Download PDF

Class 9 Mathematics Notes – Chapter 1: Real Numbers

Board: Punjab Board
Subject: Mathematics
Class: 9th
Chapter: 1

Exercise 1.1

Q1. Identify each of the following as a rational or irrational number.

(i) \(2.353535\)

Answer: Rational number.

Reason: It is a terminating decimal number.

(ii) \(0.\overline{6}\)

Answer: Rational number.

Reason: It is a non-terminating, recurring decimal number.

(iii) \(2.236067\ldots\)

Answer: Irrational number.

Reason: It is a non-terminating, non-recurring decimal number.

(iv) \(\sqrt{7}\)

Answer: Irrational number.

Reason: The square root of a non-perfect square is irrational.

(v) \(e\)

Answer: Irrational number.

Reason: Euler’s constant is approximately \(2.71828\ldots\).

(vi) \(\pi\)

Answer: Irrational number.

Reason: Pi is approximately \(3.14159\ldots\).

(vii) \(5+\sqrt{11}\)

Answer: Irrational number.

Reason: The sum of a rational number and an irrational number is irrational.

(viii) \(\sqrt{3}\)

Answer: Irrational number.

Reason: The square root of a non-perfect square is irrational.

(ix) \(\dfrac{15}{4}+\sqrt{13}\)

Answer: Irrational number.

Reason: The sum of a rational number and an irrational number is irrational.

(x) \((2-\sqrt{2})(2+\sqrt{2})\)

Answer: Rational number.

Solution:

Using the identity

\[ (a-b)(a+b)=a^2-b^2 \]

\[ (2-\sqrt{2})(2+\sqrt{2}) =2^2-(\sqrt{2})^2 \]

\[ =4-2=2=\frac{2}{1} \]

Therefore, the answer is a rational number.

Q2. Represent the following numbers on the number line.

(i) \(\sqrt{2}\)

Construction:

Draw a number line and mark the origin \(O\).
Construct a right-angled triangle with a base of \(1\) unit and a perpendicular of \(1\) unit.
By the Pythagorean theorem, the hypotenuse is

\[ h=\sqrt{1^2+1^2}=\sqrt{2} \]

4. Using the hypotenuse as a radius, draw an arc from the origin to intersect the number line. The intersection represents \(\sqrt{2}\).

(ii) \(\sqrt{3}\)

Construction:

First, construct \(\sqrt{2}\).
At the endpoint of \(\sqrt{2}\), construct a perpendicular of \(1\) unit.
The new hypotenuse is

\[ h=\sqrt{(\sqrt{2})^2+1^2} \]

\[ h=\sqrt{2+1}=\sqrt{3} \]

4. Use this hypotenuse as a radius to mark \(\sqrt{3}\) on the number line.

(iii) \(4\dfrac{1}{3}=\dfrac{13}{3}\approx4.33\)

Construction:

Mark \(4\) and \(5\) on the number line.
Divide the segment between \(4\) and \(5\) into three equal parts.
Mark the first division to the right of \(4\). This point represents \(4\dfrac{1}{3}\).

(iv) \(-2\dfrac{1}{7}=-\dfrac{15}{7}\approx-2.14\)

Construction:

Mark \(-2\) and \(-3\) on the number line.
Divide the segment between them into seven equal parts.
Mark the first division to the left of \(-2\). This point represents \(-2\dfrac{1}{7}\).

(v) \(\dfrac{5}{8}=0.625\)

Construction:

Divide the segment between \(0\) and \(1\) into eight equal parts.
Mark the fifth division to the right of \(0\). This point represents \(\dfrac{5}{8}\).

(vi) \(\dfrac{23}{4}=2.75\)

Construction:

Mark \(2\) and \(3\) on the number line.
Divide the segment between them into four equal parts.
Mark the third division to the right of \(2\). This point represents \(2\dfrac{3}{4}\).

Q3. Express the following as a rational number \(\dfrac{p}{q}\), where \(p\) and \(q\) are integers and \(q\neq0\).

