Explore topic-wise InterviewSolutions in .

Correct option is  B) -1

Correct option is (D) -5/9

Let the other numbr be x.

Then \(x\times4=-\frac{20}9\)

\(\Rightarrow\) \(x=\frac{-20}{9\times4}\) \(=\frac{-5}{9}.\)

Correct option is  D) -5/9

Correct option is  A) 1/2

The number T is multiplicative inverse of itself. 

∵ 1 × \(\frac11\) = 1 

 ⇒ 1 × 1 = 1 

∴ The multiplicative inverse of 1 is 1.

Correct option is   B) 1

Correct option is (D) -4/3

Let x be the multiplicative inverse of \(\frac{-3}{4}.\)

Then \(x\times\frac{-3}{4}=1\)

\(\Rightarrow\) \(x=\frac1{\frac{-3}4}=\frac{-4}3.\)

Correct option is  D) -4/3

Correct option is (D) -(4/5)

Multiplicative inverse of \(\frac{-9}{2}\) is \(\frac1{\frac{-9}{2}}\) \(=\frac{-2}{9}.\)

and multiplicative inverse of \(\frac{5}{18}\) is \(\frac1{\frac5{18}}=\frac{18}5.\)

\(\therefore\) Product of multiplicative inverses of \(\frac{-9}{2}\) and \(\frac{5}{18}\)

\(=\frac{-2}{9}\times\frac{18}5=\frac{-2\times2}5=\frac{-4}5.\)

Correct option is D) -(4/5)

Correct option is  C) 1

Correct option is (A) 105/34

\(\frac{-2}{7}\times \frac{-17}{15}\) \(=\frac{-2\times-17}{7\times15}\) \(=\frac{34}{105}\)

\(\therefore\) Multiplicative inverse of \(\frac{-2}{7}\times \frac{-17}{15}\) is \(\cfrac1{\frac{34}{105}}\) \(=\frac{105}{34}.\)

Correct option is   A) \(\frac{105}{34 }\)

Correct option is (D) 9/5

\(\frac{3}{5}\div\frac{1}{3}\) \(=\frac{3}{5}\times\frac31=\frac{3\times3}{5\times1}=\frac95.\)

Correct option is  D) 9/5 

Correct option is (A) 12/20

Equivalent number to \(\frac{3}{5}\) is \(\frac{3}{5}\times\frac44=\frac{3\times4}{5\times4}\) \(=\frac{12}{20}.\)

Correct option is  A) 12/20 

Correct option is (A) -1/2

\(1-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\) \(=\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\) \(=0-\frac{1}{2}=-\frac{1}{2}.\)

Correct option is  A) -1/2 

Correct option is (D) 12

\(x=\frac{-4}{9}\times-27=-4\times-3=12.\)

Correct option is  D) 12

Correct option is (D)  33/20

\(\frac{11}{2}\times\frac{3}{10}\) = \(\frac{11\times3}{2\times10}=\frac{33}{20}.\)

Correct option is    D) \(\frac{33}{20}\)

Correct option is (B)  16/75

\(\frac{-2}{5}(\frac{-7}{10}+\frac{1}{6})\) \(=\frac{-2}{5}(\frac{-7\times3+1\times5}{30})\) \(=\frac{-2}{5}\times\frac{-21+5}{30}\) \(=\frac{-2}{5}\times\frac{-16}{30}\) \(=\frac{-1\times-16}{5\times15}=\frac{16}{75}.\)

Correct option is   B) \(\frac{-16}{75}\)

Correct option is (A) 10/9

\(\because\) Krishna read in 1 hour = \(\frac{1}{3}\) part of a book

\(\therefore\) Krishna read in \(3\frac{1}{3}\) hours = \(3\frac{1}{3}\) \(\times\) \(\frac{1}{3}\) part of that book

= \(\frac{3\times3+1}{3}\) \(\times\) \(\frac{1}{3}\)

= \(\frac{10}{3}\) \(\times\) \(\frac{1}{3}\) = \(\frac{10}9\) part of the book.

Correct option is  A) 16/15

D) Not defined

Correct option is (B) \(10\frac{1}{4}\) kg

Total weight = \(5\frac{3}{4}\) + \(4\frac{1}{2}\) \(=\frac{5\times4+3}4+\frac{4\times2+1}2\)

\(=\frac{23}4+\frac92\) \(=\frac{23+9\times2}4=\frac{23+18}4\)

\(=\frac{41}4=\frac{40+1}4\) \(=10+\frac{1}{4}\)

= \(10\frac{1}{4}\) kg.

Correct option is  B) 10\(\frac14\) kg 

Correct option is (A)  −23/30

\(\frac{8}{-5}+\frac{-5}{-6}\) \(=\frac{-8}{5}+\frac{5}{6}\) \(=\frac{-8\times6+5\times5}{30}\) \(=\frac{-48+25}{30}\) \(=\frac{-23}{30}.\)

Correct option is   A) \(\frac{-23}{30}\)

Correct option is (A) -7/18

LCM of 9, 12, 18 and 3 is 36.

So, we have to find equivalent numbers to given rational numbers whose denominator is 36.

So, we can compare between these numbers.

