Class 8

Students can Download Maths Chapter 2 Life Mathematics Ex 2.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 2 Life Mathematics Ex 2.1

Question 1.
Fill in the blanks:
(i) A can finish a job in 3 days whereas B finishes it in 6 days. The time taken to complete the job together is………..days.
(ii) If 5 persons can do 5 jobs in 5 days, then 50 persons can do 50 jobs in………..days.
(iii) A can do a work in 24 days. A and B together can finish the work in 6 days. Then B alone can finish the work in…………days.
(iv) A alone can do a piece of work in 35 days. If B is 40% more efficient than A, then B will finish the work in………..days.
(v) A alone can do a work in 10 days and B alone in 15 days. They undertook the work for Rs 200000. The amount that A will get is………
Solution:
(i) 2 days
(ii) 5
(iii) 8
(iv) 25
(v) Rs 1,20,000

Question 2.
210 men working 12 hours a day can finish a job in 18 days. How many men are required to finish the job in 20 days working 14 hours a day?
Solution:
Let the required number of men be x.

More working hours ⇒ less men required.
∴ It is inverse proportion.
∴ Multiplying factor is \(\frac{12}{14}\)
Also more number of days ⇒ less men
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{18}{20}\)
∴ x = 210 × \(\frac{12}{14}\) × \(\frac{18}{20}\)= 162 men

162 men are required.

Question 3.
A cement factory makes 7000 cement bags in 12 days with the help of 36 machines. How many bags can be made in 18 days using 24 machines?
Solution:
Let the required number of cement bags be x.

Number of days more ⇒ More cement bags.
∴ It is direct variation.
∴ The multiplying factor = \(\frac{18}{12}\)
Number of machines more ⇒ More cement bags.
∴ It is direct variation.
∴ The multiplying factor = \(\frac{24}{36}\)
∴ x = 7000 × \(\frac{18}{12}\) × \(\frac{24}{36}\)

x = 7000 cement bags
7000 cement bags can be made.

Question 4.
A soap factory produces 9600 soaps in 6 days working 15 hours a day. In how many days will it produce 14400 soaps working 3 hours more a day?
Solution:
Let the required number of days be x.

To produce more soaps more days required.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{14400}{9600}\)
If more hours spend, less days required.
∴ It is indirect proportion.
∴ Multiplying factor = \(\frac{15}{18}\)
∴ x = 6 × \(\frac{14400}{9600}\) × \(\frac{15}{18}\)

x = \(\frac{15}{2}\)
\(\frac{15}{2}\) days will be needed.

Question 5.
If 6 container lorries transport 135 tonnes of goods in 5 days, how many more lorries are required to transport 180 tonnes of goods in 4 days?
Solution:
Let the number of lorries required more = x.

As the goods are more ⇒ More lorries are needed to transport.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{180}{135}\)
Again if more days ⇒ less number of lorries enough.
∴ It is direct proportion.
∴ Multiplying factor = \(\frac{5}{4}\)
∴ 6 + x = 6 × \(\frac{180}{135}\) × \(\frac{5}{4}\)

6 + x = 10
x = 10 – 6
x = 4
∴ 4 more lorries are required.

Question 6.
A can do a piece of work in 12 hours, B and C can do it 3 hours whereas A and C can do it in 6 hours. How long will B alone take to do the same work?
Solution:
Time taken by A to complete the work =12 hrs.
∴ A’s 1 hr work = \(\frac{1}{12}\)…………(1)
(B + C) complete the work in 3 hrs.
∴ (B + C)’s 1 hour work = \(\frac{1}{3}\)…………(2)
(1) + (2) ⇒
∴ (A + B + C)’s 1 hour work = \(\frac{1}{12}\) + \(\frac{1}{3}\) = \(\frac{1+4}{12}\) = \(\frac{5}{12}\)
Now (A + C) complete the work in 6 hrs.
∴(A + C)’s 1 hour work = \(\frac{1}{6}\)
∴ B’s 1 hour work = (A + B + C)’s 1 hour work – (A + C)’s 1 hour work
= \(\frac{5}{12}\) – \(\frac{1}{6}\) = \(\frac{5-2}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)
∴ B alone take 4 days to complete the work.

Question 7.
A and B can do a piece of work in 12 days, while B and C can do it in 15 days whereas A and C can do it in 20 days. How long would each take to do the same work?
Solution:
(A + B) complete the work in 12 days.
∴ (A + B)’s 1 day work = \(\frac{1}{12}\)……….(1)
(B + C) complete the work in 15 days
∴ (B + C)’s 1 day work = \(\frac{1}{15}\)……….(2)
(A + C) complete the work in 20 days
∴ (A + C)’s 1 day work = \(\frac{1}{20}\)……….(3)
Now (1) + (2) + (3) ⇒
[(A + B) + (B + C) + (A + C)]’s 1 day work = \(\frac{1}{12}\) + \(\frac{1}{15}\) + \(\frac{1}{20}\)

(2A + 2B + 2C)’s 1 day work = \(\frac{5}{60}\) + \(\frac{4}{60}\) + \(\frac{3}{60}\)
2(A + B + C)’s 1 day work = \(\frac{5+4+3}{60}\)
(A + B + C)’s 1 day work = \(\frac{12}{60×2}\)
(A + B + C)’s 1 day work = \(\frac{1}{10}\)
Now A’s 1 day’s work = (A + B + C)’s 1 day work – (B + C)’s 1 day work
\(\frac{1}{10}\) – \(\frac{1}{15}\) = \(\frac{3}{30}\) – \(\frac{2}{30}\) = \(\frac{1}{30}\)
∴ A takes 30 days to complete the work.
B’s 1 day work = (A + B + C)’s 1 day’s work – (A + C)’s 1 day’s work
= \(\frac{1}{10}\) – \(\frac{1}{20}\) = \(\frac{6}{60}\) – \(\frac{3}{60}\)
\(\frac{6-3}{60}\) = \(\frac{3}{60}\) = \(\frac{1}{20}\)
B takes 20 days to complete the work.
C’s 1 day work = (A + B + C)’s 1 day work – (A + B)’s 1 day work
\(\frac{1}{10}\) – \(\frac{1}{12}\) = \(\frac{6}{60}\) – \(\frac{5}{60}\) = \(\frac{6-5}{60}\) = \(\frac{1}{60}\)
∴ C takes 60 days to complete the work.

