Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

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

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

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

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

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

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

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

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

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

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

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.

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

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.

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

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.

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

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

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

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

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 2 Measurements Ex 2.1

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

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

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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?
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 6 Information Processing Ex 6.3 1
Solution:
(i) 12 triangles
(ii) 16 triangles
(iii) 32 triangles
(iv) 35 triangles

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

Question 2.
Find the number of dots in the tenth figure of the following patterns.
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 6 Information Processing Ex 6.3 2
Solution:
(i) 55
(ii) 100

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

Question 3.
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 6 Information Processing Ex 6.3 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:
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 6 Information Processing Ex 6.3 4
(iv) 350

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

Question 4.
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 6 Information Processing Ex 6.3 5
Solution:
(i) 20 squares
(ii) 10 squares

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

Question 5.
How many circles are there in the following figure?
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 6 Information Processing Ex 6.3 6
Solution:
7 circles

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

Question 6.
Find the minimum number of straight lines used in forming the following figures.
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 6 Information Processing Ex 6.3 7
Solution:
(i) 10
(ii) 12

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.1

Students can download Maths Chapter 3 Ratio and Proportion Ex 3.1 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 3 Ratio and Proportion Ex 3.1

Question 1.
Fill in the blanks.
(i) Ratio of ₹ 3 to ₹ 5 = ____
(ii) Ratio of 3 m to 200 cm = ______
(iii) Ratio of 5 km 400 m to 6 km = ____
(iv) Ratio of 75 paise to ₹ 2 = ____
Solution:
(i) 3 : 5
(ii) 3 : 2
Hint: 3m = 300 cm
(iii) 9 : 10
Hint: 5km 400 m = 5400m and 6 km = 6000 m
(iv) 3 : 8
Hint: ₹ 2 = 200 paise

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.1

Question 2.
Say whether the following statements are True or False.
(i) The ratio of 130 cm to 1 m is 13 : 10
(ii) One of the terms in a ratio cannot be 1
Solution:
(i) True
Hint: 1m = 100 cm
(ii) False

Question 3.
Find the simplified form of the following ratios.
(i) 15 : 20
(ii) 32 : 24
(iii) 7 : 15
(iv) 12 : 27
(v) 75 : 100
Solution:
(i) 15 : 20
= \(\frac{15}{20}\)
= \(\frac{3}{4}\)
= 3 : 4

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.1

(ii) 32 : 24
= \(\frac{32}{24}\)
= \(\frac{4}{3}\)
= 4 : 3

(iii) 7 : 15

(iv) 12 : 27
= \(\frac{12}{27}\)
= \(\frac{4}{9}\)
= 4 : 9

(v) 75 : 100
= \(\frac{75}{100}\)
= \(\frac{3}{4}\)
= 3 : 4

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.1

Question 4.
Akilan walks 10 km in an hour while Selvi walks 6 km in an hour. Find the simplest ratio of the distance covered by Akilan to that of Selvi.
Solution:
Ratio of the distance covered by Akilan to that of Selvi = 10 km : 6 km
= \(\frac{10}{6}\)
= \(\frac{5}{3}\)
= 5 : 3

Question 5.
The cost of parking a bicycle is Rs 5 and the cost of parking a scooter is Rs 15. Find the simplest ratio of the parking cost of a bicycle to that of a scooter.
Solution:
Ratio of the parking cost of a bicycle to that of a scooter = Rs 5 : Rs 15
= \(\frac{5}{15}\)
= \(\frac{1}{3}\)
= 1 : 3

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.1

Question 6.
Out of 50 students in a class, 30 are boys. Find the ratio of
(i) number of boys to the number of girls.
(ii) the number of girls to the total number of students.
(iii) the number of boys to the total number of students.
Solution:
Total no of students = 50
No of boys = 30
No of girls = 50 – 30 = 20
(i) Ratio of boys to girls = 30 : 20
= \(\frac{30}{20}\)
= \(\frac{3}{2}\)
= 3 : 2

