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Correct option is (C) \(\frac{-78}{35}\)

\(\frac{2}{5}+\frac{3}{7}-\frac{6}{5}-\frac{13}{7}\)

\(=\frac{2\times7+3\times5-6\times7-13\times5}{35}\)

\(=\frac{14+15-42-65}{35}\)

\(=\frac{-13-65}{35}\)

\(=\frac{-78}{35}.\)

Correct option is   C) \(\frac{-78}{35}\)

Correct option is (A) \(\frac{-1601}{200}\)

-8.005 \(=-\frac{8005}{1000}=-\frac{8005\div5}{1000\div5}\) = \(-\frac{1601}{200}\).

Correct option is   A) \(\frac{-1601}{200}\)

Correct option is (B) 0

\(\frac{2}{5}\times(\frac{-1}{9})+\frac{23}{180}-\frac{1}{9}\times\frac{3}{4}\)

\(=\frac{-2\times1}{5\times9}+\frac{23}{180}-\frac{1}{3\times4}\)

\(=\frac{-2}{45}+\frac{23}{180}-\frac{1}{12}\)

\(=\frac{-2\times4+23-15}{180}\)

\(=\frac{-8+23-15}{180}\)

\(=\frac{23-23}{180}\)

\(=\frac0{180}\)

= 0.

Correct option is  B) 0

Correct option is  C) 8/5

Correct option is  A) -2/3

Correct option is  C) 2

Correct option is   D) \(1.\overline{6}\)

Correct option is  A) 156

Five rational numbers greater than -2 are:
-3/2, 1, -1/2, 0, 1/2
[Other rational numbers may also be possible]

(-9/24) and (-1/18)

Take the LCM of the denominators of the given rational numbers.

LCM of 24 and 18 is 72

Express each of the given rational numbers with the above LCM as the common denominator.

Now,

(-9/24)= [(-9×3)/ (24×3)] = (-27/72)

(-1/18)= [(-1×4)/ (18×4)] = (-4/72)

Then,

= (-27/72) + (-4/72) … [∵ denominator is same in both the rational numbers]

= (-27+ (-4))/72

= (-27 – 4)/72

= (-31/72)

(i) The denominator of given rational numbers are 4 and 8 respectively

The L.C.M. of 4 and 8 is 8

Now, we rewrite the given rational numbers into forms in which both of them have the same denominator

\(\frac{3\times 2}{4\times 2} = \frac{6}{8}\) and \(\frac{-5}{8}\)

Therefore,

\(\frac{6}{8}\, -\, \frac{5}{8}\, =\, \frac{6-5}{8}\)

\(= \frac{1}{8}\)

(ii) The denominator of given rational numbers are 9 and 3 respectively

The L.C.M of 9 and 3 is 9

Now, we rewrite the given rational numbers into forms in which both of them have the same denominator

\(\frac{-5\times 1}{9\times 1} = \frac{-5}{9} \) and \(\frac{7\times 3}{3\times 3} = \frac{21}{9}\)

Therefore,

\(\frac{-5}{9} + \frac{21}{9} = \frac{-5+21}{9}\)

\(= \frac{16}{9}\)

(iii) The denominator of given rational numbers are 1 and 5 respectively 

The L.C.M of 1 and 5 is 5 

Now, we rewrite the given rational numbers into forms in which both of them have the same denominator

\(\frac{-3\times 5}{1\times 5} = \frac{-3\times 5}{5}\) and \(\frac{3}{5}\)

Therefore,

\(\frac{-15}{5} + \frac{3}{5} = \frac{3-15}{5}\)

\(= \frac{-12}{5}\)

(iv) The denominator of given rational numbers are 27 and 18 respectively

The L.C.M of 27 and 18 is 54

Now, we rewrite the given rational numbers into forms in which both of them have the same denominator

\(\frac{-7}{27} = \frac{-7\times 2}{27\times 2}\)

= \(\frac{-14}{54}\)

And,

\(\frac{11}{18} = \frac{11\times 3}{18\times 3}\)

= \(\frac{33}{54}\)

Therefore,

\((\frac{-7\times2}{27\times 2}) + \frac{33}{54}\)

= \(\frac{33}{54} - \frac{14}{54}\)

