Question 3

We will form pairs of linear equations for each problem. Then, we will solve them using the substitution method.

Step 1 — Define variables and form equations for (i)

Let the first number be $\mathbf{x}$. Let the second number be $\mathbf{y}$. The difference between the numbers is 26. One number is three times the other.

$x - y = 26 \quad \text{(Equation 1)}$

$x = 3y \quad \text{(Equation 2)}$

Step 2 — Solve for the numbers in (i)

We substitute the value of $\mathbf{x}$ from Equation 2 into Equation 1.

$3y - y = 26$

$2y = 26$

$y = \frac{26}{2}$

$\boxed{y = 13}$

Now, we substitute the value of $\mathbf{y}$ back into Equation 2.

$x = 3 \times 13$

$\boxed{x = 39}$

Step 3 — Define variables and form equations for (ii)

Let the larger angle be $\mathbf{x}$. Let the smaller angle be $\mathbf{y}$. The angles are supplementary, so their sum is 180 degrees. The larger angle exceeds the smaller by 18 degrees.

$x + y = 180 \quad \text{(Equation 1)}$

$x = y + 18 \quad \text{(Equation 2)}$

Step 4 — Solve for the angles in (ii)

We substitute the value of $\mathbf{x}$ from Equation 2 into Equation 1.

$(y + 18) + y = 180$

$2y + 18 = 180$

$2y = 180 - 18$

$2y = 162$

$y = \frac{162}{2}$

$\boxed{y = 81}$

Now, we substitute the value of $\mathbf{y}$ back into Equation 2.

$x = 81 + 18$

$\boxed{x = 99}$

Step 5 — Define variables and form equations for (iii)

Let the cost of one bat be $\mathbf{₹x}$. Let the cost of one ball be $\mathbf{₹y}$. The coach buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750.

$7x + 6y = 3800 \quad \text{(Equation 1)}$

$3x + 5y = 1750 \quad \text{(Equation 2)}$

Step 6 — Solve for the costs in (iii)

From Equation 1, we express $\mathbf{x}$ in terms of $\mathbf{y}$.

$7x = 3800 - 6y$

$x = \frac{3800 - 6y}{7} \quad \text{(Equation 3)}$

We substitute this expression for $\mathbf{x}$ into Equation 2.

$3 \left( \frac{3800 - 6y}{7} \right) + 5y = 1750$

We multiply the entire equation by 7 to clear the denominator.

$3(3800 - 6y) + 7(5y) = 1750 \times 7$

$11400 - 18y + 35y = 12250$

$11400 + 17y = 12250$

$17y = 12250 - 11400$

$17y = 850$

$y = \frac{850}{17}$

$\boxed{y = 50}$

Now, we substitute the value of $\mathbf{y}$ back into Equation 3.

$x = \frac{3800 - 6(50)}{7}$

$x = \frac{3800 - 300}{7}$

$x = \frac{3500}{7}$

$\boxed{x = 500}$

Step 7 — Define variables and form equations for (iv)

Let the fixed charges be $\mathbf{₹x}$. Let the charge per kilometre be $\mathbf{₹y}$. For 10 km, the charge paid is ₹105. For 15 km, the charge paid is ₹155.

$x + 10y = 105 \quad \text{(Equation 1)}$

$x + 15y = 155 \quad \text{(Equation 2)}$

Step 8 — Solve for charges and calculate for 25 km in (iv)

From Equation 1, we express $\mathbf{x}$ in terms of $\mathbf{y}$.

$x = 105 - 10y \quad \text{(Equation 3)}$

We substitute this expression for $\mathbf{x}$ into Equation 2.

$(105 - 10y) + 15y = 155$

$105 + 5y = 155$

$5y = 155 - 105$

$5y = 50$

$y = \frac{50}{5}$

$\boxed{y = 10}$

Now, we substitute the value of $\mathbf{y}$ back into Equation 3.

$x = 105 - 10(10)$

$x = 105 - 100$

$\boxed{x = 5}$

The charge for travelling 25 km is given by $\mathbf{x + 25y}$.

$\text{Charge for 25 km} = 5 + 25(10)$

$\text{Charge for 25 km} = 5 + 250$

$\boxed{\text{Charge for 25 km} = ₹255}$

Step 9 — Define variables and form equations for (v)

Let the numerator of the fraction be $\mathbf{x}$. Let the denominator of the fraction be $\mathbf{y}$. The fraction is $\frac{x}{y}$. If 2 is added to both, the fraction becomes $\frac{9}{11}$. If 3 is added to both, the fraction becomes $\frac{5}{6}$.

$\frac{x+2}{y+2} = \frac{9}{11}$

$11(x+2) = 9(y+2)$

$11x + 22 = 9y + 18$

$11x = 9y + 18 - 22$

$11x = 9y - 4$

$x = \frac{9y - 4}{11} \quad \text{(Equation 1)}$

$\frac{x+3}{y+3} = \frac{5}{6}$

$6(x+3) = 5(y+3)$

$6x + 18 = 5y + 15$

$6x - 5y = 15 - 18$

$6x - 5y = -3 \quad \text{(Equation 2)}$

Step 10 — Solve for the fraction in (v)

We substitute the expression for $\mathbf{x}$ from Equation 1 into Equation 2.

$6 \left( \frac{9y - 4}{11} \right) - 5y = -3$

We multiply the entire equation by 11 to clear the denominator.

$6(9y - 4) - 11(5y) = -3 \times 11$

$54y - 24 - 55y = -33$

$-y - 24 = -33$

$-y = -33 + 24$

$-y = -9$

$\boxed{y = 9}$

Now, we substitute the value of $\mathbf{y}$ back into Equation 1.

$x = \frac{9(9) - 4}{11}$

$x = \frac{81 - 4}{11}$

$x = \frac{77}{11}$

$\boxed{x = 7}$

The fraction is $\frac{x}{y}$.

$\text{Fraction} = \frac{7}{9}$

Step 11 — Define variables and form equations for (vi)

Let Jacob's present age be $\mathbf{x}$ years. Let his son's present age be $\mathbf{y}$ years. Five years hence, Jacob's age will be $\mathbf{x+5}$. Five years hence, his son's age will be $\mathbf{y+5}$. Jacob's age will be three times his son's age.

$x + 5 = 3(y + 5)$

$x + 5 = 3y + 15$

$x = 3y + 15 - 5$

$x = 3y + 10 \quad \text{(Equation 1)}$

Five years ago, Jacob's age was $\mathbf{x-5}$. Five years ago, his son's age was $\mathbf{y-5}$. Jacob's age was seven times his son's age.

$x - 5 = 7(y - 5)$

$x - 5 = 7y - 35$

$x = 7y - 35 + 5$

$x = 7y - 30 \quad \text{(Equation 2)}$

Step 12 — Solve for their present ages in (vi)

We substitute the expression for $\mathbf{x}$ from Equation 1 into Equation 2.

$3y + 10 = 7y - 30$

$10 + 30 = 7y - 3y$

$40 = 4y$

$y = \frac{40}{4}$

$\boxed{y = 10}$

Now, we substitute the value of $\mathbf{y}$ back into Equation 1.

$x = 3(10) + 10$

$x = 30 + 10$

$\boxed{x = 40}$

Answer

(i) The two numbers are 13 and 39. (ii) The two angles are 81° and 99°. (iii) The cost of one bat is ₹500 and the cost of one ball is ₹50. (iv) The fixed charge is ₹5, the charge per km is ₹10, and a person has to pay ₹255 for travelling 25 km. (v) The fraction is $\frac{7}{9}$. (vi) Jacob's present age is 40 years and his son's present age is 10 years.