(ii) Ratio of girls to the total number of students = 20 : 50
= \(\frac{20}{50}\)
= \(\frac{2}{5}\)
= 2 : 5

(iii) Ratio of boys to the total no of students = 30 : 50
= \(\frac{30}{50}\)
= \(\frac{3}{5}\)
= 3 : 5

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.1

Objective Type Questions

Question 7.
The ratio of ₹ 1 to 20 paise is _____
(a) 1 : 5
(b) 1 : 2
(c) 2 : 1
(d) 5 : 1
Solution:
(d) 5 : 1
Hint: ₹ 1 = 100 paise

Question 8.
The ratio of 1 m to 50 cm is
(a) 1 : 50
(b) 50 : 1
(c) 2 : 1
(d) 1 : 2
Solution:
(c) 2 : 1

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.1

Question 9.
The length and breadth of a window are 1 m and 70 cm respectively. The ratio of the length to the breadth is
a) 1 : 7
(b) 7 : 1
(c) 7 : 10
(d) 10 : 7
Solution:
(d) 10 : 7

Question 10.
The ratio of the number of sides of a triangle to the number of sides of a rectangle is
(a) 4 : 3
(b) 3 : 4
(c) 3 : 5
(d) 3 : 2
Solution:
(b) 3 : 4

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.1

Question 11.
If Azhagan is 50 years old and his son is 10 years old then the simplest ratio between the age of Azhagan to his son is
(a) 10 : 50
(b) 50 : 10
(c) 5 : 1
(d) 1 : 5
Solution:
(c) 5 : 1

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.3

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.3

Question 1.
Fill in the blanks.
(i) 3 : 5 :: …….. : 20
(ii) ……… : 24 :: 3 : 8
(iii) 5 : …….. :: 10 : 8 :: 15 : ………
(iv) 12 : ……… :: ………. :: 4 = 8 : 16
Solution:
(i) 12
Hint:
5x = 3 × 20 ⇒ x = 12
(ii) 9
8x = 24 × 3 ⇒ x = 9
(iii) 4, 12
Hint:
10x = 8 × 5 = 40 ⇒ x = 4
10y = 8 × 15 = 120 ⇒ y = 12
(iv) 24, 2
Hint:
16y = 8 × 4 ⇒ y = 2
12 × 4 = 2x ⇒ x = 24

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.3

Question 2.
Say True or False.
(i) 2 : 7 : : 14 : 4
(ii) 7 Persons are to 49 Persons as 11 kg is to 88 kg
(iii) 10 books are to 15 books as 3 books are to 15 books.
Solution:
(i) False
Hint:
7 × 14 ≠ 4 × 2
98 ≠ 8
(ii) False
Hint:
7 : 49 :: 11 : 48
49 × 11 ≠ 7 × 48
539 ≠ 336
(iii) False
Hint:
10 : 15 :: 3 : 5
\(\frac{10}{15}=\frac{5 \times 2}{5 \times 3}=\frac{2}{3} \Rightarrow \frac{3}{5}=\frac{3}{5}\)

Question 3.
Using the numbers 3, 9, 4, 12 write two ratios that are in proportion.
Solution:
(i) 3, 9, 4, 12
Here product of extremes = 3 × 12 = 36
Product of means = 9 × 4 = 36
3 : 9 : : 4 : 12

(ii) Also if we take 9, 3, 12, 4
Product of extremes = 9 × 4 = 36
Product of means = 3 × 12 = 36
9 : 3 : : 12 : 4

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.3

Question 4.
Find whether 12, 24,18, 36 are in order that can be expressed as two ratios that are in proportion.
Solution:
Yes, 12 : 24 : : 18 : 36
Because product of extremes 12 × 36 = 432
Product of means = 24 × 18 = 432
12 : 24 :: 18 : 36.

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.3

Question 5.
Write the mean and extreme terms in the following ratios and check whether they are in proportion.
(i) 78 liters is to 130 liters and 12 bottles are to 20 bottles
(ii) 400 gm is to 50 gm and 25 rupees is to 625 rupees
Solution:
(i) 78 : 130 :: 12 : 20
Extreme terms are 78 and 20.
Mean terms are 130 and 12.
Product of Extremes = 78 × 20 = 1560
Product of Means = 130 × 12 = 1560
Product of Extremes = Product of means
It is in proportion.

