NCERT solution for class 8 maths rational numbers ( Chapter 1)

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Solution for Exercise 1.1

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

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Answer :

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

1.Using appropriate properties find:
(i)\(-\frac{2}3×\frac{3}5+\frac{5}2-\frac{3}5×\frac{1}6 \)

(ii)\(\frac{2}5×(-\frac{3}7)-\frac{1}6×\frac{3}2+\frac{1}{14}×\frac{2}5 \)


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Answer :

(i) We have: \(-\frac{2}3×\frac{3}5+\frac{5}2-\frac{3}5×\frac{1}6 \)

\(= -\frac{2}3×\frac{3}5-\frac{3}5×\frac{1}6+\frac{5}2 (By \quad regrouping) \)

\(= \frac{3}5×(-\frac{2}3-\frac{1}6)+\frac{5}2 (by\quad distributive\quad property) \)

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

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

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

\(= -\frac{3×5}{5×6}+\frac{5}2 \)

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

so, the value obtained is =2.

(ii) we have:\(\frac{2}5×(-\frac{3}7)-\frac{1}6×\frac{3}2+\frac{1}{14}×\frac{2}5 \)

\(= \frac{2}5×(-\frac{3}7)+\frac{1}{14}×\frac{2}5 -\frac{1}6×\frac{3}2 (by \quad regrouping)\)

\(= \frac{2}5×(\frac{-3}7+\frac{1}{14})-\frac{1}6×\frac{3}2 (using \quad distributive \quad property)\)

\(= \frac{2}5×(\frac{-3×2}{7×2}+\frac{1×1}{14×1})-\frac{1}6×\frac{3}2 \)

\(= \frac{2}5×(\frac{-6}{14}+\frac{1}{14})-\frac{1}6×\frac{3}2\)

\(= \frac{2}5×(\frac{-6+1}{14})-\frac{3}{12}\)

\(= \frac{2}5×(\frac{-5}{14})-\frac{1}{4}\)

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

\(= \frac{-4}{28}-\frac{7}{28}=\frac{-4-7}{28}=\frac{-11}{28}\)

Thus, the value obtained is \(\frac{-11}{28} \).

2.Write the additive inverse of each of the following:
(i)\(\frac{2}8 \)

(ii)\(\frac{-5}9 \)

(iii)\(\frac{-6}{-5} \)

(iv)\(\frac{2}{-9} \)

(v)\(\frac{19}{-6} \)


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Answer :

We have additive inverse of a as a=-a .So, we have:

(i) Additive inverse of \(\frac{2}8 = \frac{-2}8\)

(ii)Additive inverse of \(\frac{-5}9 = -(\frac{-5}9)= \frac{5}9\)

(iii) we have \(\frac{-6}{-5}= \frac{6}5 \)

∴ the additive inverse of \(\frac{6}5 =\frac{-6}5\)

(iv)We have \(\frac{2}{-9}=\frac{-2}9\)

∴ Additive inverse of \(\frac{-2}{9} = -(\frac{-2}9)= \frac{2}9\)

(v)We have \(\frac{19}{-6}=\frac{-19}6\)

∴ Additive inverse of \(\frac{-19}{6} = -(\frac{-19}6)= \frac{19}6\)

3.verify that -(-x)=x for
(i)x=\(\frac{11}{15}\)

(ii)x=\(\frac{-13}{17}\)


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Answer :

(i) We have x=\(\frac{11}5\)

∴ -x=\(\frac{-11}{15}\)

-(-x)=\(-(\frac{-11}{15}) = \frac{11}{15}=x \quad\)[∵ (-)×(-)=(+)]

Hence verified

(ii)We have x=\(\frac{-13}{17}\)

∴ -x=\(-(\frac{-13}{17})=\frac{13}{17}\)[∵ (-)×(-)=(+)]



-(-x)=\(-(\frac{13}{17})=x (Verified)\)

4.Find the multiplicative inverse of the following :
(i)-13

(ii)\(\frac{-13}{19}\)

(iii)\(\frac{1}5\)

(iv)\(\frac{-5}8×\frac{-3}{7}\)

(v)\(-1×\frac{-2}5\)

(vi)-1


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Answer :

We know that multiplicative inverse of a is \(\frac{1}a \qquad \)[∵ \(a ×\frac{1}a=1\)]

(i) Multiplicative inverse of \(-13=\frac{-1}{13} \qquad \)[∵ \(-13 ×\frac{-1}{13}=1\)]

(ii) Multiplicative inverse of \(\frac{-13}{19}=\frac{-19}{13} \qquad \)[∵ \(\frac{-13}{19} ×\frac{-19}{13}=1\)]

(iii) Multiplicative inverse of \(\frac{1}{5}=5 \qquad \)[∵ \(\frac{1}{5} ×5=1\)]

