Bhagya

Students can download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.3

Question 1.
Observe the diagram and fill in the blanks.

(i) ‘A’, ‘O’ and ‘B’ are ……… points.
(ii) ‘A’, ‘O’ and ‘C’ are ……….. points.
(iii) ‘A’,‘B’ and ‘C’are ……… points.
(iv) ……… is the point of concurrency.
Solution:
(i) collinear points
Hint: Points on a line.
(ii) non-collinear points
Hint: Points not on a line
(iii) endpoints/non-collinear points
(iv) O is the point of concurrency.
Hint: A points where lines meet

Question 2.
Draw any line and mark any 3 points that are collinear.
Solution:

Question 3.
Draw any line and mark any 4 points that are not collinear.
Solution:

Question 4.
Draw any 3 lines to have a point of concurrency.
Solution:

Question 5.
Draw any 3 lines that are not concurrent. Find the number of points of intersection.
Solution:

Number of points of intersection = 3

Objective Type Questions

Observe the Diagram and give answers

Question 6.
A set of collinear points in the figure are
(a) A, B, C
(b) A, F, C
(c) B, C, D
(d) A, C, D
Solution:
(b) A, F, C
Hint: Collinear points are points on a line.

Question 7.
A set of non-collinear points in the figure are ……….
(a) A, F, C
(b) B, F, D
(c) E, F, G
(d)A,D,C
Solution:
(d) A, D, C

Question 8.
A point of concurrency in the figure is ______
(a) E
(b) F
(c) G
(d) H
Solution:
(b) F

Students can download Maths Chapter 1 Numbers Ex 1.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.3

Miscellaneous Practice Problems

Question 1.
Every even number greater than 2 can be expressed as the sum of two prime numbers. Verify this statement for every even number upto 16.
Solution:
Even numbers greater then 2 upto 16 are 4, 6, 8, 10, 12, 14 and 16
4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7 (or) 5 + 5
12 = 5 + 7
14 = 7 + 7 (or) 3 + 11
16 = 5 + 11 (or) 3 + 13

Question 2.
Is 173 a prime? Why?
Solution:
Yes, 173 is a prime. Because it has only 1 and itself as factors.

Question 3.
For which of the numbers, from n = 2 to 8.
Is 2n – 1 a prime?
Solution:
n = 2 ⇒ 2n – 1 = 2 × 2 – 1
= 4 – 1
= 3 (prime)
n = 3 ⇒ 2n – 1 = 2 × 3 – 1
= 6 – 1
= 5 (prime)
n = 4 ⇒ 2n – 1 = 2 × 4 – 1
= 8 – 1
= 7 (prime)
n = 5 ⇒ 2n – 1 = 2 × 5 – 1
= 10 – 1
= 9 (Not prime)
n = 6 ⇒ 2n – 1 = 2 × 6 – 1
= 12 – 1
= 11 (prime)
n = 7 ⇒ 2n – 1 = 2 × 7 – 1
= 14 – 1
= 13
n = 8 ⇒ 2n – 1 = 2 × 8 – 1
= 16 – 1
= 15 (Not prime)

Question 4.
Explain your answer with the reason for the following statements.
(a) A number is divisible by 9 if it is divisible by 3.
(b) A number is divisible by 6 if it is divisible by 12.
Solution:
(i) False, 42 is divisible by 3 but it is not divisible by 9
(ii) True, 36 is divisible by 12. Also divisible by 6.

Question 5.
Find A as required
(i) The greatest 2 digit number 9 A is divisible by 2.
(ii) The least number 567A is divisible by 3.
(iii) The greatest 3 digit number 9A6 is divisible by 6.
(iv) The number A08 is divisible by 4 and 9.
(v) The number 225A85 is divisible by 11.
Solution:
(i) 98 A = 8
(ii) 5670 A = 0
(iii) 996 A = 9
(iv) 108 A = 1
(v) 225885 A = 8

Question 6.
Numbers divisible by 4 and 6 are divisible by 24. Verify this statement and support your answer with an example.
Solution:
False 12 is divisible by both 4 and 6. But not divisible by 24

Question 7.
The sum of any two successive odd numbers is always divisible by 4. Justify this statement with an example.
Solution:
True 3 + 5 = 8 is divisible by 4.

Question 8.
Find the length of the longest rope that can be used to measure exactly the ropes of length 1 m 20cm, 3m 60 cm and 4 m.
Solution:
1 m 20 cm = 120 cm
3 m 60 cm = 360 cm
4 m = 400 cm
This is a HCF related problem. So, we need to find the HCF of 120,360 and 400.

120 = 2 × 2 × 2 × 3 × 5
360 = 2 × 2 × 2 × 3 × 3 × 5
400 = 2 × 2 × 2 × 2 × 5 × 5
HCF = 2 × 2 × 2 × 5 = 40
The length of the longest rope = 40 cm

Challenge Problems

Question 9.
The sum of three prime numbers is 80. The difference between the two of them is 4. Find the numbers.
Solution:
Given the sum of three prime numbers is 80
The numbers will be one or two-digit prime numbers. Also one of them is 2
Sum of the remaining 2 numbers = 78 [∵ one number must be even]
Also their difference = 4 (given) [Otherwise sum of three odd numbers is odd]
The numbers will be 37 and 41
The required numbers are 2, 37, 41

Question 10.
Find the sum of all the prime numbers between 10 and 20 and check whether that sum is divisible by all the single-digit numbers.
Solution:
Prime numbers between 10 and 20 are 11, 13, 17 and 19
Sum = 11 + 13 + 17 + 19 = 60
60 is divisible by 1, 2, 3, 4, 5 and 6.