(ii) 400 : 50 : : 25 : 625
Product of extremes = 400 × 625 = 2,50,000
Product of means = 50 × 25 = 1250
Here product of extremes ≠ product of means
400 : 50 and 25 : 625 are not in proportion.

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.3

Question 6.
America’s famous Golden Gate bridge is 6480 ft long with 756 ft tall towers. A model of this bridge exhibited in a fair is 60 ft long with 7 ft tall towers. Is the model in proportion to the original bridge?
Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.3 1
Solution:
6480 : 756, 60 : 7
Product of the means = 756 × 60 = 45360
Product of the extremes = 6480 × 7 = 45360
ad = bc
∴ They are in proportion

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.3

Objective Type Questions

Question 7.
Which of the following ratios are in proportion?
(a) 3 : 5, 6 : 11
(b) 2 : 3, 9 : 6
(c) 2 : 5, 10 : 25
(d) 3 : 1, 1 : 3
Solution:
(c) 2 : 5, 10 : 25
Hint:
2 × 25 = 5 × 10
⇒ 50 = 50

Question 8.
If the ratios formed using the numbers 2, 5, x, 20 in the same order are in proportion, then ‘x’ is
(a) 50
(b) 4
(c) 10
(d) 8
Solution:
(d) 8
5x = 2 × 20 ⇒ x = 8

Samacheer Kalvi 6th Maths Guide Term 1 Chapter 3 Ratio and Proportion Ex 3.3

Question 9.
If 7 : 5 is in proportion to x : 25, then ‘x’ is
(a) 27
(b) 49
(c) 35
(d) 14
Solution:
(c) 35
Hint:
5x = 7 × 25 ⇒ x = 35

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Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Students can download Maths Chapter 1 Fractions 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 3 Chapter 1 Fractions Ex 1.1

Question 1.
Fill in the blanks:
(i) 7\(\frac{3}{4}\) + 6\(\frac{1}{2}\) ………..
(ii) The sum of a whole number and a proper fraction is called ……….
(iii) 5\(\frac{1}{3}\) – 3\(\frac{1}{2}\) ………..
(iv) 8 ÷ \(\frac{1}{2}\) ………..
(v) The number which has its own reciprocal is ……….
Solution:
Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 1

(ii) Mixed Fraction

(iii) 5\(\frac{1}{3}\) – 3\(\frac{1}{2}\)
= \(\frac{16}{3}\) – \(\frac{7}{2}\) = \(\frac{32-21}{6}\) = \(\frac{11}{6}\)
= 1\(\frac{5}{6}\)

(iv) 8 ÷ \(\frac{1}{2}\)
= 8 × \(\frac{2}{1}\)
= 16

(v) 1

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

The first step would be to write the mix number 223 as an improper fraction.

Question 2.
Say True or False
(i) 3\(\frac{1}{2}\) can be written as 3 + \(\frac{1}{2}\).
(ii) The sum of any two proper fractions is always an improper fraction.
(iii) The mixed fraction of \(\frac{13}{4}\) is 3\(\frac{1}{4}\).
(iv) The reciprocal of an improper fraction is always a proper fraction.
(v) 3\(\frac{1}{4}\) × 3\(\frac{1}{4}\) = 9\(\frac{1}{16}\)
Solution:
(i) True
(ii) False
(iii) True
(iv) True
(v) False

Question 3.
Answer the following:
(i) Find the sum of \(\frac{1}{7}\) and \(\frac{3}{9}\).
(ii) What is the total of 3\(\frac{1}{3}\) and 4\(\frac{1}{6}\)?
(iii) Simplify: 1\(\frac{3}{5}\)+5\(\frac{4}{7}\).
(iv) Find the difference between \(\frac{8}{9}\) and \(\frac{2}{7}\)
(v) Subtract: 1\(\frac{3}{5}\) from 2\(\frac{1}{3}\).
(vi) Simplify: 7\(\frac{2}{7}\) – 3\(\frac{4}{21}\)
Solution:
(i) \(\frac{1}{7}\) + \(\frac{3}{9}\)
= \(\frac{9+21}{63}\) = \(\frac{30}{63}\) = \(\frac{10}{21}\)

