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#2.2

Are the following statements ‘True’ or ‘False’? Justify your answers.

(i)If the zeroes of a quadratic polynomial $ax^2 + bx + c$ are both positive, then a, b and c all have the same sign

(ii) If the graph of a polynomial intersects the x-axis at only one point, it cannot be a quadratic polynomial.

(iii) If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.

(iv) If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.

(v) If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign.

(vi) If all three zeroes of a cubic polynomial $x^3 + ax^2 - bx + c$ are positive, then at least one of a, b and c is non-negative.

(vii) The only value of k for which the quadratic polynomial $kx^2 + x + k$ has equal zeros is $\frac{1}{2}$ .

(i) Answer. [false]

Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).

Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.

If a quadratic polynomial is $p(x)=ax^2 + bx + c$ then

Sum of zeroes $=-\frac{b}{a}$

Product of zeroes $=\frac{c}{a}$

Here two possibilities can occurs:

b > 0 and a < 0, c < 0

OR b < 0 and a > 0, c > 0

Here we conclude that a, b and c all have nor same sign is given statement is false.

(ii) Answer. [False]

Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).

Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.

We know that the roots of a quadratic polynomial is almost 2, Hence the graph of a quadratic polynomial intersects the x-axis at 2 point, 1 point or 0 point.

For example :

$x^2 + 4x + 4 = 0$ (a quadratic polynomial)

$(x + 2)^2 = 0$

$x + 2 = 0$

$x =-2$

Here only one value of x exist which is –2.

Hence the graph of the quadratic polynomial $x^2 + 4x + 4 = 0$ intersect the x-axis at x = –2.

Hence, we can say that if the graph of a polynomial intersect the x-axis at only one point can be a quadratic polynomial.

Hence the given statement is false.

(iii)

Answer. [True]

Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).

Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.

Let us take an example :-

$(x -1)^2 (x -2) = 0$

It is a cubic polynomial

If we find its roots then x = 1, 2

Hence, there are only 2 roots of cubic polynomial $(x -1)^2 (x -2) = 0$ exist.

In other words we can say that the graph of this cubic polynomial intersect x-axis at two points x = 1, 2.

Hence we can say that if the graph of a polynomial intersect the x-axis at exactly two points, it need not to be a quadratic polynomial it may be a polynomial of higher degree.

Hence the given statement is true.

(iv)

Answer. [True]

Polynomial: It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).

Cubic polynomial: when the degree of polynomial is three then the polynomial is called cubic polynomial.

Let $\alpha _1 , \alpha _2 ,\alpha _3$ be the zeroes of a cubic polynomial.

It is given that two of the given zeroes have value zero.

i.e. $\alpha _1 = \alpha _2 = 0$

Let $p(y) = (y - \alpha _1)(y - \alpha _2)(y -\alpha _3)$

$p(y) = (y - 0) (y - 0) (y - \alpha _3)$

$p(y) = y^3 -y^2\alpha _3$

Here, we conclude that if two zeroes of a cubic polynomial are zero than the polynomial does not have linear and constant terms.

Hence, given statement is true.

(v)

Answer. [True]

Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).

Cubic polynomial : when the degree of polynomial is three then the polynomial is called cubic polynomial.

Let the standard equation of cubic polynomial is:

$p(x) = ax^3 + bx^2 + cx + b$

Let $\alpha ,\beta$ and $\gamma$ be the roots of p(x)

It is given that all the zeroes of a cubic polynomial are negative

i.e $\alpha = -\alpha , \beta = -\beta and \gamma = -\gamma$

Sum of zeroes $=\frac{coefficent of x^{2}}{coefficent of x^{3}}$

$\alpha +\beta +\gamma =-\frac{b}{a}$

It is given that zeroes are negative then $\alpha = -\alpha , \beta = -\beta and \gamma = -\gamma$

$\Rightarrow -\alpha +(-\beta) +(-\gamma) =-\frac{b}{a}$

$-\alpha -\beta -\gamma =-\frac{b}{a}$

$-\left (\alpha +\beta +\gamma \right ) =-\frac{b}{a}$

$\left (\alpha +\beta +\gamma \right ) =\frac{b}{a}$ …….(1)

That is $\frac{b}{a}>0$

Sum of the products of two zeroes at a time $=\frac{coefficent of x}{coefficent of x^{3}}$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$

