Saturday, September 30, 2023
Latest:

### EteacherG

a educational group

# Ncert solutions for class 8 maths chapter 1 Rational Numbers | 8th class maths EXERCISE 1

ncert solutions for class 8 maths
ncert maths class 8
class 8 maths solutions
class 8th maths

Ncert solutions for class 8 maths chapter 1 Rational Numbers. Here We learn what is in ncert maths class 8 EXERCISE 1 and how to solve questions with easiest method. ┬аIn this chapter we solve the question of NCERT class 8th maths EXERCISE 1. NCERT class 8 maths solutions Rational Numbers┬аare part of NCERT Solutions for class 8 maths chapter 1 solution PDF. Ncert solutions for class 8 EXERCISE 1 with formula and solution. 8th class maths

Ncert solutions for class 8 maths Here we solve class 8th maths ncert solutions chapter 1 concepts all questions with easy method with expert solutions. It help students in their study, home work and preparing for exam. Soon we provide ncert solutions for class 8 maths EXERCISE 1 Rational Numbers question and answers. Soon we provided ncert solutions for class 8th maths chapter 1┬аRational Numbers┬аin free PDF here. ncert solutions for class 8 maths EXERCISE 1 pdf will be provide soon. 8th class maths chapter 1 NCERT Solution and ncert solutions for class 8 maths chapter 1 pdf download book PDF.

# NCERT Solutions for Class 8 Maths EXERCISE 1

## NCERT Solutions for Class 8 Maths chapter 1class 8th mathsEx 1.1┬а

EXERCISE 1.1

1. Using appropriate properties find.
(i) $$\displaystyle -\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}$$
(ii) $$\displaystyle \left( {\frac{2}{3}} \right)\times \left( {-\frac{3}{7}} \right)-\frac{1}{6}\times \frac{3}{2}+\frac{1}{{14}}\times \frac{2}{5}$$

Solution :┬а
(i) $$\displaystyle -\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}$$
$$\displaystyle \begin{array}{l}=-\frac{2}{3}\times \frac{3}{5}-\frac{3}{5}\times \frac{1}{6}+\frac{5}{2}\\=\left( {-\frac{3}{5}} \right)\times \left( {\frac{2}{3}+\frac{1}{6}} \right)+\frac{5}{2}\\=\left( {-\frac{3}{5}} \right)\times \left( {\frac{{4+1}}{6}} \right)+\frac{5}{2}\\=\left( {-\frac{3}{5}} \right)\times \left( {\frac{5}{6}} \right)+\frac{5}{2}\\=\left( {-\frac{3}{6}} \right)+\frac{5}{2}\\=\frac{{-3+15}}{6}\\=\frac{{12}}{6}\end{array}$$
= 2

(ii) $$\displaystyle \left( {\frac{2}{3}} \right)\times \left( {-\frac{3}{7}} \right)-\frac{1}{6}\times \frac{3}{2}+\frac{1}{{14}}\times \frac{2}{5}$$
$$\displaystyle \begin{array}{l}=\left( {-\frac{6}{{35}}} \right)-\frac{3}{{12}}+\frac{2}{{70}}\\=-\frac{6}{{35}}-\frac{1}{4}+\frac{1}{{35}}\\=\frac{{-24-35+4}}{{140}}\\=\frac{{-59+4}}{{140}}\\=\frac{{-55}}{{140}}\\=\frac{{-11}}{{28}}\end{array}$$

2. Write the additive inverse of each of the following.

 (i) $$\displaystyle \frac{2}{8}$$ (ii) $$\displaystyle \frac{{-5}}{9}$$ (iii) $$\displaystyle \frac{{-6}}{{-5}}$$ (iv) $$\displaystyle \frac{2}{{-9}}$$ (v) $$\displaystyle \frac{{19}}{{-6}}$$

Solution :┬а
(i) $$\displaystyle \frac{2}{8}$$
additive inverse = $$\displaystyle \frac{-2}{8}$$

(ii) $$\displaystyle \frac{{-5}}{9}$$
additive inverse = $$\displaystyle \frac{{5}}{9}$$