= \(\frac{33-14}{54}\)

= \(\frac{19}{54}\)

(v) The denominator of given rational numbers are -4 and 8 respectively

The L.C.M of -4 and 8 is 8

Now, we rewrite the given rational numbers into forms in which both of them have the same denominator

\(\frac{31}{4} = \frac{31\times 2}{-4\times 2}\)

= \(\frac{-62}{8}\)

And,

\(\frac{5}{8}\)

Therefore,

\((\frac{-31\times 2}{4\times 2}) + \frac{-5}{8}\)

= \(\frac{-62}{8} - \frac{5}{8}\)

= \(\frac{-67}{8}\)

(vi) The denominator of given rational numbers are 36 and 12 respectively 

The L.C.M of 36 and 12 is 36 

Now, we rewrite the given rational numbers into forms in which both of them have the same denominator

\(\frac{-7\times3}{12\times 3} = \frac{-21}{36}\)

And,

\(\frac{5}{36}\)

Therefore,

\(\frac{5}{36} - \frac{21}{36}\)

= \(\frac{-16}{36}\)

= \(\frac{-4}{9}\)

(vii) The denominator of given rational numbers are 16 and 24 respectively 

The L.C.M of 16 and 24 is 48 

Now, we rewrite the given rational numbers into forms in which both of them have the same denominator

\(\frac{-5}{16} = \frac{-5\times 3}{16\times 3}\)

= \(\frac{-15}{48}\)

And,

\(\frac{7}{24} = \frac{7\times 2}{24\times 2}\)

= \(\frac{14}{48}\)

Therefore,

\(\frac{-5}{16} + \frac{7}{24}\)

= \(\frac{-15}{48} + \frac{14}{48}\)

= \(\frac{-1}{48}\)

(viii) The denominator of given rational numbers are -4 and 8 respectively

The L.C.M of 18 and 27 is 54

Now, we rewrite the given rational numbers into forms in which both of them have the same denominator

\(\frac{7}{-18} = \frac{7\times 3}{-18\times 3}\)

= \(\frac{-21}{54}\)

And,

\(\frac{8\times 2}{27\times 2}\) = \(\frac{16}{54}\)

Therefore,

\(\frac{-21}{54} + \frac{16}{54}\)

= \(\frac{16-21}{54}\)

= \(\frac{-5}{54}\)

(i) Clearly,

Denominators of the given numbers are positive

The L.C.M. of denominator 7 and 7 is 7

We have,

\(\frac{-5}{7}\, +\, \frac{3}{7}\)

\(=\, \frac{-5+3}{7}\)

\(=\, \frac{-2}{7}\)

(ii) Clearly,

Denominators of the given numbers are positive

The L.C.M. of denominators 4 and 4 is 4

We have,

\(\frac{-15}{4}\, +\, \frac{7}{4}\)

\(=\, \frac{-15+7}{4}\)

\(=\, \frac{-8}{4}\)

\(=\, -2\)

(iii) Clearly,

Denominators of the given numbers are positive

The L.C.M. of denominator 11 and 11 is 11

We have,

\(\frac{-8}{11}\, +\, \frac{-4}{11}\)

\(=\, \frac{-8-4}{11}\)

\(=\, \frac{-12}{11}\)

(iv) Clearly,

Denominators of the given numbers are positive

The L.C.M. of denominator 13 and 13 is 13

We have,

\(\frac{6}{13}\, +\, \frac{-9}{13}\)

\(=\, \frac{6-9}{13}\)

\(=\, \frac{-3}{13}\)

False.

If p/q is a rational number, p can be equal to zero (0) or any integer.

True.

If r/s is a rational number.

Then, cannot be equal to zero.

True.

Express each of the given rational numbers with 6 as the common denominator.

Now,

(2/3)= [(2×3)/ (2×2)] = (4/6)

(1)= [(1×6)/ (1×6)] = (6/6)

Then,

= 4/6 < 5/6 < 6/6

= 2/3 < 5/6 < 1

So, 5/6 lies between 2/3 and 1.

False.

X^-1 = 1/x

Then, reciprocal of 1/x = x/1 = x

False.

Express each of the given rational numbers with 4 as the common denominator.