Question 8.
Carpenter A takes 15 minutes to fit the parts of a chair while Carpenter B takes 3 more minutes than A to do the same work. Working together, how long will it take for them to fit the parts for 22 chairs?
Solution:
Time taken by A to fit a chair = 15 minutes
Time taken by B = 3 minutes more than A
= 15 + 3 = 18 minutes
∴ A’s 1 minute work = \(\frac{1}{15}\)
B’s 1 minute work = \(\frac{1}{18}\)
(A + B)’s 1 minutes work = \(\frac{1}{15}\) + \(\frac{1}{18}\)

\(\frac{12}{180}\) + \(\frac{22}{180}\) = \(\frac{22}{180}\) = \(\frac{11}{90}\)
∴ Time taken by (A + B) to fit a chair
= \(\frac{1}{\frac{11}{90}}\) = \(\frac{90}{11}\) minutes
∴ Time taken by (A + B) to fit 22 chairs
= \(\frac{90}{11}\) × 22 = 180 minutes
= \(\frac{180}{60}\) = 3 hours

Question 9.
A man takes 10 days to finish a job where as a woman takes 6 days to finish the same job. Together they worked for 3 days and then the woman left. In how many days will the man complete the remaining job?
Solution:
Man can finish the work in 10 days and women can finish the same work in 6 days.
∴ Man’s 1 day work = \(\frac{1}{10}\)
Woman’s 1 day work = \(\frac{1}{6}\)
(Man + Woman)s 1 day work = \(\frac{1}{10}\) + \(\frac{1}{6}\) = \(\frac{6}{60}\) + \(\frac{10}{60}\) = \(\frac{16}{60}\)
(Man + Woman)s 3 days work

In 3 days \(\frac{4}{5}\) th of the whole work is completed.
Remaining work = 1 – \(\frac{4}{5}\) = \(\frac{5}{5}\) – \(\frac{4}{5}\) = \(\frac{1}{5}\)
Complete work is done by the man by 10 days
∴ \(\frac{1}{5}\) of the work is done by man in \(\frac{1}{5}\) × 10 = 2 days.

Question 10.
A is thrice as fast as B. If B can do a piece of work in 24 days then, find the number of days they will take to complete the work together.
Solution:
If B does the work in 3 days, A will do it in 1 day.
B complete the work in 24 days.
∴ A complete the same work in \(\frac{24}{3}\) = 8 days.
∴ (A + B) complete the work in \(\frac{ab}{a+b}\) days = \(\frac{24×8}{24+8}\) days = \(\frac{24×8}{32}\)days = 6 days

They together complete the work in 6 days.

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 4 Statistics Additional Questions

Question 1.
The following information shows the number of students opting different subjects in a college.

Draw a pie-chart to represent the information.
Solution:

Students opting different subjects.

Question 2.
The pie chart represents the expenditures of a family on different items. Find the percentage expenditures on different items by reading the pie-chart.

Solution:
Percentage value of the component =
∴ Percentage expenditures on various items are given as :

Question 3.
The following table gives the marks of 100 students represent in the form of a histogram.

Solution:
We take marks on X axis and frequency Y axis.

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 4 Information Processing Ex 4.4

Information Processing in Maths 8th Standard  Miscellaneous Practice Problems

Information Processing Solution Question 1.
The rule of Fibonacci Sequence is F(n) = F(n – 2) + F(n – 1). Find the 11th to 20th Fibonacci numbers.
Solution:
Fibonacci series is given by F(n) = F(n – 2) + F(n – 1)
∴ F(11) = F(11 – 2) + F(11 – 1) = F(9) + F(10)
Similarly F(12) ; = F(12 – 2) + F(12 – 1) = F(11) + F(10)
F(13) = F(12) + F(11)
F(14) = F(13) + F(12)……
F(20) = F(19) + F(18)
The Fibonacci series given by

8th Maths Book Information Processing Question 2.
In a library, 385 Math books, 297 Science books and 143 Tamil books are bundled equally in numbers. What is the maximum numbers of books possible in a bundle, for all types of books? (Use repeated division method)
Solution:
In library, no. of math books – 385
Science books – 297 .
Tamil books – 143
We need to find HCF by repeated division method.
Step 1:
First let us take the bigger 2 numbers which are 385 & 297 & find their HCF using repeated division method.

Step 2:
Now let us find the HCF of 11 & 143

∴ ’11’ is the HCF of 11 & 143
∴ Therefore, the HCF of 385, 297 & 143 is 11.
So, the maximum number of books possible in a bundle, for all types of books is 11.

Repeated Subtraction Method Question 3.
Find the length of the largest which piece of wood used to measure exactly the lengths 4 m, 50 cm and 6 m 30 cm woods.( Use repeated subtraction method)
Solution:
Piece 1 of wood length = 4 m 50 cm ; 1m = 100 cm
∴ 4m 50 cm = 4 x 100 + 50 = 450 cm
Piece 2 of wood length = 6 m 30 cm = 6 x 100 + 30 = 630 cm
Now, we need to find the HCF of 450 & 630 by repeated subtraction method.
Let bigger number be m & smaller number be n
Therefore, m = 630 & n = 450
m – n = 630 – 450 = 180; now m = 450, n = 180
m – n = 450 – 180 = 270; now m = 180, n = 270
n – m = 270 – 180 = 90; now m = 180, n = 90
m – n = 180 – 90 = 90
∴ now m = n …90 is HCF
Answer:
length of larget peice of wood to measure exactly the lenth of both pieces of wood is 90 cm

Question 4.
Using both repeated division method and repeated subtraction method and find the greatest number that divides 167 and 95, leaving 5 as reminder.
We need to find a number ‘x’ that divides 167 & 95 having 5 as reminder.
Therefore, if we subtract 5 from both the numbers, there wont be any reminder when the numbers are divided by ‘x’.
∴ x is the HCF of (167 – 5) & (95 – 5)
= 162 & 90
Finding HCF of 162 & 90 by repeated division method,

∴ HCF of 162 & 90 is 18.
HCF of 162 & 90 using repeated subtraction method
Let m = 162 & n = 90
m – n = 162 – 90 = 72
m – n = 90 – 72= 18
m – n = 72 – 18 = 54
n – m = 54 – 18 = 36
n – m = 36 – 18 = 18
now m = n = 18
Answer:
The number is 18
∴ 18 is the HCF

Question 5.
Praveen recently got the registration number for his new two-wheeler. Here, the number is given in the form of mirror – image. Encode the image and find the correct registration number of praveen’s two – wheeler.

Solution:
The mirror image is

When we place an imaginary mirror & visualize the image seen in the mirror, we will get the below.

∴ The answer is Option c

Students can Download Maths Chapter 5 Information Processing Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 5 Information Processing Intext Questions

Exercise 5.1
Activity 1 (Text book Page No. 104)

Question 1.
Consider that you are going to a store with your total budget of ₹ 220 to buy things without changing the quantity of the items given in the list below with the following conditions.

Conditions :
(i) First you have to complete the price list given.
(ii) You have to buy three items as per the given price list but within your budget ₹ 220.
(iii) You won’t carry exceeding 5kg because you have to walk home carrying them, so they cannot be bulky.

Now, answer the following questions:
1. In how many ways can you buy your items? Complete the price lists given below. One is done for you.
2. Which one is the best purchase price list and why?

Answer:

Try this (Text book Page No. 106)

The teacher divides the class into four groups and setup a mock market in the class room and ask the students to involve in role play as two groups of businessmen and two groups of consumers. Consumers have to buy products at different shops and prepare a price list.

The two supermarkets in which the two groups buy are Star Food Mart and Super Provisions. Th is week they each have got a special deal on some products. At Star Food Mart, you can buy items at discount prices. At Super Provisions, there are some “BUY ONE GET ONE” deals. Flave a look at their deal:

Now, answer the following questions.

I. Here is your shopping list:
4 bottles of Protein Milk (200 ml size), 2 packets of Peanut candies(200 gm), 1 packet of Chocolate biscuits and 1 packet of Badam nuts (500 gm)
(i) If you buy all the items in one shop, where will you get the best price?
(ii) If you buy the items from different shops, how will you do it to spend the least amount of money?

II. You have ₹ 1000/- to spend to buy the following shopping list:
6 bottles of Protein Milk (200 ml size), 3 packets of Peanut candies (200 gm), 3 packets of Chocolate biscuits and 1 packet of Badam nuts (250 gm).

(i) How can you do this so that you don’t go over your budget amount ₹ 1000?
(ii) Which shop offers you the best value for money on each item?
(iii) Is the “BUY ONE GET ONE” deal at Super Provisions the same as “50% off’ deal?
Solution:
Prices of Star Food Mart.


(i) In star Food Mart
(ii) Badam nuts from super provisions and other items from Star Food Mart.

II. Comparing two stores.

(i) We can buy them from any one shop.
(ii) Star Food Mart.
(iii) Yes

Exercise 5.2
Try this (Text book Page No. 117)

Question 1.
If Kumaran cut the woods using first fit method then find the wastage pieces.
Solution:

Waste pieces = 2 feet + 4 feet + 2 feet = 8 feet

Activity 1 (Text book Page No. 117)

Question 1.
Seva Sangam has decided to deliver some aids to flood victims via lorries with a maximum capacity of 5000kg. All of these items that are given below are to be packed and sent in the lorry.

Solution:

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 4 Statistics Ex 4.3

Miscellaneous and Practice Problems

Question 1.
Draw a pie chart for the given table.

Solution:
Converting the area in percentage into components parts of 360°, we have.

Question 2.
The data on modes of transport used by the students to come to school are given below. Draw a pie chart for the data.

Solution:
Converting the percentage into components parts of 360°, we have

Mode of Transport by students.

Question 3.
Draw a histogram for the given frequency distribution.

Solution:
The given distribution is discontinuous.
Lower boundary = lower limit — \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
= 41 – \(\frac { 1 }{ 2 } \) (1) = 40.5
Upper boundary = Upper limit + \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
= 45 + \(\frac { 1 }{ 2 } \) (1) = 45.5
Now continuous frequency table is as below

Question 4.
Draw a histogram and the frequency polygon in the same diagram to represent the following data.

Solution:
The given distribution is discontinuous.
Lower boundary = lower limit – \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
= 50 – \(\frac { 1 }{ 2 } \) (1) = 49.5
Upper boundary = Upper limit + \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
= 55 + \(\frac { 1 }{ 2 } \) (1) = 55.5
∴ The continuous frequency table is as below.

Question 5.
The daily income of men and women is given below, draw a separate histogram for men and women.

Solution:
The given data is continuous frequency distribution. So we take Income in X axis and No. of men in Y axis.

Now we consider the number of women and their income we have.

Challenging Problems

Question 1.
Form a continuous frequency distribution table and draw histogram from the following data.

Solution:
Converting into continuous distribution we have

Question 2.
A rupee spent in a cloth manufacturing company is distributed as follows. Represent this in a pie chart.

Solution:
1 Rupee = 100 paise.

Expenditure of a cloth manufacturing company.

Question 3.
Draw a histogram for the following data.

Solution:
Since mid values are given, the given distributors is discontinuous.
Lower boundary = lower limit — \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
= 15 – \(\frac { 1 }{ 2 } \) (10) = 10
Upper boundary = Upper limit + \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
15 + \(\frac { 1 }{ 2 } \) (10) = 20
The continuous distribution will be as follows.

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 3 Geometry Ex 3.2

Miscellaneous and Practice Problems

Question 1.
Identify the centroid of ΔPQR.

Solution:
In ΔPQR, PT = TR ⇒ QT is a median from vertex Q.
QS = SR ⇒ PS is a median from vertex P.
QT and PS meet at W and therefore W is the centroid of ΔPQR.

Question 2.
Name the orthocentre of ΔPQR.

Solution:
∠P = 90°
This is a right triangle
∴ orthocentre = P [∴ In right triangle orthocentre is the vertex containing 90°]

Question 3.
In the given figure, A is the midpoint of YZ and G is the centroid of the triangle XYZ. If the length of GA is 3 cm, find XA.

Solution:
Given A is the midpoint of YZ.
∴ ZA = AY
G is the centroid of ΔXYZ centroid divides each median in a ratio 2 : 1 ⇒ XG : GA = 2 : 1
\(\frac{XG}{GA}\) = \(\frac{2}{1}\)
\(\frac{XG}{3}\) = \(\frac{2}{1}\)
XG = 2 × 3
XG = 6 cm
XA = XG + GA
= 6 + 3 ⇒ XA = 9 cm

Challenging Problems

Question 4.
Find the length of an altitude on the hypotenuse of a right angled triangle of legs of length 15 feet and 20 feet.
Solution:

Since ∠B = 90°
Using pythagoras theorem
AC² = AB² + BC² = 20² + 15²
= 400 + 225
AC² = 62⁵
AC² = 25⁵
AC = 25
Area of a triangle ΔABC = \(\frac{1}{2}\) × 15 × 20 = 150 feet²
Again Area of ΔABC = \(\frac{1}{2}\) × AC × BD
150 = \(\frac{1}{2}\) × 25 × BD
BD = \(\frac{2 × 150}{25}\) = \(\frac{300}{25}\)
BD = 12 feet
∴ Length of the altitude on the hypotenuse of the right angled triangle is 12 feet.

Question 5.
If I is the incentre of ΔXYZ, ∠IYZ = 30° and ∠IZY = 40°, find ∠YXZ.

Solution:
Since I is the incentre of ΔXYZ
∠IYZ = 30° ⇒ ∠IYX = 30°
∠IZY = 40° ⇒ ∠IZX = 40°
∴ ∠XYZ = ∠XYI + ∠IYZ = 30° + 30°
∠XYZ = 60°
lll ly ∠XYZ = ∠XZI + ∠IZY = 40° + 40°
∠XYZ = 80°
By angle sum property of a triangle
∠XZY + ∠XYZ + ∠YXZ = 180°
80° + 60° + ∠YXZ = 180°
140° + ∠YXZ = 180°
∠YXZ = 180° – 140°
∠YXZ = 40°

Question 6.
In ΔDEF, DN, EO, FM are medians and point P is the centroid. Find the following.
(i) If DE = 44, then DM = ?
(ii) If PD = 12, then PN = ?
(iii) If DO = 8, then FD = ?
(iv) If OE = 36, then EP = ?

