Bhagya

Students can download Maths Chapter 5 Information Processing Ex 5.1 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 5 Information Processing Ex 5.1

Question 1.
Convert the following numerical expressions into Tree diagrams
(i) 8 + (6 × 2)
(ii) 9 – (2 × 3)
(iii) (3 × 5) – (4 – 2)
(iv) [(2 × 4) + 2] × (8 – 2)]
(v) [(6 + 4) × 7] – [2 × (10 – 5)]
(vi) [(4 × 3) – 2] + [8 × (5 – 3)]
Solution:

Question 2.
Convert the following tree diagrams into numerical expressions.

Solution:
(i) The numerical Expression is 9 × 8
(ii) The numerical expression is (7 + 6) – 5
(iii) The numerical expression is (8 + 2) – (6 + 1)
(iv) The numerical expression is (5 × 6) – (10 ÷ 2)

Question 3.
Convert the following algebraic expressions into tree diagrams.
(i) 10 v
(ii) 3a – b
(iii) 5x + y
(iv) 20t × p
(v) 2(a + b)
(vi) (x × y) – (y × z)
(vii) 4x + 5y
(viii) (Im – n) ÷ (pq + r)
Solution:

Question 4.
Convert Tree diagrams into Algebraic expressions.

Solution:
(i) Algebraic Expression is p + q
(ii) Algebraic Expression is l – m
(iii) Algebraic Expression is (a × b) – c (or) (ab) – c
(iv) Algebraic Expression is (a + b) – (c + d)
(v) Algebraic Expression is (8 ÷ a) + [ (6 ÷ 4) + 3]

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 2 Chapter 4 Geometry Ex 4.3

Miscellaneous Practice Problems

Question 1.
What are the angles of an isosceles right-angled triangle?
Solution:
Since it is a right-angled triangle
One of the angles is 90°
The other two angles are equal because it is an isosceles triangle.
The other two angles must be 45° and 45°
Angles are 90°, 45°, 45°.

Question 2.
Which of the following correctly describes the given triangle?

(a) It is a right isosceles triangle
(b) It is an acute isosceles triangle
(c) It is an obtuse isosceles triangle
(d) It is an obtuse scalene triangle
Solution:
(c) It is an obtuse isosceles triangle

Question 3.
Which of the following is not possible?
(a) An obtuse isosceles triangle.
(b) An acute isosceles triangle.
(c) An obtuse equilateral triangle.
(d) An acute equilateral triangle.
Solution:
(c) An obtuse equilateral triangle.

Question 4.
If one angle of an isosceles triangle is 124°, then find the other angles
Solution:
In an isosceles triangle, any two sides are equal. Also, the two angles are equal.
Sum of three angles of a triangle = 180°
Given one angle = 124°
Sum of other two angles = 180° – 124° = 56°
Other angles are = \(\frac{56}{2}\) = 28°
28° and 28°.

Question 5.
The diagram shows a square ABCD. If the line segment joins A and C, then mention the type of triangle so formed.
Solution:

Isosceles right-angled triangles

Question 6.
Draw a line segment AB of length 6 cm. At each end of this line segment AB, draw a line perpendicular to the line segment AB. Are these lines parallel?
Solution:

yes, they are parallel

Challenge Problems

Question 7.
Is a triangle possible with the angles 90°, 90°, and 0°, Why?
Solution:
No, a triangle cannot have more than one right angle.

Question 8.
Which of the following statements is true? Why?
(a) Every equilateral triangle is an isosceles triangle.
(b) Every isosceles triangle is an equilateral triangle
Solution:
“(a)” is true, because an isosceles triangle need not have three equal sides

Question 9.
If one angle of an isosceles triangle is 70°, then find the possibilities for the other two angles.
Solution:
(i) Given one angle = 70°
Also, it is an isosceles triangle.
Another one angle also can be 70°.
Sum of these two angles = 70° + 70° = 140°
We know that the sum of three angles in a triangle = 180°.
Third angle = 180° – 140° = 40°
One possibility is 70°, 70°, and 40°
(ii) Also if one angle is 70°
Sum of other two angles = 180° – 70° = 110°
Both are equal. They are \(\frac{110}{2}\) = 55°.
Another possibility is 70°, 55° and 55°.

Question 10.
Which of the following can be the sides of an isosceles triangle?
(a) 6 cm, 3 cm, 3 cm
(b) 5 cm, 2 cm, 2 cm
(c) 6 cm, 6 cm, 7 cm
(d) 4 cm, 4 cm, 8 cm
Solution:
(c) 6 cm, 6 cm, 7 cm

Question 11.
Study the given figure and identify the following triangles,

(a) equilateral triangle
(b) isosceles triangle
(c) scalene triangles
(d) acute triangle
(e) obtuse triangle
(f) right triangle
Solution:(a) BC = 1 + 1 + 1 + 1 = 4 cm
AB = AC = 4 cm
∆ABC is an equilateral triangle.
(b) ∆ABC and ∆AEF are isosceles triangles.
Since AB = AC = 4 cm Also AE = AF.
(c) In a scalene triangle, no two sides are equal.
∆AEB, ∆AED, ∆ADF, ∆AFC, ∆ABD, ∆ADC, ∆ABF, and ∆AEC are scalene triangles.
(d) In an acute-angled triangle all the three angles are less than 90°.
∆ABC, ∆AEF, ∆ABF, and ∆AEC are acute-angled triangles.
(e) In an obtuse-angled triangle any one of the angles is greater than 90°.
∆AEB and ∆AFC are obtuse angled triangles.
(f) In a right triangle, one of the angles is 90°.
∆ADB, ∆ADC, ∆ADE, and ∆ADF are right-angled triangles.