Question 11.
Find the smallest number which is exactly divisible by all the numbers from 1 to 9.
Solution:
2520

Question 12.
The product of any three consecutive numbers is always divisible by 6. Justify this statement with an example.
Solution:
Yes. Because one of every two consecutive integers is even and so the product of three consecutive integers is even and divisible by 2.
Also one of every 3 consecutive integers is divisible by 3.
Product of any three consecutive integers is divisible by 6.
Example: 5 × 6 × 7

Question 13.
Malarvizhi, Karthiga, and Anjali are friends and natives of the same village. They work in different places. Malarvizhi comes to her home once in 5 days. Similarly, Karthiga and Anjali come to their homes once in 6 days and 10 days respectively, Assuming that they met each other on the 1st of October, when will all the three meet again?
Solution:
This is an LCM related problem. So, we need to find the LCM of 5, 6, and 10.

LCM = 5 × 2 × 1 × 3 × 1
= 30
All the three will meet again once in 30 days.

Question 14.
In an apartment consisting of 108 floors, two lifts A & B starting from the ground floor, stop at every 3rd and 5th floors respectively. On which floors, will both of them stop together?
Solution:
LCM of 3 and 5 = 3 × 5 = 15
The lifts stop together at floors 15, 30, 45, 60, 75, 90, and 105.

Question 15.
The product of 2 two-digit numbers is 300 and their HCF is 5. What are the numbers?
Solution:
15 × 20 = 300
HCF of 15 and 20 is 5
The numbers are 15 and 20

Question 16.
Find whether the number 564872 is divisible by 88. (use of the test of divisibility rule for 8 and 11 will help)
Solution:
564872 Divisibility by 8
564872 It is divisible by 8
Divisibility by 11
5 + 4 + 7 = 16
6 + 8 + 2 = 16
16 – 16 = 0
It is divisible by both 8 and 11 and hence divisible by 88.

Question 17.
Wilson, Mathan, and Guna can complete one round of a circular track in 10, 15, and 20 minutes respectively. If they start together from at the starting point of 7 am, at what time will they meet together again at the same starting point?
Solution:
This is an LCM related problem. So, we need to find the LCM of 10, 15, and 20.

LCM = 5 × 2 × 1 × 3 × 2 = 60 min
They will meet together again after 60 minutes.
ie. 7.am + 60 minutes = 8 am.

Students can download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.1

Question 1.
Fill in the blanks.
(i) A line through two endpoints ‘A’ and ‘B’ is denoted by ______
(ii) Aline segment from point ‘B’ to point ‘A’ is denoted by ______
(iii) A ray has ______ endpoint(s).
Solution:
(i) \(\overleftrightarrow { AB }\)
(ii) \(\bar { BA } \)
(iii) one

Question 2.
How many line segments are there in the given line? Name them.

Solution:
10, \(\overline { PQ } \), \(\overline { PA } \), \(\overline { PB } \), \(\overline { PC } \), \(\overline { AB } \), \(\overline { BC } \), \(\overline {CQ} \), \(\overline { AQ } \), \(\overline { BQ } \), \(\overline { AC } \).

Question 3.
Measure the following line segments.

Solution:
\(\overline { XY } \) = 2.4 cm, \(\overline { AB } \) = 3.4 cm, \(\overline { EF } \) = 4 cm, \(\overline { PQ } \) = 3 cm.

Question 4.
Construct a line segment using a ruler and compass.
(1) \(\overline { AB } \) = 7.5 cm
(2) \(\overline { CD } \) = 3.6 cm
(3) \(\overline { QR } \) = 10 cm
Solution:
(1)
(i) Draw line 1 and mark a point A on it.
(ii) Measure 7.5 cm using a compass, placing the pointer at ‘O’ and the pencil pointer at 7.5 cm.
(iii) Place the pointer of the compass at A then draw a small arc on the line 1 with the pencil pointer. It cuts line 1 at a point and name that point as B.
(iv) Now \(\overline { AB } \) is the required line segment of length 7.5 cm.

(2)
(i) Draw a line 1 and mark a point C on it.
(ii) Measure 3.6 cm using a compass, placing the pointer at O and the pencil pointer at 3.6 cm.
(iii) Place the pointer of the compass at C then draw a small arc on the line 1 with the pencil pointer. It cuts line 1 at a point and names the point as D.
(iv) Now \(\overline { CD } \) is the required line segment of length 3.6 cm.

(3)
(i) Draw a line 1 and mark a point Q on it.
(ii) Measure 10 cm using compass placing the pointer at O and the pencil pointer at 10 cm.
(iii) Place the pointer of the compass at Q then draw a small arc on the line 1 with the pencil pointer. It cuts the line 1 at a point and name that point as R.
(iv) Now \(\overline { QR }\) is the required line segment of length 10 cm.

