## NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots Exercise 6.4

NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots Exercise 6.4 are comprehensive and exhaustive solutions. They have been meticulously prepared by Vedantu’s subject matter experts who have vast experience in the education industry. The Class 8 Maths Exercise 6.4 Solution has been prepared according to the guidelines of CBSE. NCERT Solution for Class 8 Maths Chapter 6 Exercise 6.4 can be conveniently downloaded in free PDF formats. that will allow you to study when you are offline as well.

Science Students who are looking for NCERT Solutions for Class 8 Science will also find the Solutions curated by our Master Teachers really Helpful.

## Access NCERT Solutions for Mathematics Chapter 6 – Squares and Square Roots

### Exercise 6.4

1. Find the square root of each of the following numbers by division method.

2304

Ans: The square root of the given number is

$\quad\quad 48 \quad$ | |

$\quad 4\quad$ | $\overline{23}\:\overline{04}\\$ $ -16\qquad $ |

$\quad 88\quad$ | ${\;}704\\ $ $ {\;}704 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{2304} = 48$

4489

Ans: The square root of the given number is

$\quad\quad 67 \quad$ | |

$\quad 6\quad$ | $ \overline{44}\:\overline{89}\\ $ $ -36\qquad $ |

$\quad 127\quad$ | ${\;} 889\\ $ $ {\;} 889 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{4489} = 67$.

3481

Ans: The square root of the given number is

$\quad\quad 59 \quad$ | |

$\quad 5\quad$ | $\overline{34}\:\overline{81}\\ $ $ -25\qquad $ |

$\quad 109\quad$ | ${\;} 981\\ $ $ {\;} 981 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{3481} = 59$

529

Ans: The square root of the given number is

$\quad \quad 23 \quad$ | |

$\quad 2\quad$ | $\overline{5}\:\overline{29}\\ $ $ -4\qquad $ |

$\quad 43\quad$ | $ {} 129\\ $ $ {} 129 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{529} = 23$.

3249

Ans: The square root of the given number is

$\quad\quad 57 \quad$ | |

$\quad 5\quad$ | $ \overline{32}\:\overline{49}\\ $ $ -25\qquad $ |

$\quad 107\quad$ | ${\;} 749\\$ $ {\;} 749 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{3249} = 57$

1369

Ans: The square root of the given number is

$\quad\quad 37 \quad$ | |

$\quad 3\quad$ | $ \overline{13}\:\overline{69}\\ $ $ -9\qquad $ |

$\quad 67\quad$ | $ {\;} 469\\ $ $ {\;} 469 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{1369} = 37$

5776

Ans: The square root of the given number is

$\quad\quad 76 \quad$ | |

$\quad 7\quad$ | $ \overline{57}\:\overline{76}\\ $ $ -49\qquad $ |

$\quad 146\quad$ | ${\;} 876\\ $ $ {\;} 876 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{5776} = 76$.

7921

Ans: The square root of the given number is

$\quad\quad 30 \quad$ | |

$\quad 8\quad$ | $ \overline{79}\:\overline{21}\\ $ $ -64\qquad $ |

$\quad 169\quad$ | $ {\;} 1521\\ $ $ {\;} 1521 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{7921} = 89$

576

Ans: The square root of the given number is

$\quad\quad 24 \quad$ | |

$\quad 2\quad$ | $\overline{5}\:\overline{76}\\ $ $ -4\qquad $ |

$\quad 44\quad$ | $ {\;} 176\\ $ $ {\;} 176 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{576} = 24$

1024

Ans: The square root of the given number is

$\quad\quad 32 \quad$ | |

$\quad 3\quad$ | $\overline{10}\:\overline{24}\\$ $ -9\qquad $ |

$\quad 62\quad$ | $ {\;} 124\\ $ $ {\;} 124 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{1024} = 32$

3136

Ans:The square root of the given number is

$\quad\quad 56 \quad$ | |

$\quad 5\quad$ | $ \overline{31}\:\overline{36}\\ $ $ -25\qquad $ |

$\quad 106\quad$ | $ {\;} 636\\ $ $ {\;} 636 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{3136} = 56$