Question 12.
Two sides of the triangle are given in the table. Find the third side of the triangle.

Solution:
(i) between 3 and 11
(ii) between 0 and 16
(iii) between 4 and 11
(iv) between 4 and 24

Question 13.
Complete the following table:

Solution:
(i) Always acute angles
(ii) Acute angle
(iii) Obtuse angle

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 2 Chapter 4 Geometry Ex 4.1

Question 1.
Fill in the blanks.
(a) Every triangle has at least _____ acute angles.
(b) A triangle in which none of the sides equal is called a _____.
(c) In an isosceles triangle ______ angles are equal.
(d) The sum of three angles of a triangle is ______.
(e) A right-angled triangle with two equal sides is called ______.
Solution:
(a) Two
(b) Scalene Triangle
(c) Two
(d) 180°
(e) Isosceles right-angled triangle

Question 2.
Match the following

Solution:
(i) Scalene triangle
(ii) Right-angled triangle
(iii) Obtuse angled triangle
(iv) Isosceles triangle
(v) Equilateral triangle

Question 3.
In ΔABC, name the

a) Three sides: ……….., ………., ……….
b) Three Angles: ………., ………, ……….
c) Three Vertices: ………., ………., ……….
Solution:
(a) \(\overline{\mathrm{AB}}, \overline{\mathrm{BC}}, \overline{\mathrm{CA}}\)
(b) ∠ABC, ∠BCA, ∠CAB or ∠A, ∠B, ∠C
(c) A, B, C

Question 4.
Classify the given triangles based on its sides as scalene, isosceles or equilateral.

Solution:
(i) Equilateral triangle
(ii) Scalene triangle
(iii) Isosceles triangle
(iv) Scalene triangle

Question 5.
Classify the given triangles based on their angles as acute-angled, right-angled, or obtuse-angled.

Solution:
(i) Acute angled triangle
(ii) Right-angled triangle
(iii) Obtuse angled triangle
(iv) Acute angled triangle

Question 6.
Classify the following triangles based on its sides and angles?

Solution:
(i) Isosceles Acute angled triangle
(ii) Scalene Right-angled triangle
(iii) Isosceles Obtuse angled triangle
(iv) Isosceles Right-angled triangle
(v) Equilateral Acute angled triangle
(vi) Scalene Obtuse angled triangle

Question 7.
Can a triangle be formed with the following sides? If yes, name the type of triangle.
(i) 8 cm, 6 cm, 4 cm
(ii) 10 cm, 8 cm, 5 cm
(iii) 6.2 cm, 1.3 cm, 3.5 cm
(iv) 6 cm, 6 cm, 4 cm
(v) 3.5 cm, 3.5 cm, 3.5 cm
(vi) 9 cm, 4 cm, 5 cm
Solution:
(i) Sum of two smaller sides of the triangle
= 6 + 4 = 10 cm > 8 cm
It is greater than the third side. So, a triangle can be formed a scalene triangle.

(ii) Sum of two smaller sides of the triangle
= 8 + 5 = 13 cm > 10 cm
It is greater than the third side. So, a triangle can be formed a scalene triangle.

(iii) Sum of two smaller sides of the triangle
= 1.3 + 3.5 = 4.8 cm < 6.2 cm
It is not greater than the third side. So, a triangle cannot be formed.

(iv) Two sides are equal.
So, a triangle can be formed. Isosceles triangle.

(v) Three sides are equal.
So, a triangle can be formed equilateral triangle.

(vi) Sum of two smaller sides of the triangle
= 4 + 5 = 9 cm = 9 cm
It is equal to the third side. No, a triangle cannot be formed.

Question 8.
Can a triangle be formed with the following angles? If yes, name the type of triangle.
(i) 60°, 60°, 60°
(ii) 60°, 40°, 42°
(iii) 90°, 55°, 35°
(iv) 60°, 90°, 90°
(v) 70°, 60°, 50°
(vi) 100°, 50°, 30°
Solution:
(i) 60°, 60°, 60°
Sum of the angles = 60°+ 60°+ 60° = 180°
A triangle can be formed as an acute-angled triangle.

(ii) 60°, 40°, 42°
Sum of the angles = 60° + 40° + 42° = 142°
A triangle cannot be formed.

(iii) 90°, 50°, 35°
Sum of the angles = 90° + 55° + 35° = 180°
A triangle can be formed Right-angled triangle.

(iv) 60°, 90°, 90°
No, a triangle can not be formed.
A triangle cannot have more than one right angle.

(v) 70°, 60°, 50°
Sum of the angles = 70° + 60° + 50° = 180°
A triangle can be formed as an acute-angled triangle.

