Exercise 1.3 — Class 9 Mathematics Solutions

Solution

1. Define the algebraic variables.

   Let the three consecutive integers be:

   • First integer $= x$
   • Second integer $= x + 1$
   • Third integer $= x + 2$

2. Set up the linear equation according to the problem statement.

   $$x + (x + 1) + (x + 2) = 42$$

3. Solve for $x$.

   $$3x + 3 = 42$$

   Subtract $3$ from both sides:

   $$3x = 42 - 3$$ $$3x = 39 \implies x = \frac{39}{3} = 13$$

4. Calculate the three integers.

   • First integer $= x = 13$
   • Second integer $= x + 1 = 13 + 1 = 14$
   • Third integer $= x + 2 = 13 + 2 = 15$

5. Check and verify.

   $$13 + 14 + 15 = 42 \quad \checkmark$$

Solution

1. State the formula for the area of a right-angled triangle.

   For a right-angled triangle with perpendicular sides $\overline{AC}$ and $\overline{AB}$:

   $$\text{Area of } \Delta ABC = \frac{1}{2} \times \overline{AC} \times \overline{AB}$$

2. Substitute the given values into the formula.

   $$1 + \sqrt{15} = \frac{1}{2} \times (\sqrt{3} + \sqrt{5}) \times \overline{AB}$$

3. Isolate $\overline{AB}$.

   Multiply both sides by $2$:

   $$2(1 + \sqrt{15}) = (\sqrt{5} + \sqrt{3}) \times \overline{AB}$$ $$\overline{AB} = \frac{2(1 + \sqrt{15})}{\sqrt{5} + \sqrt{3}}$$

4. Rationalize the denominator by multiplying by conjugate $(\sqrt{5} - \sqrt{3})$.

   $$\overline{AB} = \frac{2(1 + \sqrt{15})(\sqrt{5} - \sqrt{3})}{(\sqrt{5} + \sqrt{3})(\sqrt{5} - \sqrt{3})} = \frac{2(1 + \sqrt{15})(\sqrt{5} - \sqrt{3})}{(\sqrt{5})^2 - (\sqrt{3})^2}$$ $$\text{Denominator} = 5 - 3 = 2$$ $$\overline{AB} = \frac{2(1 + \sqrt{15})(\sqrt{5} - \sqrt{3})}{2} = (1 + \sqrt{15})(\sqrt{5} - \sqrt{3})$$

5. Expand the product.

   $$(1)(\sqrt{5} - \sqrt{3}) + \sqrt{15}(\sqrt{5} - \sqrt{3}) = \sqrt{5} - \sqrt{3} + \sqrt{75} - \sqrt{45}$$

   Simplifying radicals $\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}$ and $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$:

   $$= \sqrt{5} - \sqrt{3} + 5\sqrt{3} - 3\sqrt{5}$$ $$= (5\sqrt{3} - \sqrt{3}) + (\sqrt{5} - 3\sqrt{5}) = 4\sqrt{3} - 2\sqrt{5}$$

6. Match with the form $a\sqrt{3} + b\sqrt{5}$.

   $$\overline{AB} = (4\sqrt{3} - 2\sqrt{5})\text{ cm} \implies a = 4, \; b = -2$$

Solution

1. Simplify the dimensions of the rectangle.

   • Length $L = 2 + \sqrt{18} = 2 + \sqrt{9 \times 2} = 2 + 3\sqrt{2}\text{ m}$
   • Width $W = 5 - \frac{4}{\sqrt{2}} = 5 - \frac{4\sqrt{2}}{2} = 5 - 2\sqrt{2}\text{ m}$

2. Apply the formula for the area of a rectangle.

   $$\text{Area} = L \times W = (2 + 3\sqrt{2})(5 - 2\sqrt{2})$$

3. Expand the algebraic binomial product.

   $$\text{Area} = 2(5) - 2(2\sqrt{2}) + 3\sqrt{2}(5) - 3\sqrt{2}(2\sqrt{2})$$ $$= 10 - 4\sqrt{2} + 15\sqrt{2} - 6(2)$$ $$= 10 + 11\sqrt{2} - 12$$ $$= -2 + 11\sqrt{2}\text{ m}^2$$

4. State integers $a$ and $b$.

   Comparing with $a + b\sqrt{2}$:

   $$a = -2, \quad b = 11$$

Solution

1. Define the two unknown numbers.

   Let:

