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Let the required number be x

(-33/16) ÷ x = (-11/4)

x = (-33/16) ÷ (-11/4)

x = (-33/16) × (4/-11)

x = (3/4)

Given sum of two numbers is -8

One of them is (-15/7)

Let the required number be x

x + (-15/7) = -8

The LCM of 7 and 1 is 7

Consider (-8/1) = (-8/1) × (7/7) = (-56/7)

On substituting

x + (-15/7) = (-56/7)

x = (-56/7) – (-15/7)

x = (-56/7) + (15/7)

x = (-56 + 15)/7

x = (-41/7)

Given (3/7)

Let the required number be x

(3/7) – x = (5/4)

– x = (5/4) – (3/7)

The LCM of 4 and 7 is 28

Consider (5/4) = (5/4) × (7/7) = (35/28)

Again (3/7) = (3/7) × (4/4) = (12/28)

On substituting

-x = (35/28) – (12/28)

– x = (35 -12)/28

– x = (23/28)

x = (-23/28)

Let the second number be ‘x’ say

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

\(\Rightarrow 8\,+\,\frac{5}6=\frac{48\,+\,5}{6}=\frac{53}{6}\)

∴ The other number (x) = \(\frac{53}{6}\)

Let the number = x 

Now, 

According to question,

\(-1+\) x \(=\frac{5}{7}\)

\(\Rightarrow\) x \(=\frac{5}{7}-(-1)\)

\(\Rightarrow\) x \(=\frac{5-(-7)}{7}\)

\(\Rightarrow\) x \(=\frac{5+7}{7}\)

\(\Rightarrow\) x \(=\frac{12}{7}\)

Therefore, \(\frac{12}{7}\) should be added to \(-1\) so as to get \(\frac{5}{7}\)

Given (-7/8)

Let the required number be x

x + (-7/8) = (5/9)

The LCM of 8 and 9 is 72

x = (5/9) – (-7/8)

x = (5/9) + (7/8)

Consider (5/9) = (5/9) × (8/8) = (40/72)

Again (7/8) = (7/8) × (9/8) = (63/72)

On substituting

x = (40/72) + (63/72)

x = (40 + 63)/72

x = (103/72)

Given (-5/11)

Let the required number be x

x + (-5/11) = (26/33)

x = (26/33) – (-5/11)

x = (26/33) + (5/11)

Consider (5/11) = (5/11) × (3/3) = (15/33)

On substituting

x = (26/33) + (15/33)

x = (41/33)

Correct option is   D) 3\(\frac{1}{2}\)

Rational number between \(\frac{-1}{3}\) and \(\frac{1}{2}\)

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

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

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

= \(\frac{1}{2}\times\frac{1}{6}\)

= \(\frac{1}{12}\)

Now,

\(=\frac{1}{2}(\frac{1}{12}+\frac{1}{2})\)

\(=\frac{1}{2}(\frac{1+6}{12})\)

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

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

Given ((2/3) + (3/5))

Let the required number be x

((2/3) + (3/5)) + x = (-2/15)

Consider (2/3) = (2/3) × (5/5) = (10/15)

Again (3/5) = (3/5) × (3/3) = (9/15)

On substituting

((10/15) + (9/15)) + x = (-2/15)

x = (-2/15) – ((10/15) + (9/15))

x = (-2/15) – (19/15)

x = (-2 -19)/15

x = (-21/15)

x = (- 7/5)

Let the number added be x 

Then,

\(\frac{-3}{5}+\text{x}=\frac{-1}{3}\)

\(\Rightarrow\) \(\text{x}=\frac{1}{3}-\frac{-3}{5}\)

\(\Rightarrow\) \(\text{x}=\frac{-1\times5-(-3)\times3}{15}\)

\(\Rightarrow\) \(\text{x}=\frac{-5+9}{15}\)

\(\Rightarrow\) \(\text{x}=\frac{4}{15}\)

Given ((1/2) + (1/3) + (1/5))

Let the required number be x

((1/2) + (1/3) + (1/5)) + x = 3

x = 3 – ((1/2) + (1/3) + (1/5))

LCM of 2, 3 and 5 is 30

Consider (1/2) = (1/2) × (15/15) = (15/30)

(1/3) = (1/3) × (10/10) = (10/30)

(1/5) = (1/5) × (6/6) = (6/30)

On substituting

x = 3 – ((15/30) + (10/30) + (6/30))

x = 3 – (31/30)

(3/1) = (3/1) × (30/30) = (90/30)

x = (90/30) – (31/30)

x = (90 – 31)/30

x = (59/30)

