Bhagya

Students can download Maths Chapter 1 Numbers Ex 1.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Ex 1.2

Question 1.
Fill in the blanks with > or < or =.

1. 48792 ………. 48972
2. 1248654 ……… 1246854
3. 658794 ……… 658794

Solution:

1. <
2. >
3. =

Question 2.
Say True or False.

1. The difference between the smallest number of seven digits and the largest number of six digits is 10.
2. The largest 4 digit number formed by the digits 8, 6, 0, 9 using each digit only once is 9086
3. The total number of 4 digit number is 9000

Solution:

1. False
   Hint: 1000000 – 999999 = 1
2. False
   Hint: 9999 – 999 = 9000
3. True

Question 3.
Of the numbers 1386787215, 137698890, 86720560, which one is the largest? Which one is the smallest?
Solution:
We know that the number with more digits is greater.
The greatest number is 1386787215
The smallest number is 86720560

Question 4.
Arrange the following numbers in the descending order:
128435, 10835, 21354, 6348, 25840
Solution:
128435 > 25840 > 21354 > 10835 > 6348

Question 5.
Write any eight-digit number with 6 in ten lakhs place and 9 in ten-thousandth place.
Solution:
76594231
86493725

Question 6.
Rajan writes a 3-digit number, using the digits 4, 7, and 9. What are the possible numbers he can write?
Solution:
974, 947, 479, 497, 749, 794

Question 7.
The password to access my ATM card includes the digits 9, 4, 6, and 8. It is the smallest 4 digit even number. Find the password of my ATM card.
Solution:
Given digits are 9, 4, 6, and 8.
The smallest number with these digits is 4689 Given that it is an even number.
It may be 4698.
So password of the ATM card is 4698.

Question 8.
Postal Index Number consists of six digits. The first three digits are 6, 3, and 1. Make the largest and the smallest Postal Index Number by using the digits 0, 3, and 6 each only once.
Solution:
Largest Postal Index Number = 631603
Smallest Postal Index Number = 631036

Question 9.
The heights (in meters) of the mountains in Tamil Nadu are as follows:

(i) Which is the highest mountain listed above?
(ii) Order the mountains from the highest to the lowest.
(iii) What is the difference between the heights of the mountains Anaimudi and Mahendragiri?
Solution:
(i) Anaimudi (2695 m)
(ii) Anaimudi, Doddabetta, Velliangiri, Mahendragiri
(iii) 2695 m – 1647 m = 1048 m

Objective Type Questions

Question 10.
Which list of numbers is in order from the smallest to the largest?
(a) 1468, 1486, 1484
(b) 2345, 2435, 2235
(c) 134205, 134208, 154203
(d) 383553, 383548, 383642
Solution:
(c) 134205, 134208, 154203

Question 11.
The Arabian Sea has an area of 1491000 square miles. This area lies between which two numbers?
(a) 1489000 and 1492540
(b) 1489000 and 1490540
(c) 1490000 and 1490100
(d) 1480000 and 1490000
Solution:
(a) 1489000 and 1492540

Question 12.
The chart below shows the number of newspapers sold as per the Indian Readership Survey in 2018. Which could be the missing number in the table?

(a) 8
(b) 52
(c) 77
(d) 26
Solution:
(d) 26

Students can download Maths Chapter 1 Numbers Ex 1.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 1 Numbers Ex 1.1

Question 1.
Fill in the blanks.

1. The smallest 7 digit number is _______
2. The largest 8 digit number is _______
3. The place value of 5 in 7005380 is ________
4. The expanded form of the number 76,70,905 is _______

Solution:

1. 10,00,000
2. 9,99,99,999
3. 5 × 1000 = 5000
4. 7 × 10,00,000 + 6 × 1,00,000 + 7 × 10,000 + 0 + 9 × 100 + 0 + 5 × 1
   (or)
   70,00,000 + 6,00,000 + 70,000 + 900 + 5

Question 2.
Say True or False.

1. In the Indian System of Numeration, the number 67999037 is written as 6,79,99,037
2. The successor of a one-digit number is always a one-digit number 9 + 1 = 10
3. The predecessor of a 3-digit number is always a 3 or 4 digit number 100 – 1 = 99
4. 88888 = 8 × 10000 + 8 × 100 + 8 × 10 + 8 × 1

Solution:

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

Question 3.
Complete the given order.
Ten crores, crore, ten lakh, ………, ……….., ……….., ………, …….., ……..
Solution:
Ten crores, Crore, Ten lakh, Lakh, Ten Thousand, Thousand, Hundred, Ten, One

Question 4.
How many ten thousands are there in the smallest 6 digit number?
Solution:
Smallest 6 digit number = 1,00,000
= \(\frac{100000}{10000}\) = 10
1 Lakh = 10 Ten Thousands

Question 5.
Using the digits 5, 2, 0, 7, 3 forms the largest 5 digit number and the smallest 5 digit number.
Solution:
Given digits = 5, 2, 0, 7, 3
Largest 5 digit number – 75320
Smallest 5 digit number – 20357

Question 6.
Observe the commas and write down the place value of 7.
(i) 56,74,56,345
(ii) 567,456,345
Solution:
(i) 70,00,000
(ii) 7,000,000

Question 7.
Write the following numbers in the International System by using commas.

1. 347056
2. 7345671
3. 634567105
4. 1234567890

Solution:

1. 347,056
2. 7,345,671
3. 634,567,105
4. 1,234,567,890

Question 8.
Write the largest six-digit number and put commas in the Indian and the International Systems.
Solution:
Largest 6 digit number = 9,99,999
Indian System: 9,99,999 (Nine Lakh Ninety-Nine Thousand Nine Hundred Ninety-Nine)
International System: 999,999 (Nine Hundred Ninety-Nine Thousand, Nine Hundred Ninety-Nine)

Question 9.
Write the number names of the following numerals in the Indian System.
(i) 75,32,105
(ii) 9,75,63,453
Solution:
(i) Seventy-five lakh thirty-two thousand one hundred five.
(ii) Nine crore seventy-five lakh sixty-three thousand four hundred fifty-three.

Question 10.
Write the number of names in words using the International System.

1. 345,678
2. 8,343,710
3. 103,456,789

Solution:

1. Three hundred forty-five thousand six hundred seventy-eight.
2. Eight million three hundred forty-three thousand seven hundred ten.
3. One hundred three million four hundred fifty-six thousand seven hundred eighty-nine.

Question 11.
Write the number name in numerals.

1. Two crores thirty lakhs fifty-one thousand nine hundred eighty.
2. Sixty-six million three hundred forty-five thousand twenty-seven.
3. Seven hundred eighty-nine million, two hundred thirteen thousand four hundred fifty-six.

