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 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.

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 Ex 6.3

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

Solution:
(i) The single portions A, B, C, D, E, F, G and H are 8 triangles.
Taking the combination of 2 E & F, F & G & H and E & 11 are 4 triangles.
There are 8 + 4 = 12 triangles.

(ii) The smaller portions A, B, C, D, E, F, G and H are 8 triangles.
Combining 2 at a time B & C, D & F, F & G, A & H are 4 triangles.
Taking 4 at a time
Combination of B, C, D and E; D, E, F and G; F, G, H and A; H, A, B and C are 4 triangles.
Total triangles = 8 + 4 + 4 = 16.

(iii) Every part A, B, C, D, E, F, G, H, I, J, K, L, M, N, O and P are 16 triangles.
Combining two at a time. A & B, B & C, C & D, D & A, J & K, L & M, N & O, P & I are 8 triangles.
Taking the combination of 4, we have I, J, E & A; K, L, F and B, M, N, G and C, P, O, H and D are 4 triangles.
Taking 8 parts together we have QRS, RST, STQ, and TRQ are 4 triangles.
Total triangles = 16 + 8 + 4 + 4 = 32 triangles.

(iv)

Taking smaller triangles we have AFB, BFG, BGC, CGH, HCD, DHI, DIE, EIJ, EJAand AJF ⇒ 10 triangles.
Taking the combination of 2.
CGD, CID, DJE, DHE, EIA, EFA, AGB, AJB, BHC and BFC are 10 triangles.
Combining 3 at a time we have CED, DAE, EBA, ACB, BDC, CEF, DBJ, GDA, EBH, ACI ⇒ 10 triangles.
Other triangles are BCE, CAD, DBE, ADB and ACE are 5 triangles.
Total 10 + 10 + 10 + 5 = 35 triangles.

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

Solution:
(i) The number of dots given are 1, 3, 6, 10, 15,…

1st number = 1
2nd number = 1 + 2 = 3
3rd number = 3 + 3 = 6
4th number = 6 + 4 = 10
5th number = 10 + 5 = 15
6th number will be 15 + 6 = 21
7th number will be 21 + 7 = 28
8th number will be 28 + 8 = 36
9th number will be 36 + 9 = 45
10th number will be 45 + 10 = 55
Number of dots in the 10th figure = 55
(ii) The number of dots given are 1, 4, 9, 11, 16, 25,…
The no. of dots are in the pattern 1 × 1, 2 × 2, 3 × 3, 4 × 4, 5 × 5, …… 10 × 10
The number of dots in the 10th figure = 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:

First number is 2
2nd number is 2 + 3 = 5
3rd number is 5 + 4 = 9
4th number is 9 + 5 = 14
5th number is 14 + 6 = 20
(iv) Number of dots in the 25th pattern is

The number of dots in the 25th pattern = 350.

Question 4.
Count the number of squares in each of the following figures?

Solution:

Smaller squares 4 × 4 = 16 (As numbered).
As a whole bigger = 4
Total squares = 20

As we see in the figure in the middle 1, 2, 3, and 4 are 4 small squares.
Also, we have 9 and 10 = 2 big squares.
Outer squares 5, 6, 7, & 8 = 4.
Total = 4 + 4 + 1 + 1 = 10 squares.

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

Solution:

There are 7 Circles.

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

Solution:

There are 5 horizontal lines and 5 vertical lines.
Total of 10 lines minimum needed.

There are 12 straight lines used which is minimum.

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 Ex 6.2

Question 1.
In the following magic triangle, arrange the numbers from 1 to 6, so that you get the same sum on all its sides.

Solution:
Step 1: Complete the corners with smaller numbers 1, 2 and 3.

Step 2: The side having smallest numbers 1 & 2 are to be filled with the greatest number 6, the second smallest 1 & 3 side to be filled with the second largest 5 at the middle and so on.

The magic sum is 1 + 6 + 2 = 2 + 4 + 3 = 3 + 5 + 1 = 9. Some other ways are given below.

The magic sum = 1 + 6 + 3 = 3 + 2 + 5 = 5 + 4 + 1 = 10.

The magic sum 6 + 1 + 4 = 4 + 5 + 2 = 2 + 3 + 6 = 11.

The magic sum 4 + 3 + 5 = 5 + 1 + 6 = 6 + 2 + 4 = 12.

Question 2.
Using the numbers from 1 to 9
(i) Can you form a magic triangle?
(ii) How many magic triangles can be formed?
(iii) What are the sums of the sides of the magic triangle?

