# NCERT Solutions Class 8 Maths Chapter 12 Exercise 12.1

Class 8 is a crucial academic year in a student’s life. In Class 8, students are introduced to a variety of new concepts. When students progress from Class 7 to Class 8, the level of difficulty increases. As a result, students must prepare more and devote more time to their studies. The numerous concepts taught in Class 8 are quite useful in further education. Class 8 topics serve as a basis for higher levels. As the difficulty level of Class 8 Mathematics grows, students may find Mathematics to be a tough subject. To become proficient at solving mathematics problems, students must practise a large number of questions.They must complete all NCERT exercise questions from all Class 8 Mathematics chapters.

Mathematics is generally viewed as difficult by students. In order to make Mathematics easy and simple to learn, students must practise a significant number of questions. Furthermore, consistency is required in Mathematics for effective learning. Students must solve questions of varying difficulty levels in order to be fully prepared for Mathematics examinations. Students should begin the chapter by answering basic questions and working their way up to questions that require them to apply their thinking and analytical skills. Students may often face questions demanding analytical ability during exams; therefore, students must be prepared for all sorts of questions. Students will be able to answer questions quickly and within the time limit with enough practice.

Mathematics is a very essential topic for all students. Mathematics is a fascinating subject which introduces new laws, formulas, and concepts to students. Studying Mathematics also develops the students’ cognitive and reasoning abilities. Mathematics can help students develop their critical thinking skills. It is a significant subject for developing mental discipline. Mathematics has several applications. In some way or another, Mathematics is used in everyday life. As a result, Mathematics is introduced to students at an early age. Moreover, other Science subjects also make use of Mathematical concepts.

To make Mathematics simple and easy to learn, students must practise a large number of questions. Additionally, Mathematics requires consistency for the best understanding. Students should be ready for various types of questions because they may regularly encounter questions in exams that call for analytical thinking. With adequate practice, students will be able to respond to questions quickly and within the allotted time.Students must finish problems of varying levels of complexity in order to be fully prepared for Mathematics exams. Simple questions should be answered at the start of the chapter, and when they are answered, students should gradually move on to questions that requirethem to apply their analytical and reasoning skills.

Chapter 12 of Class 8 Mathematics is on Exponents and Powers. In this chapter, students learn about How to write large numbers easily using Exponents and Powers, how to express numbers in standard form, how to deal with negative Exponents, numerous laws related to Exponents and much more. To study this chapter effectively, students can take help from the resources offered by Extramarks. To find the most accurate solutions for Exercise 12.1, students can use the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1. The NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 are available to download from the website and mobile application of Extramarks. The NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 can be downloaded in PDF format by students. The Class 8 Maths Chapter 12 Exercise 12.1 can be easily solved with the help of the solutions provided by Extramarks.

## NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers (EX 12.1) Exercise 12.1

To score well in Mathematics, students must solve a lot of questions. With adequate practice, students will be able to respond to questions quickly and within the allotted time.The NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 are offered by Extramarks for the benefit of students. Students can download the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 from the website and mobile application of Extramarks. The NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 can be downloaded in PDF format for easy access. Students can always refer to the Extramarks’ study materials for a thorough understanding of concepts.

With the help of the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 students will be able to solve  NCERT Class 8 Maths Chapter 12 Exercise 12.1 thoroughly. The NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 written in a stepwise manner for the students’ clarity.The NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 can help students with self-study and enhance their Mathematics skills.

### Access NCERT Solution for Class 8 Maths Chapter 12- Exponents and Powers Exercise 12.1

With the help of the link provided, students can access the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1. Students are advised to download the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 in PDF format. The NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 will help students score well in the Mathematics exams. Students must download the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 to write the most precise solutions for Class 8 Maths Exercise 12.1.

### NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers Exercise 12.1

In Chapter 12 Class 8 Mathematics there are two exercises. Both exercises must be solved by students. The Class 8 Maths Exercise 12.1 Solution is offered by Extramarks. The NCERT Solutions for Class 8 Maths, Chapter 12, Exercise 12.1 can be downloaded from Extramarks’ website and mobile app.The NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 are written by expert Mathematics teachers for the benefit of students. Students will be able to easily solve all questions with the help of the NCERT Solutions for Class 8 Maths Chapter 12 Exercise 12.1.NCERT Solutions for Class 8

Class 8 students study a wide range of subjects. The core subjects in Class 8 include Mathematics, Social Science, Science, Sanskrit/Hindi, and English. Students must thoroughly study all subjects since they are important. Extramarks resources can help students prepare for these subjects. Extramarks provides students with past years’ papers and the solutions, NCERT Solutions, and revision notes to help them prepare for their Class 8 examinations efficiently. Students must use the offered materials to enhance their Class 8 scores. Extramarks’ materials are excellent for self-study. Students may rely on Extramarks’ materials. Subject-matter experts curate and proofread all materials on a regular basis. In addition, the study materials are frequently updated to incorporate the most recent CBSE guidelines.

