Category: Class 6

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 2 Integers Ex 2.1

Question 1.
Fill in the blanks.
(i) The potable water available at 100m below the ground level is denoted as ______ m.
Hint: Below ground level – negative; ground level – 0; above ground level – positive.
(ii) A swimmer dives to a depth of 7 feet from the ground into the swimming pool. The integer that represents this, is ______ feet.
Hint: Below ground level – negative numbers.
(iii) -46 is to the _____ of -35 on the number line.
Hint: -46 <-35
(iv) There are _____ integers from -5 to +5 (both inclusive).

(v) ______ is an integer which is neither positive nor negative.

Solutions:
(i) -100
(ii) -7
(iii) Left
(iv) 11
(v) 0

Question 2.
Say True or False.
(i) Each of the integers -18, 6, -12, 0 is greater than -20.
(ii) -1 is to the right of 0.
(iii) -10 and 10 are at equal distance from 1.
(iv) All negative integers are greater than zero.
(v) All whole numbers are integers.
Solution:
(i) True
(ii) False
(iii) False
(iv) False
(v) True

Question 3.
Mark the numbers 4, -3, 6, -1 and -5 on the number line.
Solution:

Question 4.
On the number line, which number is
i) 4 units to the right of-7?
ii) 5 units to the left of 3?
Solution:

(i) – 3 is 4 units to the right of -7
(ii) – 2 is 5 units to the left of 3

Question 5.
Find the opposite of the following numbers.
(i) 44
(ii) -19
(iii) 0
(iv) -312
(v) 789
Solution:
(i) Opposite of 44 is – 44
(ii) Opposite of-19 is + 19 or 19
(iii) Opposite of 0 is 0
(iv) Opposite of-312 is + 312 or 312
(v) Opposite of 789 is – 789.

Question 6.
If 15 km east of a place is denoted as + 15 km, what is the integer that represents 15 km west of it?
Solution:
Opposite of east is west.
∴ If 15 km east is + 15 km, then 15 km west is – 15 km.

Question 7.
From the following number lines, identify the correct and the wrong representations with reason?

Solution:
(i) This representation is wrong. Because the numbers are not continuously marked as -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6,
(ii) This is the correct representation of Integers
(iii) The representation is wrong. – 2 is misplaced in the number line.
(iv) This representation is correct as the numbers are marked at equal intervals in the correct order.
(v) Here negative integers are marked wrongly to the left of 0

Question 8.
Write all the integers between the given numbers.
(a) 7 and 10
(b) -5 and 4
(c) -3 and 3
(d) -5 and 0
Solution:

Question 9.
Put the appropriate signs as <, > or = in the box?

Solution:
(i) We know that -7 is to the left of 8.
(ii) -8 is to the left of -7,
(iii) -1000 is to the left of -999,
(iv) -111 and-111 coincides in the number line.
(v) 0 is to the right of -200,

Question 10.
Arrange the following integers in ascending order.
i) -11, 12, -13, 14, -15, 16, -17, 18, -19, -20
ii) -28, 6, -5, -40, 8, 0, 12, -1, 4, 22
iii) -100, 10, -1000, 100,0, -1, 1000, 1, -10
Solution:
(i) -11, 12, -13, 14, -15, 16, -17, 18, -19, -20

• First separating the positive integers 12, 14, 16, 18 and the negative integers -11,-13,-15,-17,-19,-20.
• Then arranging the positive integers in ascending order we get 12, 14, 16, 18 and negative integers in ascending order as -20, -19, -17, -15, -13, -11 4
• Now the ascending order : -20, -19, -17, -15, -13, -11, 12, 14, 16, 18.

(ii) -28, 6, -5, -40, 8, 0, 12, -1, 4, 22

• Positive integers are 6, 8, 12, 4, 22 Negative integers are -28, -5, -40, -1
• Arranging the positive integers in ascending order we get 4, 6, 8, 12, 22 and the negative integers in ascending order -40, -28, -5, -1
• The number 0 is neither positive nor negative and stand in the middle.
• In ascending order : -40, -28, -5, -1, 0, 4, 6, 8,12, 22

(iii) -100, 10, -1000, 100, 0, -1, 1000, 1, -10

• Separating positive integers 10, 100, 1000, 1 and negative integers -100, -1000, -1, -10.
• Now the positive integers in ascending order 1,10,100,1000 and the negative integers in ascending order. -1000, -100, -10, -1
• Also ‘0’ stand in the middle as its is neither positive nor negative.
• ∴ The numbers in ascending order: -1000, -100, -10, -1, 0, 1, 10, 100, 1000.