(iv) Multiplicative inverse of \(\frac{-5}8×\frac{-3}{7}=\frac{-8}5×\frac{-7}{3}\)

we have \(\frac{-5}8×\frac{-3}{7}=\frac{15}{56}\)

and we have the multiplicative inverse as \(\frac{-8}5×\frac{-7}{3}=\frac{56}{15}\)

∴ \(\frac{15}{56} ×\frac{56}{15}=1\)

(v) We have \(-1\frac{-2}5=\frac{2}5\)

∴Multiplicative inverse of \(\frac{2}5=\frac{5}2 \qquad \)[∵ \(\frac{2}{5} ×\frac{5}{2}=1\)]

(vi)Multiplicative inverse of \(-1=\frac{1}{-1}=-1 \qquad \)[∵ \(-1 ×\frac{1}{-1}=1\)]

5.Name the property under multiplication used in each of the following:
(i)\(\frac{-4}5×1=1×\frac{-4}5=\frac{-4}5\)

(ii)\(\frac{-13}{17}×\frac{-2}7=\frac{-2}7×\frac{-13}{17}\)

(iii)\(\frac{-19}{29}×\frac{29}{-19}=1\)


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Answer :

(i)Commutative property of multiplication.

(ii)Commutative property of multiplication.

(iii)Multiplicative inverse property.

6.Multiply \(\frac{6}{13}\) by the reciprocal of \(\frac{-7}{16}\)

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Answer :

We have reciprocal of \(\frac{-7}{16}=\frac{16}{-7}=\frac{-16}7\)

∴ \(\frac{6}{13}×\frac{-16}{7}=\frac{6×(-16)}{13×7}=-\frac{96}{91}\)

7.Tell what property allows you to compute
\(\frac{1}3×(6×\frac{4}3)as (\frac{1}3×6)×\frac{4}3\)

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Answer :

We know that a × (b× c) = (a × b) × c shows the associative property of multiplications.

∴\(\frac{1}3× (6× \frac{4}3)=(\frac{1}3× 6)× \frac{4}3\) shows the associative property of multiplication.

8.Is \(\frac{8}9\) the multiplicative inverse of \(-1\frac{1}8\)? Why or why not?

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Answer :

We have \(-1\frac{1}8=\frac{-9}{8}\)

Since multiplication inverse of \(\frac{8}9 is \frac{9}8 \) but not \(\frac{-9}8\)

Because, \(\frac { 8 }{ 9 } \times -1\frac { 1 }{ 8 } =\frac { 8 }{ 9 } \times \frac { -9 }{ 8 } =-1\neq 1\)

So, \(\frac{8}9 \) is not the multiplicative inverse of \(-1\frac{1}8\).

9. Is 0.3 the multiplicative inverse of \(3\frac { 1 }{ 3 }\)? Why or why not?

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Answer :

Yes, 0.3 is multiplicative inverse of \(3\frac { 1 }{ 3 } \).
Because \(0.3\times 3\frac { 1 }{ 3 } =\frac { 3 }{ 10 } \times \frac { 10 }{ 3 } =1\).

10.Write:
(i) The rational number that does not have a reciprocal.

(ii) The rational numbers that are equal to their reciprocals.

(iii) The rational number that is equal to its negative.


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Answer :

(i) The rational number 0 has no reciprocal because there is no rational number which when multiplied with 0, gives 1.

(ii) We know that the reciprocal of 1 is 1 and the reciprocal of -1 is -1. Thus,1 and -1 are the only rational numbers which are their own reciprocals.

(iii) 0 is the rational number which equal to its negative.

11.Fill in the blanks :

(i) Zero has ……………. reciprocal.

(ii) The numbers ……………. and ………. are their own reciprocals.

(iii) The reciprocal of – 5 is …………….

(iv) Reciprocal of \(\frac { 1 }{ x }\) , where \(x\neq 0\) is ………

(v) The product of two rational numbers is always a ………….

(vi) The reciprocal of a positive rational number is ………..


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Answer :

(i) No
(ii) 1, -1

(iii)\(\frac { -1 }{ 5 }\)

(iv) x

(v) rational number

(vi) positive

Solution for Exercise 1.2

aasa

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Answer :

assss

1.Represent these numbers on the number line.
(i) \(\frac { 7 }{ 4 } \)

(ii)\( \frac { -5 }{ 6 } \)


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Answer :

(i)For \(\frac { 7 }{ 4 }\) ,as we have the positive number, we make 7 markings of distance each on the right of zero and starting from 0. The seven markings are


Here, point P represents \(\frac { 7 }{ 4 } \)on the number line.
(ii) For \(\frac { -5 }{ 6 }\) , as the number is negative,we make 5 markings of distance \(\frac { 1 }{ 6 }\) each on the left and starting from 0. The fifth marking \(\frac { -5 }{ 6 } \).