Students must prepare diligently for Class 8.A thorough preparation for Class 8will ensure that students understand the basic concepts of the core subjects. Students who excel in Class 8 will be able to study more successfully in higher classes. Students in Class 8 must complete all of the NCERT practise questions and example questions mentioned in the NCERT textbooks. In this way, students will be able to prepare effectively. If students face any difficulties, they can use the various study materials provided by Extramarks.

Q.1

$\begin{array}{l}\text{Evaluate.}\\ {\text{(i) 3}}^{\text{-2}}{\text{(ii) (-4)}}^{\text{-2}}\text{(iii)}{\left(\frac{\text{1}}{\text{2}}\right)}^{\text{-5}}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right){3}^{–2}=\frac{1}{{3}^{2}}=\frac{1}{9}\text{\hspace{0.17em}}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ \\ \left(\mathrm{ii}\right){\text{(-4)}}^{\text{-2}}=\frac{1}{{\left(-4\right)}^{2}}=\frac{1}{16}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ \\ \left(\mathrm{iii}\right){\left(\frac{1}{2}\right)}^{–5}=\frac{1}{{2}^{-5}}={2}^{5}=2×2×2×2×2=32\end{array}$

Q.2

$\begin{array}{l}\text{Simplify and express the result in power notation}\\ \text{with positive exponent}\text{.}\\ {\text{(i) (- 4)}}^{\text{5}}{\text{÷ (- 4)}}^{\text{8}}\text{(ii)}{\left(\frac{\text{1}}{{\text{2}}^{\text{3}}}\right)}^{\text{2}}\text{(iii)}{\left(\text{– 3}\right)}^{\text{4}}\text{×}{\left(\frac{\text{5}}{\text{3}}\right)}^{\text{4}}\text{}\\ {\text{(iv) (3}}^{\text{– 7}}{\text{÷ 3}}^{\text{– 10}}{\text{) × 3}}^{\text{– 5}}{\text{(v) 2}}^{\text{– 3}}{\text{× (- 7)}}^{\text{– 3}}\end{array}$

Ans

$\begin{array}{l}{\text{(i) (- 4)}}^{\text{5}}{\text{÷ (- 4)}}^{\text{8}}\\ {\text{=(- 4)}}^{\text{5-8}}\text{\hspace{0.17em}}\left[{\text{By using a}}^{\text{m}}{\text{÷a}}^{\text{n}}{\text{=a}}^{\text{m-n}}\right]\\ {\text{=(- 4)}}^{\text{-3}}\\ \text{=}\frac{\text{1}}{{\left(\text{-4}\right)}^{\text{3}}}\text{\hspace{0.17em}}\left[{\text{By using a}}^{\text{-m}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{m}}}\right]\\ \\ \text{(ii)}{\left(\frac{\text{1}}{{\text{2}}^{\text{3}}}\right)}^{\text{2}}\\ \text{=}\frac{\text{1}}{{\left({\text{2}}^{\text{3}}\right)}^{\text{2}}}\\ \text{=}\frac{\text{1}}{{\text{2}}^{\text{6}}}\text{}\left[\text{By using}{\left({\text{a}}^{\text{m}}\right)}^{\text{n}}{\text{=a}}^{\text{mn}}\right]\end{array}$

$\begin{array}{l}\text{(iii)}{\left(-3\right)}^{4}×{\left(\frac{5}{3}\right)}^{4}\\ ={\left(-1\right)}^{4}×{3}^{4}×\frac{{5}^{4}}{{3}^{4}}\\ ={\left(-1\right)}^{4}×{5}^{4}\\ ={5}^{4}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\left(-1\right)}^{4}=1\right]\\ \end{array}$