Question 5.
From the given figure
(i) identify the parallel lines
(ii) identify the intersecting lines
(iii) name the points of intersection.

Solution:
(i) Parallel lines:
(a) \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{AB}}\)
(b) \(\overrightarrow{\mathrm{EF}} \text { and } \overrightarrow{\mathrm{GH}}\)
(ii) Intersecting lines:
(a) \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{EF}}\)
(b) \(\overrightarrow{\mathrm{CD}} \text { and } \overrightarrow{\mathrm{GH}}\)
(c) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{EF}}\)
(d) \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{GH}}\)
(iii) Point of Intersection:
P, Q, R and S are the points of Intersection.
(iii) P, Q, R and S

Question 6.
From the given figure, name the
(i) parallel lines
(ii) intersecting lines
(iii) points of intersection.

Solution:
(i) \(\overleftrightarrow { CD } \) and \(\overleftrightarrow { EF } \), \(\overleftrightarrow { CD } \) and \(\overleftrightarrow { IJ } \), \(\overleftrightarrow { EF } \) and \(\overleftrightarrow { IJ } \)
(ii) \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { CD } \), \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { EF } \), \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { IJ } \), \(\overleftrightarrow { GH } \) and \(\overleftrightarrow { IJ } \), \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { GH } \)
(iii) P, Q and R

Question 7.
From the given figure, name
(i) all pairs of parallel lines.
(ii) all pairs of intersecting lines.
(iii) pair of lines whose point of intersection is ‘V’.
(iv) point of intersection of the lines ‘l₂‘ and ‘l₃‘.
(v) point of intersection of the lines ‘l₁‘, and ‘l₅

Solution:
(i) Pairs of parallel lines:

• l₃ and l₄
• l₄ and l₅
• l₃ and l₅

(ii) Pairs of intersecting lines:

• l₁ and l₂
• l₁ and l₃
• l₁ and l₄
• l₁ and l₅
• l₂ and l₃
• l₂ and l₄
• l₂ and l₅

(iii) l₁ and l₂ intersect at ‘V’
(iv) point of intersection of the lines ‘l₂‘ and; l₅‘ is ‘Q’
(v) point of intersection of the lines ‘l₁‘ and ‘l₅‘ is ‘U’

Objective Type Questions

Question 8.
The number of line segments in
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

Question 9.
A line is denoted as __________
(a) AB
(b) \(\overrightarrow{AB}\)
(c) \(\overleftrightarrow {AB} \)
(d) \(\overline { AB }\)
Solution:
(c) \(\overleftrightarrow {AB} \)

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

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.1

Question 1.
Fill in the blanks.
(i) The number of prime numbers between 11 and 60 is _______
(ii) The numbers 29 and _______ are twin primes.
(iii) 3753 is divisible by 9 and hence divisible by _______
(iv) The number of distinct prime factors of the smallest 4 digit number is______
(v) The sum of distinct prime factors of 30 is ________
Solution:
(i) 12
(ii) 31
(iii) 3
(iv) 2
(v) 10

Question 2.
Say True or False.
(i) The sum of any number of odd numbers is always even.
(ii) Every natural number is either prime or composite.
(iii) If a number is divisible by 6, then it must be divisible by 3.
(iv) 16254 is divisible by 2, 3, 6, and 9.
(v) The number of distinct prime factors of 105 is 3.
Solution:
(i) False
(ii) False
(iii) True
(iv) True
(v) True

Question 3.
Write the smallest and the biggest two-digit prime number.
Solution:
Smallest two-digit prime number – 11
Biggest two-digit prime number – 97.

Question 4.
Write the smallest and the biggest three-digit composite number.
Solution:
Smallest three-digit composite number – 100
Biggest three-digit composite number – 999

Question 5.
The sum of any three odd natural number is odd. Justify this statement with an example.
Solution:
True.
1 + 3 + 5 = 9 (Odd)

Question 6.
The digits of the prime number 13 can be reversed to get another prime number 31. Find if any such pair exists up to 100.
Solution:
(17, 71), (37, 73) and (79, 97)

Question 7.
Your friend says that every odd number is prime. Give an example to prove him/her wrong.
Solution:
False. 15 is an odd number. But not prime.

Question 8.
Each of the composite numbers has at least three factors. Justify this statement with an example.
Solution:
Factors of 4 are 1, 2, 4

Question 9.
Find the dates of any month in a calendar which are divisible by both 2 and 3.
Solution:
Every month the dates 6, 12, 18, 24, and 30 (excluding February) are divisible by both 2 and 3.

Question 10.
I am a two-digit prime number and the sum of my digits is 10.1 am also one of the factors of 57. Who am I?
Solution:
Two-digit prime numbers with a sum of digits 10 are 19, 37, 73.
of these factors of 57 is 19. the number is 19.