Question 11.
Arrange the following integers in descending order.
i) 14, 27, 15, -14, -9, 0, 11, -17
ii) -99, -120, 65, -46, 78, 400, -600
iii) 111, -222, 333, -444, 555, -666, 7777, -888
Solution:
(i) 14, 27, 15, -14, -9, 0,11, -17

• Separating the positive integers 14, 27, 15, 11 and negative integers -14, -9, -17
• Arranging in descending order we get the positive integers 27,15,14,11 and the negative integers -9, -14, -17.
• ‘0’ is neither positive nor negative and so it stand in middle.
• ∴ The numbers in descending order : 27, 15, 14, 11, 0, -9, -14, -17

(ii) -99, -120, 65, -46, 78, 400, -600

• Separating the positive integers 65, 78, 400 and negative integers -99, -120, -46, -600
• Arranging the positive integers in descending order as 400, 78, 65 and the negative integers in descending order -46, -99, -120, -600.
• The numbers in descending order : 400, 78, 65, -46, -99, -120, -600.

(iii) 111, -222, 333, -444, 555, -666, 7777, -888

• Separating the positive integers 111, 333, 555, 7777 and negative integers -222, -444, -666, -888
• Arranging the positive integers in descending order as 7777, 555, 333, 111 and negative integers in descending order as -222, -444, -666, -888
• The numbers in descending order : 7777, 555, 333, 111, -222, -444, -666, -888

Objective Type Questions

Question 12.
There are _______ positive integers from – 5 to 6.
(a) 5
(b) 6
(c) 7
(d) 11
Solution:
(b) 6
Hint:

Question 13.
The opposite of 20 units to the left of 0 is
(a) 20
(b) 0
(c) -20
(d) 40
Solution:
(a) 20
Hint:

Question 14.
One unit to the right of -7 is
(a) +1
(b) -8
(c) -7
(d) -6
Solution:
(d) -6
Hint:

Question 15.
3 units to the left of 1 is
(a) -4
(b) -3
(c) -2
(d) 3
Solution:
(c) -2
Hint:

Question 16.
The number which determines marking the position of any number to its opposite on a number line is?
(a) -1
(b) 0
(c) 1
(d) 10
Solution:
(b) 0
Hint:

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 1 Fractions Intext Questions

(Try These Textbook Page No. 2)

Question 1.
If all the three cakes are divided among the total participants of the function what would be each one’s share? Discuss.
Solution:
Total participants of the function = 9
Total number of cakes = 3
∴ Each cake should be divided into 3 equal parts.
∴ Total number of equal parts of cake = 9
∴ Each one’s share may be \(\frac{1}{9}\) of total cakes or \(\frac{1}{3}\) of a cake.

Question 2.
Observe the following and represent the shaded parts as fraction.

Solution:
(i) Total number of equal parts = 8
Shaded Parts = 3
Fraction representing the shaded parts = \(\frac{3}{8}\)

(ii) Total number of equal parts = 15
Shaded parts = 5
∴ Fraction representing the shaded parts = \(\frac{5}{15}\)

(iii) Total number of equal parts = 9
Shaded parts = 3
∴ Fraction representing the shaded parts = \(\frac{3}{9}\)

(iv) Total number of equal parts = 9
Shaded parts = 5
Fraction representing the shaded parts = \(\frac{5}{9}\)

Question 3.
Look at the following beakers, express the quantity of water as fraction and arrange them in ascending order:

Solution:
Quantity of water in the first beaker = 1 full
Quantity of water in the second beaker = \(\frac{1}{4}\)
Quantity of water in the third beaker = \(\frac{3}{4}\)
Quantity of water in the fourth beaker = \(\frac{1}{2}\)
Ascending order \(\frac{1}{4}\) < \(\frac{1}{2}\) < \(\frac{3}{4}\) < 1

Question 4.
Write the fraction of shaded part in the following.

Solution:
(i) Total number of equal parts = 3
Shaded parts = 2
Fraction representing the shaded portion = \(\frac{2}{3}\)

(ii) Total number of equal parts = 4
Shaded parts = 3
Fraction representing the shaded parts = \(\frac{3}{4}\)

(iii) Total number of equal parts = 5
Shaded parts = 4
Fraction representing the shaded parts = \(\frac{4}{5}\)

Question 5.
Write the fraction that represents the dots In the triangle.

Solution:
Total number of dots = 24
Number of dots in the triangle = 6
∴ Fraction represents the dots in the triangle = \(\frac{6}{24}\)

Question 6.
Find the fractions of the shaded and unshaded portions in the following.

Solution:
(a) Total number of equal parts = 8
Shaded parts = 2
∴ Fraction representing shaded parts = \(\frac{2}{8}\)

(b) Total number of equal parts = 8
Unshaded parts = 6
∴ Fraction of unshaded portion = \(\frac{6}{8}\)

(Activity Textbook Page No. 3)

Take a rectangular paper. Fold it into two equal parts. Shade one part, write the fraction. Again fold it into two halves. Write the fraction for the shaded part. Continue this process 5 times and write the fraction of the shaded part. Establish the equivalent fractions of \(\frac{1}{2}\) in the folded paper to your friends.