\(\frac{-4}{9}=\frac{-4}{9}\times\frac44=\frac{-4\times4}{9\times4}=\frac{-16}{36},\)

\(\frac{-5}{12}=\frac{-5}{12}\times\frac33=\frac{-5\times3}{12\times3}=\frac{-15}{36},\)

\(\frac{-7}{18}=\frac{-7}{18}\times\frac22=\frac{-7\times2}{18\times2}=\frac{-14}{36}\) and

\(\frac{-2}3=\frac{-2}3\times\frac{12}{12}=\frac{-2\times12}{3\times12}=\frac{-24}{36}.\)

Since, -24 < -16 < -15 < -14

\(\Rightarrow\) \(\frac{-24}{36}<\frac{-16}{36}<\frac{-15}{36}<\frac{-14}{36}\)

\(\Rightarrow\) \(\frac{-2}{3}<\frac{-4}{9}<\frac{-5}{12}<\frac{-7}{18}.\)

Thus, biggest number among all given rational numbers is \(\frac{-7}{18}.\)

Correct option is  A) -7/18

(i)

Since the denominators of both the numbers are different 

therefore, we will take their LCM 0f 3 and 4 = 12

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

And,

\(\frac{1}{3}= \frac{1\times4}{3\times4}=\frac{4}{12}\)

Therefore,

= \(\frac{1}{3}-\frac{3}{4}\)

= \(\frac{4}{12}-\frac{9}{12}\)

= \(\frac{4-9}{12}\)

= \(\frac{-5}{12}\)

(ii)

Since the denominators of both the numbers are different 

therefore, we will take their LCM 0f 6 and 3 = 6

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

And,

\(\frac{1}{3}= \frac{1\times2}{3\times2}=\frac{2}{6}\)

Therefore,

\(\frac{1}{3}-(\frac{-5}{6})\)

= \(\frac{2}{6}-(\frac{-5}{6})\)

= \(\frac{2-(-5)}{6}\)

= \(\frac{2+5}{6}\)

= \(\frac{7}{6}\)

(iii)

Since the denominators of both the numbers are different 

therefore, we will take their LCM 0f 9 and 5 = 45

\(\frac{-8}{9}= \frac{-8\times5}{9\times5}=\frac{-40}{45}\)

And,

\(\frac{-3}{5}= \frac{-3\times9}{5\times9}=\frac{-27}{45}\)

Therefore,

\(\frac{-3}{5}-(\frac{-8}{9})\)

= \(\frac{-27}{45}-(\frac{-40}{45})\) 

= \(\frac{-27-(-40)}{45}\) 

= \(\frac{-27+40}{45}\)

= \(\frac{13}{45}\)

(iv)

We can write, -1 = \(\frac{-1}{1}\)

Since the denominators of both the numbers are different 

therefore, we will take their LCM 0f 1 and 7 =7

\(\frac{-1}{1}= \frac{-1\times7}{1\times7}=\frac{-7}{7}\)

And,

\(\frac{-9}{7}= \frac{-9\times1}{7\times1}=\frac{-9}{7}\)

Therefore,

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

= \(\frac{-7}{7}-(\frac{-9}{7})\)

= \(\frac{-7+9}{7}\)

= \(\frac{2}{7}\) 

Correct option is (B) 1/20

\(\frac{3}{8}+\frac{-2}{5}+\frac{7}{8}-\frac{4}{5}\) \(=\frac{3\times5-2\times8+7\times5-4\times8}{40}\) \(=\frac{15-16+35-32}{40}\) \(=\frac{-1+3}{40}=\frac{2}{40}=\frac{1}{20}.\)

Correct option is  B) 1/20 

Correct option is (A) -1/4

\(\frac{7}{20}\) - \(\frac{3}{5}\) = \(\frac{7-3\times4}{20}=\frac{7-12}{20}\) \(=\frac{-5}{20}=\frac{-1}{4}.\)

Correct option is  A) -1/4 

Correct option is (C)  −19/12

Let we subtract x from \(\frac{-3}{4}\) to get \(\frac{5}{6}\).

Thus, \(\frac{-3}{4}\) - x = \(\frac{5}{6}\)

\(\Rightarrow\) x = \(\frac{-3}{4}\) - \(\frac{5}{6}\) \(=\frac{-3\times3-5\times2}{12}=\frac{-9-10}{12}=\frac{-19}{12}.\)

Correct option is   C) \(\frac{-19}{12}\)

−2 can be represented as -14/7.

Therefore, five rational numbers greater than −2 are

-13/7, -12/7, -11/7 -10/7, -9/7

i) The additive inverse of \(\frac{-2}{17}\)   is \(\frac{2}{17}\)

ii) The additive inverse of \(\frac{3}{-11 }\) is \(\frac{3}{11}\)

iii) The additive inverse of \(\frac{-17}{5}\) is \(\frac{17}{5}\)

iv) The additive inverse of \(\frac{-11}{-25}\) is \(\frac{-11}{25}\)

(i) Rearranging and grouping the numbers in pairs in such a way that each group contains a pair of rational numbers with equal denominator

We have,

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

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

= \(\frac{-6}{15}+\frac{30}{15}\)

= \(\frac{24}{15}\)

= \(\frac{8}{5}\)

(ii) Rearranging and grouping the numbers in pairs in such a way that each group contains a pair of rational numbers with equal denominator

We have,

\(\frac{3}{7}+\frac{-4}{9}+\frac{-11}{7}+\frac{7}{9}\)

\(\frac{3}{7}-\frac{11}{7}+\frac{7}{9}-\frac{4}{9}\) = \(\frac{-8}{7}+\frac{3}{9}\)

= \(\frac{-8}{7}+\frac{1}{3}\)

= \(\frac{-24}{21}+\frac{7}{21}\)

= \(\frac{-17}{21}\)

(iii) Rearranging and grouping the numbers in pairs in such a way that each group contains a pair of rational numbers with equal denominator