(iii) $$\displaystyle \frac{{-6}}{{-5}}$$
additive inverse = $$\displaystyle \frac{{-6}}{{5}}$$

(iv) $$\displaystyle \frac{2}{{-9}}$$
additive inverse = $$\displaystyle \frac{2}{{9}}$$

(v) $$\displaystyle \frac{{19}}{{-6}}$$
additive inverse = $$\displaystyle \frac{{19}}{{6}}$$

3. Verify that тИТ (тИТ x) = x
(i) $$\displaystyle x=\frac{{11}}{{15}}$$┬а ┬а (ii) $$\displaystyle x=-\frac{{13}}{{17}}$$

Solution :┬а┬а
(i) $$\displaystyle x=\frac{{11}}{{15}}$$
given $$\displaystyle x=\frac{{11}}{{15}}$$
additive inverse of $$\displaystyle x=\frac{{11}}{{15}}$$ is $$\displaystyle -x=\frac{{-11}}{{15}}$$ ,
because $$\displaystyle \frac{{11}}{{15}}+\left( {\frac{{-11}}{{15}}} \right)=0$$

the same equality $$\displaystyle \frac{{11}}{{15}}+\left( {\frac{{-11}}{{15}}} \right)=0$$ , shows that the additive inverse of $$\displaystyle -x=\frac{{-11}}{{15}}$$ is $$\displaystyle x=\frac{{11}}{{15}}$$┬а
so, $$\displaystyle -\left( {\frac{{-11}}{{15}}} \right)=\frac{{11}}{{15}}$$
so, тИТ (тИТ x) = x

(ii) $$\displaystyle x=-\frac{{13}}{{17}}$$
given $$\displaystyle x=-\frac{{13}}{{17}}$$
additive inverse of $$\displaystyle x=-\frac{{13}}{{17}}$$ is $$\displaystyle x=\frac{{13}}{{17}}$$ рд╣реИ,┬а
because $$\displaystyle \left( {\frac{{-13}}{{17}}} \right)+\frac{{13}}{{17}}=0$$ рд╣реИред

рд╕рдорд┐рдХрд╛ $$\displaystyle \left( {\frac{{-13}}{{17}}} \right)+\frac{{13}}{{17}}=0$$ , shows that the additive inverse of $$\displaystyle x=\frac{{13}}{{17}}$$ is $$\displaystyle x=-\frac{{13}}{{17}}$$┬а
so, $$\displaystyle -\left( {\frac{{-13}}{{17}}} \right)=\frac{{13}}{{17}}$$
so, тИТ (тИТ x) = x

4. Find the multiplicative inverse of the following.

 (i) тИТ 13 (ii) $$\displaystyle \frac{{-13}}{{19}}$$ (iii) $$\displaystyle \frac{{1}}{{5}}$$ (iv) $$\displaystyle \frac{{-5}}{8}\times \left( {\frac{{-3}}{7}} \right)$$ (v) $$\displaystyle -1\times \left( {\frac{{-2}}{5}} \right)$$ (vi) тИТ 1 ┬а ┬а

Solution :
(i) тИТ 13
multiplicative inverse = $$\displaystyle \frac{{-1}}{{13}}$$

(ii) $$\displaystyle \frac{{-13}}{{19}}$$
multiplicative inverse = $$\displaystyle \frac{{-19}}{{13}}$$

(iii) $$\displaystyle \frac{{1}}{{5}}$$
multiplicative inverse = $$\displaystyle \frac{{5}}{{1}}$$ рдпрд╛ 5

(iv) $$\displaystyle \frac{{-5}}{8}\times \left( {\frac{{-3}}{7}} \right)$$
multiplicative inverse = $$\displaystyle \frac{{-56}}{{15}}$$

(v) $$\displaystyle -1\times \left( {\frac{{-2}}{5}} \right)$$
multiplicative inverse = $$\displaystyle \frac{{-5}}{{2}}$$

