SolveItClass 9 · NCERT

NCERT Solutions · Class 9 Mathematics Introduction to Linear Polynomials

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Exercise Set 2.5 1–3 (part 5 of 8)

  1. Exercise 1

    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\displaystyle 10 modules, her bill was ₹400. When she accessed 14\displaystyle 14 modules, her bill was ₹500. If the monthly bill y\displaystyle y depends on the number of modules accessed, x\displaystyle x, according to the relation y=ax+b\displaystyle y=a x+b, find the values of a\displaystyle a and b\displaystyle b.

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    Two readings give two equations; subtract to remove the fixed fee.Here \(\displaystyle y = ax + b\), where \(\displaystyle x\) is the number of modules accessed, \(\displaystyle a\) is the cost of one module and \(\displaystyle b\) is the fixed monthly fee.Put the two observations into the relation.
    $\displaystyle 10$ modules cost ₹$\displaystyle 400$: \(\displaystyle 10a + b = 400 \qquad \ldots (1)\)
    $\displaystyle 14$ modules cost ₹$\displaystyle 500$: \(\displaystyle 14a + b = 500 \qquad \ldots (2)\)
    Both bills contain the same fixed fee \(\displaystyle b\). So if we subtract equation ($\displaystyle 1$) from equation ($\displaystyle 2$), the \(\displaystyle b\) cancels and only the cost of the extra modules is left: \[(14a + b) - (10a + b) = 500 - 400 \] \[4a = 100 \quad \Longrightarrow \quad a = 25 \]That makes sense on its own: $\displaystyle 4$ extra modules cost ₹$\displaystyle 100$ extra, so one module costs ₹25.Now substitute \(\displaystyle a = 25\) into equation ($\displaystyle 1$) to get \(\displaystyle b\): \[10(25) + b = 400 \quad \Longrightarrow \quad 250 + b = 400 \quad \Longrightarrow \quad b = 150 \]Check in the equation we did not use for the substitution, equation ($\displaystyle 2$): \[14(25) + 150 = 350 + 150 = 500 \quad \checkmark \]So the rule for the bill is \(\displaystyle y = 25x + 150\): ₹$\displaystyle 25$ per module plus a fixed ₹$\displaystyle 150$ a month.Answer: \(\displaystyle a = 25\) and \(\displaystyle b = 150\), so \(\displaystyle y = 25x + 150\).

  2. Exercise 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\displaystyle 10 hours, her bill was ₹800. When she used it for 15\displaystyle 15 hours, her bill was ₹1100. If the monthly bill y\displaystyle y depends on the hours of the use of the badminton court, x\displaystyle x, according to the relation y=ax+b\displaystyle y=a x+b, find the values of a\displaystyle a and b\displaystyle b.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Two readings give two equations; subtract to remove the fixed fee.In \(\displaystyle y = ax + b\), \(\displaystyle x\) is the number of hours of court use, \(\displaystyle a\) is the charge per hour and \(\displaystyle b\) is the fixed monthly fee.
    $\displaystyle 10$ hours cost ₹$\displaystyle 800$: \(\displaystyle 10a + b = 800 \qquad \ldots (1)\)
    $\displaystyle 15$ hours cost ₹$\displaystyle 1100$: \(\displaystyle 15a + b = 1100 \qquad \ldots (2)\)
    The fixed fee \(\displaystyle b\) is the same in both bills, so subtracting ($\displaystyle 1$) from ($\displaystyle 2$) removes it and leaves only what the extra hours cost: \[(15a + b) - (10a + b) = 1100 - 800 \] \[5a = 300 \quad \Longrightarrow \quad a = 60 \]Sense check: $\displaystyle 5$ extra hours added ₹$\displaystyle 300$, so one hour costs ₹60.Substitute \(\displaystyle a = 60\) in equation ($\displaystyle 1$): \[10(60) + b = 800 \quad \Longrightarrow \quad 600 + b = 800 \quad \Longrightarrow \quad b = 200 \]Check with equation ($\displaystyle 2$): \[15(60) + 200 = 900 + 200 = 1100 \quad \checkmark \]So the bill follows \(\displaystyle y = 60x + 200\): ₹$\displaystyle 60$ per hour on the court plus a fixed ₹$\displaystyle 200$ a month.Answer: \(\displaystyle a = 60\) and \(\displaystyle b = 200\), so \(\displaystyle y = 60x + 200\).

  3. Exercise 3

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

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    This solution has not been cross-checked against the answer printed in NCERT.

    Two known temperature pairs, two unknowns.We are told the relation has the form \[{}^{\circ}\mathrm{C} = a \cdot {}^{\circ}\mathrm{F} + b \] Write \(\displaystyle C\) for the Celsius reading and \(\displaystyle F\) for the Fahrenheit reading, so \(\displaystyle C = aF + b\).Substitute the two known pairs.Melting ice: \(\displaystyle C = 0\) when \(\displaystyle F = 32\): \[0 = 32a + b \qquad \ldots (1) \] Boiling water: \(\displaystyle C = 100\) when \(\displaystyle F = 212\): \[100 = 212a + b \qquad \ldots (2) \]Subtract ($\displaystyle 1$) from ($\displaystyle 2$) so that \(\displaystyle b\) cancels: \[100 - 0 = (212a + b) - (32a + b) \] \[100 = 180a \quad \Longrightarrow \quad a = \frac{100}{180} = \frac{5}{9} \]This number has a meaning: between freezing and boiling there are $\displaystyle 100$ Celsius degrees but $\displaystyle 180$ Fahrenheit degrees, so one Fahrenheit degree is only \(\displaystyle \tfrac{5}{9}\) of a Celsius degree.Find \(\displaystyle b\) from equation ($\displaystyle 1$): \[0 = 32 \cdot \frac{5}{9} + b \quad \Longrightarrow \quad b = -\frac{160}{9} \]Check with equation ($\displaystyle 2$): \[212 \cdot \frac{5}{9} - \frac{160}{9} = \frac{1060 - 160}{9} = \frac{900}{9} = 100 \quad \checkmark \]The relationship. Putting the two values back, \[C = \frac{5}{9}F - \frac{160}{9} = \frac{5}{9}\left(F - 32\right) \] which is the familiar conversion rule: subtract $\displaystyle 32$, then take five-ninths.Answer: \(\displaystyle a = \dfrac{5}{9}\) and \(\displaystyle b = -\dfrac{160}{9}\), so \(\displaystyle {}^{\circ}\mathrm{C} = \dfrac{5}{9}\left({}^{\circ}\mathrm{F} - 32\right)\).