We know,For any real number a≠0, \(a^{-1}=\frac{1}{a}\,\,so,(\frac{-3}{7})^{-1}\) \(=\frac{7}{-3}=\frac{7\times-1}{-3\times-1}=\frac{-7}{3}\)

Given ((3/4) – (2/3))

Let the required number be x

((3/4) – (2/3)) – x = (-1/6)

– x = (-1/6) – ((3/4) – (2/3))

Consider (3/4) = (3/4) × (3/3) = (9/12)

(2/3) = (2/3) × (4/4) = (8/12)

On substituting

– x = (-1/6) – ((9/12) – ((8/12))

– x = (-1/6) – (1/12)

(1/6) = (1/6) × (2/2) = (2/12)

– x = (-2/12) – (1/12)

– x = (-2 – 1)/12

– x = (-3/12)

x = (3/12)

x = (1/4)

Let the number subtracted be x

Then,

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

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

\(\Rightarrow\) \(\text{x}=\frac{-3\times1-(-1)\times2}{4}\)

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

\(\Rightarrow\) \(\text{x}=\frac{-1}{4}\)

Let the number subtracted be x.

Then,

\(\frac{-5}{3}-\text{x}=\frac{5}{6}\)

\(\Rightarrow\) \(\text{x}=\frac{-5}{3}-\frac{5}{6}\)

\(=\frac{-5\times2-5\times1}{6}\)

\(=\frac{-10-5}{6}\)

\(=\frac{-15}{6}=\frac{-15\div3}{6\div3}=\frac{-5}{2}\)

(i) Given (3/8), (5/8)

(5/8) – (3/8) = (5 – 3)/8

= (2/8)

= (1/4)

(ii) Given (-7/9), (4/9)

(4/9) – (-7/9) = (4/9) + (7/9)

= (4 + 7)/9

= (11/9)

(iii) Given (-2/11), (-9/11)

(-9/11) – (-2/11) = (-9/11) + (2/11)

= (-9 + 2)/ 11

= (-7/11)

(iv) Given (11/13), (-4/13)

(-4/13) – (11/13) = (-4 – 11)/13

= (-15/13)

Correct option is  D) none

Cost of \(3\frac{1}{2}\,m\) of cloth \(=Rs\,166\frac{1}{4}\)

Cost of \(1\,m\) of cloth \(=\) Cost of \(3\frac{1}{2}\,m\) of cloth \(\div\) \(3\frac{1}{2}\)

\(=Rs\,166\frac{1}{4}\div3\frac{1}{2}\)

\(=Rs\,\frac{665}{4}\div\frac{7}{2}\)

\(=Rs\,\frac{665}{4}\times\frac{2}{7}\)

\(=Rs\,\frac{1330}{28}\)

\(=Rs\,47\frac{1}{2}\)

Cost of \(1\,m\) of cloth \(=Rs\,47\frac{1}{2}\)

Sum of \(\frac{13}{5}\) and \(\frac{-12}{7}\)

\(\frac{13}{5}+\frac{-12}{7}\)

\(=\frac{13\times7+(-12)\times5}{35}\)

= \(\frac{91-60}{35}\)

= \(\frac{31}{35}\)

Product of \(\frac{-31}{7}\) and \(\frac{1}{-2}\)

\(\frac{-31}{7}\times\frac{1}{-2}\)

= \(\frac{-31\times1}{7\times-2}\)

= \(\frac{-31}{-14}\)

= \(\frac{31}{14}\)

Now,

According to the question,

\(\frac{31}{35}\div\frac{31}{14}\)

= \(\frac{31}{35}\times\frac{14}{31}\)

= \(\frac{2}{5}\)

Given sum of two numbers = (-1/3)

One of them is (-12/3)

Let the required number be x

x + (-12/3) = (-1/3)

x = (-1/3) – (-12/3)

x = (-1/3) + (12/3)

x = (-1 + 12)/3

x = (11/3)

Correct option is   C) \(\frac{177}{20}\)

Correct option is   A) \(\frac{a+b}{2}\)

Correct option is  D) 1/6

First we have to find the sum of (-36/11) and (49/22)

= (-36/11) + (49/22)

LCM of 11 and 22 is 22

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

Now,

= [(-36×2)/ (11×2)] = (-72/22)

= [(49×1)/ (22×1)] = (49/22)

Then,

= (-72/22) + (49/22)

= (-72+49)/22

= (-23/22)

Now we have to find the sum of (33/8) and (-19/4)

