NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.5) Exercise 13.5
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For students, Class 10 is an important academic year. It has an impact on their academic and professional endeavours. However, a student’s success is not solely based on how well they perform throughout the course of one academic year. Since Class 9 and Class 10 are two sides of the same coin, the academic session of Class 9 is seen as a turning moment in students’ academic careers. Before taking the Class 10 board exams, a student must pass the annual exams for Class 9. It also provides the framework for upcoming classes. Students who are adept at the Class 9 curriculum have a higher chance of doing well in the Class 10 board exams.
The possibility of questions from topics taught in Class 9 appearing in competitive exams is high. Students must therefore reinforce their understanding of the curriculum of Class 9. The subjects of Science and Mathematics require a lot of practice. One of the greatest resources for learning the content are the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5. This resource can help students examine how well they comprehend key concepts in class 9 Mathematics and develop their skills.
The importance of the Mathematics curriculum in Class 9 is evident. This is an essential resource that outlines the subjects studied and lays the foundation for the Class 10 board exams. Research requirements and tasks are outlined in the curriculum. The Class 9 syllabus is very important for students who wish to excel in their future board examinations. Students are advised to take their annual exams Class 9 seriously as they can affect their academic career in the future.
Students can find the Class 9 syllabus on the Extramarks website along with resources like the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5 can be a great tool to prepare for Class 9 Mathematics examinations. However, students are advised to go through the syllabus first before starting the preparation for any subject. This can help students plan their preparation efficiently and save a lot of time by doing so. Time management is crucial when it comes to preparation for examinations. Making use of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5 along with other resources and being thorough with the syllabus can help students manage their time efficiently.
The Mathematics textbook for Class 9 compiled by the NCERT can be efficiently studied and prepared with the help of tools like the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5. By covering the key concepts presented throughout the course, NCERT books provide students with a firm conceptual foundation in the subject. This enables students to score better on exams. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5, when used in conjunction with the NCERT textbook, may increase students’ chances of performing well in their exams.
To aid students in better understanding challenging concepts and computations, subjectmatter experts of Extramarks have created the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5 would assist students in moving closer to their goals.
For students in elementary and middle schools, the National Council for Educational Research and Training publishes and distributes textbooks. What one can find most amazing while considering the value of NCERT books for CBSE board exam preparation is that these books concentrate on solidifying conceptual understanding to assist students in understanding basic ideas. To optimally utilise the learning potential of NCERT books, students are advised by the experts at Extramarks to utilise the NCERTrelated resources that are available on the Extramarks website, such as the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5. Both test preparation and classroom instruction can be completed efficiently with the use of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5, and other similar tools.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.5) Exercise 13.5
Highly qualified mentors of Mathematics at Extramarks have developed the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5. Students can also find resources for elementary and grade levels similar to the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5 on the Extramarks website. Students are encouraged to use these learning assets in order to fully understand everything that is being discussed in class. Without a concrete conceptual foundation in elementary school classroom learning materials, students are unable to comprehend the middle and high school curriculum. Without a reliable foundation, students may lose interest in the subject. Students are encouraged to use tools created by exceptionally knowledgeable mentors at Extramarks. Numerous elearning resources, such as the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5, are readily available on Extramarks.
Using the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5 is a great way to prepare for the topics presented in Maths Class 9 Chapter 13 Exercise 13.5. Students can successfully complete this chapter using the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5 and other tools especially targeted towards this chapter available on the Extramarks website. Students can find detailed answers to all questions in the Class 9 Maths Chapter 13 Exercise 13.5 in the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5. Mentors with expertise in Mathematics prepare the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5. Students can understand the material better if they spend more time studying quality reference materials such as NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5.
The Maths Class 9 Chapter 13 Exercise 13.5 is just one of the many exercises from this chapter. Exercise 13.5 Class 9 Maths specifically can be studied with the help of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5 but there are similar solutions to all the other exercises as well available on the Extramarks learning portal.
Access NCERT solutions Class 9 Maths Chapter 13 – Surface Areas and Volumes
A degree in Mathematics can provide students with a wide range of intriguing and rewarding career prospects. Students can become more adept at problemsolving as a result of studying Mathematics. It equips the students with abilities that they can employ in a variety of fields and professional responsibilities. There are many benefits to pursuing a degree in Mathematics. Students who are interested in doing so can benefit greatly from the NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.5 and other similar tools.
