Review Exercise 1 — Class 9 Mathematics Solutions

Solution

1. Identify the radicand.

   The number under the radical is $7$, which is a prime number and not a perfect square.

2. Apply the radical property of real numbers.

   The square root of any positive integer that is not a perfect square has a non-terminating and non-recurring decimal representation:

   $$\sqrt{7}\approx \mathrm{2.6457513...}$$

   Since it cannot be expressed in the form $\frac{p}{q}$ where $p,q\in \mathbb{Z}$ and $q\ne 0$, $\sqrt{7}$ is an irrational number.

Solution

1. Identify the mathematical constants.

   • $\pi$ represents the ratio of a circle's circumference to its diameter ($\pi \approx \mathrm{3.14159...}$).
   • $e$ (Euler's constant) is the base of natural logarithms ($e\approx \mathrm{2.71828...}$).

2. Classify their decimal expansions.

   Both $\pi$ and $e$ are transcendental numbers with non-terminating, non-repeating decimal expansions. Therefore, both are irrational numbers.

Solution

1. Recall the definition of square roots.

   If $n$ is a perfect square (e.g. $4,9,16$), then $\sqrt{n}$ is an integer and hence rational.

2. Analyze non-perfect squares.

   If $n$ is not a perfect square, its decimal form is non-terminating and non-recurring, which means $\sqrt{n}$ cannot be written as $\frac{p}{q}$. Thus, $\sqrt{n}$ is an irrational number.

Solution

1. Analyze the sum of surds.

   Both $\sqrt{3}$ and $\sqrt{5}$ are distinct irrational numbers.

2. Verify by contradiction.

   Assume $x=\sqrt{3}+\sqrt{5}$ were rational. Squaring both sides:

   $$\begin{array}{cc}{x}^2& =(\sqrt{3}+\sqrt{5}{)}^2\\ & =3+2\sqrt{15}+5\\ & =8+2\sqrt{15}\\ \sqrt{15}& =\frac{x^2-8}{2}\end{array}$$

   If $x$ were rational, then $\sqrt{15}$ would also be rational, which is a contradiction since $15$ is not a perfect square. Hence, $\sqrt{3}+\sqrt{5}$ is irrational.

Solution

1. Review equality properties.

   • Reflexive Property: $\forall x\in ℝ,x=x$.
   • Symmetric Property: $x=y⟹y=x$.
   • Transitive Property: $x=y\text{ and }y=z⟹x=z$.
   • Trichotomy Property: Exactly one of $x<y$, $x=y$, or $x>y$ holds.

2. Conclude the property.

   The equation $x=x$ describes the reflexive property of equality.

Solution

Solution

1. Express all bases as powers of $2$.

   $$\begin{array}{cc}8& =2^3\\ 64& =2^6\end{array}$$

2. Substitute and simplify exponents.

   $$\begin{array}{cc}{2}^x\times (2^3{)}^x& =2^6\\ 2^x\times 2^{3x}& =2^6\\ 2^{x+3x}& =2^6\\ 2^{4x}& =2^6\end{array}$$

3. Equate exponents and solve for $x$.

   $$\begin{array}{cc}4x& =6\\ x& =\frac{6}{4}\\ x& =\frac{3}{2}\end{array}$$

Solution

Solution

1. Factorize radicands into perfect square components.

   $$\begin{array}{cc}\sqrt{75}& =\sqrt{25\times 3}=5\sqrt{3}\\ \sqrt{27}& =\sqrt{9\times 3}=3\sqrt{3}\end{array}$$

2. Add the like surds.

   $$\begin{array}{cc}\sqrt{75}+\sqrt{27}& =5\sqrt{3}+3\sqrt{3}\\ & =(5+3)\sqrt{3}\\ & =8\sqrt{3}\end{array}$$

Solution

1. Apply difference of squares identity $(a+b)(a-b)=a^2-b^2$.

   $$\begin{array}{cc}(3+\sqrt{5})(3-\sqrt{5})& =(3{)}^2-(\sqrt{5}{)}^2\\ & =9-5\\ & =4\end{array}$$

2. Classify the result.

   The integer $4=\frac{4}{1}$ is an even composite number, which is a rational number ($4\in \mathbb{Q}$).

