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 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:

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 4 Geometry Intext Questions

Try These (Textbook Page No. 80, 81)

Question 1.
Name all the line segments.

Solution:
\(\overline{\mathrm{AB}}, \overline{\mathrm{AE}}, \overline{\mathrm{EB}}, \overline{\mathrm{CD}}, \overline{\mathrm{CE}} \text { and } \overline{\mathrm{ED}}\)

Question 2.
If AB = 5 cm, say which of the following measures are correct in fig 4.9.

Solution:
Fig 4.9(i) and fig 4.9(ii) measures are correct.

Try These (Textbook Page No. 85)

Question 1.

1. Name the rays in the given figure.
2. What is the common point of all these rays?
Solution:
1. \(\overrightarrow{\mathrm{TA}}, \overrightarrow{\mathrm{TB}}, \overrightarrow{\mathrm{TC}} \text { and } \overrightarrow{\mathrm{TD}}\) are the rays given
2. Point T is the common point of all these rays.

Try These (Textbook Page No. 90, 95)

Question 1.
Which direction will you face if you start facing West and take three right turns clockwise?
Solution:
Will be facing South.

Question 2.
Which direction will you face if you start facing North and take two right turns anticlockwise?
Solution:
Will be facing South.

Question 3.
Adjust the hands of the clock for following time, note the angle made between the hour hand and the minute hand and write the type of angle.


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 1 Chapter 4 Geometry Additional Questions

Question 1.
From the given figure, name the parallel lines

Solution:
(i) Parallel Lines:
\(\overrightarrow{\mathrm{CD}}\) and \(\overrightarrow{\mathrm{EF}}\) ; \(\overrightarrow{\mathrm{CD}}\) and \(\overrightarrow{\mathrm{IJ}}\) ; \(\overrightarrow{\mathrm{EF}}\) and \(\overrightarrow{\mathrm{IJ}}\) are parallel lines.
(ii) Intersecting lines:
(a) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{CD}}\)
(b) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{EF}}\)
(c) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{GH}}\)
(d) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{IJ}}\)
(e) \(\overrightarrow{\mathrm{GH}} \text { and } \overrightarrow{\mathrm{IJ}}\)
(iii) Points of Intersection:
P, Q and R are the points of intersection.

Question 2.
(a) Name the line segments in the figure.
(b) Is Q, the endpoint of each line segment?

Solution:
(a) \(\overline{\mathrm{QP}} \text { and } \overline{\mathrm{QR}}\) are the line segments
(b) Yes, Q is the end point of each line segment

Question 3.
How many lines can pass through
(a) one given point
(b) two given points.
Solution:

(a) An infinite number of lines can pass through one given point.
(b) Exactly one and only one line can pass through two given points.

Question 4.
A line contains how many points?
(a) minimum?
(b) maximum?
Solution:
(a) A line contains a minimum of two points.
(b) A line contain a maximum of infinitely many points.

Question 5.
Write the (a) maximum and (b) the minimum number of point of intersection of three lines.
Solution:

Maximum – 3 points of intersection
Minimum – No point of intersection

Fill in the blanks.

Question 6.
Complementary angle of 20° is _____
Solution:
70°

Question 7.
The supplementary angle of 90° is _____
Solution:
90°

Question 8.
78°, 12°, ______
Solution:
Complementary angle

Answer the following question.

Question 9.
∠ABD =?

Solution:
On Sum of complementary angles = 90°
∠ABC = 90°
∠CBD = 30°
∠ABD = ∠ABC – ∠DBC = 90° – 30° = 60°
∠ABD = 60°
Complementary angle of 30° = 60°

Question 10.
In the following figure, name the angles.

Solution:
∠AOB, ∠BOZ, ∠AOZ

Question 11.
Write the alternate name of the angle ∠XYZ in the given figure.

Solution:
∠Y or ∠ZYX

Question 12.
Draw the diagram of two angles having only one common point.

Solution:
∠COD and ∠AOB have the point ‘O’ in common

Question 13.
What are the supplementary and complementary angles of 60°?
Solution:
Supplementary angle is 120°
Complementary angle is 30°

Question 14.
How many lines can you draw passing through three collinear points? Draw the figure also.
Solution:
Only one.

Question 15.
Write the maximum number of lines that can pass through a single point.
Solution:
Infinite.

Question 16.
Use a protractor to draw an angle 45°.
Solution:

Construction:
1. Drawn the base ray PQ.
2. Placed the centre of the protractor at the vertex P. Lined up the ray \(\overrightarrow{\mathrm{PQ}}\) with the 0° line. Then drawn and labelled a pointed (R) at the 45° mark on the inner scale (a) anticlockwise and (b) outer scale (clockwise)
3. Removed the protractor and drawn at \(\overrightarrow{\mathrm{PR}}\) to complete the angle
Now ∠P = ∠QPR = ∠RPQ = 45°.

