Laxmi

Students can Download Maths Chapter 5 Information Processing Ex 5.1 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 5 Information Processing Ex 5.1

I. Choose the correct answer:

Question 1.
Online or television advertisements influence on spending decisions by
(a) using special music
(b) making me think I need the item
(c) using attractive pictures
(d) all the above
Answer:
(d) all the above

Question 2.
When I go shopping, I will buy
(a) something that looks attractive
(b) something my friend has
(c) something that I need to purchase
(d) the first thing I see in the store
Answer:
(c) something that I need to purchase

Question 3.
The best shopping choice is
(a) always shop at brand name stores
(b) compare the choices before buying
(c) the same thing my friends bought
(d) always buy at a regular shop
Answer:
(b) compare the choices before buying

II. Say true or false:

Question 1.
Wise consumers take time to compare two or three shops before spending money.
Answer:
True

Question 2.
Taking time to analyse advertisements cannot save money when shopping.
Answer:
False

Question 3.
One cannot shop on double and triple coupon discounts in available days.
Answer:
True

Question 4.
Every time one must make a shopping list and stick on within his/her budget.
Answer:
False

III. Find the best buy of the following purchases:

(i) A pack of 5 chocolate bars for ₹ 175 or 3 chocolate bars for ₹ 114?
(ii) Basker buy 1 1/2 dozen of eggs for ₹ 81 and Aruna buy 15 eggs for ₹ 64.5?
Solution:
(i)

Best buy is a pack of 5 chocolate bars for ₹ 175.

(ii)

Best buy is 15 eggs for ₹ 64.5

Question 4.
Using the given picture find the total special offer price of fresh sweets and bakery products to buy 1/2 kg laddu, 1 kg cake, 6 pockets of bread.

Solution:

Question 5.
Using given picture prepare a price list.
Suppose you plan to buy 1 1/2 kg of apple, 2 kg of pomegranate, 2 kg of banana, 3 kg of mango, 1/2 kg of papaya, 3 kg of onion, 1 1/2 kg of tomato, and 1 kg of carrot in shop 1, how much will you save compared to shop 2.

Solution:

Price in shop 2 – Shop 1 = 715 – 675.75 = ₹ 39.25
We can save ₹ 39.25 in shop 1 compared to shop 2.

Question 6.
When you plan to buy a shirt, one shop offers a discount of ₹ 200 on MRP ₹ 1000 and another shop offers 15% discount on the same MRP. Where would you buy?
Solution:
Price in I^st shop = ₹ 1000 – ₹ 200 = ₹ 800

Question 7.
Amazing park is offers a package deal of 5 entrance passes for ₹ 130. If one entrance pass normally costs ₹ 30, how much will you save by taking advantage of this special deal?
Solution:
Cost of one entrance pass = ₹ 30
∴ Cost of 5 entrance passes = 5 × 30 = ₹ 150
But special deal price = ₹ 130
Amount of saving = 150 – 130 = ₹ 20

Question 8.
Consider that you are going to buy the toys that are given below in the price list. Prepare a comparison price chart table and find where will you get the best offer price and also find how much you save?

[The amount to the nearest rupee as ₹ 789.84 is rounded off to ₹ 790]
Solution:

By shopping at Toys & Trades mart we can save ₹ 512 (4328 – 3816) then Golden toys in Toys & Trades mart we get the best price.

Students can Download Maths Chapter 4 Statistics 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 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.

Students can Download Maths Chapter 4 Statistics Ex 4.3 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 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.

Students can Download Maths Chapter 1 Number System Ex 1.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th 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 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.2

Question 1.
Add by using grid 0.51 + 0.25.
Solution:
Here 0.51 = \(\frac { 51 }{ 100 } \) and 0.25 = \(\frac { 25 }{ 100 } \)

First we shade the region 0.51 and then 0.25.
The sum is the total shaded area. 0.51 + 0.25 = 0.76

Question 2.
Add the following by using place value grid.
(i) 25.8 + 18.53
(ii) 17.4 + 23.435
Solution:
(i) 25.8 + 18.53.
Using place value grid.

Therefore 25.8 + 18.53 = 44.33

(ii) 17.4 + 23.435
Lets use the place value grid.

Therefore 17.4 + 23.435 = 40.835

Question 3.
Find the value of 0.46 – 0.13 by grid model.
Solution:
Here 0.46 = \(\frac { 46 }{ 100 } \) and 0.13 = \(\frac { 13 }{ 100 } \)

Shading the region 0.46 and then crossing out 0.13 from the shaded area. The left out shaded region without cross marks is the difference. So 0.46 – 0.13 = 0.33

Question 4.
Subtract the following by using place value grid, (i) 6.567 from 9.231 (ii) 3.235 from 7
Solution:
(i) Let as use place value grid

Therefore 9.231 – 6.567 = 2.664

(ii) Let as use place value grid.

Therefore 7 – 3.235 = 3.765

Question 5.
Simplify: 23.5 – 27.89 + 35.4 – 17.
Solution:
23.5 – 27.89 + 35.4 – 17 = 14.01

Question 6.
Sulaiman bought 3.350 kg of Potato, 2.250 kg of Tomato and some onions. If the weight of the total items are 10.250 kg, then find the weight of onions?
Solution:
Weight of Potato = 3.350 kg
Weight of Tomato = 2.250 kg
Total weight of Potato and Tomato = (3.350 + 2.250 kg)
= 5.600 kg
Weight of potato, tomato and onions = 10.250
Weight of potato and tomato = 5.600
∴ Weight of onions = (10.250 – 5.600) Kg = 4.650 Kg
Weight of onions = 4.650 Kg

Question 7.
What should be subtracted from 7.1 to get 0.713?
Solution:
To get the number to be subtracted

We have 7.1 – 0.713 = 6.387
∴ The number to be subtracted = 6.387

Question 8.
How much is 35.6 km less than 53.7 km?
Solution:
To get the answer we must subtract 53.7km – 35.6 km = 18.1 km

So 35.6 km is 18.1 km less than 53.7 km.

Question 9.
Akilan purchased a geometry box for ₹ 25.75, a pencil for ₹ 3.75 and a pen for ₹ 17.90. He gave ₹ 50 to the shopkeeper. What amount did he get back?
Solution:
Cost of geometry box = ₹ 25.75 (+)
Cost of Pencil box = ₹ 3.75

Question 10.
Find the perimeter of an equilateral triangle with a side measuring 3.8 cm.
Solution:
Perimeter of an equilateral triangle = (Side + Side + Side) Sq. units.

Given side = 3.8
∴ Perimeter = 3.8 + 3.8 + 3.8
Perimeter of the triangle = 11.4 cm

Objective Type Questions

Question 1.
1.0 + 0.83 = ?
(i) 0.17
(ii) 0.71
(iii) 1.83
(iv) 1.38
Answer:

(iii) 1.83

Question 2.
7.0 – 2.83 = ?
(i) 3.47
(ii) 4.17
(iii) 7.34
(iv) 4.73
Answer:

(ii) 4.17

Question 3.
Subtract 1.35 from 3.51
(i) 6.21
(ii) 4.86
(iii) 8.64
(iv) 2.16
Answer:

(iv) 2.16

Question 4.
Sum of two decimals is 4.78 and one decimal is 3.21 then the other one is
(i) 1.57
(ii) 1.75
(iii) 1.59
(iv) 1.58
Answer:

(i) 1.57

Question 15.
The difference of two decimals is 86.58 and one of the decimal is 42.31 Find the other one
(i) 128.89
(ii) 128.69
(iii) 128.36
(iv) 128.39
Answer:

(i) 128.89

Students can Download Maths Chapter 4 Statistics Ex 4.2 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 Ex 4.2

Question 1.
Which of the following data can be represented in a histogram?