(i) \(0.\overline{4}\)

Solution:

Let

\[ x=0.\overline{4}=0.4444\ldots \tag{1} \]

Multiplying both sides by \(10\),

\[ 10x=4.4444\ldots \tag{2} \]

Subtracting equation (1) from equation (2),

\[ 10x-x=4.4444\ldots-0.4444\ldots \]

\[ 9x=4 \]

\[ \boxed{x=\frac{4}{9}} \]

(ii) \(0.\overline{37}\)

Solution:

Let

\[ x=0.\overline{37}=0.373737\ldots \tag{1} \]

Multiplying both sides by \(100\),

\[ 100x=37.373737\ldots \tag{2} \]

Subtracting equation (1) from equation (2),

\[ 100x-x=37.373737\ldots-0.373737\ldots \]

\[ 99x=37 \]

\[ \boxed{x=\frac{37}{99}} \]

(iii) \(0.\overline{21}\)

Solution:

Let

\[ x=0.\overline{21}=0.212121\ldots \tag{1} \]

Multiplying both sides by \(100\),

\[ 100x=21.212121\ldots \tag{2} \]

Subtracting equation (1) from equation (2),

\[ 100x-x=21 \]

\[ 99x=21 \]

\[ x=\frac{21}{99} \]

Dividing the numerator and denominator by \(3\),

\[ \boxed{x=\frac{7}{33}} \]

Q4. Name the property used in the following.

(i) \((a+4)+b=a+(4+b)\)

Answer: Associative property of addition.

(ii) \(\sqrt{2}+\sqrt{3}=\sqrt{3}+\sqrt{2}\)

Answer: Commutative property of addition.

(iii) \(x-x=0\)

Answer: Additive inverse property.

(iv) \(a(b+c)=ab+ac\)

Answer: Distributive property of multiplication over addition.

(v) \(16+0=16\)

Answer: Additive identity property.

(vi) \(100\times1=100\)

Answer: Multiplicative identity property.

(vii) \(4\times(5\times8)=(4\times5)\times8\)

Answer: Associative property of multiplication.

(viii) \(ab=ba\)

Answer: Commutative property of multiplication.

Q5. Name the property used in the following.

(i) \(-3<-1\Rightarrow0<2\)

Answer: Additive property of order.

Adding \(3\) to both sides gives

\[ -3+3<-1+3 \]

\[ 0<2 \]

(ii) If \(a<b\), then \(\dfrac{1}{a}>\dfrac{1}{b}\).

Answer: Reciprocal property of order, provided \(a\) and \(b\) are positive.

Note: The positivity condition is necessary for this reciprocal inequality.

(iii) If \(a<b\), then \(a+c<b+c\).

Answer: Additive property of order.

(iv) If \(ac<bc\) and \(c>0\), then \(a<b\).

Answer: Multiplicative property of order for a positive number.

(v) If \(ac<bc\) and \(c<0\), then \(a>b\).

Answer: Multiplicative property of order for a negative number. Multiplication or division by a negative number reverses the inequality sign.

(vi) Either \(a>b\), \(a=b\), or \(a<b\).

Answer: Trichotomy property.

Q6. Insert two rational numbers between the following numbers.

(i) \(\dfrac{1}{3}\) and \(\dfrac{1}{4}\)

Solution:

Find the average of the two numbers:

\[ \frac{\frac13+\frac14}{2} =\frac{\frac7{12}}{2} =\frac7{24} \]

Now find the average of \(\dfrac14\) and \(\dfrac7{24}\):

\[ \frac{\frac14+\frac7{24}}{2} =\frac{\frac{13}{24}}{2} =\frac{13}{48} \]

Answer:

\[ \boxed{\frac{13}{48},\ \frac7{24}} \]

(ii) \(3\) and \(4\)

Solution:

First average:

\[ \frac{3+4}{2}=\frac72 \]

Second average:

\[ \frac{3+\frac72}{2} =\frac{\frac{13}{2}}{2} =\frac{13}{4} \]

Answer:

\[ \boxed{\frac{13}{4},\ \frac72} \]

(iii) \(\dfrac35\) and \(\dfrac45\)

Solution:

Express the fractions with denominator \(50\):

\[ \frac35=\frac{30}{50},\qquad \frac45=\frac{40}{50} \]

Two rational numbers between them are

\[ \boxed{\frac{31}{50},\ \frac{32}{50}} \]

Exercise 1.2

Q1. Rationalize the denominator of the following.