Replace $\alpha = -\alpha , \beta = -\beta and \gamma = -\gamma$

$(-\alpha ) (-\beta) +(-\beta) (-\gamma) +(-\gamma) (-\alpha )=\frac{c}{a}$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$ …..(2)

That is $\frac{c}{a}>0$

Product of all zeroes $=\frac{-constant terms}{coefficent of x^{3}}$

$\alpha \beta \gamma =-\frac{d}{a}$

Replace $\alpha = -\alpha , \beta = -\beta and \gamma = -\gamma$

$(-\alpha) (-\beta) (-\gamma) =-\frac{d}{a}$

$-(\alpha \beta \gamma)=-\frac{d}{a}$

$\alpha \beta \gamma =\frac{d}{a}$ ……(3)

That is $\frac{d}{a}>0$

From equation (1), (2) and (3) we conclude that all the coefficient and the constant term of the polynomial have the same sign.

Hence, given statement is true.

(vi)

Answer. [False]

Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).

Cubic polynomial : when the degree of polynomial is three then the polynomial is called cubic polynomial.

The given cubic polynomial is

$p(x)=x^3 + ax^2 - bx + c$

Let a, b and g are the roots of the given polynomial

Sum of zeroes $=\frac{(coefficent of x^{2} )}{(coefficent of x^{3})}$

$\alpha +\beta +\gamma =-\frac{a}{1}=-a$

We know that when all zeroes of a given polynomial are positive then their sum is also positive

But here a is negative

Sum of the product of two zeroes at a time $=\frac{(coefficent of x)}{(coefficent of x^{3})}$

$\alpha \beta +\beta \gamma +\gamma \alpha =-\frac{b}{1}=-b$

Also here b is negative

Product of all zeroes $=\frac{-(constant terms)}{(coefficent of x^{3})}$

$\alpha \beta \gamma =\frac{-c}{1}=-c$

Also c is negative

Hence if all three zeroes of a cubic polynomial $x^3 + ax^2 - bx + c$ are positive then a, b and c must be negative.

Hence given statement is false.

(vii)

Answer. [False]

Polynomial : It is an expression of more than two algebraic terms, especially then sum of several terms that contains different powers of the same variable(s).

Quadratic polynomial : when the degree of polynomial is two then the polynomial is called cubic polynomial.

Let $p(x)=kx^2 + x + k$

Here it is gives that zeroes of p(x) are equal and we know that when any polynomial having equal zeroes then their discriminate will be equal to zero.

i.e. d = 0

$b^2 - 4ac = 0$ (Q d = b2 – 4ac)

$p(x)=kx^2 + x + k$

Here, a = k, b = 1, c = k

$b^2 - 4ac = 0$

$(1)2 -4 \times k \times k = 0$

$1 - 4k^2 = 0$

$1 = 4k^2$

$\frac{1 }{4}= k^2$

$k=\pm \frac{1}{2}$

$\therefore$ When $+\frac{1}{2}$ and $-\frac{1}{2}$ then the given quadratic polynomial has equal zeroes.

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#2.2

(i)Answer the following and justify:

Can $x^2 - 1$ be the quotient on division of $x^6 + 2x^3 + x - 1$ by a polynomial in x of degree 5?

(ii)Answer the following and justify:

What will the quotient and remainder be on division of $ax^2 + bx + c \ by \ px^3 + qx^2 + rx+ s, p\neq 0?$ $ax^2 + bx + c \ by \ px^3 + qx^2 + rx+ s, p\neq 0?$

(iii)Answer the following and justify:

If on division of a polynomial p (x) by a polynomial g (x), the quotients zero, what is the relation between the degrees of p (x) and g (x)?

(iv) Answer the following and justify:

If on division of a non-zero polynomial p (x) by a polynomial g (x), the remainder is zero, what is the relation between the degrees of p (x) and g (x)?

(v) Answer the following and justify:

Can the quadratic polynomial $x^2 + kx + k$ have equal zeroes for some odd integer k > 1?

(i) Answer. [false]

Solution.

Let divisor of a polynomial in x of degree 5 is = $x^5 + x + 1$

Quotient = $x^2 - 1$

Dividend = $x^6 + 2x^3 + x - 1$

According to division algorithm if one polynomial p(x) is divided by the other polynomial $g(x) \neq 0$ , then the relation among p(x), g(x), quotient q(x) and remainder r(x) is given by

$p(x) = g(x) \times q(x) + r(x)$

i,e. Dividend = Divisor × Quotient + Remainder

$= (x^5 + x + 1)(x^2 - 1) +$ Remainder

$= x^7 - x^5 + x^3 + x + x^2 - 1 +$ Remainder

Here it is of degree seven but given dividend is of degree six.