Solution:
(i) Yes, we can form
(ii) 5
(iii)


Sums are 17, 19, 20, 21 and 23.

Question 3.
Arrange the odd numbers from 1 to 17 without repetition to get a sum of 30 on each side of the magic triangle.

Solution:
The odd numbers between 1 to 17 are 1, 3, 5,7,9,11,13,15,17.
Step 1: Place the smaller numbers 1, 3, 5 on the comers.

Step 2: Arrange another set of smaller numbers 7, 9 and 11 on each side.

Step 3: Arrange the remaining numbers 13,15,17 to give the total 30.

Magic sum = 30.

Question 4.
Put the numbers 1, 2, 3, 4, 5, 6 and 7 in the circles so that each straight line of three numbers add up to the same total.

Solution:
Here there is even number of terms. Also we know that 1 + 6 = 7, 2 + 5 = 7, 3 + 4 = 7; so placing 7 at the centre, and the pairs (1, 6) (2, 5) and (3, 4) at. the opposite ends we get, the answer.

Question 5.
Place the number 1 to 12 in the 12 circles so that the sum of the numbers in each of the six lines of the star is 26. Use each number from 1 to 12 exactly once. Find more possible ways.

Solution:
The given star can be viewed as two magical triangular as.

Now the required arrangement is

Some other arrangements are

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 5 Statistics Intext Questions

Recap (Textbook Page No. 108)

Question 1.
Count the objects in the following figure and complete the table that follows :

From Fig 5.1 and the table, answer the following question.
(i) The total number of objects in the above picture is ______
(ii) The difference between the number of squares and the number of bats is ______
(iii) The ratio of the number of balls to the number of bats is _______
(iv) What are the objects equal in number?
(v) How many more balls are there than the number of bats?
Solution:
(i) 24
(ii) 0
(iii) \(\frac{8}{6}=\frac{4}{3}\)
(iv) Bat and Square
(v) 2 balls more

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 5 Statistics Ex 5.4

Miscellaneous Practice Problems

Question 1.
The heights (in centimetres) of 40 children are

Prepare a tally mark table.
Solution:

Question 2.
There are 1000 students in a school. Data regarding the mode of transport of the students as given below. Draw a pictograph to represent the data.

Solution:
The pictograph for the mode of transport of students.

Question 3.
The following pictograph shows the total savings of a group of friends in a year. Each picture represents a saving of Rs. 100. Answer the following questions.

(i) What is the ratio of Ruby’s saving to that of Thasnim’s?
(ii) What is the ratio of Kuzhali’s savings to that of others?
(iii) How much is Iniya’s savings?
(iv) Find the total amount of savings of all your friends?
(v) Ruby and Kuzhali save the same amount. Say True or False.
Solution:
(i) Ratio of Ruby’s saving to that of Thasnim’s
\(=\frac{\text { Ruby’s saving }}{\text { Thasnim’s saving }}=\frac{5 \times 100}{4 \times 100}=\frac{5}{4}=5: 4\)
Ratio of Ruby’s saving to that of Thasnims = 5 : 4
(ii) Ratio of Kuzhali’s savings to that of others
\(=\frac{\text { kuzhali’s saving }}{\text { others saving }}=\frac{5 \times 100}{19 \times 100}=\frac{5}{19}=5: 19\)
Ratio of Kuzhali’s saving to that of others = 5 : 19
(iii) Iniya’s saving = 3 × 100 = ₹ 300
(iv) Saving of all the friends = (5 + 7 + 4 + 5 + 3) × 100 = 24 × 100 = ₹ 2400.
Total savings = ₹ 2400
(v) True.

Challenging Problems

Question 4.
The table shows the number of moons that orbit each of the planets in our solar system.

Make a Bar graph for the above data.
Solution:

Question 5.
The predictions of Weather in the month of September is given below:

(i) Make a frequency table of the types of weather by reading the calender.
(ii) How many days are either cloudy or partly cloudy
(iii) How many days do not have rain? Give two ways to find the answer?
(iv) Find the ratio of the number of sunny days to Rainy days.
Solution:
Frequency Table for the Type of weather for the month of September

(ii) 14 days.
(iii) For 24 days there is no rain.
(a) From the total 30 days, we subtract the rainy days i.e 30 – 6 = 24 days (from the frequency table)
(b) From the picture, we can count the non-rainy days.
(iv) Ratio of number of Sunny day to Rainy days
\(=\frac{\text { Number of Sunny days }}{\text { Number of Rainy days }}=\frac{10}{6}=\frac{5}{3}=5: 3\)
The ratio of a number of Sunny days to Rainy days = 5 : 3.