(i) \(\dfrac{13}{4+\sqrt5}\)

Solution:

Multiply the numerator and denominator by the conjugate \(4-\sqrt5\):

\[ \frac{13}{4+\sqrt5} \times\frac{4-\sqrt5}{4-\sqrt5} \]

\[ =\frac{13(4-\sqrt5)}{4^2-(\sqrt5)^2} \]

\[ =\frac{13(4-\sqrt5)}{16-5} \]

\[ =\frac{52-13\sqrt5}{11} \]

Answer:

\[ \boxed{\frac{52-13\sqrt5}{11}} \]

(ii) \(\dfrac{\sqrt2+\sqrt5}{\sqrt3}\)

Solution:

Multiply the numerator and denominator by \(\sqrt3\):

\[ \frac{(\sqrt2+\sqrt5)\sqrt3}{\sqrt3\cdot\sqrt3} \]

\[ =\frac{\sqrt6+\sqrt{15}}{3} \]

Answer:

\[ \boxed{\frac{\sqrt6+\sqrt{15}}{3}} \]

(iii) \(\dfrac{\sqrt2-1}{\sqrt5}\)

Solution:

Multiply the numerator and denominator by \(\sqrt5\):

\[ \frac{(\sqrt2-1)\sqrt5}{\sqrt5\cdot\sqrt5} \]

\[ =\frac{\sqrt{10}-\sqrt5}{5} \]

Answer:

\[ \boxed{\frac{\sqrt{10}-\sqrt5}{5}} \]

(iv) \(\dfrac{6-4\sqrt2}{6+4\sqrt2}\)

Solution:

Multiply the numerator and denominator by \(6-4\sqrt2\):

\[ \frac{(6-4\sqrt2)^2} {(6+4\sqrt2)(6-4\sqrt2)} \]

\[ =\frac{36-48\sqrt2+32}{36-32} \]

\[ =\frac{68-48\sqrt2}{4} \]

Answer:

\[ \boxed{17-12\sqrt2} \]

(v) \(\dfrac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\)

Solution:

Multiply the numerator and denominator by \(\sqrt3-\sqrt2\):

\[ \frac{(\sqrt3-\sqrt2)^2} {(\sqrt3+\sqrt2)(\sqrt3-\sqrt2)} \]

\[ =\frac{3-2\sqrt6+2}{3-2} \]

Answer:

\[ \boxed{5-2\sqrt6} \]

(vi) \(\dfrac{4\sqrt3}{\sqrt7+\sqrt5}\)

Solution:

Multiply the numerator and denominator by \(\sqrt7-\sqrt5\):

\[ \frac{4\sqrt3(\sqrt7-\sqrt5)} {(\sqrt7+\sqrt5)(\sqrt7-\sqrt5)} \]

\[ =\frac{4(\sqrt{21}-\sqrt{15})}{7-5} \]

\[ =2\sqrt{21}-2\sqrt{15} \]

Answer:

\[ \boxed{2\sqrt{21}-2\sqrt{15}} \]

Q2. Simplify the following.

(i) \(\left(\dfrac{81}{16}\right)^{-\frac34}\)

Solution:

Using the negative exponent rule,

\[ \left(\frac{81}{16}\right)^{-\frac34} =\left(\frac{16}{81}\right)^{\frac34} \]

\[ =\left[\left(\frac23\right)^4\right]^{\frac34} \]

\[ =\left(\frac23\right)^3 \]

Answer:

\[ \boxed{\frac8{27}} \]

(ii) \(\left(\dfrac34\right)^{-2}\div\left(\dfrac49\right)^3\times\dfrac{16}{27}\)

Solution:

\[ =\frac{16}{9}\div\frac{64}{729}\times\frac{16}{27} \]

\[ =\frac{16}{9}\times\frac{729}{64}\times\frac{16}{27} \]