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 1 Numbers Additional Questions

Answer the following questions.

Question 1.
How many thousands are there in 1 lakhs?
Solution:
\(\frac{1,00,0000}{1000}\) = 100 Thousands

Question 2.
The difference between successor and predecessor of any number is 2. Is it true? Justify your answer.
Solution:
It is true that the difference between successor and predecessor of any number is 2.
Because the difference between any number and its predecessor is 1.
Also the difference between the number and its successor is 1.
The total difference is 2.

Question 3.
The expanded form of the number 6,00,001 is given as 6 × 100000 + 1 × 1. Can you write like this Comment.
Solution:
Yes. We can write the expansion of the number 600001 as 6 × 100000 + 1 × 1.
Because 6 × 100000 + 1 × 1 = 600000 + 1 = 600001

Question 4.
Write the greatest five digit number using the digits 2, 3, 4, 0 and 7.
Solution:
Greatest five digit number = 74320

Question 5.
Can you write the least five digit number using the digits 2,3,4,0 and 7 as 02347. Why? What will be the correct number?
Solution:
No, we cannot write the least five digit number using the digits 2, 3, 4, 0 and 7 as 02347. If it is 02347, the left most zero has no value. It becomes 4 digit number 2347.
The correct number will be 20347.

Question 6.
Write the relation between Largest two digit number and Smallest three digit number.
Solution:
Largest two digit number + 1 = Smallest three digit number.
99 + 1 = 100

Question 7.
Name the property being illustrated in each of the cases.

1. (30 + 20) + 10 = 30 + (20 + 10)
2. 10 × 35 = (10 × 30) + (10 × 5)

Solution:

1. Associativity
2. Distribution of multiplication over addition.

Question 8.
10 crore = ____
Solution:
100 million

Question 9.
The heights of five boys in class VI are 135, 141, 129, 132, 145 (in centimetres) in height. Arrange their heights as how they stand in the assembly?
Solution:
129 cm < 132 cm < 135 cm < 141 cm < 145 cm

Question 10.
The number lock has the password number with 3 digits. The number is the least even number and less than 200. Middle digit has no value separately. Find the password. The digits are used only once.
Solution:
102

Question 11.
Arrange in ascending order. 123456, 123546, 123623, 123511
Solution:
123456 < 123511 < 123546 < 123623 Question 12. Arrange in descending order. 8461, 7535, 2943, 6214 Solution: 8461 > 7535 > 6214 > 2943

Question 13.
Find the numbers between 572634 and 562634 which is approximated to ten thousand place.
Solution:
562634 < 570000 < 572634

Question 14.
Evaluate the following:
(a) 44 ÷ 2 + (7 + 80 ÷ 10) – 14 + 23
(b) 17 × 6 – 4 – 2 + 20 – (22 + 18)
(c) 16 × 144 ÷ 16 ÷ 9 + 16 + 15 – 20
(d) 12 × 36 ÷ 12 ÷ 3 + 5 + 6 – 2
(e) 15 – [17 + 30 ÷ 6 – (6 + 6) + 7]
Solution:
(a) 44 ÷ 2 + (7 + 80 ÷ 10) – 14 + 23 (Given)
= 44 ÷ 2 + (7 + 8) – 14 + 23 (To complete the bracket ÷ done first)
= 44 ÷ 2 + 15 – 14 + 23 (Bracket completed second)
= 22 + 15 – 14 + 23 (÷ completed third)
= 37 – 37 (+ completed fourth)
= 0 (- completed last)
∴ 44 ÷ 2 + (7 + 80 ÷ 10) – 14+ 23 = 0.

(b) 17 × 6 – 4 – 2 + 20 – (22 + 18) (Given)
= 17 × 6 – 4 – 2 + 20 – 40 (Bracket completed first)
= 102 – 4 – 2 + 20 – 40 (× completed second)
= 102 – 4 – 22 – 40 (+ completed third)
= 98 – 22 – 40 (÷ completed one by one)
= 76 – 40
= 36
∴ 17 × 6 – 4 – 2+ 20 – (22 + 18) = 36

(c) 16 × 144 ÷ 16 ÷ 9 + 16 + 15 – 20 (Given)
= 16 × 9 ÷ 9+16 + 15 – 20 (÷ completed first)
= 16 × 1 + 16 + 15 – 20 (÷ completed second)
= 16 + 16 + 15 – 20 (× completed third)
= 32 + 15 – 20 (+ completed fourth)
= 47 – 20 (+ completed fifth)
= 27 (- completed last)
∴ 16 × 144 ÷ 16 ÷ 9 + 16 + 15 – 20 = 27