= (33/8) + (-19/4)

LCM of 8 and 4 is 8

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

Now,

= [(33×1)/ (8×1)] = (33/8)

= [(-19×2)/ (4×2)] = (-38/8)

Then,

= (33/8) + (-38/8)

= (33-38)/8

= (-5/8)

Now,

= (-5/8) – (-23/22)

We have:

= (-5/8) – (-23/22)

= (-5/8) + (additive inverse of -23/22)

= (-5/8) + (23/22)

LCM of 8 and 22 is 88

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

Now,

= [(-5×11)/ (8×11)] = (-55/88)

= [(23×4)/ (22×4)] = (92/88)

Then,

= (-55/88) + (92/88)

= (-55+92)/88

= (37/88)

(-12/5) or -3

The LCM of the denominators 5 and 1 is 5

∴ (-12/5) = [(-12×1)/ (5×1)] = (-12/5)

and (-3/1) = [(-3×5)/ (1×5)] = (-15/5)

Now, -12 > -15

⇒ (-12/5) > (-15/5)

Hence, (-12/5) > (-3)

∴ -3 is greater.

(5/9) or (-3/8)

The LCM of the denominators 9 and 8 is 72

∴ (5/9) = [(5×8)/ (9×8)] = (40/72)

and (-3/8) = [(-3×9)/ (8×9)] = (-27/72)

Now, 40 > -27

⇒ (40/72) > (-27/72)

Hence, (5/9) > (-3/8)

∴ 5/9 is greater.

(i)\(\frac{3}{8}\) is a positive number and all positive numbers are greater than 0. 

Therefore,\(\frac{3}{8}\)> 0

(ii) \(\frac{-2}{9}\) is a negative number and all negative numbers are less than 0. 

Therefore, 0 >\(\frac{-2}{9}\)

(iii) Both \(\frac{-3}{4}\)and \(\frac{1}{4}\) have the same denominator 4. Therefore, we can directly compare both the numbers. Since, 1 > -3

Therefore, \(\frac{-3}{4}\)>\(\frac{1}{4}\)

(iv) Both \(\frac{-3}{4}\) and\(\frac{-4}{7}\) have the same denominator 7. Therefore, we can directly compare both the numbers. Since, -4 > -5

Therefore, \(\frac{-4}{7}\)>\(\frac{-5}{7}\)

(v)\(\frac{-3}{4}\) and \(\frac{-1}{4}\)have different denominators. Therefore, we take LCM of 3 and 4 that is 12.

 Now,

\( \frac{2}{3}=\frac{2\times4}{3\times4}=\frac{8}{12}\)

And,

\( \frac{3}{4}=\frac{3\times3}{4\times3}=\frac{9}{12}\)

Since, 9 > 8

\(\frac{9}{12}\)>\(\frac{8}{12}\)

Hence, \(\frac{2}{3}\)>\(\frac{3}{4}\)

(vi) We can write -1=\(\frac{-1}{1}\)

\(\frac{1}{2}\)and \(\frac{-1}{1}\)have different denominators.

Therefore, we take LCM of 1 and 2 that is 2.

 Now,

\( \frac{-1}{2}=\frac{-1\times1}{2\times1}=\frac{-1}{2}\)

And,

\( \frac{-1}{1}=\frac{-1\times2}{1\times2}=\frac{-2}{2}\)

Since, -1 > -2

Therefore,  \(\frac{-1}{2}\)>\(\frac{-2}{2}\)

Hence, \(\frac{-1}{2}\)>-1

(i) (7/9) or (-5/9)

Since both denominators are same therefore compare the numerators.

We have,

= 7 > -5

∴ (7/9) > (-5/9)

(ii) (-6/11) or (5/11)

Since both denominators are same therefore compare the numerators.

We have,

= -6 < 5

∴ (-6/11) < (5/11)

(iii) (-15/4) or (-17/4)

Since both denominators are same therefore compare the numerators.

We have,

= -15 > -17

∴ (-15/4) > (-17/4)

(i) Given (-3/8), 0

We know that every positive rational number is greater than zero and every negative

rational number is smaller than zero. Thus, – (3/8) > 0

(ii)  Given (5/2), 0

We know that every positive rational number is greater than zero and every negative rational number is smaller than zero. Thus, (5/2) > 0

(iii) Given (– 4/11), (3/11)

We know that every positive rational number is greater than zero and every negative rational number is smaller than zero, also the denominator is same in given question now we have to compare the numerator, thus – 4/11 < 3/11.