We have,

\(\frac{2}{5}+\frac{8}{3}+\frac{-11}{15}+\frac{4}{5}+\frac{-2}{3}\)

\(\frac{2}{5}+\frac{4}{5}+\frac{8}{3}-\frac{2}{3}-\frac{11}{15}\) = \(\frac{6}{5}+\frac{8-2}{3}-\frac{11}{15}\)

= \(\frac{6}{5}+\frac{6}{3}+\frac{-11}{15}\)

= \(\frac{18}{15}+\frac{30}{15}-\frac{11}{15}\)

= \(\frac{18+30-11}{15}\)

= \(\frac{37}{15}\)

(iv) Rearranging and grouping the numbers in pairs in such a way that each group contains a pair of rational numbers with equal denominator

We have,

\(\frac{4}{7}+0+\frac{-8}{9}+\frac{-13}{7}+\frac{17}{21}\)

\(\frac{4}{7}-\frac{13}{7}+0+\frac{-8}{9}+\frac{17}{21}\) = \(\frac{4-13}{7}+\frac{17}{21}-\frac{8}{9}\)

= \(\frac{-9}{7}+\frac{17}{21}-\frac{8}{9}\)

= \(\frac{-27+17}{21}-\frac{8}{9}\)

= \(\frac{-10\times9}{21\times 9}-\frac{8\times 21}{9\times 21}\)

= \(\frac{-90}{189}-\frac{168}{189}\)

= \(\frac{-258}{189}\)

i) As the property states \((x+y)+z=x+(y+z)\)

We use value as

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

Let us consider L.H.S  \((\frac{1}{2}+\frac{2}{3})+\frac{-1}{5}\\\)

L.C.M for 2 and 3 is 6

\(\frac{(1\times3)}{(2\times3)}+\frac{(2\times2)}{(3\times2)}\\=\frac{3}{6}+\frac{4}{6}\\=\frac{7}{6}\\=\frac{7}{6}+\frac{-1}{5}\) (Since the denominators are same we add it)
L.C.M for 6 and 5 is 30

\(\frac{(7\times5)}{(6\times5)}+\frac{(-1\times6)}{(5\times6)}\\=\frac{35}{30}+\frac{-6}{30}\\=\frac{(35+(-6))}{30}\\=\frac{(35-6)}{30}\\=\frac{29}{30}\)(Since denominators are same we add it)

Let us consider R.H.S \(=\frac{1}{2}+(\frac{2}{3}+\frac{-1}{5})\)

L.C.M for 3 and 5 is 15

\(\frac{2}{3}+\frac{-1}{5}\\=\frac{(2\times5)}{(3\times5)}+\frac{(-1\times3)}{(5\times3)}\\=\frac{10}{15}+\frac{-3}{15}\)

\(\frac{10}{15}+\frac{-3}{15}\\=\frac{(10-3)}{15}\\=\frac{7}{15}\)    (Since the denominators are same we add it)

L.C.M for 2 and 15 is 30

\(\frac{1}{2}+\frac{7}{15}\\=\frac{(1\times15)}{(2\times15)}+\frac{(7\times2)}{(15\times2)}\\=\frac{15}{30}+\frac{14}{30}\\=\frac{29}{30}\)   (Since denominator is same we add it) 

\(\therefore\) L.H.S = R.H.S (associativity of addition of rational number is verified.

ii) As the property states \((x+y)+z=x+(y+z)\)

We use value as

\((\frac{-2}{5}+\frac{4}{3})+\frac{-7}{10}\\=\frac{-2}{5}+(\frac{4}{3 }+\frac{-7}{10})\)

Let us consider L.H.S \((\frac{-2}{5}+\frac{4}{3})+\frac{-7}{10}\\ \)

 LCM for 5 and 3 is 15

\(\frac{(-2\times3)}{(5\times3)}+\frac{(4\times5)}{(3\times5)}\\=\frac{-6}{15}+\frac{20}{15}\\=\frac{-6}{15}+\frac{20}{15}\\=\frac{(-6+20)}{15}\\=\frac{14}{15}\) ( Since denominators are same we add it)

Now,

\(\frac{14}{15}+\frac{-7}{10}\)

L.C.M for 15 and 30 is 30

\(\frac{(14\times2)}{(15\times2)}+\frac{(-7\times3)}{(10\times3)}\\=\frac{28}{30}+\frac{-21}{30}\\=\frac{(28+(-21))}{30}\\=\frac{(28-21)}{30}\\=\frac{7}{30}\) (Since the denominators are same we add it)
Let us take R.H.S \(\frac{-2}{5}+(\frac{4}{3 }+\frac{-7}{10})\)

L.C.M for 3 and 10 is 30

\(\frac{4}{3}+\frac{-7}{10}\\=\frac{(4\times10)}{(3\times10)}+\frac{(-7\times3)}{(10\times3)}\\=\frac{40}{30}+\frac{-21}{30}\\=\frac{40}{30}+\frac{-21}{30}\\=\frac{(40-21)}{30}\\=\frac{19}{30}\) ( Since the denominators are same we add it)

Now,

\(\frac{-2}{5}+\frac{19}{30}\)

LCM for 5 and 30 is 30

\(\frac{-2}{5}+\frac{19}{30}\\=\frac{(-2\times6)}{(5\times6)}+\frac{(19\times1)}{(30\times1)}\\=\frac{-12}{30}+\frac{19}{30}\\=\frac{(-12+19)}{30}\\=\frac{7}{30}\) (Since denominators are same we add it)

 \(\therefore\) L.H.S = R.H.S (associativity of addition of rational number is verified.