Question 11.
Find the prime factorisation of each number by factor tree method and division method.
(i) 60
(ii) 128
(iii) 144
(iv) 198
(v) 420
(vi) 999
Solution:

60 = 2 × 30 = 2 × 2 × 15
= 2 × 2 × 3 × 5

128 = 2 × 2 × 2 × 2 × 2 ×2 × 2
128 = 2 × 64 = 2 × 2 × 32
= 2 × 2 × 2 × 16
= 2 × 2 × 2 × 2 × 8
128 = 2 × 2 × 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2 × 2 × 2

144 = 2 × 2 × 2 × 2 × 3 × 3
144 = 2 × 72 = 2 × 2 × 36
= 2 × 2 × 2 × 18 = 2 × 2 × 2 × 2 × 9
= 2 × 2 × 2 × 2 × 3 × 3

198 = 2 × 3 × 3 × 11
198 = 2 × 99 = 2 × 3 × 33 = 2 × 3 × 3 × 11

420 = 2 × 210
= 2 × 2 × 105
= 2 × 2 × 3 × 35 = 2 × 2 × 3 × 5 × 7

999 = 3 × 333 = 3 × 3 × 111
= 3 × 3 × 3 × 37

Question 12.
If there are 143 math books to be arranged in equal numbers in all the stacks, then find the number of books in each stack and also the number of stacks.
Solution:

143 = 11 × 13 (11, 13) or (13, 11)
143 = 11 × 13

Question 13.
The difference between two successive odd numbers is
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(b) 2

Question 14.
The only even prime number is
(a) 4
(b) 6
(c) 2
(d) 0
Solution:
(c) 2

Question 15.
Which of the following numbers is not a prime?
(a) 53
(b) 92
(c) 97
(d) 71
Solution:
(b) 92

Question 16.
The sum of the factors of 27 is
(a) 28
(b) 37
(c) 40
(d) 31
Solution:
(c) 40

Question 17.
The factors of the number are 1, 2, 4, 5, 8, 10, 16, 20, 40, and 80. What is the number?
(a) 80
(b) 100
(c) 128
(d) 160
Solution:
(a) 80

Question 18.
The prime factorisation of 60 is 2 × 2 × 3 × 5. Any other number which has the same prime factorisation as 60 is
(a) 30
(b) 120
(c) 90
(d) Impossible
Solution:
(d) Impossible

Question 19.
If the number 6354*97 is divisible by 9 then the value of * is
(a) 2
(b) 4
(c) 6
(d) 7
Solution:
(a) 2

Question 20.
The number 87846 is divisible by
(a) 2 only
(b) 3 only
(c) 11 only
(d) all of these
Solution:
(d) all of these

Students can download Maths Chapter 4 Geometry Ex 4.4 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.4

Miscellaneous Practise Problems

Question 1.
Find the type of lines marked in thick lines (Parallel, intersecting or perpendicular)

Solution:
(i) Parallel lines
(ii) Parallel lines
(iii) Parallel and Perpendicular lines
(iv) Intersecting lines

Question 2.
Find the parallel and intersecting line segments in the picture given below.

Solution:
(a) Parallel line segments

• \(\overline{\mathrm{YZ}} \text { and } \overline{\mathrm{DE}}\)
• \(\overline{\mathrm{EA}} \text { and } \overline{\mathrm{ZV}}\)
• \(\overline{\mathrm{VW}} \text { and } \overline{\mathrm{AB}}\)
• \(\overline{\mathrm{WX}} \text { and } \overline{\mathrm{BC}}\)
• \(\overline{\mathrm{YX}} \text { and } \overline{\mathrm{DC}}\)
• \(\overline{\mathrm{YD}} \text { and } \overline{\mathrm{XC}}\)
• \(\overline{\mathrm{XC}} \text { and } \overline{\mathrm{WB}}\)
• \(\overline{\mathrm{WB}} \text { and } \overline{\mathrm{VA}}\)
• \(\overline{\mathrm{VA}} \text { and } \overline{\mathrm{ZE}}\)
• \(\overline{\mathrm{ZE}} \text { and } \overline{\mathrm{YD}}\)

(b) Intersecting line segments

• DE and ZV
• WX and DC

Question 3.
Name the following angles as shown in the figure.

(i) ∠1 =
(ii) ∠2 =
(iii) ∠3 =
(iv) ∠1 + ∠2 =
(v) ∠2 + ∠3 =
(vi) ∠1 + ∠2 + ∠3 =
Solution:
(i) ∠1 = ∠DBC or ∠CBD
(ii) ∠2 = ∠DBE or ∠EBD
(iii) ∠3 = ∠ABE or ∠EBA
(iv) ∠1 + ∠2 = ∠EBC or ∠CBE
(v) ∠2 + ∠3 = ∠ABD or ∠DHA
(vi) ∠1 + ∠2 + ∠3 = ∠ABC or ∠B or ∠CBA

Question 4.
Measure the angles of the given figures using a protractor and identify the type of angle as acute, obtuse, right or straight.

Solution:
(i) right angle
(ii) acute angle
(iii) straight angle
(iv) obtuse angle

Question 5.
Draw the following angles using the protractor.
(i) 45°
(ii) 120°
(iii) 65°
(iv) 135°
(v) 0°
(vi) 180°
(vii) 38°
(viii) 90°
Solution:

Question 6.
From the figures given below, classify the following pairs of angles into complementary and non-complementary.

Solution:
We know that the two angles are complementary if they add up to 90°.
Therefore (a) (i) is complementary.
In (v) ∠ABC and ∠CBD are complementary
(b) (ii), (iii), (iv) and (v) are non-complementary

Question 7.
From the figures given below, classify the following pairs of angles into supplementary and non supplementary.