Solution:

1. 2,30,51,980
2. 66,345,027
3. 789,213,456

Question 12.
Tamil Nadu has about twenty-six thousand three hundred forty-five square kilometres of Forest land. Write the number mentioned in the statement in the Indian System.
Solution:
26,345

Question 13.
The number of employees in the Indian Railways is about 10 lakh. Write this in the International System of numeration.
Solution:
1,000,000 (One Million)

Objective Type Questions

Question 14.
1 billion is equal to
(a) 100 crore
(b) 100 million
(c) 100 lakh
(d) 10000 lakh
Solution:
(a) 100 crore

Question 15.
The successor of 10 million is
(a) 1000001
(b) 10000001
(c) 9999999
(d) 100001
Solution:
(b) 10000001

Question 16.
The difference between successor and predecessor of 99999 is
(a) 90000
(b) 1
(c) 2
(d) 99001
Solution:
(c) 2

Question 17.
The expanded form of the number 6,70,905 is
(a) 6 × 10000 + 7 × 1000 + 9 × 100 + 5 × 1
(b) 6 × 10000 + 7 × 1000 + 0 × 100 + 9 × 100 + 0 × 10 + 5 × 1
(c) 6 × 1000000 + 7 × 10000 + 0 × 1000 + 9 × 100 + 0 × 10 + 5 × 1
(d) 6 × 100000 + 7 × 10000 + 0 × 1000 + 9 × 100 + 0 × 10 + 5 × 1
Solution:
(d) 6 × 100000 + 7 × 10000 + 0 × 1000 + 9 × 100 + 0 × 10 + 5 × 1

Students can download Maths Chapter 2 Measurements Ex 2.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2

Question 1.
Say the time in two ways:

Solution:
(i) 10 : 15 hours; quarter past 10; 45 minutes to 11
(ii) 6 : 45 hours; quarter to 7; 45 minutes past 6
(iii) 4 : 10 hours; 10 minutes past 4; 50 minutes to 5
(iv) 3 : 30 hours; half-past 3; 30 minutes to 4
(v) 9 : 40 hours; 20 minutes to 10; 40 minutes past 9.

Question 2.
Match the following:

Solution:
(i) d
(ii) e
(iii) b
(iv) c
(v) a

Question 3.
Convert the following:
(i) 20 minutes into seconds
(ii) 5 hours 35 minutes 40 seconds into seconds
(iii) 3 ½ hours into minutes
(iv) 580 minutes into hours
(v) 25200 seconds into hours
Solution:
(i) 20 minutes into seconds:
1 min = 60 seconds
20 min = 20 × 60 seconds
= 1200 seconds

(ii) 5 hours 35 min 40 seconds into seconds
Solution:
1 hour = 60 min
1 min = 60 seconds
1 hour = 3600 seconds
5 hours = 5 × 3600 seconds
= 18000 seconds
35 min = 35 × 60 seconds
= 2100 seconds
5 hours 35 minutes 40 seconds
= (18000 + 2100 + 40) seconds
= 20140 seconds

(iii) 3 ½ hours into minutes
Solution:
1 hour = 60 minutes
3 ½ hours = 3 hours + 30 min
= (3 × 60 + 30) min
= (180 + 30)min
= 210 min

(iv) 580 minutes into hours
Solution:
1 hour = 60 min
580 min
= \(\frac{580}{60}\) hours
= \(\frac{290}{30}\) hours
= \(\frac{29}{3}\)
= 9 \(\frac{2}{3}\)
= 9 hours 40 min

(v) 25200 seconds into hours:
Solution:
25200 seconds = \(\frac{25200}{3600}\)
= \(\frac{126}{18}\) hours
= \(\frac{63}{9}\) hours
= 7 hours

Question 4.
The duration of electricity consumed by the farmer for his pump set on Monday and Tuesday was 7 hours 20 minutes 35 seconds and 3 hours 44 minutes 50 seconds respectively. Find the total duration of consumption of electricity.
Solution:
The total duration of electricity consumed on both days
= 7 hours 20 min 35 sec + 3 hours 44 min 50 sec
= (7 + 3) hours (20 + 44) min (35 + 50) sec
= 10 hours 64 min 85 seconds
= 11 hours 5 min 25 seconds

Question 5.
Subtract 10 hours 20 min 35 seconds from 12 hours 18 min 40 seconds.
Solution:
12 hours 18 min 40 seconds
= (12 × 3600) + (18 × 60) + 40 seconds
= 43200 + 1080 + 40 seconds
= 44320 seconds
10 hours 20 min 35 seconds
= (10 × 3600) + (20 × 60) + 35 seconds
= 36000 + 1200 + 35 seconds
= 37235 seconds
Difference:
44320 – 37235
= 7085
7085 seconds = (1 × 3600) + 3480 + 5 seconds
= 1 hour 58 minutes 5 seconds

Question 6.
Change the following into 12 hour format
(i) 02:00 hours
(ii) 08:45 hours
(iii) 21:10 hours
(iv) 11:20 hours
(v) 00:00 hours
Solution:
(i) 2 am
(ii) 08:45 am
(iii) 9:10 pm
(iv) 11:20 am
(v) 12 midrid

Question 7.
Change the following into 24-hour format.
(i) 3.15 am
( ii) 12.35 pm
(iii) 12.00 noon
(iv) 12.00 mid night
Solution:
(i) 03.15 hours
(ii) 12.35 hours
(iii) 12.00 hours
(iv) 24.00 hours

Question 8.
Calculate the duration of time
(i) from 5.30 am to 12.40 pm
(ii) from 1.30 pm to 10.25 pm
(iii) from 20.00 hours to 4.00 hours
(iv) from 17.00 hours to 5.15 hours
Solution:
(i) from 5.30 a.m. to 12 .40 p.m.
Duration of time from 5.30 a.m. to noon = 12 : 00 – 5 : 30 = 6 : 30 i.e 6 hours 30 minutes
From noon to 12.40 p.m the duration = 00 hours 40 minutes
Total duration = 6 hours 30 minutes + 00 hours 40 minutes
= 6 hours 70 minutes
= 6 hours + (60 + 10) minutes
= 6 hours + 1 hr 10 minutes
= 7 hours 10 minutes
∴ Duration of time from 5.30 am to 12.40 pm = 7 hours 10 minutes

(ii) From 1.30 pm to 10.25 pm
= (1.30 pm to 10.00 pm) + 25 min
= 8 hrs 30 min + 25 min
= 8 hrs 55 min

(iii) From 20.00 hours to 4.00 hours
= (20.00 hrs to 24.00 hrs) + (24.00 hrs to 4.00 hrs)
= 4 hrs + 4 hrs
= 8 hours

(iv) From 17.00 hrs to 5.15 hours
= (17.00 hrs to 05.00 hrs) + 15 min
= 12 hours + 15 min
= 12 hours 1.5 min

Question 9.
The departure and arrival timing of the Vaigai Superfast Express (No. 12635) from Chennai Egmore to Madurai Junction are given. Read the details and answer the following.

(i) At what time does the Vaigai Express start from Chennai and arrive at Madurai?
(ii) How many halts are there between Chennai and Madurai?
(iii) How long does the train halt at the Villupuram Junction?
(iv) At what time does the train come to Sholavandan?
(v) Find the journey time from Chennai Egmore to Madurai?
Solution:
(i) 13.40 hours – 21.20 hours
(ii) 8 halts
(iii) 5 minutes
(iv) 20.34 hours
(v) 7 hours 40 minutes

Question 10.
Manickam joined a chess class on 20.02.2017 and due to an exam, he left practice after 20 days. Again he continued to practice from 10.07.2017 to 31.03.2018. Calculate how many days did he practice?
Solution:
From the date of joining = 20 days From 10.07.2017 to 31.03.2018
July – 22
Aug – 31
Sep – 30
Oct – 31
Nov – 30
Dec – 31
Jan – 31
Feb – 28
Mar – 31
Total – 265
Total no of practice days = 265 + 20 = 285 days

Question 11.
A clock gains 3 minutes every hour. If the clock is set correctly at 5 am, find the time shown by the clock at 7 p.m?
Solution:
Time gained for 1 hour = 3 min
Time duration from 5 am to 7 pm = 14 hours
Time gained for 14 hours = 1 4 × 3 minutes
= 42 minutes
So, at 7 pm, the clock shows 7 hrs 42 minutes

Question 12.
Find the number of days between Republic day and Kalvi Valarchi Day in 2020.
Solution:
In 2020 Republic Day will be celebrated on 26th January and Kalvi Valarchi Day will be celebrated on 15th July.
Number of days between 26.01.2020 and 15.07.2020
January – 6 Days (from 26.01.2020)
February – 29 Days (2020 is a leap year)
March – 31 Days
April – 30 Days
May – 31 Days
June – 30 Days
July – 15 Days (upto 15.07.2020)
Total – 172 Days.
∴ Total number of days = 172

Question 13.
If the 11th of Jan 2018 is Thursday, what is the day on 20th July of the same year?
Solution:
Jan – 21
Feb – 28
Mar – 31
April – 30
May – 31
June – 30
July – 19
Total – 190 days
190 days = 27 weeks + 1 day
The required day is the first day after Thursday.
Therefore 20th July 2018 is Friday.