Answer:

\[ \boxed{12} \]

(iii) \((0.027)^{-\frac13}\)

Solution:

\[ (0.027)^{-\frac13} =\left(\frac{27}{1000}\right)^{-\frac13} \]

\[ =\left[\left(\frac3{10}\right)^3\right]^{-\frac13} \]

\[ =\left(\frac3{10}\right)^{-1} \]

Answer:

\[ \boxed{\frac{10}{3}} \]

(iv) \(\sqrt[4]{\dfrac{x^{14}y^{21}z^{35}}{y^{14}z^7}}\)

Solution:

Simplify the expression inside the radical:

\[ \sqrt[4]{x^{14}y^7z^{28}} \]

Using fractional exponents,

\[ =x^{\frac{14}{4}}y^{\frac74}z^{\frac{28}{4}} \]

Answer:

\[ \boxed{x^{\frac72}y^{\frac74}z^7} \]

Note: This transcription follows the exponents visible in the supplied text.

(v) \(\dfrac{5(25)^{n+1}-25(5)^{2n}}{5(5)^{2n+3}-(25)^{n+1}}\)

Solution:

Express all terms using base \(5\):

\[ =\frac{5\cdot5^{2n+2}-5^2\cdot5^{2n}} {5\cdot5^{2n+3}-5^{2n+2}} \]

\[ =\frac{5^{2n+3}-5^{2n+2}} {5^{2n+4}-5^{2n+2}} \]

Factor out \(5^{2n+2}\):

\[ =\frac{5^{2n+2}(5-1)} {5^{2n+2}(25-1)} \]

\[ =\frac4{24} \]

Answer:

\[ \boxed{\frac16} \]

(vi) \(\dfrac{16^{n+1}+20(4^{2n})}{2^3\times8^{n-1}}\)

Solution:

Express all terms using base \(2\):

\[ =\frac{2^{4n+4}+20(2^{4n})} {2^3\times2^{3n-3}} \]

\[ =\frac{16(2^{4n})+20(2^{4n})}{2^{3n}} \]

\[ =\frac{36(2^{4n})}{2^{3n}} \]

Answer:

\[ \boxed{36\cdot2^n} \]

(vii) \(64^{-\frac23}\div9^{-\frac32}\)

Solution:

\[ = (4^3)^{-\frac23}\div(3^2)^{-\frac32} \]

\[ =4^{-2}\div3^{-3} \]

\[ =\frac1{16}\div\frac1{27} \]

Answer:

\[ \boxed{\frac{27}{16}} \]

(viii) \(\dfrac{3^n\times9^{n+1}}{3^{n-1}\times9^{n-1}}\)

Solution:

Express \(9\) as \(3^2\):

\[ =\frac{3^n\times3^{2n+2}} {3^{n-1}\times3^{2n-2}} \]

\[ =\frac{3^{3n+2}}{3^{3n-3}} \]

\[ =3^5 \]

Answer:

\[ \boxed{243} \]

(ix) \(\dfrac{5^{n+3}-6\cdot5^{n+1}}{9\cdot5^n-2^n\cdot5^n}\)

Solution:

Factor out \(5^{n+1}\) from the numerator and \(5^n\) from the denominator:

\[ =\frac{5^{n+1}(25-6)} {5^n(9-2^n)} \]

\[ =\frac{5\cdot19}{9-2^n} \]

Answer:

\[ \boxed{\frac{95}{9-2^n}} \]

Q3. If \(x=3+\sqrt8\), find the value of the following.

