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:

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

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.

   $$13+14+15=42✓$$

Solution

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

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

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

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 $a\sqrt{3}+b\sqrt{5}$.

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

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 $a+b\sqrt{2}$:

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

Solution

1. Define the two unknown numbers.

   Let:

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

2. Formulate the system of equations.

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

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

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

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

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

5. Verification.

   • Sum: $45+23=68✓$
   • Difference: $45-23=22✓$

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:

   $$F+S=72⟹F=72-S$$

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 $S$.

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

   • 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: $46+26=72\text{ years}✓$
   • Ratio 6 years ago: $\frac{\text{Father’s age}}{\text{Son’s age}}=\frac{40}{20}=2✓$

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

2. Calculate the Taxable Income.

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

3. Calculate the tax payable.

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

Solution

1. Write the given parameters.

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

2. Apply the Compound Amount formula.

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

3. Calculate the compound markup.

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