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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, (-2/3) × ((-4/6) + (-7/9)) = ((-2/3) × (-4/6)) + ((-2/3) × (-7/9))

LHS = (-2/3) × ((-4/6) + (-7/9))

= (-2/3) × ((-12 – 14)/18)

= – (2/3) × (-26/18)

= – (1/3) × (-26/9)

= 26/27

RHS = ((-2/3) × (-4/6)) + ((-2/3) × (-7/9))

= ((-1/3) × (-4/3)) + ((-2/3) × (-7/9))

= (4/9) + (14/27)

= (12 + 14)/27

= 26/27

By comparing LHS and RHS

LHS = RHS

∴ 26/27 = 26/27

Hence x × (y + z) = x × y + x × z

False.

Between any two rational numbers there are infinite rational numbers.

(14/15) – (13/20)

We have:

= (14/15) – (13/20)

= (14/15) + (additive inverse of 13/20)

= (14/15) + (-13/20)

LCM of 15 and 20 is 60

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

Now,

= [(14×4)/ (15×4)] = (56/60)

= [(-13×3)/ (20×3)] = (-39/60)

Then,

= (56/60) + (-39/60)

= (56-39)/60

= (17/60)

Correct answer is (a) –1

Correct answer is (d) -1

Correct answer is (c) 1/2

(a)[-8(1/9)]

Because,

= [-2(1/9)]-6 = (-19/9) – (6)

= (-19-54)/9

= (-73/9)

= [-8(1/9)]

Multiplicative inverse of (-2/3) is (b) (-3/2)

Reciprocal of -6 is (c) (-1/6)

Let the number added be x.

Then,

\(\frac{7}{12}+\text{x}=\frac{-4}{15}\)

\(\Rightarrow\) \(\text{x}=\frac{-4}{15}-\frac{7}{12}\)

\(=\frac{-4\times4-7\times5}{60}\)

\(=\frac{-16-35}{8}\)

\(=\frac{-51}{60}=\frac{-51\div3}{60\div3}=\frac{-17}{20}\)

(a)(2/-3)

Because,

First we write (2/-3) in standard form

= (2×-1)/ (-3×-1)

= (-2/3)

Then,

LCM of 3 and 5 is 15

= (-2 × 5)/ (3×5) = (-10/15)

= (-4×3) / (5×3) = (-12/15)

Clearly,

= (-10) > (-12)

Hence,

(-2/3)> (-4/5)

∴ (2/-3) > (-4/5)

\(\frac{2}{3}+\frac{-4}{5}+\frac{7}{15}+\frac{-11}{20}\)

LCM of 3, 5, 15, 20

\(=\frac{2\times20+(-4)\times12+7\times4+(-11)\times3}{60}\)

\(=\frac{40-48+28-33}{60}\)

\(=\frac{68-81}{60}\)

\(=\frac{-31}{60}\)

(a)(5/-6)

Because,

First we write (5/-6) in standard form

= (5×-1)/ (-6×-1)

= (-5/6)

Then,

LCM of 6 and 12 is 12

= (-5 × 2)/ (6×2) = (-10/12)

= (-7×1) / (12×1) = (-7/12)

Clearly,

= (-10) < (-7)

Hence,

(-5/6)< (-7/12)

∴ (5/-6) < (-7/12)

Let the number added be x.

Then,

\(\frac{5}{6}+\text{x}=\frac{-4}{7}\)

\(\Rightarrow\) \(\text{x}=\frac{-4}{7}-\frac{5}{6}\)

\(=\frac{-4\times6-5\times7}{42}\)

\(=\frac{-24-35}{42}\)

\(=\frac{-59}{42}\)

Let the number be x

It is given that,

x + (\(\frac{2}{3}+\frac{3}{5}\)) = \(\frac{-2}{15}\)

\(\frac{15x+10+9}{15}\) = \(\frac{-2}{15}\)

15x = -2-19

x = \(\frac{-21}{15}\)

x = \(\frac{-7}{5}\)

Let the number be x

It is given that,

(\(\frac{3}{4}\)-\(\frac{2}{3}\)) - x = \(\frac{-1}{6}\)