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 2
Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 3

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Question 4.
Convert mixed fractions into improper fractions and vice versa:
(i) 3\(\frac{7}{18}\)
(ii) \(\frac{99}{7}\)
(iii) \(\frac{47}{6}\)
(iv) 12\(\frac{1}{9}\)
Solution:
(i) 3\(\frac{7}{18}\) = \(\frac{61}{18}\)
(ii) \(\frac{99}{7}\) = 14\(\frac{1}{7}\)
(iii) \(\frac{47}{6}\) = 7\(\frac{5}{6}\)
(iv) 12\(\frac{1}{9}\) = \(\frac{109}{9}\)

Question 5.
Multiply the following:
(i) \(\frac{2}{3}\) × 6
(ii) 8\(\frac{1}{3}\) × 5
(iii) \(\frac{3}{8}\) × \(\frac{4}{5}\)
(iv) 3\(\frac{5}{7}\) × 1\(\frac{1}{13}\)
Solution:
(i) \(\frac{2}{3}\) × 6 = 4

(ii) 8\(\frac{1}{3}\) × 5
= \(\frac{25}{3}\) × 5
= \(\frac{125}{3}\)
= 41\(\frac{2}{3}\)

(iii) \(\frac{3}{8}\) × \(\frac{4}{5}\)
= \(\frac{12}{40}\) = \(\frac{3}{10}\)

(iv) 3\(\frac{5}{7}\) × 1\(\frac{1}{13}\)
= \(\frac{26}{7}\) × \(\frac{14}{13}\)
= \(\frac{4}{1}\)
= 4

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Question 6.
Divide the following
(i) \(\frac{3}{7}\) ÷ 4
(ii) \(\frac{4}{3}\) ÷ \(\frac{5}{9}\)
(iii) 4\(\frac{1}{5}\) ÷ 3\(\frac{3}{4}\)
(iv) 9\(\frac{2}{3}\) ÷ 1\(\frac{2}{3}\)
Solution:
(i) \(\frac{3}{7}\) ÷ 4
= \(\frac{3}{7}\) × \(\frac{1}{4}\) = \(\frac{3}{28}\)

(ii) \(\frac{4}{3}\) ÷ \(\frac{5}{9}\)
= \(\frac{4}{3}\) × \(\frac{9}{5}\)
= \(\frac{12}{5}\)
= 2\(\frac{2}{5}\)

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 4

Question 7.
Gowri purchased 3\(\frac{1}{2}\) kg of tomatoes, \(\frac{3}{4}\) kg of brinjal and 1\(\frac{1}{4}\) kg of onion. What is the total weight of the vegetables she bought?
Solution:
Total weight of vegetables bought
Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 5

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Question 8.
An oil tin contains 3\(\frac{3}{4}\) litres of oil of which 2\(\frac{1}{2}\) litres of oil is used. How much oil is left over?
Solution:
Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 6

Question 9.
Nilavan can walk 4\(\frac{1}{2}\) km in an hour. How much distance will he cover in 3\(\frac{1}{2}\) hours?
Solution:
Distance walked in an hour = 4\(\frac{1}{2}\) km
Distance covered in 3\(\frac{1}{2}\)
Hours
Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 7

Question 10.
Ravi bought a curtain of length 15\(\frac{3}{4}\) m. If he cut the curtain into small pieces each of length 2\(\frac{1}{4}\) m, then how many small curtains will he get?
Solution:
Total length = 15\(\frac{3}{4}\)
Length of the small pieces = 2\(\frac{1}{4}\)
Small curtains obtained
Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1 8
= 7