Solution:
First time = \(\frac{1}{2}\)
Second time = \(\frac{2}{4}\)
Third time = \(\frac{4}{8}\)
Fourth time = \(\frac{8}{16}\)
Fifth time = \(\frac{16}{32}\)

(Try These Textbook Page No. 4)

Question 1.
Find the unknown in the following equivalent fractions.

Solution:

(Try These Textbook Page No. 7)

Question 1.
Shade the rectangle for the given pair of fractions and say which is greater among them.

Solution:

Question 2.
Which is greater \(\frac{3}{8}\) or \(\frac{3}{5}\) ?
Solution:
LCM of the denominators 8 and 5 is 40.
Finding the equivalent fractions.

Question 3.
Arrange the fractions in ascending order : \(\frac{3}{5}, \frac{9}{10}, \frac{11}{15}\)
Solution:

Question 4.
Arrange the fractions in descending order : \(\frac{9}{20}, \frac{3}{4}, \frac{7}{12}\)
Solution:

(Try These Textbook Page No. 9)

(i) \(\frac{2}{3}+\frac{5}{7}\)
Solution:
By cross multiplication technique

(ii) \(\frac{3}{5}-\frac{3}{8}\)
Solution:
By cross multiplication technique

(Activity Textbook Page No. 10)

Question 1.
Using the given fractions \(\frac{1}{5}, \frac{1}{6}, \frac{1}{10}, \frac{1}{15}, \frac{2}{15}, \frac{4}{15}, \frac{1}{30}, \frac{7}{30} \text { and } \frac{9}{30}\) fill in the missing ones in the given 3 × 3 square in such a way that the addition of fractions through rows, columns and diagonals give the same total \(\frac{1}{2}\)

Solution:

(Try These Textbook Page No. 10)

Question 1.
Complete the following table. The first one is done for you.

(Try These Textbook Page No. 11)

Question 1.
(i) Are 5\(\frac{2}{3}\) and 5\(\frac{4}{6}\) equal?
Solution:

(ii) \(\frac{3}{2} \neq 3 \frac{1}{2}\) why?
Solution:

Question 2.
Convert 3\(\frac{1}{3}\) into improper fraction.
Solution:

Question 3.
Convert \(\frac{45}{7}\) into mixed fraction.
Solution:

(Try These Textbook Page No. 13)

Question 1.
Find the sum of 5\(\frac{4}{9}\) and 3\(\frac{1}{6}\).
Solution:

Question 2.
Subtract 7\(\frac{1}{6}\) and 12\(\frac{3}{8}\)
Solution:

Question 3.
Subtract the sum of 6\(\frac{1}{6}\) and 3\(\frac{1}{5}\) from the sum of 9\(\frac{2}{3}\) and 2\(\frac{1}{2}\).
Solution:

(Try These Textbook Page No. 15)

Question 1.
2\(\frac{1}{4}\) × 3 is not equal to 6\(\frac{1}{4}\). Why?
Solution:

Question 2.
Simplify : 35 × \(\frac{3}{7}\).
Solution:

Question 3.
Find the value of \(\frac{1}{5}\) of 15.
Solution:
\(\frac{1}{5}\) of 15 = \(\frac{1}{5}\) × 15 = \(\frac{15}{5}\) = 3

Question 4.
Find the value of \(\frac{1}{3}\) of \(\frac{3}{4}\)
Solution:

Question 5.
Multiply 7\(\frac{3}{4}\) by 5\(\frac{1}{2}\).
Solution:

(Activity Textbook Page No. 15)

Take a paper. Fold it into 4 parts vertically of equal width. Shade one part of it with red. Then, fold it into 3 parts horizontally of equal width. Shade two parts of it with blue. Now, you count the number of shaded grids which have both the colours. (Hint: The total number of grids is the product of \(\frac{2}{3}\) and \(\frac{1}{4}\))
Solution:
Activity to be done by the students themselves

(Try These Textbook Page No. 17)

Question 1.
(i) How many 6s are there in 18?
Solution:
Number of 6s in 18 are \(\frac{18}{6}\) = 3

(ii) How many \(\frac{1}{4}\)s are there in 5?
Solution:
Number of \(\frac{1}{4}\) s in 5 are 5 ÷ \(\frac{1}{4}\) = 5 × \(\frac{4}{1}\) = 20

(iii) \(\frac{1}{3}\) ÷ 5 = ?
Solution:
\(\frac{1}{3}\) ÷ 5 = \(\frac{1}{3} \times \frac{1}{5}\) = \(\frac{1}{15}\)

(Try These Textbook Page No. 18)

Question (i)
Find the value of 5 ÷ 2\(\frac{1}{2}\).
Solution:

Question (ii)
Simplify: \(1 \frac{1}{2} \div \frac{1}{2}\)
Solution:

Question (iii)
Divide 8 \(\frac{1}{2}\) by 4\(\frac{1}{4}\).
Solution:

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 5 Information Processing Intext Questions

Exercise 5.1

Try These (Text book Page No. 74)

Question 1.
Find the 10th term of the Fibonacci sequence
Solution:
We have the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,….
The 10th term is 55.