Question 14.
(i) Convert 480 days into years.
(ii) Convert 38 months into years
Solution:
(i) 480 days = \(\frac{480}{365}\)
= 1 year 115 days
= 1 year 3 months 25 days

(ii) 38 months = \(\frac{38}{12}\)
= 3 years 2 months

Question 15.
Calculate your age as on 01.06.2018
Solution:
My date of birth 20.11.1999
Convert in the format yyyy/mm/dd

My age is 18 yrs 6 months 11 days

Objective Type Questions

Question 16.
2 days = _____ hours.
(a) 38
(b) 48
(c) 28
(d) 40
Solution:
(b) 48

Question 17.
3 weeks = ……… days
(i) 21
(ii) 7
(iii) 14
(iv) 28
Solution:
(i) 21

Question 18.
The number of ordinary years between two consecutive leap years is _____.
(a) 4 years
(b) 2 years
(c) 1 year
(d) 3 years
Solution:
(d) 3 years

Question 19.
What time will it be 5 hours after 22:35 hours?
(i) 2:30 hours
(ii) 3:35 hours
(iii) 4:35 hours
(iv) 5:35 hours
Solution:
(ii) 3:35 hours

Question 20.
2\(\frac { 1 }{ 2 }\) years is equal to ______ months.
(a) 25
(b) 30
(c) 24
(d) 5
Solution:
(b) 30

Students can download Maths Chapter 2 Measurements Ex 2.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.1

Question 1.
Fill in the blanks.
(i) 250 ml + \(\frac{1}{2}\) ml = _____ l.
(ii) 150 kg 200 g + 55 kg 750 g = ____ kg ____ g.
(iii) 20 l – 1 l 500 ml = ____ l ___ ml
(iv) 450 ml × 5 = ____ l ____ ml.
(v) 50 Kg ÷ 100 g = ______
Solution:
(i) \(\frac{3}{4}\) l
(ii) 205 kg 950 g
(iii) 18 l 500 ml
(iv) 2l 250 ml
(v) 500

Question 2.
True or False.
(i) Pugazhenthi ate 100 g of nuts which is equal to 0.1 kg.
(ii) Meena bought 250 ml of buttermilk which is equal to 2.5 l.
(iii) Karkuzhali’s bag 1 kg 250 g and Poongkodi’s bag 2 kg 750 g. The total weight of their bags 4 kg.
(iv) Vanmathi bought 4 books each weighing 500 g. Total weight of 4 books is 2 kg.
(v) Gayathri bought 1 kg of birthday cake. She shared 450 g with her friends. The weight of cake remaining is 650 g.
Solution:
(i) True
(ii) False
(iii) True
(iv) True
(v) False

Question 3.
Convert into indicated units:
(i) 10 l and 50 ml into ml
(ii) 4 km and 300 m into m
(iii) 300 mg into g
Solution:
(i) 10 l and 50 ml
= 10 × 1000 ml + 50 ml
= (10000 + 50)ml
= 10050 ml

(ii) 4 km and 300 m
= 4 × 1000 + 300 m
= (4000 + 300) m
= 4300 m

(iii) 300 mg
= \(\frac{300}{1000}\)g
= 0.3 g

Question 4.
Convert into higher units:
(i) 13000 mm
(km, m, cm)
Solution:
13000 mm
= \(\frac{13000}{10}\) cm
= 1300 cm
= \(\frac{13000}{1000}\) m = 13000 mm
= 13 m
= \(\frac{13000}{1000000}\) km = 13000 mm
= 0.013 km

(ii) 8257 ml (kl, l)
Solution:
8257 ml
= \(\frac{8257}{1000}\) l
= 8.257 l
= 8257 ml
= \(\frac{8257}{1000000}\) kl
= 0.008257 kl

Question 5.
Convert into lower units:
(i) 15 km (m, cm, mm)
(ii) 12 kg (g, mg)
Solution:
(i) 15 km = 15 × 1000 m = 15000 m
15 km = 15 × 100000 cm
= 1500000 cm
15 km = 15 × 1000000 mm
= 15000000 mm

(ii) 12 kg (g, mg)
Solution:
12 kg = 12 × 1000 g
= 12000 g
12 kg = 12 × 1000000 mg
= 12000000 mg

Question 6.
Compare and put > or < or = in the following:
(i) 800 g + 150 g ____ 1 kg
(ii) 600 ml + 400 ml ____ 1 l
(iii) 6 m 25 cm ____ 600 cm + 25 cm
(iv) 88 cm ____ 8 m 8 cm
(v) 55 g ____ 550 mg
Solution:
(i) 800 g + 150g < 3kg
(ii) 600 ml + 400 ml = 1 l
(iii) 6 m 25 cm = 600 cm + 25 cm
(iv) 88 cm < 8 m 8 cm
(v) 55 g > 550 mg

Question 7.
Geetha brought 2 l and 250 ml of water in a bottle. Her friend drank 300 ml from it. How much of water is remaining in the bottle?
Solution:
Quantity of water Geetha brought = 2 l 250 ml
= 2 × 1000 + 250 ml
= 2000 + 250 ml
= 2,250 ml.
Quantity of water her friend drank = 300 ml
Remaining water = 2250 – 300 = 1950 ml. = 1 litre 950 ml.
Remaining water = 1 litre 950 ml.

Question 8.
Thenmozhi’s height is 1.25 m now she grows 5 cm every year. What would be her height after 6 years?
Solution:
Thenmozhi’s present height = 1.25 m
Rate of growth per year = 5 cm
Her growth in 6 years = 5 cm × 6 = 30 cm.
After 6 years her height = 1.25 m + 30 cm
= 1.25 × 100 + 30 cm
= 125 + 30 cm
= 155 cm.
∴ After 6 years Thenmozhi’s height will be 155 cm.

Question 9.
Priya bought 22\(\frac { 1 }{ 2 }\) kg of onion, Krishna bought 18\(\frac { 3 }{ 4 }\) kg of onion and Sethu bought 9 kg 250 g of onion. What is the total weight of onion did they buy?
Solution:
Priya’s weight = 22 kg 500 g
Krishna’s weight = 18 kg 750 g
Sethu’s weight = 9 kg 250 g
Total weight = 49 kg 1500 g = 49 kg + 1 kg 500 g = 50 kg + 500 g.
Their total weight = 50 kg 500 g.