$\begin{array}{l}{\text{(iv) (3}}^{\text{– 7}}{\text{÷ 3}}^{\text{– 10}}{\text{) × 3}}^{\text{– 5}}\\ {\text{=3}}^{\left(\text{– 7}\right)\text{–}\left(\text{-10}\right)}{\text{× 3}}^{\text{– 5}}\text{}\left[{\text{By using a}}^{\text{m}}{\text{÷a}}^{\text{n}}{\text{=a}}^{\text{m-n}}\right]\\ {\text{=3}}^{\text{3}}{\text{× 3}}^{\text{– 5}}\\ {\text{=3}}^{\left(\text{3}\right)\text{+}\left(\text{-5}\right)}\text{}\left[{\text{By using a}}^{\text{m}}{\text{×a}}^{\text{n}}{\text{=a}}^{\text{m+n}}\right]\\ {\text{=3}}^{\text{-2}}\\ \text{=}\frac{\text{1}}{{\text{3}}^{\text{2}}}\text{}\left[{\text{By using a}}^{\text{-m}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{m}}}\right]\\ \\ {\text{(v) 2}}^{\text{– 3}}{\text{× (- 7)}}^{\text{-3}}\\ \text{=}\frac{\text{1}}{{\text{2}}^{\text{3}}}\text{×}\frac{\text{1}}{{\left(\text{-7}\right)}^{\text{3}}}\\ \text{=}\frac{\text{1}}{{\left[\text{2×}\left(\text{-7}\right)\right]}^{\text{3}}}\text{}\left[{\text{By using a}}^{\text{m}}{\text{×b}}^{\text{m}}{\text{=ab}}^{\text{m}}\right]\\ \text{=}\frac{\text{1}}{{\left(\text{-14}\right)}^{\text{3}}}\end{array}$

Q.3

$\begin{array}{l}\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{value}\text{}\mathrm{of}:\\ \left(\mathrm{i}\right)\text{}\left({3}^{0}\text{}+\text{}{4}^{–1}\right)\text{}×\text{}{2}^{2}\\ \left(\mathrm{ii}\right)\text{ }\left({2}^{–1}\text{}×\text{}{4}^{–1}\right)\text{}÷\text{}{2}^{–2}\\ \left(\mathrm{iii}\right)\text{ }{\left(\frac{1}{2}\right)}^{–2}+{\left(\frac{1}{3}\right)}^{–2}+{\left(\frac{1}{4}\right)}^{–2}\\ {\text{(iv) (3}}^{\text{-1}}{\text{+ 4}}^{\text{-1}}{\text{+ 5}}^{\text{-1}}{\text{)}}^{\text{0}}\text{}\\ \left(\mathrm{v}\right)\text{}{\left\{{\left(\frac{–2}{3}\right)}^{–2}\right\}}^{2}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }\left(3\mathrm{°}+{4}^{-1}\right)×{2}^{2}\\ =\left(1+\frac{1}{4}\right)×{2}^{2}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{0}\text{=1,}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ =\frac{5}{4}×4\\ =5\\ \\ \left(\mathrm{ii}\right)\text{ }\left({2}^{-1}×{4}^{-1}\right)÷{2}^{-2}\\ =\left({2}^{-1}×{\left({2}^{2}\right)}^{-1}\right)÷{2}^{-2}\text{}\\ ={2}^{-1}×\left({2}^{-2}\right)÷{2}^{-2}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\text{n}}{\text{=a}}^{\text{mn}}\right]\\ ={2}^{-1+\left(-2\right)}÷{2}^{-2}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{\mathrm{m}}×{\mathrm{a}}^{\text{n}}{\text{=a}}^{\text{m+n}}\right]\end{array}$

$\begin{array}{l}{\text{=2}}^{\text{-3}}{\text{÷2}}^{\text{-2}}\\ {\text{=2}}^{\text{-3-(-2)}}\text{}\left[{\text{By using a}}^{\text{m}}{\text{÷a}}^{\text{n}}{\text{=a}}^{\text{m-n}}\right]\\ {\text{=2}}^{\text{-3+2}}\\ {\text{=2}}^{\text{-1}}\\ \text{=}\frac{\text{1}}{\text{2}}\\ \\ \text{(iii) }{\left(\frac{\text{1}}{\text{2}}\right)}^{\text{-2}}\text{+}{\left(\frac{\text{1}}{\text{3}}\right)}^{\text{-2}}\text{+}{\left(\frac{\text{1}}{\text{4}}\right)}^{\text{-2}}\\ \text{=}{\left(\frac{\text{2}}{\text{1}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{3}}{\text{1}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{4}}{\text{1}}\right)}^{\text{2}}\text{ }\left[{\text{By using a}}^{\text{-m}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{m}}}\right]\\ {\text{=2}}^{\text{2}}{\text{+3}}^{\text{2}}{\text{+4}}^{\text{2}}\\ \text{=4+9+16}\\ \text{=29}\\ \\ {\text{(iv) (3}}^{\text{-1}}{\text{+ 4}}^{\text{-1}}{\text{+ 5}}^{\text{-1}}{\text{)}}^{\text{0}}\text{}\\ \text{=}{\left(\frac{\text{1}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}}\text{+}\frac{\text{1}}{\text{5}}\right)}^{\text{0}}\text{}\left[{\text{By using a}}^{\text{-m}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{m}}}\right]\\ \text{=1}\left[{\text{By using a}}^{\text{0}}\text{=1}\right]\end{array}$