iii) As the property states \((x+y)+z=x+(y+z)\)

We use value as

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

Let us consider L.H.S

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

 LCM for 1 and 5 is 5

\(\frac{(-2\times5)}{(1\times5)}+\frac{(3\times1)}{(5\times1)}\\=\frac{-10}{5}+\frac{3}{5}\\=\frac{(-10+3)}{5}\\=\frac{-7}{5}\)

Now,

\(\frac{-7}{5}+\frac{-4}{3}\)

LCM for 5 and 3 is 15

\(\frac{(-7\times3)}{(5\times3)}+\frac{(-4\times5)}{(3\times5)}\\=\frac{-21}{15}+\frac{-20}{15}\\=\frac{(-21+(-20))}{15}\\=\frac{-21-20}{15}\\=\frac{-41}{15}\) (Since the denominators are same)

Let us consider R.H.S

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

LCM for 5 and 3 is 15

\(\frac{-3}{5}+\frac{-4}{3}\\=\frac{(3\times3)}{(5\times3)}+\frac{(-4\times5)}{(3\times5)}\\=\frac{9}{15}+\frac{-20}{15}\\=\frac{(9-20)}{15}\\=\frac{-11}{15}\) (Since denominators are same)

Now,

\(\frac{-2}{1}+\frac{-11}{15}\\=\frac{(-2\times15)}{(1\times15)}+\frac{(-11\times1)}{(15\times1)}\\=\frac{-30}{15}+\frac{-11}{15}\\=\frac{-41}{15}\)

 \(\therefore\) L.H.S = R.H.S (associativity of addition of rational number is verified.

i) We classified the rational number with same denominators.

= (2+(-4))/5 + (7+(-1))/3
= (2-4)/5 + (7-1)/3
= -2/5 + 6/3

L.C.M for 5 and 3 is 15

\(= \frac{(-2\times3)}{(5\times3)}+\frac{(6\times5)}{(3\times5)}\\=\frac{-6}{15}+\frac{30}{15}\)

Now the given denominators are same, we add it.

\(\frac{(-6+30)}{15}\\=\frac{24}{15}\\=\frac{8}{5}\)

ii) We classified the rational number with same denominators.

 = 3/7 + (-11/7) + (-4/9) + 7/9

\(=\frac{(3+(-11))}{7}+\frac{(-4+7)}{9}\\=\frac{-8}{7}+\frac{3}{9}\\=\frac{-8}{7}+\frac{1}{3}\)

L.C.M for 7 and 3 is 21

\(=\frac{(-8\times3)}{(7\times3)}+\frac{(1\times7)}{(3\times7)}\\=\frac{-24}{21}+\frac{7}{21}\\=\frac{(-24+7)}{21}\\=\frac{-17}{21}\)

iii) We classified the rational number with same denominators.

2/5 + 4/5 + 8/3 + (-2/3) + (-11/15)

Now the given denominators are same, we add it.

\(=\frac{(2+4)}{5}+\frac{(8+(-2))}{3}+\frac{-11}{15}\\=\frac{6}{5}+\frac{(8-2)}{3}+\frac{-11}{15}\\=\frac{6}{5}+\frac{2}{1}+\frac{-11}{15}\)

L.C.M for 5,1 and 15 is 15

\(=\frac{(6\times3)}{(5\times3)}+\frac{(2\times15)}{(1\times15)}+\frac{(-11\times1)}{(15\times1)}\\=\frac{-18}{15}+\frac{30}{15}+\frac{-11}{15}\\=\frac{(18+30+(-11))}{15}\\=\frac{(18+30-11)}{15}\\=\frac{37}{15}\)

(i) In order to verify this property, let us consider the following expressions:

Verification:

\(\frac{1}{2}+[\frac{2}{3}+(\frac{-1}{5})] = \frac{1}{2}+[\frac{10}{15}-\frac{3}{15}]\)

= \(\frac{1}{2}+\frac{7}{15}\)

= \(\frac{15+14}{30}\)

= \(\frac{29}{30}\)

And,

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

= \(\frac{7}{6}-\frac{1}{5}\)

= \(\frac{35-6}{30}\)

= \(\frac{29}{30}\)

Therefore,

The associative property of additional of rational numbers has been verified

(ii) In order to verify this property, let us consider the following expressions:

Verification:
\(\frac{-2}{5}+[\frac{4}{3}+(\frac{-7}{10})]=\frac{-2}{5}+[\frac{40}{30}-\frac{21}{30}]\)

= \(\frac{-12}{30}+\frac{19}{30}\)

= \(\frac{-12+19}{30}\)

= \(\frac{7}{30}\)

And,

\((\frac{-2}{5}+\frac{4}{3})+(\frac{-7}{10})=(\frac{-6}{15}+\frac{20}{15})-\frac{7}{10}\)

= \(\frac{14}{15}-\frac{7}{10}\)

= \(\frac{28-21}{30}\)

= \(\frac{7}{30}\)

Therefore,

The associative property of additional of rational numbers has been verified

(iii) In order to verify this property, let us consider the following expressions:

Verification:

\(\frac{-7}{11}+[\frac{2}{-5}+(\frac{-3}{22})]=\frac{-7}{11}+[\frac{44}{-110}-\frac{15}{110}]\):

= \(\frac{-7}{11}-\frac{29}{110}\)

= \(\frac{-70-29}{110}\)

= \(\frac{-99}{110}\)

And,

\((\frac{-7}{11}+\frac{2}{-5})+(\frac{-3}{22})\) \(=(\frac{-35}{55}-\frac{22}{35})-\frac{3}{22}\)