Question 10.
Maran walks 1.5 km every day to reach the school while Mahizhan walks 1400 m. Who walks more distance and by how much?
Solution:
Distance which Maran walks = 1.5 km = 1.5 × 1000 m = 1500 m
The distance which Mahizhan walks = 1400 m.
Here 1500 > 1400
∴ Difference = 1500 – 1400 = 100 m.
∴ Maran walks more distance = 100 m.

Question 11.
In a JRC one day camp, 150 gm of rice and 15 ml oil are needed for a student. If there are 40 students to attend the camp how much rice and oil are needed?
Solution:
Rice needed for one student = 150 g
Rice needed for 40 students = 150 g × 40 = 6000 g. = \(\frac{6000}{1000}\) kg = 6 kg.
Oil needed for one student = 15 ml
Oil needed for 40 students = 15 ml × 40 = 600 ml. = \(\frac{600}{1000}\) l = 0.6 l
∴ For the camp 6 kg of rice and 0.6 l of oil needed.

Question 12.
In a school, 200 litres of lemon juice is prepared. If 250 ml lemon juice is given to each student, how many students get the juice?
Solution:
Total lemon juice prepared = 200 l = 200 × 1000 ml = 2,00,000 ml.
∴ Quantity of Lemon juice given to one student = 250 ml.
∴ Number of students can get = \(\frac{2,00,000}{250}\) = 800
∴ 800 students can get the lemon juice.

Question 13.
How many glasses of the given capacity will fill a 2 litre jug?
(i) 100 ml ___
(ii) 50 ml ____
(iii) 500 ml ____
(iv) 1 l ____
(v) 250 ml ____
Solution:
2 litre = 2 × 1000 ml = 2000 ml.
(i) 100 ml
\(\frac{2000}{100}=20\)
20 glasses of 100 ml.
(ii) 50 ml
\(\frac{2000}{50}=40\)
40 glasses of 50 ml
(iii) 500 ml
\(\frac{2000}{500}=4\)
4 glasses of 500 ml
(iv) 1 l
\(\frac{2 l}{1 l}=2\)
2 glasses of 1 l.
(v) 250 ml
\(\frac{2000}{250}=8\)
8 glasses of 250 ml can fill the jug.

Objective Type Questions

Question 14.
9 m 4 cm is equal to ……..
(i) 94 cm
(ii) 904 cm
(iii) 9.4 cm
(iv) 0.94 cm
Solution:
(ii) 904 cm

Question 15.
1006 g is equal to ____
(a) 1 kg 6 g
(b) 10 kg 6 g
(c) 100 kg 6 g
(d) 1 kg 600 g
Solution:
(a) 1 kg 6 g

Question 16.
Every day 150 l of water is sprayed in the garden. Water sprayed in a week is ……
(i) 700 l
(ii) 1000 l
(iii) 950 l
(iv) 1050 l
Solution:
(iv) 1050 l

Question 17.
Which is the greatest 0.007 g, 70 mg, 0.07 cg?
(a) 0.07 cg
(b) 0.007 g
(c) 70 mg
(d) all are equal
Solution:
(d) all are equal

Question 18.
7 km – 4200 m is equal to ……….
(i) 3 km 800 m
(ii) 2 km 800 m
(iii) 3 km 200 m
(iv) 2 km 200 m
Solution:
(ii) 2 km 800 m

Students can download Maths Chapter 8 Statistics Additional Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Additional Questions

I. Choose the Best Answer

Question 1.
The Arithmetic mean of all the factors of 10 is ………
(a) 4.5
(b) 5.5
(c) 10
(d) 55
Solution:
(a) 4.5

Question 2.
The mean of five numbers is 27, if one number is excluded, then mean is 25. Then the excluded number is ……..
(a) 0
(b) 15
(c) 25
(d) 35
Solution:
(d) 35

Question 3.
The mean of 8 numbers is 15. If each number is multiplied by 2, then the new mean will be ……..
(a) 7.5
(b) 30
(c) 10
(d) 25
Solution:
(b) 30

Question 4.
The median of 11, 8, 4, 9, 7, 5, 2, 4, 10 is ……..
(a) 1
(b) 8
(c) 4
(d) 11
Solution:
(a) 1

Question 5.
Median is …….
(a) the most frequent value
(b) the least frequent value
(c) middle most value
(d) mean of first and last values
Solution:
(c) middle most value

Question 6.
The mode of the distribution is …….

(a) 3
(b) 4
(c) 6
(d) 14
Solution:
(b) 4

Question 7.
Mode is …….
(a) the middle value
(b) extreme value
(c) minimum value
(d) the most repeated value
Solution:
(d) the most repeated value

Question 8.
The mode of the data 72, 33, 44, 72, 81, 72, 15 is ……..
(a) 12
(b) 33
(c) 81
(d) 15
Solution:
(a) 12

Question 9.
The Arithmetic mean of 10 number is -7. If 5 is added to every number, then the new arithmetic mean is ……..
(a) 17
(b) 12
(c) -2
(d) -7
Solution:
(c) -2

Question 10.
The Arithmetic mean of integers from -5 to 5 is ……..
(a) 25
(b) 10
(c) 3
(d) 0
Solution:
(d) 0

II. Answer the Following Questions

Question 1.
Find the Arithmetic mean for the following data.

Solution:

Arithmetic mean (\(\bar { X }\)) = \(\frac{Σfx}{Σf}\)
= \(\frac{5540}{96}\)
= 57.7
∴ Arithmetic mean = 57.7

Question 2.
Calculate the Arithmetic mean of the following data using step deviation method.

Solution:
Assumed mean = 35

Arithmetic mean (\(\bar { X }\)) = A + \(\frac{Σfd}{Σf}\) × c
= 35 + \(\frac{(-20)}{100}\) × 10
= 35 – 2
= 33
∴ Arithmetic mean = 33

Question 3.
Find the median for the following data.

Solution:
The given class intervals is inclusive type we convert it into exclusive type.

\(\frac{N}{2}\) = \(\frac{70}{2}\)
= 35
Median class is 25.5 – 30.5
Here l = 25.5, f = 26, m = 30, c = 5
Median = l + \(\frac{\frac{N}{2}-m}{f}\) × c
= 25.5 + \(\frac{35-30}{26}\) × 5
= 25.5 + \(\frac{5×5}{26}\)
= 25.5 + 0.96
= 26.46
∴ Median = 26.46

Question 4.
Calculate the mode of the following data.

Solution:

The highest frequency is 30. Corresponding class interval is the modal class.
Here l = 25, f = 30, f₁ = 18, f₂ = 20 and c = 5
Mode

∴ Arithmetic mean = 25 + 2.727
= 25 + 2.73
= 27.73
mode = 27.73

Question 5.
Find the mean, median and mode of marks obtained by 20 students in an examination. The marks are given below.