Solution:
Ans: (ii) and (iv) are supplementary angles.
(i), and (iii) non-supplementary angles.

Question 8.
From the figure

(i) name a pair of complementary angles.
(ii) name a pair of supplementary angles.
Solution:
(i) ∠FAE; ∠EAD
(ii) ∠FAD; ∠DAC
∠BAC; ∠CAE
∠FAB; ∠BAC
∠FAB; ∠FAE

Question 9.
Find the complementary angle of
(i) 30°
(ii) 26°
(iii) 85°
(iv) 0°
(v) 90°
Solution:

Question 10.
Find the supplementary angle of
(i) 70°
(ii) 35°
(iii) 165°
(iv) 90°
(v) 0°
(vi) 180°
(vii) 95°
Solution:

Challenging Problems

Question 11.
Think and write an object having
(i) Parallel lines (1) ……….. (2) ………. (3) ………..
(ii) Perpendicular lines (1) ……… (2) ……… (3) ………..
(iii) Intersecting lines (1) ……….. (2) ………. (3) ……….
Solution:
(i) Legs of the table, railway tracks, edges of the scale
(ii) Adjacent sides of a Board, Crossbars of windows, Adjacent sides of the textbook
(iii) Crossbars of windows, Ladder, blades of a scissor.

Question 12.
Which angle is equal to twice its complement.
Solution:
Let the angle be x
According to the problem, x = 2 × (90 – x)
x = 180 – 2x
x + 2x = 180
3x = 180
x = \(\frac{180}{3}\)
x = 60
∴ The angle is 60°

Question 13.
Which angle is equal to two-thirds of its supplement.
Solution:
Let the angle be x
According to the problem,
x = \(\frac{2}{3}\) × (180° – x)
3x = 2(180 – x)
3x = 360 – 2x
3x + 2x = 360°
5x = 360°
x = \(\frac{360°}{5}\)
x = 72°
∴ The angle is 72°

Question 14.
Given two angles are supplementary and one angle is 20° more than the other. Find the two angles.
Solution:
Let the angles be x and x + 20°
According to the problem,
x + x + 20 = 180°
2x + 20° = 180°
2x = 180° – 20°
2x = 160°
x = \(\frac{160°}{2}\)
x = 80°
x + 20 = 80° + 20°
= 100°
∴ The two angles are 80° and 100°

Question 15.
Two complementary angles are in ratio 7 : 2. Find the angles.
Solution:
Let the angles be 7x, 2x
According to the problem,
7x + 2x = 90
9x = 90
x = \(\frac{90}{9}\)
x = 10
7x = 7 × 10
= 70
2x = 2 × 10
= 20
∴ Two angles are 70° and 20°

Question 16.
Two supplementary angles are in ratio 5 : 4. Find the angles.
Solution:
Total of two supplementary angles = 180°
Given they are in the ratio 5 : 4
Dividing total angles to 5 + 4 = 9 equal parts.
One angle \(=\frac{5}{9} \times 180=100^{\circ}\)
Another angle \(=\frac{4}{9} \times 180=80^{\circ}\)
Two angles are 100° and 80°.

Students can download Maths Chapter 5 Statistics Ex 5.4 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.4

Miscellaneous Practice Problems

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

Prepare a tally marks table.
Solution:

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

Solution:
Mode of transport of the students
Scale: 1 Unit=100 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 the 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 numbers of moons that orbit each of the planets in oar solar system.

Make a Bar graph for the above data.
Solution:
Number of moons that orbit each of the planets in our solar system
Scale: 1 Unit = 2 moons

Question 5.
The prediction of the weather in the month of September is given below.

(i) Make a frequency table of the types of weather by reading the calendar.
(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:
(i)
(ii) 14 days
(iii) 24 days (30 – 6 = 24 days)
(iv) 10 : 6

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

Represent this data using pictograph.
Solution:
Students responses in an interview about their future profession
Scale: 1 Unit = 1 student

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 a 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 one hour.

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

Solution:
Vans = 50; Buses = 40; Cars = 65; Others = 15
Total Vehicles = 245

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

Draw the bar graph showing the same information.
Solution:
The lengths (in nearest cm) of drumsticks
Scale : 1 Unit = 1 drumstick

Students can download Maths Chapter 4 Geometry Ex 4.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.2

Question 1.
Use any number of the given dots to make different angles.
(i) An Accute angle

(ii) An Obtuse Angle

(iii) A Right Angle

(iv) A Straight Angle

Solution:
(i) An Accute angle

(ii) An Obtuse Angle

(iii) A Right Angle

(iv) A Straight Angle

Question 2.
Name the vertex and sides that form each angle.

Solution:
(i) D, DE and DF
(ii) D, DE and DC
(iii) P, PQ and PR
(iv) S, SV and ST

Question 3.
Pick out the Right angles from the given figures.

Solution:
(i), (iii), (v)

Question 4.
Pick out the Accute angles from the given figures.

Solution:
(i), (iii), (iv) are the Acute Angles.

Question 5.
Pick out the Obtuse angles from the given figures.

Solution:
(i) and (ii) are the Obtuse Angles.

Question 6.
Name the angle in each figure given below in all the possible ways.