Solution:
Given DN, EO, FM are medians.
∴ FN = EN
DO = FO
EM = DM
(i) If DE = 44, then
DM = \(\frac{44}{2}\) = 22
DM = 22

(ii) If PD = 12, PN = ?
\(\frac{PD}{PN}\) = \(\frac{2}{1}\)
\(\frac{12 }{PN}\) = \(\frac{2}{1}\) ⇒ PN = \(\frac{12}{2}\) = 6
PN = 6

(iii) If DO = 8, then
FD = DO + OF = 8 + 8
FD = 16

(iv) If OE = 36,
then \(\frac{EP}{PO}\) = \(\frac{2}{1}\)
\(\frac{EP}{2}\) = PO
OE = OP + PE
36 = \(\frac{PE}{2}\) + PE
36 = \(\frac{PE}{2}\) + \(\frac{2PE}{2}\)
36 = \(\frac{3PE}{2}\)
PE = \(\frac{36 × 2}{3}\)
PE = 24

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 2 Life Mathematics Intext Questions

Exercise 2.1
Try These (Text book Page No. 33)

Classify the given examples as direct or inverse proportion:
(i) Weight of pulses to their cost.
Solution:
As weight increases cost also increases.
∴ Weight and cost are direct proportion.

(ii) Distance travelled by bus to the price of ticket.
Solution:
As the distance increases price to travel also increases.
∴ Distance and price are direct proportion.

(iii) Speed of the athelete to cover a certain distance.
Solution:
As the speed increases, the time to cover the distance become less.
So speed and time are in indirect proportion.

(iv) Number of workers employed to complete a construction in a specified time.
Solution:
As the number of workers increases, the amount of work become less, so they are in indirect proportion.

(v) Volume of water flown through a pipe to its pressure.
Solution:
As the pressure increases, volume also increases.
∴ They are direct proportions.

(vi) Area of a circle to its radius.
Solution:
If the radius of the circle increases its area also increases.
∴ Area and radius of circles are direct proportion.

Use the concept of direct and inverse proportions and try to answer the following questions:
Question 1.
A student can type 21 pages in 15 minutes. At the same rate, how long will it take the student to type 84 pages?
Solution:
Direct proportion
No. of minutes = x
k = \(\frac{21}{15}\)
\(\frac{21}{15}\) = \(\frac{84}{x}\)

Question 2.
The weight of an iron pipe varies directly with its length. If 8 feet of an iron pipe weighs 3.2 kg, find the proportionality constant k and determine the weight of a 36 feet iron pipe.
Solution:

Weight of 36 feet iron pipe = x
\(\frac{36}{x}\) = 2.5

Question 3.
A car covers a distance of 765 km in 51 litres of petrol. How much distance would it cover in 30 litres of petrol?
Solution:
Direct proportion
k = \(\frac{51}{765}\)
Distance cover = x km
\(\frac{30}{x}\) = \(\frac{51}{765}\)

Question 4.
If x and y vary inversely and x = 24 when y = 8, find x when y = 12.
Solution:
k = xy = 24 × 8 = 192
∴ 12 × x = 192

Question 5.
If 35 women can do a piece of work in 16 days, in how many days will 28 women do the same work?
Solution:
Inverse proportion
No. of days = x
k = 35 × 16
∴ 28 × x = 35 × 16

Question 6.
A farmer has food for 14 cows which can last for 39 days. How long would the food last, if 7 more cows join his cattle?
Solution:
Inverse variation
k = xy = 14 × 39
No. of cow = 14 + 7 = 21
No. of days = x
21 × x = 14 × 39

Question 7.
Identify the type of proportion and fill in the blank boxes:

Solution:
Direct proportion
\(\frac{x}{y}\) = k = \(\frac{1}{20}\)
(i) x = 2; y = ?
\(\frac{2}{y}\) = \(\frac{1}{20}\) ⇒ y = 2 × 20 = 40

(ii) x = ?; y = 60
\(\frac{x}{60}\) = \(\frac{1}{20}\) ⇒ x = \(\frac{60}{20}\) = 3

(iii) x = 4; y = ?
\(\frac{4}{y}\) = \(\frac{1}{20}\) ⇒ y = 80

(iv) x = 4; y = ?
\(\frac{8}{y}\) = \(\frac{1}{20}\) ⇒ y = 20 × 8 = 160

(v) x = ?; y = 180
\(\frac{x}{180}\) = \(\frac{1}{20}\)
x = \(\frac{180}{20}\) = 9

(vi) x = 12; y = ?
\(\frac{12}{y}\) = \(\frac{1}{20}\)
y = 12 × 20 = 240

(vii) x = ?; y = 360
\(\frac{x}{360}\) = \(\frac{1}{20}\) ⇒ x = \(\frac{360}{20}\) = 18

(viii) x = 24; y = ?
\(\frac{24}{y}\) = \(\frac{1}{20}\) ⇒ y = 24 × 20 = 480

Question 8.
Identify the type of proportion and fill in the blank boxes:

Solution:
Inverse proportion
k = xy = 1 × 144 = 144
(i) x = 2; y = ?
2y = 144
y = 72

(ii) X = ?; y = 48
48x = 144
x = \(\frac{144}{48}\) = 3

(iii) x = 4; y = ?
4y = 144
y = \(\frac{144}{4}\) = 36

(iv) x = 8; y = ?
8 y = 144
y = \(\frac{144}{8}\) = 18

(v) x = ?; y = 16
16x = 144
y = \(\frac{144}{16}\) = 9

(vi) x = 12; y = ?
12y = 144
y = \(\frac{144}{12}\) = 12

(vii) x = ?; y = 9
9x = 144
x = \(\frac{144}{9}\) = 16

(viii) x = 24; y = ?
24y = 144
y = \(\frac{144}{24}\) = 6

Try These (Text book Page No. 38)

Question 1.
When x = 5 and y = 5 find k, if x and y vary directly.
Solution:
If x and y vary directly then \(\frac{x}{y}\) = k
Here x = 5; y = 5
∴ k = \(\frac{5}{5}\)
k = 1

Question 2.
When x and y vary inversely, find the constant of variation when x = 64 and y = 0.75
Solution:
Given
x =64, y = 0.75
and also given x and y vary inversely.
∴ xy = k. the constant of variation.
∴ Constant = 64 × 0.75
Constant of variation = 48

Think (Text book Page No. 38)

(i) When x and y are in direct proportion and if y is doubled, then what happens to x?
Solution:
If x and y are in direct proportion \(\frac{x}{y}\) = k, constant.
if y is doubled, then \(\frac{x}{2}\) must be equal to k. So x also to be doubled.

(ii) if \(\frac{x}{y-x}\) = \(\frac{6}{7}\) What is \(\frac{x}{y}\)?
Solution:
if \(\frac{x}{y-x}\) = \(\frac{6}{7}\)
\(\frac{y-x}{x}\) = \(\frac{7}{6}\)
\(\frac{y}{x}\) – \(\frac{x}{x}\) = \(\frac{7}{6}\)
\(\frac{y}{x}\) = \(\frac{7}{6}\) + \(\frac{x}{x}\)
\(\frac{y}{x}\) = \(\frac{7}{6}\) + 1
\(\frac{y}{x}\) = \(\frac{7+6}{6}\)
\(\frac{y}{x}\) = \(\frac{13}{6}\)
\(\frac{x}{y}\) = \(\frac{6}{13}\)

Try These (Text book Page No. 40)

Identify the different variations present in the following questions:
Question 1.
24 men can make 48 articles in 12 days. Then, 6 men can make …………. articles in 6 days.
Solution:
Let the required no. of articles be x