= \(\frac{-57}{55}-\frac{3}{22}\)

= \(\frac{-114+15}{110}\)

= \(\frac{-99}{110}\)

Therefore,

The associative property of additional of rational numbers has been verified

(iv) In order to verify this property, let us consider the following expressions:

Verification:

\(-2+[\frac{3}{5}+(\frac{-4}{3})]\) = \(-2+[\frac{9}{15}-\frac{20}{15}]\)

= \(-2-\frac{11}{15}\)

= \(\frac{-30-11}{15}\)

= \(\frac{-41}{15}\)

And,

\((-2+\frac{3}{5})+(\frac{-4}{3})\) = \(=(\frac{-10}{5}+\frac{3}{35})-\frac{4}{3}\)

= \(\frac{-7}{5}-\frac{4}{3}\)

= \(\frac{-21-20}{15}\)

= \(\frac{-41}{15}\)

Therefore, The associative property of additional of rational numbers has been verified

i) The LCM for 9 and 6 is 18.

\(\frac{(8\times2)}{(9\times2)}+\frac{(-11\times3)}{(6\times3)}\\\frac{16}{18}+\frac{-33}{18}\)

Since the denominators are same we can add them directly.We get

\(\frac{(16-33)}{18}=\frac{-17}{18 }\)

ii)  Firstly convert the denominator to positive number.

\(\frac{5}{-7}=\frac{(5\times-1)}{(-7\times-1)}\\\frac{-5}{7}\\=\frac{3}{1}+\frac{-5}{7}\)

The LCM for 1 and 7 which is 7

\(\frac{(3\times7)}{(1\times7)} +\frac{(-5\times1)}{(7\times1)}\\=\frac{21}{7}+\frac{-5}{7}\)

Since the denominators are same we can add them directly. We get

\(\frac{(21-5)}{7}=\frac{16}{7}\)

iii) Firstly convert the denominator to positive number then,

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

Now,

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

So, \(\frac{-1}{12}+\frac{-2}{15}\)

The LCM for 12 and 15 which is 60.Now

\(\frac{(-1\times5)}{(12\times5)}+\frac{(-2\times4)}{(15\times4)}\\=\frac{-5}{60}+\frac{-8}{60}\)

Since the denominators are same we can add them.

\(\frac{(-5-8)}{60}=\frac{-13}{60}\)

iv) As we know that anything is added to 0 results is the same.

\(0+\frac{-3}{5}\\=\frac{-3}{5}\)

v) The LCM for 1 and 5  is 5

\(\frac{(1\times5)}{(1\times5)}+\frac{(-4\times1)}{(5\times1)}\\=\frac{5}{5}+\frac{-4}{5}\)

Since the denominators are same we can add them directly.We get

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

i) Using the commutativity law, the addition of rational numbers is commutative.

\(\therefore\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}\)

let us consider the given fraction

\(\frac{-11}{5}\) and \(\frac{4}{7}\) as

\(\frac{-11}{5}\)+ \(\frac{4}{7}\)and \(\frac{4}{7}\)+\(\frac{-11}{5}\)

The denominators are 5 and 7.The LCM of 5 and 7 is 35
We rewrite the given fraction in order to get the same denominator.We get

Now,\(\frac{-11}{5}\) = \(\frac{(11\times7)}{(5\times7)}\)

 \(=\frac{(-11\times7)}{(5\times7)}\\=\frac{-77}{35}\\\frac{4}{7}=\frac{(4\times5)}{(7\times5)}\\=\frac{20}{35}\)

Since the denominators are same we can add them directly

\(\frac{-77}{35}+\frac{20}{35}\\=\frac{(-77+20)}{35}\\=\frac{-50}{35}\)

ii) Firstly we need to convert the denominators to positive numbers.We get

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

Using the commutativity law, the addition of rational numbers is commutative.

\(\therefore\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}\)

In order to verify the above property let us consider the given fraction.

\(\frac{4}{9}\) and \(\frac{-7}{12}\) as

\(\frac{4}{9}\)+\(\frac{-7}{12}\) and \(\frac{-7}{12}\)+ \(\frac{4}{9}\)

The denominators are 9 and 12. By taking LCM for 9 and 12 is 36
We rewrite the given fraction in order to get the same denominator.We get

Now , \(\frac{4}{9}=\frac{(4\times4)}{(9\times4)}\\=\frac{16}{36}\)

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

= \(\frac{-7}{12}\)+\(\frac{4}{9}\)

The denominators are 12 and 9 By taking LCM for 12 and 9 is 36.
We rewrite the given fraction in order to get the same denominator.We get

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

\(\frac{4}{9}=\frac{(4\times4)}{9\times4)}\\=\frac{16}{36}\)

Since the denominators are same we can add them directly

\(\frac{-21}{36}+\frac{16}{36}=\frac{(-21+16)}{36}\\=\frac{-5}{36}\)

\(\therefore\frac{4}{9}+\frac{-7}{12}=\frac{-7}{12}+\frac{4}{9}\)  is satisfied

iii) Firstly we need to convert the denominators to positive numbers.