(d) 12 × 36 ÷ 12 ÷ 3 + 5 + 6 – 2 (Given)
= 12 × 3 ÷ 3 + 5 + 6 – 2 (÷ completed first)
= 12 × 1 + 5 + 6 – 2 (÷ completed second)
= 12 + 5 + 6 – 2 (× completed third)
= 17 + 6 – 2 (+ completed forth)
= 23 – 2 (+ completed fifth)
= 21 (- completed last)
∴ 12 × 36 ÷ 12 ÷ 3 + 5 + 6 – 2 = 21

(e) 15 – [17 + 30 ÷ 6 – (6 + 6) + 7] (Given)
= 15 – [17 + 30 ÷ 6 – 12 + 7] (Inner bracket completed first)
= 15 – [17 + 5 – 12 + 7] (÷ completed second)
= 15 – [22 – 19] (+ completed third)
= 15 – 3 (bracket completed forth)
= 12 (- completed last)
∴ 15 – [17 + 30 ÷ 6 – (6 + 6) + 7] = 12.

Question 15.
An export company produced 235219 shirts, 158342 trousers and 11704 jackets in a year. What is the total production of all the three items in that year?
Solution:
Number of shirts produced = 235219
Number of trousers produced = 158342
Number of jackets produced = 11704
Total production of all items = 405265
Total production of all items in that year = 4,05,265

Question 16.
India’s population has been steadily increasing from 439 million in 1961 to 1028 million in 2001. Find the total increase in population from 1961 to 2001. Write the increase in population in the Indian system of Numeration using commas suitably.
Solution:
Population of India in 1961 = 439 millions = 439,000,000
Population of India in 2001 = 1028 millions = 1,028,000,000
Increase in population from 1961 to 2001 = Population in 2001 – Population in 1961
= 1028000000 – 439000000
= 589000000
= 589 million.
Increase in population in Indian System = 58,90,00,000

Question 17.
A person had ₹ 10,00,000 with him. He purchased a flat for ₹ 8,70,000. With the remaining money, he has to buy a T.V. for 1 lakh. How much money was left with him to buy a T.V?
Solution:
Total money the person had = ₹ 10,00,000
Cost of flat = ₹ 8,70,000
Remaining money = ₹ 1,30,000
Now he has ₹ 1,30,000. So it is enough to buy a TV for ₹ 1,00,000.

Question 18.
A box contains 50 packets of biscuits, each weighing 120g. How many such boxes can be loaded in a van, which cannot carry more than 900 kg?
Solution:
Given: Total number of packets = 50.
Weight of each packet = 120 g
Weight of a box = 50 × 120 g = 6000 g = 6 kg [∵ 1000 g = 1 kg]
Required number of boxes = \(\frac{900}{6}\) = 150.
150 boxes are required.

Question 19.
How much money was collected from 5342 students for a charity show, if each student contributed ₹ 670?
Solution:
Total number of students = 5342
Contribution of each student = ₹ 670
Total money collected = 5342 × 670 = ₹ 35,79,140
Total money collected = ₹ 35,79,140

Question 20.
Estimate the following to the nearest hundreds
(a) 439 + 334 + 4317
(b) 1,08,734 – 47,599
(c) 8325 – 491
(d) 4,89,348 – 48,365
Solution:
(a) 439 + 334 + 4317
439 ⇒ 400
334 ⇒ 300
4317 ⇒ 4300
Sum = 5,000

(b) 1,08,734 – 47,599
1,08,734 ⇒ 1,08,700
47,599 ⇒ 47,600
Difference = 61,100

(c) 8325 – 491
8325 ⇒ 8300
491 ⇒ 500
Differences = 7,800

(d) 4,89,348 – 48,365
4,89,348 ⇒ 4,89,300
48,365 ⇒ 48,400
Difference = 4,40,900

Question 21.
Estimate the following products:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Solution:
(a) 578 × 161
578 ⇒ 600
161 ⇒ 200
Estimated product is 600 × 200 = 1,20,000

(b) 5281 × 3491
5281 ⇒ 5000
3491 ⇒ 3500
Estimated Product = 5000 × 3500 = 1,75,00,000

(c) 1291 × 592
1291 ⇒ 1300
592 ⇒ 600
Estimated Product is = 1300 × 600 = 7,80,000

(d) 9250 × 29
9250 ⇒ 9000
29 ⇒ 30
Estimated Product is 9000 × 30 = 2,70,000

Question 22.
Are all whole numbers are natural numbers? Justify your answer?
Solution:
No, all whole numbers are not natural numbers.
Because ‘0’ belongs to the whole number system. But it is not in a natural number system.
All whole numbers except ‘0’ are natural numbers.