$\begin{array}{l}\left(\mathrm{v}\right){\left\{{\left(\frac{-2}{3}\right)}^{-2}\right\}}^{2}\\ ={\left\{{\left(\frac{3}{-2}\right)}^{2}\right\}}^{2}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ ={\left\{\frac{{3}^{2}}{{\left(-2\right)}^{2}}\right\}}^{2}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{m}}=\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{b}}^{\mathrm{m}}}\right]\\ ={\left(\frac{9}{4}\right)}^{2}\\ =\frac{81}{16}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Evaluate}\\ \left(\mathrm{i}\right)\text{ }\frac{{8}^{–1}×{5}^{3}}{{2}^{–4}}\text{}\left(\mathrm{ii}\right)\text{ }\left({5}^{–1}×{2}^{–1}\right)×{6}^{–1}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\frac{{8}^{-1}×{5}^{3}}{{2}^{-4}}\\ =\frac{{2}^{4}×{5}^{3}}{{8}^{1}}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ =\frac{{2}^{4}×{5}^{3}}{{2}^{3}}\end{array}$

$\begin{array}{l}={2}^{4-3}×{5}^{3}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]\text{}\\ =2×125\\ =250\text{}\end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\text{ }\left({5}^{–1}×{2}^{–1}\right)×{6}^{–1}\\ =\left(\frac{1}{5}×\frac{1}{2}\right)×\frac{1}{6}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ =\frac{1}{10}×\frac{1}{6}\\ =\frac{1}{60}\end{array}$

Q.5

$\mathrm{Find}\text{}\mathrm{the}\text{}\mathrm{value}\text{}\mathrm{of}\text{}\mathrm{m}\text{}\mathrm{for}\text{}\mathrm{which}\text{}{5}^{\mathrm{m}}\text{}÷\text{}{5}^{–3}\text{}=\text{}{5}^{5}.$

Ans

$\begin{array}{l}{\text{5}}^{\text{m}}{\text{÷ 5}}^{\text{-3}}{\text{= 5}}^{\text{5}}\\ ⇒{\text{5}}^{\text{m-(-3)}}{\text{= 5}}^{\text{5}}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\text{n}}{\text{=a}}^{\text{m-n}}\right]\\ ⇒{\text{5}}^{\text{m+3}}{\text{= 5}}^{\text{5}}\\ ⇒\mathrm{m}+3=5\text{}\left(\text{Since the bases are same,\hspace{0.17em}\hspace{0.17em}their}\\ \text{powers must be equal.)}\\ ⇒\mathrm{m}=5-3\\ ⇒\mathrm{m}=2\end{array}$

Q.6 Evaluate

$\left(\mathrm{i}\right){\left\{{\left(\frac{1}{3}\right)}^{–1}–{\left(\frac{1}{4}\right)}^{–1}\right\}}^{–1}\left(\mathrm{ii}\right){\left(\frac{5}{8}\right)}^{–7}×{\left(\frac{8}{5}\right)}^{–4}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\text{}{\left\{{\left(\frac{\text{1}}{\text{3}}\right)}^{\text{-1}}\text{–}{\left(\frac{\text{1}}{\text{4}}\right)}^{\text{-1}}\right\}}^{\text{-1}}\\ \text{=}{\left\{\left(\frac{\text{3}}{\text{1}}\right)\text{–}\left(\frac{\text{4}}{\text{1}}\right)\right\}}^{\text{-1}}\text{}\left[{\text{By using a}}^{\text{-m}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{m}}}\right]\\ \text{=}{\left\{\text{3-4}\right\}}^{\text{-1}}\\ {\text{= (-1)}}^{\text{-1}}\\ \text{=}\frac{\text{1}}{\left(\text{-1}\right)}\\ \text{= -1}\end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\text{}{\left(\frac{5}{8}\right)}^{–7}×{\left(\frac{8}{5}\right)}^{–4}\\ =\frac{{5}^{-7}}{{8}^{-7}}×\frac{{8}^{-4}}{{5}^{-4}}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{m}}=\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{b}}^{\mathrm{m}}}\right]\end{array}$