Solution

1. Evaluate the Left Hand Side ($\text{L.H.S.}$).

   $$\begin{array}{cc}\text{L.H.S.}& =a(b+c)\\ & =\frac{3}{2}(\frac{5}{3}+\frac{7}{5})\\ & =\frac{3}{2}\left(\frac{5\times 5+7\times 3}{15}\right)\\ & =\frac{3}{2}\left(\frac{25+21}{15}\right)\\ & =\frac{3}{2}\left(\frac{46}{15}\right)\\ & =\frac{3\times 46}{2\times 15}\\ & =\frac{138}{30}\\ & =\frac{23}{5}\end{array}$$

2. Evaluate the Right Hand Side ($\text{R.H.S.}$).

   $$\begin{array}{cc}\text{R.H.S.}& =ab+ac\\ & =(\frac{3}{2}\times \frac{5}{3})+(\frac{3}{2}\times \frac{7}{5})\\ & =\frac{15}{6}+\frac{21}{10}\\ & =\frac{5}{2}+\frac{21}{10}\\ & =\frac{5\times 5+21\times 1}{10}\\ & =\frac{25+21}{10}\\ & =\frac{46}{10}\\ & =\frac{23}{5}\end{array}$$

3. Conclusion.

   $$\text{L.H.S.}=\text{R.H.S.}=\frac{23}{5}$$

Solution

1. Evaluate the Left Hand Side ($\text{L.H.S.}$).

   $$\begin{array}{cc}\text{L.H.S.}& =(a+b)c\\ & =(\frac{3}{2}+\frac{5}{3})\times \frac{7}{5}\\ & =\left(\frac{3\times 3+5\times 2}{6}\right)\times \frac{7}{5}\\ & =\left(\frac{9+10}{6}\right)\times \frac{7}{5}\\ & =\left(\frac{19}{6}\right)\times \frac{7}{5}\\ & =\frac{19\times 7}{6\times 5}\\ & =\frac{133}{30}\end{array}$$

2. Evaluate the Right Hand Side ($\text{R.H.S.}$).

   $$\begin{array}{cc}\text{R.H.S.}& =ac+bc\\ & =(\frac{3}{2}\times \frac{7}{5})+(\frac{5}{3}\times \frac{7}{5})\\ & =\frac{21}{10}+\frac{35}{15}\\ & =\frac{21}{10}+\frac{7}{3}\\ & =\frac{21\times 3+7\times 10}{30}\\ & =\frac{63+70}{30}\\ & =\frac{133}{30}\end{array}$$

3. Conclusion.

   $$\text{L.H.S.}=\text{R.H.S.}=\frac{133}{30}$$

Solution

1. Evaluate the Left Hand Side ($\text{L.H.S.}$).

   $$\begin{array}{cc}\text{L.H.S.}& =a+(b+c)\\ & =\frac{4}{3}+(\frac{5}{2}+\frac{7}{4})\\ & =\frac{4}{3}+\left(\frac{5\times 2+7\times 1}{4}\right)\\ & =\frac{4}{3}+\frac{17}{4}\\ & =\frac{4\times 4+17\times 3}{12}\\ & =\frac{16+51}{12}\\ & =\frac{67}{12}\end{array}$$

2. Evaluate the Right Hand Side ($\text{R.H.S.}$).

   $$\begin{array}{cc}\text{R.H.S.}& =(a+b)+c\\ & =(\frac{4}{3}+\frac{5}{2})+\frac{7}{4}\\ & =\left(\frac{4\times 2+5\times 3}{6}\right)+\frac{7}{4}\\ & =\frac{23}{6}+\frac{7}{4}\\ & =\frac{23\times 2+7\times 3}{12}\\ & =\frac{46+21}{12}\\ & =\frac{67}{12}\end{array}$$

3. Conclusion.

   $$\text{L.H.S.}=\text{R.H.S.}=\frac{67}{12}$$

Solution

1. Evaluate the Left Hand Side ($\text{L.H.S.}$).

   $$\begin{array}{cc}\text{L.H.S.}& =a\times (b\times c)\\ & =\frac{4}{3}\times (\frac{5}{2}\times \frac{7}{4})\\ & =\frac{4}{3}\times \frac{35}{8}\\ & =\frac{4\times 35}{3\times 8}\\ & =\frac{140}{24}\\ & =\frac{35}{6}\end{array}$$

