Review Exercise 1

Question 1: Choose the Correct Option

(i) $\sqrt{7}$ is:

• (a) Integer
• (b) Rational number
• © Irrational number ✅
• (d) Natural number

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

• (a) Natural numbers
• (b) Integers
• © Rational numbers
• (d) Irrational numbers ✅

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(iii)

If $n$ is not a perfect square, then $\sqrt{n}$ is:

• (a) Rational number
• (b) Natural number
• © Integer
• (d) Irrational number ✅

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(iv)

$\sqrt3+\sqrt5$ is:

• (a) Whole number
• (b) Integer
• © Rational number
• (d) Irrational number ✅

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(v)

For all $x\in\mathbb{R}$,

$$x=x$$

is called:

• (a) Reflexive property ✅
• (b) Transitive property
• © Symmetric property
• (d) Trichotomy property

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(vi)

Let

$$a,b,c\in\mathbb{R}$$

If

$$a>b \quad \text{and} \quad b>c \Rightarrow a>c$$

then this is called:

• (a) Trichotomy property
• (b) Transitive property ✅
• © Additive property
• (d) Multiplicative property

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(vii)

If

$$2^x\times8^x=64$$

then

$$x=$$

• (a) $\dfrac{3}{2}$ ✅
• (b) $\dfrac{3}{4}$
• © $\dfrac{5}{6}$
• (d) $\dfrac{2}{3}$

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(viii)

Let

$$a,b\in\mathbb{R}$$

If

$$a=b \quad \text{and} \quad b=a$$

then this is called:

• (a) Reflexive property
• (b) Symmetric property ✅
• © Transitive property
• (d) Additive property

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(ix)

$$\sqrt{75}+\sqrt{27}=$$

• (a) $\sqrt{102}$
• (b) $8\sqrt3$ ✅
• © $5\sqrt3$
• (d) $9\sqrt3$

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(x)

The product of

$$(3+\sqrt5)(3-\sqrt5)$$

is:

• (a) Prime number
• (b) Odd number
• © Rational number ✅
• (d) Irrational number

Question 2: Verify the Distributive Property

Given

$$a=\frac{3}{2},\qquad b=\frac{5}{3},\qquad c=\frac{7}{5}$$

Verify:

$$a(b+c)=ab+ac$$

$$(a+b)c=ac+bc$$

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(i) Verify

$$a(b+c)=ab+ac$$

Left-Hand Side (LHS)

First, find

$$b+c$$

$$=\frac53+\frac75$$

Taking the LCM (15),

$$=\frac{25+21}{15} =\frac{46}{15}$$

Now,

$$a(b+c) =\frac32\times\frac{46}{15} =\frac{138}{30} =\frac{23}{5}$$

Right-Hand Side (RHS)

Find

$$ab$$

$$=\frac32\times\frac53 =\frac52$$

Find

$$ac$$

$$=\frac32\times\frac75 =\frac{21}{10}$$

Now add them.

$$\frac52+\frac{21}{10} =\frac{25+21}{10} =\frac{46}{10} =\frac{23}{5}$$

Since

$$\text{LHS}=\text{RHS}=\frac{23}{5},$$

the property is verified.

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Final Answer

$$\boxed{a(b+c)=ab+ac}$$

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(ii) Verify

$$(a+b)c=ac+bc$$

Left-Hand Side (LHS)

First,

$$a+b =\frac32+\frac53$$

LCM (6),

$$=\frac{9+10}{6} =\frac{19}{6}$$

Multiply by (c).

$$\frac{19}{6}\times\frac75 =\frac{133}{30}$$

Right-Hand Side (RHS)

We already know

$$ac=\frac{21}{10}$$

Now,

$$bc =\frac53\times\frac75 =\frac73$$

Add them.

$$\frac{21}{10}+\frac73$$

LCM (30),

$$=\frac{63+70}{30} =\frac{133}{30}$$

Hence,

$$\text{LHS}=\text{RHS} =\frac{133}{30}$$

The property is verified.

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Final Answer

$$\boxed{(a+b)c=ac+bc}$$

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Question 3: Verify the Associative Property

Given

$$a=\frac43,\qquad b=\frac52,\qquad c=\frac74$$

Verify the associative property for:

• Addition
• Multiplication

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(i) Addition

Verify

$$(a+b)+c=a+(b+c)$$

Left-Hand Side

$$a+b =\frac43+\frac52 =\frac{8+15}{6} =\frac{23}{6}$$

Now,

$$\frac{23}{6}+\frac74$$

LCM (12),

$$=\frac{46+21}{12} =\frac{67}{12}$$

Right-Hand Side

First,

$$b+c =\frac52+\frac74 =\frac{10+7}{4} =\frac{17}{4}$$

Now,

$$\frac43+\frac{17}{4}$$

LCM (12),

$$=\frac{16+51}{12} =\frac{67}{12}$$

Since

$$\text{LHS}=\text{RHS},$$

the associative property of addition is verified.

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(ii) Multiplication

Verify

$$(ab)c=a(bc)$$

Left-Hand Side

$$ab =\frac43\times\frac52 =\frac{10}{3}$$

Now,

$$\frac{10}{3}\times\frac74 =\frac{70}{12} =\frac{35}{6}$$

Right-Hand Side

First,

$$bc =\frac52\times\frac74 =\frac{35}{8}$$

Now,

$$\frac43\times\frac{35}{8} =\frac{140}{24} =\frac{35}{6}$$

Since

$$\text{LHS}=\text{RHS},$$

the associative property of multiplication is verified.