Given:

\[ x=3+\sqrt8 \]

Rationalizing the reciprocal,

\[ \frac1x=\frac1{3+\sqrt8} \]

\[ =\frac{3-\sqrt8}{(3+\sqrt8)(3-\sqrt8)} \]

\[ =\frac{3-\sqrt8}{9-8}=3-\sqrt8 \]

Therefore,

\[ \boxed{\frac1x=3-\sqrt8} \]

(i) \(x+\dfrac1x\)

\[ =(3+\sqrt8)+(3-\sqrt8) \]

Answer:

\[ \boxed{6} \]

(ii) \(x-\dfrac1x\)

\[ =(3+\sqrt8)-(3-\sqrt8) \]

\[ =2\sqrt8=4\sqrt2 \]

Answer:

\[ \boxed{4\sqrt2} \]

(iii) \(x^2+\dfrac1{x^2}\)

Using the identity

\[ \left(x+\frac1x\right)^2 =x^2+\frac1{x^2}+2 \]

\[ x^2+\frac1{x^2}=6^2-2 \]

Answer:

\[ \boxed{34} \]

(iv) \(x^2-\dfrac1{x^2}\)

Using the identity

\[ x^2-\frac1{x^2} =\left(x+\frac1x\right)\left(x-\frac1x\right) \]

\[ =6\times4\sqrt2 \]

Answer:

\[ \boxed{24\sqrt2} \]

(v) \(x^4+\dfrac1{x^4}\)

\[ x^4+\frac1{x^4} =\left(x^2+\frac1{x^2}\right)^2-2 \]

\[ =34^2-2=1156-2 \]

Answer:

\[ \boxed{1154} \]

(vi) \(\left(x-\dfrac1x\right)^2\)

\[ =(4\sqrt2)^2 \]

\[ =16\times2 \]

Answer:

\[ \boxed{32} \]

Q4. Find rational numbers \(p\) and \(q\) such that

\[ \frac{8-3\sqrt2}{4+3\sqrt2}=p+q\sqrt2 \]

Solution:

Rationalize the denominator by multiplying by \(4-3\sqrt2\):

\[ \frac{(8-3\sqrt2)(4-3\sqrt2)} {(4+3\sqrt2)(4-3\sqrt2)} \]

Expanding the numerator,

\[ = \frac{32-24\sqrt2-12\sqrt2+18} {16-18} \]

\[ =\frac{50-36\sqrt2}{-2} \]

\[ =-25+18\sqrt2 \]

Comparing with \(p+q\sqrt2\),

Answer:

\[ \boxed{p=-25,\quad q=18} \]

Q5. Simplify the following.

(i) The supplied expression contains unclear radical indices and exponents. The source gives the following result:

\[ \boxed{\frac{75\sqrt3}{1024}} \]

Note: The original expression should be checked against the textbook before this result is used, because its radical notation is not fully recoverable from the supplied text.

(ii) The supplied expression is partially corrupted. The source gives the following result:

\[ \boxed{\frac23} \]

Note: Verify the original exponents and fraction layout in the textbook.

(iii) The supplied expression is partially corrupted. The source gives the following result:

\[ \boxed{\frac65} \]

Note: Verify the original radical indices and denominator in the textbook.

(iv) \((a^{1/3}+b^{1/3})(a^{2/3}-a^{1/3}b^{1/3}+b^{2/3})\)

Solution:

Using the sum of cubes identity,

\[ (x+y)(x^2-xy+y^2)=x^3+y^3 \]

Let

\[ x=a^{1/3},\qquad y=b^{1/3} \]

Then

\[ (x+y)(x^2-xy+y^2)=x^3+y^3 \]

\[ =a+b \]

Answer:

\[ \boxed{a+b} \]

Exercise 1.3

Q1. The sum of three consecutive integers is forty-two. Find the three integers.

Solution:

Let the three consecutive integers be

\[ x,\quad x+1,\quad x+2 \]

According to the question,

\[ x+(x+1)+(x+2)=42 \]

\[ 3x+3=42 \]

\[ 3x=39 \]

\[ x=13 \]

Therefore, the three integers are

\[ \boxed{13,\ 14,\ 15} \]

Q2. The diagram shows a right-angled triangle \(ABC\), where \(AC=(\sqrt3+\sqrt5)\) cm and the area is \((1+\sqrt{15})\text{ cm}^2\). Find \(AB\) in the form \((a\sqrt3+b\sqrt5)\) cm.