Therefore $x^2 - 1$ can not be the quotient of $x^6 + 2x^3 + x - 1$ because division algorithm is not satisfied.

Hence, given statement is false.

(ii) Here dividend is $ax^2 + bx + c$

and divisor is $px^3 + qx^2 + rx + s$

According to division algorithm if one polynomial p(x) is divided by the other polynomial g(x) ¹ 0 then the relation among p(x), g(x) quotient q(x) and remainder r(x) is given by

$p(x) = g(x) \times q(x) + r(x)$

where degree of r(x) < degree of g(x).

i.e. Dividend = Devisor × Quotient + Remainder

Here degree of divisor is greater than degree of dividend therefore.

According to division algorithm theorem $ax^2 + bx + c$ is the remainder and quotient will be zero.

That is remainder = $ax^2 + bx + c$

Quotient = 0

(iii) Division algorithm theorem :- According to division algorithm if one polynomial p(x) is divided by the other polynomial g(x) is then the relation among p(x), g(x), quotient q(x) and remainder r(x) is given by

p(x) = g(x) × q(x) + r(x)

where degree of r(x) < degree of g(x)

i.e. Dividend = Division × Quotient + Remainder

In the given statement it is given that on division of a polynomial p(x) by a polynomial g(x), the quotient is zero.

The given condition is possible only when degree of divisor is greater than degree of dividend

i.e. degree of g(x) > degree of p(x).

(iv)

Division algorithm theorem :- According to division algorithm if one polynomial p(x) is divided by the other polynomial g(x) is then the relation among p(x), g(x), quotient q(x) and remainder r(x) is given by

p(x) = g(x) × q(x) + r(x) ….(1)

where degree of r(x) < degree of g(x)

According to given statement on division of a non-zero polynomial p(x) by a polynomial g(x) then remainder r(x) is zero then equation (1) becomes

p(x) = g(x) × q(x) ….(2)

From equation (2) we can say g(x) is a factor of p(x) and degree of g(x) may be less than or equal to p(x).

(v)

Answer. [false]

Polynomial : It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s)

Quadratic polynomial : when the degree of polynomial is two then the polynomial is called quadratic polynomial.

Let $p(x)=x^2 + kx + k$

It is given that zeroes of p(x) has equal and we know that when any polynomial having equal zeroes than their discriminate is equal to zero

i.e. $d = b^2 - 4ac = 0$

$p(x)=x^2 + kx + k$

here, a = 1, b = k, c = k

$\therefore d = (k)^2 - 4 \times 1 \times k = 0$
$k^2 - 4k = 0$

$k(k -4) = 0$

$k = 0, k = 4$

Hence, the given quadratic polynomial $x^2 + kx + k$ have equal zeroes only when the values of k will be 0 and 4.

Hence, given statement is not correct.

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infoexpert21

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#2.2

Write whether the following statements are True or False. Justify your answer.

i. A binomial can have at most two terms

ii. Every polynomial is a binomial.

iii. A binomial may have degree 5.

iv. Zero of a polynomial is always 0

v.A polynomial cannot have more than one zero.

vi. The degree of the sum of two polynomials each of degree 5 is always 5

i. False

Solution :- Binomial: A binomial is an expression that has two numbers, terms or letters joined by the sign + or –.

Binomial necessarily means consisting of two terms only. These terms should not be like terms.

For example: $x^{2}+1$ is a binomial

x + 2x is not a binomial as these are like terms.

So the given statement is false, as a binomial has exactly two terms (not at most two terms).

ii. False

Solution:- Binomial: A binomial is an expression that has two numbers, terms or letters joined by the sign + or –.

Binomial necessarily means consisting of two terms only. These terms should not be like terms.

For example: $x^{2}+1$ is a binomial

x + 2x is not a binomial as these are like terms.

Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s)

Its degree is always a whole number.

For example: $x^{0},x^{2}+2$ etc.

Because a binomial has exactly two terms but a polynomial can be monomial (single term), binomial (two terms), trinomial (three terms) etc.

$\therefore$ The given statement is False.

iii. True

Solution:- Binomial: A binomial is an expression that has two numbers, terms or letters joined by the sign + or –.