   • Greater number $= x$
   • Smaller number $= y$

2. Formulate the system of equations.

   $$x + y = 68 \quad \text{--- (1)}$$ $$x - y = 22 \quad \text{--- (2)}$$

3. Add equations (1) and (2) to eliminate $y$.

   $$(x + y) + (x - y) = 68 + 22$$ $$2x = 90 \implies x = \frac{90}{2} = 45$$

4. Substitute $x = 45$ into equation (1) to find $y$.

   $$45 + y = 68$$ $$y = 68 - 45 = 23$$

5. Verification.

   • Sum: $45 + 23 = 68 \quad \checkmark$
   • Difference: $45 - 23 = 22 \quad \checkmark$

Solution

1. Write the given data and formula.

   • Temperature in Celsius $C = 48^\circ\text{C}$
   • Conversion formula:

   $$F = \frac{9}{5}C + 32$$

2. Substitute $C = 48$ into the formula.

   $$F = \frac{9}{5}(48) + 32$$

3. Multiply and calculate decimal fraction.

   $$\frac{9 \times 48}{5} = \frac{432}{5} = 86.4$$

4. Add $32$ to get final temperature in Fahrenheit.

   $$F = 86.4 + 32 = 118.4^\circ\text{F}$$

Solution

1. Define variables for present ages.

   Let:

   • Present age of son $= S$
   • Present age of father $= F$

   Given that the sum of their present ages is $72$ years:

   $$F + S = 72 \implies F = 72 - S$$

2. Express ages six years ago.

   • Son's age six years ago $= S - 6$
   • Father's age six years ago $= F - 6 = (72 - S) - 6 = 66 - S$

3. Formulate equation based on the condition given for 6 years ago.

   "Father's age was 2 times the age of the son":

   $$(F - 6) = 2(S - 6)$$ $$66 - S = 2(S - 6)$$

4. Solve for the son's present age $S$.

   $$66 - S = 2S - 12$$ $$66 + 12 = 2S + S$$ $$78 = 3S \implies S = \frac{78}{3} = 26\text{ years}$$

   • Present age of Son $= 26\text{ years}$
   • Present age of Father $= 72 - 26 = 46\text{ years}$

5. Find ages six years ago.

   • Son's age 6 years ago $= 26 - 6 = 20\text{ years}$
   • Father's age 6 years ago $= 46 - 6 = 40\text{ years}$

6. Verification.

   • Sum of present ages: $46 + 26 = 72\text{ years} \quad \checkmark$
   • Ratio 6 years ago: $\frac{\text{Father's age}}{\text{Son's age}} = \frac{40}{20} = 2 \quad \checkmark$

Solution

1. Write the given financial values.

   • Cost Price (C.P.) $= \text{Rs. } 1500$
   • Selling Price (S.P.) $= \text{Rs. } 1520$

2. Calculate total profit.

   $$\text{Profit} = \text{S.P.} - \text{C.P.} = 1520 - 1500 = \text{Rs. } 20$$

3. Apply profit percentage formula.

   $$\text{Profit \%} = \left(\frac{\text{Profit}}{\text{C.P.}}\right) \times 100$$ $$\text{Profit \%} = \left(\frac{20}{1500}\right) \times 100 = \frac{20}{15} = \frac{4}{3}\% = 1\frac{1}{3}\% \approx 1.33\%$$

Solution

1. State the given income details.

   • Total Annual Income $= \text{Rs. } 9,60,000$
   • Exempted Amount $= \text{Rs. } 1,30,000$
   • Tax Rate $= 0.75\%$

2. Calculate the Taxable Income.

   $$\text{Taxable Income} = \text{Total Income} - \text{Exempted Amount}$$ $$\text{Taxable Income} = 9,60,000 - 1,30,000 = \text{Rs. } 8,30,000$$

3. Calculate the tax payable.

   $$\text{Income Tax} = \text{Taxable Income} \times \text{Tax Rate}$$ $$\text{Income Tax} = 8,30,000 \times \frac{0.75}{100} = 8,300 \times 0.75 = \text{Rs. } 6,225$$

Solution

1. Write the given parameters.

   • Principal Amount $P = \text{Rs. } 3,75,000$
   • Markup Rate $R = 14\%$ per annum
   • Time Period $n = 1$ year

2. Apply the Compound Amount formula.

   $$A = P \left(1 + \frac{R}{100}\right)^n$$ $$A = 3,75,000 \left(1 + \frac{14}{100}\right)^1 = 3,75,000 \times 1.14 = \text{Rs. } 4,27,500$$

3. Calculate the compound markup.

   $$\text{Compound Markup} = A - P = 4,27,500 - 3,75,000 = \text{Rs. } 52,500$$