(vi) тИТ 1
multiplicative inverse = тИТ 1

5. Name the property under multiplication used in each of the following.

 (i) $$\displaystyle \frac{{-4}}{5}\times 1=1\times \left( {\frac{{-4}}{5}} \right)=-\frac{4}{5}$$ (ii) $$\displaystyle -\frac{{13}}{{17}}\times \left( {-\frac{2}{7}} \right)=\frac{{-2}}{7}\times \left( {\frac{{-13}}{{17}}} \right)$$ (iii) $$\displaystyle \frac{{-19}}{{29}}\times \frac{{29}}{{-19}}=1$$ ┬а

Solution :┬а
(i) $$\displaystyle \frac{{-4}}{5}\times 1=1\times \left( {\frac{{-4}}{5}} \right)=-\frac{4}{5}$$
Answer – 1 is the multiplicative identity

(ii) $$\displaystyle -\frac{{13}}{{17}}\times \left( {-\frac{2}{7}} \right)=\frac{{-2}}{7}\times \left( {\frac{{-13}}{{17}}} \right)$$

(iii) $$\displaystyle \frac{{-19}}{{29}}\times \frac{{29}}{{-19}}=1$$

6. Multiply $$\displaystyle \frac{6}{{13}}$$ by the reciprocal of $$\displaystyle \frac{{-7}}{{16}}$$
Solution :┬а
Multiplicative inverse $$\displaystyle \frac{{-7}}{{16}}$$ is $$\displaystyle \frac{{-16}}{{7}}$$
so, $$\displaystyle \frac{6}{{13}}\times \frac{{-16}}{7}$$
$$\displaystyle =\frac{{-96}}{{91}}$$

7. Tell what property allows you to compute $$\displaystyle \frac{1}{3}\times \left( {\frac{{-2}}{7}} \right)=\frac{{-2}}{7}\times \left( {\frac{{-13}}{{17}}} \right)$$
\displaystyle \frac{{-7}}{{16}}\$
Solution :
Associativity property

8. Is $$\displaystyle -1\frac{1}{8}$$ the multiplicative inverse of $$\displaystyle \frac{{8}}{{9}}$$ ? Why or why not?
Solution :┬а
converting the $$\displaystyle -1\frac{1}{8}$$ in simple fraction we get $$\displaystyle \frac{{-9}}{8}$$
Multiplicative inverse of $$\displaystyle \frac{{-9}}{8}$$ is $$\displaystyle \frac{{-8}}{9}$$ ,
which does not correspond to the given number.

so, Multiplicative inverse of $$\displaystyle -1\frac{1}{8}$$ is $$\displaystyle \frac{{-8}}{9}$$ , not $$\displaystyle \frac{{8}}{{9}}$$

9. Is 0.3 the multiplicative inverse of $$\displaystyle 3\frac{1}{3}$$ ? Why or why not?
Solution :
converting the $$\displaystyle 3\frac{1}{3}$$ in simple fraction we get $$\displaystyle \frac{{10}}{3}$$┬а
so, Multiplicative inverse of $$\displaystyle \frac{{10}}{3}$$ is $$\displaystyle \frac{{3}}{10}$$┬а
$$\displaystyle \frac{{3}}{10}$$ = 0.3
So, multiplicative inverse of $$\displaystyle 3\frac{1}{3}$$ is 0.3

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

(ii) The rational numbers that are equal to their reciprocals.
Answer – 1 and тИТ 1

(iii) The rational number that is 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 тАУ 5 is ________.
(iv) Reciprocal of $$\displaystyle \frac{{1}}{x}$$ , where (x тЙа 0) is _______________ .
(v) The product of two rational numbers is always a _______.
(vi) The reciprocal of a positive rational number is ________.

Solution :
(i) Zero has no reciprocal.
(ii) The numbers 1 and тИТ 1 are their own reciprocals.
(iii) The reciprocal of тАУ 5 is $$\displaystyle \underline{{\frac{{-1}}{5}}}$$ .
(iv) Reciprocal of $$\displaystyle \frac{{1}}{x}$$ , where (x тЙа 0) is x.
(v) The product of two rational numbers is always a rational number.
(vi) The reciprocal of a positive rational number is positive.

error: Content is protected !!