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Objective Type Questions

Question 11.
Which of the following statement is incorrect?
(a) \(\frac{1}{2}\) > \(\frac{1}{3}\)
(b) \(\frac{7}{8}\) > \(\frac{6}{7}\)
(c) \(\frac{8}{9}\) < \(\frac{9}{10}\)
(d) \(\frac{10}{11}\) > \(\frac{9}{10}\)
Solution:
(d) \(\frac{10}{11}\) > \(\frac{9}{10}\)

Question 12.
The difference between \(\frac{3}{7}\) and \(\frac{2}{9}\) is
(a) \(\frac{13}{63}\)
(b) \(\frac{1}{9}\)
(c) \(\frac{1}{7}\)
(d) \(\frac{9}{16}\)
Solution:
(a) \(\frac{13}{63}\)

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Question 13.
The reciprocal of \(\frac{53}{17}\) is
(a) \(\frac{53}{17}\)
(b) 5\(\frac{3}{17}\)
(c) \(\frac{17}{53}\)
(d) 3\(\frac{5}{17}\)
Solution:
(c) \(\frac{17}{53}\)

Question 14.
If \(\frac{6}{7}=\frac{\mathbf{A}}{49}\), then the value of A is
(a) 42
(b) 36
(c) 25
(d) 48
Solution:
(a) 42
Hint: \(\frac{6 \times 49}{7}=42\)

Samacheer Kalvi 6th Maths Guide Term 3 Chapter 1 Fractions Ex 1.1

Question 15.
Pugazh has been given four choices for his pocket money by his father. Which of the choices should he take in order to get the maximum money?
(a) \(\frac{2}{3}\) of Rs 150
(b) \(\frac{3}{5}\) of Rs 150
(c) \(\frac{4}{5}\) of Rs 150
(d) \(\frac{1}{5}\) of Rs 150
Solution:
(c) \(\frac{4}{5}\) of Rs 150

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

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 2 Chapter 1 Numbers Ex 1.2

Question 1.
Fill in the blanks
(i) The HCF of 45 and 75 is …….
(ii) The HCF of two successive even numbers is …….
(iii) If the LCM of 3 and 9 is 9, then their HCF is ………
(iv) The LCM of 26, 39 and 52 is ……..
(v) The least number that should be added to 57 so that the sum is exactly divisible by 2, 3, 4 and 5 is ………
Solution:
(i) 15
(ii) 2
(iii) 3
(iv) 156
(v) 3

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

The Least common multiple (LCM) of 15 and 20 is 60.

Question 2.
Say True or False
(i) The numbers 57 and 69 are co-primes.
(ii) The HCF of 17 and 18 is 1.
(iii) The LCM of two successive numbers is the product of the numbers.
(iv) The LCM of two co-primes is the sum of the numbers.
(v) The HCF of two numbers is always a factor of their LCM.
Solution:
(i) False
(ii) True
(iii) True
(iv) False
(v) True

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

So, now the prime factorization of 84 with the upside-down division method is 2 × 2 × 3 × 7.

Question 3.
Find the HCF of each set of numbers using the prime factorisation method.
(i) 18, 24
(ii) 51, 85
(iii) 61, 76
(iv) 84, 120
(v) 27, 45, 81
(vi) 45, 55, 95
Solution:
(i) 18, 24
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 1
HCF = 2 × 3 = 6

(ii) 51, 85
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 2
HCF = 17

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

(iii) 61, 76
61 = 1 × 61
For 61, 76, the common factor is 1
HCF = 1

(iv) 84, 120
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 3
HCF = 2 × 2 × 3 = 12

(v) 27, 45, 81
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 4
HCF = 3 × 3 = 9

(vi) 45, 55, 95
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 5
Factors of 55 = 5 × 11
Factors of 45 = 3 × 3 × 5
Factors of 95 = 5 × 19
HCF = 5

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Question 4.
Find the LCM of each set of numbers using the prime factorisation method.
(i) 6, 9
(ii) 8, 12
(iii) 10, 15
(iv) 14, 42
(v) 30, 40, 60
(vi) 15, 25, 75
Solution:
(i) 6, 9
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 6
Factors of 6 = 2 × 3
Factors of 9 = 3 × 3
LCM of 6 and 9 = 3 × 2 × 3 = 18