The point P represents \(\frac { -5 }{ 6 }\) on the number line.

2.Represent \(\frac { -2 }{ 11 } ,\frac { -5 }{ 11 } ,\frac { -9 }{ 11 }\) on the number line.

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Answer :

As the given numbers are negative so mark a point 0 and mark eleven points till the point -1 on the left of 0 taking each division of \(\frac{1}{11}\) distance.The points marked are as follows.


Hence, point A represents \(\frac { -2 }{ 11 } \), point B represents \(\frac { -5 }{ 11 }\), point C represents \(\frac { -9 }{ 11 }\)

3.Write five rational numbers which are smaller than 2.

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Answer :

There can be many rational numbers as example. Few of them are:1, \(\frac{1}2\),0,\(\frac{1}3\),\(\frac{1}6\).

4.Find ten rational numbers between \(\frac { -2 }{ 5 }\) and \(\frac { 1 }{ 2 }\).

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Answer :

We have rational numbers as \(\frac{-2}{5} \) and \(\frac{1}{2} \)
Making same denominator:

⇒ \(\frac{-2}5 =\frac{-2×4}{5×4}=\frac{-8}{20}\)

and \(\frac{1}2 =\frac{1×10}{2×10}=\frac{10}{20}\)

Now, we have to find ten rational numbers between \(\frac{-8}{20}\) and\(\frac{10}{20}\)

So,these rational numbers can be \(\frac{-7}{20},\frac{-6}{20},\frac{-5}{20},\frac{-4}{20},\frac{-3}{20},\frac{-2}{20},\frac{-1}{20},0,\frac{1}{20},\frac{2}{20}\)

5.Find five rational numbers between
(i)\(\frac { 2 }{ 3 }\) and \(\frac { 4 }{ 5 }\)

(ii) \(\frac { -3 }{ 2 } \)and \(\frac { 5 }{ 3 }\)

(iii) \(\frac { 1 }{ 4 }\) and \(\frac { 1 }{ 2 }\)

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Answer :

(i) We have rational numbers as: \(\frac{2}3\) and\(\frac{4}5\)

Making the denominator same for both the rational numbers:

\(\frac{2}3=\frac{2}3×\frac{20}{20}=\frac{40}{60}\) and \(\frac{4}5=\frac{4×12}{5×12}=\frac{48}{60}\)

So now we have to see five rational number between \(\frac{40}{60}\) and \(\frac{48}{60}\)

These can be, \(\frac{41}{60},\frac{42}{60},\frac{43}{60},\frac{44}{60},\frac{46}{60}\)

(ii) We have rational numbers as: \(\frac{-3}2\) and\(\frac{5}3\)

Making the denominator same for both the rational numbers:

\(\frac{-3}2=\frac{-3×3}{2×3}=\frac{-9}{6}\) and \(\frac{5}3=\frac{5×2}{3×2}=\frac{10}{6}\)

So now we have to see five rational number between \(\frac{-9}{6}\) and \(\frac{10}{6}\)

These can be, \(\frac{-8}{6},\frac{-7}{6},\frac{-6}{6},\frac{3}{6},\frac{4}{6}\)

(iii) We have rational numbers as: \(\frac{1}4\) and\(\frac{1}2\)

Making the denominator same for both the rational numbers:

\(\frac{1}4=\frac{1×8}{2×8}=\frac{8}{32}\) and \(\frac{1}2=\frac{1×16}{2×16}=\frac{16}{32}\)

So now we have to see five rational number between \(\frac{8}{32}\) and \(\frac{16}{32}\)

These can be, \(\frac{9}{32},\frac{10}{32},\frac{11}{32},\frac{12}{32},\frac{13}{32}\)

6.Write five rational numbers greater than -2.

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Answer :

Rational numbers greater than -2 are many and few of them are: -1, 0, 1, \(\frac{1}4\),\(\frac{3}4\).

7.Find ten rational numbers between \(\frac { 3 }{ 5 }\) and \(\frac { 3 }{ 4 }\).

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Answer :

We have rational numbers as: \(\frac{3}5\) and\(\frac{3}4\)

Making the denominator same for both the rational numbers:

\(\frac{3}5=\frac{3×20}{5×20}=\frac{60}{100}\) and \(\frac{3}4=\frac{3×25}{4×25}=\frac{75}{100}\)

So now we have to see ten rational number between \(\frac{60}{100}\) and \(\frac{75}{100}\)

So, these can be, \(\frac{61}{100},\frac{62}{100},\frac{63}{100},\frac{64}{100},\frac{65}{100},\frac{66}{100},\frac{67}{100},\frac{68}{100},\frac{69}{100},\frac{70}{100}\)