Question 23.
Use associative property of addition to add 847 + 306 + 453
Solution:
847 + 306 + 453
= (847 + 453) + 306
= 1300 + 306
= 1606
∴ 847 + 306 + 453 = 1606

Question 24.
Find the value of (1063 × 127) – (1063 × 27)
Solution:
(1063 × 127) – (1063 × 27)
= 1063 (127 – 27) [Taking 1063 as common]
= 1063 × 100
= 106300.
i.e (1063 × 127) – (1063 × 27) = 106300

Question 25.
Find the product using suitable properties
(a) 738 × 103
(b) 1005 × 168
Solution:
(a) We have 738 × 103
= 738 × (100 + 3)
= 738 × 100 + 738 × 3 [By distributive property of multiplication over addition]
= 73800 + 2214
= 76014

(b) 1005 × 168
= (1000 + 5) × 168
= (168 × (1000 + 5) (By commutative property)
= (168 × 1000) + (168 × 5)
= 1,68,000 + 840
= 1,68,840

Question 26.
Write the largest six-digit number and write the number names in words using the Indian and International system.
Solution:
The largest six-digit number is 999999
Number names are nine lakh ninety-nine thousand nine hundred and ninety-nine

Question 27.
In a mobile store, the number of mobiles sold during a month is 1250, Assuming that the same number of mobiles are sold every month, find the number of mobiles sold in 2 years.
Solution:
Number of mobiles sold in 1 month = 1250
1 year = 12 months
2 years = 2 × 12 = 24 months
Number of mobiles sold in 24 months = 1250 × 24= 30,000
Number of mobiles sold in 2 years = 30,000

Question 28.
Simplify 24 + 2 × 8 ÷ 2 – 1
Solution:
24 + 2 × 8 ÷ 2 – 1
= 24 + 2 × 4 – 1
= 24 + 8 – 1
= 32 – 1
= 31

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 2 Introduction to Algebra Additional

Questions

Question 1.
Additive identity ____
Solution:
0

Question 2.
Multiplicative identity ____
Solution:
1

Question 3.
Express to an algebraic statement.
(i) ‘t’ is added to 100
(ii) 4 less to 9 times of y.
Solution:
(i) t + 100
(ii) 9y – 4

Question 4.
Find the rule which gives the number of sticks in the following pattern.

Solution:
Let ‘x’ be the no. of R’s formed.

The rule is 6x.
Let y ’ be the no. of S’s formed.

The rule is 5y.

Question 5.
How old was Suja 6 years from now?
Solution:
Let Suja’s present age be ‘a’ years.
6 years from now Suja will be (a + 6) years old.

Question 6.
Price of Apple per kg is ₹ 50 more than price of orange per kg. What is the cost of Apple per kg?
Solution:
Let the price of orange be ₹ b
Price of Apple will be ₹ (b + 50)

Question 7.
Given ‘n’ students like ice cream. What may 2n show?
Solution:
2n shows double the number of students who like ice cream.

Question 8.
Price of oil per litre is ₹ 5 more than three times the price of cool drinks ₹ ‘p’ Express algebraically.
Solution:
Price of cool drinks per kg = ₹ p
Three times = 3p
5 Rs. more = 3p + 5
Price of oil per kg = ₹ (3p + 5)

Question 9.
Complete the table and by inspection of the table find the value of m when m + 10 = 16.

Solution:

From the table m+ 10 = 16 when m = 6.

Question 10.
Express algebraically (a) y divided by r (b) double times x is subtracted from 10
Solution:
(a) \(\frac{y}{r}\)
(b) 10 – 2x

Question 11.
Give verbal expression of
(a) 7x + 18
(b) \(\frac{4 x}{3}\)
Solution:
(a) 18 added to 7 times x
(b) 4 times x divided by 3.

Question 12.
Rajini’s Father’s age is 5 years more than 3 times Rajini’s age. What is her father’s age?
Solution:
3x + 5

Question 13.

Find the rule for the above pattern.
Solution:
2p

Question 14.
Prepare a table for 3x + 10. From the table find the value of x when 3x + 10 = 25.
Solution:
5

Question 15.
Complete the table and find the solution of the equation \(\frac{z}{3}=4\) using the table.

Solution:

Question 16.
Form the expression for which Ramu is 3 years younger than Mathu.
Solution:
m – 3

Question 17.
A tap is to be pasted along the edges of a square shaped gift box. Its length is 4 cm. What is the length of tap needed for one side.
Solution:
\(\frac{4 p}{4}=p\)

Question 18.
The value of y in 7y – 20 = 99.
Solution:
y = 17

Question 19.
Nine added to two times x gives 301. Find the value of x.
Solution:
x = 146

Question 20.
Aarthi is 3 years younger to Harini. If the sum of their ages is 23, how old is Harini?
Solution:
Let Harini’s age be x years
Aarthi’s age is x – 3 years
Given sum of their ages is 23.
i.e., x + (x – 3) = 23