(iv) Given (– 7/12), (5/- 8)

Consider (– 7/12)

Multiply both numerator and denominator by 2 then we get

(-7/12) × (2/2) = (-14/24)…… (1)

Now consider (5/-8)

Multiply both numerator and denominator by 3 we get

(5/-8) × (3/3) = (15/-24)…… (2)

The denominator is same in equation (1) and (2) now we have to compare the numerator, thus (– 7/12) > (5/- 8)

(v) Given (4/-9), (– 3/- 7)

Consider (4/-9)

Multiply both numerator and denominator by 7 then we get

(4/-9) × (7/7) = (28/-63)…… (1)

Now consider (-3/-7)

Multiply both numerator and denominator by 9 we get

(-3/-7) × (9/9) = (-27/-63)…… (2)

The denominator is same in equation (1) and (2) now we have to compare the numerator, thus (4/-9) < (– 3/- 7)

(vi) Given (– 5/8), (3/- 4)

Now consider (3/-4)

Multiply both numerator and denominator by 2 we get

(3/-4) × (2/2) = (6/-8)

The denominator is same in above equation now we have to compare the numerator, thus (– 5/8) > (3/- 4)

(vii) Given (5/9), (-3/- 8)

Consider (5/9)

Multiply both numerator and denominator by 8 then we get

(5/9) × (8/8) = (40/72)…… (1)

Now consider (5/-8)

Multiply both numerator and denominator by 9 we get

(-3/-8) × (9/9) = (-27/-72)…… (2)

The denominator is same in equation (1) and (2) now we have to compare the numerator, thus (5/9) > (-3/- 8)

(viii) Given (5/- 8), (-7/12)

Consider (5/-8)

Multiply both numerator and denominator by 3 then we get

(5/-8) × (3/3) = (15/-24)…… (1)

Now consider (-7/12)

Multiply both numerator and denominator by 2 we get

(-7/12) × (2/2) = (-14/24)…… (2)

The denominator is same in equation (1) and (2) now we have to compare the numerator, thus (5/- 8) < (-7/12)

(i)\(\frac{-4}{3}\) and\( \frac{-8}{7}\) have different denominators. Therefore, we take LCM of 3 and 7 that is 21. 

Now,

\(\frac{-4}{3}=\frac{-4\times7}{3\times7}=\frac{-28}{21}\)

And,

\(\frac{-8}{7}=\frac{-8\times3}{7\times3}=\frac{-24}{21}\)

Since, -24 > -28

Therefore,\(\frac{-24}{21}\)>\(\frac{-28}{21}\)

Hence,\(\frac{-8}{7}\)>\(\frac{-4}{3}\)

(ii)

\(\frac{-7}{9}=\frac{-7\times-1}{-9\times-1}=\frac{-7}{9}\)

\(\frac{-7}{9}\)and\(\frac{-5}{8}\) have different denominators. Therefore, we take LCM of 9 and 8 that is 72.

 Now,

\(\frac{-7}{9}=\frac{-7\times8}{9\times8}=\frac{-56}{72}\)

And,

\(\frac{-5}{8}=\frac{-5\times9}{8\times9}=\frac{-45}{72}\)

Since, -45 > -56

Therefore, \(\frac{-45}{72}\)>\(\frac{-56}{72}\)

Hence, \(\frac{-5}{8}\)>\(\frac{-7}{9}\)

(iii)

\(\frac{-4}{5}=\frac{-4\times-1}{-5\times-1}=\frac{-4}{5}\)

\(\frac{-1}{3}\) and\(\frac{-4}{5}\) have different denominators. Therefore, we take LCM of 3 and 5 that is 15. 

Now,

\(\frac{-1}{3}=\frac{-1\times5}{3\times5}=\frac{-5}{15}\)

And,

\(\frac{-4}{5}=\frac{-4\times3}{5\times3}=\frac{-12}{15}\)

Since, -5 > -12

Therefore,  \(\frac{-5}{15}\)>\(\frac{-12}{15}\)

Hence, \(\frac{-1}{3}\)>\(\frac{-4}{5}\)

(iv)

\(\frac{-9}{-13}=\frac{9\times-1}{-13\times-1}=\frac{-9}{13}\)

And,

\(\frac{7}{-12}=\frac{7\times-1}{-12\times-1}=\frac{-7}{12}\)

\(\frac{-9}{13}\)and \(\frac{-7}{12}\)have different denominators.