Question 6.
26 students were interviewed to find out what they want to become in future. Their responses are given in the following table.

Represent this data using the pictograph
Solution:

Question 7.
Yasmin of class VI was given a task to count the number of books which are biographies, in her school library. The information collected by her is represented as follows.

Observe the pictograph and answer the following questions.
(i) Which title has the maximum number of biographies?
(ii) Which title has the minimum number of biographies?
(iii) Which title has exactly half the number of biographies as Novelists?
(iv) How many biographies are there on the title of Sportspersons?
(v) What is the total number of biographies in the library?
Solution:
(i) ‘The title Novelists’ have the maximum number of biographies
(ii) ‘The title Scientists’ have the minimum number of biographies.
(iii) ‘Sportspersons’ title has exactly half the number of biographies as Novelist.
(iv) \((1 \times 20)+\frac{20}{4}=20+5=25\) biographies are there in the title sportsperson.
(v) 8 × 20 = 160 biographies are there in the library.

Question 8.
The bar graph illustrates the results of a survey conducted on vehicles crossing over a Toll Plaza in on hour.

Observe the bar graph carefully and fill up the following table.

Solution:

Question 9.
The lengths (in the nearest centimetre) of 30 drumsticks are given as follows.

Draw the bar graph showing the same information.
Solution:
The bar graph showing the same information is given below:

Question 10.
Given two angles are supplementary i.e. their sum = 180°.
Solution:
Let the angle be x.
Then anothcf angle = x + 20 (given)

The two angles are 80° and 100°

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 5 Statistics Additional Questions

Question 1.
A _____ is a collection of numbers gathered to give some meaningful information.
Solution:
Data

Question 2.
The data can be arranged in a tabular form using ____ marks
Solution:
Tally

Question 3.
A observation occurring 8 times in a data is represented as _____ using tally marks.
Solution:

Question 4.
Tally marks  represents _____
Solution:
6

Question 5.
Following are the choice of sports for 20 students of class VI.

Arrange the names of sports in a table using tally marks.
Solution:

Question 6.
Mariana threw a die 40 times and noted the number appearing each time as shown below.

Make a table and enter the data using tally marks.
Find the number that appeared
(a) the minimum number of times
(b) the maximum number of times
(c) Find those numbers that appear an equal number of times.
Solution:
Table using Tally Marks

(a) The number ‘4’ appeared the minimum number of times
(b) The number ‘5’ appeared the maximum number of times
(c) Numbers ‘1’ and ‘6’ appeared the equal number of times.

Question 7.
What is the advantage of using pictograph?
Solution:
The data can be analyzed and interpreted. The pictures and symbols help us to understand better.

Question 8.
Total number of students of a school in different years is shown in the following table

(A) Prepare a pictograph of students using one symbol to represent 100 students and answer the following questions
(a) How many symbols represent the total number of students in the year 2000?
(b) How many symbols represent the total number of students in the year 1997?
Solution:
(A) The pictograph is given by

(a) 9 Full pictures and a half picture represent the number of students in the year 2000
(b) 5 symbols.

Question 9.
The number of bottles of honey sold by three different shops are given below.

(A) Draw a pictograph and answer the following questions (use the scale – 20 bottles)
(a) What is the total profit of shop A, if the profit gained on each bottle is ‘E’150?
(b) If the total number of bottles sold is 400 how many figures must be drawn to shops?
(c) Find the difference between the number of bottles sold by shop B and the number of bottles sold by shop ‘C’.
Solution:
(A) Pictograph of number of bottles sold

(a) Profit per bottle = ₹ 150
Shop A’s total profit = 80 × 150 = ₹ 12,000
(b) For 400 bottles, the figures to be drawn is \(\frac{400}{20}\) = 20 bottles for shop ‘C ’
(c) Difference = No. of bottles sold by shop ‘C’ – No. of bottles sold by shop ‘B’ = 180 – 140 = 40 bottles

Question 10.
The following data is the total savings of a group of 5 friends in a year.

Draw a pictograph to represent the data.
Solution:

Question 11.
The number of mathematics books sold by a shopkeeper on six consecutive days is shown below.

Draw a bar graph to represent the above information choosing the scale of your choice.
Solution:

Question 12.
The following bar graph shows the number of people visited Mahabalipuram over a period of 5 months.