Solution:
(i) ∠M or ∠LMN or ∠NML
(ii) ∠Q or ∠PQR or ∠RQP
(iii) ∠N or ∠MNO or ∠ONM
(iv) ∠A or ∠TAS or ∠SAT
(v) ∠Y or ∠XYZ or ∠ZYX
(vi) There are 3 angles in (vi)

• ∠ADC or ∠CDA
• ∠ CDB or ∠BDC
• ∠D or ∠ADB or ∠BDA

Question 7.
Say True or False.
(i) 20° and 70° are complementary.
(ii) 88° and 12° are complementary.
(iii) 80° and 180° are supplementary.
(iv) 0° and 180° are supplementary.
Solution:
(i) True
Hint: 20°+ 70° = 90°
(ii) False
Hint: 88° + 180° = 260° ≠ 1
(iii) False
Hint: 80° + 180° = 260° ≠ 1
(iv) True
Hint: 0° + 180° = 180°

Question 8.
Draw and label each of the angles.
(i) ∠NAS = 90°n
(ii) ∠BIG = 35°
(iii) ∠SMC = 145°
Solution:

Question 9.
Identify the types of angles shown by the hands of the given clock.

Solution:
(i) Obtuse angle
(ii) Zero angle
(iii) Straight angle
(iv) Acute angle
(v) Right angle

Question 10.
Find the supplementary/complementary angles in each case.

Solution:

Objective Type Questions

Question 11.
In this figure, which is not the correct way of naming an angle?

(a) ∠Y
(b) ∠ZXY
(c) ∠ZYX
(d) ∠XYZ
Solution:
(b) ∠ZXY

Question 12.
In this figure, ∠AYZ = 45. If point ‘A’ is shifted to point ‘B’ along the ray, then the measure of ∠BYZ is

(a) more than 45°
(b) 45°
(c) less than 45°
(d) 90°
Solution:
(b) 45°

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.2

Question 1.
What is the slope of a line whose inclination with positive direction of x -axis is
(i) 90°
(ii) 0°
Solution:
Here θ = 90°
Slope (m) = tan θ
Slope = tan 90°
= undefined.

(ii) Here θ = 0°
Slope (m) = tan θ
Slope = tan 0°
= 0

Question 2.
What is the inclination of a line whose slope is
(i) 0
(ii) 1
Solution:
(i) m = 0
tan θ = 0 ⇒ θ = 0°
(ii) m = 1 ⇒ tan θ = tan 45° ⇒ 0 = 45°

Question 3.
Find the slope of a line joining the points
(i) (5,\(\sqrt { 5 }\)) with the origin
(ii) (sin θ, -cos θ) and (-sin θ, cos θ)
Solution:
(i) The given points is (5,\(\sqrt { 5 }\)) and (0, 0)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) = \(\frac{0-\sqrt{5}}{0-5}\)
= \(\frac{\sqrt{5}}{5}=\frac{1}{\sqrt{5}}\)

(ii) The given points is (sin θ, -cos θ) and (-sin θ, cos θ)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{\cos \theta+\cos \theta}{-\sin \theta-\sin \theta}\)
= \(\frac{2 \cos \theta}{-2 \sin \theta}\) = – cot θ

Question 4.
What is the slope of a line perpendicular to the line joining A(5,1) and P where P is the mid-point of the segment joining (4,2) and (-6,4).
Solution:
Mid point of XY = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\) = (\(\frac { 4-6 }{ 2 } \),\(\frac { 2+4 }{ 2 } \))
= (\(\frac { -2 }{ 2 } \),\(\frac { 6 }{ 2 } \)) = (-1, 3)

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\) = (\(\frac { 3-1 }{ -1-5 } \))
= \(\frac { 2 }{ -6 } \) = – \(\frac { 1 }{ 3 } \)

Question 5.
Show that the given points are collinear: (-3, -4), (7,2) and (12, 5)
Solution:
The vertices are A(-3, -4), B(7, 2) and C(12, 5)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { 2+4 }{ 7+3 } \) = \(\frac { 6 }{ 10 } \) = \(\frac { 3 }{ 5 } \)
Slope of BC = \(\frac { 5-2 }{ 12-7 } \) = \(\frac { 3 }{ 5 } \)
Slope of AB = Slope of BC = \(\frac { 3 }{ 5 } \)
∴ The three points A,B,C are collinear.

Question 6.
If the three points (3, -1), (a, 3) and (1, -3) are collinear, find the value of a.
Solution:
The vertices are A(3, -1), B(a, 3) and C(1, -3)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { 3+1 }{ a-3 } \) = \(\frac { 4 }{ a-3 } \)
Slope of BC = \(\frac { 3+3 }{ a-1 } \) = \(\frac { 6 }{ a-1 } \)
Since the three points are collinear.
Slope of AB = Slope BC
\(\frac { 4 }{ a-3 } \) = \(\frac { 6 }{ a-1 } \)
6 (a – 3) = 4 (a – 1)
6a – 18 = 4a – 4
6a – 4a = -4 + 18
2a = 14 ⇒ a = \(\frac { 14 }{ 2 } \) = 7
The value of a = 7