2. Evaluate the Right Hand Side ($\text{R.H.S.}$).

   $$\begin{array}{cc}\text{R.H.S.}& =(a\times b)\times c\\ & =(\frac{4}{3}\times \frac{5}{2})\times \frac{7}{4}\\ & =\frac{20}{6}\times \frac{7}{4}\\ & =\frac{10}{3}\times \frac{7}{4}\\ & =\frac{70}{12}\\ & =\frac{35}{6}\end{array}$$

3. Conclusion.

   $$\text{L.H.S.}=\text{R.H.S.}=\frac{35}{6}$$

Solution

1. Recall the definition of a rational number.

   A real number is rational if it can be expressed in the quotient form:

   $$x=\frac{p}{q},\text{where }p,q\in \mathbb{Z}\text{ and }q\ne 0$$

2. Express $0$ as a fraction.

   The number $0$ can be written by dividing $0$ by any non-zero integer $q$ (such as $1,2,3,-5$):

   $$\begin{array}{cc}0& =\frac{0}{1}=\frac{0}{2}=\frac{0}{-5}\end{array}$$

3. Verify the conditions.

   • Numerator $p=0$: $0\in \mathbb{Z}$ (is an integer).
   • Denominator $q=1$: $1\in \mathbb{Z}$ and $q\ne 0$.

4. Conclusion.

   Since $0$ satisfies all requirements of $\frac{p}{q}$, $0$ is a rational number.

Solution

1. Understand ordering on the real number line.

   For any two real numbers $a$ and $b$, their relative positions on the number line must satisfy one and only one order relation.

2. Formal Statement of Trichotomy Property.

   For all $a,b\in ℝ$, exactly one of the following three conditions is true:

   $$\begin{array}{cc}1.& a<b(a\text{ is strictly less than }b)\\ 2.& a=b(a\text{ is equal to }b)\\ 3.& a>b(a\text{ is strictly greater than }b)\end{array}$$

Solution

1. Find the first rational number ($q_1$) as the midpoint of $4$ and $5$.

   $$\begin{array}{cc}{q}_1& =\frac{4+5}{2}\\ & =\frac{9}{2}(=4.5)\end{array}$$

   Since $4<\frac{9}{2}<5$, $\frac{9}{2}$ lies strictly between $4$ and $5$.

2. Find the second rational number ($q_2$) between $4$ and $\frac{9}{2}$.

   $$\begin{array}{cc}{q}_2& =\frac{4+\frac{9}{2}}{2}\\ & =\frac{\frac{8+9}{2}}{2}\\ & =\frac{17}{4}(=4.25)\end{array}$$

3. Verify the ordering.

   $$4<\frac{17}{4}<\frac{9}{2}<5(4<4.25<4.5<5)$$

Solution

1. Convert radical form to exponential form.

   Using $\sqrt[n]{a}=a^{\frac{1}{n}}$:

   $$\begin{array}{cc}\sqrt[5]{\frac{x^{15}{y}^{35}}{z^{20}}}& ={\left(\frac{x^{15}{y}^{35}}{z^{20}}\right)}^{\frac{1}{5}}\end{array}$$

2. Distribute exponent $\frac{1}{5}$ to all numerator and denominator factors.

   $$\begin{array}{cc}& =\frac{(x^{15}{)}^{\frac{1}{5}}(y^{35}{)}^{\frac{1}{5}}}{(z^{20}{)}^{\frac{1}{5}}}\end{array}$$

3. Multiply exponents using $(a^m{)}^n=a^{mn}$.

   $$\begin{array}{cc}& =\frac{x^{15\times \frac{1}{5}}\cdot y^{35\times \frac{1}{5}}}{z^{20\times \frac{1}{5}}}\\ & =\frac{x^3{y}^7}{z^4}\end{array}$$

Solution

1. Express $27$ as $3^3$.

   $$\begin{array}{cc}\sqrt[3]{(27{)}^{2x}}& =\sqrt[3]{(3^3{)}^{2x}}\\ & =\sqrt[3]{3^{6x}}\end{array}$$