900

Ans: The square root of the given number is

$\quad \quad 30 \quad$ | |

$\quad 3\quad$ | $\overline{9}\:\overline{00}\\ $ $-9\qquad $ |

$\quad 60\quad$ | $ {\;} 00\\ $ $ {\;} 00 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{900} = 30$

2. Find the number of digits in the square root of each of the following numbers (without any calculation).

64

Ans: Starting from the right side, take the numbers into pairs and place the bar above the number, then we get

$64 = \overline{64}$

There is only one bar (one pair) in the given number, so the number of digits in the square root will be one digit.

144

Ans: Starting from the right side, take the numbers into pairs and place the bar above the number, then we get

\[144 = \overline{1}\overline{44}\]

There are only two bar (two pair) in the given number, so the number of digits in the square root will be two-digit.

4489

Ans: Starting from the right side, take the numbers into pairs and place the bar above the number, then we get

$4489 = \overline{44}\overline{89}$

There is only two bar (two pair) in the given number, so the number of digits in the square root will be two-digit.

27225

$27225 = \overline{2}\overline{72}\overline{25}$

There is only three bar (three pair) in the given number, so the number of digits in the square root will be three-digit.

390625

$390625 = \overline{39}\overline{06}\overline{25}$

There is only three bar (three pair) in the given number, so the number of digits in the square root will be three-digit.

3. Find the square root of the following decimal numbers.

\[~\mathbf{2}.\mathbf{56}\]

Ans: The square root of the given number is

$\quad\quad 1.6 \quad$ | |

$\quad 1\quad$ | $\overline{2.}\:\overline{56}\\ $ $ -1\qquad $ |

$\quad 26\quad$ | ${\;} 156\\ $ $ {\;} 156 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{2.56} = 1.6$.

\[\mathbf{7}.\mathbf{29}\]

Ans: The square root of the given number is

$\quad\quad 2.7 \quad$ | |

$\quad 2\quad$ | $ \overline{7.}\:\overline{29}\\ $ $ -4\qquad $ |

$\quad 47\quad$ | ${\;} 329\\ $ $ {\;} 329 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{7.29} = 2.7$.

51.84

Ans: The square root of the given number is

$\quad\quad 7.2 \quad$ | |

$\quad 7\quad$ | $\overline{51.}\:\overline{84}\\ $ $ -49\qquad $ |

$\quad 142 \quad$ | ${\;} 284\\ $ $ {\;} 284 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{51.84} = 7.2$.

42.25

Ans: The square root of the given number is

$\quad\quad 6.5 \quad$ | |

$\quad 6\quad$ | $\overline{42}\:\overline{.25}\\ $ $ -36\qquad $ |

$\quad 125\quad$ | $ {\;} 625\\ $ $ {\;} 625 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{42.25} = 6.5$.

31.36

Ans: The square root of the given number is

$\quad\quad 5.6 \quad$ | |

$\quad 5\quad$ | $ \overline{31}\:\overline{.36}\\ $ $ -9\qquad $ |

$\quad 106\quad$ | $ {\;} 636\\$ $ {\;} 636 $ |

$\qquad\quad 0\quad$ |

$\Rightarrow \text{ }\sqrt{31.36} = 5.6$.

4. Find the least number, which must be subtracted from each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained.

$402$

Ans: The square root of the given number is

$\quad\quad 20 \quad$ | |

$\quad 2\quad$ | $\overline{4}\:\overline{02}\\$ $-4\qquad $ |

$\quad 40\quad$ | ${\;} 002\\$ $ {\;} 002 $ |

$\qquad\quad 0\quad$ |

The remainder of the long division method is$2$.

It shows that the square of $20$ is less than the square root of $402$ .

Therefore a perfect square will be get by subtracting $2$ from the$402$.

$\therefore \Rightarrow $The required perfect square $= 402 - 2 = 400$

The square root of the perfect square$\sqrt{400} = 20$.

1989

Ans: The square root of the given number is

$\quad\quad 44 \quad$ | |

$\quad 4\quad$ | $\overline{19}\:\overline{89}\\ $ $ -16\qquad $ |

$\quad 84\quad$ | $ {\;} 389\\ $ $ {\;} 336 $ |

$\qquad\quad 53\quad$ |

The remainder of the long division method is $53$.

It shows that the square of $44$ is less than the square root of $1989$ .

Therefore a perfect square will be obtained by subtracting $53$ from the$1989$.

$\therefore \Rightarrow $The required perfect square $= 1989 - 53 = 1936$

The square root of the perfect square$\sqrt{1989} = 44$.

3250

Ans: The square root of the given number is

$\quad\quad 57 \quad$ | |

$\quad 5\quad$ | $ \overline{32}\:\overline{50}\\ $ $ -25\qquad $ |

$\quad 107\quad$ | $ {\;} 750\\ $ $ {\;} 749 $ |

$\qquad\quad 1\quad$ |

The remainder of the long division method is $1$.

It shows that the square of $57$ is less than the square root of $3250$ .

Therefore a perfect square will be get by subtracting $2$ from the $3250$.

$\therefore \Rightarrow $The required perfect square $= 3250 - 1 = 3249$

The square root of the perfect square $\sqrt{3249} = 57$.

825

Ans: The square root of the given number is

$\quad\quad 28 \quad$ | |

$\quad 2\quad$ | $\overline{8}\:\overline{25}\\ $ $ -4\qquad $ |

$\quad 48\quad$ | $ {\;} 425\\ $ $ {\;} 384 $ |

$\qquad\quad 41\quad$ |

The remainder of the long division method is $41$.

It shows that the square of $28$ is less than the square root of $825$ .

Therefore a perfect square will be get by subtracting $41$ from the $825$.

$\therefore \Rightarrow $The required perfect square $= 825 - 41 = 784$

The square root of the perfect square $\sqrt{784} = 28$.

4000

Ans: The square root of the given number is

$\quad\quad 63 \quad$ | |

$\quad 6\quad$ | $ \overline{40}\:\overline{00}\\ $ $ -36\qquad $ |

$\quad 123\quad$ | $ {\;} 400\\ $ $ {\;} 369 $ |

$\qquad\quad 31\quad$ |

The remainder of the long division method is $31$.

It shows that the square of $63$ is less than the square root of $4000$ .

Therefore a perfect square will be get by subtracting $31$ from the $4000$.

$\therefore \Rightarrow $The required perfect square $= 4000 - 31 = 3969$

The square root of the perfect square $\sqrt{3969} = 63$.

5. Find the least number which must be added to each of the following numbers so as to get a perfect square. Also, find the square root of the perfect square so obtained.

525

Ans: The square root of the given number is

$\quad\quad 22 \quad$ | |

$\quad 2\quad$ | $ \overline{5}\:\overline{25}\\ $ $ -4\qquad $ |

$\quad 42\quad$ | $ {\;} 125\\ $ $ {\;} 84 $ |

$\qquad\quad 41\quad$ |

The remainder of the long division method is $41$.

It shows that the square of $22$ is less than the square root of $525$ .

${{22}^{2}} = 484 \And {{23}^{2}} = 529$

We need to find the least number, which must be added to the given number to get a perfect square.

Therefore a perfect square will be get by subtracting ${{23}^{2}}$ from the $525$.

$\therefore \Rightarrow $The required perfect square $= 529 - 525 = 4$

The square root of the perfect square $\sqrt{529} = 23$.

1750

Ans: The square root of the given number is

$\quad\quad 41 \quad$ | |

$\quad 4\quad$ | $ \overline{17}\:\overline{50}\\ $ $ -16\qquad $ |

$\quad 51\quad$ | $ {\;} 150\\ $ $ {\;} 81 $ |

$\qquad\quad 69\quad$ |

The remainder of the long division method is \[69\].

It shows that the square of $41$ is less than the square root of $1750$ .

${{41}^{2}} = 1681 \And {{42}^{2}} = 1764$

We need to find the least number, which must be added to the given number so as to get a perfect square.

Therefore a perfect square will be get by subtracting ${{42}^{2}}$ from the $1750$.

$\therefore \Rightarrow $The required perfect square $= 1764 - 1750 = 14$

The square root of the perfect square $\sqrt{1764} = 42$.

252

Ans: The square root of the given number is

$\quad\quad 15 \quad$ | |

$\quad 1\quad$ | $ \overline{2}\:\overline{52}\\ $ $ -1\qquad $ |

$\quad 25\quad$ | $ {\;} 152\\ $ $ {\;} 125 $ |

$\qquad\quad 27\quad$ |

The remainder of the long division method is \[27\].

It shows that the square of $15$ is less than the square root of $252$ .

${{15}^{2}} = 225 \And {{16}^{2}} = 256$

We need to find the least number, which must be added to the given number so as to get a perfect square.

Therefore a perfect square will be get by subtracting ${{16}^{2}}$ from the $252$.

$\therefore \Rightarrow $The required perfect square $= 256 - 252 = 4$

The square root of the perfect square $\sqrt{1256} = 16$.

1825

Ans: The square root of the given number is

$\quad\quad 42 \quad$ | |

$\quad 4\quad$ | $ \overline{18}\:\overline{25}\\ $ $ -16\qquad $ |

$\quad 82\quad$ | $ {\;} 225\\ $ $ {\;} 164 $ |

$\qquad\quad 61\quad$ |

The remainder of the long division method is \[61\].

It shows that the square of $42$ is less than the square root of $1825$ .

${{42}^{2}} = 1764 \And {{43}^{2}} = 1849$

We need to find the least number, which must be added to the given number so as to get a perfect square.

Therefore a perfect square will be get by subtracting ${{43}^{2}}$ from the $1825$.

$\therefore \Rightarrow $The required perfect square $= 1849 - 1825 = 24$

The square root of the perfect square $\sqrt{1849} = 43$.

6412

Ans: The square root of the given number is

$\quad\quad 80 \quad$ | |

$\quad 8\quad$ | $ \overline{64}\:\overline{12}\\ $ $ -64\qquad $ |

$\quad 160\quad$ | $ {\;} 012\\$ $ {\;} 0 $ |

$\qquad\quad 12\quad$ |

The remainder of the long division method is \[12\].

It shows that the square of $15$ is less than the square root of $6412$ .

${{80}^{2}} = 6400 \And {{81}^{2}} = 6561$

Therefore a perfect square will be get by subtracting ${{81}^{2}}$ from the $6412$.

$\therefore \Rightarrow $The required perfect square $= 6561 - 6412 = 149$

The square root of the perfect square $\sqrt{6561} = 81$.

6. Find the length of the side of a square whose area is $441 {{m}^{2}}$.

Ans: Let us take the length of the side of the square to be $x m$.

Area of Square $= {{\left( x \right)}^{2}} = 441 {{m}^{2}}$.

$x = \sqrt{441}$

The square root of $x $is

$\quad\quad 21 \quad$ | |

$\quad 2\quad$ | $ \overline{4}\:\overline{41}\\ $ $ -4\qquad $ |

$\quad 41\quad$ | $ {\;} 041\\ $ $ {\;} 041 $ |

$\qquad\quad 0\quad$ |

$\therefore \Rightarrow x = 21 m$.

So, The length of the side of a square is $21 m$.

7. In a right triangle

$ ABC, \therefore B\text{ }=\text{ }90{}^\circ . \\ $

If \[AB = 6 cm , BC = 8 cm, \]find \[AC\]

Ans:

\[\Delta ABC\]is a right angle at \[B.\]

By Pythagoras theorem,

\[A{{C}^{2}} = A{{B}^{2}} + B{{C}^{2}}\]

\[A{{C}^{2}} = {{\left( 6 cm \right)}^{2}} + {{\left( 8 cm \right)}^{2}}\]

\[A{{C}^{2}} = \left( 36 + 64 \right) c{{m}^{2}}\]

\[A{{C}^{2}} = \left( 100 \right)c{{m}^{2}}\]

\[AC = \left( \sqrt{100} \right)cm = \left( \sqrt{10 \times 10} \right)cm\]

\[AC = 10cm\].

If \[AC = 13 cm, BC = 5 cm,\] find \[AB\]

Ans:

\[\Delta ABC\]is a right angle at \[B.\]

By Pythagoras theorem,

\[A{{C}^{2}} = A{{B}^{2}} + B{{C}^{2}}\]

\[{{\left( 13 \right)}^{2}} = A{{B}^{2}} + {{\left( 5 cm \right)}^{2}}\]

\[A{{B}^{2}} = {{\left( 13 cm \right)}^{2}} - {{\left( 5 cm \right)}^{2}}\]

\[A{{B}^{2}} = \left( 169 - 25 \right)c{{m}^{2}}\]

\[A{{B}^{2}} = \left( 144 \right)c{{m}^{2}}\]

\[AC = \left( \sqrt{144} \right)cm = \left( \sqrt{12 \times 12} \right)cm\]

\[AB = 12 cm\]

8. A gardener has $1000$plants. He wants to plant these in such a way that the number of rows and the number of columns remain the same. Find the minimum number of plants he needs more for this.

Ans:

The gardener has $1000$ plants and it has the same number of rows and the same number of columns.

The square root of 1000 is

$\quad\quad 31 \quad$ | |

$\quad 3\quad$ | $ \overline{10}\:\overline{00}\\ $ $ -9\qquad $ |

$\quad 61\quad$ | $ {\;} 100\\ $ $ {\;} 061 $ |

$\qquad\quad 39\quad$ |

The remainder of the long division method is \[39\].

It shows that the square of $31$ is less than the square root of $1000$ .

${{31}^{2}} = 961 \And {{32}^{2}} = 1024$

Therefore a perfect square will be get by subtracting ${{32}^{2}}$ from the$1000$.

$\therefore \Rightarrow $The required perfect square $= 1024 - 1000 = 24$

Hence, The required number of plants is $24$.

9. These are $500$ children in a school. For a P.T. drill, they have to stand in such a manner that the number of rows is equal to the number of columns. How many children would be left out in this arrangement?

Ans:

It is given that there are $500$ children in a school.

For a P.T. drill, they have to stand in such a manner that the number of rows is equal to the number of columns.

To find that the number of children left in this arrangement,

$\quad\quad 22 \quad$ | |

$\quad 2\quad$ | $ \overline{5}\:\overline{00}\\ $ $ -4\qquad $ |

$\quad 42\quad$ | $ {\;} 100\\ $ $ {\;} 084 $ |

$\qquad\quad 16\quad$ |

The remainder of the long division method is$16$.

It shows that the square of $22$ is less than the square root of $500$ .

Therefore a perfect square will be get by subtracting $16$ from the $500$.

$\therefore \Rightarrow $ The required perfect square $= 500 - 16 = 484$

The square root of the perfect square $16$.

Then the number of children would be left out in this arrangement is $16$

### Maths NCERT Solutions Class 8 Chapter 6 Exercise 6.4

Class 8 is one of the most conceptually vital formative years for a student. It is important to have a good grip on the topics and subtopics learned in class 8 to understand more complex concepts in subsequent years. Concepts that you learn in Class 8 Maths will help you build a strong conceptual foundation. This is the right time to focus on learning Maths fundamentals the most. Class 8 is pretty much the building block of your future. Class 8 Maths Chapter 6- Squares and Square Roots is a very important chapter from the exam point of view. Prior knowledge of exponents shall help you learn this chapter well.

Chapter 6 is quite lengthy, and it is not easy to learn this chapter all on your own. You are bound to face challenges while solving NCERT Practice problems on your own. Hence it is advisable that you learn to solve all the solved examples, practice exercises, and relevant questions with the help of NCERT Solution of Class 8 Maths Chapter 6. This way you will rid yourself of all the ambiguity related to squares and square roots.