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Final Answer

$$\boxed{(a+b)+c=a+(b+c)}$$

and

$$\boxed{(ab)c=a(bc)}$$

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Question 4: Is Zero a Rational Number?

Solution

A rational number is any number that can be written in the form

$$\frac{p}{q},$$

where

• (p) and (q) are integers, and
• (q\neq0).

Zero can be written as

$$0=\frac01$$

where

• (0) is an integer,
• (1) is a non-zero integer.

Since zero satisfies the definition of a rational number, it is a rational number.

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Final Answer

$$\boxed{\text{Yes, }0\text{ is a rational number because }0=\frac01.}$$

Question 5: State the Trichotomy Property of Real Numbers

Statement

For any two real numbers

$$a \text{ and } b,$$

exactly one of the following statements is true:

$$a<b,$$

or

$$a=b,$$

or

$$a>b.$$

Only one of these relations can hold at a time.

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Final Answer

$$\boxed{\text{For any two real numbers, exactly one of }a<b,\;a=b,\;\text{or }a>b\text{ is true.}}$$

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Question 6: Find Two Rational Numbers Between 4 and 5

Solution

There are infinitely many rational numbers between 4 and 5.

Two examples are

$$4.2$$

and

$$4.8$$

Since

$$4<4.2<5$$

and

$$4<4.8<5,$$

both are rational numbers lying between 4 and 5.

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Final Answer

$$\boxed{4.2\text{ and }4.8}$$

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Question 7: Simplify the Following

(i)

Given

$$\sqrt[5]{\frac{x^{15}y^{35}}{z^{20}}}$$

Formula

$$\sqrt[n]{a^m}=a^{\frac{m}{n}}$$

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Solution

Apply the fifth root to each factor.

$$=\frac{x^{15/5}y^{35/5}}{z^{20/5}}$$

$$=\frac{x^3y^7}{z^4}$$

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Final Answer

$$\boxed{\frac{x^3y^7}{z^4}}$$

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(ii)

Given

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

Solution

Since

$$27=3^3,$$

we have

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

$$=\sqrt[3]{3^{6x}}$$

$$=3^{6x/3}$$

$$=3^{2x}$$

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Final Answer

$$\boxed{3^{2x}}$$

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(iii)

Given

$$\frac{6(3)^{n+2}}{3^{n+1}-3^n}$$

Solution

Factor the denominator.

$$3^{n+1}-3^n = 3^n(3-1)$$

$$=2\cdot3^n$$

Substitute into the expression.

$$=\frac{6\cdot3^{n+2}}{2\cdot3^n}$$

Simplify.

$$=\frac62\cdot3^{(n+2)-n}$$

$$=3\cdot3^2$$

$$=3\cdot9$$

$$=27$$

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Final Answer

$$\boxed{27}$$

Question 8: Find Three Consecutive Odd Integers

Given

The sum of three consecutive odd integers is

$$51.$$

Find the three integers.

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Solution

Let the three consecutive odd integers be

$$x,\;x+2,\;x+4.$$

According to the given information,

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

Combine like terms.

$$3x+6=51$$

Subtract 6 from both sides.

$$3x=45$$

Divide both sides by 3.

$$x=15$$

Therefore, the three consecutive odd integers are

$$15,\;17,\;19.$$

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Verification

$$15+17+19=51$$

Hence, the answer is correct.

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Final Answer

$$\boxed{15,\;17,\;19}$$

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Question 9: Balls in Two Buckets

Given

Abdullah picked up

$$96$$

balls and placed them into two buckets.

One bucket has

$$28$$

more balls than the other.

Find the number of balls in each bucket.

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Solution

Let the smaller bucket contain

$$x$$

balls.

Then the larger bucket contains

$$x+28$$

balls.

According to the question,

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

Combine like terms.

$$2x+28=96$$

Subtract 28 from both sides.

$$2x=68$$

Divide by 2.

$$x=34$$

Therefore,

Larger bucket:

$$34+28=62$$

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Verification

$$34+62=96$$

and

$$62-34=28$$

Both conditions are satisfied.

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Final Answer

$$\boxed{34\text{ balls and }62\text{ balls}}$$

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Question 10: Simple Profit

Given

Principal Amount

$$Rs.\,350,000$$

Rate for first 2 years

$$7\frac14\%$$

Rate for next 5 years

$$8\%$$

Find the total amount after

$$7$$

years.

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Formula

Simple Profit:

$$SI=\frac{PRT}{100}$$

Amount:

$$A=P+SI$$

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Solution

Step 1: Profit for the First 2 Years

Convert the mixed percentage.

$$7\frac14\%=7.25\%$$

Simple profit:

$$SI_1 = \frac{350000\times7.25\times2}{100}$$

$$=Rs.\,50,750$$

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Step 2: Profit for the Next 5 Years

$$SI_2 = \frac{350000\times8\times5}{100}$$

$$=Rs.\,140,000$$

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Step 3: Total Profit

$$50,750+140,000 = Rs.\,190,750$$

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Step 4: Total Amount

$$350,000+190,750 = Rs.\,540,750$$

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Final Answer

$$\boxed{Rs.\,540,750}$$

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Summary

Question	Answer
8	$$\boxed{15,\;17,\;19}$$
9	$$\boxed{34\text{ balls and }62\text{ balls}}$$
10	$$\boxed{Rs.\,540,750}$$