Solution:

Arithmetic mean:
Here Σfx = 560, Σf = 20
\(\bar { X }\) = \(\frac{Σfx}{Σf}\)
= \(\frac{560}{20}\)

Median:
Median class is 20 – 30
Here l = 20, \(\frac{N}{2}\) = 10, m = 5, c = 10, f = 5
Median

= 20 + 10
= 30

Mode:
The highest frequency is 8
Modal class is 30 – 40
Here, l = 30, f = 8, f₁ = 5, f₂ = 2, c = 10
Mode

= 33.33
Mean = 28, median = 30, mode = 33.33

Students can download Maths Chapter 8 Statistics Ex 8.4 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Ex 8.4

Question 1.
Let m be the mid point and b be the upper limit of a class in a continuous frequency distribution. The lower limit of the class is …….
(a) 2m – b
(b) 2m + b
(c) m – b
(d) m – 2b
Solution:
(a) 2m – b

Question 2.
The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is ……..
(a) 101
(b) 100
(c) 99
(d) 98
Solution:
(c) 99
Hint:
Total of 8 numbers = 81 × 7 = 567
Total of 7 numbers = 78 × 6 = 468
The number is = 567 – 468
= 99

Question 3.
A particular observation which occurs maximum number of times in a given data is called its ………
(a) frequency
(b) range
(c) mode
(d) median
Solution:
(c) mode

Question 4.
For which set of numbers do the mean, median and mode all have the same values?
(a) 2, 2, 2, 4
(b) 1, 3, 3, 3, 5
(c) 1, 1, 2, 5, 6
(d) 1, 1, 2, 1, 5
Solution:
(b) 1, 3, 3, 3, 5

Question 5.
The algebraic sum of the deviations of a set of n values from their mean is ……..
(a) 0
(b) n – 1
(c) n
(d) n + 1
Solution:
(a) 0

Question 6.
The mean of a, b, c, d and e is 28. If the mean of a, c and e is 24, then mean of b and d is ……..
(a) 24
(b) 36
(c) 26
(d) 34
Solution:
(d) 34
Hint:
Mean = 28
a + b + c + d + e = 28 × 5 = 140
= 140 …… (1)
But \(\frac{a+c+e}{3}\) = 24
a + c + e = 72
a + b + c + d + e = 140
b + d + 72 = 140
b + d = 140 – 72
= 68
Mean = \(\frac{68}{2}\)
= 34

Question 7.
If the mean of five observations x, x + 2, x + 4, x + 6, x + 8, is 11, then the mean of first three observations is …….
(a) 9
(b) 11
(c) 13
(d) 15
Solution:
(a) 9
Hint:
Mean = \(\frac{x+x+2+x+4+x+6+x+8}{5}\)
11 = \(\frac{5x+20}{5}\)
5x + 20 = 55
5x = 55 – 20
= 35
x = \(\frac{35}{5}\)
= 7
Mean of first 3 observation is = \(\frac{7+9+11}{3}\)
= 9

Question 8.
The mean of 5, 9, x, 17 and 21 is 13, then find the value of x.
(a) 9
(b) 13
(c) 17
(d) 21
Solution:
(b) 13
Hint:
Mean = \(\frac{5+9+x+17+21}{5}\)
13 = \(\frac{52+x}{5}\)
65 = 52 + x
x = 65 – 52
= 13

Question 9.
The mean of the square of first 11 natural numbers is
(a) 26
(b) 46
(c) 48
(d) 52
Solution:
(b) 46
Hint:

= 46

Question 10.
The mean of a set of numbers is \(\bar { X }\). If each number is multiplied by z, the mean is ……..
(a) \(\bar { X }\) + z
(b) \(\bar { X }\) – z
(c) z\(\bar { X }\)
(d) \(\bar { X }\)
Solution:
(c) z\(\bar { X }\)
Hint:
If each observation is multiplied by k, k ≠ 0 then the arithmetic mean is also multiplied by the same quantity.

Students can download Maths Chapter 4 Geometry Ex 4.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.4

Question 1.
The length of the tangent to a circle from a point P, which is 25 cm away from the centre is 24 cm. What is the radius of the circle?
Solution:
Let the radius AB be r. In the right ∆ ABO,
OB² = OA² + AB²
25² = 24² + r²

25² – 24² = r²
(25 + 24) (25 – 24) = r²
r = \(\sqrt { 49 }\) =7
Radius of the circle = 7 cm

Question 2.
∆ LMN is a right angled triangle with ∠L = 90°. A circle is inscribed in it. The lengths of the sides containing the right angle are 6 cm and 8 cm. Find the radius of the circle.
Solution:
LN = 6; ML = 8. In the right ∆ LMN,
MN² = LN² + LM²
= 6² + 8² = 36 + 64 = 100
MN = \(\sqrt { 100 }\) = 10

OA= OB = OC = r
AN = CN (Tangent of the circle)
LN – AL= CN
LN – r = CN
8 – r = CN ……(1)
MC = MB (tangent of the circle)
MC = ML – LB
MC = 6 – r …….(2)

Add (1) and (2)
MC + CN = (6 – r) + (8 – r)
MN = 14 – 2r
10 = 14 – 2r
2r = 14 – 10 = 4
r = \(\frac { 4 }{ 2 } \) = 2 cm
radius of the circle = 2 cm

Question 3.
A circle is inscribed in ∆ ABC having sides 8 cm, 10 cm and 12 cm as shown in figure, Find AD, BE and CF.

Solution:
AD = AF = x (tangent of the circle)
BD = BE = y (tangent of the circle)
CE = CF = z (tangent of the circle)
AB = AD + DB
x + y = 12 ……(1)
BC = BE + EC
y + z= 8 …….(2)
AC = AF + FC
x + z = 10 ……(3)
Add (1) (2) and (3)
2x + 2y + 2z = 12 + 8 + 10
x + y + z = \(\frac { 30 }{ 2 } \) = 15 …….(4)
By x + y = 12 in (4)
z = 3
y + z = 8 in (4)
x = 7
x + z = 10 in (4)
y = 5
AD = 7 cm; BE = 5 cm and CF = 3 cm

Question 4.
PQ is a tangent drawn from a point P to a circle with centre O and QOR is a diameter of the circle such that ∠POR = 120° . Find ∠OPQ.
Solution:
∠POQ = 180° – 120° = 60°
In ∆OPQ, we know

∠POQ + ∠OQP + ∠OPQ = 180°
(Sum of the angles of a ∆ is 180°)
60° + 90° + ∠OPQ = 180°
∠OPQ = 180° – 150° = 30°
∠OPQ = 30°

Question 5.
A tangent ST to a circle touches it at B. AB is a chord such that ∠ABT = 65°. Find ∠AOB, where “O” is the centre of the circle.
Solution:
Given ∠ABT = 65°
∠OBT = 90°(TB is the tangent of the circle)
∠ABO = 90° – 65° = 25°
∠ABO + ∠BOA + ∠OAB = 180°
25° + x + 25° = 180° (Sum of the angles of a ∆)

(OA and OB are the radius of the circle.
∴ ∠ABO = ∠BAO = 25°
x + 50 = 180°
x = 180° – 50° = 130°
∴ ∠BOA = 130°

Question 6.
In figure, O is the centre of the circle with radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle E, if AB is the tangent to the circle at E, find the length of AB.

Solution:
In the right ∆ OTP,
PT² = OT² – OP²
= 13² – 5²
= 169 – 25
= 144
PT = \(\sqrt { 144 }\) = 12 cm
Since lengths of tangent drawn from a point to circle are equal.
∴ AP = AE = x
AT = PT – AP
= (12 – x) cm
Since AB is the tangent to the circle E.
∴ OE ⊥ AB
∠OEA = 90°
∠AET = 90°
In ∆AET, AT² = AE² + ET²
In the right triangle AET,
AT² = AE² + ET²
(12 – x)² = x² + (13 – 5)²
144 – 24x + x² = x² + 64
24x = 80 ⇒ x = \(\frac { 80 }{ 24 } \) = \(\frac { 20 }{ 6 } \) = \(\frac { 10 }{ 3 } \)
BE = \(\frac { 10 }{ 3 } \) cm
AB = AE + BE
= \(\frac { 10 }{ 3 } \) + \(\frac { 10 }{ 3 } \) = \(\frac { 20 }{ 3 } \)
Lenght of AB = \(\frac { 20 }{ 3 } \) cm

Question 7.
In two concentric circles, a chord of length 16 cm of larger circle becomes a tangent to the smaller circle whose radius is 6 cm. Find the radius of the larger circle.
Solution:
Here AP = PB = 8 cm
In ∆OPA,
OA² = OP² + AP²

= 6² + 8²
= 36 + 64
= 100
OA = \(\sqrt { 100 }\) = 10 cm
Radius of the larger circle = 10 cm

Question 8.
Two circles with centres O and O’ of radii 3 cm and 4 cm, respectively intersect at two points P and Q, such that OP and O’ P are tangents to the two circles. Find the length of the common chord PQ.
Solution:
In ∆ OO’P
(O’O)² = OP² + O’P²
= 3² + 4²

= 9 + 16
(OO’)² = 25
∴ OO’ = 5cm
Since the line joining the centres of two intersecting circles is perpendicular bisector of their common chord.
OR ⊥ PQ and PR = RQ
Let OR be x, then O’R = 5 – x again Let PR = RQ = y cm
In ∆ ORP, OP² = OR² + PR²
9 = x² + y² …(1)
In ∆ O’RP, O’P² = O’R² + PR²
16 = (5 – x)² + y²
16 = 25 + x² – 10x + y²
16 = x² + y² + 25 – 10x
16 = 9 + 25 – 10x (from 1)
16 = 34 – 10x
10x = 34 – 16 = 18
x = \(\frac { 18 }{ 10 } \) = 1.8 cm
Substitute the value of x = 1.8 in (1)
9 = (1.8)² + y²
y² = 9 – 3.24
y² = 5.76
y = \(\sqrt { 5.76 }\) = 2.4 cm
Hence PQ = 2 (2.4) = 4.8 cm
Length of the common chord PQ = 4.8 cm

Question 9.
Show that the angle bisectors of a triangle are concurrent.
Solution:
Given: ABC is a triangle. AD, BE and CF are the angle bisector of ∠A, ∠B, and ∠C.
To Prove: Bisector AD, BE and CF intersect
Proof: The angle bisectors AD and BE meet at O. Assume CF does not pass through O. By angle bisector theorem.
AD is the angle bisector of ∠A
\(\frac { BD }{ DC } \) = \(\frac { AB }{ AC } \) …..(1)
BE is the angle bisector of ∠B

\(\frac { CE }{ EA } \) = \(\frac { BC }{ AB } \) …….(2)
CF is the angle bisector ∠C
\(\frac { AF }{ FB } \) = \(\frac { AC }{ BC } \) …….(3)
Multiply (1) (2) and (3)
\(\frac { BD }{ DC } \) × \(\frac { CE }{ EA } \) × \(\frac { AF }{ FB } \) = \(\frac { AB }{ AC } \) × \(\frac { BC }{ AB } \) × \(\frac { AC }{ BC } \)
So by Ceva’s theorem.
The bisector AD, BE and CF are concurrent.

Question 10.
In ∆ABC , with ∠B = 90° , BC = 6 cm and AB = 8 cm, D is a point on AC such that AD = 2 cm and E is the midpoint of AB. Join D to E and extend it to meet at F. Find BF.
Solution:
Consider ∆ABC. Then D, E and F are respective points on the sides AC, AB and BC.
By construction D, E, F are collinear.
By Menelaus’ theorem \(\frac { AE }{ EB } \) × \(\frac { BF }{ FC } \) × \(\frac { CD }{ DA } \) = 1 ……(1)
AD = 2 cm; AE = EB = 4 cm; BC = 6 cm; FC = FB + BC = x + 6
In ∆ABC, By Pythagoras theorem.

AC²= AB² + BC²
AC² = 8² + 6² = 64 + 36 = 100
AC = \(\sqrt { 100 }\) = 10
CD = AC – AD
= 10 – 2 = 8 cm
Substituting the values in (1) we get
\(\frac { 4 }{ 4 } \) × \(\frac { x }{ x+6 } \) × \(\frac { 8 }{ 2 } \) = 1
\(\frac { x }{ x+6 } \) × 4 = 1
4x = x + 6
3x = 6 ⇒ x = \(\frac { 6 }{ 3 } \) = 2
The value of BF = 2 cm

Question 11.
An artist has created a triangular stained glass window and has one strip of small length left before completing the window. She needs to figure out the length of left out portion based on the lengths of the other sides as shown in the figure.

Solution:
Given that AE = 3 cm, EC = 4 cm, CD = 10 cm, DB = 3 cm, AF = 5 cm.
Let FB be x
Using Ceva’s theorem we have
\(\frac { AE }{ EC } \) × \(\frac { CD }{ DB } \) × \(\frac { BF }{ AF } \) = 1
\(\frac { 3 }{ 4 } \) × \(\frac { 10 }{ 3 } \) × \(\frac { x }{ 5 } \) = 1
\(\frac { 2x }{ 4 } \) = 1
2x = 4 ⇒ x = \(\frac { 4 }{ 2 } \) = 2
The value of BF = 2

Question 12.
Draw a tangent at any point R on the circle of radius 3.4 cm and centre at P ?
Answer:
Given Radius = 3.4 cm


Steps of construction:

1. Draw a circle with centre “O” of radius 3.4 cm.
2. Take a point P on the circle Join OP.
3. Draw a perpendicular line TT’ to OP which passes through P.
4. TT’ is the required tangent.

Question 13.
Draw a circle of radius 4.5 cm. Take a point on the circle. Draw the tangent at that point using the alternate segment theorem.
Answer:
Radius of the circle = 4.5 cm


Steps of construction:

1. With O as centre, draw a circle of radius 4.5 cm.
2. Take a point L on the circle. Through L draw any chord LM.
3. Take a point M distinct from L and N on the circle, so that L, M, N are in anti-clockwise direction. Join LN and NM.
4. Through “L” draw tangent TT’such that ∠TLM = ∠MNL
5. TT’ is the required tangent.

Question 14.
Draw the two tangents from a point which is 10 cm away from the centre of a circle of radius 5 cm. Also, measure the lengths of the tangents.
Answer:
Radius = 5 cm; Distance = 10 cm


Steps of construction:

1. With O as centre, draw a circle of radius 5 cm.
2. Draw a line OP =10 cm.
3. Draw a perpendicular bisector of OP, which cuts OP at M.
4. With M as centre and MO as radius draw a circle which cuts previous circle at A and B.
5. Join AP and BP. AP and BP are the required tangents.

Verification: In the right ∆ OAP
PA² = OP² – OA²
= 10² – 5² = \(\sqrt { 100-25 }\) = \(\sqrt { 75 }\) = 8.7 cm
Lenght of the tangent is = 8.7 cm

Question 15.
Take a point which is 11 cm away from the centre of a circle of radius 4 cm and draw the two tangents to the circle from that point.
Answer:
Radius = 4 cm; Distance = 11 cm


Steps of construction:

1. With O as centre, draw a circle of radius 4 cm.
2. Draw a line OP = 11 cm.
3. Draw a perpendicular bisector of OP, which cuts OP at M.
4. With M as centre and MO as radius, draw a circle which cuts previous circle A and B.
5. Join AP and BP. AP and BP are the required tangents.