Use the graph to answer the following questions
(a) How many people visited Mahabalipuram in April?
(b) How many more people visited Mahabalipuram in May than in January?
(c) In which month, were there as many visitors as in March?
(d) In which month was there 1500 more visitors than in January?
(e) How many visitors were there in 5 months?
Solution:
(a) 3500 people visited Mahabalipuram in April
(b) 500 more people
(c) January
(d) April
(e) 11,500 people

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 3 Ratio and Proportion Intext Questions

Recap (Textbook Page No. 55)

Question 1.
Which of the following fractions is not a proper fraction?
(a) \(\frac{1}{3}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{5}{10}\)
(d) \(\frac{10}{5}\)
Solution:
(d) \(\frac{10}{5}\)

Question 2.
The equivalent fraction of \(\frac{1}{7}\) is ____
(a) \(\frac{2}{15}\)
(b) \(\frac{1}{49}\)
(c) \(\frac{7}{49}\)
(d) \(\frac{100}{7}\)
Solution:
(c) \(\frac{7}{49}\)

Question 3.
Write >, < or = in the box.
(i) \(\frac{5}{8}\) ____ \(\frac{1}{10}\)
(ii) \(\frac{9}{12}\) _____ \(\frac{3}{4}\)
Solution:

Question 4.
Arrange these fractions from the least to the greatest: \(\frac{1}{2}, \frac{1}{4}, \frac{6}{8}, \frac{1}{8}\)
Solution:
\(\begin{array}{l}{\frac{1}{2}=\frac{1 \times 4}{2 \times 4}=\frac{4}{8}} \\ {\frac{1}{4}=\frac{1 \times 2}{4 \times 2}=\frac{2}{8}}\end{array}\)
Comparing \(\frac{4}{8}, \frac{2}{8}, \frac{6}{8} \text { and } \frac{1}{8}\). we have \(\frac{1}{8}<\frac{2}{8}<\frac{4}{8}<\frac{6}{8}\)
i.e, \(\frac{1}{8}<\frac{1}{4}<\frac{1}{2}<\frac{6}{8}\)

Question 5.
Annan says the \(\frac{2}{6}\) th of the group of triangles given below are blue. Is he correct?

Solution:
No, he is not correct because of the total of 6 triangles 4 are blue, i.e \(\frac{4}{6}^{\text {th }}\) triangles are blue.

Question 6.
Joseph has a flower garden. Draw a picture which shows that \(\frac{2}{10}\) th of the flowers are red and the rest of them are yellow.
Solution:

Question 7.
Malarkodi has 10 oranges. If she ate 4 oranges, what fraction of oranges she was not eaten by her?
Solution:
\(\frac{\text { Oranges not eaten }}{\text { Total oranges }}=\frac{10-4}{10}=\frac{6}{10}\)

Question 8.
After sowing seeds on day one, Muthu observes the growth of two plants and records it. In 10 days, if the first plant grew \(\frac{1}{4}\) th of an inch and the second plant grew \(\frac{3}{8}\) th of an inch, then which plant grew more?
Solution:
Comparing \(\frac{1}{4}\) th of an inch and \(\frac{3}{8}\) th of an inch.
\(\frac{1}{4}=\frac{2}{8}<\frac{3}{8}\)
Second plant grew more.

Try These (Textbook Page No. 57 to 60)

Question 1.
Write the ratio of red tiles to blue tiles and yellow tiles to red tiles.

Solution:
(i) Red tiles to blue tiles = 2 : 3
(ii) Yellow tiles to red tiles = 2 : 2

Question 2.
Write the ratio of blue tiles to that of red tiles and red tiles to that of total tiles.

Solution:
The ratio of blue tiles to red tiles = 3 : 5
The ratio of red tiles to total tiles = 5 : 8

Question 3.
Write the ratio of shaded portion to the unshaded portions in the following shapes.

Solution:
(i) Ratio = 1 : 2
(ii) Ratio = 5 : 4

Question 4.
If the given quantity is in the same unit, put ‘✓’ otherwise put ‘ ✗’ in the table below.

Solution:

Question 5.
Write the ratios in the simplest form and fill in the table.

Solution:

Try These (Textbook Page No. 64)

Question 1.
For the given ratios, find two equivalent ratios and complete the table.

Solution:

Question 2.
Write three equivalent ratios and fill in the boxes.

Solution:

Question 3.
Find the given ratios, find their simplest form and complete the table.

Solution:

Try These (Textbook Page No. 70)

Question 1.
Fill the box by using cross product rule of two ratios
Solution:

By cross product rule we have 1 × □ = 5 × 8
1 × 40 = 40
∴ \(\frac{1}{8}=\frac{5}{40}\)

Question 2.
Use the digits 1 to 9 only once and write as many ratios that are in proportion as possible (For example \(\frac{2}{4}=\frac{3}{6}\))
Solution: