# NCERT Solutions for Class 9 Maths Exercise 13.5 Chapter 13 Surface Area And Volume

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5 have been provided here for students to prepare well for the second term and competitive examinations. NCERT Solutions for Class 9 Maths has been designed by experienced subject experts in accordance with the NCERT syllabus and guidelines as directed by CBSE. The questions with solutions of 9th class Maths are helpful for students, to do their homework and aids as an instant reference guide.

### Access Other Exercise Solutions of Class 9 Maths Chapter 13 Surface Areas and Volumes

Exercise 13.1 solution (9 questions)
Exercise 13.2 solution (8 questions)
Exercise 13.3 solution (9 questions)
Exercise 13.4 solution (5 questions)
Exercise 13.6 solution (8 questions)
Exercise 13.7 solution (9 questions)
Exercise 13.8 solution (10 questions)
Exercise 13.9 solution (3 questions)

### Access Answers to NCERT Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5

1. A matchbox measures 4 cmÃ—2.5cmÃ—1.5cm. What will be the volume of a packet containing 12 such boxes?

Solution:

Dimensions of a matchbox (a cuboid) are lÃ—bÃ—h = 4 cmÃ—2.5 cmÃ—1.5 cm respectively

Formula to find the volume of matchbox = lÃ—bÃ—h = (4Ã—2.5Ã—1.5) = 15

Volume of matchbox = 15 cm3

Now, volume of 12 such matchboxes = (15Ã—12) cm3 = 180 cm3

Therefore, the volume of 12 matchboxes is 180cm3.

2. A cuboidal water tank is 6m long, 5m wide and 4.5m deep. How many litres of water can it hold? (1 m3= 1000 l)

Solution:

Dimensions of a cuboidal water tank are: l = 6 m and b = 5 m and h = 4.5 m

Formula to find volume of tank, V = lÃ—bÃ—h

Put the values, we get

V = (6Ã—5Ã—4.5) = 135

Volume of water tank is 135 m3

Again,

We are given that, amount of water that 1m3 volume can hold = 1000 l

Amount of water, 135 m3volume hold = (135Ã—1000) litres = 135000 litres

Therefore, given cuboidal water tank can hold up to135000 litres of water.

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

Solution:

Given:

Length of cuboidal vessel, l = 10 m

Width of cuboidal vessel, b = 8m

Volume of cuboidal vessel, V = 380 m3

Let the height of the given vessel be h.

Formula for Volume of a cuboid, V = lÃ—bÃ—h

Using formula, we have

lÃ—bÃ—h = 380

10Ã—8Ã—h= 380

Or h = 4.75

Therefore, the height of the vessels is 4.75 m.

4. Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3.

Solution:

The given pit has its length(l) as 8m, width (b)as 6m and depth (h)as 3 m.

Volume of cuboidal pit = lÃ—bÃ—h = (8Ã—6Ã—3) = 144 (using formula)

Required Volume is 144 m3

Now,

Cost of digging per m3 volume = Rs 30

Cost of digging 144 m3 volume = Rs (144Ã—30) = Rs 4320

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.

Solution:

Length (l) and depth (h) of tank is 2.5 m and 10 m respectively.

To find: The value of breadth, say b.

Formula to find the volume of a tank = lÃ—bÃ—h = (2.5Ã— bÃ—10) m3= 25b m3

Capacity of tank= 25b m3, which is equal to 25000b litres

Also, capacity of a cuboidal tank is 50000 litres of water (Given)

Therefore, 25000 b = 50000

This implies, b = 2

Therefore, the breadth of the tank is 2 m.

6. A village, having a population of 4000, requires 150 litres of water per head per day.

It has a tank measuring 20 mÃ—15 mÃ—6 m. For how many days will the water of this tank last?

Solution:

Length of the tank = l = 20 m

Breadth of the tank = b = 15 m

Height of the tank = h = 6 m

Total population of a village = 4000

Consumption of the water per head per day = 150 litres

Water consumed by the people in 1 day = (4000Ã—150) litres = 600000 litres â€¦(1)

Formula to find the capacity of tank, C = lÃ—bÃ—h

Using given data, we have

C = (20Ã—15Ã—6) m3= 1800 m3

Or C = 1800000 litres

Let water in this tank last for d days.

Water consumed by all people in d days = Capacity of tank (using equation (1))

600000 d =1800000

d = 3

Therefore, the water of this tank will last for 3 days.

7. A godown measures 40 mÃ—25mÃ—15 m. Find the maximum number of wooden crates each

measuring 1.5mÃ—1.25 mÃ—0.5 m that can be stored in the godown.

Solution:

From statement, we have

Length of the godown = 40 m

Height = 15 m

Whereas,

Length of the wooden crate = 1.5 m

Height = 0.5 m

Since godown and wooden crate are in cuboidal shape. Find the volume of each using formula, V = lbh.

Now,

Volume of godown = (40Ã—25Ã—15) m3 = 15000 m3

Volume of a wooden crate = (1.5Ã—1.25Ã—0.5) m3 = 0.9375 m3

Let us consider that, n wooden crates can be stored in the godown, then

The volume of n wooden crates = Volume of godown

0.9375Ã—n =15000

Or n= 15000/0.9375 = 16000

Hence, the number of wooden crates that can be stored in the godown is 16,000.

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.

Solution:

Side of a cube = 12cm (Given)

Find the volume of cube:

Volume of cube = (Side)3 = (12)3cm3= 1728cm3

Surface area of a cube with side 12 cm = 6a2 = 6(12) 2 cm2 â€¦(1)

Cube is cut into eight small cubes of equal volume, say side of each cube is p.

Volume of a small cube = p3

Surface area = 6p2 â€¦(2)

Volume of each small cube = (1728/8) cm3 = 216 cm3

Or (p)3 = 216 cm3

Or p = 6 cm

Now, Surface areas of the cubes ratios = (Surface area of bigger cube)/(Surface area of smaller cubes)

From equation (1) and (2), we get

Surface areas of the cubes ratios = (6a2)/(6p2) = a2/p2 = 122/62 = 4

Therefore, the required ratio is 4 : 1.

9. A river 3m deep and 40m wide is flowing at the rate of 2km per hour. How much water will fall into the sea in a minute?

Solution:

Given:

Depth of river, h = 3 m

Width of river, b = 40 m

Rate of water flow = 2km per hour = 2000m/60min = 100/3 m/min

Now, Volume of water flowed in 1 min = (100/3) Ã— 40 Ã— 3 = 4000m3

Therefore, 4000 m3water will fall into the sea in a minute.

By practising problems in NCERT Solutions for Class 9 Maths Chapter 13 students can score well in second term examination and can know the easy way to solve the problems. All the 9 questions included in this exercise help students analyze the situation and apply the formula. It includes easy to solve questions which can be expected in examinations. Solve NCERT solutions, where problems are explained in a detailed way following each and every step.
NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes exercise consists of application-level questions that help to determine the volume of the cuboid. It also helps in finding the length and breadth of the cuboid by applying the formula.

### Key Features of NCERT Solutions for Class 9 Maths Chapter 13 â€“ Surface Areas and Volume Exercise 13.5

Solving these solutions help students to:

• Students can self-assess their learning abilities about the chapter and topics
• Students can analyze the type of questions that appear for the exams
• Improve their efficiency and speed in solving the problems
• Remember the formulas in an easy way and apply relevantly, as necessary