$\begin{array}{l}=\frac{{8}^{7}}{{5}^{7}}×\frac{{5}^{4}}{{8}^{4}}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ =\frac{{8}^{7-4}}{{5}^{7-4}}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]\\ =\frac{{8}^{3}}{{5}^{3}}\\ =\frac{512}{125}\end{array}$

Q.7 Simplify

$\text{(i)}\frac{{\text{25×t}}^{\text{-4}}}{{\text{5}}^{\text{-3}}{\text{×10×t}}^{\text{-8}}}\text{(t ≠ 0) (ii)}\frac{{\text{3}}^{\text{-5}}{\text{×10}}^{\text{-5}}\text{×125}}{{\text{5}}^{\text{-7}}{\text{×6}}^{\text{-5}}}\text{}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\frac{25×{\mathrm{t}}^{-4}}{{5}^{–3}×10×{\mathrm{t}}^{–8}}\left(\mathrm{t}\ne 0\right)\\ =\frac{{5}^{2}×{\mathrm{t}}^{-4}}{{5}^{–3}×5×2×{\mathrm{t}}^{–8}}\\ =\frac{{5}^{2}×{\mathrm{t}}^{-4}}{{5}^{–3+1}×2×{\mathrm{t}}^{–8}}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{:}{\mathrm{a}}^{\mathrm{m}}×{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\right]\\ =\frac{{5}^{2}×{\mathrm{t}}^{-4}}{{5}^{–2}×2×{\mathrm{t}}^{–8}}\end{array}$

$\begin{array}{l}=\frac{{5}^{2+2}×{\mathrm{t}}^{-4+8}\text{}}{2}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{:}{\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]\\ =\frac{{5}^{4}×{\mathrm{t}}^{4}\text{}}{2}\\ =\frac{625{\mathrm{t}}^{4}}{2}\end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\frac{{3}^{–5}×{10}^{–5}×125}{{5}^{–7}×{6}^{–5}}\\ =\frac{{3}^{–5}×{\left(2×5\right)}^{–5}×{5}^{3}}{{5}^{–7}×{\left(2×3\right)}^{–5}}\\ ={3}^{–5+5}×{2}^{-5+5}×{5}^{-5+3+7}\left[\mathrm{By}\text{}\mathrm{using}\text{:}{\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]\\ ={3}^{0}×{2}^{0}×{5}^{5}\left[\mathrm{By}\text{}\mathrm{using}\text{:}{\mathrm{a}}^{0}={\mathrm{a}}^{1}\right]\\ ={5}^{5}\end{array}$

## 1. How can NCERT Solutions in clarifying doubts?

It is true that the difficulty level for all subjects increases in Class 8, however, students may make it easy on themselves by studying consistently and solving questions from NCERT exercises. Class 8 Mathematics must also be prepared in this manner. Extramarks’ materials can assist students in completing NCERT exercises. To solve the questions of Chapter 12 Class 8, students can take the help of the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1.

## 2. From where can students download the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1?

Students can download the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 from the website and mobile application of Extramarks.

## 3. Can the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 be downloaded in PDF format?

Yes, the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 are available in PDF format. Students are advised to download the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 in PDF format for easy access.

## 4. Will the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 help students to solve mock papers?

Yes, the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 will help students in solving the mock examination questions. They can also refer to Extramarks’ past years’ papers and its solutions to find authentic solutions. Students must download the  NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 for preparing Chapter 12 efficiently. The study resources provided by Extramarks are reliable and authentic. The study resources provided by Extramarks cover the whole syllabus holistically and can be used by students to prepare for their exams. Further, the study materials are updated regularly according to the latest CBSE guidelines.

## 5. Are solutions of other Chapter 6 Exercises also provided by Extramarks?

Yes, along with the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1, solutions for Exercise 12.2 are also available on the website and mobile application of Extramarks for the benefit of students. Students can refer to the solutions in case of any challenges.

## 6. Are the NCERT Solutions For Class 8 Maths Chapter 12 Exercise 12.1 also provided in Hindi by Extramarks?

Extramarks does provide study materials in Hindi for students. They can visit the website and mobile application to download the study materials in Hindi.