Class 8 Maths Chapter 6 deals with a number of important concepts such as Square Number, Square Roots, Properties of Square Numbers, Numbers between Square Numbers, Adding Odd Numbers, Sum of Consecutive Natural Numbers, Product of Two Consecutive Odd or Even Numbers, Finding the Square of A Number, Pythagorean Triplets, Finding Square Root Through Repeated Subtraction, Finding Square Root Through Prime Factorization, Finding Square Root Through Division Method, Square Roots of Decimals, and Estimating Square Roots

Class 8 Maths Chapter 6 Exercise 6.4 deals with Finding the Square Root Through Division Method, Finding Square Roots of Decimals, and Estimation of Square Roots. NCERT exercises mostly follow the same pattern as the CBSE question papers. Exercise 6.4 Class 8 also follows the same pattern. It includes all possible question types from this section that can be asked in your CBSE Class 8 Maths exams. They also include questions that cover all the topics in this section to facilitate splendid conceptual understanding.

### Advantages of CBSE Maths NCERT Solutions Class 8 Chapter 6 Exercise 6.4

NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.4 include step by step solutions to all the exercise questions. This makes understanding this chapter a less complicated task. Vedantu’s NCERT Solution of Class 8 Maths Chapter 6 Exercise 6.4 is comprehensive, extensive, yet crisp, and concise. They contain the shortest simplest and most logical answers to all questions facilitating effective learning. Vedantu’s NCERT Solutions have been devised to get you accustomed to tackling various question types with ease and confidence.

There are a plethora of advantages of using Vedantu as your learning hub. Some of them have been sighted below

Vedantu’s NCERT Solution of Maths Class 8 Chapter 6 Exercise 6.4 and all other NCERT Solutions are of exceptional quality. They have been devised to give you an all-inclusive learning experience that is based on the NCERT and CBSE guidelines.

Vedantu’s has all your course content in one place. All the course content is very well- structured and laid out in an organized fashion class wise, subject wise, and chapter wise. This makes the Vedantu App a very user-friendly and student-friendly App.

With the Vedantu App, you can study anywhere and anytime as per your discretion. Wherever you go, you can take your studies there. With the option of free PDF downloads, you can easily study offline. You can study without the internet, on the go, in a low network zone, and even when your data has expired.

Vedantu’s Class 8 Maths NCERT Solutions Chapter 6 Exercise 6.4 and all other solutions are a hundred percent accurate.

Exercise 6.4 Class 8 Maths NCERT Solution is comprehensive, extensive, easy to understand, yet to the point. These solutions are time and effort saving solutions.

Vedantu’s NCERT Solutions have been devised based on proven study strategies that help you remember better, learn faster, and attain more marks in your exams.

These solutions have been designed to break down complex concepts into simpler and step by step solutions to facilitate full conceptual clarity.

On the Vedantu App, you can Watch Lectures and LIVE Quizzes, attend LIVE Master Classes, Watch Free Conceptual Videos, participate in Ongoing Competitions and take Online Tests. These help you maintain conceptual clarity always and enhance your capabilities. LIVE courses and LIVE Master Classes teach you effective shortcuts and hacks that help you solve problems sums quickly, efficiently and effectively. Taking Vedantu’s Online Tests makes you accustomed to the exam pattern and exam environment. It helps you estimate the quality of your preparation.

### About Vedantu

Vedantu is an online learning platform that has brought about a significant improvement in the outcome of the exam scores of students of all classes all across the nation. It has substantially revolutionized the education space of the nation. Vedantu’s one of a kind one-on-one tutorials ensure that all students get the personal attention they need to unleash their full potential and perform well in their exams. On the Vedantu App, you can choose your tutor and start learning immediately. Vedantu aims to impart excellent quality education to students worldwide. It uses advanced technologies and modern and innovative teaching methodologies to facilitate effective learning. Vedantu is India’s pioneer e-Learning Platform that has changed the way students study. It has made learning a more interesting, engaging, exciting and stress-free process. All its efforts are directed towards making learning a more straightforward and natural process of a student’s life. It has a very futuristic and modern outlook towards education. The Vedantu App addresses all the anomalies of the existing education system of the country. All its course material is developed by proficient subject experts in the best of your interests. All NCERT Solutions are available on the Vedantu App as free PDF Downloads.