1. The number of mountain climbers in the age group 20 to 60.
2. Production of cycles in different years.
3. The number of students in each class of a school.
4. The number votes polled from 7 am to 6 pm in an election.
5. The wickets fallen from 1 over to 50th over in a one day cricket match.

Answers:

1. Yes
2. No
3. No
4. Yes
5. Yes

Question 2.
Fill in the blanks

1. The area of the rectangles are proportional to the _______ given.
2. The total area of the histogram is _______ to the total frequency of the given data.
3. _______ is a graphical representation of continuous frequency distribution with rectangles.
4. Histogram is a graphical representation of _______ data.

Answers:

1. frequency
2. proportional
3. Histogram
4. grouped

Question 3.
In a village, there are 570 people who have cell phones. An NGO survey their cell phone usage. Based on this survey a histogram is drawn. Answer the following questions.

(i) How many people use the cell phone for less than 3 hours?
Solution:
330 people (110+ 220)

(ii) How many of them use the cell phone for more than 5 hours?
Solution:
150 of them (100 + 50)

(iii) Are people using cell phone for less than 1 hour?
Solution:
No.

(iv) Give your suggestions on the data.
Solution:
People should minimise the usage of cell phones.

Question 4.
Draw a histogram for the following data.

Solution:
The given data is continuous frequency distribution taking class intervals on X axis and No. of students on Y-axis, the histogram is given below.

Question 5.
Construct a histogram from the following distribution of total marks of 40 students in a class.

Solution:
The given distribution is continuous taking marks on X axis and No. of students on Y-axis the histogram is constructed.

Question 6.
The distribution of heights (in cm) of 100 people is given below. Construct a histogram and the frequency polygon imposed on it.

Solution:
The given distribution is discontinuous.
Converting into continuous distribution we have
Lower boundary = lower limit – \(\frac { 1 }{ 2 } \) (gap between the adjacent class interval)
= 125 – \(\frac { 1 }{ 2 } \) (1) = 124.5
Upper boundary = Upper limit + \(\frac { 1 }{ 2 } \) (gap between adjacent class interval)
= 135 + \(\frac { 1 }{ 2 } \) = 135.5
∴ The new frequency table is

+

Question 7.
In a study of dental problem, the following data were obtained.

Represent the above data by a frequency polygon.
Solution:
Finding the midpoints of the class interval we get.

The points to be plotted are A (5,5), B (15, 13), C (25, 25), D (35, 14), E (45, 30), G (65, 43.), H (75, 50) to obtain the frequency polygon ZABCDEFGHI.
Where I imagined class between 80 and 90.

Question 8.
The marks obtained by 50 students in Mathematics are given below (i) Make a frequency distribution table taking a class size of 10 marks (ii) Draw a histogram and a frequency polygon.

Solution:
Maximum marks obtained = 89
Minimum marks obtained = 08
Range = Maximum marks – Minimum marks = 89 – 08 = 81
Taking the class size = 10, then

Now we have the continuous frequency table.

We will draw the histogram taking class interval in x axis and frequency in y axis as follows.

Objective Type Questions

Question 1.
Data is a collection of _______
(a) numbers
(b) words
(c) measurements
(d) all the three
Answer:
(d) all the three

Question 2.
The number of times an observation occurs in the given data is called _______
(a) tally marks
(b) data
(c) frequency
(d) none of these
Answer:
(c) frequency

Question 3.
The difference between the largest value and the smallest value of the given data is _______
(a) range
(b) frequency
(c) variable
(d) none of these
Answer:
(a) range

Question 4.
The data that can take values between a certain range is called _______
(a) ungrouped
(b) grouped
(c) frequency
(d) none of these
Answer:
(b) grouped

Question 5.
Inclusive series is a _______ series
(a) continuous
(b) discontinuous
(c) both
(d) none of these
Answer:
(b) discontinuous

Question 6.
In a class interval the upper limit of one class is the lower limit of the other class. This is _______ series.
(a) Inclusive
(b) exclusive
(c) ungrouped
(d) none of these
Answer:
(b) exclusive

Question 7.
The graphical representation of un grouped data is _______
(a) histogram
(b) frequency polygon
(c) pie chart
(d) all the three

Question 8.
Histogram is a graph of a _______ frequency distribution.
(a) continuous
(b) discontinuous
(c) discrete
(d) none of these
Answer:
(a) continuous

Question 9.
A _______ is a line graph for the graphical representation of the continuous frequency distribution.
(a) frequency polygon
(b) histogram
(c) pie chart
(d) bar graph
Answer:
(a) frequency polygon

Question 10.
The graphical representation of grouped data is _______
(a) bar graph
(b) pictograph
(c) pie chart
(d) histogram
Answer:
(d) histogram

Students can Download Maths Chapter 1 Number System Ex 1.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th 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 7th Maths Solutions Term 3 Chapter 1 Number System Ex 1.1

Question 1.
Round each of the following decimals to the nearest whole number.
(i) 8.71
(ii) 26.01
(iii) 69.48
(iv) 103.72
(v) 49.84
(vi) 101.35
(vii) 39.814
(viii) 1.23
Solution.
(i) 8.71
Underlining the digit to be rounded 8.71. Since the digit next to the underlined digit, 7 which is greater than 5, adding 1 to the underlined digit.
Hence the nearest whole number 8.71 rounds to is 9.

(ii) 26.01
Underlining the digit to be rounded 26.01. Since the digit next to the underlined digit, 0 which is less than 5, the underlined digit 6 remains the same.
∴ The nearest whole number 26.01 rounds to is 26.

(iii) 69.48
Underlining the digit to be rounded 69.48. Since the digit next to the underlined digit, 4 which is less than 5, the underlined digit 9 remains the same.
∴ The whole number is 69.48 rounds to is 69.

(iv) 103.72
Underlining the digit to be rounded 103.72 since the digit next to the underlined digit, 7 which is greater than 5, we add 1 to the under lined digit.
Hence the nearest whole number 103.72 rounds to is 104.

(v) 49.84
Underlining the digit to be rounded 49.84. Since the digit next to the underlined digit 8 which is greater than 5, we add 1 to the underlined digit.
Hence the nearest whole number 49.84 rounds to 50.

(vi) 101.35
Underlining the digit to be rounded 101.35. Since the digit next to the underlined digit 3 is less than 5, the underlined digit 1 remains the same.
Hence the nearest whole number 101.35 rounds to is 101.

(vii) 39.814
Underlining the digit to be rounded 39.814. Since the digit next to the underlined digit 8 is greater than 5, we add 1 to the underlined digit.
Hence the nearest whole number 39.814 rounds to is 40.