2. Convert cube root to exponential form.

   $$\begin{array}{cc}& =(3^{6x}{)}^{\frac{1}{3}}\\ & =3^{6x\times \frac{1}{3}}\\ & =3^{2x}(\text{or }{9}^x)\end{array}$$

Solution

1. Expand powers with sum exponents using $a^{m+n}=a^m{a}^n$.

   $$\begin{array}{cc}\frac{6(3{)}^{n+2}}{3^{n+1}-3^n}& =\frac{6\cdot (3^n\cdot 3^2)}{(3^n\cdot 3^1)-3^n}\end{array}$$

2. Factor out $3^n$ from numerator and denominator.

   $$\begin{array}{cc}& =\frac{6\cdot 9\cdot 3^n}{3^n(3-1)}\end{array}$$

3. Cancel common factor $3^n$ and evaluate.

   $$\begin{array}{cc}& =\frac{54\cdot 3^n}{2\cdot 3^n}\\ & =\frac{54}{2}\\ & =27\end{array}$$

Solution

1. Define the algebraic variables.

   Let the three consecutive odd integers be:

   • First integer $=n$
   • Second integer $=n+2$
   • Third integer $=n+4$

2. Set up the linear equation.

   $$n+(n+2)+(n+4)=51$$

3. Solve for $n$.

   $$\begin{array}{cc}3n+6& =51\\ 3n& =51-6\\ 3n& =45\\ n& =\frac{45}{3}\\ n& =15\end{array}$$

4. Determine the integers.

   • First integer $=15$
   • Second integer $=15+2=17$
   • Third integer $=15+4=19$

5. Verification.

   $$15+17+19=51✓$$

Solution

1. Define the variables.

   Let:

   • Balls in first (smaller) bucket $=x$
   • Balls in second (larger) bucket $=x+28$

2. Set up the linear equation.

   $$x+(x+28)=96$$

3. Solve for $x$.

   $$\begin{array}{cc}2x+28& =96\\ 2x& =96-28\\ 2x& =68\\ x& =\frac{68}{2}\\ x& =34\end{array}$$

4. Calculate balls in each bucket.

   • First bucket: $x=34\text{ balls}$
   • Second bucket: $34+28=62\text{ balls}$

5. Verification.

   $$\begin{array}{cc}34+62& =96✓\\ 62-34& =28✓\end{array}$$

Solution

1. Identify the given financial parameters.

   • Principal invested ($P$) $=\text{Rs. }\mathrm{350,000}$
   • Total investment duration $=7\text{ years}$

2. Calculate simple profit for first 2 years ($I_1$).

   • Rate $R_1=7\frac{1}{4}\%=\frac{29}{4}\%=7.25\%$
   • Time $T_1=2\text{ years}$

   $$\begin{array}{cc}{I}_1& =\frac{P\times R_1\times T_1}{100}\\ & =\frac{\mathrm{350,000}\times \frac{29}{4}\times 2}{100}\\ & =\frac{3500\times 29\times 2}{4}\\ & =\frac{3500\times 29}{2}\\ & =1750\times 29\\ & =\text{Rs. }\mathrm{50,750}\end{array}$$

3. Calculate simple profit for remaining 5 years ($I_2$).

   • Rate $R_2=8\%$
   • Time $T_2=7-2=5\text{ years}$

   $$\begin{array}{cc}{I}_2& =\frac{P\times R_2\times T_2}{100}\\ & =\frac{\mathrm{350,000}\times 8\times 5}{100}\\ & =3500\times 40\\ & =\text{Rs. }\mathrm{140,000}\end{array}$$

4. Calculate total profit ($I$).

   $$\begin{array}{cc}I& =I_1+I_2\\ & =\mathrm{50,750}+\mathrm{140,000}\\ & =\text{Rs. }\mathrm{190,750}\end{array}$$

5. Calculate total accumulated amount ($A$).

   $$\begin{array}{cc}A& =P+I\\ & =\mathrm{350,000}+\mathrm{190,750}\\ & =\text{Rs. }\mathrm{540,750}\end{array}$$