Question 2.
If the 11^th term of the Fibonacci sequence is 89 and 13^th term is 233 then, what is the 12^th term?
Solution:
Given 11^th term of the Fibonacci sequence is 89 and 13^th term of is 233
We know that a number in Fibonacci sequence is got by adding the previous two consecutive numbers.
11^th term + 12^th term = 13^th term
89 +12^th term =233
∴ 12^th term = 233 – 89 = 144
∴ 12^th term in Fibonacci sequence = 144.

Try this (Text book Page No.75)

Question 1.
Are two consecutive Fibonacci numbers relatively prime?
Solution:
GCD of two consecutive numbers in Fibonacci sequence is 1.
∴ They are relatively prime.

Try These (Text book Page No. 78)

Question 1.
The teacher should give oral instructions to the students to draw the geometrical figure already drawn by him / her.
i) Draw a square. In the middle of the square, draw a circle in such a way that the circle does not touch any side of the square. Divide the circle into four equal parts. Shade the bottom right part of the circle. Ask the students to show that figure drawn by them.
ii) Draw a triangle on a piece of paper. Make it crazy looking by adding some features. Give instructions to your friend to draw it exactly the same.

Solution:
i) The figure will be as drawn below.
ii) The instructions may be:

1. Draw a square of side 3 cm.
2. Draw an equilateral triangle of side 2 cm inside the square without touching the sides of the square.
3. Draw a small circle at the middle of the triangle.
4. Draw two eyes above the circle.
5. Draw a mouth below the circle.
6. Draw two legs at the base of the triangle.
7. Draw hands to the other to sides of the triangle stretching inside the square. Draw two ears above the hands.

Question 2.
Suppose your friend wants to come to your house from his/her house. Give clear instructions in order, to reach your house.
Solution:
Activity to be done by the students themselves

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 1 Fractions Additional Questions

Question 1.
Color the part according to the given fraction.

Solution:
(i) Here \(\frac{3}{4}\) shows out of 4 parts 3 parts are shaded

(ii) Here \(\frac{2}{4}\) shows out of 4 parts 2 parts are shaded

Question 2.
Identify the error if any

Solution:
In the given figure, shaded portion is not equal to unshaded portion. So the fraction is not equal to \(\frac{1}{2}\).

Question 3.
What fraction of an hour is 20 minutes?
Solution:
We know that total minutes in an hour = 60 min
∴ Required fraction = \(\frac{20 \min }{60 \min }=\frac{20}{60}=\frac{1}{3}\)

Question 4.
Write a fraction equivalent to \(\frac{3}{5}\) with numerator 15.
Solution:
Given, numerator of an equivalent fraction = 15
Equivalent fraction of \(\frac{3}{5}=\frac{3 \times 5}{5 \times 5}=\frac{15}{25}\)

Question 5.
Which is the larger fraction \(\frac{6}{10}\) or \(\frac{7}{10}\)?
Solution:
Here the denominators, of both fractions are same.
Also 7 > 6 So \(\frac{7}{10}>\frac{6}{10}\)

Question 6.
Sona got one-fifth of the total marks and Mala got one-third of the total marks. Who got more?
Solution:
We know that if the numerators are same in two fractions, the fraction with smaller denominator is greater.
∴ \(\frac{1}{3}>\frac{1}{5}\)
∴ Mala got more marks.

Question 7.
A piece of rope \(\frac{7}{8}\) metre long is cut into two pieces. One piece was \(\frac{1}{4}\) m long. How long is the other?
Solution:

Question 8.
Meena travelled 3\(\frac{1}{2}\) km by bus, then she walked 1\(\frac{1}{8}\) km to reach a town. How much she travelled to reach-the town?
Solution:

She travelled 4\(\frac{5}{8}\) km

Question 9.
What should be subtracted from the sum of 2\(\frac{1}{4}\) and 3\(\frac{1}{7}\) to get 2\(\frac{3}{28}\) ?
Solution:

Question 10.
Compare 4\(\frac{2}{3}\) and 5\(\frac{3}{7}\)?
Solution:

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 5 Information Processing Additional Questions

Question 1.
Find the figure which is different from others.