(viii) 1.23
Underlining the digit to be rounded 1.23. Since the digit next to the underlined digit 2, is less than 5, the underlined digit 1 remains the same.
Hence the nearest whole number 1.23 rounds to is 1.

Question 2.
Round each decimal number to the given place value.
(i) 5.992; tenths place
(ii) 21.805; hundredth place
(iii) 35.0014; thousandth place
Solution:
(i) 992; tenths place
Underlining the digit to be rounded 5.992. Since the digit next to the underlined digit is 9 greater than 5, we add 1 to the underlined digit.
Hence the rounded number is 6.0.

(ii) 21.805; hundredth place
Underlining the digit to be rounded 21.805 since the digit next to the underlined digit is 5, we add 1 to the underlined digit.
Hence the rounded number is 21.81.

(iii) 35.0014; thousandth place
Underlining the digit to be rounded 35.0014. Since the digit next to the underlined digit is 4 less than 5 the underlined digit remains the same.
Hence the rounded number is 35.001.

Question 3.
Round the following decimal numbers upto 1 places of decimal.
(i) 123.37
(ii) 19.99
(iii) 910.546
Solution:
(i) 123.37
Rounding 123.37 upto one places of decimal means round to the nearest tenths place. Underling the digit in the tenths place of 123.37 gives 123.37. Since the digit next to the tenth place value is 7 which is greater than 5, we add 1 to the underlined digit to get 123.4. Hence the rounded value of 123.37 upto one places of decimal is 123.4.

(ii) 19.99
Rounding 19.99 upto one places of decimal means round to the nearest tenth place. Underling the digit in the tenths place of 19.99 gives 19.99. Since the digit next to the tenth place value is 9 which is greater than 5, we add 1 to the underlined digit to get 20.
Hence the rounded value of 19.99 upto one places of decimal is 20.0.

(iii) 910.546
Rounding 910.546 upto one places of decimal means round to the nearest tenths place underlining the digit in the tenths place of 910.546 gives 910.546. Since the digit next to the tenth place value is 4, which is less than 5 the underlined digit remains the same. Hence the rounded value of 910.546 upto one places of decimal is 910.5.

Question 4.
Round the following decimal numbers upto 2 places of decimal.
(i) 87.755
(ii) 301.513
(iii) 79.997
Solution:
(i) 87.755
Rounding 87.755 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 87.755 gives 87.755. Since the digit next to the hundredth place value is 5, we add 1 to the underlined digit.

Hence the rounded value of 87.755 upto two places of decimal is 87.76.

(ii) 301.513
Rounding 301.51 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 301.513 gives 301.513. Since the digit next to the underlined digit 3 is less than 5, the underlined digit remains the same.

∴ The rounded value of 301.513 upto 2 places of decimal is 301.51.

(iii) 79.997
Rounding 79.997 upto 2 places of decimal means round to the nearest hundredths place. Underlining the digit in the hundredth place of 79.997 gives 79.997. Since the digit next to the underlined digit 7 is greater than 5, we add 1 to the underlined number.

Hence the rounded value of 79.997 upto 2 places of decimal is 80.00.

Question 5.
Round the following decimal numbers upto 3 place of decimal
(a) 24.4003
(b) 1251.2345
(c) 61.00203
Solution:
(a) 24.4003
Rounding 24.4003 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandths place of 24.4003 gives 24.4003. In 24.4003 the digit next to the thousandths value is 3 which is less than 5.

∴ The underlined digit remains the same. So the rounded value of24.4003 upto 3 places of decimal is 24.400.

(b) 1251.2345
Rounding 1251.2345 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandths place of 1251.2345 gives 1251.2345, the digit next to the thousandths place value is 5 and so we add 1 to the underlined digit. So the rounded value of 1251.2345 upto 3 places of decimal is 1251.235.

(c) 61.00203
Rounding 61.00203 upto 3 places of decimal means rounding to the nearest thousandths place. Underlining the digit in the thousandth place of 61.00203 gives 61.00203. In 61.00203, the digit next to the thousandths place value is 0, which is less than 5.

Hence the underlined digit remains the same. So the rounded value of 61.00203 upto 3 places of decimal is 61.002.

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 12th English Solutions Supplementary Chapter 2 Life of Pi

Warm Up

Imagine you are going on a trek or an adventure. It requires grit and a strong will to survive the odds and emerge unscathed. Most importantly, you should carry an emergency kit.

Choose the ten most essential items from the box below and complete the table.

Answer:

Samacheer Kalvi 12th English Life of Pi Textual Questions

1. Answer the following questions in a sentence or two each, based on your understanding of the story.

Question (a)
Describe the pathetic condition of Pi in the middle of the ocean.
Answer:
Pi was alone and orphaned. He was drifting in the middle of pacific ocean hanging on to the oar. An adult tiger was in front of him. Sharks were beneath his boat. A storm was raging about him.

Question (b)
Who was Richard Parker?
Answer:
Richard Parker was a Bengal tiger. It stayed in the same boat. Each was cautious of the other.

Question (c)
Richard Parker’s survival seemed incredible to Pi. Why?
Answer:
The elements of nature which threatened his survival the previous night had calmed down. The life boat did not sink. The shark prowled but did not attack. The waves did not pull him off. Thus Richard Parker’s survival seemed incredible.

Question (d)
Why was the great beast not behaving naturally?
Answer:
It was unnatural of 450 pound heavy tiger to lie quietly for two and half days unnoticed by Pi in the lower deck of the boat. It was suffering from sea sickness and sedation. Pi’s father had sedated it early to lessen the stress of sea voyage on the animal. That is why it behaved unnaturally.

Question (e)
Why was Pi not afraid of the hyena?
Answer:
The presence of Richard Parker had intimidated the hyena to such an extent that he hid himself behind the Zebra. So, Pi wasn’t afraid of the hyena.

Question (f)
Describe Pi’s struggle to find drinking water.
Answer:
Pi was just two feet above Richard Parker. He moved the yellow tarpaulin. Thirst egged him on to unroll the tight tarpaulin. He saw the bow and it had an end bench. Upon it, a harp glittered like a diamond. He found a lid shaped like a rounded-out triangle. He found a pile of orange life jackets. He undid the harp and opened the lid. It opened on to a locker. The open locker shined with new things. He found water cans piled up.

Question (g)
What was Pi’s reaction when he discovered drinking water?
Answer:
Hitting the lid against the tarpaulin hook, Pi opened the water can and started drinking. To the gurgling beat of his greedy throat, pure delicious, beautiful, crystalline water flowed into his system. It was liquid life. He drained that golden cup to the very last drop sucking at the hole to catch any remaining moisture.

Question (h)
Why did Pi want to tame Parker?
Answer:
A part of Pi was glad about Richard Parker. A part of Pi did not want Richard Parker to die because if he died, Pi would be left alone to deal with despair, a foe is much more formidable than a tiger. So, he wanted to tame Richard Parker.

Question (i)
What saved Pi’s life?
Answer:
Parker kept Pi from thinking too much about his family and his tragic circumstances. He pushed him on to go on living. It was Richard Parker who pushed him to go on living.