Solution:

The area of a right-angled triangle is

\[ \text{Area}=\frac12\times\text{base}\times\text{height} \]

Therefore,

\[ 1+\sqrt{15} =\frac12\times AB\times(\sqrt3+\sqrt5) \]

\[ AB=\frac{2(1+\sqrt{15})}{\sqrt3+\sqrt5} \]

Rationalizing the denominator,

\[ AB=\frac{2(1+\sqrt{15})(\sqrt5-\sqrt3)} {(\sqrt3+\sqrt5)(\sqrt5-\sqrt3)} \]

Since the denominator equals \(5-3=2\),

\[ AB=(1+\sqrt{15})(\sqrt5-\sqrt3) \]

Expanding,

\[ AB=\sqrt5-\sqrt3+\sqrt{75}-\sqrt{45} \]

\[ =\sqrt5-\sqrt3+5\sqrt3-3\sqrt5 \]

\[ =4\sqrt3-2\sqrt5 \]

Answer:

\[ \boxed{AB=(4\sqrt3-2\sqrt5)\text{ cm}} \]

Q3. A rectangle has sides of length \((2+\sqrt{18})\) m and \(\left(5-\dfrac4{\sqrt2}\right)\) m. Express its area in the form \(a+b\sqrt2\).

Solution:

Simplify the sides:

\[ 2+\sqrt{18}=2+3\sqrt2 \]

\[ 5-\frac4{\sqrt2}=5-2\sqrt2 \]

Area of a rectangle:

\[ A=\text{length}\times\text{width} \]

\[ A=(2+3\sqrt2)(5-2\sqrt2) \]

Expanding,

\[ A=10-4\sqrt2+15\sqrt2-12 \]

\[ A=-2+11\sqrt2 \]

Answer:

\[ \boxed{A=(-2+11\sqrt2)\text{ m}^2} \]

Q4. Find two numbers whose sum is \(68\) and difference is \(22\).

Solution:

Let the numbers be \(x\) and \(y\).

\[ x+y=68 \]

\[ x-y=22 \]

Adding the equations,

\[ 2x=90 \]

\[ x=45 \]

Substituting into \(x+y=68\),

\[ 45+y=68 \]

\[ y=23 \]

Answer:

\[ \boxed{45\text{ and }23} \]

Q5. Convert \(48^\circ\text{C}\) to Fahrenheit using \(F=\dfrac95C+32\).

Solution:

\[ F=\frac95(48)+32 \]

\[ F=86.4+32 \]

Answer:

\[ \boxed{118.4^\circ\text{F}} \]

Q6. The sum of the ages of a father and son is \(72\) years. Six years ago, the father’s age was twice the son’s age. What was the son’s age six years ago?

Solution:

Let the present ages of the father and son be \(F\) and \(S\).

\[ F+S=72 \]

Therefore,

\[ F=72-S \]

Six years ago,

\[ F-6=2(S-6) \]

Substitute \(F=72-S\):

\[ 72-S-6=2S-12 \]

\[ 66-S=2S-12 \]

\[ 78=3S \]

\[ S=26 \]

The son’s age six years ago was

\[ 26-6=20 \]

Answer:

\[ \boxed{20\text{ years}} \]

Q7. Mirha sells a toy for Rs. 1,520. What will the selling price be to get a \(15\%\) profit?

Assumption given in the source: Cost price = Rs. 1,520.

Solution:

Profit:

\[ \text{Profit}=\frac{15}{100}\times1520 \]

\[ = \text{Rs. }228 \]

Selling price:

\[ \text{SP}=\text{CP}+\text{Profit} \]

\[ =1520+228 \]

Answer:

\[ \boxed{\text{Rs. }1748} \]

Q8. Tayyab’s annual income is Rs. 9,60,000, with an exempted amount of Rs. 1,30,000. Find the tax at \(0.75\%\).

Solution:

Taxable income:

\[ 960000-130000=830000 \]

Tax:

\[ \text{Tax}=\frac{0.75}{100}\times830000 \]

Answer:

\[ \boxed{\text{Rs. }6225} \]

Q9. Find the compound markup on Rs. 3,75,000 for one year at \(14\%\), compounded annually.