Therefore, we take LCM of 13 and 12 that is 156. Now,

\(\frac{-9}{13}=\frac{-9\times-12}{13\times12}=\frac{-108}{156}\)

And,

\(\frac{-7}{12}=\frac{-7\times-13}{12\times13}=\frac{-91}{156}\)

Since, -91 > -108

Therefore, \(\frac{-91}{156}\)>\(\frac{-108}{156}\)

Hence,  \(\frac{-7}{12}\)>\(\frac{-9}{13}\)

(v)

\(\frac{-4}{5}=\frac{4\times-1}{-5\times-1}=\frac{-4}{5}\)

\(\frac{-7}{10}\)and\(\frac{-4}{5}\) have different denominators.

Therefore, we take LCM of 10 and 5 that is 10. Now,

\(\frac{-7}{10}=\frac{-7\times1}{10\times1}=\frac{-7}{10}\)

And,

\(\frac{-4}{5}=\frac{-4\times2}{5\times2}=\frac{-8}{10}\)

Since, -7 > -8

Therefore, \(\frac{-7}{10}\)>\(\frac{-8}{10}\)

Hence, \(\frac{-7}{10}\)>\(\frac{-4}{5}\)

(vi)

 We can write-3=\(\frac{-3}{1}\)

\(\frac{-3}{1}\) and \(\frac{-12}{5}\) have different denominators. Therefore, we take LCM of 1 and 5 that is 5. 

Now,

\(\frac{-12}{5}=\frac{-12\times1}{5\times1}=\frac{-12}{5}\)

And,

\(\frac{-3}{1}=\frac{-3\times5}{1\times5}=\frac{-15}{5}\)

Since, -12 > -15

Therefore, \(\frac{-12}{5}\)>\(\frac{-15}{5}\)

Hence,  \(\frac{-12}{5}\)>-3

-3/5 is smaller than 0.

Since, -3/5 lies on the left side of zero(0). On the number line, -3/5 is smaller than 0 i.e. -3/5 < 0.

-7/11 = -70/110; 6/-11 = -60/110

We have -61/110, -62/110, ......., -69/110 between -7/11 and 6/-11.

7/−5 is a rational number

Because of rational number is a rational number which is of the form p/q , q ≠ 0 and p and q are integers.

But fraction is part of a whole.

0, −1/2, 1/2, 3/4, 6/7, -5, 6

(i) \(\frac{-8}{3}+(\frac{-1}{4})+(\frac{-11}{6})+\frac{3}{8}-3\)

= \(\frac{-8}{3}-\frac{1}{4}-\frac{11}{6}+\frac{3}{8}-3\)

= \(\frac{-8\times 8}{3\times 8}-\frac{1\times 6}{4\times 6}-\frac{11\times 4}{6\times 4}+\frac{3\times 3}{8\times 3}-\frac{3\times 24}{1\times 24}\)

= \(\frac{-64-6-44+9-72}{24}\)

 = \(\frac{-59\times 3}{8\times 3}\)

 = \(\frac{-59}{8}\)

(ii) \(\frac{6}{7}+1+(\frac{-7}{9})+(\frac{19}{21})+\frac{-12}{7}\)

= \(\frac{6\times 9}{7\times 9}+\frac{63}{63}-\frac{7\times 7}{9\times 7}-\frac{19\times 3}{21\times 3}-\frac{12\times 9}{7\times 9}\) 

= \(\frac{54}{63}+\frac{63}{63}-\frac{49}{63}+\frac{57}{63}-\frac{108}{63}\)

= \(\frac{54+63-49+57-108}{63}\)

= \(\frac{17}{63}\)

(iii) \(\frac{15}{2}+(\frac{9}{8})+(\frac{-11}{3})+6+\frac{-7}{6}\)

 = \(\frac{15}{2}+\frac{9}{8}-\frac{11}{3}+6-\frac{7}{6}\)

 = \(\frac{15\times 12}{2\times 12}+\frac{9\times 3}{8\times 3}-\frac{11\times 8}{3\times 8}+\frac{6\times 24}{1\times 24}-\frac{7\times 4}{6\times 4}\)

 = \(\frac{180+27-88+144-28}{24}\)

  = \(\frac{235}{24}\)

(iv) \(\frac{-7}{4}+0+(\frac{-9}{5})+(\frac{19}{10})+\frac{11}{14}\)

 = \(\frac{-7}{4}+0-\frac{9}{5}+\frac{19}{10}+\frac{11}{14}\)

= \(\frac{-7\times 35}{4\times 35}-\frac{9\times 28}{5\times 28}+\frac{19\times 14}{10\times 14}+\frac{11\times 10}{14\times 10}\)

= \(\frac{-245-252+266+110}{140}\)

= \(\frac{-121}{140}\)

(v) \(\frac{-7}{4}+(\frac{5}{3})+(\frac{-1}{2})+\frac{-5}{6}+2\)

 = \(\frac{-7}{4}+\frac{5}{3}-\frac{1}{2}-\frac{5}{6}+2\)

 = \(\frac{-7\times 6}{4\times 6}+\frac{5\times 8}{3\times 8}-\frac{1\times 12}{2\times 12}-\frac{5\times 4}{6\times 4}+\frac{2\times 24}{1\times 24}\)

 = \(\frac{-42+40-12-20+48}{24}\)

  = \(\frac{14}{24}\)

   = \(\frac{7}{12}\)

i) (-16/21) × (14/5)

\(=\frac{-16}{21}\times\frac{14}{5}\\=\frac{-16}{3}\times\frac{2}{5}\\=\frac{-16\times2}{3\times5}\\=\frac{-32}{15}\) 

ii) (7/6) × (-3/28)

\(=\frac{7}{6}\times\frac{-3}{28}\\=\frac{1}{2}\times\frac{-1}{4}\\=\frac{-1}{8}\)

iii) (-19/36) × 16

\(=\frac{-19}{36}\times16\\=\frac{-19}{9}\times4\\=\frac{-19\times4}{9}\\=\frac{-76}{9}\)

iv) (-13/9) × (27/-26)

\(=\frac{-13}{9}\times\frac{27}{-26}\\=\frac{-1}{1}\times\frac{-3}{2}\\=\frac{-3}{-2}\\=\frac{3}{2}\)

v) (-5/9) × (72/-25)

\(=\frac{-5}{9}\times\frac{72}{-25}\\=\frac{-1}{1}\times\frac{8}{-5}\\=\frac{-1\times8}{1\times-5}\\=\frac{-8}{-5}\\=\frac{8}{5}\)

(-5/3) × (-3/5) = 1.

= (-5 × -3)/ (3 × 5)

= (-1 × -1)/ (1 × 1)

= 1

We know 5 = 5/1, y where Nr and Dr both are positive

∴ 5 is a positive, rational number.

(i) Given (-9/12) and (8/-12)

The standard form of (-9/12) is (-3/4) [on diving the numerator and denominator of given number by their HCF i.e. by 3]

The standard form of (8/-12) = (-2/3) [on diving the numerator and denominator of given number by their HCF i.e. by 4]

Since, the standard forms of two rational numbers are not same. Hence, they are not equal.

(ii) Given (-16/20) and (20/-25)

Multiplying numerator and denominator of (-16/20) by the denominator of (20/-25)

i.e. -25.

(-16/20) × (-25/-25) = (400/-500)

Now multiply the numerator and denominator of (20/-25) by the denominator of

(-16/20) i.e. 20

(20/-25) × (20/20) = (400/-500)

Clearly, the numerators of the above obtained rational numbers are equal.

Hence, the given rational numbers are equal

(iii) Given (-7/21) and (3/-9)

Multiplying numerator and denominator of (-7/21) by the denominator of (3/-9)

i.e. -9.

(-7/21) × (-9/-9) = (63/-189)

Now multiply the numerator and denominator of (3/-9) by the denominator of

(-7/21) i.e. 21

(3/-9) × (21/21) = (63/-189)

Clearly, the numerators of the above obtained rational numbers are equal.

Hence, the given rational numbers are equal

(iv) Given (-8/-14) and (13/21)

Multiplying numerator and denominator of (-8/-14) by the denominator of (13/21)

i.e. 21

(-8/-14) × (21/21) = (-168/-294)

Now multiply the numerator and denominator of (13/21) by the denominator of

(-8/-14) i.e. -14

(13/21) × (-14/-14) = (-182/-294)

Clearly, the numerators of the above obtained rational numbers are not equal.

Hence, the given rational numbers are also not equal.

(i) Rational number

(ii) Standard form

(iii) Standard

(iv) b ÷ m

(v) Positive, negative

(vi) (-1/1)

(vii) Zero

(viii) Ratio

 (i) \(\frac{8}{-5}\) and \(\frac{-24}{15}\)

\(\frac{8}{-5}\times\frac{-3}{-3} = \frac{-24}{15}\)

∴ The above pair of numbers represent the same rational number.

(ii) \(\frac{1}{3}\) and \(\frac{-1}{9}\)

\(\frac{1}{3}\) is a positive rational number and \(\frac{-1}{9}\) is a negative Rational Number.

Therefore this pair does not represent the same rational number.

(iii) \(\frac{-5}{-9}\) and \(\frac{5}{-9}\)

\(\frac{-5}{-9}\times\frac{-1}{-1} = \frac{5}{9}\)  is a positive rational number and \(\frac{5}{-9}\) is a negative rational number.

(i) Given (2/3) and (5/x)

Also given that they are equivalent rational number so (2/3) = (5/x)

x = (5 × 3)/2

x = (15/2)

(ii) Given (-3/7) and (x/4)

Also given that they are equivalent rational number so (-3/7) = (x/4)

x = (-3 × 4)/7

x = (-12/7)

(iii) Given (3/5) and (x/-25)

Also given that they are equivalent rational number so (3/5) = (x/-25)

x = (3 × -25)/5

x = (-75)/5

x = -15

(iv) Given (13/6) and (-65/x)

Also given that they are equivalent rational number so (13/6) = (-65/x)

x = 6/13 x (- 65)

x = 6 x (-5)

x = -30

(i) 7/12 and 28/48

Now, first rational number is 7/12 and it is already in the standard form because there is no common factor in 7 and 12 other than 1. 

So, 7/12 is in its standard form ......(a)

Now, Consider 28/48

28 = 2 × 2 × 7 

48 = 2 × 2 × 2 × 2 × 3 

HCF = 2 × 2 = 4

Now, to reduce the rational numbers to its standard form, we divide the numerator and denominator by their HCF. First we take HCF of 28 and 48: 

Now, 28/48 = (28 ÷ 4)/(48 ÷ 4) = 7/12 .......(b)

From (a) and (b), we can say that the rational numbers 7/12 and 28/48 are equivalent.

(ii) -2/-3 and -16/24

First we multiply the numerator and denominator of –2/–3 by (–1), we get

-2/-3 = (-2) x (-1)/ (-3) x (-1) = 2/3  .....(a)

Now it is in its standard form.

Now, Consider 16/24

HCF of 16 and 24 is 2 × 2 × 2 = 8 

16 = 2 × 2 × 2 × 2 

24 = 2 × 2 × 2 × 3 

HCF = 2 × 2 × 2 = 8

So, -16/24 = (-16 ÷ 8)/(24 ÷ 8) = -2/3  .......(b)

From (a) and (b), we can say that the rational numbers -2/3-3 and -16/24 are not equivalent.

(-19/36) × (16)

The product of two rational numbers = (product of their numerator)/ (product of their denominator)

We have,

= (-19×16)/ (36×1)

On simplifying,

= (-19×4)/ (9×1)

= (-76/9)

(i) Given (8/9) + (-11/6)

The LCM of 9 and 6 is 18

(8/9) = (8/9) × (2/2) = (16/18)

(11/6) = (11/6) × (3/3) = (33/18)

= (16 – 33)/18

= (-17/18)

(ii) Given (-5/16) + (7/24)

The LCM of 16 and 24 is 48

Now (-5/16) = (-5/16) × (3/3) = (-15/48)

Consider (7/24) = (7/24) × (2/2) = (14/48)

(-5/16) + (7/24) = (-5/48) + (14/48)

= (14 – 15) /48

= (-1/48)

(iii) Given (1/-12) + (2/-15)

The LCM of 12 and 15 is 60

Consider (-1/12) = (-1/12) × (5/5) = (-5/60)

Now (2/-15) = (-2/15) × (4/4) = (-8/60)

(1/-12) + (2/-15) = (-5/60) + (-8/60)

= (-5 – 8)/60

= (-13/60)

(iv) Given (-8/19) + (-4/57)

The LCM of 19 and 57 is 57

Consider (-8/57) = (-8/57) × (3/3) = (-24/57)

(-8/19) + (-4/57) = (-24/57) + (-4/57)

= (-24 – 4)/57

= (-28/57)

(i) Given (3/4) and (-3/5)

If p/q and r/s are two rational numbers such that q and s do not have a common factor other than one, then

(p/q) + (r/s) = (p × s + r × q)/ (q × s)

(3/4) + (-3/5) = (3 × 5 + (-3) × 4)/ (4 × 5)

= (15 – 12)/ 20

= (3/20)

(ii) Given -3 and (3/5)

If p/q and r/s are two rational numbers such that q and s do not have a common factor other than one, then

(p/q) + (r/s) = (p × s + r × q)/ (q × s)

(-3/1) + (3/5) = (-3 × 5 + 3 × 1)/ (1 × 5)

= (-15 + 3)/ 5

= (-12/5)

(iii) Given (-7/27) and (11/18)

LCM of 27 and 18 is 54

(-7/27) = (-7/27) × (2/2) = (-14/54)

(11/18) = (11/18) × (3/3) = (33/54)

(-7/27) + (11/18) = (-14 + 33)/54

= (19/54)

(iv) Given (31/-4) and (-5/8)

LCM of -4 and 8 is 8

(31/-4) = (31/-4) × (2/2) = (62/-8)

(31/-4) + (-5/8) = (-62 – 5)/8

= (-67/8)

(i) Since, the denominators of given rational numbers are negative therefore, we will make them positive.

\(\frac{7}{-18} = \frac{7\times-1}{-18\times-1} = \frac{-7}{18}\)

Now, 

since, the denominators of given rational numbers are different therefore, we take their LCM.

LCM of 18 and 27 = 54

\(\frac{-7}{18} = \frac{-7\times3}{18\times3} = \frac{-21}{54}\)

And

\(\frac{8}{27} = \frac{8\times2}{27\times2} = \frac{16}{54}\)

Now,

\(\frac{-7}{18}+\frac{8}{27}\)

= \(\frac{-21}{54}+\frac{16}{54}\)

= \(\frac{-21+16}{54}\)

= \(\frac{-5}{54}\)

(ii) Since, the denominators of given rational numbers are negative therefore, we will make them positive.

\(\frac{1}{-12} = \frac{1\times-1}{-12\times-1} = \frac{-1}{12}\)

And,

\(\frac{2}{-15} = \frac{2\times-1}{-15\times-1} = \frac{-2}{15}\)

Now, 

since, the denominators of given rational numbers are different therefore, we take their LCM. LCM of 12 and 15 = 60

\(\frac{-1}{12} = \frac{-1\times5}{12\times5} = \frac{-5}{60}\)

And

\(\frac{-2}{15} = \frac{-2\times4}{15\times4} = \frac{-8}{60}\)

Now,

\(\frac{-5}{60}+\frac{8}{60}\)

= \(\frac{-5+(8)}{60}\)

= \(\frac{-5-8}{60}\)

= \(\frac{-13}{60}\)

(iii) We can write \(-1\) as \(\frac{-1}{1}.\)

Now,

since, the denominators of given rational numbers are different therefore, we take their LCM. 

LCM of 1 and 4 = 4

\(\frac{-1}{1} = \frac{-1\times4}{1\times4} = \frac{-4}{4}\)

And

\(\frac{3}{4} = \frac{3\times1}{4\times1} = \frac{3}{4}\)

Now,

\(-1+\frac{3}{4}\)

= \(\frac{-4+3}{4}\)

= \(\frac{-1}{4}\) 

(iv) We can write 2 as \(\frac{2}{1.}\)

Now, 

since, the denominators of given rational numbers are different therefore, we take their 

LCM. LCM of 1 and 4 = 4

\(\frac{2}{1} = \frac{2\times4}{1\times4} = \frac{8}{4}\)

And

\(\frac{-5}{4} = \frac{-5\times1}{\times1} = \frac{-5}{4}\)

Now,

\(2+\frac{-5}{4}\)

= \(\frac{8+(-5)}{4}\)

= \(\frac{8-5}{4}\)

= \(\frac{3}{4}\)

(v) \(0+\frac{-2}{5}\)

On adding, any number to 0 we get the same number. 

Therefore,

\(0+\frac{-2}{5}= \frac{-2}{5}\)

(9/-13) and (-11/-13)

First we write each of the given numbers with a positive denominator.

(9/-13) = [(9× (-1))/ (-13×-1)]

= (-9/13)

(-11/-13) = [(-11× (-1))/ (-13×-1)]

= (11/13)

Then,

(-9/13)+ (11/13)

We have:

= [(-9 + 11)/13] … [∵ denominator is same in both the rational numbers]

= (2/13)

In the question is given to verify the property x × (y + z) = x × y + x × z

The arrangement of the given rational number is as per the rule of distributive property of multiplication over addition.

Then, (-½) × (¾ + ¼) = (-½ × ¾) + (-½ × ¼)

LHS = (-½) × (¾ + ¼)

= (-½) × ((3 + 1)/4)

= -½ × (4/4)

= -½ × 1

= -½

RHS = (-½ × ¾) + (-½ × ¼)

= (-3/8) + (-1/8)

= (-3 – 1)/8

= -4/8

= -½

By comparing LHS and RHS

LHS = RHS

∴ -½ = -½

Hence x × (y + z) = x × y + x × z