Question (j)
How did Pi reunite with this family?
Answer:
Richard Parker and Pi reached an island. Parker left him without even looking back at him. A ship did find Pi and he got reunited with his family.

Additional Questions

Question (a)
How many days did Pi drift on a lifeboat in the pacific ocean?
Answer:
Pi drifted on a boat in the Pacific ocean for 227 days.

Question (b)
What happened to the ship Pi sailed with his family?
Answer:
The ship disappeared with much burbling and belching lights flickered and went out.

Question (c)
What did Pi look for after the ship sank?
Answer:
Pi looked about for his family, for survivors, for another boat and for anything that might bring him hope.

Question (d)
Why did Pi find himself very weak?
Answer:
For about three days Pi had not had a drop of water to drink. He did not have a minute of sleep in three days. So, he found himself very weak.

Question (e)
What appeared to be a conundrum for Pi?
Answer:
Pi had failed to notice the presence of a 450 pound Bengal Tiger in his lifeboat for about two and a half days. The boat was just twenty six feet long. So, his failure to notice the big beast in a small boat appeared to be a conundrum.

Question (f)
Why did Richard Parker seem to be the largest stowaway in the history of navigation?
Answer:
Richard Parker was 450 pound heavy. He was covering from the tip of his nose to the tip of his tail over a third of the length of the ship he was on. So, the feat surely made Richard Parker the largest stowaway proportionately speaking in the history of navigation.

2. Answer the following questions in about 50 words.

Question (a)
How did the presence of Richard Parker help Pi?
Answer:
Pi had to be alert to stay alive in the presence of a 450 pound heavy tiger whose moods may swing. The tiger’s presence did not allow him to think much about his family and the tragic circumstances he was placed in life. He had to keep alert to stay alive. He hated Richard Parker but was grateful to the animal for keeping him out of despair, an enemy much more formidable than the Bengal tiger.

Question (b)
Describe the lifeboat.
Answer:
The life-boat was three and a half feet deep, eight feet wide and twenty six feet long exactly. It was designed to accomodate a maximum of thirty two people. But just the presence of three (i.e.) Zebra, hyena and the tiger along with him made it appear that the boat was already crowded. In the lower part there were many orange life boats and packaged drinking water kept in a closed box.

Question (c)
How did Pi feel after drinking water?
After drinking water, he felt great. A sense of well being quickly overcame him. His mouth became moist and soft. His skin relaxed. His joints moved with greater ease. His heart began to beat like a merry drum. Blood started flowing through his veins like cars from a wedding party honking their way through town. Strength and suppleness came back to his muscles. His head became clearer. In fact, he felt like returning from the world of dead to life.

Question (d)
Did Pi want Richard Parker to die? Answer, giving reasons.
Answer:
No, Pi did not want Richard Parker to die. He was in fact glad about Parker. If he still had the will to live, it was because of Parker. He kept him from thinking too much about his family and the tragic circumstances. Though he hated him for his capability to strike terror in his heart, he wanted him to live. If Parker died, Pi will have to live alone with despair which is a much more formidable enemy than Parker himself.

Additional Questions

Question (a)
Why did Pi compare himself to a losing Tennis champion who bounces back towards the end?
Answer:
Pi was pitted against two foes. The first was a hyena and the second was Richard Parker, the Bengal tiger. The presence of tiger made hyena very quiet. Though he was scared of hyena, the presence of Parker outmatched him. Thirst made him overcome the fear of both foes. The debilitating thirst made him look for water inside the boat. This was like a losing Tennis champion who attacks the confident foe fearing nothing.

3. Answer in a paragraph.

Question (a)
How did the presence of Richard Parker influence the attitude of Pi during his stay on the boat?
Answer:
It was Richard Parker who calmed down Pi. The irony of life is that the animal that scared him witless to start with was the same who brought him peace, purpose and even wholeness. They were literally and figuratively in the same boat. They had to live together. A part of Pi’s personality was glad about Parker’s presence, who gave him the will to hang on to life under the toughest conditions.

He realized danger and peace can co-exist when one is aware of one’s innate strength. He realizes the fact that staying with a tiger is less tormenting than living alone with despair. Loneliness and despair can easily kill a person. He decided to focus on the chances of survival and shut off the tragic circumstances he was placed in a lonely boat in the middle of Pacific Ocean. In short, he became optimistic, balanced and stoical in his attitude to life thanks to the presence of Richard Parker.

Question (b)
Water is the elixir of life. Substantiate the statement with reference to the story of ‘Life Of Pi’.
Answer:
The author calls it ‘liquid life’. For a thirsty man, a cup of water is more precious than gold. He had not drunk water for two and a half days after the ship wreck. Pi drank two litres of the most exquisite nectar (i.e.) water. Suddenly Pi’s forehead was wet with fresh perspiration. Everything in him right down to the pores of his skin was expressing joy. A sense of well-being quickly overcame him. His skin relaxed, his joints moved with greater ease.

His heart began to beat like a merry drum. Blood started flowing through his veins like cars from a wedding party honking their way through the town. Strength and suppleness came back to his muscles. His head became clearer. Truly he was returning to life from the world of dead. After being thirsty for a while, to be drunk on water is noble and ecstatic basked in bliss and plentitude for several minutes. These experiences of Pi highlight the indisputable fact that water is the elixir of life.

Question (c)
If you were lost at sea for as long as Pi was, what is the one item you would want with you? Write a diary entry in which you identify the item and explain why it is the one thing you would want with you.
Answer:
If I were lost in the seas similar to Pi, I wouldn’t have to worry about water because I will carry enough water to last for 140 days. Lifejackets would be already available. A whistle could be my choice of item. Even at night if a ship is passing by I can whistle and attract the attention of the ship. When I feel lonely I can whistle. I can scare away prowling sharks too with my whistle as it would sound like the sound made by dolphins in deep waters.

Occasionally fishermen also come deep into the ocean with their trawlers or fishing boats in early evenings or dawn. I could easily invite their attention. Inspite of having a whistle, if no boat or ship comes that side, it would be difficult to survive. But I can whistle as well as wave the orange life jackets to attract possible rescuers.

Additional Questions

Question (а)
Why does Pi say that thirst is worse than hunger?
Answer:
Prolonged hunger may result in the discomfort of death due to lack of oxygen. But thirst is a long drawn affair. Lord Jesus Christ died on the cross due to suffocation. But his only desire during the last moments of his life was water to drink. If thirst can be so taxing even God incarnate desires of, one can imagine how an ordinary human could drive a thirsty man to go raving mad in thirst. Pi had never before experienced physical hell than that putrid taste and pasty feeling in the mouth. It was an unbearable pressure at the back of his throat. He had a sensation that his blood was turning into a thick syrup that barely flowed through his veins.

Question (b)
How did Pi convince himself that there must be water on board?
Answer:
The divining rod in Pi’s mind dipped sharply and spring gushed forth. He remembered that he was on a genuine regulation lifeboat and such a lifeboat was surely outfitted with supplies. A captain would never fail in so elementary way of preserving water to ensure safety and survival of his crew in the event of a disaster. Besides, it is natural that a ship chandler would think of making a little extra money under the noble pretext of saving lives. It was settled beyond doubt that there was water on board.