Solution:

For one year, the compound markup is equal to the simple markup.

\[ \text{Markup}=\frac{P\times R\times T}{100} \]

\[ =\frac{375000\times14\times1}{100} \]

Answer:

\[ \boxed{\text{Rs. }52500} \]

Review Exercise 1

Q1. Multiple Choice Questions

(i) \(\sqrt7\) is:

Correct Answer: (c) Irrational number

(ii) \(\pi\) and \(e\) are:

Correct Answer: (d) Irrational numbers

(iii) If \(n\) is not a perfect square, then \(\sqrt n\) is:

Correct Answer: (d) Irrational number

(iv) \(\sqrt3+\sqrt5\) is:

Correct Answer: (d) Irrational number

(v) For all \(x\in\mathbb R\), \(x=x\) is called:

Correct Answer: (a) Reflexive property

(vi) Let \(a,b,c\in\mathbb R\). Then \(a>b\) and \(b>c\Rightarrow a>c\) is called:

Correct Answer: (b) Transitive property

(vii) \(2^x\times8^x=64\), then \(x=\)

Correct Answer: (a) \(\dfrac32\)

(viii) Let \(a,b\in\mathbb R\). Then \(a=b\) and \(b=a\) is called:

Correct Answer: (b) Symmetric property

(ix) \(\sqrt{75}+\sqrt{27}=\)

Simplifying,

\[ \sqrt{75}+\sqrt{27} =5\sqrt3+3\sqrt3 =8\sqrt3 \]

Correct Answer: (d) \(8\sqrt3\)

(x) The product \((3+\sqrt5)(3-\sqrt5)\) is:

\[ (3+\sqrt5)(3-\sqrt5)=9-5=4 \]

Correct Answer: (d) Rational number

Q2. If \(a=\dfrac32\), \(b=\dfrac53\), and \(c=\dfrac75\), verify the following.

(i) \(a(b+c)=ab+ac\)

LHS:

\[ a(b+c)=\frac32\left(\frac53+\frac75\right) \]

\[ =\frac32\left(\frac{25+21}{15}\right) \]

\[ =\frac32\times\frac{46}{15} =\frac{23}{5} \]

RHS:

\[ ab+ac =\frac32\times\frac53+\frac32\times\frac75 \]

\[ =\frac52+\frac{21}{10} \]

\[ =\frac{25+21}{10} =\frac{23}{5} \]

Since LHS = RHS,

Answer: Verified.

(ii) \((a+b)c=ac+bc\)

LHS:

\[ (a+b)c =\left(\frac32+\frac53\right)\frac75 \]

\[ =\frac{19}{6}\times\frac75 =\frac{133}{30} \]

RHS:

\[ ac+bc =\frac32\times\frac75+\frac53\times\frac75 \]

\[ =\frac{21}{10}+\frac73 \]

\[ =\frac{63+70}{30} =\frac{133}{30} \]

Since LHS = RHS,

Answer: Verified.

Q3. If \(a=\dfrac43\), \(b=\dfrac52\), and \(c=\dfrac74\), verify the associative properties of addition and multiplication.

(i) Associative property of addition

\[ a+(b+c)=(a+b)+c \]

LHS:

\[ \frac43+\left(\frac52+\frac74\right) \]

\[ =\frac43+\frac{17}{4} =\frac{16+51}{12} =\frac{67}{12} \]

RHS:

\[ \left(\frac43+\frac52\right)+\frac74 \]

\[ =\frac{23}{6}+\frac74 =\frac{46+21}{12} =\frac{67}{12} \]

Since LHS = RHS,

Answer: The associative property of addition is verified.

(ii) Associative property of multiplication

\[ a(bc)=(ab)c \]

LHS:

\[ \frac43\left(\frac52\times\frac74\right) =\frac43\times\frac{35}{8} =\frac{35}{6} \]

RHS:

\[ \left(\frac43\times\frac52\right)\frac74 \]

\[ =\frac{10}{3}\times\frac74 =\frac{35}{6} \]

Since LHS = RHS,

Answer: The associative property of multiplication is verified.

Q4. Is \(0\) a rational number? Explain.

