NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes (Ex 13.3) Exercise 13.3

VSAT 2022

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes (Ex 13.3) Exercise 13.3

Free PDF download of NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.3 (Ex 13.3) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.3 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all NCERT Book Solutions in your emails. Download maths NCERT solutions class 9 prepared by Master teacher at Vedantu. Students can also avail of NCERT Solutions for Class 9 Science from our website.

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Access NCERT solutions Class 9 Maths Chapter 13 - Surface Areas and Volumes

Exercise (13.3)

1. Diameter of the base of a cone is $\text{10}\text{.5 cm}$ and its slant height is $\text{10 cm}$. Find its curved surface area. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

We are Given the Following:

The slant height $\left( \text{l} \right)$ of the cone $\text{= 10 cm}$

The diameter of the base of cone $\text{= 10}\text{.5 cm}$

So, the radius $\left( \text{r} \right)$ of the base of cone $\text{= }\frac{\text{10}\text{.5}}{\text{2}}\text{ cm = 5}\text{.25 cm}$

The curved surface area of cone, $\text{A =  }\!\!\pi\!\!\text{ rl}$

$\Rightarrow \text{A = }\left( \frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ 5}\text{.25 }\!\!\times\!\!\text{ 10} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \text{22 }\!\!\times\!\!\text{ 0}\text{.75 }\!\!\times\!\!\text{ 10} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 165 c}{{\text{m}}^{\text{2}}}$

Therefore, the curved surface area of the cone is $\text{165 c}{{\text{m}}^{\text{2}}}$.


2. Find the total surface area of a cone, if its slant height is $\text{21 m}$and diameter of its base is $\text{24 m}$. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

We are Given the Following:

The slant height $\left( \text{l} \right)$ of the cone $\text{= 21 m}$

The diameter of the base of cone $\text{= 24 m}$

So, the radius $\left( \text{r} \right)$ of the base of cone $\text{= }\frac{\text{24}}{\text{2}}\text{ m = 12 m}$

The total surface area of cone, $\text{A =  }\!\!\pi\!\!\text{ r}\left( \text{l + r} \right)$

$\Rightarrow \text{A = }\left( \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  12  }\!\!\times\!\!\text{  }\left( \text{21 + 12} \right) \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  12  }\!\!\times\!\!\text{  33} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 1244}\text{.57 }{{\text{m}}^{\text{2}}}$

Therefore, the total surface area of the cone is $\text{1244}\text{.57 }{{\text{m}}^{\text{2}}}$.


3. Curved surface area of a cone is $\text{308 c}{{\text{m}}^{\text{2}}}$ and its slant height is $\text{14 cm}$. Find

i. Radius of the Base

Ans:

It is given that the slant height $\left( \text{l} \right)$ of the cone $\text{= 14 cm}$

The curved surface area of the cone $\text{= 308 c}{{\text{m}}^{\text{2}}}$

Let us assume the radius of base of cone be $\text{r}$.

We know that curved surface area of the cone $\text{=  }\!\!\pi\!\!\text{ rl}$

$\therefore \text{ }\!\!\pi\!\!\text{ rl = 308 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \left( \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  r  }\!\!\times\!\!\text{  14} \right)\text{ cm = 308 c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{r = }\frac{\text{308}}{\text{44}}\text{ cm}$

$\Rightarrow \text{r = 7 cm}$

Hence, the radius of the base is $\text{7 cm}$.


ii. Total Surface Area of the Cone. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

The total surface area of the cone is the sum of its curved surface area and the area of the base.

Total surface area of cone, $\text{A =  }\!\!\pi\!\!\text{ rl +  }\!\!\pi\!\!\text{ }{{\text{r}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{308 + }\frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  }{{\left( \text{7} \right)}^{\text{2}}} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left[ \text{308 + 154} \right]\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 462 c}{{\text{m}}^{\text{2}}}$

Hence, the total surface area of the cone is $\text{462 c}{{\text{m}}^{\text{2}}}$.