Hint:
D is different from other figures.
All other figures has 5 lines and D has 6 lines in it.
Solution:
(d)

Question 2.
If GARIMA is coded as 725432 and TINA as 6482, how will MARTINA be coded?
(a) 3256482
(b) 3265842
(c) 3645862
(d) 3658426
Hint: GARIMA and TINA both have the lettersT and A and they are coded as 4 and 2 repectively.
∴ In MARTINA, I coded as 4 and A coded as 2 in (A) 3256482
Solution:
(a) 3256482

Question 3.
If ‘+’ is ‘×’, is ‘+% ‘×’ is ‘÷’ and ‘÷’ is then answer the following 21 ÷ 8 8 + 2 – 12 × 3 = ?
(a) 14
(b) 9
(c) 13
(d) 11

Solution:
(a) 14

Question 4.
Which number will replace the question mark?

(a) 7
(b) 14
(c) 48
(d) 49
Hint: [∴ Left side numbers are square numbers] .
Solution:
(d) 49.

Question 5.
Find the correct figure which replaces the ‘?

Hint: Upside down
Solution:

Question 6.
Which comes next ?

Hint : Here the figures rotate clockwise and anticlockwise alternatively making an angle 90° and one arrow deleted in every successive block.
Solution:

Question 7.
Find the number which replaces the question mark in the following series.

(a) 25
(b) 49
(c) 97
(d) 193

Solution:
(d) 193

Question 8.
Identify the number which replaces the question mark?

(a) 82
(b) 124
(c) 100
(d) 64
Hint: (2 + 8)² = 10² = 100
Solution:
(c) 100

Question 9.

Hint: In each step each one of the existing elements moves to the clockwise adjacent
comer. Also in one step, the element that reaches the upper-right comer gets replaced by a new element and in the next step, the element that reaches the lower left comer gets replaced by a new element.
Solution:
(d) 5

Question 10.
1, 6, 15, ?, 45, 66, 91. Find the missing term.
(a) 25
(b) 28
(c) 33
(d) 38
Hint: 1 + 5 = 6, 6 + 9 = 15, so missing term = 15 + 13 = 28
Solution:
(b) 28

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 1 Fractions Ex 1.2

Miscellaneous Practice Problems

Question 1.
Sankari purchased 2\(\frac{1}{2}\) m cloth to stitch a long skirt and 1\(\frac{3}{4}\) m cloth to stitch blouse. If the cost is ₹ 120 per metre then find the cost of cloth purchased by her
Solution:
Cloth to stitch a long skirt = 2\(\frac{1}{2}\) m
Cloth to stitch a blouse = 1\(\frac{3}{4}\) m
Total length of the cloth = \(2 \frac{1}{2}+1 \frac{3}{4} m=2+\frac{1}{2}+1+\frac{3}{4}\) m

Cost of cloth purchased by Sankari = ₹ 510

Question 2.
From his office, a person wants to reach his house on foot which is at a distance of 5\(\frac{3}{4}\) km. If he had walked 2\(\frac{1}{2}\) km, how much distance still he has to walk to reach his house?
Solution:

Distance still he has to be walked = 3\(\frac{1}{4}\) km

Question 3.
Which is smaller? The difference between \(\frac{1}{2}\) and 3\(\frac{2}{3}\) or the sum of 1\(\frac{1}{2}\) and 2\(\frac{1}{4}\)
Solution:

Question 4.
Mangai bought 6\(\frac{3}{4}\) kg of apples. If Kalai bought 1\(\frac{1}{2}\) times as Mangai bought, then how many kilograms of apples did Kalai buy?
Solution:
Weight of apples Mangai bought = 6\(\frac{3}{4}\) kg

Weight of apples Kalai bought = 10\(\frac{1}{8}\) kg

Question 5.
The length of the staircase is 5 \(\frac{1}{2}\) m. If one step is set at \(\frac{1}{4}\) m, then how many steps will be there in the staircase?
Solution:
Length of the staircase = 5\(\frac{1}{2}\) m
Distance between each step = \(\frac{1}{4}\) m
∴ Number of steps in the staircase = \(5 \frac{1}{2} \div \frac{1}{4}=\frac{11}{2} \div \frac{1}{4}=\frac{11}{2} \times \frac{4}{1}\) = 22
There will be 22 steps in the staircase

Challenge Problems

Question 6.
By using the following clues, find who am I?
(i) Each of my numerator and denominator is a single digit number.
(ii) The sum of my numerator and denominator is a multiple of 3.
(iii) The product of my numerator and denominator is a multiple of 4
Solution:
The numerator may be any one of!, 2, 3,4, 5, 6, 7, 8, 9 and the denominator may be any one of 1, 2,3,4, 5,6, 7,8,9. Sum of numerator and denominator is a multiple of 3.
∴ Possible proper fractions are \(\frac{1}{2}, \frac{1}{5}, \frac{1}{8}, \frac{2}{4}, \frac{2}{7}, \frac{3}{6}, \frac{3}{9}, \frac{4}{5}, \frac{4}{8}, \frac{5}{7}, \frac{6}{9}\)
Also given the product of numerator and denominator is a multiple of 4.
∴ Possible fractions are \(\frac{1}{8}, \frac{2}{4}, \frac{4}{5}, \frac{4}{8}\)

Question 7.
Add the difference between 1\(\frac{1}{3}\) and 3\(\frac{1}{6}\) and the difference between 4\(\frac{1}{6}\) and 2\(\frac{1}{3}\)
Solution:

Question 8.
What fraction is to be subtracted from 9\(\frac{3}{7}\) to get 3\(\frac{1}{5}\) ?
Solution:

Question 9.
The sum of two fractions is 5\(\frac{3}{9}\). If one of the fractions is 2\(\frac{3}{4}\), find the other fraction.
Solution:

The other number is 2\(\frac{7}{12}\)

Question 10.
By what number should 3\(\frac{1}{16}\) be multiplied to get 9\(\frac{3}{16}\) ?
Solution:

The number to be multiplied is 3.

Question 11.
Complete the fifth row in the Leibnitz triangle which is based on subtraction.

Solution:

Question 12.
A painted \(\frac{3}{8}\) of the wall of which one third is painted in yellow colour. What fraction is the yellow colour of the entire wall.

Solution:

\(\frac{1}{8}\) of the wall is painted yellow.

Question 13.
A rabbit has to cover 26\(\frac{1}{4}\) m to fetch its food. If it covers 1\(\frac{3}{4}\) m in one jump, thenhow many jumps will it take to fetch its food?

Solution:
Total distance to be covered by the rabbit = 26\(\frac{1}{4}\)m
Distance covered in one jump = 1\(\frac{3}{4}\) m

∴ The rabbit jumps 15 times to fetch its food.

Question 14.
Look at the picture and answer the following questions :
(i) What is the distance from School to library via bus stop?
(ii) What is the distance between school and library via Hospital?
(iii) Which is the shortest distance between (i) and (ii)?
(iv) The distance between School and Hospital is ____ times the distance between school and Bus stop.

Solution:


The distance between school and Hospital is 6 times the distance between school and bus stop.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Intext Questions

Try These (Textbook Page No. 1)

Question 1.
(i) Observe and complete:
1 + 3 = ?
5 + 11 = ?
21 + 47 = ?
___ + ____ = ?
From this observation, we conclude that “the sum of any two odd numbers is always an _____”
(ii) Observe and complete:
5 × 3 = ?
7 × 9 = ?
11 × 13 = ?
_____ × ____ = ?
From this observation, we conclude that “the product of any two odd numbers is always an _____”
Justify the following statements with appropriate examples:
(iii) The sum of an odd number and an even number is always an odd number.
(iv) The product of an odd and an even number is always an even number.
(v) The product of only three odd numbers is always an odd number.
Solution:
(i) 1 + 3 = 4
5 + 11 = 16
21 + 47 = 68
An odd number + another odd number = An Even number
From this observation, we conclude that the sum of any two odd numbers is always an even number.
(ii) 5 × 3 = 15
7 × 9 = 63
11 × 13 = 143
An odd number × Another odd number = An odd number
From this observation, we conclude that “the product of any two odd numbers is always an odd number.”
(iii) Take the odd number 5 and the even number 10
Their sum = 5 + 10 = 15, which is odd.
∴ Sum of an odd number and an even number is always an odd number.
(iv) Take the odd number 5 and the even number 10.
Their product = 5 × 10 = 50, which is even
Thus the product of an odd and an even number is always an even number.
(v) Consider 7 × 5 × 3
We know that the product of any two odd numbers is an odd number
7 × 5 = 35, odd number.
Also we have 35 × 3 = 105
∴ 7 × 5 × 3 = 105, an odd number.
So the product of three odd numbers is always an odd number.

Try These (Textbook Page No. 3)

Question 1.
(i) Say True or False
(a) The smallest odd natural number is 1.
(b) 2 is the smallest even whole number.
(c) 12345 + 5063 is an odd number.
(d) Every number is a factor of itself.
(e) A number which is a multiple of 6 is also a multiple of 2 and 3.
(ii) Is 7, a factor of 27?
(iii) Is 12, a factor or a multiple of 12?
(iv) Is 30, a factor or a multiple of 10?
(v) Which of the following numbers has 3 as a factor?
(a) 8
(b) 10
(c) 12
(d) 14
(vi) The factors of 24 are 1, 2, 3, __, 6, ___, 12 and 24. Find the missing ones.
(vii) Look at the following numbers carefully and find the missing multiples.

Solution:
(i) (a) True
(b) False
(c) False
(d) True
(e) True
(ii) No, 7 is not a factor of 27. Because 7 does not divide 27 exactly
(iii) 12 is both a factor and a multiple of 12
(iv) 30 is a multiple of 10
(v) (a) Factors of 8 are 1, 2, 4, 8
(b) Factors of 10 are 1, 2, 5, 10
(c) Factors of 2 are 1, 2, 3, 4, 6, 12
(d) Factors of 14 are 1, 2, 7, 14
∴ The number 12 has 3 as a factor
(vi) Factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24.
Missing Factors 4, 8.
(vii)

Try These (Textbook Page No. 6)

Question 1.
Express 68 and 128 as the sum of two consecutive primes.
Solution:
68 = 31 + 37
128 = 61 + 67

Question 2.
Express 79 and 104 as the sum of any three odd primes.
Solution:
79 = 37 + 31 + 11
79 = 41 + 31 + 7
79 = 61 + 11 + 7
79 = 59 + 13 + 7
79 = 53 + 19 + 7 and so on.
104 cannot be expressed as the sum of three odd primes.
Because we know that “ the sum of any two odd numbers is an even number”.
Also the sum of an odd and even number is always an odd number.
104 = 61 + 41 + 2
104 = 97 + 5 + 2
104 = 89 + 13 + 2 and so on.

Try These (Textbook Page No. 8)

Question 1.
Are the leap years divisible by 2?
Solution:
Leap years are divisible by 4.
Leap years are divisible by 2.

Question 2.
Is the first 4 digit number divisible by 3?
Solution:
The first four-digit number is 1000.
Sum of the digits is 1 + 0 + 0 + 0 = 1, not divisible by 3.
1000 is not divisible by 3.

Question 3.
Is your date of birth (DDMMYYYY) divisible by 3?
Solution:
Date of birth 25.05.2007
Sum of digits = 2 + 5 + 0 + 5 + 2 + 0 + 0 + 7 = 21
Again 2 + 1 = 3, divisible by 3.
My date of birth is divisible by 3.

Question 4.
Identify the numbers in the sequence 2000, 2006, 2010, 2015, 2019, 2025 that are divisible by both 2 and 5.
Solution:
We know that a number is divisible by both 2 and 5 if it is divisible by 10. 2000 and 2010 are divisible by 10.

Question 5.
Check whether the sum of 5 consecutive numbers is divisible by 5.
Solution:
Take the first five consecutive natural numbers 1, 2, 3, 4 and 5.
Their sum 1 + 2 + 3 + 4 + 5 = 15, divisible by 5.
Also, 2 + 3 + 4 + 5 + 6 = 20, divisible by 5.
3 + 4 + 5 + 6 + 7 = 25, divisible by 5.
Generally, the sum of 5 consecutive natural numbers is divisible by 5.

Try These (Textbook Page No. 19)

A small boy went to a town to sell a basket of wood apples. On the way, some robbers grabbed the fruits from him and ate them! The small boy went to the King and complained. The King asked him, “How many wood apples did you bring?”. The boy replied, “Your Majesty! I didn’t know, but I knew that if you divided my fruits into groups of 2, one fruit would be left in the basket”. He continued saying that if the fruits were divided into groups of 3, 4, 5 and 6, the fruits left in the basket would be 2, 3, 4 and 5 respectively. Also, if the fruits were divided into groups of 7, no fruit would be there in the basket. Can you find the number of fruits, the small boy had initially?
(This problem is taken from the famous Mathematics problems collection book in Tamil called “Kanakkathikaram” under the heading of “Wood Apple Problem”)
Solution:
The total number of fruits, when divided by 2, 3, 4, 5 and 6, leaves the remainders 1, 2, 3, 4 and 5 respectively.
Here (2 – 1) = (3 – 2) = (4 – 3) = (5 – 4) = (6 – 5) = 1.
∴ The required no. of fruits will be LCM (2, 3, 4, 5, 6) – 1
L CM (2, 3, 4, 5, 6) = 2 × 3 × 2 × 5 = 60

Now take the multiples of 60 and subtract 1 from it.
Also checking the conditions, multiplies of 60 are 60, 120, 180, …..
The multiple -1
59, 119, 179, ……
The required number = 119
∴ The total number of fruits = 119.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Additional Questions

Question 1.
A pair of prime numbers whose difference is 2, is called _____
Solution:
Twin primes

Question 2.
The first 4 digit number divisible by 3 ______
Solution:
1002

Question 3.
Prime triplet ______
Solution:
(3, 5, 7)

Question 4.
100 years = ______
Solution:
100 years = 1 Century

Question 5.
Loss = ____
Solution:
Loss = CP – SP

Question 6.
526 ml _____
Solution:
526 ml = 0.526 L

Question 7.
No sides are equal _____
Solution:
Scalene Triangle

Question 8.
What is the total number of primes up to 100?
Solution:
25

Question 9.
Check whether (37, 39) is a twin prime?
Solution:
No, because 39 is not a prime number.

Question 10.
Check the divisibility by 11 of 684398?
Solution:
In 684398
Sum of digits in odd places = 8 + 3 + 8 = 19
Sum of digits in even places = 6 + 4 + 9 = 19.
Difference = 19 – 19 = 0.
684398 is divisible by 11.

Question 11.
Is 53249624 is divisible by 8? How?
Solution:
In 53249624, consider the last three digits 624, which is divisible by 8.
53249624 is divisible by 8.

Question 12.
Factorise 1056
Solution:

1056 = 2 × 2 × 2 × 2 × 2 × 3 × 11

Question 13.
Express 42 and 100 as the sum of two consecutive primes?
Solution:
42 = 19 + 23
100 = 47 + 53

Question 14.
A heap of stones can be made up into groups of 21. When made up into groups of 16, 20, 25 and 45 there are 3 stones left in each case. How many stones at least can there be in the heap?
Solution:
LCM of 16, 20, 25, 45 = 2 × 5 × 2 × 5 × 2 × 2 × 3 × 3 = 3600

The heap contain 3600 + 3 = 3603 stones at least.
3603 stones at least can there be in the heap.

Question 15.
Find the largest number of four digits exactly divisible by 12, 15, 18 and 27
Solution:
The largest number of four digits = 9999
lcm of 12, 15, 18, 27 is 540
Dividing 9999 by 540
We get 279 as the remainder


LCM = 2 × 3 × 3 × 2 × 5 × 3 = 540
Required number = 9999 – 279 = 9720

Question 16.
Find the least number which when divided by 6, 7, 8, 9 and 12 leaves the same remainder 1 in each case.
Solution:
Required number = [LCM (6, 7, 8, 9, 12)] + 1
LCM (6, 7, 8, 9, 12) = 3 × 2 × 2 × 2 × 3 × 7 = 504
Required number = 504 + 1 = 505

Question 17.
Product of two co-prime numbers is 117. Then what will be their LCM?
Solution:
We know that LCM × HCF = Product of two numbers
Also, we know that HCF of two co-primes = 1
LCM × 1 = 117
LCM = 117

Question 18.
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
Solution:
LCM of 2, 4, 6, 8, 10 and 12 is 120
So the bell will toll together after every 120 seconds i.e 2 minutes.
In 30 minutes, they will toll together \(\frac{30}{2}\) + 1 = 16 times.
LCM = 2 × 2 × 3 × 2 × 5 = 120

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Intext Questions

Try These (Textbook Page No. 138)

Question 1.
Now drop the condition that each digit must be used exactly once. List the numbers that are possible now and find the numbers that were not listed above.
Solution:
Here repetition of digits allowed.
Fixing 9 for a thousand places we have

Fixing 6 for thousand places we have

Fixing 5 for thousands place

Fixing 3 for thousands place, we get

Thus we have 4 × 4 × 4 × 4 = 256 numbers.
(ii) The underlined 24 numbers were listed as the numbers with non-repeated digits. Remaining numbers are not listed in the previous list.

Question 2.
Mother had a lot of wooden pieces in different shapes with her. She gave 6 triangles to Kannagi and 6 circles to Madhan and asked them to create different figures using them. They tried and got some interesting figures. Can you create figures like them that are nice and interesting?

Solution:

Try These (Textbook Page No. 139)

Question 1.
Form a group with two of your friends and try this. All three of you together have to draw a scene. First, one of your friends should draw one part, next the other friend has to continue it, and finally you have to complete it. No discussion or any other communication is allowed. Finally, each person tells what (s)he actually intended to draw the full picture.
Solution:
Activity to be done by the students themselves.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Additional Questions

Question 1.
In a meeting, breakfast is served with coffee, Tea and Milk. Breakfast items are Idly, Dosai and Pongal. Each one can have a drink and breakfast. How many such combinations can be there?
Solution:

So they have 9 combinations such as

1. Coffee and Idly
2. Coffee and Dosai
3. Coffee and Pongal
4. Milk and Idly
5. Milk and Dosai
6. Milk and Pongal
7. Tea and Idly
8. Tea and Dosai
9. Tea and Pongal.

Question 2.
Place the numbers from 1 to 12 exactly once in the circles so that the sum of numbers in each of the 5 lines of the star is 24.

Solution:
By trial and error, we arrange the numbers to get the sum 24.

Question 3.
Place the numbers from 1 to 12 in the circles without repetition to get a sum of 28, 29, 30 and 31.

Solution:

Question 4.
Find the minimum number of straight lines used in forming the figure.

Solution:

The straight lines may be labelled as
Vertical lines = 5
Horizontal lines = 3
Standing lines = 6
Total of 14 lines are needed.

Question 5.
Find the dots in the 12th figure of the following:

Solution:

10th figure will have 10 × 3 = 30 dots.
And 12th figure will have 12 × 3 = 36 dots.