.5 feet0.5\text{ feet} per month.
  • The total height hh after tt months is calculated as: Total Height=Initial Height+(Monthly Growth×t)\text{Total Height} = \text{Initial Height} + (\text{Monthly Growth} \times t)

(i) Find the height after 7 months.

Step 1 · Calculate Height after 7 Months

Given growth per month =0.5 feet= 0.5\text{ feet} and initial height =1.75 feet= 1.75\text{ feet}.

Growth in 7 months=0.5×7=3.5 feet\begin{aligned} \text{Growth in 7 months} &= 0.5 \times 7 \\[0.6em] &= 3.5\text{ feet} \end{aligned} Height after 7 months=1.75+3.5=5.25 feet\begin{aligned} \text{Height after 7 months} &= 1.75 + 3.5 \\[0.6em] &= 5.25\text{ feet} \end{aligned}
Answer

(i) 5.25 feet5.25\text{ feet}

(ii) Make a table of values for tt varying from 0 to 10 months and show how the height, hh, increases every month.

Step 1 · Construct the Table of Values

Starting at t=0t = 0 with h=1.75 feeth = 1.75\text{ feet}, add 0.5 feet0.5\text{ feet} for each subsequent month:

t (months)h (feet)01.7512.2522.7533.2543.7554.2564.7575.2585.7596.25106.75\begin{array}{|c|c|} \hline t\text{ (months)} & h\text{ (feet)} \\ \hline 0 & 1.75 \\ \hline 1 & 2.25 \\ \hline 2 & 2.75 \\ \hline 3 & 3.25 \\ \hline 4 & 3.75 \\ \hline 5 & 4.25 \\ \hline 6 & 4.75 \\ \hline 7 & 5.25 \\ \hline 8 & 5.75 \\ \hline 9 & 6.25 \\ \hline 10 & 6.75 \\ \hline \end{array}
Answer

(ii) $$ \begin{array}{|c|c|} \hline t\text{ (months)} & h\text{ (feet)} \ \hline 0 & 1.75 \ \hline 1 & 2.25 \ \hline 2 & 2.75 \ \hline 3 & 3.25 \ \hline 4 & 3.75 \ \hline 5 & 4.25 \ \hline 6 & 4.75 \ \hline 7 & 5.25 \ \hline 8 & 5.75 \ \hline 9 & 6.25 \ \hline 10 & 6.75 \ \hline \end{array}

(iii) Find an expression that relates hh and tt, and explain why it represents linear growth.

Step 1 · Formulate Expression and Explain Linear Growth

Let hh be the height in feet and tt be the time in months.

h=1.75+0.5th = 1.75 + 0.5t

This represents linear growth because:

  • The rate of change is constant (0.5 feet0.5\text{ feet} per month).
  • The variable tt is of degree 11 in the equation.
Answer

(iii) h=1.75+0.5th = 1.75 + 0.5t; it represents linear growth because the height increases at a constant rate of 0.5 feet0.5\text{ feet} each month.

Common Mistakes
  • Forgetting Initial Height: Calculating only the growth (0.5t0.5t) and forgetting to add the initial height of 1.75 feet1.75\text{ feet}.
  • Off-by-One in Table Starting Value: Beginning the table at t=1t = 1 with 1.75 feet1.75\text{ feet} instead of at t=0t = 0 (the initial state).

More questions in Exercise 2.4

Q1

Suppose a plant has height 1.75 feet and it grows by 0.5 feet each month.

(i) Find the height after 7 months.

(ii) Make a table of values for tt varying from 0 to 10 months and show how the height, hh, increases every month.

(iii) Find an expression that relates hh and tt, and explain why it represents linear growth.

Q2

A mobile phone is bought for ₹10,000. Its value decreases by ₹800 every year.

(i) Find the value of the phone after 3 years.

(ii) Make a table of values for tt varying from 0 to 8 years and show how the value of the phone, vv, depreciates with time.

(iii) Find an expression that relates vv and tt, and explain why it represents linear decay.

Q3

The initial population of a village is 750. Every year, 50 people move from a nearby city to the village.

(i) Find the population of the village after 6 years.

(ii) Make a table of values for tt varying from 0 to 10 years and show how the population, PP, increases every year.

(iii) Find an expression that relates PP and tt, and explain why it represents linear growth.

Q4

A telecom company charges ₹600 for a certain recharge scheme. This prepaid balance is reduced by ₹15 each day after the recharge.

(i) Write an equation that models the remaining balance b(x)b(x) after using the scheme for xx days. Explain why it represents linear decay.

(ii) After how many days will the balance run out?

(iii) Make a table of values for xx varying from 1 to 10 days and show how the balance b(x)b(x) reduces with time.

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