Binomial necessarily means consisting of two terms only. These terms should not be like terms.

For example: $x^{2}+1$ is a binomial

x + 2x is not a binomial as these are like terms.

Degree of polynomial: Degree of polynomial is the highest power of the polynomial’s monomials with non-zero coefficient.

For any binomial of the form $x^{5}+2$ , we can see that the degree is 5.

So, a binomial may have degree 5.

Therefore the given statement is True.

iv. False

Solution :- Polynomial:- It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s)

Its degree is always a whole number.

For example: $x^{0},x^{2}+2$ etc.

We know that for finding the zero of a polynomial, we need to find a value of x for which the polynomial will be zero

i.e., p(x)=0

Let us consider an example:

p(x) = x - 2

Now to find the zero of this polynomial, we have:

x - 2 =0

x = 2 (which is not zero)

Hence the given statement is false because zero of a polynomial can be any real number.

v. False

Solution :- Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s).

Its degree is always a whole number.

For example: $x^{0},x+2,x^{3}+1,x^{4}+x^{3}+x^{2}+1$ etc.

We know that for finding the zero of a polynomial, we need to find a value of x for which the polynomial will be zero

i.e., p(x)=0

So, a polynomial can have any number of zeroes. It depends upon the degree of polynomial.

$\therefore$ The given statement is False.

vi. False

Solution :- Degree of polynomial: Degree of polynomial is the highest power of the polynomial’s monomials with non-zero coefficient.

The degree of the sum of two polynomials may be less than or equal to 5.

For example: $x^{5}+1$ and $-x^{5}+2x^{3}+1$ are two polynomials of degree 5 but the degree of the sum of the two polynomials $2x^{3}+2$ is 3.

Hence the given statement is False.

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#2.2

Which of the following expressions are polynomials? Justify your answer

(i) 8

(ii) $\sqrt{3}x^{2}-2x$

(iii) $1-\sqrt{5x}$

(iv) $\frac{1}{5x^{-2}}+5x+7$

(v) $\frac{(x-2)(x-4)}{x}$

(vi) $\frac{x}{x+1}$

(vii) $\frac{1}{7}a^{3}-\frac{2}{\sqrt{3}}a^{2}+4a-7$

(viii) $\frac{1}{2x}$

(i, ii, iv, vii)

Solution

Polynomial:- It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s)

Its degree is always a whole number.

For example: $x^{0},x+2$ etc.

(i) Here 8 is a polynomial because it can also be written as $8.x^{0}$ i.e., multiply by $x^{0}$ .

(ii) $\sqrt{3}x^{2}-2x$ is also a polynomial having degree two.

(iii) $1-\sqrt{5x}$ is not a polynomial because its exponent is in fraction.

(iv) $\frac{1}{5x^{-2}}+5x+7$ can be written as $5x^{2}+5x+7$ and it is a polynomial having degree two.

(v) $\frac{(x-2)(x-4)}{x}$ is not polynomial because it has negative exponent.

(vi) $\frac{1}{x+1}$ is not a polynomial because it have negative exponent.

(vii) $\frac{1}{7}a^{3}-\frac{2}{\sqrt{3}}a^{2}+4a-7$ is a polynomial of degree three.

(viii) $\frac{1}{2x}$ is not a polynomial because it have negative exponent.

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#2.2

8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed $\frac{2}{5}$ of the water. Pratap consumed the remaining water.

(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?

Given

Total water = 5 litre.

i) The amount of water vidya consumed :

$=\frac{2}{5}\:\:o\!f 5\:liter=\frac{2}{5}\times5=2\:liter$

Hence vidya consumed 2 liters of water from the bottle.

ii) The amount of water Pratap consumed :

$=\left (1-\frac{2}{5} \right )\:\:o\!f 5\:liter=\frac{3}{5}\times5=3\:liter$

Hence, Pratap consumed 3 liters of water from the bottle.

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#2.2

7. Find:

$(a)\frac{1}{2} \; \; o\! f\; \; (i)\; \; 2\frac{3}{4} \; \; \;(ii)\; 4\frac{2}{9}$ $(b)\frac{5}{8} \; \; o\! f\; \; (i)\; \; 3\frac{5}{6} \; \; \;(ii)\; 9\frac{2}{3}$

$(a) (i)\frac{1}{2} \; \; o\! f\; \;\; \; 2\frac{3}{4} \;$

As we know that of is equivalent to multiply,

$\frac{1}{2} \; \; o\! f\; \;\; \; 2\frac{3}{4} \;=\frac{1}{2}\times2\frac{3}{4}=\frac{1}{2}\times\frac{11}{4}=\frac{11}{8}=1\frac{3}{8}$

$(a)(ii)\frac{1}{2} \; \; o\! f\;\; 4\frac{2}{9}$

As we know that of is equivalent to multiply,

$\frac{1}{2} \; \; o\! f\;\; 4\frac{2}{9}=\frac{1}{2}\times4\frac{2}{9}=\frac{1}{2}\times\frac{38}{9}=\frac{38}{18}=\frac{19}{9}=2\frac{1}{9}$

$(b)(i)\frac{5}{8} \; \; o\! f\; \; 3\frac{5}{6} \; \;$

As we know that of is equivalent to multiplication, so

$\frac{5}{8} \; \; o\! f\; \; 3\frac{5}{6} \; =\frac{5}{8}\times3\frac{5}{6}=\frac{5}{8}\times\frac{23}{6}=\frac{115}{48}=2\frac{19}{48}$

$(b)(ii)\frac{5}{8} \; \; o\! f\; \;\; 9\frac{2}{3}$

As we know that of is equivalent to multiplication, so

$\frac{5}{8} \; \; o\! f\; \;\; 9\frac{2}{3}=\frac{5}{8}\times \frac{29}{3}=\frac{145}{24}=6\frac{1}{24}$

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#2.2

6. Multiply and express as a mixed fraction :

$(a) 3\times 5\frac{1}{5}$ $(b) 5\times 6\frac{3}{4}$ $(c) 7\times 2\frac{1}{4}$

$(d) 4\times 6\frac{1}{3}$ $(e) 3\frac{1}{4}\times 6$ $(f) 3\frac{2}{5}\times 8$

$(a) 3\times 5\frac{1}{5}$

On Multiplying, we get

$\Rightarrow 3\times\frac{5\times5+1}{5}=3\times\frac{26}{5}=\frac{3\times26}{5}=\frac{78}{5}$

Converting This into Mixed Fraction,

$\Rightarrow \frac{78}{5}=\frac{75+3}{5}=\frac{75}{5}+\frac{3}{5}=15+\frac{3}{5}=15\frac{3}{5}$

$(b) 5\times 6\frac{3}{4}$

On Multiplying, we get

$\Rightarrow 5\times\frac{6\times4+3}{4}=5\times\frac{27}{4}=\frac{5\times27}{4}=\frac{135}{4}$

Converting This into Mixed Fraction,

$\Rightarrow \frac{135}{4}=\frac{132+3}{4}=\frac{132}{4}+\frac{3}{4}=33+\frac{3}{4}=33\frac{3}{4}$

$(c) 7\times 2\frac{1}{4}$

On multiplying, we get

$7\times 2\frac{1}{4}=7\times \frac{4\times2+1}{4}=7\times\frac{9}{4}=\frac{7\times9}{4}=\frac{63}{4}$

Converting it into a mixed fraction, we get

$\frac{63}{4}=\frac{60+3}{4}=\frac{60}{4}+\frac{3}{4}=15+\frac{3}{4}=15\frac{3}{4}$

$(d) 4\times 6\frac{1}{3}$

On multiplying, we get

$4\times 6\frac{1}{3}=4\times\frac{3\times6+1}{3}=4\times\frac{19}{3}=\frac{4\times 19}{3}=\frac{76}{3}$

Converting it into a mixed fraction,

$\frac{76}{3}=\frac{75+1}{3}=\frac{75}{3}+\frac{1}{3}=25+\frac{1}{3}=25\frac{1}{3}$

$(e) 3\frac{1}{4}\times 6$

Multiplying them, we get

$3\frac{1}{4}\times 6=\frac{4\times3+1}{4}\times6=\frac{13}{4}\times6=\frac{13\times6}{4}=\frac{78}{4}$

Now, converting the result fraction we got to mixed fraction,

$\frac{78}{4}=\frac{76+2}{4}=\frac{76}{4}+\frac{2}{4}=19+\frac{1}{2}=19\frac{1}{2}$

$(f) 3\frac{2}{5}\times 8$

On multiplying, we get

$3\frac{2}{5}\times 8=\frac{5\times3+2}{5}\times8=\frac{17}{5}\times8=\frac{17\times8}{5}=\frac{136}{5}$

Converting this into a mixed fraction, we get

$\frac{136}{5}=\frac{135+1}{5}=\frac{135}{5}+\frac{1}{5}=27+\frac{1}{5}=27\frac{1}{5}$

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#2.2

4. Shade:

$(i) \frac{1}{2}$ of the circles in box (a)                 $(ii) \frac{2}{3}$ of the triangles in box (b)

$(iii) \frac{3}{5}$    of the squares in box (c).

1) In figure a there are 12 circles: half of 12 = 6

2) In figure b there are 9 triangles: 2/3 of 9 = 6

3) In figure c there are 15 triangles: 3/5 of 15 = 9

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#2.2

5. Find:

             $(a)\; \frac{1}{2}\;\; of\; (i)24\; \; (ii)46$               $(b)\; \frac{2}{3}\;\; of\; (i)18\; \; (ii)27$

              $(c)\; \frac{3}{4}\;\; of\; (i)16\; \; (ii)36$                 $(d)\; \frac{4}{5}\;\; of\; (i)20\; \; (ii)35$

$(a)(i)\; \frac{1}{2}\;\; of\; 24$

On Multiplying we get,

$\; \frac{1}{2}\;\; of\; 24=\frac{1}{2}\times24=\frac{1\times24}{2}=12$

$(a)\;ii) \frac{1}{2}\;\; of\;46$

On multiplying, we get

$\Rightarrow \frac{1}{2}\;\; of\;46=\frac{1}{2}\times46=\frac{1\times46}{2}=23.$

$(b)(i)\; \frac{2}{3}\;\; of\;18$

On Multiplying, we get

$\Rightarrow \frac{2}{3}\;\; of\;18=\frac{2}{3}\times18=\frac{2\times18}{3}=\frac{36}{3}=12.$

$(b)(ii)\; \frac{2}{3}\;\; of\;27$

On multiplying, we get

$\Rightarrow \frac{2}{3}\:\:of\:\:27=\frac{2}{3}\times27=\frac{2\times27}{3}=\frac{54}{3}=18$

$(c)(i)\; \frac{3}{4}\;\; of\; 16$

On multiplying, we get

$\Rightarrow \frac{3}{4}\:\:of\:\:16=\frac{3}{4}\times16=\frac{3\times16}{4}=\frac{48}{4}=12.$

$(c)(ii)\; \frac{3}{4}\;\; of\; 36$

$\Rightarrow \frac{3}{4}\:\:of\:\:36=\frac{3}{4}\times36=\frac{3\times36}{4}=\frac{108}{4}=27$

$(d)(i)\; \frac{4}{5}\;\; of\; 20\;$

On multiplying, we get

$\frac{4}{5}\;\; of\; 20\;=\frac{4}{5}\times20=\frac{4\times20}{5}=\frac{80}{5}=16$

$(d)(ii)\; \frac{4}{5}\;\; of\; 35\;$

On Multiplying, we get,

$\; \frac{4}{5}\;\; of\; 35\;=\frac{4}{5}\times35=\frac{4\times35}{5}=\frac{140}{4}=28$

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#2.2

3. Multiply and reduce to lowest form and convert into a mixed fraction:

$(i) 7\times \frac{3}{5}$ $(ii) 4\times \frac{1}{3}$ $(iii) 2\times \frac{6}{7}$ $(iv) 5\times \frac{2}{9}$ $(v) \frac{2}{3}\times 4$

$(vi) \frac{5}{2}\times 6$ $(vii) 11\times \frac{4}{7}$ $(viii) 20\times \frac{4}{5}$ $(ix) 13\times \frac{1}{3}$ $(x) 15\times \frac{3}{5}$

$(i) 7\times \frac{3}{5}$

On Multiplying, we get

$\Rightarrow \frac{7\times3}{5}=\frac{21}{5}=\frac{20+1}{5}=\frac{20}{5}+\frac{1}{5}=4+\frac{1}{5}=4\frac{1}{5}$

$(ii) 4\times \frac{1}{3}$

On Multiplying, we get

$\Rightarrow \frac{4\times1}{3}=\frac{4}{3}=\frac{3+1}{3}=\frac{3}{3}+\frac{1}{3}=1+\frac{1}{3}=1\frac{1}{3}$

$(iii) 2\times \frac{6}{7}$

On Multiplying, we get

$\Rightarrow \frac{2\times6}{7}=\frac{12}{7}=\frac{7+5}{7}=\frac{7}{7}+\frac{5}{7}=1+\frac{5}{7}=1\frac{5}{7}$

$(iv) 5\times \frac{2}{9}$

On Multiplying, we get

$\Rightarrow \frac{5\times2}{9}=\frac{10}{9}=\frac{9+1}{9}=\frac{9}{9}+\frac{1}{9}=1+\frac{1}{9}=1\frac{1}{9}$

$(v) \frac{2}{3}\times 4$

On Multiplying, we get

$\Rightarrow \frac{2\times4}{3}=\frac{8}{3}=\frac{6+2}{3}=\frac{6}{3}+\frac{2}{3}=2+\frac{2}{3}=2\frac{2}{3}$

$(vi) \frac{5}{2}\times 6$

On Multiplying, we get

$\Rightarrow \frac{5\times6}{2}=\frac{30}{2}=15$

$(vii) 11\times \frac{4}{7}$

On Multiplying, we get

$\Rightarrow \frac{11\times4}{7}=\frac{44}{7}$

Converting this into a mixed fraction, we get

$\Rightarrow \frac{44}{7}=\frac{42+2}{7}=\frac{42}{7}+\frac{2}{7}=6+\frac{2}{7}=6\frac{2}{7}.$

$(viii) 20\times \frac{4}{5}$

On Multiplying, we get

$\Rightarrow \frac{20\times4}{5}=\frac{80}{5}=16$

$(ix) 13\times \frac{1}{3}$

On multiplying, we get

$\Rightarrow \frac{13\times1}{3}=\frac{13}{3}$

Converting this into a mixed fraction,

$\Rightarrow\frac{13}{3}=\frac{12+1}{3}=\frac{12}{3}+\frac{1}{3}=4+\frac{1}{3}=4\frac{1}{3}$ .

$(x) 15\times \frac{3}{5}$

On multiplying, we get,

$\Rightarrow \frac{15\times3}{5}=\frac{45}{5}=9$ .

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Pankaj Sanodiya

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Which of the following expressions are polynomials? Justify your answer (i) 8 (ii) (iii) (iv) (v) (vi) (vii) (viii)

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8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed of the water. Pratap consumed the remaining water. (i) How much water did Vidya drink? (ii) What fraction of the total quantity of water did Pratap drink?

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Pankaj Sanodiya

7. Find:

Which of the following expressions are polynomials? Justify your answer

(i) 8

(ii) $\sqrt{3}x^{2}-2x$

(iii) $1-\sqrt{5x}$

(iv) $\frac{1}{5x^{-2}}+5x+7$

(v) $\frac{(x-2)(x-4)}{x}$

(vi) $\frac{x}{x+1}$

(vii) $\frac{1}{7}a^{3}-\frac{2}{\sqrt{3}}a^{2}+4a-7$

(viii) $\frac{1}{2x}$

8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed $\frac{2}{5}$ of the water. Pratap consumed the remaining water.

(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?

7. Find:

$(a)\frac{1}{2} \; \; o\! f\; \; (i)\; \; 2\frac{3}{4} \; \; \;(ii)\; 4\frac{2}{9}$ $(b)\frac{5}{8} \; \; o\! f\; \; (i)\; \; 3\frac{5}{6} \; \; \;(ii)\; 9\frac{2}{3}$

6. Multiply and express as a mixed fraction :

$(a) 3\times 5\frac{1}{5}$ $(b) 5\times 6\frac{3}{4}$ $(c) 7\times 2\frac{1}{4}$

$(d) 4\times 6\frac{1}{3}$ $(e) 3\frac{1}{4}\times 6$ $(f) 3\frac{2}{5}\times 8$

4. Shade:

$(i) \frac{1}{2}$ of the circles in box (a) $(ii) \frac{2}{3}$ of the triangles in box (b)

$(iii) \frac{3}{5}$ of the squares in box (c).

5. Find:

$(a)\; \frac{1}{2}\;\; of\; (i)24\; \; (ii)46$ $(b)\; \frac{2}{3}\;\; of\; (i)18\; \; (ii)27$

$(c)\; \frac{3}{4}\;\; of\; (i)16\; \; (ii)36$ $(d)\; \frac{4}{5}\;\; of\; (i)20\; \; (ii)35$