4. Sequence the following incidents logically to write the summary of the story ‘Life of Pi’.

• As he looked around, he was shocked to find Richard Parker on board.
• His search for water took him dangerously close to Richard Parker but nothing could stop him neither Richard Parker nor the hyena
• Pi left Richard Parker in a jungle and reunited with his family.
• Pi came back to life and his senses after drinking the elixir of life.
• He understood that it was Richard Parker who helped him survive for 227 days.
• He was pinned by weakness having had no food, water or even sleep for nearly three days.
• Strangely his thirst overpowered his fear of Richard Parker and he went about exploring for ‘ fresh water.
• A little later, he succeeded in his search, when he found stacks of cans of drinking water.
• Pi was stranded in the Pacific on a lifeboat.
• Then, he realized that Parker who scared him earlier brought him peace, purpose and wholeness.

Answer:

1. Pi was stranded in the Pacific on a lifeboat.
2. He was pinned by weakness having had no food, water or even sleep for nearly three days.
3. As he looked around, he was shocked to find Richard Parker on board.
4. Strangely his thirst overpowered his fear of Richard Parker and he went about exploring for fresh water.
5. His search for water took him dangerously close to Richard Parker but nothing could stop him neither Richard Parker nor the hyena.
6. A little later, he succeeded in his search, when he found stacks of cans of drinking water.
7. Pi came back to life and his senses after drinking the elixir of life.
8. He understood that it was Richard Parker who helped him survive for 227 days.
9. Then, he realized that Parker who scared him earlier brought him peace, purpose and wholeness.
10. Pi left Richard Parker in a jungle and reunited with his family.

Life of Pi About The Author

Yann Martel was born in Spain to French Canadian parents. Martel’s father worked as a diplomat and the family moved to Costa Rica, France, Mexico and Canada during Martel’s childhood. He grew up speaking both French and English. Martel studied philosophy at Trent University in Ontario, and later spent a year in India visiting religious sisters and zoos.
His first three books received little critical or popular attention but with the publication of Life of Pi in 2001, Martel became internationally famous and he was awarded the Man Booker Prize in 2002.

Life of Pi Summary in English

Introduction
Pi recounts the ordeal he faced in the middle of the pacific ocean after the ship wreck. Francis Adirubasamy, a close business associate of his father, a swimming champion taught him to swim. He also gave him his unusual name Piscine Molitor, a Persian swimming club he used to frequent. Pi is named after that swimming club (i.e) Piscine Molitor. Pi’s father once ran a zoo in Pondicherry. As the political situation in India was volatile, he decided to shift the family to Canada. They set sail in a cargo ship on June 21,1977. A crew and many cages full of animals sailed along with them in that cargo ship.

Crisis in the mid ocean
When the cargo ship was sailing in the middle of the Pacific Ocean, there was a huge storm. The ship started sinking. Pi jumped into a lifeboat. Very soon he found himself in the company of Richard Parker, a tiger, a zebra, and a hyena, all in a state of shock. Initially, the hyena keeps quiet and hides behind the zebra unwilling to confront the tiger. After two and a half days of drifting in the lifeboat, Pi discovered that Richard Parker was on board just below him. He was unusually quiet because of sea-sickness and the impact of heavy sedation given to him by Pis father.

Water, water everywhere but no water to drink

Pi felt very thirsty. His blood had thickened. His .legs had become stiff. He was afraid that he may die if he did not drink liquid life.

Search for drinking water
When he lifted the yellow tarpaulin, he found Richard Parker just two feet below his legs. Initially, he was scared of seeking the attention of Richard Parker. But his thirst was so much that he preferred to risk his life to find water. The narrator says, ‘thirst is worse than hunger’. Even Jesus Christ who died on the cross yearned for water only before dying. He opened the tarpaulin still further and found hyena looking at him. But Pi wasn’t afraid as much more formidable 450-pound heavy beast was there between him and the hyena

Analysis of the boat
Pi found that the boat was three and a half feet deep and eight feet wide and twenty-six feet long.lt was designed to accommodate 32 people. But it appeared to be cramped due to the presence of three animals (i.e.) zebra, Richard Parker, and the hyena. Tarpaulin, life jackets the lifeboat and the oars were also orange in color. Even the plastic headless whistle was orange in colour.

Discovery of drinking water

The narrator felt like a dried out lizard. He undid the hasp quietly. He pulled on the lid. It opened on to a locker. He looked down between his legs and got delighted because new things glistened. He experienced a heady mixture of hope, surprise, disbelief, thrill, gratitude all crushed into one. He was positively giddy with happiness. The wine of life was found in pale golden cans. Each can contained 500 ml. He felt that he was about to be released from the hell of thirst. He had water stock to last for 124 days.

Returning from the world of dead

He hit the other end of the can hard against the hook.At a stretch he gulped down 4 cans (i.e.) 2 litres of water. It was the most exquisite of nectars. His forehead was wet with fresh, clean refreshing perspiration, his skin relaxed. His joints moved with greater ease. His heart began to beat like a merry drum, blood flowed through his veins like cars from a wedding party honking their way through the tour. Strength and suppleness came back to his muscles. Truly, he was coming back from the world of the dead to life.

Parker and Pi
Richard Parker had finished off the hyena and the zebra. So, Richard Parker and Pi were literally and figuratively in the same boat. He wanted to tame Parker to stay alive.

Richard parker’s help
It was Richard Parker’s presence that shut off all the other memories of family tragedy and the tragic circumstances he was in. Richard Parker egged him to go on living. He hated Parker yet felt grateful to him. He admits that without Parker, he wouldn’t have been alive to tell the story of his survival.

Inevitable separation
At last Pi’s boat reached an island. Richard Parker jumped and walked away without even looking back at Pi. Pi was found by a ship. He got reunited with the other family members.

Conclusion
The life of Pi teaches us the great lesson that one should focus on living even in the most adverse situation in life.

Life of Pi Summary in Tamil

முன்னுரை:
பசுபிக் கடலின் நடுவே கப்பல் முறிந்ததும் தான் எதிர்கொண்ட சோதனையை பை விவரிக்கிறார். பிரான்சிஸ் அதிருபசாமி தன் தந்தையின் நெருங்கிய தொழில் ரீதியான நண்பனும், நீச்சல் வீரனும் ஆவார். அவர் தான் பைக்கு நீச்சல் கற்றுக் கொடுத்தார். அவர் தான் அடிக்கடி, செல்லும் கிளப் ஒன்றின் பேரான *Pisune molitor’ என்ற வழக்கத்தில் இல்லாத பெயரை இவருக்கு சூட்டினார். பெயரை உச்சரிக்க எளிமையாக அதைச் சுருக்கி “பை” என்று அவர் அழைக்கப்பட்டார். ஒருபோது பை யின் அப்பா பாண்டிச்சேரியில் விலங்கியல் பூங்கா ஒன்றை நடத்தினார். இந்தியாவில் அரசியல் சூழ்நிலை நிச்சயமற்றதாக இருந்ததால் அவர் தம் குடும்பத்தை கனடாவிற்கு மாற்ற முடிவு செய்தார். அவர்கள் ஒரு சரக்குக் கப்பலில் ஜுன் 21, 1977 ஆம் ஆண்டு பயணத்தை மேற்கொண்டனர். கூண்டு நிறைய விலங்குகளும், ஒரு பணியாள் சகிதமாய் கப்பல் பயணத்தை மேற்கொண்டனர்.

நடுக்கடலில் நடந்த விபத்து:
சரக்குக் கப்பல் பசுபிக் கடன் நடுவில் சென்று கொண்டிருந்த போது திடீரென்று புயல் காற்று வீசியது. கப்பல் மூழ்கத் தொடங்கியது. பை மீட்பு படகு ஒன்றில் தாவி ஏறிக் கொண்டார். அங்கு, ரிச்சர்ட் பார்க்கர் எனும் புலி, ஒரு வரிக்குதிரை, ஒரு கழுதைப்புலி உடன் காணப்பட்டது. அவை யாவும் அதிர்ச்சிக்கு உள்ளாகி | இருந்தன. தொடக்கத்தில் புலியுடன் சண்டையிடும் | நோக்கம் இல்லாமல் வரிக்குதிரைக்குப் பின்னால் கழுதைப்புலி ஒளிந்து கொண்டது. இரண்டரை நாள் நிதானமான பயணத்திற்கு பின்னால் ரிச்சர்ட் பார்க்கர் தனக்கு கீழ்த்தளத்தில் இருக்கிறது என்பதை அறிந்தார் பை. கடல் வியாதி காரணமாகவும், ‘பை’ யின் தகப்பனார் அதற்கு உடலில் செலுத்திய மயக்க மருந்தின் காரணமாகவும் அது அமைதியாகக் காணப்பட்டது.

எங்கும் தண்ணீர், ஆனால் குடிப்பதற்கு தண்ணீர் இல்லை :

‘பை’ க்கு மிகவும் தண்ணீ ர் தாகம் ஏற்பட்டது. இரத்தம் கட்டியாக மாறியது போல அவர் உணர்ந்தார். கால்கள் விறைப்பாக மாறின. உயிர் தரும் நீரைக் குடிக்காவிட்டால் இறந்து விடுவோமோ என்ற ஐயம் அவருக்குத் தோன்றியது.

குடிநீரைத் தேடும் படலம்:
மஞ்சள் நிற தார்ச்சீலையை அகற்றிப் பார்த்தால் இரண்டடி கீழே ரிச்சர்ட் பார்க்கர் இருந்தது. ஆனால் தாக மிகுதியால் அதைப் பொருட்படுத்தாமல் நீரைத் தேடலானார். கதையாசிரியர் தாகம் பசியைவிடக் கொடுமையானது என்கிறார். ஏசு கிறிஸ்துவும் இறப்பதற்கு முன்னர் ஏங்கியது தண்ணீருக்காகத்தான். தார்ச்சீலையை மேலும் நீக்கிப் பார்த்தால் அங்கு கழுதைப்புலி பையை நோக்கியது. ஆனால், பை பயப்படவில்லை . ஏனெனில், அதைவிட பலமிக்க 450 பவுண்ட் கனம் கொண்ட மிருகம் அவருக்கும் கழுதைப் புலிக்கும் இடையே இருந்தது.

படகை ஆராய்தல்:
படகானது 3 ஆழமாகவும், 8 அடி அகலமாகவும், 26 அடி நீளமாகவும் இருந்தது. அது 32 ஆட்களைக் கொள்ளும் அளவுக்கு வடிவமைக்கப்பட்டு இருந்தது. அது வரிக்குதிரை. ரிச்சர்ட் பார்க்கர் மற்றும் கழுதைப் புலி இருந்ததால் குறுகியதாகக் காணப்பட்டது.) தார்ச்சீலை. உயிர்க் கவசம். மீட்பு படகு மற்றும் துடுப்பும் ஆரஞ்சு நிறத்தில் இருந்தன. பிளாஸ்டிக்காலான தலையற்ற விசிலும் ஆரஞ்சு நிறத்தில் இருந்தது.

குடிநீரைக் கண்டுபிடிக்கும் படலம்;

கதையாசிரியர் உலர்ந்து போன பல்லியைப் போல் உணர்ந்தார். கொக்கியை அகற்றி, மூடியைக் கழற்றினார். ஒரு அலமாரி தென்பட்டது. கால்களுக்கு நடுவே புதிய பொருட்கள் மின்னின. கதையாசிரியருக்கு தன்னம்பிக்கை, ஆச்சரியம், அவநம்பிக்கை, சிலிர்ப்பு, நன்றியுணர்வு என அனைத்தும் கலந்த கலவையான ஒரு உணர்வு ஏற்ப்பட்டது. சந்தோசத்தால் தலை கிறுகிறுத்துப் போயிற்று. தங்க நிறக் குவளைகளில் தண்ணீர் என்ற பானம் தென்பட்டது. ஒவ்வொரு குவளையிலும் 500மிலி, தண்ணீ ர் இருந்தது. தாகம் எனும் நரகத்தில் இருந்து தான் விடுதலை ஆகப் போவதை உணர்ந்தார். அவருக்கு 124 நாட்களுக்கு தேவையான நீர் இருந்தது.

இறந்தவர்களின் உலகத்தில் இருந்து மீண்டு வருதல்;

குவளையைக் கூர்மையான கொக்கியால் துளைத்தார். முத்தாக நீர் தென்பட்டது. குவளையை கொக்கியை நோக்கி இடித்தார். ஒரே மூச்சில் 2 லிட்டர் தண்ணீரைக் குடித்தார். அருமையான தேனைப் போல் நீர் இருந்தது. நெற்றி வியர்வைத் |துளியால் நனைந்தது. சதை தளர்ந்தது. மூட்டு இலேசாக நகர்ந்தது. நெஞ்சு படபட என அடித்தது. சத்தத்துடன் கல்யாண கேளிக்கையில் நுழையும் மகிழுந்துகளைப் போல் இரத்தம் நாடியில் ஓடியது. பலமும், மிருதுவான | தன்மையும் சதைக்கு மீண்டும் வந்தது. இறந்தவர்கள் உலகத்தில் இருந்து உண்மையிலேயே மீண்டு வந்தது போல் உணர்ந்தார்.

பார்க்கரும், பையும்:
ரிச்சர்ட் பார்க்கர் கழுதைப் புலியையும், வரிக் குதிரையையும் கொன்று தின்று முடித்துவிட்டது. ஆதலால், ரிச்சர்ட் பார்க்கரும், பையும் எதிர் எதிராக ஒரே படகில் இருந்தனர். பார்க்கரை பழக்கப்படுத்தி தன் வசப்படுத்திக் கொண்டால் தான் அவர் உயிர் தப்பமுடியும் என பை எண்ணினார்.

ரிச்சர்ட் பார்க்கரின் உதவி:
ரிச்சர்ட் பார்க்கர் அங்கு இருந்தது தான், அவர்தன் குடும்ப கவலையையும், தான் மாட்டிக் கொண்டிருக்கும் துயரமான சூழ்நிலையையும் மறக்கும் வண்ணம் செய்தது என்றார். ரிச்சர்ட் பார்க்கர் கதாசிரியர் உயிர் வாழத் தூண்டுதலாக இருந்தது. பார்க்கர் இல்லாவிடில் தான் இந்தக் கதையைக் கூறுவதற்கு உயிருடன் இருந்திருக்க இயலாது என்பதை அவர் ஒப்புக் கொண்டார்.

நிச்சயம் நேரிடக்கூடிய பிரிவு:
கடைசியில் ‘பைன் படகு ஒரு தீவை அடைந்தது. ரிச்சர்ட் பார்க்கர் படகில் இருந்து குதித்துக் கதையாசிரியரைத் திரும்பிக் கூடப் பார்க்காமல் சென்றுவிட்டது. ‘பையை ஒரு கப்பல் கண்டுகொண்டது. அவர் மீட்கப்பட்ட பின் இதர குடும்பத்தினருடன் அவர் சேர்ந்து கொண்டார்.

முடிவுரை:
பையின் வாழ்க்கை நமக்கு கற்றுத் தரும் பாடம் என்னவென்றால், எப்பேர்ப்பட்ட சூழ்நிலையிலும் | மனிதன் வாழவேண்டும் என்ற குறிக்கோளுடன் இருக்க வேண்டும் என்பதே ஆகும்.

Life of Pi Glossary

Textual:

Additional:

Students can Download Maths Chapter 4 Statistics Ex 4.1 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 4 Statistics Ex 4.1

Question 1.
Fill in the blanks

1. Data has already been collected by some other person is _______ data.
2. The upper limit of the class interval (25-35) is _______
3. The range of the data 200, 15, 20, 103, 3, 197, is _______
4. If a class size is 10 and range is 80 then the number of classes are _______
5. Pie chart is a _______ graph.

Answers:

1. Secondary
2. 35
3. 197
4. 8
5. circular

Question 2.
Say True or False

1. Inclusive series is a continuous series.
2. Pie charts are easy to understand.
3. Same pie chart can be used for different samples.
4. Media and business people use pie charts.
5. A pie diagram is a circle broken down into component sectors.

Answers:

1. False
2. True
3. False
4. True
5. True

Question 3.
The continuous series of

Solution:
Difference in the gap = 24 – 20 = 4
Here half of the gap = \(\frac { 1 }{ 2 } \) (4) = 2
∴ 2 is the adjustment factor. So we subtract 2 from the lower limit and add 2 to the upper limit to make it as a continuous series.

Question 4.
Represent the following data in ungrouped frequency table which gives the number of children in 25 families.
1, 3, 0, 2, 5, 2, 3, 4, 1, 0, 5, 4, 3, 1, 3, 2, 5, 2, 1, 1, 2, 6, 2, 1, 4
Solution:
The data given is raw data. Ascending order : 0, 1, 2, 3, 4, 5, 6

∴ Tabulating in frequency distribution table we get

Question 5.
Form a continuous frequency distribution table for the marks obtained by 30 students in a X std public examination.
328, 470, 405, 375, 298, 326, 276, 362, 410, 255, 391, 370, 455, 229, 300, 183, 283, 366, 400, 495, 215, 157, 374, 306, 280, 409, 321, 269, 398, 200.
Solution:
Maximum mark obtained = 495
Minimum marks obtained = 157
Range = Maximum value – Minimum value
Range = 495 – 157 = 338
If we take the class size as 50 then the number of class intervals possible

Question 6.
Apaint company asked a group of students about their favourite colours and made a pie chart of their iindings. Use the information to answer the following questions.

(i) What percentage of the students like red colour?
(ii) How many students liked green colour?
(iii) What fraction of the students liked blue?
(iv) How many students did not like red colour?
(v) How many students liked pink or blue?
(vi) How many students were asked about their favourite colours?
Solution:
Total percentage of students = 100%
∴ 50 students = 100 % – (30% + 20% + 25% + 15%)
= 100 % – 90%
50 students = 10%
of total students = 50
∴ \(\frac { 10 }{ 100 } \) (Total students) = 50
Total students = \(\frac{50 \times 100}{10}\) = 500.
Total students = 500

(i) 20% of the students like red colour.

(ii) 15% of the students liked green colour.
\(\frac { 15 }{ 100 } \) × 500 = 75 students liked green colour.

(iii) 25% students liked blue ⇒ \(\frac { 25 }{ 100 } \) students liked blue.
⇒ \(\frac { 1 }{ 4 } \) students liked blue.

(iv) Percentage of students liked red colour = 20 %
Percentage of students did not like red colour
= 100% – 20% = 80%
∴ Number of students did not like red colour
= 80% of 500 = \(\frac { 80 }{ 100 } \) × 500 = 400
400 students did not like red colour.

(v) Students liked pink or blue = students liked pink + students liked blue.
= 30% of 500 + 25% of 500
= \(\frac { 30 }{ 100 } \) × 500 + × 500 = 150 + 125 = 275

(vi) Total number of students = 500.
500 students were asked about their favourite colour.

Question 7.
Write any five points from the given pie chart information regarding pollutants entering in the oceans.

Solution:

1. The pie chart gives the information about the amount of pollutants entering in oceans from various sources.
2. 30% of ocean is polluted by sewage.
3. 20% of ocean is polluted by air.
4. Farm run off equally contribute as much as air in polluting ocean.
5. The observation is about off shore oil, litter, Industrial waste water, Maritime transportation, Air pollution, Farm run off and sewage.

Question 8.
A survey gives the following information of food items preferred by people. Draw a Pie chart.

Solution:
Total number of people = 160 + 90 + 80 + 50 + 30 + 40 = 450
Converting the number of people prefer various food items into components part of 360°


Food items preferred by people.

Question 9.
Draw a pie chart for the following information.

Solution:


Water in Ocean

Question 10.
Income from various sources for Government of India from a rupee is given below. Draw a pie chart.

Solution:

Income from various sources for Government of India in a rupee.

Question 11.
Monthly expenditure of Kumaran’s family is given below. Draw a suitable Pie chart.

Also
1. Find the amount spent for education if Kumaran spends ₹ 6000 for Rent.
2. What is the total salary of Kumaran?
3. How much did he spend more for food than education?
Solution:

Monthly expenditure of Kumaran’s family

1. Given Kumaran spends ₹ 6000 for Rent.
∴ 15% oftotal expenditure
\(\frac { 15 }{ 100 } \) (Total Expenditure) = 6000
Total Expenditure = \(\frac{6000 \times 100}{15}\)
Total Expenditure = ₹ 40,000
Amount spend for education = 20% of total expenditure.
\(\frac { 20 }{ 100 } \) × 40,000 = ₹ 8000

2. Total salary of Kumaran = ₹ 40,000

3. Amount spend for food = 50% of (40,000)
Amount spend for the food than education = 20,000 – 8,000 = ₹ 12,000

Students can Download Maths Chapter 1 Life Mathematics 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 2 Chapter 1 Life Mathematics Intext Questions

Exercise 1.1
Try This (Text book Page no. 1)

Question 1.
Find the indicated percentage value of the given numbers
Solution:

Think (Text book page no. 6)

Question 1.
An increase from 200 to 600 ¡s clearly a 200%increase. Isn’t it? (check!). With a lot of pride, the traffic police commiuêener of a city reported that the accidents had decreased by 200% ¡n one year. He cameupwith this number stating that the accidents had gone down from 600 last year to 2this year. Is the decrease from 600 to 200, the same 200% as above? Justify.
Solution:
Increase from original value 200 to 600

Decrease from original value 600 to 200

here original value is 600
% decrease = \(\frac{600-200}{600}\) x 100 = \(\frac{400}{600}\) x 100 = 66.67% decrease
Increase from 200 → 600 and % decrease from 600 → 200 are not the same

Try This (Text book page no. 7)

Question 1.
What percent of a day is 10 hours?
Solution:
In a day, there are 24 hours
10 hrs out of 24 hrs is \(\frac{10}{24}\)
As a percentage, we need to multiply by 100
∴ Percentage = \(\frac{10}{24}\) x 100 = 41.67%

Question 2.
Divide ₹ 350 among P, Q and R such that P gets 50% of what Q gets and Q gets 50% of what R gets.
Solution:
Let R get x, Q gets 50% of what R gets
∴ Q gets = \(\frac{50}{100}\) × x = \(\frac{x}{2}\)
P gets 50% of what Q gets
∴ P gets = \(\frac{50}{100}\) x \(\frac{x}{2}\) = \(\frac{x}{4}\)
Since 350 is divided among the three
∴ 350 = x + \(\frac{x}{2}\) + \(\frac{x}{4}\)
350 = \(\frac{4x+2x+x}{4}\) = \(\frac{7x}{4}\) = 350
x = \(\frac{350×4}{7}\) = 200
Q gets = \(\frac{x}{2}\) = \(\frac{200}{2}\) = 100,
P gets = \(\frac{x}{4}\) = \(\frac{200}{4}\) = 50
∴ P = 50
Q = 100
R = 200

Exercise 1.2
Think (Text book Page No. 13)

Question 1.
A shopkeeper marks the price of a marker board 15% above the cost price and then allows a discount of 15% on the marked price. Does he gain or lose in the transaction?
Solution:
Let cost price of marker board be 100
CP = 100 Marks it 15% above CP
∴ Marked price = MP = \(\frac{15}{100}\) x CP + CP = \(\frac{15}{100}\) 100 + 100 = 15 + 100 = 115
Discount % = 15%

∴ He sells it 97.75 which is less than his cost price. Therefore he loses
Loss = 97.75 – 100 = – 2.25

Try This (Text book Page No. 14)

Question 1.
By selling 5 articles, a man gains the cost price of 1 article. Find his gain percentage.
Solution:
Let cost price of article be C.P. Let S.P of 1 articles be SP by selling 5 articles at SP he makes a gain of cost price of one article.
Gain on 1 article = SP – CP; ⇒ Gain% = \(\frac{SP-CP}{CP}\) x 100
Gain on 2 articles = 2 x (SP – CP)
Gain on 5 articles = 5 x (SP – CP)
Given than gain on 5 articles is CP of 1 article
∴ 5(SP – CP) = CP
\(\frac{SP-CP}{CP}\) = \(\frac{1}{5}\)
Gain percentage \(\frac{SP-CP}{CP}\) x 100 = \(\frac{1}{5}\) x 100 % = 20%

Question 2.
By selling 8 articles, a shopkeeper gains the selling price of 3 articles. Find his gain percentage.
Solution:
Let cost price of 1 article be CP. Let selling price of 1 article be SP.
Gain on 1 article = SP – CP
Gain on 8 articles = 8 x (SP – CP)
Given that gain on 8 articles in selling price of 3 articles
8(SP – CP) = 3 x SP
∴ \(\frac{SP-CP}{CP}\) = \(\frac{3}{8}\)
∴ \(\frac{SP}{SP-CP}\) = \(\frac{8}{3}\)
Subtracting 1 on both sides
\(\frac{SP}{SP-CP}\) – 1 = \(\frac{8}{3}\) – 1 = \(\frac{SP-(SP-CP)}{SP-CP}\)= \(\frac{8-3}{3}\)
\(\frac{SP-SP-CP}{SP-CP}\) = \(\frac{5}{3}\) ⇒ \(\frac{CP}{SP-CP}\) = \(\frac{5}{3}\)
\(\frac{SP-CP}{CP}\) = \(\frac{3}{5}\) (taking reciprocals on both sides)
Gain% \(\frac{SP-CP}{CP}\) x 100 = \(\frac{3}{5}\) x 100 = 3 x 20 = 60%

Question 3.
If the C.P of 20 articles is equal to the S.P of 15 articles, find the profit or loss percentage.
Solution:
Given CP of 20 article = SP of 15 articles
∴ SP of 15 articles = CP of 20 articles
∴ SP of 1 article = \(\frac{1}{15}\) x CP of 20 articles
SP = \(\frac{1}{15}\) x 20 CP = \(\frac{20}{15}\) CP = \(\frac{4}{3}\) CP
∴ SP = \(\frac{4}{3}\) CP ⇒ SP is greater than CP
It is a profit.

Exercise 1.3
Try This (Text book Page No. 23)

Question 1.
Find the principal which gives ₹ 420 as C.I @ 20% p.a compounded half yearly for one year.
Solution:
CI = ₹420
Rate = ₹ 20% p.a
Principle = ₹ [required to find] Time period (n) = 1 year.
However, let us value of r to be 20% p.a so for half yearly, r is \(\frac{20}{2}\) = 10%
Formula for Amount (A) when compounded half yearly is

Question 2.
The price of a laptop depreciates @ 4% p.a. If its present price is ₹ 24,000, find its price after 3 years.
Solution:
Let original price of laptop be ‘P’, Rate of depreciation is 4% p.a,
Present price is ₹ 24,000 (D).
Formula for depreciation is

Price after 3 years from now is
P(1 – \(\frac{4}{100}\))ⁿ⁺³ = ? ⇒ (1 – \(\frac{4}{100}\))ⁿ (1 – \(\frac{4}{100}\))³
From (1)
24,000 x (1 – \(\frac{4}{100}\))³
24,000 x \(\frac{96}{100}\) x \(\frac{96}{100}\) x \(\frac{96}{100}\) = 21233.66

Actvity I (text book Page No. 23)

Question 1.
Mukunthan invests 30,000/- for 3 months in a bank which gives C.I at the rate of 12% compounded monthly. A private company offers his S.l at the rate of 12% p.a What is the difference in the interests received by Mukunthan? Do by traditional method and verify your answer by calculator.
Solution:
Principal = 30,000
Time period = 3 months
In Bank rate of interest for CI = 12% compounded monthly
∴ A = (1 + \(\frac{r}{100}\))ⁿ = 30,000(1 + \(\frac{12}{100}\))³
30,000 x \(\frac{112}{100}\) x \(\frac{112}{100}\) x \(\frac{112}{100}\) = 42147.84
∴ CI = A – P = 42147.84 – 30,000
CI = 12147.84
In private company,
Rate of single Interest SI = 12% p.a
So, for 3 months, i.e \(\frac{3}{12}\) = \(\frac{1}{4}\) year,

∴ Difference in interest = CI – SI = 12,147.84 – 900 = 11247.84

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