Four of the top advantages of studying Mathematics are:
 A better understanding of the world acquired.: Numerous fields of study and industries use Mathematics. Both realworld problems and solutions are affected by it. Humans use mathematical concepts in everything, from simple operations like counting and purchasing to more complex issues like data analysis. To master these skills, students are advised to make use of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5, and other such tools available on the Extramarks website.
 Gaining a foundational understanding of the subject with a Mathematics degree enables students to contribute to the future development of numerous sectors. The worldview that the students receive from a Mathematics degree can help them succeed in whichever route they choose, whether they decide to specialise in Mathematics or use the information they gain in another academic field. No matter the future plans of the students, indepth knowledge of Mathematics always proves to be beneficial and that can be attained by resources like the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5 .
 The core skill prerequisite to learning Mathematics is problemsolving. Along with teaching students how to solve difficult mathematical equations, working on mathematical functions will also help the students become more adept at other types of problemsolving, such as:
 Pattern recognition
 working in reverse
 Visualising
 systematically working
 utilising logical thinking
These skills will be useful in a wide range of occupations and circumstances, regardless of whether students decide to pursue a career in Mathematics or in a completely different area. Hence, it is advised that students take full advantage of tools like the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5 to master the subject of Mathematics.
 Students can utilise the abilities they learn in Mathematics in a variety of professional and social contexts. In this way, the Mathematics degree will teach them abilities that will serve them well throughout their lives. The following are just a few examples of the transferable talents that a Mathematics degree can give students:
 Data Evaluation
 Structure Critical thinking
 management of time
 Decisionmaking in communications
Exercise (13.5)
A majority of students perceive Mathematics to be a challenging subject, but all students must sincerely study it. However, Mathematics is an important, fascinating, and extraordinary aspect of human reality. Cuttingedge developments are made possible in a number of crucial sectors, including Engineering, Science, Business, and Technology with the assistance of concepts embedded in Mathematics. With the aid of helpful resources available on the Extramarks website like the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5, students can master the themes encapsulated in Exercise 13.5.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5
Attempting to retain the steps involved in a mathematical problem is not recommended, but it is important to keep this in mind. This will turn out to be ineffective. It is far better to focus attention on comprehending the reasoning and method involved. Students are advised to be aware of the main concepts in order to grasp the topic well. They will benefit in the long run. For instance: Algebra and Geometry serve as the foundation for later, more complex Arithmetic topics. Students can take help from the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5, and other similar tools to achieve this.
Selfstudy is the key to mastering Mathematics. This is another extremely effective method for understanding Mathematics. Selfstudy can improve the students’ comprehension of the concepts and familiarise them with the challenging ones. Students are encouraged to spend time attempting to comprehend the steps involved in solving a question if they run into difficulty. It can be simpler to go on to the remaining question after they have a firm grasp of the original problem. Students can take help from the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5, for selfguided learning.
NCERT Solutions for Class 9
The use of the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5 here can help students study on a regular basis. Mathematics cannot be learned well through reading and listening alone. Practice will aid in solidifying the ideas and fundamentals in their mind. Each problem has unique properties, thus it is crucial to have found several solutions to it before taking the examination. The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5 are the best tool in this regard.
CBSE Study Materials for Class 9
The NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5, are just one of several helpful tools that the students can refer to for their examination preparation. Students can easily access these tools by visiting the Extramarks website. These tools are not only easily accessible but also extremely easy to comprehend.
CBSE Study Materials
The more students use tools such as the NCERT Solutions For Class 9 Maths Chapter 13 Exercise 13.5, the more they can get revision done for mathematical questions that are given within their NCERT Mathematics book. It is recommended that students spend adequate time practising and revising with the solutions for Class 9 Chapter 13 Maths Exercise 13.5. This can help the students achieve great scores in the examinations.
Q.1 A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes?
Ans
Volume of a matchbox = 4 cm × 2.5 cm × 1.5 cm
= 15 cm^{3}
Volume of a packet = Volume of 12 matchboxes
= 12 x 15 cm^{3}
= 180 cm^{3}
Thus, the volume of a packet containing 12 matchboxes is 180 cm^{3}.
Q.2 A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m^{3} = 1000 l)
Ans
Volume of cuboidal water tank = 6 x 5 x 4.5 m^{3}
= 135 m^{3}
Since, 1 m^{3} = 1000 litres
So, volume of water in tank = 135 x 1000 litres
= 135000 litres
Q.3 A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?
Ans
$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{\hspace{0.17em}Volume of a liquid = 380 m}}^{\text{3}}\\ \text{\hspace{0.17em}Length of cuboidal vessel = 10 m}\\ \text{Breadth of cuboidal vessel = 8 m}\\ \text{Let height of vessel be h m}\text{.}\\ \text{Then,}\\ \text{volume of cuboidal vessel}=\text{\hspace{0.17em}}l\times b\times h\\ {\text{380 m}}^{\text{3}}=10\times 8\times h\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}h=\frac{380}{80}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4.75\text{\hspace{0.17em}}m\\ \text{Thus, cuboidal vessel should be 4}\text{.75 m high}\text{.}\end{array}$
Q.4 Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs 30 per m^{3}.
Ans
Volume of cuboidal pit = 8 x 6 x 3
= 144 m^{3}
Cost of digging 1 m^{3} pit = Rs. 30
Then, the cost of digging cuboidal pit
= Rs. 30 x 144
= Rs. 4320
Thus, the cost of digging a cuboidal pit is Rs. 4320.
Q.5 The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Ans
Capacity of cuboidal tank = 50,000 litres
Length of cuboidal tank = 2.5 m
Depth of cuboidal tank = 10 m
Let breadth of cuboidal tank = p m
Since, 1000 litres = 1 m^{3}
So, 50,000 litres = 50 m^{3}
Then, Volume of tank = lbh
50 m^{3}= 2.5 x p x 10
p = 2 m
Thus, the breadth of the tank is 2 m.
Q.6 A village, having a population of 4000, approximately requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?
Ans
Volume of water tank = 20 m × 15 m × 6 m
= 1800 m^{3}
= 1800000 litres
[Since, 1 m^{3} = 1000 litres]
Water require for a person = 150 litres
Water require for 4000 persons
= 150 x 4000 litres
= 600000 litres
Number of days to last the water of this tank
= (1800000/600000) days
= 3 days
Thus, the water of tank will last in 3 days in a village of 4000 people.
Q.7 A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
Ans
$\begin{array}{l}\text{Number of wooden crates in godown}\\ =\frac{\text{Volume of godown}}{\text{Volume of wooden crates}}\\ =\frac{\text{4}0\text{m}\times \text{25 m}\times \text{15 m}}{\text{1}.\text{5 m}\times \text{1}.\text{25 m}\times \text{}0.\text{5 m}}\\ =\frac{\text{4}00\text{m}\times \text{2500 m}\times \text{150 m}}{\text{15 m}\times \text{125 m}\times \text{5 m}}\\ =80\times 20\times 10\\ =16000\\ \text{Thus},\text{16000 wooden crates can be stored in the godown}\text{.}\end{array}$
Q.8 A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
Ans
$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Side of solid cube}=\text{12 cm}\\ \text{Number of cubes of equal volume}=\text{8}\\ \text{Side of each cube of equal volume}=\frac{\text{Volume of solid cube}}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{12\times 12\times 12}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=216\text{\hspace{0.17em}}c{m}^{3}\\ \text{Side\hspace{0.17em}}\text{of each cube of equal volume}=\sqrt[3]{216}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}6\text{\hspace{0.17em}}cm\\ \text{Ratio of sufaces area of both cubes}=\frac{4{\left(12\right)}^{2}}{4{\left(6\right)}^{2}}\\ =\frac{12\times 12}{6\times 6}\\ =4:1\end{array}$
$\text{Thus},\text{the required ratio of surface areas of both cubes is 4:1}\text{.}$
Q.9 A part of a river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
Ans
$\begin{array}{l}\text{Speed of river = 2 km per hour}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\left(\text{2000/60}\right)\text{m per minute}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\frac{\text{100}}{\text{3}}\text{m per min}\\ \text{Then,}\\ \text{Volume of flowing water in one minute}\\ \text{\hspace{0.17em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}}\left(\frac{\text{100}}{\text{3}}\right)\left(\text{3\hspace{0.17em}}\right)\left(\text{40}\right)\\ {\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.17em}=\hspace{0.33em}4000\hspace{0.17em}m}}^{\text{3}}\\ {\text{Thus, 4000 m}}^{\text{3}}\text{\hspace{0.17em}water will fall into the sea in a minute.}\end{array}$
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FAQs (Frequently Asked Questions)
1. Is it important to retain the theorems in Mathematics?
It may seem like retaining the theorms is the only way to prepare for them. However, students are advised to practice these theorms enough times so that it becomes easier to actually understand them. When students truly understand the theorems, they would not feel the need to retain them.