Now,

-¾ = [(-3×1)/ (4×1)] = (-3/4)

-2/1 = [(-2×4)/ (1×4)] = (-8/4)

Then,

-3/4 > -8/4

Hence,

-¾ > -2

So, -¾ is greater than -2.

False.

Express each of the given rational numbers with 10 as the common denominator.

Now,

½ = [(1×5)/ (2×5)] = (5/10)

(1)= [(1×10)/ (1×10)] = (10/10)

Then,

½ is equal to 5/10

So, 5/6 does not lies between ½ and 1.

True

We Know that there are infinite rational numbers lie between two rational numbers.

True.

Express each of the given rational numbers with 2 as the common denominator.

Now,

-3/1 = [(-3×2)/ (1×2)] = (-6/2)

-4/1 = [(-4×2)/ (1×2)] = (-8/2)

Then,

-8/2 > -7/2 > -6/2

-4 > -7/2 > -3

So, -7/2 is lies between -3 and -4.

False.

We know that rational numbers can be expressed in the form of p/q.

True.

Because, 0/1 is a rational number.

False.

The correct form is for any rational number x, (x) × (-1) = – x.

True.

Express each of the given rational numbers with 6 as the common denominator.

Now,

1/1 = [(1×6)/ (1×6)] = (6/6)

2/1 = [(2×6)/ (1×6)] = (12/6)

Then,

6/6 < 9/6 < 12/6

1 < 9/6 < 2

So, 9/6 is lies between 1 and 2.

False.

There are infinite positive rational number on the right side of 0 on the number line.

True

If a = 0, then multiplicative inverse of a/b is not defined.

So, if a ≠ 0, then multiplicative inverse of a/b is b/a.

False.

Because, for rational numbers x and y, if x < y then x – y is a negative rational number.

Example, let x = 2 and y = 3

Then,

= X – y

= 2 – 3

= -1

False.

Because, the correct answer is reciprocal of a negative rational number is negative rational number. i.e. reciprocal of -3/5 is -5/3.

False.

The additive inverse of ½ is -½.

True.

Let x/y = ½ and its additive inverse c/d = -1/2

Then, (x/y) + (c/d)

= ½ + (-½ )

= ½ – ½

= 0

False.

Let x = 3

Then, 3 + 1 = 4

3 ≠ 4

So, it is clear that x + 1 ≠ x

To find a rational number x between two rational numbers \(\frac{a}{b}\) and \(\frac{c}{d},\) we use

\(\text{x}=\frac{1}{2}(\frac{a}{b}+\frac{c}{d})\)

Therefore, 

to find rational number x (let) between \(\frac{2}{3}\) and \(\frac{3}{4}\)

\(\text{x}=\frac{1}{2}(\frac{2}{3}+\frac{3}{4})\)

\(\Rightarrow\) \(\text{x}=\frac{1}{2}(\frac{8+9}{12})\)

\(\Rightarrow\) \(\text{x}=\frac{1}{2}\times\frac{17}{12}\)

\(\Rightarrow\) \(\text{x}=\frac{17}{24}\)

Now if we find a rational number between \(\frac{2}{3}\) and \(\frac{17}{24}\) it will also be between \(\frac{2}{3}\) and \(\frac{3}{4}\)since \(\frac{17}{24}\) lies between \(\frac{2}{3}\) and \(\frac{3}{4}\) 

Therefore, 

to find rational number y (let) between \(\frac{2}{3}\)  and \(\frac{17}{24}\)

\(\text{y}=\frac{1}{2}(\frac{2}{3}+\frac{17}{24})\)

\(\Rightarrow\) \(\text{y}=\frac{1}{2}(\frac{16+17}{24})\)

\(\Rightarrow\) \(\text{y}=\frac{1}{2}\times\frac{33}{24}\)

\(\Rightarrow\) \(\text{y}=\frac{33}{48}\)

Now if we find a rational number between \(\frac{17}{24}\) and \(\frac{3}{4}\) it will also be between \(\frac{2}{3}\) and \(\frac{3}{4}\)since \(\frac{17}{24}\) lies between \(\frac{2}{3}\) and \(\frac{3}{4}\) 

Therefore, 

to find rational number z (let) between \(\frac{17}{24}\) and \(\frac{3}{4}\)

\(\text{z}=\frac{1}{2}(\frac{17}{24}+\frac{3}{4})\)

\(\Rightarrow\) \(\text{z}=\frac{1}{2}(\frac{17+18}{24})\)

\(\Rightarrow\) \(\text{x}=\frac{1}{2}\times\frac{35}{24}\)

\(\Rightarrow\) \(\text{z}=\frac{35}{48}\) 

False.

Let x/y = 2/3 and its additive inverse c/d = -2/3

Then, (x/y) – (c/d)

= (2/3) – (-2/3)

= (2/3) + (2/3)

= 4/3

To find a rational number x between two rational numbers \(\frac{a}{b}\) and \(\frac{c}{d},\) we use

\(\text{x}=\frac{1}{2}(\frac{a}{b}+\frac{c}{d})\)

Therefore, 

to find rational number x (let) between 4 and 5

\(\text{x}=\frac{1}{2}(4+5)\)

\(\Rightarrow\) \(\text{x}=\frac{1}{2}\times9\)

\(\Rightarrow\) \(\text{x}=\frac{9}{2}\)

Now if we find a rational number between 4 and \(\frac{9}{2}\) it will also be between 4 and 5 since \(\frac{9}{2}\) lies between 4 and 5 

Therefore, 

to find rational number y (let) between 4 and \(\frac{9}{2}\)

\(\text{y}=\frac{1}{2}(4+\frac{9}{2})\)

\(\Rightarrow\) \(\text{y}=\frac{1}{2}(\frac{8+9}{2})\) 

\(\Rightarrow\) \(\text{y}=\frac{1}{2}\times\frac{17}{2}\) 

\(\Rightarrow\) \(\text{y}=\frac{17}{4}\) 

Now if we find a rational number between \(\frac{9}{2}\) and 5 it will also be between 4 and 5 since \(\frac{9}{2}\) lies between 4 and 5 

Therefore, 

to find rational number z (let) between \(\frac{9}{2}\)and 5

\(\text{z} =\frac{1}{2}(\frac{9}{2}+5)\)

\(\Rightarrow\) \(\text{z}= \frac{1}{2}(\frac{9+10}{2})\)

\(\Rightarrow\) \(\text{z}=\frac{1}{2}\times\frac{19}{2}\)

\(\Rightarrow\) \(\text{z}=\frac{19}{4}\)

False.

Reciprocal of non-zero rational number q/p is p/q.

To find a rational number x between two rational numbers \(\frac{a}{b}\) and \(\frac{c}{d},\) we use

\(\text{x}=\frac{1}{2}(\frac{a}{b}+\frac{c}{d})\)

Therefore, 

to find rational number x (let) between -3 and -2

\(\text{x}= \frac{1}{2}(-3+(-2))\)

\(\Rightarrow\) \(\text{x}=\frac{1}{2}(-3-2)\)

\(\Rightarrow\) \(\text{x}= \frac{-5}{2}\)

Now if we find a rational number between \(\frac{-5}{2}\) and d -2 it will also be between -3 and -2 since \(\frac{-5}{2}\) ies between -3 and -2 

Therefore,

to find rational number y (let) between \(\frac{-5}{2}\) and -2

True.

For example,

Let x = – 1/3 and y = -2/3

Then,

= x + y

= (-1/3) + (-2/3)

= -1/3 – 2/3

= -3/3

= – 1

False.

If x and y are rational numbers, then y is known as the negative of x.

To find a rational number x between two rational numbers \(\frac{a}{b}\) and \(\frac{c}{d},\) we use

\(\text{x}= \frac{1}{2}(\frac{a}{b}+\frac{c}{d})\)

Therefore, 

to find rational number x (let) between \(\frac{-1}{3}\) and \(\frac{1}{2}\)

\(\text{x}= \frac{-1}{2}(\frac{-1}{3}+\frac{1}{2})\)

\(\Rightarrow\) \(\text{x}=\frac{1}{2}(\frac{-1\times2+1\times3}{6})\)

\(\Rightarrow\) \(\text{x}= \frac{1}{2}(\frac{-2+3}{6})\)

\(\Rightarrow\) \(\text{x}=\frac{1}{2}\times\frac{1}{6}\)

\(\Rightarrow\) \(\text{x}=\frac{1}{12}\)

True.

Let y be a positive rational number.

Then,

The negative of the negative of y is = – (- y)

= y

True

Since, zero is neither a positive integer nor a negative integer.

To find a rational number x between two rational numbers \(\frac{a}{b}\) and \(\frac{c}{d},\) we use

\(\text{x} =\frac{1}{2}(\frac{a}{b}+\frac{c}{d})\)

Therefore, to find rational number x (let) between 2 and 3

\(\text{x}=\frac{1}{2}(2+3)\)

\(\Rightarrow\) \(\text{x}=\frac{1}{2}\times5\)

\(\Rightarrow\) \(\text{x}=\frac{5}{2}\)

False.

The negative of 1 = -1

To find a rational number x between two rational numbers \(\frac{a}{b}\) and \(\frac{c}{d},\) we use

\(\text{x}=\frac{1}{2}(\frac{a}{b}+\frac{c}{d})\)

Therefore, to find rational number \(\text{x}\) (let) between \(\frac{1}{4}\) and \(\frac{1}{3}\)

\(\text{x}= \frac{1}{2}(\frac{1}{4}+\frac{1}{3})\)

\(\Rightarrow\) \(\text{x}=\frac{1}{2}(\frac{1\times3+1\times4}{12})\)

\(\Rightarrow\) \(\text{x}=\frac{1}{2}(\frac{3+4}{12})\)

\(\Rightarrow\) \(\text{x}=\frac{1}{2}\times\frac{7}{12}\)

\(\Rightarrow\) \(\text{x}=\frac{7}{24}\)

False.

Let x = 2, y = 3

Then,

LHS = x – y

= 2 – 3

= -1

RHS = y – x

= 3 – 2

= 1

By comparing LHS and RHS

-1 ≠ 1

LHS ≠ RHS

(i) No rational numbers are not always closed under division,

Since, \(\frac{a}{0}\) ∞ which is not a rational number

(ii) No rational numbers are not always commutative under division,

Let \(\frac{a}{b}\) and \(\frac{c}{d}\) be two rational numbers.

\(\frac{a}{b}\div\frac{c}{d}=\frac{ad}{bc}\)

And

\(\frac{c}{d}\div\frac{a}{b}=\frac{bc}{ad}\)

Therefore,

\(\frac{a}{b}\div\frac{c}{d}\neq\frac{c}{d}\div\frac{a}{b}\)

Hence, rational numbers are not always commutative under division

(iii) No rational numbers are not always associative under division,

Let \(\frac{a}{b},\frac{c}{d}\) and \(\frac{e}{f}\) be two rational numbers.

\(\frac{a}{b}\div(\frac{c}{d}\div\frac{e}{f})= \frac{ade}{bcf}\)

And,

\((\frac{a}{b}\div\frac{c}{d})=\frac{e}{f}=\frac{adf}{bce}\)

Therefore,

\(\frac{a}{b}\div(\frac{c}{d}\div\frac{e}{f})\neq(\frac{a}{b}\div\frac{c}{d})\div\frac{e}{f}\)

Hence, rational numbers are not always associative under division.

(iv) No we cannot divide 1 by 0.

Since, \(\frac{a}{0}= ∞\) which is not defined.

True.

Let x = 2, y = 3

Then,

LHS = 2 × 3

= 6

RHS = 3 × 2

= 6

By comparing LHS and RHS

6 = 6

LHS = RHS

(b) (x + y)/2 is a rational number between x and y

Let us assume the value of x and y is 6 and 9 respectively

Then,

= (6 + 9)/ 2

= 14/2

= 7

Hence, the value 7 is lies between 6 and 9.

False.

Let x = 2

Then,

For every rational number x

(x) × (0) = 0

2 × 0 = 0

Form the question it is given that equivalent of 5/7 = 45/denominator

To get 45 in the numerator multiply both numerator and denominator by 9

Then,

= (5 × 9)/ (7 × 9)

= 45/63

So, the equivalent of 5/7, whose numerator is 45 is (45/63)