Question 7.
The line through the points (-2, a) and (9,3) has slope –\(\frac { 1 }{ 2 } \) Find the value of a.
Solution:
The given points are (-2, a) and (9, 3)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
– \(\frac { 1 }{ 2 } \) = \(\frac { 3-a }{ 9+2 } \) ⇒ – \(\frac { 1 }{ 2 } \) = \(\frac { 3-a }{ 11 } \)
2(3 – a) = -11 ⇒ 6 – 2a = -11
-2a = -11 – 6 ⇒ -2a = -17 ⇒ a = – \(\frac { 17 }{ 2 } \)
∴ The value of a = \(\frac { 17 }{ 2 } \)

Question 8.
The line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8,12) and (x, 24). Find the value of x.
Solution:
Find the slope of the line joining the point (-2, 6) and (4, 8)
Slope of line (m1) = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 8-6 }{ 4+2 } \) = \(\frac { 2 }{ 6 } \) = \(\frac { 1 }{ 3 } \)
Find the slope of the line joining the points (8, 12) and (x, 24)
Slope of a line (m2) = \(\frac { 24-12 }{ x-8 } \) = \(\frac { 12 }{ x-8 } \)
Since the two lines are perpendicular.
m₁ × m₂ = -1
\(\frac { 1 }{ 3 } \) × \(\frac { 12 }{ x-8 } \) = -1 ⇒ \(\frac{12}{3(x-8)}=-1\)
-1 × 3 (x – 8) = 12
-3x + 24 = 12 ⇒ – 3x = 12 -24
-3x = -12 ⇒ x = \(\frac { 12 }{ 3 } \) = 4
∴ The value of x = 4

Question 9.
Show that the given points form a right angled triangle and check whether they satisfies Pythagoras theorem.
(i) A(1, -4) , B(2, -3) and C(4, -7)
(ii) L(0, 5), M(9,12) and N(3,14)
Solution:
(i) The vertices are A(1, -4), B(2, -3) and C(4, -7)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3+4 }{ 2-1 } \) = \(\frac { 1 }{ 1 } \) = 1
Slope of BC = \(\frac { -7+3 }{ 4-2 } \) = \(\frac { -4 }{ 2 } \) = -2
Slope of AC = \(\frac { -7+4 }{ 4-1 } \) = – \(\frac { 3 }{ 3 } \) = -1
Slope of AB × Slope of AC = 1 × -1 = -1

∴ AB is ⊥^r to AC
∠A = 90°
∴ ABC is a right angle triangle
Verification:

20 = 2 + 18
20 = 20 ⇒ Pythagoras theorem verified

(ii) The vertices are L(0, 5), M(9, 12) and N(3, 14)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of LM = \(\frac { 12-5 }{ 9-0 } \) = \(\frac { 7 }{ 9 } \)

Slope of MN = \(\frac { 14-12 }{ 3-9 } \) = \(\frac { 2 }{ -6 } \) = – \(\frac { 1 }{ 3 } \)
Slope of LN = \(\frac { 14-5 }{ 3-0 } \) = \(\frac { 9 }{ 3 } \) = 3
Slope of MN × Slope of LN = – \(\frac { 1 }{ 3 } \) × 3 = -1
∴ MN ⊥ LN
∠N = 90°
∴ LMN is a right angle triangle
Verification:


130 = 90 + 40
130 = 130 ⇒ Pythagoras theorem is verified

Question 10.
Show that the given points form a parallelogram:
A (2.5,3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5).
Solution:
Let A(2.5, 3.5), B(10, -4), C(2.5, -2.5) and D(-5, 5) are the vertices of a parallelogram.


Slope of AB = Slope of CD = -1
∴ AB is Parallel to CD ……(1)

Slope of BC = Slope of AD
∴ BC is parallel to AD
From (1) and (2) we get ABCD is a parallelogram.

Question 11.
If the points A(2, 2), B(-2, -3), C(1, -3) and D(x, y) form a parallelogram then find the value of x and y.
Solution:
Let A(2, 2), B(-2, -3), C(1, -3) and D(x, y) are the vertices of a parallelogram.

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -3-2 }{ -2-2 } \) = \(\frac { -5 }{ -4 } \) = \(\frac { 5 }{ 4 } \)
Slope of BC = \(\frac { -3+3 }{ -2-1 } \) = \(\frac { 0 }{ -3 } \) = 0
Slope of CD = \(\frac { y+3 }{ x-1 } \)
Slope of AD = \(\frac { y-2 }{ x-2 } \)
Since ABCD is a parallelogram
Slope of AB = Slope of CD
\(\frac { 5 }{ 4 } \) = \(\frac { y+3 }{ x-1 } \)
5(x – 1) = 4 (y + 3)
5x – 5 = 4y + 12
5x – 4y = 12 + 5
5x – 4y = 17 ……(1)
Slope of BC = Slope of AD
0 = \(\frac { y-2 }{ x-2 } \)
y – 2 = 0
y = 2
Substitute the value of y = 2 in (1)
5x – 4(2) = 17
5x -8 = 17 ⇒ 5x = 17 + 18
5x = 25 ⇒ x = \(\frac { 25 }{ 5 } \) = 5
The value of x = 5 and y = 2.

Question 12.
Let A(3, -4), B(9, -4) , C(5, -7) and D(7, -7). Show that ABCD is a trapezium.
Solution:
Let A(3, -4), B(9, -4), C(5, -7) and D(7, -7) are the vertices of a quadrilateral.

Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB = \(\frac { -4+4 }{ 9-3 } \) = \(\frac { 0 }{ 6 } \) = 0
Slope of BC = \(\frac { -7+4 }{ 5-9 } \) = \(\frac { -3 }{ -4 } \) = \(\frac { 3 }{ 4 } \)
Slope of CD = \(\frac { -7+7 }{ 7-5 } \) = \(\frac { 0 }{ 2 } \) = 0
Slope of AD = \(\frac { -7+4 }{ 7-3 } \) = \(\frac { -3 }{ 4 } \) = – \(\frac { 3 }{ 4 } \)
The slope of AB and CD are equal.
∴ AB is parallel to CD. Similarly the slope of AD and BC are not equal.
∴ AD and BC are not parallel.
∴ The Quadrilateral ABCD is a trapezium.

Question 13.
A quadrilateral has vertices at A(-4, -2), B(5, -1) , C(6, 5) and D(-7, 6). Show that the mid-points of its sides form a parallelogram.
Solution:
Let A(-4, -2), B(5, -1), C(6, 5) and D(-7, 6) are the vertices of a quadrilateral.





Slope of EF = Slope of GH = \(\frac { 7 }{ 10 } \)
∴ EF || GH …….(1)
Slope of FG= Slope of EH = – \(\frac { 7 }{ 12 } \)
∴ FG || EH ……(2)
From (1) and (2) we get EFGH is a parallelogram.
The mid point of the sides of the Quadrilateral ABCD is a Parallelogram.

Question 14.
PQRS is a rhombus. Its diagonals PR and QS intersect at the point M and satisfy QS = 2PR. If the coordinates of S and M are (1, 1) and (2, -1) respectively, find the coordinates of P.
Solution:
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of SM = \(\frac { 1+1 }{ 1-2 } \) = \(\frac { 2 }{ -1 } \) = -2
Slope of PM = \(\frac { 1 }{ 2 } \) (Since SM and PM are ⊥^r)
Let the point p be (a,b)
Slope of PM = \(\frac { 1 }{ 2 } \)
\(\frac { b+1 }{ a-2 } \) = \(\frac { 1 }{ 2 } \) ⇒ a – 2 = 2b + 2

a – 2b = 4
a = 4 + 2b ……(1)
Given QS = 2PR
\(\frac { QS }{ 2 } \) = PR
∴ SM = PR
SM = 2PM (PR = 2PM)

Squaring on both sides

∴ (b + 1)² = \(\frac { 1 }{ 4 } \) ⇒ b + 1 = ± \(\frac { 1 }{ 2 } \)
b = \(\frac { 1 }{ 2 } \) – 1 (or) b = – \(\frac { 1 }{ 2 } \) – 1
= – \(\frac { 1 }{ 2 } \) – 1 (or) b = –\(\frac { 1 }{ 2 } \) – 1
= – \(\frac { 1 }{ 2 } \) (or) – \(\frac { 3 }{ 2 } \)
a = 4 + 2b
a = 4 + 2 (\(\frac { -1 }{ 2 } \))
a = 3
a = 4 + 2 (\(\frac { -3 }{ 2 } \))
a = 4 – 3
a = 1
The point of p is (3,\(\frac { -1 }{ 2 } \)) (or) (1,\(\frac { -3 }{ 2 } \))

Students can download Maths Chapter 5 Information Processing Ex 5.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 5 Information Processing Ex 5.2

Miscellaneous Practice Problems

Question 1.
Write the missing numbers in the trees.

Solution:

Question 2.
Write the missing operations in the trees.

Solution:

Question 3.
Check whether the Tree diagrams are equal or not.

Solution:
c ÷ (a ÷ b), a ÷ (b ÷ c) Not equal

Challenge Problems

Question 4.
Convert ti e following questions into tree diagrams:
(i) The number of people who visited a library in the last 5 months were 1210, 2100, 2550, 3160 and 3310. Draw the tree diagram of the total number of people who had used the library for the 5 months.
(ii) Ram had a bank deposit of Rs. 7,55,250 and he had withdrawn Rs. 5,34,500 for educational purpose. Find the amount left in his account. Draw a tree diagram for this.
(iii) In a cycle factory, 1,600 bicycles were manufactured on a day. Draw tree diagram to find the number of bicycle produced in 20 days.
(iv) A company with 30 employees decided to distribute Rs. 90, 000 as a special bonus equally among its employees. Draw tree diagram to show how much will each receive?
Solution:

Question 5.
Write the numerical expression which gives the answer 10 and also convert into tree diagram.
Solution:

Question 6.
Use brackets in appropriate place to the expression 3 x 8 – 5 which gives 19 and convert it into tree diagram for it.
Solution:

= 3 × 8 – 5
= (3 × 8) – 5 = 24 – 5
= 19

Question 7.
A football team gains 3 and 4 points for successive 2 days and loses 5 points on the third day. Find the total points scored by the team and also represent this in tree diagram.
Solution:

Students can download Maths Chapter 6 Information Processing Ex 6.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 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:
One of the answers is.

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
(ii) 5
(iii) 17, 19, 20, 21, 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:

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

Solution:

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: