Class 8 Maths

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2.

• Mensuration Class 8 Ex 11.1
• Mensuration Class 8 Ex 11.3
• Mensuration Class 8 Ex 11.4

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 11
Chapter Name	Mensuration
Exercise	Ex 11.2
Number of Questions Solved	11
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2

Ex 11.2 Class 8 Maths Question 1.
The shape of the top surface of a table is trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Solution:
Area of top surface of a table
= Area of the trapezium

Ex 11.2 Class 8 Maths Question 2.
The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.
Solution:
Let the required side be x cm.

Ex 11.2 Class 8 Maths Question 3.
Length of the fence of a trapezium A shaped field ABCD is 120 m. IF BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD B and BC.

Solution:
Let ABCD be the given trapezium in which BC = 48 m, CD = 17 m and
AD = 40 m.

Ex 11.2 Class 8 Maths Question 4.
The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Solution:
Let ABCD be the given quadrilateral in which BE ⊥ AC and DF ⊥ AC.
It is given that
AC =24m, BE = 8m and DF = 13 m.
Now, area of quad. ABCD
= area of ∆ABC + area of ∆ACD

Ex 11.2 Class 8 Maths Question 5.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
Area of a rhombus = \(\frac { 1 }{ 2 } \) x (product of diagonals)
\(=\left( \frac { 1 }{ 2 } \times 7.5\times 12 \right) { cm }^{ 2 }=45{ cm }^{ 2 }\)

Ex 11.2 Class 8 Maths Question 6.
Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:
Let ABCD be a rhombus of side 6 cm and whose altitude DF = 4 cm. Also, one of its diagonals, BD =8 cm.
Area of the rhombus ABCD

Ex 11.2 Class 8 Maths Question 7.
The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m² is ₹ 4.
Solution:
Area of the floor = 3000 x Area of one tile

Ex 11.2 Class 8 Maths Question 8.
Mohan wants to buy a trapezium shaped field. Its side along the. river is parallel to and twice the side along the road. If the area of this field is 10500 m² and the perpendicular distance between the two parallel sides is 100m, find the length of the side along the river.

Solution:
Let the parallel sides of the trapezium shaped field berm and 2x m. Then, its area

∴ The length of the side along the river is 2 x 70, i.e., 140 metres.

Ex 11.2 Class 8 Maths Question 9.
Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
Solution:
Area of the octagonal surface ABCDEFGH = Area (trap. ABCH) + Area (rect. HCDG) +Area (trap. GDEF)

Ex 11.2 Class 8 Maths Question 10.
There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?
Solution:
Taking Jyoti’s diagram :

Ex 11.2 Class 8 Maths Question 11.
Diagram of the adjacent picture frame has outer dimensions = 24 cm x 28 cm and inner dimensions 16 cm x 20 cm. Find the area of each section of the frame, if the width of each section is same.

Solution:

We hope the NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4.

• Mensuration Class 8 Ex 11.1
• Mensuration Class 8 Ex 11.2
• Mensuration Class 8 Ex 11.3

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 11
Chapter Name	Mensuration
Exercise	Ex 11.4
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4

Ex 11.4 Class 8 Maths Question 1.
Given a cylindrical tank, in which situation will you find surface area and in which situation volume.
(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.
Solution:
(a) Volume
(b) Surface area
(c) Volume

Ex 11.4 Class 8 Maths Question 2.
Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?

Solution:
Cylinder B has greater volume.
Surface area of cylinder A
= (2πrh + 2πr² sq. units,

Thus, cylinder having greater volume also has greater surface area.

Ex 11.4 Class 8 Maths Question 3.
Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?
Solution:
Volume of the cuboid =900 cm³
⇒ (Area of the base) x Height = 900 cm³
⇒ 180 x Height = 900
⇒ Height = \(\frac { 900 }{ 180 } \) = 5
Hence, the height of the cuboid is 5 cm.

Ex 11.4 Class 8 Maths Question 4.
A cuboid is of dimensions 60 cm x 54 cm x 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?
Solution:

Ex 11.4 Class 8 Maths Question 5.
Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base is 140 cm?
Solution:
Let the h be the height of cylinder whose radius, r = \(\frac { 140 }{ 2 } \) cm = 70 cm = \(\frac { 70 }{ 100 } \) m and volume =1.54 m³.

Ex 11.4 Class 8 Maths Question 6.
A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in liters that can be stored in the tank?

Solution:
Quantity of milk that can be stored in the tank = Volume of the tank

Ex 11.4 Class 8 Maths Question 7.
If each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?
Solution:
Let x units be the edge of the cube. Then, its surface area = 6x² and its volume = x³.
When its edge is doubled,
(i) Its surface area = 6 (2x)² = 6 x 4 x 2 = 24 x 2
⇒ The surface area of the new cube will be 4 times that of the original cube.
(ii) Its volume = (2x)³ = 8x³.
⇒ The volume of the new cube will be 8 times that of the original cube.

Ex 11.4 Class 8 Maths Question 8.
Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute. If the volume of reservoir is 108 m3, find the number of hours it will take to fill the reservoir.

Solution:
Volume of the reservoir =108 m³
= 108 x 1000 litres
= 108000 litres
Since water is pouring into reservoir @ 60 litres per minute
∴ Time taken to fill the reservoir = \(\frac { 108000 }{ 60\times 60 } \) hours
= 30 hours

We hope the NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3.

• Comparing Quantities Class 8 Ex 8.1
• Comparing Quantities Class 8 Ex 8.2

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 8
Chapter Name	Comparing Quantities
Exercise	Ex 8.3
Number of Questions Solved	12
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3

Ex 8.3 Class 8 Maths Question 1.
Calculate the amount and compound interest on
(a) ₹ 10,800 for 3 years at 12 \(\frac { 1 }{ 2 } \) % per annum compounded annually.
(b) ₹ 18,000 for 2 \(\frac { 1 }{ 2 } \) years at 10% per annum compounded annually.
(c) ₹ 62,500 for 1 \(\frac { 1 }{ 2 } \) years at 8% per annum compounded half yearly.
(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify)
(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.
Solution:

Ex 8.3 Class 8 Maths Question 2.
Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
[Hint : Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for \(\frac { 4 }{ 12 } \) years)
Solution:
Here, P = ₹ 26400, R =15% per annum
and n = 2 years 4 months =2\(\frac { 1 }{ 3 } \) years.

Hence, Kamala will pay ₹ 36659.70 to the bank.

Ex 8.3 Class 8 Maths Question 3.
Fabina borrows ? 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much? .
Solution:
In case of Fabina :
P = ₹ 12500, R =12% per annum and T =3 years. Then,

Hence, Fabina pays 362.50 more as interest ₹ (4500 – 4137.50), i.e., ₹ 362.50 more as interest.

Ex 8.3 Class 8 Maths Question 4.
I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?
Solution:
Here, P = ₹ 12000, R = 6% per annum and T = 2 years.

So, I have to pay ₹ (1483.20 -1440), i.e., ₹ 43.20 in excess.

Ex 8.3 Class 8 Maths Question 5.
Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half-yearly. What amount would he get
(i) after 6 months?
(ii) after 1 year.
Solution:
Here, Principal =₹ 60000, Rate = 12% per annum = 6%per half-year.
(i) Time = 6 months = 1 half-year

Ex 8.3 Class 8 Maths Question 6.
Arif took a loan of ? 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1\(\frac { 1 }{ 2 } \) years if the interest is 2
(i) compounded annually
(ii) compounded half-yearly.
Solution:
Here, P = ₹ 80000
Rate = 10% per annum = 5% per half-year,
Time = 1\(\frac { 1 }{ 2 } \) years = 3 half-years.

Ex 8.3 Class 8 Maths Question 7.
Maria invested ? 8,000 in a business. She would he paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.
Solution:

Ex 8.3 Class 8 Maths Question 8.
Find the amount and the compound interest on ? 10,000 for 1\(\frac { 1 }{ 2 } \) years at 10% per annum, compounded half-yearly. Would this interest be more than the interest he would get if it was compounded annually?
Solution:
Here, Principal = ? 10000
Time = 1\(\frac { 1 }{ 2 } \) years = 3 half years,

This interest is more than the interest that he would get if it was compounded annually.

Ex 8.3 Class 8 Maths Question 9.
Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 12\(\frac { 1 }{ 2 } \) % per annum, interest being compounded half yearly.
Solution:
Here, Principal = ₹ 4096,
Time = 18 months = 3 half years

Ex 8.3 Class 8 Maths Question 10.
The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005?
Solution:

Ex 8.3 Class 8 Maths Question 11.
In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Solution:
We have, P = Original count of bacteria = 506000;
Rate of increase = R = 2.5% per hour, Time = 2 hours.

Ex 8.3 Class 8 Maths Question 12.
A scooter was bought at ? 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Solution:
We have, V₀ = Initial value = ₹ 42000
R = Rate of depreciation = 8% p.a.

We hope the NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2.

• Direct and Inverse Proportions Class 8 Ex 13.1

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 13
Chapter Name	Direct and Inverse Proportions
Exercise	Ex 13.2
Number of Questions Solved	11
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2

Ex 13.2 Class 8 Maths Question 1.
Which of the following are in inverse proportion?
(i) The number of workers on a joh and the time to complete the job.
(ii) The time taken for a journey and the distance travelled in a uniform speed.
(iii) Area of cultivated land and the crop harvested.
(iv) The time taken for a fixed journey and the speed of the vehicle.
(v) The population of a country and the area of land per person.
Solution:
(i) We know that more is the number of workers to do a job, less is the time taken to finish the job.
So, it is a case of inverse proportion.
(ii) Clearly, more is the time taken, more is the distance travelled in a uniform speed.
So, it is a case of direct proportion.
(iii) Clearly, more is the area cultivated land, more is the crop harvested. So, it is a case of direct proportion.
(iv) We know that more is the speed of the vehicle, less is the time to cover a fixed distance.
So, it is a case of inverse proportion.
(v) Clearly, more is the population, less is the area of land per person in a country.
So, it is a case of inverse proportion.

Ex 13.2 Class 8 Maths Question 2.
In a Television game show, the prize money of ₹ 1,00,000 is to be divided equally amongst the winners. Complete the following table and find whether the prize money given to an individual winner is directly or inversely proportional to the number of winners?

Solution:
Clearly, more the number of winners, less is the prize for each winner. So, it
∴ 4 x x = 1 x 100000

Ex 13.2 Class 8 Maths Question 3.
Rehman is making a wheel using spokes. He wants to fix equal spokes in such a way that the angles between any pair of consective spokes are equal. Help him by completing the following table.

(i) Are the number of spokes and the angles formed between the pairs of consecutive spokes in inverse proportion?
(ii) Calculate the angle between a pair of consecutive spokes on a wheel with 15 spokes.
(iii) How many spokes would be needed, if the angle between a pair of consecutive spokes is 40°?
Solution:
Clearly, more is the number of spokes, less is the measure of angle between a pair of consecutive spokes.
Also, we observe here
4 x 90° = 6 x 60° (= 360°)
So, it is a case of inverse proportion.

(i) Yes, the number of spokes and the angles formed between the pairs of consecutive spokes are in inverse proportion.
(ii) Let x° be the angle between a pair of consecutive spokes on a wheel with 15 spokes.
∴ 15 x x = 4 x 90 ⇒ x= \(\frac { 4\times 90 }{ 15 } \) = 24
Thus, the required angle is 24°.
(iii) Let x spokes are needed on a wheel if the angle between a pair of consecutive spokes is 40°.
∴ x x 40 = 4 x 90 ⇒ x= \(\frac { 4\times 90 }{ 40 } \) = 9
Thus, the required number of spokes is 9.

Ex 13.2 Class 8 Maths Question 4.
If a box of sweets is divided among 24 children, they will get 5 sweets each. How many would each get, if the number of the children is reduced by 4?
Solution:
Since the number of the children were reduced by 4. Therefore, their number becomes 24 – 4 = 20.
Let each child get x sweets when their number is 20.
Thus, we have the following table :

Since less children will get more sweets. So, it is a case of inverse proportion.
∴ 24 x 5 = 20 x x ⇒ x= \(\frac { 24\times 5 }{ 20 } \) = 6
Hence, each child will get 6 sweets.

Ex 13.2 Class 8 Maths Question 5.
A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would the food last if there were 10 more animals in his cattle?
Solution:
Let the food will now last for x days. Then,

Clearly, more is the number of animals, less will be the number of days for food to last. .
So, it is a case of inverse proportion.
∴ 24 x 6 = 30 x x ⇒ x= \(\frac { 20\times 6 }{ 30 } \) = 4
Hence, the food will now last for 4 days.

Ex 13.2 Class 8 Maths Question 6.
A contractor estimates that 3 persons could rewire Jasminder’s house in 4 days. If, he uses 4 persons instead of three, how long should they take to complete the job?
Solution:
Let the rewiring be completed by 4 persons in x days.

Clearly, more is the number of persons, less is the number of days required.
So, it is a case of inverse proportion.
∴ 4 x 3 = x x 4 ⇒ x= \(\frac { 4\times 3 }{ 3 } \) = 3
Hence, the job will be completed in 3 days.

Ex 13.2 Class 8 Maths Question 7.
A batch of bottles were packed in 25 boxes with 12 bottles in each box. If the same batch is packed using 20 bottles in each box, how many boxes would be filled?
Solution:
Let JC boxes be needed when 20 bottles are packed in each box. Then,

Clearly, more is the number of bottles, less will be the number of boxes needed for packing.
So, it is a case of inverse proportion.
∴ 25 x 12 = x x 20 ⇒ x= \(\frac { 25\times 12 }{ 20 } \) = 15
Hence, the boxes needed for packing is 15.

Ex 13.2 Class 8 Maths Question 8.
A factory requires 42 machines to produce a given number of articles in 63 days. HoW many machines would be required to produce the same number of articles in 54 days?
Solution:
Let x machines are required to produce a fixed number of articles in 54 days. Then,

Clearly, less is the number of days to produce a fixed number of articles, more machines are required.
So, it is a case of inverse proportion.
∴ 42 x 63 = x x 54 ⇒ x= \(\frac { 42\times 63 }{ 54 } \) = 49
Hence, the number of machines needed is 49.

Ex 13.2 Class 8 Maths Question 9.
A car takes 2 hours to reach a destination by travelling at the speed of 60 km/h. How long will it take when the car travels at the speed of 80 km/h?
Solution:
Let the car takes x hours to reach a destination by travelling at the speed of 80 km/h. Then,

Clearly, more the speed, less will be the time taken. So, it is a case of inverse proportion.
∴ 60 x 2 = 80 x x ⇒ x= \(\frac { 60\times 2 }{ 80 } \) = \(\frac { 3 }{ 2 } \)
Hence, the time taken will be 1\(\frac { 1 }{ 2 } \) hours.

Ex 13.2 Class 8 Maths Question 10.
Two persons could fit new windows in a house in 3 days.
(i) One of the persons fell ill before the work started. How long would the job take now?
(ii) How many persons would be needed to fit the windows in one day?
Solution:
(i) Let x days be taken to fix the windows. Then

Clearly, less the number of persons, more will be the number of days to fix the windows.
So, it is a case of inverse proportion.
∴ 2 x 3 = 1 x x ⇒ x = 6
Hence, one person will finish the given work in 6 days.
(ii) Let x persons will fix the windows in 1 day. Then,

Clearly, less the number of days, more will be the number of persons to fix the windows.
So, it is a case of inverse proportion
∴ 2 x 3 = x x 1
⇒ x = 6
Hence, 6 persons will finish the given work in 1 day.

Ex 13.2 Class 8 Maths Question 11.
A school has 8 periods a day each of 45 minutes duration. How long would each period he, if the school has 9 periods a day, assuming the number of school hours to be the same?
Solution:
Let x minutes be the duration of period when the school has 9 periods a day. Then

Clearly, more the periods, less will be the duration of the period.
So, it is a case of inverse proportion.
∴ 8 x 45 = 9 x x ⇒ x = \(\frac { 8\times 45 }{ 9 } \) = 40
Hence, the duration of period is 40 minutes.

We hope the NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 13 Direct and Inverse Proportions Ex 13.2, drop a comment below and we will get back to you at the earliest.