(i) Mens and days are Indirect variables.
(ii) Men and Articles are direct variables
(iii) Days and articles are also direct variables using formula.
Let P₁ = 24, D₁ = 12, W₁ = 48
P₂ = 6, D₂ = 6, W₂ = x

Question 2.
15 workers can lay a road of length 4 km In 4 hours. Then, …………. workers can lay a road of length 8 km in 8 hours.
Solution:
Let the required number of workers be x

Length and workers are direct variable as more length need more workers.
The proportion is 4 : 8 : : 15 : x ……….(1)
Hours and workers are indirect variables as more working hours need less men.
∴ The proportion is 8 : 4 : : 15 : x ………..(2)
Combining (1) and (2)

Product of the extremes = product of the means
4 × 8 × x = 8 × 4 × 15
x = \(\frac{8×4×15}{4×8}\)
x = 15 workers

Question 3.
25 women working 12 hours a day can complete a work in 36 days. Then, 20 women must ……….. work hours to complete the same work in 30 days.
Solution:
Let the required hours be x

As women increases hours to work decreases
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{25}{20}\)
As days increases hours needed become less
∴ It is also an indirect variation.
∴ Multiplying factor is \(\frac{36}{30}\)
∴ x = 12 × \(\frac{25}{20}\) × \(\frac{36}{30}\)
x = 18 hours

Question 4.
In a camp, there are 420 kg of rice sufficient for 98 persons for 45 days. The number of days that 60 kg of rice will last for 42 persons is…………
Solution:
Let the required number of days be x.

If amount of rice is more it will last for more days.
∴ It is Direct Proportion.
∴ Multiplying factor is \(\frac{60}{420}\)
If men increases number of days the rice lasts decreases
∴ It is an inverse proportion.
∴ Multiplying factor is \(\frac{98}{42}\)
x = 45 × \(\frac{60}{420}\) × \(\frac{98}{42}\)

x = 15 days

Try These (Text book Page No. 44)

Question 1.
Vikram can do one-third of work in p days. He can do \(\frac{3}{4}\)th of work in ………… days.
Solution:
\(\frac{1}{3}\) of the work will be done in p days
∴ Full work will be completed in 3p days
\(\frac{3}{4}\)th of the work will be done in = 3p × \(\frac{3}{4}\) = \(\frac{9}{4}\)p = 2\(\frac{1}{4}\)p days

Question 2.
If m persons can complete a work in n days, then 4m persons can complete the same work in ……….. days and \(\frac{m}{4}\) persons can complete the same work in…….. days
Solution:
Given m persons complete a work in n days
(i) Then work measured in terms of Man days = mn
4 m men do the work it will be completed in \(\frac{mn}{4m}\) days = \(\frac{n}{4}\) days.
(ii) \(\frac{m}{4}\) persons can complete the same work in \(\frac{mn}{\frac{m}{4}}\) days = \(\frac{4mn}{m}\) = 4n days

Students can Download Maths Chapter 4 Statistics Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 4 Statistics Intext Questions

Exercise 4.1
Try These (Text book Page no. 77)

Question 1.
Arrange the given data in ascending and descending order:
9, 34, 4, 13, 42, 10, 25, 7, 31, 4, 40
Solution:
Ascending order: 4, 4, 7, 9, 10, 13, 25, 31, 34, 40, 42.
Descending order : 42, 40, 34, 31, 25, 13, 10, 9, 7, 4, 4

Question 2.
Find the range of the given data : 53, 42, 61, 9, 39, 63, 14, 20, 06, 26, 31, 4, 57
Solution:
Ascending order of the given data:
4, 6, 9, 14, 20, 26, 31, 39, 42, 53, 57, 61, 63
Here largest value = 63
Smallest value = 4
∴ Range = Largest value – smallest value = 63 – 4 = 59

Think (Text book Page no. 79)

How will you change the given series as continuous series
15 – 25
28 – 38
41 – 51
54 – 64
Solution:
Given series
15 – 25
28 – 38
41 – 51
54 – 64
Difference in the gap = 28 – 25 = 3
Here half of the gap = \(\frac{1}{2}\)(3) = 1.5
∴ 1.5 is the adjustment factor. So we subtract 1.5 from the lower limit and add 1.5 to the upper limit to make it as a continuous series.

Discontinuous series	Continuous series
15-25	13.5-26.5
28-38	26.5-39.5
41-51	39.5-52.5
54 – 64	52.5-65.5

Think (Text book Page no. 80)

If we want to represent the given data by 5 classes, then how shall we find the interval?
Solution:
We can find the class size by the formula
Number of class intervals = \(\frac{Range}{Class size}\)

Try These (Text book Page no. 82)

Question 1.
Prepare a frequency table for the data : 3, 4, 2, 4, 5, 6, 1, 3, 2, 1, 5, 3, 6, 2, 1, 3, 2, 4
Solution:
Ascending order of the given data.
1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 6
The distribution table:

∴ Frequency Table:

Question 2.
Prepare a grouped frequency table for the data :
10, 9, 3, 29, 17, 34, 23, 20, 39, 42, 5, 12, 19, 47, 18, 19, 27, 7, 13, 40, 38, 24, 34, 15, 40
Largest value = 47
Smallest value = 3
Range = Largest value – Smallest value = 47 – 3 = 44
Suppose we take class size as 10, then Number of class intervals possible
= \(\frac{Range}{Class size}\) = \(\frac{44}{10}\) = 4.4
\(\tilde { – } \) 5

Exercise 4.2
Think (Text book Page no. 94)

When joining two adjacent midpoints w ithout using a ruler, can you get a polygon?
Solution:
No, because it may be curved lines and they are not considered as polygons.

Students can Download Maths Chapter 2 Life Mathematics Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 3 Chapter 2 Life Mathematics Additional Questions

Question 1.
120 men had food for 200 days. After 5 days 30 men left the camp. How long will the remaining food last.
Solution:
Since 30 men left after 5 days, the remaining food is sufficient for 120 men for 195 days. Suppose the remaining food lasts for x days for the remaining 90 men.
We have

More men means less days the food lasts
∴ It is inverse proportion
120 : 90 = x : 195
Product of extremes = Product of means
120 × 195 = 90 × x
x = \(\frac{120×195}{90}\)
x = 90
x = 260.
∴ Remaining food last for 260 days.

Question 2.
15 men earn Rs 900 in 5 days, how much will 20 men earn in 7 days?
Solution:
In one day 15 men earn Rs 900
In one day 15 men earn \(\frac{900}{5}\) = Rs 180
In one day 1 men earn \(\frac{180}{5}\) = Rs 12
∴ 1 men earn in 7 days = 12 × 7 = Rs 84
∴ 20 men earn in 7 days = 84 × 20 = 1680

Question 3.
A and B together can do a piece of work in 10 days, B and C can do the same work together in 12 days, A and C can do together in 15 days. How long will it take to complete the work working three of them altogether?
Solution:
(A + B)’s 1 day’s work = \(\frac{1}{10}\)……….(1)
(B + C)’s 1 day’s work = \(\frac{1}{12}\)……….(2)
(A + C)’s 1 day’s work = \(\frac{1}{15}\)……….(3)
(l) + (2) + (3) ⇒
[A + B + B + C + A + C]’s 1 day work = \(\frac{1}{10}\) + \(\frac{1}{12}\) + \(\frac{1}{15}\)
(2A + 2B + 2C)’s 1 day work = \(\frac{6 + 5 + 4}{60}\)
2(A + B + C)’s 1 day work = \(\frac{15}{60}\)
(A + B + C)’s 1 day’s work = \(\frac{1}{4 × 2}\) = \(\frac{1}{8}\)
∴ A + B + C work together to finish the work in 8 days.

Students can Download Science Term 3 Chapter 4 Water Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Science Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Science Solutions Term 3 Chapter 4 Water

Samacheer Kalvi 8th Science Water Text Book Exercise

I. Choose the best answer:

Question 1.
Water changes to ice at ………….
(a) 0°C
(b) 100°C
(c) 102°C
(d) 98°C
Answer:
(a) 0°C

Question 2.
Solubility of carbon dioxide in water is high when the ………….
(a) pressure is low
(b) pressure is high
(c) temperature is high
(d) None of the above
Answer:
(b) pressure is high

Question 3.
The gas collected at the cathode on electrolysis of water is ………….
(a) oxygen
(b) hydrogen
(c) nitrogen
(d) carbon dioxide
Answer:
(b) hydrogen

Question 4.
Which of the following is a water pollutant?
(a) Lead
(b) Alum
(c) Oxygen
(d) Chlorine
Answer:
(a) Lead

Question 5.
Permanent hardness of water is due to the presence of ………….
(a) Sulphates and Chlorides
(b) Dust particles
(c) Carbonates and Bicarbonates
(d) Other soluble particles
Answer:
(a) Sulphates and Chlorides

II. Fill in the blanks:

1. Water is colourless, odourless and ………….
2. The boiling point of water is ………….
3. Temporary hardness of water can be removed by …………. of water
4. The density of water is maximum at …………..
5. Loading speeds up the process of …………..

Answer:

1. tasteless
2. 100°C
3. boiling
4. 4°C
5. Sedimentation

III. State True or False. If false, correct the statement:

Question 1.
Sewage should be treated well before being discharged it into water bodies.
Answer:
True.

Question 2.
Sea water is suitable for irrigation as it contains dissolved salts.
Answer:
False.

Correct statement:
Sea water is not suitable for irrigation as it has high salinity.

Question 3.
Excessive use of chemical fertilizers depletes the soil and causes water pollution.
Answer:
True.

Question 4.
Water unfit for drinking is called potable water.
Answer:
False

Correct statement:
Water suitable for drinking is called potable water.

Question 5.
Soap lathers well in hard water.
Answer:
False

Correct statement:
Soap lathers well in soft water.

IV. Match the following:

1. Universal solvent – Water Pollutant
2. Hard water – Kills germs
3. Boiling – Ozonisation
4. Sterilization – Water
5. Sewage – Stomach ailments

Answer:

1. Universal solvent
2. Hard water
3. Boiling
4. Sterilization
5. Sewage

V. Give reasons for the following:

Question 1.
Alum is added to water in sedimentation tank.
Reason:

1. Chemical substance potash alum is added to water to speed up the process of sedimentation.
2. This process is called loading.
3. The particles of potash alum combine with the suspended impurities and make them settle down at a faster rate.

Question 2.
Water is a universal solvent.
Reason:

1. Water has a unique property to dissolve more substances than any other liquids.
2. It can dissolve solids such as salt and sugar, liquids such as honey and milk and gases such as oxygen and carbon dioxide in it.
3. Water can dissolve more number of substances than any other solvent.
4. Therefore, it is called as universal solvent.

Question 3.
Ice floats on water.
Reason:

1. This is because ice is lighter than water.
2. It means that the density of ice is lower than that of water.
3. Since ice is a bad conductor of heat it does not allow heat to pass through it.
4. So, the water below the ice remains in liquid form, where most of the aquatic life lives.

Question 4.
Aquatic animals can breathe in water.
Reason:

1. Air dissolved in water is important for the aquatic animals to survive.
2. Aquatic animals extracts the oxygen form the water and expels water.
3. They can breathe in water only through the dissolved oxygen present in water.

Question 5.
Sea water is unfit for drinking.
Reason:

1. Every litre of sea water contains 35 grams of dissolved salts most commonly known as sodium chloride (NaCl).
2. Such water is called saline water.
3. It is not suitable for drinking and is said to be non-potable water.

Question 6.
Hard water is not good for washing utensils.
Reason:
Hard water damages the utensils and containers in which it is stored and forms a hard layer.

VI. Define the following:

Question 1.
Freezing Point.
Answer:
The temperature at which a liquid turns into solid when cooled is known as freezing point. The freezing point of water is 0°C.

Question 2.
Boiling point.
Answer:
The temperature at which a water boils and changes to steam is called as boiling point. The boiling point of water is 100°C at atmospheric pressure.

Question 3.
Specific heat capacity.
Answer:
Amount of heat that is needed to raise the temperature of a unit mass of a substance by 1°C is called specific heat capacity.

Question 4.
Latent heat of fusion.
Answer:
The amount of heat energy required by ice to change into water is called latent heat of fusion of ice.

Question 5.
Potable water.
Answer:
The water suitable for drinking is called potable water.

VII. Answer in brief:

Question 1.
Name the gas evolved at cathode and anode when water is electrolysis. State their ratio by volume.
Answer:

1. The gas which is evolved at cathode: Hydrogen (H₂).
2. The gas which is evolved at anode: Oxygen (O₂)
3. The ratio of H₂ and O₂ = 2 : 1.

Question 2.
State the importance of dissolved oxygen and carbon dioxide in water.
Answer:

1. Fish extracts the oxygen from the water and expels water through the gills. Fish can survive in water only through the dissolved oxygen present in water.
2. Aquatic plants make use of dissolved carbon dioxide for photosynthesis.
3. Carbon dioxide dissolved in water reacts with limestone to form calcium bicarbonate.
4. Marine organisms such as snails, oysters, etc., extract calcium carbonate from calcium bicarbonate to build their shells.

Question 3.
What are the causes of temporary hardness and permanent hardness of water?
Answer:

1. Temporary hardness is due to the presence of carbonate and bicarbonate salts of calcium and magnesium.
2. Permanent hardness is due to the presence of chloride and sulphate salts of calcium and magnesium.

Question 4.
Define specific latent heat of vaporization of water.
Answer:

1. When water attains the temperature of 100°C, it starts changing its state from liquid to gaseous state, however, the temperature of water does not rise above 100°C.
2. It is because the heat energy supplied only changes the state of the boiling water.
3. This heat energy is stored in steam and is commonly called latent heat of vaporization of steam.

Question 5.
What are the methods of removing hardness of water?
Answer:

1. Boiling – Temporary hardness is easily removed from water by boiling.
2. Adding washing soda – Washing soda is used to remove permanent hardness of water.
3. Ion-exchange – This converts hard water into soft water.
4. Distillation – Temporary and permanent hardness both can be removed by the method of distillation.

VIII. Answer in detail:

Question 1.
How is water purified at a water purification plant?
Answer:
In conventional water treatment plant, water is subjected to different process. They are:
Sedimentation:

1. Water from lakes or rivers is collected in large sedimentation tanks.
2. There, it is allowed to stand undisturbed so that suspended impurities settle down at the bottom of the tank.
3. Sometimes, a chemical substance such as potash alum is added to water to speed up the process of sedimentation, this is called loading.
4. The particles of potash alum combine with the suspended impurities and make them settle down at a faster rate.

Filtration:

1. Water from the sedimentation tanks is then pumped to the filtration tanks.
2. Filtration tanks contain filter beds made up of gravel, sand, pebbles, activated charcoal and concrete.
3. Water passes through these layers and becomes free from any remaining dissolved or suspended impurities completely.

sterilisation:

1. The filtered water is treated chemically to remove the remaining germs or bacteria, this process is called sterilisation.
2. The chemicals that are used in this process are chlorine and ozone.
3. The process of adding chlorine in adequate amounts to water is called chlorination.
4. The water from filtration tanks is pumped into chlorination tanks, where chlorine is added to remove harmful bacteria and other germs.
5. Ozonisation is a process in which water is treated with ozone gas to kill the germs present in it.
6. The sterilisation of water can also be done by exposing it to air and sunlight.
7. Oxygen from the air and sunlight destroy the germs present in water.
8. Aeration is the process in which air under pressure is blown into filtered water, this also helps to kill the germs.

Question 2.
What is permanent hardness of water? How can it be removed?
Answer:
The hardness due to the presence of chloride and sulphate salts of calcium and magnesium is known as permanent hardness of water.
Removal of hardness:
1. Adding washing soda.

1. Washing soda is used to remove permanent hardness of water.
2. It converts chlorides and sulphates into insoluble carbonates.
3. These insoluble carbonates are removed by filtration.

2. Distillation.

1. Temporary and permanent hardness both can be removed by the method of distillation.
2. The water obtained after distillation is called distilled water.
3. It is the purest form of water.

Question 3.
What is Electrolysis? Explain the electrolysis of water.
Answer:
The process of breaking down of water molecules by the passage of electric current is known as electrolysis of water.
Electrolysis of Water:

1. A glass beaker is fixed with two carbon electrodes and it is filled with water up to one third of its volume.
2. The positive carbon electrode acts as anode and the negative carbon electrode acts as cathode.
3. Two test tubes are placed on the electrodes.
4. The electrodes are connected to a battery’ and current is passed until the test tubes are filled with a particular gas.
5. If the gas collected is tested using a burning splint we can notice that the gas in cathode side bums with a popping sound when the burning splint is brought near the mouth of the test tube.
6. This property is usually shown by hydrogen gas and so it is confirmed that the gas inside the test tube is hydrogen.
7. The burning splint placed near the anode side bums more brightly confirming that it is oxygen gas. This experiment shows that water is made up of hydrogen and oxygen.
8. The ratio of hydrogen and oxygen is 2:1. Hence, for every two volumes of hydrogen collected at the cathode, there is one volume of oxygen collected at the anode.

Question 4.
Explain the different ways by which water gets polluted.
Answer:
1. Domestic Sewage.
Untreated sewage contains impurities such as organic matter from food waste, toxic chemicals from household products and it may also contain disease-causing microbes.

2. Domestic waste and plastics.
Plastics block drains spreading vector borne diseases such as malaria and dengue. Waste in water bodies negatively impact aquatic life.

3. Agricultural activities

• Fertilizers, pesticides and insecticides used in agriculture can dissolve in rainwater and flow into water bodies such as rivers and lakes.
• This causes an excess of nutrients such as nitrates and phosphates as well as toxic chemicals into the water bodies and they can be harmful to aquatic life.

4. Industrial waste.

• Many industries release toxic waste such as lead, mercury, cyanides, cadmium, etc.
• If this waste is unregulated and is released into water bodies, it negatively impacts humans, plants, animals and aquatic life.

5. Oil spills.
Oil spills cause water pollution which is harmful to aquatic life.

6. Thermal pollution.
Water used for cooling purposes is discharged back to a river or to original water source at a raised temperature and sometimes with chemicals. This rise in temperature decreases the amount of oxygen dissolved in water which adversely affects the aquatic life.

Samacheer Kalvi 8th Science Water Additional Questions

I. Choose the correct Answer:

Question 1.
What is the chemical formula of water?
(a) HO₂
(b) H₂O
(C) H₂O₂
(d) HO
Answer:
(b) H₂O

Question 2.
In which one of the following states water is present on Earth?
(a) solid
(b) liquid
(c) gas
(d) All of these
Answer:
(d) All of these

Question 3.
In the process of electrolysis of water, the positive carbon electrode acts as ……………..
(a) cathode
(b) anode
(c) both a and b
(d) none
Answer:
(b) anode

Question 4.
The boiling point of water increases with …………….. in pressure.
(a) decrease
(b) increase
(c) increase or decrease
(d) none
Answer:
(b) increase

Question 5.
The freezing point of water …………….. with increase in pressure.
(a) increase
(b) decreases
(c) remains same
(d) none
Answer:
(b) decreases

Question 6.
Every litre of sea water contains …………….. grams of dissolved salts.
(a) 40
(b) 70
(c) 35
(d) 10
Answer:
(c)35

Question 7.
The process of adding chlorine in adequate amounts to water is called ……………..
(a) Sterilisation
(b) Ozonisation
(c) Aeration
(d) Chlorination
Answer:
(d) Chlorination

Question 8.
…………….. from the air and sunlight destroy the germs present in water.
(a) Oxygen
(b) Hydrogen
(c) Nitrogen
(d) none
Answer:
(a) Oxygen

Question 9.
Aquatic plants make use of dissolved …………….. for photosynthesis.
(a) O₂
(b) CO₂
(c) N₂
(d) H₂
Answer:
(b)CO₂

Question 10.
Which one of the following has the highest latent heat of vaporization?
(a) Ice
(b) Water
(c) Steam
(d) Metal
Answer:
(c) Steam

II. Fill in the blanks:

1. The chemical name of water is …………….
2. …………….
3. Water was first prepared by ……………..
4. Pure water boils at ……………. °C at one atmospheric pressure.
5. Freezing of water will cause an ……………. is the volume.
6. ……………. has the highest latent heat of fusion.
7. One gram of water requires ……………. of heat to raise its temperature by l°C.
8. Water is circulated around car engine using the ……………. pump and the heat is absorbed.
9. Pure water is ……………. and it shows no action towards litmus paper
10. ……………. does not react with water and any temperature.

Answer:

1. Dihydrogen monoxide
2. O₂
3. Henry Cavendish
4. 100
5. expansion
6. Ice
7. 1 Calorie
8. radiator
9. neutral
10. Copper

III. True or False. If false, give the correct statement:

Question 1.
Washing soda is used to remove permanent hardness of water.
Answer:
True.

Question 2.
Temporary and permanent hardness both can be removed by the method of boiling.
Answer:
False

Correct statement:
Temporary and permanent hardness both can be removed by the method of distillation.

Question 3.
The pleasant taste of drinking water is due to the presence of dissolved substance which include air, CO₂ and minerals.
Answer:
True.

Question 4.
Micro-plastics can be found in almost every freshwater source.
Answer:
True.

Question 5.
During electrolysis, hydrogen and oxygen are obtained in the ratio 1 : 2.
Answer:
False

Correct statement:
During electrolysis, hydrogen and oxygen are obtained in the ratio 2:1.

IV. Match the following:

1. Density of pure water – (a) Inflammable air
2. Henry Cavendish – (b) Cathode
3. Hydrogen – (c) 1 g/cm³
4. Negative carbon electrode – (d) Hydrogen

Answer:

1. c
2. d
3. a
4. b

V. Very short answer questions:

Question 1.
Name the process by which temporary hardness is easily removed from water.
Answer:
Boiling.

Question 2.
Name the chemical which converts chlorides and sulphates into insoluble carbonates.
Answer:
Washing soda.

Question 3.
What is the taste of distilled water & boiled water?
Answer:
No taste.

Question 4.
When did Henry Cavendish discover Water?
Answer:
Water was first prepared in 1781 by an English scientist Henry Cavendish.

VI. Short Answer Questions:

Question 1.
What are the sources of water pollution?
Answer:

1. House hold detergents
2. Domestic sewage
3. Domestic waste and plastics
4. Agricultural activities
5. Oil spills
6. Industrial waste.

Question 2.
How plastic sheets affect the soil?
Answer:

1. Plastic sheets are used in agriculture to grow vegetables.
2. At the end of the season, these plastic sheets are ploughed back into the soil.
3. The plastic sheets break into tiny pieces and get eaten by earthworms, which is harmful to their health and that of soil.

Question 3.
Write a note on ion-exchange.
Answer:

1. Another method used to remove the hardness of water is to pass it through a column of ion-exchange resins where calcium and magnesium ions get replaced by sodium ions.
2. This converts hard water into soft water.

Question 4.
What is the reason for hardness of water?
Answer:
The hardness of water is due to the presence of dissolved salts of calcium and magnesium.

Question 5.
Mention the disadvantages of hard water.
Answer:

1. It is not good for washing clothes. It forms scum with soap and detergents, which makes the soap ineffective and also spoils the clothes further.
2. It damages the utensils and containers in which it is stored and forms a hard layer.
3. It forms scales on the machine parts used in industries and decreases their efficiency.
4. It results in stomach ailments if consumed for a long period.

Question 6.
What is soft water?
Answer:
Water contains a number of dissolved salts and minerals. When these salts are present in very small quantities in water, it is called soft water.

Question 7.
Write the characteristics of potable water.
Answer:

1. Potable water should be colourless and odourless.
2. II should be transparent.
3. It should be free from harmful micro-organisms such as bacteria, virus and protozoa.
4. It should be free from suspended impurities.
5. I should contain some minerals and salts, necessary for our body and some dissolved gases to add taste.

Question 8.
Write a short note on catalytic nature of water.
Answer:
Water acts as a catalyst in a number of reactions. Perfectly dry hydrogen and chlorine gases do not react in the presence of sunlight. However in the presence of traces of water, the reaction takes place with explosion to produce hydrogen chloride.

Question 9.
What are the physical properties of pure water?
Answer:
Pure water is a clear and transparent liquid:

1. Pure water boils at 100° C at one atmospheric pressure.
2. Pure water freezes at exactly 0°C at one atmospheric pressure.
3. Pure water has a density of 1 gm/cm₃.

Question 10.
Write the chemical equation for the production of water when zinc reacts with sulphuric acid.
Answer:
Zn + H₂SO₄ → ZnSO₄ + H₂
2H₂+ O₂ → 2H₂O

VII. Answer in detail:

Question 1.
Tabulate the sources and effects of domestic and industrial pollutants.
Answer:

Question 2.
Mention some simple ideas to avoid water pollution.
Answer:

1. Use detergents that are biodegradable and avoid those that contain toxic chemicals,
2. Wear clothing that is made from natural fibres such as cotton and avoid wearing synthetic fibres such as nylon.
3. Do not throw waste such as plastics into water bodies. Always separate your waste into recyclable, non-recyclable and biodegradable so that it does not cause pollution.
4. Domestic waste water should be treated properly, and all harmful substances should be removed from it, so it can be reused for flushing toilets and gardening.
5. Use bio-pesticides (natural pest control) instead of chemical pest control.
6. Use compost made from cow dung, garden waste and kitchen waste as a fertiliser.
7. Water released from industries should be treated before being discharged.

Question 3.
Explain about oil spills.
Answer:

1. There are large crude oil and natural gas reserves below the sea bed.
2. With the increasing exploration of crude oil in the oceans, accidents in drilling and transporting oil have also increased.
3. Oil spills cause water pollution which is harmful to aquatic life.
4. The oil which remains floating on the water surface blocks sunshine, reduces the oxygen dissolved in water and suffocates marine organisms.

Question 4.
Explain the importance of dissolved salts in water.
Answer:

1. They are essential for the growth and development of plants.
2. They add taste to water.
3. They supply the essential minerals needed for our bodies.
4. Most of the chemical reactions important for our living take place in the cells of our body with the help of water.

Samacheer Kalvi 8th Science Water Intext Activities

Activity – 1

Question 1.
Take some anhydrous copper (II) sulphate powder and place it in a watch glass. Add water drop by drop to the anhydrous copper (II) sulphate. Do you notice any colour change in the powder? You can notice the powder turning blue. It is a test for water.

Answer:

1. The reaction between anhydrous copper (II) sulphate and water is used as a test for water.
2. The white solid turns blue in the presence of water.

Activity -3

Question 1.
Place a sample of tap water on a clean watch glass and place it over a beaker containing water, as shown in Tap water the figure. Boil the water in the beaker. When all the water has evaporated from the watch glass, remove it from the burner and let it cool. What do you see on the watch glass?

Answer:

1. We can observe a number of concentric rings of solid matter deposited on the watch glass.
2. These are the dissolved solids left behind after the evaporation of water Salts, minerals and impurities are the solids dissolved in water.

Activity -5

Question 1.
Take two pots with similar plants. Water one of the plants with tap water and the other with sea water. Record your findings and note the difference observed.
Answer:
Observation:

1. Due to the high salinity of sea water, the plant starts to droop
2. The plant which is watered with tap water grows well.

Activity – 6

Question 1.
Take samples of water from different sources (like a tube well, a lake, a pond or a river) and pour equal quantities of each sample of water into different test tubes. Measure the height of water in each test tube with a scale. Add one or two drops of liquid soap to each test tube. Shake each test tube five times and observe the height of the lather in each sample. Record your observations in the table. Which water is soft ? Which water is hard? Can you say why?
Answer:
Samples of water (Source):

1. Tap water
2. Well water
3. Pond water
4. River water

Height of lather:

1. Less
2. Less
3. More
4. More

Question 2.
Which water is soft?
Answer:

1. Pond and river water are considered as soft due to presence of very small quantities of salts.
2. Salts are present in very small quantities.

Question 3.
Which water is hard?
Answer:
Tap and well water are considered as hard due to presence of large quantity of minerals and salts.