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

= \(\frac{-4}{7}\)

By using the commutativity law, the addition of rational numbers is commutative

\(\therefore\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}\)

In order to verify the above property let us consider the given fraction

\(\frac{-4}{1}\) and \(\frac{-4}{7}\) as

\(\frac{-4}{1}\)+ \(\frac{-4}{7}\) and \(\frac{-4}{7}\) + \(\frac{-4}{1}\)

The denominators are 1 and 7 By taking LCM for 1 and 7 is 7.
We rewrite the given fraction in order to get the same denominator.We get

Now,

\(\frac{-4}{1}=\frac{(-4\times7)}{(1\times7)}\\=\frac{-28}{7}\)

\(\frac{-4}{7}=\frac{(-4\times1)}{(7\times1)}\\=\frac{-4}{7}\)

Since the denominators are same we can add them directly
\(\frac{-28}{7}+\frac{-4}{7}=\frac{(-28+(-4))}{7}\\=\frac{(-28-7)}{7}\\=\frac{-32}{7}\)

\(\frac{-4}{7}+-\frac{4}{1}\)

The denominators are 7 and 1. By taking LCM for 7 and 1 is 7. We rewrite the given fraction in order to get the same denominator.

Now,

\(\frac{-4}{7}=\frac{(-4\times1)}{(7\times1)}\\=\frac{-4}{7}\)

\(\frac{-4}{1}=\frac{(-4\times7)}{(1\times7)}\\=\frac{-28}{7}\)

Since the denominators are same we can add them directly.We get

\(\frac{-4}{7}+\frac{-28}{7}=\frac{(-4+(-28))}{7}\\=\frac{(-4-28)}{7}\\=\frac{-32}{7}\)

\(\therefore \frac{-4}{1}+\frac{-4}{7}=\frac{-4}{7}+\frac{-4}{1}\) is satisfied

(i) The addition of rational number is commutative

i.e, if \(\frac{a}{b}\) and \(\frac{c}{d}\) are any two rational numbers, then

\(\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}\)

Verification: In order to verify this property,

Let us consider two expressions:

\(\frac{-11}{5}+\frac{4}{7}\)

And,

\(\frac{4}{7}+\frac{-11}{5}\)

We have:

\(\frac{-11}{5}+\frac{4}{7}=\frac{-77}{35}+\frac{20}{35}\)

= \(\frac{-77+20}{35}\)

= \(\frac{-57}{35}\)

And,

\(\frac{4}{7}+\frac{-11}{5}=\frac{20}{35}+\frac{-77}{35}\)

= \(\frac{20-77}{35}\)

= \(\frac{-57}{35}\)

Therefore,

\(\frac{-11}{5}+\frac{4}{7}=\frac{4}{7}+\frac{-11}{5}\)

(ii) The addition of rational number is commutative i.e, if \(\frac{a}{b}\) and \(\frac{c}{d}\) are any two rational numbers, then

\(\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b}\)

Verification: In order to verify this property,

Let us consider two expressions:

\(\frac{4}{9}+\frac{-7}{12}\)

And,

\(\frac{-7}{12}+\frac{4}{9}\)

We have:

\(\frac{4}{9}+\frac{-7}{12}=\frac{16}{36}+\frac{-21}{36}\)

= \(\frac{16-21}{36}\)

= \(\frac{-5}{36}\)

And,

\(\frac{-7}{12}+​​\frac{4}{9}=\frac{-21}{36}+\frac{16}{36}\)

\(=\frac{-21+16}{36}\)

= \(\frac{-5}{36}\)

Therefore,

\(\frac{4}{9}+\frac{-7}{12}=\frac{-7}{12}+\frac{4}{9}\)

(iii) The addition of rational number is commutative i.e, if \(\frac{a}{b}\) and \(\frac{c}{d}\)are any two rational numbers, then

\(\frac{a}{b}+​​\frac{c}{d}=\frac{c}{d}+\frac{a}{b}\)

Verification: In order to verify this property,

Let us consider two expressions:

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

And,

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

We have:

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

= \(\frac{-9+2}{15}\)

= \(\frac{-7}{15}\)

And,

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

= \(\frac{2-9}{15}\)

= \(\frac{-7}{15}\)

Therefore,

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

(iv) The addition of rational number is commutative i.e, if \(\frac{a}{b}\) and \(\frac{c}{d}\) are any two rational numbers, then

\(\frac{a}{b}+​​\frac{c}{d}=\frac{c}{d}+\frac{a}{b}\)

Verification: In order to verify this property,

Let us consider two expressions:

\(\frac{2}{-7}+\frac{12}{-35}\)

And,

\(\frac{12}{-35}+\frac{2}{-7}\)

We have:

\(\frac{2}{-7}+​​\frac{12}{-35}=\frac{-10}{35}+\frac{-12}{35}\)

= \(\frac{-10-12}{35}\)

= \(\frac{-22}{35}\)

And,

\(\frac{12}{-35}+​​\frac{2}{7}=\frac{-12}{35}+\frac{-10}{35}\)

= \(\frac{-12-10}{35}\)

= \(\frac{-22}{35}\)

Therefore,

\(\frac{2}{-7}+​​\frac{12}{-35}=\frac{12}{-35}+\frac{2}{-7}\)

(v) The addition of rational number is commutative i.e, if \(\frac{a}{b}\) and \(\frac{c}{d}\) are any two rational numbers, then

\(\frac{a}{b}+​​\frac{c}{d}=\frac{c}{d}+\frac{a}{b}\)

Verification: In order to verify this property, Let us consider two expressions:

\(4+\frac{-3}{5}\)

And,

\(\frac{-3}{5}+4\)

We have:

\(4+\frac{-3}{5}\)=\(\frac{20}{5}-\frac{3}{5}\)

= \(\frac{20-3}{5}\)

= \(\frac{17}{5}\)

And,

\(\frac{-3}{5}+4\)= \(\frac{-3}{5}+\frac{20}{5}\)

= \(\frac{-3+20}{5}\)

= \(\frac{17}{5}\)

