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# NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.5) Exercise 13.5

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.5) Exercise 13.5

Last updated date: 30th Jan 2023
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Free PDF download of NCERT Solutions for Class 9 Maths Chapter 13 Exercise 13.5 (Ex 13.5) 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 13.5 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise NCERT Solution in your emails. You can also download NCERT Solutions for Class 9 Maths to help you to revise complete syllabus and score more marks in your examinations. Students can also avail of NCERT Solutions Class 9 Science from our website.

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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Ex 13.5) Exercise 13.5
Surface Area and Volume L-1 | Surface Area & Volume of Cuboid & Cube | CBSE 9 Maths Ch 13 | Term 2
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## Exercise (13.5)

1. A matchbox measures $\text{4 cm }\!\!\times\!\!\text{ 2}\text{.5 cm }\!\!\times\!\!\text{ 1}\text{.5 cm}$. What will be the volume of a packet containing $\text{12}$ such boxes?

Ans:

We are given the following:

Length of matchbox $\text{l = 4 cm}$

Breadth of matchbox $\text{b = 2}\text{.5 cm}$

Height of matchbox $\text{h = 1}\text{.5 cm}$

The volume of single matchbox $\text{V = l }\!\!\times\!\!\text{ b }\!\!\times\!\!\text{ h}$

$\Rightarrow \text{V = }\left( \text{4 }\!\!\times\!\!\text{ 2}\text{.5 }\!\!\times\!\!\text{ 1}\text{.5} \right)\text{ c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 15 c}{{\text{m}}^{\text{3}}}$

So, the volume of $\text{12}$ matchboxes $\text{= }\left( \text{15 }\!\!\times\!\!\text{ 12} \right)\text{ c}{{\text{m}}^{\text{3}}}\text{ = 180 c}{{\text{m}}^{\text{3}}}$

Hence, the volume of $\text{12}$ matchboxes is $\text{180 c}{{\text{m}}^{\text{3}}}$.

2. A cuboidal water tank is $\text{6 m}$ long, $\text{5 m}$ wide and $\text{4}\text{.5 m}$ deep. How many litres of water can it hold? $\left( \text{1 }{{\text{m}}^{\text{3}}}\text{ = 1000 litre} \right)$

Ans:

We are given the following:

Length of tank $\text{l = 6 m}$

Breadth of tank $\text{b = 5 m}$

Height of tank $\text{h = 4}\text{.5 m}$

The volume of tank $\text{V = l }\!\!\times\!\!\text{ b }\!\!\times\!\!\text{ h}$

$\Rightarrow \text{V = }\left( \text{6 }\!\!\times\!\!\text{ 5 }\!\!\times\!\!\text{ 4}\text{.5} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 135 }{{\text{m}}^{\text{3}}}$

We know the amount of water with volume $\text{1 }{{\text{m}}^{\text{3}}}$ can hold $\text{= 1000 litres}$

So, the amount of water with volume $\text{135 }{{\text{m}}^{\text{3}}}$ can hold $\text{= }\left( \text{1000 }\!\!\times\!\!\text{ 135} \right)\text{ litres = 135000 litres}$

Therefore, the tank can contain $\text{135000 litres}$ of water.

3. A cuboidal vessel is $\text{10 m}$ long and $\text{8 m}$ wide. How high must it be made to hold $\text{380}$ cubic metres of a liquid?

Ans:

We are given the following:

Length of vessel $\text{l = 10 m}$

Width of vessel $\text{b = 8 m}$

Volume of vessel $\text{V = 380 }{{\text{m}}^{\text{3}}}$

Let $\text{h}$ be the height of the vessel.

We know the volume of cuboidal vessel $\text{= l }\!\!\times\!\!\text{ b }\!\!\times\!\!\text{ h}$

$\therefore \text{l }\!\!\times\!\!\text{ b }\!\!\times\!\!\text{ h = 380 }{{\text{m}}^{\text{3}}}$

$\Rightarrow \left[ \text{10 }\!\!\times\!\!\text{ 8 }\!\!\times\!\!\text{ h} \right]\text{ }{{\text{m}}^{2}}\text{ = 380 }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{h = 4}\text{.75 m}$

Therefore, the height of the vessel is $\text{4}\text{.75 m}$.

4. Find the cost of digging a cuboidal pit $\text{8 m}$ long, $\text{6 m}$ broad and $\text{3 m}$deep at the rate of $\text{Rs}\text{. 30}$ per ${{\text{m}}^{\text{3}}}$.

Ans:

We are given the following:

Length of tank $\text{l = 8 m}$

Breadth of tank $\text{b = 6 m}$

Depth of tank $\text{h = 3 m}$

The volume of tank $\text{V = l }\!\!\times\!\!\text{ b }\!\!\times\!\!\text{ h}$

$\Rightarrow \text{V = }\left( \text{8 }\!\!\times\!\!\text{ 6 }\!\!\times\!\!\text{ 3} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 144 }{{\text{m}}^{\text{3}}}$

It is given that the cost of digging per ${{\text{m}}^{\text{3}}}$ volume $\text{= Rs}\text{. 30}$

So, the cost of digging per $\text{144 }{{\text{m}}^{\text{3}}}$ volume $\text{= Rs}\text{. }\left( \text{30 }\!\!\times\!\!\text{ 144} \right)\text{ = Rs}\text{. 4320}$

Therefore, the required cost of digging is $\text{Rs}\text{. 4320}$.

5. The capacity of a cuboidal tank is $\text{5000 litres}$ of water. Find the breadth of the tank, if its length and depth are respectively $\text{2}\text{.5 m}$ and $\text{10 m}$.

Ans:

We are given the following:

Length of tank $\text{l = 2}\text{.5 m}$

Depth of tank $\text{h = 10 m}$

Let us assume the breadth of the tank to be $\text{b m}$.

The volume of tank $\text{V = l }\!\!\times\!\!\text{ b }\!\!\times\!\!\text{ h}$

$\Rightarrow \text{V = }\left( \text{2}\text{.5 }\!\!\times\!\!\text{ b }\!\!\times\!\!\text{ 10} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 25b }{{\text{m}}^{\text{3}}}$

We can write the capacity of the tank $\text{= 25b }{{\text{m}}^{\text{3}}}\text{ = 25000b litres}$

But it is given that the capacity of the tank is $\text{50000 litres}$

$\therefore \text{25000b = 50000}$

$\Rightarrow \text{b = 2}$

Therefore, the breadth of the cuboidal tank is $\text{2 m}$.

6. A village, having a population of $\text{4000}$, requires $\text{150 litres}$ of water per head per day. It has a tank measuring $\text{20 m }\!\!\times\!\!\text{ 15 m }\!\!\times\!\!\text{ 6 m}$. For how many days will the water of this tank last?

Ans:

We are given the following:

Length of tank $\text{l = 20 m}$

Breadth of tank $\text{b = 15 m}$

Height of tank $\text{h = 6 m}$

The capacity of tank $\text{V = l }\!\!\times\!\!\text{ b }\!\!\times\!\!\text{ h}$

$\Rightarrow \text{V = }\left( \text{20 }\!\!\times\!\!\text{ 15 }\!\!\times\!\!\text{ 6} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 1800 }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 1800000 litres}$

The total water consumed by the people in a day $\text{= }\left( \text{4000 }\!\!\times\!\!\text{ 150} \right)\text{ litres = 600000 litres}$

Let us assume that the water in the tank will last for $\text{n}$ days.

$\therefore \text{n }\!\!\times\!\!\text{ 600000 = 1800000}$

$\Rightarrow \text{n = 3}$

Therefore, the water in the tank will last $\text{3}$ days.

7. A godown measures $\text{40 m }\!\!\times\!\!\text{ 25 m }\!\!\times\!\!\text{ 15 m}$. Find the maximum number of wooden crates each measuring $\text{1}\text{.5 m }\!\!\times\!\!\text{ 1}\text{.25 m }\!\!\times\!\!\text{ 0}\text{.5 m}$ that can be stored in the godown.

Ans:

We are given the following:

Length of godown $\text{L= 40 m}$

Breadth of godown $\text{B = 25 m}$

Height of godown $\text{H = 15 m}$

Length of wooden crate $\text{l = 1}\text{.5 m}$

Breadth of wooden crate $\text{b = 1}\text{.25 m}$

Height of wooden crate $\text{h = 0}\text{.5 m}$

The volume of godown ${{\text{V}}_{\text{g}}}\text{ = L }\!\!\times\!\!\text{ B }\!\!\times\!\!\text{ H}$

$\Rightarrow {{\text{V}}_{\text{g}}}\text{ = }\left( \text{40 }\!\!\times\!\!\text{ 25 }\!\!\times\!\!\text{ 15} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow {{\text{V}}_{\text{g}}}\text{ = 15000 }{{\text{m}}^{\text{3}}}$

The volume of a wooden crate ${{\text{V}}_{\text{c}}}\text{ = l }\!\!\times\!\!\text{ b }\!\!\times\!\!\text{ h}$

$\Rightarrow {{\text{V}}_{\text{c}}}\text{ = }\left( \text{1}\text{.5 }\!\!\times\!\!\text{ 1}\text{.25 }\!\!\times\!\!\text{ 0}\text{.5} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow {{\text{V}}_{\text{c}}}\text{ = 0}\text{.9375 }{{\text{m}}^{\text{3}}}$

Let us assume that $\text{n}$ wooden crates can be stored in the godown.

So, we can say that the volume of $\text{n}$ wooden crates will be the same as the volume of godown.

$\therefore \text{0}\text{.9375 }\!\!\times\!\!\text{ n = 15000}$

$\Rightarrow \text{n = 16000}$

Therefore, we can easily store $\text{16000}$ wooden crates in the godown.

8. A solid cube of side $\text{12 cm}$ 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:

Given side of cube $\text{a = 12 cm}$

Volume of this cube $\text{V = }{{\left( \text{a} \right)}^{\text{3}}}$

$\Rightarrow \text{V = }{{\left( \text{12 cm} \right)}^{\text{3}}}$

$\Rightarrow \text{V = 1728 c}{{\text{m}}^{\text{3}}}$

Let us assume the side of the smaller cube is $\text{x}$.

So, the volume of smaller cube $\text{= }\frac{\text{1728}}{\text{8}}\text{ = 216 c}{{\text{m}}^{\text{3}}}$

$\therefore {{\text{x}}^{\text{3}}}\text{ = 216 c}{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{x = 6 cm}$

Hence, the side of the smaller cubes will be $\text{6 cm}$.

The ratio of the surface areas $\text{R = }\frac{\text{6}{{\text{a}}^{\text{2}}}}{\text{6}{{\text{x}}^{\text{2}}}}$

$\Rightarrow \text{R = }\frac{{{\left( \text{12} \right)}^{\text{2}}}}{{{\left( \text{6} \right)}^{\text{2}}}}$

$\Rightarrow \text{R = }\frac{\text{4}}{\text{1}}$

Therefore the ratio between the surface areas of the cubes is $\text{4 : 1}$.

9. A river $\text{3 m}$ deep and $\text{40 m}$ wide is flowing at the rate of $\text{3}$ km per hour. How much water will fall into the sea in a minute?

Ans:

Given the rate of flow of water $\text{R = 2 km/hr}$

$\Rightarrow \text{R = }\left( \frac{\text{2 }\!\!\times\!\!\text{ 1000}}{\text{60}} \right)\text{ m/min}$

$\Rightarrow \text{R = }\left( \frac{\text{100}}{3} \right)\text{ m/min}$

Given the depth of the river $\text{h = 3 m}$

Width of the river $\text{b = 40 m}$

The volume of water that will flow in $\text{1 min}$ $\text{V = }\left( \frac{\text{100}}{\text{3}}\text{ }\!\!\times\!\!\text{ 40 }\!\!\times\!\!\text{ 3} \right)\text{ }{{\text{m}}^{\text{3}}}$

$\Rightarrow \text{V = 4000 }{{\text{m}}^{\text{3}}}$

Therefore, $\text{4000 }{{\text{m}}^{\text{3}}}$ of water will fall into the sea in a minute.

## NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5

Opting for the NCERT solutions for Ex 13.5 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 13.5 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 Subject Surface Areas and Volumes 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 13.5 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 13.5, 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 13.5 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.