4. A conical tent is $\text{10 m}$ high and the radius of its base is $\text{24 m}$. Find

i. slant height of the tent

Ans:


(image will be uploaded soon)


From the figure we can say that $\text{ABC}$ is a conical tent.

It is given that the height $\left( \text{h} \right)$ of conical tent $\text{= 10 m}$

The radius $\left( \text{r} \right)$ of conical tent $\text{= 24 m}$

Let us assume the slant height as $\text{l}$.

In $\text{ }\!\!\Delta\!\!\text{ ABD}$, we will use Pythagorean Theorem.

$\therefore \text{A}{{\text{B}}^{\text{2}}}\text{ = AD}{{\text{ }}^{\text{2}}}\text{ + B}{{\text{D}}^{\text{2}}}$

$\Rightarrow {{\text{l}}^{\text{2}}}\text{ = }{{\text{h}}^{\text{2}}}\text{ + }{{\text{r}}^{\text{2}}}$

\[\Rightarrow {{\text{l}}^{\text{2}}}\text{ = }{{\left( \text{10 m} \right)}^{\text{2}}}\text{ + }{{\left( \text{24 m} \right)}^{\text{2}}}\]

$\Rightarrow {{\text{l}}^{\text{2}}}\text{ = 676 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{l = 26 m}$

The slant height of the tent is $\text{26 m}$.


ii. cost of canvas required to make the tent, if cost of $\text{1 }{{\text{m}}^{\text{2}}}$ canvas is $\text{Rs}\text{. 70}$. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

The curved surface area of the tent, $\text{A =  }\!\!\pi\!\!\text{ rl}$

$\Rightarrow \text{A = }\left( \frac{\text{22}}{\text{7}}\text{  }\!\!\times\!\!\text{  24  }\!\!\times\!\!\text{  26} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = }\left( \frac{13728}{7} \right)\text{ }{{\text{m}}^{\text{2}}}$

It is given that the cost of $\text{1 }{{\text{m}}^{\text{2}}}$ of canvas $\text{= Rs}\text{. 70}$

So, the cost of $\frac{13728}{7}\text{ }{{\text{m}}^{\text{2}}}$ canvas $\text{= Rs}\text{. }\left( \frac{\text{13728}}{\text{7}}\text{  }\!\!\times\!\!\text{  70} \right)\text{ = Rs}\text{. 137280}$

Hence, the cost of canvas required to make the tent is $\text{Rs}\text{. 137280}$.


5. What length of tarpaulin $\text{3 m}$ wide will be required to make conical tent of height $\text{8 m}$ and base radius $\text{6 m}$? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately $\text{20 cm}$. $\left[ \text{Use  }\!\!\pi\!\!\text{  = 3}\text{.14} \right]$

Ans:

We are Given the Following:

The base radius $\left( \text{r} \right)$ of tent $\text{= 6 m}$

The height $\left( \text{h} \right)$ of tent $\text{= 8 m}$

So the slant height of the tent, $\text{l = }\sqrt{{{\text{r}}^{\text{2}}}\text{ + }{{\text{h}}^{\text{2}}}}$ 

$\Rightarrow \text{l = }\left( \sqrt{{{\text{6}}^{\text{2}}}\text{ + }{{\text{8}}^{\text{2}}}} \right)\text{ m}$

$\Rightarrow \text{l = }\left( \sqrt{\text{100}} \right)\text{ m}$

$\Rightarrow \text{l = 10 m}$

The curved surface area of the tent, $\text{A =  }\!\!\pi\!\!\text{ rl}$

$\Rightarrow \text{A = }\left( \text{3}\text{.14  }\!\!\times\!\!\text{  6  }\!\!\times\!\!\text{  10} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 188}\text{.4 }{{\text{m}}^{\text{2}}}$

It is give the width of tarpaulin $\text{= 3 m}$

Let us assume the length of the tarpaulin sheet required be $\text{x}$.

It is given that there will be a wastage of $\text{20 cm}$.

So, the new length of the sheet $\text{=}\left( \text{x - 0}\text{.2} \right)\text{ m}$

We know that the area of the rectangular sheet required will be the same as the curved surface area of the tent.

$\therefore \left[ \left( \text{x - 0}\text{.2} \right)\text{  }\!\!\times\!\!\text{  3} \right]\text{ m = 188}\text{.4 }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{x - 0}\text{.2 m = 62}\text{.8 m}$

$\Rightarrow \text{x = 63 m}$

The length of tarpaulin sheet required is $\text{63 m}$.


6. The slant height and base diameter of a conical tomb are $\text{25 m}$ and $\text{14 m}$ respectively. Find the cost of white-washing its curved surface at the rate of $\text{Rs}\text{. 210}$ per $\text{100 }{{\text{m}}^{\text{2}}}$. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

We are Given the Following:

The base radius $\left( \text{r} \right)$ of tomb $\text{= 7 m}$

The slant height $\left( \text{l} \right)$ of tomb $\text{= 25 m}$

The curved surface area of the conical tomb, $\text{A =  }\!\!\pi\!\!\text{ rl}$

$\Rightarrow \text{A = }\left( \frac{22}{7}\text{  }\!\!\times\!\!\text{  7  }\!\!\times\!\!\text{  25} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 550 }{{\text{m}}^{\text{2}}}$

It is given that the cost of white-washing $\text{1 }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 210}$

So, the cost of white-washing $550\text{ }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \frac{\text{210}}{\text{100}}\text{  }\!\!\times\!\!\text{  550} \right)\text{ = Rs}\text{. 1155}$

Hence, the cost of white-washing the curved surface area of a conical tomb is $\text{Rs}\text{. 1155}$.


7. A joker’s cap is in the form of a right circular cone of base radius $\text{7 cm}$ and the height $\text{24 cm}$. Find the area of the sheet required to make $\text{10}$ such caps. $\left[ \text{Assume  }\!\!\pi\!\!\text{  = }\frac{\text{22}}{\text{7}} \right]$

Ans:

We are Given the Following:

The base radius $\left( \text{r} \right)$ of conical cap $\text{= 7 cm}$

The height $\left( \text{h} \right)$ of conical cap $\text{= 24 cm}$

So the slant height of the tent, $\text{l = }\sqrt{{{\text{r}}^{\text{2}}}\text{ + }{{\text{h}}^{\text{2}}}}$ 

$\Rightarrow \text{l = }\left( \sqrt{{{\text{7}}^{\text{2}}}\text{ + 2}{{\text{4}}^{\text{2}}}} \right)\text{ cm}$

$\Rightarrow \text{l = }\left( \sqrt{625} \right)\text{ cm}$

$\Rightarrow \text{l = 25 cm}$

The curved surface area of one conical cap, $\text{A =  }\!\!\pi\!\!\text{ rl}$

$\Rightarrow \text{A = }\left( \frac{22}{7}\text{  }\!\!\times\!\!\text{  7  }\!\!\times\!\!\text{  25} \right)\text{ c}{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 550 c}{{\text{m}}^{\text{2}}}$

So, the curved surface area of $\text{10}$ conical caps $\text{= }\left( \text{550  }\!\!\times\!\!\text{  10} \right)\text{ c}{{\text{m}}^{\text{2}}}\text{ = 5500 c}{{\text{m}}^{\text{2}}}$

Therefore, the total area of the sheet required is $\text{5500 c}{{\text{m}}^{\text{2}}}$.


8. A bus stop is barricaded from the remaining part of the road, by using $\text{50}$ hollow cones made of recycled cardboard. Each cone has a base diameter of $\text{40 cm}$ and height $\text{1 m}$. If the outer side of each of the cones is to be painted and the cost of painting is $\text{Rs}\text{. 12}$ per ${{\text{m}}^{\text{2}}}$, what will be the cost of painting all these cones? 

$\left[ \text{Use  }\!\!\pi\!\!\text{  = 3}\text{.14 and take }\sqrt{\text{1}\text{.02}}\text{=1}\text{.02} \right]$

Ans:

We are Given the Following:

The base radius $\left( \text{r} \right)$ of cone $\text{= }\frac{\text{40}}{\text{2}}\text{ = 20 cm = 0}\text{.2 m}$

The height $\left( \text{h} \right)$ of cone $\text{= 1 m}$

So the slant height of the cone, $\text{l = }\sqrt{{{\text{r}}^{\text{2}}}\text{ + }{{\text{h}}^{\text{2}}}}$ 

$\Rightarrow \text{l = }\left( \sqrt{{{\left( \text{0}\text{.2} \right)}^{\text{2}}}\text{ + }{{\left( \text{1} \right)}^{\text{2}}}} \right)\text{ m}$

$\Rightarrow \text{l = }\left( \sqrt{1.04} \right)\text{ m}$

$\Rightarrow \text{l = 1}\text{.02 m}$

The curved surface area of one cone, $\text{A =  }\!\!\pi\!\!\text{ rl}$

$\Rightarrow \text{A = }\left( \text{3}\text{.14  }\!\!\times\!\!\text{  0}\text{.2  }\!\!\times\!\!\text{  1}\text{.02} \right)\text{ }{{\text{m}}^{\text{2}}}$

$\Rightarrow \text{A = 0}\text{.64056 c}{{\text{m}}^{\text{2}}}$

So, the curved surface area of $\text{50}$ cones $\text{= }\left( \text{50  }\!\!\times\!\!\text{  0}\text{.64056} \right)\text{ }{{\text{m}}^{\text{2}}}\text{ = 32}\text{.028 }{{\text{m}}^{\text{2}}}$

It is given that the cost of painting $\text{1 }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. 12}$

So, the cost of painting $32.028\text{ }{{\text{m}}^{\text{2}}}$ area $\text{= Rs}\text{. }\left( \text{32}\text{.028  }\!\!\times\!\!\text{  12} \right)\text{ = Rs}\text{. 384}\text{.336}$

We can also write the cost approximately as $\text{Rs}\text{. 384}\text{.34}$.

Therefore, the cost of painting all the hollow cones is $\text{Rs}\text{. 384}\text{.34}$.


NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas And Volumes Exercise 13.3

Opting for the NCERT solutions for Ex 13.3 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 13.3 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 13 Exercise 13.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 9 Maths Chapter 13 Exercise 13.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 9 Maths Chapter 13 Exercise 13.3 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.


Important Topics Covered in Exercise 13.3 of Class 9 Maths NCERT Solutions

NCERT Solutions Class 9 Maths Exercise 13.3 is based on the following topics.

  • Image of a three-dimensional geometric shape, i.e., the right circular cone (students can easily relate to it with an ice cream cone)

  • Determining the surface area of the right-angular cone


Exercise 13.3 explains the step-by-step derivation of the formula to calculate the curved surface area of the right circular cone so that the students can understand the concepts in a simplified way.


The key takeaways from this exercise are the formulas of the surface area and volume of the right circular cone which are given below.

  • Curved Surface Area of a Cone = 1/2 × l × 2πr = πrl,

where ‘r’ is its radius and ‘l’ is its slant height. So, ‘l’ = (r2 + h2)

  • Total Surface Area of a Cone = πrl + πr 2 = πr(l + r)

Students are advised to memorise the formulas for finding the surface areas and volumes of various figures as it will help them while solving the sum given in this exercise as well as in the exams.


Chapter wise NCERT Solutions for Class 9 Maths


CBSE Class 9 Maths Chapter 13 Other Exercises

Chapter 13 Surface Areas and Volumes All Exercises in PDF Format

Exercise 13.1

8 Questions & Solutions (4 Short Answers, 4 Long Answers)

Exercise 13.2

11 Questions & Solutions (5 Short Answers, 6 Long Answers)

Exercise 13.4

9 Questions & Solutions (4 Short Answers, 5 Long Answers)

Exercise 13.5

9 Questions & Solutions (4 Short Answers, 5 Long Answers)

Exercise 13.6

8 Questions & Solutions (8 Short Answers)

Exercise 13.7

9 Questions & Solutions (9 Long Answers)

Exercise 13.8

10 Questions & Solutions (5 Short Answers, 5 Long Answers)

Exercise 13.9

3 Questions & Solutions (3 Long Answers)

FAQs (Frequently Asked Questions)

1. What are the benefits of referring to NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3?

NCERT Solutions are the best study materials to find answers to all the textbook exercise problems at one place. Using NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3, students will be able to find out surface areas and volumes of a cone. If they have any doubts regarding any question, they can refer to the solutions provided by subject matter experts. E-learning platforms like Vedantu provide authentic and well-researched NCERT Solutions designed as per the latest guidelines and syllabus to help students in learning a chapter better. Students can refer to exercise-wise NCERT Solutions for Class 9 Maths Chapter 13 to score well in the exams.