Therefore,

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

(vi) The addition of rational number is commutative i.e, if \(\frac{a}{b}\) and \(\frac{c}{d}\) are any two rational numbers, then

\(\frac{a}{b}+​​\frac{c}{d}=\frac{c}{d}+\frac{a}{b}\)

Verification: In order to verify this property,

Let us consider two expressions:

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

And,

\(\frac{4}{-7}-4\)

We have:

\(-4+\frac{4}{-7}\)=\(\frac{-28}{7}-\frac{4}{7}\)

= \(\frac{-28-4}{7}\)

= \(\frac{-32}{7}\)

And,

\(\frac{4}{-7}-4\)=\(\frac{-4}{7}-\frac{28}{7}\)

= \(\frac{-4-28}{7}\)

= \(\frac{-32}{7}\)

Therefore,

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

i)  We add the given fraction

\(\frac{12}{5}+\frac{43}{10}\)

The L.C.M of 5 and 10  is 10

\(\frac{(-12\times2)}{(5\times2)}+\frac{(43\times1)}{(10\times1)}\\=\frac{-24}{10}+\frac{43}{10}\)

The denominators are same we can add them directly.We get

\(\frac{(-24+43)}{10}=\frac{19}{10}\)

\(\therefore\)  \(\frac{19}{10}\)can be written in mixed fraction as \(\frac{9}{10}{1}\)

ii)  We add the given fraction

\(\frac{-31}{6}+\frac{-27}{8}\)

The L.C.M of  6 and 8  is 24

\(\frac{(-31\times4)}{(6\times4)}+\frac{(-27\times3)}{(8\times3)}\\=\frac{-124}{24}+\frac{-81}{24}\)

The denominators are same we can add them directly.We get

\(\frac{(-124-81)}{24}=\frac{-205}{24}\)

\(\frac{-205}{24}\) can be written in mixed fraction as \(-8\frac{13}{24}\)

(i) The denominators of the given rational numbers 5 and 10 respectively.

The L.C.M of 5 and 10 is 10

Now,

We write the given rational numbers into forms in which both of them have the same denominator

\(\frac{-12\times2}{5\times2}=\frac{-24}{10}\)

And,

\(\frac{43\times1}{10\times1}=\frac{43}{10}\)

Therefore,

\(\frac{-24}{10}+\frac{43}{10}=\frac{-24+43}{10}\)

= \(\frac{19}{10}\)

= \(1\frac{9}{10}\)

(ii) The denominators of the given rational numbers 7 and 4 respectively.

The L.C.M of 7 and 4 is 28

Now,

We write the given rational numbers into forms in which both of them have the same denominator

\(\frac{24\times4}{7\times4}=\frac{96}{28}\)

And,

\(\frac{-11\times7}{4\times7}=\frac{-77}{28}\)

Therefore,

\(\frac{96}{28}-\frac{77}{28}=\frac{96-77}{28}\)

= \(\frac{19}{28}\)

= \(1\frac{9}{10}\)

(iii) The denominators of the given rational numbers 6 and 8 respectively.

The L.C.M of 6 and 8 is 24

Now,

We write the given rational numbers into forms in which both of them have the same denominator

\(\frac{-31\times4}{6\times4}=\frac{-124}{24}\)

And,

\(\frac{-27\times3}{8\times3}=\frac{-81}{24}\)

Therefore,

\(\frac{-124}{24}-\frac{81}{24}=\frac{-124-81}{24}\)

= \(\frac{-205}{24}\)

 \(-8\frac{13}{24}\)

(iv) The denominators of the given rational numbers 6 and 8 respectively.

The L.C.M of 6 and 8 is 24

Now,

We write the given rational numbers into forms in which both of them have the same denominator

\(\frac{101\times4}{6\times4}=\frac{404}{24}\)

And,

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

Therefore,

\(\frac{404}{24}+\frac{21}{24}=\frac{404+21}{24}\)

= \(\frac{425}{24}\)

= \(17\frac{17}{24}\)

(d) Commutative

(a) Associative under multiplication

 (i) 315+115 = 430

Closure property of addition

(ii) \(\frac{3}{4}\times \frac{9}{5} = \frac{27}{20}\)

Closure property of multiplication.

(iii) 5 + 0 = 0 + 5 = 5

0 is the additive identity.

(iv) \(\frac{8}{9}\times1 = \frac{8}{9}\)

1 is the multiplicative identity

(v) \(\frac{8}{17} + \frac{-8}{17} = 0\)

\(\frac{-8}{17}\)

(vi) \(\frac{22}{23}\times\frac{23}{22} = 1\)

\(\frac{23}{22}\) is the multiplicative identity

(i) true 

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

Then,

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

which is also a rational number

Hence, 

Rational numbers are always closed under subtraction.

(ii) false

\(\frac{a}{0}=\infty\)

Hence, 

Rational numbers are not always closed under division.

(iii) false

\(\frac{1}{0}=\infty\)

Hence, \(\frac{1}{0}=\infty\) 

(iv) false

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

Then,

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

And

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

Therefore,

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

Hence, 

Subtraction is not commutative on rational numbers.