This the length of the tangents PA = PB = 10.2 cm
Verification: In the right angle triangle OAP
PA² = OP² – OA²
= 11² – 4² = 121 – 16 = 105
PA = \(\sqrt { 105 }\) = 10.2 cm
Length of the tangents = 10.2 cm

Question 16.
Draw the two tangents from a point which is 5 cm away from the centre of a circle of diameter 6 cm. Also, measure the lengths of the tangents.
Answer:
Radius = 3cm; Distance = 5cm.


Steps of construction:

1. With O as centre, draw a circle of radius 3 cm.
2. Draw a line OP = 5 cm.
3. Draw a perpendicular bisector of OP, which cuts OP at M.
4. With M as centre and MO as radius draw a circle which cuts previous circles at A and B.
5. Join AP and BP, AP and BP are the required tangents.

The length of the tangent PA = PB = 4 cm
Verification: In the right angle triangle OAP
PA² = OP² – OA²
= 5² – 3²
= 25 – 9
= 16 PA
= \(\sqrt { 16 }\) = 4 cm
Length of the tangent = 4 cm

Question 17.
Draw a tangent to the circle from the point P having radius 3.6 cm, and centre at O. Point P is at a distance 7.2 cm from the centre.
Answer:
Radius = 3.6; Distance = 7.2 cm.


Steps of construction:

1. With O as centre, draw a circle of radius 3.6 cm.
2. Draw a line OP = 7.2 cm.
3. Draw a perpendicular bisector of OP which cuts OP at M.
4. With M as centre and MO as radius draw a circle which cuts the previous circle at A and B.
5. Join AP and BP, AP and BP are the required tangents.

Length of the tangents PA = PB = 6.26 cm
Verification: In the right triangle ∆OAP
PA² = OP² – OA²
= 7.22 – 3.62 =(7.2 + 3.6) (7.2 – 3.6)
PA² = 10.8 × 3.6 = \(\sqrt { 38.88 }\)
PA = 6.2 cm
Length of the tangent = 6.2 cm

Students can download Maths Chapter 8 Statistics Ex 8.3 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 8 Statistics Ex 8.3

Question 1.
The monthly salary of 10 employees in a factory are given below:
Rs 5000, Rs 7000, Rs 5000, Rs 7000, Rs 8000, Rs 7000, Rs 7000, Rs 8000, Rs 7000, Rs 5000
Find the mean, median and mode.
Solution:

Mean = 6600
Median:
Arrange in ascending order we get.
5000, 5000, 5000, 7000, 7000, 7000, 7000, 7000, 8000, 8000
The number of values = 10
Median = Average of (\(\frac{10}{2})^{th}\) and (\(\frac{10}{2}+1)^{th}\)
= Average of 5^th value and 6^th value
= \(\frac{7000+7000}{2}\)
∴ Median = 7000
Mode: 7000 repeated 5 times
∴ Mode = 7000

Question 2.
Find the mode of the given data: 3.1, 3.2, 3.3, 2.1, 1.3, 3.3, 3.1
Solution:
3.1 occuring two times
3.3 occuring two times
∴ 3.1 and 3.3 are the mode (bimodal)

Question 3.
For the data 11, 15, 17, x + 1, 19, x – 2, 3 if the mean is 14, find the value of x. Also find the mode of the data.
Solution:
Arithmetic mean

∴ 2x + 64 = 14 × 7
2x = 98 – 64
2x = 34
x = \(\frac{34}{2}\)
= 17
The given numbers are 11, 15, 17, 18, 19, 15 and 3
15 occuring two times
∴ Mode = 15
The value of x = 17 and mode = 15

Question 4.
The demand of track suit of different sizes as obtained by a survey is given below:

Which size is demanded more?
Solution:
The highest frequency is 37
The corresponding value is the mode
∴ Mode = 40
Size 40 is demanded more.

Question 5.
Find the mode of the following data:

Solution:

The highest frequency is 46
20 – 30 is the modal class
Here l = 20, f = 46, f₁ = 38, f₂ = 34 and c = 10
Mode

= 20 + 4
= 24
∴ Mode = 24

Question 6.
Find the mode of the following distribution

Solution:
In the given table the class intervals are in inclusive form; convert them into exclusive form.

The highest frequency is 14
Modal class is 54.5 – 64.5
Here l = 54.5, f = 14, f₁ = 10, f₂ = 8 and c = 10
mode

= 58.5
∴ Mode = 58.5

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.1

Question 1.
Find the area of the triangle formed by the points
(i) (1,-1), (-4, 6) and (-3, -5)
(ii) (-10, -4), (-8, -1) and (-3, -5)
Solution:
(i) Let the vertices A (1, -1), B (-4, 6) and C (-3, -5)


= \(\frac { 1 }{ 2 } \) [(6 + 20 + 3) – (4 – 18 – 5)] = \(\frac { 1 }{ 2 } \) [29 – (-19)] = \(\frac { 1 }{ 2 } \) [29 + 19]
= \(\frac { 1 }{ 2 } \) × 48 = 24 sq. units.
Area of ∆ABC = 24 sq. units

(ii) Let the vertices be A(-10, -4), B(-8 -1) and C(-3, -5)

Area of ∆ABC = \(\frac { 1 }{ 2 } \)[(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
= \(\frac { 1 }{ 2 } \) [(50 + 3 + 32) – (12 + 40 + 10)]

= \(\frac { 1 }{ 2 } \) [85 – (62)] = \(\frac { 1 }{ 2 } \) [23] = 11.5
Area of ∆ACB = 11.5 sq.units

Question 2.
Determine whether the sets of points are collinear?
(i) (-\(\frac { 1 }{ 2 } \),3)
(ii) (a,b + c), (b,c + a) and (c,a + b)
Solution:
(i) Let the points be A (-\(\frac { 1 }{ 2 } \),3), B (-5, 6) and C(-8, 8)
Area of ∆ABC = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₃ + x₁y₃)]
= \(\frac { 1 }{ 2 } \) [(- 3 – 40 – 24) – (-15 – 48 – 4)]

= \(\frac { 1 }{ 2 } \) [-67 + 67] = \(\frac { 1 }{ 2 } \) × 0 = 0
Area of a ∆ is 0.
∴ The three points are collinear.

(ii) Let the points be A (a, b + c), B (b, c + a) and C (c, a + b)
Area of the triangle = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]

Since the area of a triangle is 0.
∴ The given points are collinear.