2. How to calculate the surface area and volume of a cone?

The surface area and volume of a cone can be calculated with the help of the formulas given below:

  • Curved Surface Area of a Cone= πrl

  • Total Surface Area of a Cone=  πrl + πr2 = πr (l + r)

  • The volume of a Cone= 1/3 πr2h

Where r is the base radius, h is the height and l is the slant height of the cone.

3. Where can I avail exercise-wise NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes?

Vedantu provides well-prepared NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes for all the exercises. Chapter 13 of CBSE Class 9 Maths is an exhaustive chapter as it includes a total of nine exercises. Students can find solutions to all the exercises’ questions on Vedantu’s site provided by expert tutors. This is an important chapter from the examination point of view. Hence, students must solve each and every problem and refer to Vedantu’s site for the needed solutions. They can also register for master classes for better doubt clearance.

4. What are the salient features of Vedantu’s NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.3?

NCERT Solutions by Vedantu are the most comprehensive study resource available online. Class 9 Mathematics NCERT Solutions for Chapter 13 Surface Areas and Volumes Exercise 13.3 as well as other exercises include step-by-step explanations of the problems. Students must practice all the questions of the exercise as it is really important from the exam point of view. Some of the salient features of this study material curated by Vedantu are:

  • It helps in self-assessment

  • Provides effective revision and practice

  • Prepared by subject experts

5. What are the topics covered in Class 9 Maths Chapter 13 Exercise 13.3?

Class 9 Maths Chapter 13 Exercise 13.3 covers the following important topics.

  • Diameter of a cone

  • Surface area of a cone

  • Radius of a base 

6. How could I learn Chapter 13 Exercise 13.3 of Class 9 Maths in the efficient and fastest way?

Fastest way doesn’t mean that you take shortcuts in solving the answers. It means the learning time. Suppose you are not having much time for preparation for exams, and you want to learn the concepts in a faster yet effective way, then in such a case you can visit Vedantu, which has all the concepts for you in a concise form. The PDF of NCERT solutions for Class 9 Maths Chapter 13 Exercise 13.3 carries proper explanations for the same.

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It is easy to download the latest NCERT Solutions for Chapter 13 Exercise 13.3 of Class 9 Maths just by clicking on one specific link. How to approach that correct link is very important. Firstly, visit Vedantu website (vedantu.com). Then click on the page NCERT Class 9 Maths book, where you will be re-directed towards the download of the latest NCERT Solutions for Class 9 Maths. Select the relevant chapter and the exercise for the latest NCERT Solutions for Chapter 13, Exercise 13.3. 

8. Can I take the help of the Maths experts at Vedantu for Exercise 13.3 of Chapter 13 of Class 9 Maths?

Yes, you can take the help of Maths experts at Vedantu. This facility is available to everyone who wants to get the right guidance regarding their subjects. They will help you by clearing all your doubts. By getting help from the Maths experts, you can solve the questions properly. They are ready to help you out till you are satisfied and understand the concepts thoroughly. You can also download the PDF of NCERT solutions for Class 9 Maths free of cost. These solutions are available at free of cost on Vedantu’s website(vedantu.com) and mobile app.

9. How can I improve in Chapter 13 Exercise 13.3 of Class 9 Maths?

If you are not able to improve your Maths for Class 9 Chapter 13 Exercise 13.3, then don’t simply jump to solving exercises . First understand the concept and try solving example sums. Then proceed to solving the sums from 13.3. However, solve all the problems to improve yourself in the said exercise.

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