Answer: Yes, \(0\) is a rational number because it can be expressed as \(\dfrac pq\), where \(p\) and \(q\) are integers and \(q\neq0\).

For example,

\[ 0=\frac01 \]

Q5. State the trichotomy property of real numbers.

Answer: For any two real numbers \(a\) and \(b\), exactly one of the following relations holds:

\[ \boxed{a<b,\quad a=b,\quad\text{or}\quad a>b} \]

Q6. Find two rational numbers between \(4\) and \(5\).

Solution:

First average:

\[ \frac{4+5}{2}=\frac92 \]

Second average:

\[ \frac{4+\frac92}{2} =\frac{\frac{17}{2}}2 =\frac{17}{4} \]

Answer:

\[ \boxed{\frac{17}{4},\ \frac92} \]

Q7. Simplify the following.

(i) \(\sqrt{\dfrac{x^{15}y^{35}}{z^{20}}}\)

Using fractional exponents,

\[ \sqrt{\frac{x^{15}y^{35}}{z^{20}}} =\frac{x^{15/2}y^{35/2}}{z^{10}} \]

Answer:

\[ \boxed{\frac{x^{15/2}y^{35/2}}{z^{10}}} \]

(ii) \(\sqrt[3]{(27)^{2x}}\)

Solution:

\[ \sqrt[3]{(27)^{2x}} =\left((3^3)^{2x}\right)^{1/3} \]

\[ =3^{6x/3} \]

Answer:

\[ \boxed{3^{2x}} \]

(iii) \(\dfrac{6(3)^{n+2}}{3^{n+1}-3^n}\)

Solution:

Factor out \(3^n\):

\[ =\frac{6\cdot3^n\cdot3^2}{3^n(3-1)} \]

\[ =\frac{54\cdot3^n}{2\cdot3^n} \]

Answer:

\[ \boxed{27} \]

Q8. The sum of three consecutive odd integers is \(51\). Find the integers.

Solution:

Let the integers be

\[ x,\quad x+2,\quad x+4 \]

According to the question,

\[ x+(x+2)+(x+4)=51 \]

\[ 3x+6=51 \]

\[ 3x=45 \]

\[ x=15 \]

Answer:

\[ \boxed{15,\ 17,\ 19} \]

Q9. Abdullah picked up \(96\) balls into two buckets. One bucket has \(28\) more balls than the other. Find the number of balls in each bucket.

Solution:

Let the smaller number of balls be \(x\). The larger number is \(x+28\).

\[ x+(x+28)=96 \]

\[ 2x+28=96 \]

\[ 2x=68 \]

\[ x=34 \]

The larger number is

\[ 34+28=62 \]

Answer:

\[ \boxed{34\text{ balls and }62\text{ balls}} \]

Q10. Salma invested Rs. 3,50,000 at \(7\dfrac14\%\) per annum simple profit for two years, then at \(8\%\) per annum for the remaining years. Find the amount at the end of seven years.

Solution:

Principal:

\[ P=\text{Rs. }350000 \]

For the first two years, the rate is

\[ R_1=7.25\% \]

Step 1: Calculate the profit for the first two years.

\[ I_1=\frac{P\times R_1\times T_1}{100} \]

\[ =\frac{350000\times7.25\times2}{100} \]

\[ I_1=\text{Rs. }50750 \]

Step 2: Calculate the profit for the next five years.

The rate is \(R_2=8\%\).

\[ I_2=\frac{P\times R_2\times T_2}{100} \]

\[ =\frac{350000\times8\times5}{100} \]

\[ I_2=\text{Rs. }140000 \]

Step 3: Calculate the total profit.

\[ I=I_1+I_2 \]

\[ =50750+140000 \]

\[ I=\text{Rs. }190750 \]

Step 4: Calculate the total amount.

\[ A=P+I \]

\[ =350000+190750 \]

Answer:

\[ \boxed{\text{Rs. }540750} \]

Note: The source text’s second interest rate is corrupted in one place; the calculation above uses the \(8\%\) rate shown in its working.

Notes by Sir Dilawar

For more notes, visit QuranEducationPoint.com