Question 3.
Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p’

Solution:
(i) Let the vertices be A (0,0) B (p, 8), c (6, 2)
Area of a triangle = 20 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 20

\(\frac { 1 }{ 2 } \) [(0 + 2p + 0) – (0 + 48 + 0)] = 20
\(\frac { 1 }{ 2 } \) [2p – 48] = 20
2p – 48 = 40 ⇒ 2p = 40 + 48
p = \(\frac { 88 }{ 2 } \) = 44
The value of p = 44

(ii) Let the vertices be A (p, p), B (5, 6) and C (5, -2)
Area of a triangle = 32 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 32

\(\frac { 1 }{ 2 } \) [6p – 10 + 5p) – (5p + 30 – 2p)] = 32
\(\frac { 1 }{ 2 } \) [11 p – 10 – 3p – 30] = 32
11p – 10 – 3p – 30 = 64
8p – 40 = 64
8p = 64 + 40 ⇒ 8p = 104
p = \(\frac { 104 }{ 8 } \) = 13
The value of p = 13

Question 4.
In each of the following, find the value of ‘a’ for which the given points are collinear.
(i) (2,3), (4, a) and (6, -3)
(ii) (a, 2 – 2a), (-a + 1, 2a) and (-4 -a, 6 – 2a).
Solution:
(i) Let the points be A (2, 3), B(4, a) and C(6, -3).
Since the given points are collinear.
Area of a triangle = 0
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 0)

\(\frac { 1 }{ 2 } \) [(2a – 12 + 18) – (12 + 6a – 6)] = 0
2a + 6 – (6 + 6a) = 0
2a + 6 – 6 – 6a = 0
-4a = 0 ⇒ a = \(\frac { 0 }{ 4 } \) = 0
The value of a = 0

(ii) Let the points be A (a, 2 – 2a), B (-a + 1, 2a) C (-4 -a, 6 – 2a).
Since the given points are collinear.
Area of a ∆ = 0

6a² – 2a – 2 – (-2a² – 6a + 2) = 0
6a² – 2a – 2 + 2a² + 6a – 2 = 0
8a² + 4a – 4 = 0 (Divided by 4)
2a² + a – 1 = 0
2a² + 2a – a – 1 = 0
2a (a + 1) – 1 (a + 1) = 0

(a + 1) (2a – 1) = 0
a + 1 = 0 (or) 2a – 1 = 0
a = -1 (or) 2a = 1 ⇒ a = \(\frac { 1 }{ 2 } \)
The value of a = -1 (or) \(\frac { 1 }{ 2 } \)

Question 5.
Find the area of the quadrilateral whose vertices are at
(i) (-9, -2), (-8, -4), (2, 2) and (1, -3)
(ii) (-9, 0), (-8,6), (-1, -2) and (-6, -3)
Solution:
(i) Let the vertices A (-9, -2), B(-8, -4), C(2, 2) and D(1, -3).
Plot the vertices in a graph.

[Note: Consider the points in counter clock wise order]

Area of the Quadrilateral ABDC = \(\frac { 1 }{ 2 } \) [36 + 24 + 2 – 4 – (16 – 4 – 6 – 18)]
= \(\frac { 1 }{ 2 } \) [58 – (-12)] – \(\frac { 1 }{ 2 } \)[58 + 12]
= \(\frac { 1 }{ 2 } \) × 70 = 35 sq. units 2
Area of the Quadrilateral = 35 sq. units

(ii) Let the vertices A(-9, 0), B(-8, 6), C(-1, -2) and D(-6, -3)
Plot the vertices in a graph and take them in counter – clock wise order.
Area of the Quadrilateral DCB

= \(\frac { 1 }{ 2 } \) [33 + 35] = \(\frac { 1 }{ 2 } \) × 68 = 34 sq. units
Area of the Quadrilateral = 34 sq. units

Question 6.
Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are (-4, -2), (-3, k), (3, -2) and (2, 3)
Solution:
Let the vertices A (-A, -2), B (-3, k), C (3, -2) and D (2, 3)
Area of the Quadrilateral = 28 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)] = 28

-7k + 21 = 56
-7k = 56 – 21
-7k = 35 ⇒ 7k = – 35
k = – \(\frac { 35 }{ 7 } \) = -5
The value of k = -5

Question 7.
If the points A(-3, 9), B(a, b) and C(4, -5) are collinear and if a + b = 1, then find a and b.
Solution:
Since the three points are collinear
Area of a ∆ = 0

-3b – 5a + 36 – 9a – 4b – 15 = 0
-7b – 14a + 21 = 0
(÷ by 7) – b – 2a + 3 = 0
2a + b – 3 = 0

Substitute the value of a = 2 in (2) ⇒ 2 + b = 1
b = 1 – 2 = -1
The value of a = 2 and b = -1

Question 8.
Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC . Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.
Solution:
Let the vertices of the ∆ABC be A(x₁,y₁), B(x₂,y₂), C(x₃,y₃)

Question 9.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.

Solution:

= \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)]
= \(\frac { 1 }{ 2 } \) [(16 + 80 + 36 + 80) – (-64 – 24 – 100 – 24)]
= \(\frac { 1 }{ 2 } \) [212 – (-212)]

= \(\frac { 1 }{ 2 } \) [212 + 212] = \(\frac { 1 }{ 2 } \) [424] = 212 sq. units

= \(\frac { 1 }{ 2 } \) [90 – (-90)]

= \(\frac { 1 }{ 2 } \) [90 + 90]
= \(\frac { 1 }{ 2 } \) × 180 = 90 sq. units
Area of the patio = Area of the Quadrilateral ABCD – Area of the Quadrilateral EFGH
= (212 – 90) sq. units
Area of the patio = 122 sq. units

Question 10.
A triangular shaped glass with vertices at A(-5, -4), B(l, 6) and C(7, -4) has to be painted. If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.
Solution:
Given the vertices of the triangular glass is A (-5, -4), B (1, 6), and C (7, -4)


= \(\frac { 1 }{ 2 } \) [(20 + 42 – 4) – (-28 – 4 – 30)]
= \(\frac { 1 }{ 2 } \) [58 – (-62)]
= \(\frac { 1 }{ 2 } \) [58 + 62]
= \(\frac { 1 }{ 2 } \) × 120 = 60 sq. feet
Number of cans to paint 6 square feet = 1
∴ Number of cans = \(\frac { 60 }{ 6 } \) = 10 ⇒ Number of cans = 10

Question 11.
In the figure, find the area of
(i) triangle AGF
(ii) triangle FED
(iii) quadrilateral BCEG.

Solution:
Area of a triangle = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
(i) Area of ∆AGF = \(\frac { 1 }{ 2 } \) [(-2.5 – 13.5 – 6) – (-13.5 – 1 – 15)]
= \(\frac { 1 }{ 2 } \) [-22 – (-29.5)]

= \(\frac { 1 }{ 2 } \) [-22 + 29.5]
= \(\frac { 1 }{ 2 } \) × 7.5 = 3.75 sq.units

(ii) Area of ∆FED = \(\frac { 1 }{ 2 } \) [(-2 + 4.5 + 3) – (4.5 + 1 – 6)]
= \(\frac { 1 }{ 2 } \) [5.5 – (-0.5)]

= \(\frac { 1 }{ 2 } \) [5.5 + 0.5] = \(\frac { 1 }{ 2 } \) × 6 = 3 sq.units

(iii)

= \(\frac { 1 }{ 2 } \) [(4 + 2 + 0.75 + 9) – (-4 -1.5 – 4.5 -2)]
= \(\frac { 1 }{ 2 } \) [15.75 + 12]
= \(\frac { 1 }{ 2 } \) [27.75] = 13.875
= 13.88 sq. units

Students can download Maths Chapter 6 Information Processing Ex 6.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.3

Question 1.
How many Triangles are there in each of the following figures?

Solution:
(i) 12 triangles
(ii) 16 triangles
(iii) 32 triangles
(iv) 35 triangles

Question 2.
Find the number of dots in the tenth figure of the following patterns.

Solution:
(i) 55
(ii) 100

Question 3.

(i) Draw the next pattern.
(ii) Prepare a table for the number of dots used for each pattern.
(iii) Explain the pattern.
(iv) Find the number of dots in the 25th pattern.
Solution:

(iv) 350

Question 4.

Solution:
(i) 20 squares
(ii) 10 squares

Question 5.
How many circles are there in the following figure?

Solution:
7 circles

Question 6.
Find the minimum number of straight lines used in forming the following figures.

Solution:
(i) 10
(ii) 12