(v) true

\((\frac{-7}{8})=-1\times\frac{-7}{8}=\frac{7}{8}\)

The L.C.M of denominators 5 and 4 is 20

Converting the given rational numbers into equivalent rational number having common denominator 20

we get:

\(\frac{3}{5}\times \frac{20}{20}\) = \(\frac{60}{100}\)

 \(\frac{3}{4}\times \frac{25}{25}\) = \(\frac{75}{100}\)

Clearly,

61, 62, 63,…, 74 are integers between numerators 60 and 75 of these equivalent rational numbers

Thus we have

\(\frac{61}{100}\),\(\frac{62}{100}\),.......,\(\frac{74}{100}\)

As rational number between \(\frac{3}{5}\) and \(\frac{3}{4}\)

We can take only 10 of these as required rational numbers

\(\frac{61}{100}\),\(\frac{62}{100}\),\(\frac{63}{100}\),...........,\(\frac{73}{100}\),\(\frac{74}{100}\)

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, (-1/5) × ((2/15) + (-3/10)) = ((-1/5) × (2/15)) + ((-1/5) × (-3/10))

LHS = (-1/5) × ((2/15) + (-3/10))

= (-1/5) × ((4 – 9)/30)

= (-1/5) × (-5/30)

= (-1/1) × (-1/30)

= 1/30

RHS = ((-1/5) × (2/15)) + ((-1/5) × (-3/10))

= (-2/75) + (3/50)

= (-4 + 9)/150

= 5/150

= 1/30

By comparing LHS and RHS

LHS = RHS

∴ 1/30 = 1/30

Hence x × (y + z) = x × y + x × z

(i) We have,

x = \(\frac{-7}{3}\), y = \(\frac{12}{5}\) and z = \(\frac{4}{9}\)

= x x (y x z) = \(\frac{-7}{3}\times (\frac{12}{5}\times \frac{4}{9})\)

= \(\frac{-7}{3}(\frac{48}{45})\)

= \(\frac{-112}{45}\)

(x x y) x z = \((\frac{-7}{3}\times \frac{12}{5})\times \frac{4}{9}\)

= \(\frac{-7}{3}(\frac{48}{45})\)

= \(\frac{-112}{45}\)

(ii) We have,

 x = 0, y = \(\frac{-3}{5}\) and z = \(\frac{-9}{4}\)

= x x (y x z) = \(0\times (\frac{-3}{5}\times \frac{-9}{4})\)

 = 0

 (x x y) x z = \((0\times \frac{-3}{5})\times \frac{-9}{4}\)

 = 0

(iii) We have,

 x = \(\frac{1}{2}\), y = \(\frac{5}{-4}\) and z = \(\frac{-7}{5}\)

= x x (y x z) = \(\frac{1}{2}\times (\frac{5}{-4}\times \frac{-7}{5})\)

= \(\frac{1}{2}(\frac{7}{4})\)

= \(\frac{7}{8}\)

 (x x y) x z = \((\frac{1}{2}\times \frac{5}{-4})\times \frac{-7}{5}\)

= \(\frac{-5}{8}(\frac{-7}{5})\)

= \(\frac{7}{8}\)

(iv) We have,

  x = \(\frac{5}{7}\), y = \(\frac{-12}{13}\) and z = \(\frac{-7}{18}\)

= x x (y x z) = \(\frac{5}{7}\times (\frac{-12}{13}\times \frac{-7}{18})\)

 = \(\frac{10}{39}\)

  (x x y) x z = \((\frac{5}{7}\times \frac{-12}{13})\times \frac{-7}{8}\)

= \(\frac{60}{91}(\frac{-7}{18})\)

= \(\frac{10}{39}\)

Correct answer is (c) 4/6

36 trousers of equal sizes can he stitched with 64 mts of cloth, then the length of the cloth ¡s required for each trouser

= 64 ÷ 36

= \(\frac{64}{36}=\frac{16}9= 1\frac79\)

No. of trousers = 24

Total length of cloth = 54

Length of cloth required for each trousers = \(\frac{Total\,length\,of\,cloth}{No.\,of\,trousers}\)

= \(\frac{54}{12}\)

 = \(\frac{9}{2}\)m

Given

((-13/5) + (12/7)) ÷ (-31/7) x (-1/2)

= ((-13/5) × (7/7) + (12/7) × (5/5)) ÷ (31/14)

= ((-91/35) + (60/35)) ÷ (31/14)

= (-31/35) ÷ (31/14)

= (-31/35) × (14/31)

= (-14/35)

= (-2/5)

The sum is given -13/5 + 12/7

= -13/5 + 12/7

((-13×7) + (12×5)) /35

(-91+60)/35

= -31/35

Now, Product of -31/7 and -1/2

= -31/7 × -1/2

= (-31×-1)/(7×2)

= 31/14

∴ by dividing the sum and the product we get the value as,

= (-31/35) / (31/14)

= (-31/35) × (14/31)

= (-31×14)/(35×31)

= -14/35

= -2/5

Suppose x be the rational number to be subtracted to \(\frac{-5}{3}\) to get \(\frac{5}{6}\)

Then,

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

\(\frac{-5}{3}\) - \(\frac{x\times3}{3}\) = \(\frac{5}{6}\)

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

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

-5-3x =  \(\frac{15}{6}\)

-3x = \(\frac{-15}{6}\)+ 5

-3x = \(\frac{15\times 1+5\times 6}{6}\)

-3x = \(\frac{15+30}{6}\)

-3x = \(\frac{45}{6}\)

-18x = 45

x = \(\frac{-45}{18}\)

x = \(\frac{-5}{2}\)

Therefore,

The required number x = \(\frac{-5}{2}\)

Given sum of two numbers = (-4/3)

One of them is -5

Let the required number be x

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

LCM of 1 and 3 is 3

(-5/1) = (-5/1) × (3/3) = (-15/3)

On substituting

x + (-15/3) = (-4/3)

x = (-4/3) – (-15/3)

x = (-4/3) + (15/3)

x = (-4 + 15)/3

x = (11/3)