- x + [ \(\frac{3\times 3}{4\times 3}\) - \(\frac{2\times 4}{4\times 3}\)] =  \(\frac{-1}{6}\)

- x +  [\(\frac{9}{12}\)-\(\frac{8}{12}\)]  =  \(\frac{-1}{6}\)

\(\frac{-12x+9-8}{12}\) = \(\frac{-1}{6}\)

-12x-1 = -2-12x = -2-1-12x = -3

x = \(\frac{-3}{-12}\)

x = \(\frac{1}{4}\)

(b)(-19/12)

Because,

Let the missing number be X,

(-3/4) – (X) = (5/6)

By sending (-X) from the left hand side to the right side it becomes (X) and (5/6) from right hand side to left hand side it becomes (-5/6)

= (X) = (-3/4) – (5/6)

LCM of 4 and 6 is 12

(-3 × 3)/ (4×3) = (-9/12)

(-5×2) / (6×2) = (-10/12)

(X) = (-9-10)/12

(X) = (-19/12)

Let the number added be x.

Then,

\(\frac{-5}{7}+\text{x}=\frac{-2}{3}\)

\(\Rightarrow\) \(\text{x}=\frac{-2}{3}-\frac{-5}{7}\)

\(=\frac{-2\times7-(-5)\times3}{21}\)

\(=\frac{-14+15}{21}\)

\(=\frac{1}{21}\)

Let the number added be x 

Then,

\(\frac{-3}{5}+\text{x}=\frac{2}{3}\)

\(\Rightarrow\) \(\text{x} =\frac{2}{3}-\frac{-3}{5}\)

\(\Rightarrow\) \(\text{x}=\frac{2\times5-(-3)\times3}{15}\)

\(\Rightarrow\) \(\text{x}=\frac{10+9}{15}\)

\(\Rightarrow\) \(\text{x}=\frac{19}{15}\)

Multiplicative inverse of a rational number \(\frac{a}{b}\) = \(\frac{b}{a}\)

Therefore,

(i) Negative inverse of \(\frac{-3}{4}\) = \(\frac{4}{-3}\) 

\(\frac{4}{-3}=\frac{4\times-1}{-3\times-1}=\frac{-4}{3}\)

(ii) Negative inverse of \(\frac{11}{4}\) = \(\frac{4}{11}\)

i) 2/3 – 3/5

\(=\frac{(2\times5-3\times3)}{15}\\=\frac{1}{15}\)

ii) (-4/7) – (2/-3)

= (2/-3) = (-2/3)  (convert the denominator to positive number by multiplying by -1)
= -4/7 – (-2/3)

\(=\frac{(-4\times3-(-2\times7))}{21}\\=\frac{(-12+14)}{21}\\=\frac{2}{21}\)  

iii)  4/7 - -5/-7

 convert the denominator to positive number by multiplying by -1
= (-5/-7) = (5/7)
= 4/7 – 5/7

\(=\frac{(4-5)}{7}\\=\frac{-1}{7}\)

iv)  (-2) – (5/9)

\(=\frac{-2}{1}-\frac{5}{9}\\=\frac{(-2\times9-5\times1)}{9}\\=\frac{(-18-5)}{9}\\=\frac{-23}{9}\)

v) (-3/-8) - (-2/7)
 convert the denominator to positive number by multiplying by -1
(-3/-8) = 3/8
= 3/8 – (-2/7)
\(=\frac{(3\times7-(-2\times8))}{56}\\=\frac{(21+16)}{56}\\=\frac{37}{56}\)

vi) 5/63 - (-8/21)
\(=\frac{(5\times1-(-8\times3))}{63}\\=\frac{(5+24)}{63}\\=\frac{29}{63}\)

vii) 7/24 – 19/36

\(=\frac{(7\times3-19\times2)}{72}\\=\frac{(21-38)}{72}\\=\frac{-17}{72}\)

Given that,

Sum of two numbers = 5/9
One of the number = 1/3
By using the formula,
Other number = sum of number – given number

= 5/9 – 1/3

\(=\frac{(5\times1-1\times3)}{9}\\=\frac{(5-3)}{9}\\=\frac{2}{9}\)

\(\therefore\) the other number is \(\frac{2}{9}\)

Correct option is (C) 4

\(3\frac{1}{5}-\frac{1}{5}+1\) \(=\frac{3\times5+1}{5}-\frac{1}{5}+1\) \(=\frac{15+1-1}{5}+1\) \(=\frac{15}{5}+1\) = 3+1 = 4.

Correct option is  C) 4

Correct option is (A) 25/9

Let number to be added is x.

\(\therefore\) x + (-2) = \(\frac{7}{9}\)

\(\Rightarrow\) x = \(\frac{7}{9}\) -(-2) = \(\frac{7}{9}\) + 2 = \(\frac{7+9\times2}{9}\)

\(\Rightarrow\) x = \(\frac{7+18}{9}\) = \(\frac{25}{9}\).

Correct option is  A) 25/9

Correct option is (B) -28/15

\(4-(2\frac{2}{3}+3\frac{1}{5})\) \(=4-(\frac{2\times3+2}{3}+\frac{3\times5+1}{5})\)

\(=4-(\frac{8}{3}+\frac{16}{5})\)

\(=4-\frac{8\times5+16\times3}{15}\)

\(=4-\frac{40+48}{15}\)

\(=\frac{15\times4-88}{15}\)

\(=\frac{60-88}{15}=\frac{-28}{15}.\)

Correct option is  B) -28/15 

Correct option is (B) \(11\frac{1}{3}\)

\(7\frac{5}{9}\) \(\times\) \(\frac{3}{2}\) = \(\frac{7\times9+5}{9}\) \(\times\) \(\frac{3}{2}\) = \(\frac{63+5}{9}\) \(\times\) \(\frac{3}{2}\) \(=\frac{68}{3\times2}\) \(=\frac{34}3=\frac{33+1}3=11+\frac13\) \(=11\frac13.\)

Correct option is  B) 11\(\frac{1}{3}\) 

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

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

Let the number added be x

Then,

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

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

\(\Rightarrow\) \(\text{x}=\frac{3\times3-(-2)\times4}{12}\) 

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

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

Given that,

Sum of two numbers = -1/3
One of the number = -12/3
By using the formula,
Other number = sum of number – given number

= (-1/3) – (-12/3)

\(=\frac{(-1+12)}{3}\\=\frac{11}{3}\)

\(\therefore\) the other number is \(\frac{11}{3}\)

Correct option is  D) 0

i)  By using the property

x × y = y × x

\(\frac{-1}{3}\times\frac{2}{7}=\frac{2}{7}\times\frac{-1}{3}\\=\frac{(-1\times2)}{(3\times7)}=\frac{(2\times-1)}{(7\times3)}\\=\frac{-2}{21}=\frac{-2}{21}\)

\(\therefore\) the property is satisfied.

ii) By using the property

    x × y = y × x

\(=2\times\frac{7}{-8}=\frac{7}{-8}\times2\\=\frac{(2\times7)}{-8}=\frac{(7\times2)}{-8}\\=\frac{14}{-8}=\frac{14}{-8}\)

\(\therefore\) the property is satisfied.

iii) By using the property

   x × y = y × x

\(0\times\frac{-15}{8}=\frac{-15}{8}\times0\\=0=0\)

\(\therefore\)  the property is satisfied.

Correct option is   A) \(\frac{43}{9}\)

Correct option is (A)  43/9

Let x = \(4.\bar{7}\) 

\(\Rightarrow\) x = 4.777...     ______(1)

Multiply equation (1) by 10, we obtain

10x = 47.777...    ______(2)

Subtract equation (1) from (2), we get

10x - x = 47.777.... - 4.777....

\(\Rightarrow\) 9x = 43

\(\Rightarrow\) x = \(\frac{43}{9}\).

Correct option is  D) 1

Correct option is (A) 48

\(\frac{4}{5}\) of 1 hour = \(\frac{4}{5}\) \(\times\) 1 hour

= \(\frac{4}{5}\) \(\times\) 60 minutes

= 4 \(\times\) 12 minutes

= 48 minutes.

Correct option is A) 48

Correct option is (C) 0

\(\frac{a}{y-z}=\frac{b}{z-x}=\frac{c}{x-y}\) \(=\frac{ax+by+cz}{x(y-z)+y(z-x)+z(x-y)}\)

\(=\frac{ax+by+cz}{xy-xz+yz-xy+xz-yz}\) \(=\frac{ax+by+cz}0\)

\(\Rightarrow\) ax+by+cz = 0.

Correct option is  C) 0

Correct option is (C) 1.5625

\(\frac{25}{16}=\frac{5^2}{2^4}=\frac{5^2\times5^4}{2^4\times5^4}=\frac{5^6}{10^4}\) \(=\frac{15625}{10000}\) = 1.5625

Correct option is  C) 1.5625

Correct option is (A) –9

\(-2-\frac{1}{2}+\frac{1}{2}-7\) = -2 - 7 = -9.

Correct option is  A) – 9

D) All the above

Correct option is (D) All of the above

\(\frac{10}{2}\) = 5

\(\frac{15}{3}\) = 5

5 = \(5\times\frac{10}{10}=\frac{50}{10}.\)

Given that,

Sum of two numbers = -4/3
One of the number = -5/1
By using the formula,
Other number = sum of number – given number

=(-4/3)–(-5/1)

\(=\frac{(-4\times1-(-5\times3))}{3}\\=\frac{(-4+15)}{3}\\=\frac{11}{3}\)

\(\therefore\) the other number is \(\frac{11}{3}\)

Correct option is  B) – 3

Correct option is (C) 775

Area = \(41\frac{2}{3}\) \(\times\) \(18\frac{3}{5}\) \(=\frac{41\times3+2}3\times\frac{18\times5+3}5\) \(=\frac{123+2}3\times\frac{90+3}5\) \(=\frac{125}3\times\frac{93}5\) \(=25\times31\) \(=775\,m^2.\)

Correct option is  C) 775

Correct option is (B) 17/36

\(\frac{5}{9}-\frac{7}{12}+\frac{1}{2}\) \(=\frac{5\times4-7\times3+1\times18}{36}\) \(=\frac{20-21+18}{36}=\frac{18-1}{36}\) \(=\frac{17}{36}.\)

Correct option is  B) \(\frac{17}{36}\)

(i) x = 3/5

– x = -3/5

– (-x) = – (-3/5)

x = 3/5

(ii) x = -7/9

– x = – (-7/9)

-x = 7/9

– (-x) = – (7/9)

x = – 7/9

(iii) x = 13/-15

– x = – (-13/15)

-x = 13/15

– (-x) = – (13/15)

x = -13/15

Given that,

Sum of two rational numbers = -8/1
One of the number = -15/7
Let the other number as x

\(x+\frac{-15}{7}=-8\\\frac{(7x-15)}{7}=-8\\=7x-15=-8\times7\\=7x-15=-56\\=7x=-56+15\\=x=\frac{-41}{7}\)

\(\therefore\) the other number is \(\frac{-41}{7}\)

Let the number as x to be added to -5/11 to get 26/33
So,

\(\frac{-5}{11}+x=\frac{26}{33}\\x=\frac{26}{33}+\frac{5}{11}\\x=\frac{(26\times1+5\times3)}{33}\\=\frac{41}{33}\)

\(\therefore\) The required number is \(\frac{41}{33}\)

Correct option is  B) 1

Correct option is  C) ba – ca

 Let the number x to be added to -7/8 to get 5/9

\(\frac{-7}{8}+x=\frac{5}{9}\\\frac{(-7+8x)}{8}=\frac{5}{9}\\(-7+8x)\times9=5\times8\\-63+72x=40\\72x=40+73\\x=\frac{103}{72}\)

\(\therefore\) the required number is \(\frac{103}{72}\)  

Let the number be x to subtracted from -5/3 to get 5/6
So,

\(\frac{-5}{3}-x=\frac{5}{6}\\x=\frac{-5}{3}-\frac{5}{6}\\x=\frac{(-5\times2-5\times1)}{6}\\=\frac{(10-5)}{6}\\=\frac{-15}{6}\\=\frac{-5}{2}\)

\(\therefore\) the required number is \(\frac{-5}{2}\)

-333 < -210 < -26 < -8 < 0 < 21 < 33 < 85 < 2011. 

-333, -210, -26, -8, 0, 21, 33, 85, 2011