(ii) 8, 12
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 7

(iii) 10, 15
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 8
Factors of 10 = 2 × 5
Factors of 15 = 3 × 5
LCM of 10 and 15 = 5 × 2 × 3 = 30

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

(iv) 14, 42
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 9
Factors of 14 = 2 × 7
Factors of 42 = 2 × 3 × 7
LCM of 14 and 42 = 7 × 2 × 3 = 42

(v) 30, 40, 60
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 10
Factors of 30 = 2 × 3 × 5
Factors of 40 = 2 × 2 × 2 × 5
Factors of 60 = 2 × 2 × 3 × 5
LCM of 30, 40, 60 = 2 × 5 × 3 × 2 × 2 = 120

(vi) 15, 25, 75
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 11
Factors of 15 = 3 × 5
Factors of 25 =5 × 5
Factors of 75 = 3 × 5 × 5
LCM of 15, 25, 75 = 5 × 3 × 5 = 75

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Question 5.
Find the HCF and the LCM of the numbers 154, 198, 286
Solution:
HCF
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 12
Factors of 154 = 2 × 7 × 11
Factors of 198 = 2 × 3 × 3 × 11
Factors of 286 = 2 × 13 × 11
HCF of 154, 198, 286 = 11 × 2 = 22
LCM
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 13
LCM = 2 × 11 × 7 × 9 × 13
= 22 × 63 × 13
= 18018

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Question 6.
What is the greatest possible volume of a vessel that can be used to measure exactly the volume of milk in cans (in full capacity) of 80 litres, 100 litres and 120 litres?
Solution:
This is an HCF related problem. So, we need to find the HCF of 80, 100, 120
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 14
80 = 2 × 2 × 2 × 2 × 5
100 = 2 × 2 × 5 × 5
120 = 2 × 2 × 2 × 3 × 5
HCF of 80, 100 and 120 = 2 × 2 × 5 = 20
Volume of the vessel = 20 litres

Question 7.
The traffic lights at three different road junctions change after every 40 seconds, 60 seconds, and 72 seconds respectively. If they changed simultaneously together at 8 a.m at the junctions, at what time will they simultaneously change together again?
Solution:
This is an LCM related problem. So, we need to find the LCM of 40, 60, 72
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 15
LCM of 40, 60 and 72
= 2 × 2 × 3 × 5 × 2 × 1 × 1 × 3
= 360 seconds
= 6 minutes
The traffic lights will change simultane¬ously again at 8 : 06 am.

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Question 8.
The LCM of two numbers is 210 and their HCF is 14. How many such pairs are possible.
Solution:
Product of two numbers = LCM × HCF
x × y = 210 × 14
x × y = 2940
(12, 245), (20, 147)
Two pairs are possible.
Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2 16

Question 9.
The LCM of two numbers is 6 times their HCF. If the HCF is 12 and one of the numbers is 36, then find the other number.
Solution:
36x = 6 × 12 × 12
⇒ \(x=\frac{6 \times 12 \times 12}{36}=24\)

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Objective Type Questions

Question 10.
Which of the following pairs is co-prime?
(a) 51, 63
(b) 52, 91
(c) 71, 81
(d) 81, 99
Solution:
(c) 71, 81

Question 11.
The greatest four-digit number which is exactly divisible by 8, 9, and 12 is
(a) 9999
(b) 9996
(c) 9696
(d) 9936
Solution:
(d) 9936

Samacheer Kalvi 6th Maths Guide Term 2 Chapter 1 Numbers Ex 1.2

Question 12.
The HCF of two numbers is 2 and their LCM is 154. If the difference between the numbers is 8, then the sum is
(a) 26
(b) 36
(c) 46
(d) 56
Solution:
(b) 36

Question 13.
Which of the following cannot be the HCF of two numbers whose LCM is 120?
(a) 60
(b) 40
(c) 80
(d) 30
Solution:
(c) 80