(vi) 100°, 50°, 30°
Sum of the angles = 100° + 50° + 30° = 180°
A triangle can be formed as an obtuse-angled triangle.

Question 9.
Two angles of the triangles are given. Find the third angle
(i) 80°, 60°
(ii) 75°, 35°
(iii) 52°, 68°
(iv) 50°, 90°
(v) 120°, 30°
(vi) 55°, 85°
Solution:
(i) 80°, 60°
Let the third angle be x.
Sum of the angles = 180°
80° + 60° + x = 180°
140 + x = 180°
x = 180°- 140°
x = 40°
Third angle = 40°

(ii) 52°, 68°
Let the third angle be x.
Sum of the angles = 180°
52° + 68° + x = 180°
120 + x = 180°
180° – 120°
x = 60°
Third angle = 60°

(iii) 75°, 35°
Let the third angle be x.
Sum of the angles 180°
75° + 35° + x = 180°
110 + x = 180°
x = 180° – 110°
x = 70°
Third angle = 70°

(iv) 50°, 90°
Let the third angle be x. Sum of the angles = 180°
50° + 90° + x = 180°
140 + x = 180°
x = 180° – 140°
x = 40°
Third angle = 40°

(v) 120°, 30°
Let the third angle be x.
Sum of the angles = 180°
120° + 30° + x = 180°
150 + x = 180°
x = 180° – 150°
x = 30°
Third angle = 30°

(vi) 55°, 85°
Let the third angle be x.
Sum of the angles = 180°
55° + 85° + x = 180°
140 + x = 180°
x = 180° – 140°
x = 40°
Third angle = 40°

Question 10.
I am a closed figure with each of my three angles is 60°. Who am I?
Solution:
Equilateral triangle

Question 11.
Using the given information, write the type of triangle in the table given below.

Solution:
(ii) Acute angled triangle, Isosceles triangle
(iii) Right-angled triangle, Isosceles triangle
(iv) Acute angled triangle, Scalene triangle
(v) Acute angled triangle, Scalene triangle
(vi) Right-angled triangle. Scalene triangle
(vii) Obtuse angled triangle, Scalene triangle
(viii) Obtuse angled triangle, Isosceles triangle

Objective Type Questions

Question 12.
The given triangle is ……….

(a) a right-angled triangle
b) an equilateral triangle
(c) a scalene triangle
(d) an obtuse-angled triangle
Solution:
(b) an equilateral triangle

Question 13.
If all angles of a triangle are less than a right angle, then it is called _____.
(a) an obtuse-angled triangle
(b) a right-angled triangle
(c) an isosceles right-angled triangle
(d) an acute-angled triangle
Solution:
(d) an acute-angled triangle

Question 14.
If two sides of a triangle are 5 cm and 9 cm, then the third side is
(a) 5 cm
(b) 3 cm
(c) 4 cm
(d) 14 cm
Solution:
(a) 5 cm

Question 15.
The angles of a right-angled triangle are
(a) acute, acute, obtuse
(c) acute, right, right
(c) right, obtuse, acute
(d) acute, acute, right
Solution:
(d) acute, acute, right

Question 16.
An equilateral triangle is
(a) an obtuse-angled triangle
(b) a right-angled triangle
(c) an acute-angled triangle
(d) a scalene triangle
Solution:
(c) an acute-angled triangle

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Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 3 Bill, Profit and Loss Ex 3.2

Miscellaneous Practice Problems

Question 1.
A Shopkeeper buys three articles for Rs 325, Rs 450, and Rs 510. He is able to sell them for Rs 350, Rs 425, and Rs 525 respectively. Find the gain or loss to the shopkeeper on the whole.
Solution:
C.P of three articles = 325 + 450 + 510 = ₹ 1285
S.P of three articles = 350 + 425 + 525 = ₹ 1,300
Here S.P > C.P
Profit = S.P – C.P = 1,300 – 1285 = ₹ 15
The shopkeeper gained = ₹ 15

Question 2.
A stationery shop owner bought a scientific calculator for ₹ 750. He had put a battery worth ₹ 100 in it. He had spent ₹ 50 for its outer pouch. He was able to sell it at ₹ 850. Find his profit or loss.
Solution:
Cost of the scientific calculator = ₹ 750
Cost of its battery = ₹ 100
Cost of outer pouch = ₹ 50
Cost Price of the calculator = ₹ 750 + ₹ 100 + ₹ 50 = ₹ 900
S.P = ₹ 850
Here S.P > C.P
Loss = C.P – S.P = 900 – 850 = ₹ 50
Loss = ₹ 50

Question 3.
Nathan paid Rs 800 and bought 10 bottles of honey from a village vendor. He sold them in a gain for Rs 100 per bottle. Find his profit or loss.
Solution:
C.P of 10 bottles of honey = ₹ 800
C.P of 1 bottle honey = 800/10 = ₹ 80
S.P of a bottle honey = ₹ 100
Here S.P > C.P
Profit per bottle = ₹ 100 – ₹ 80 = ₹ 20
Profit for 10 bottles = 20 × 10 = ₹ 200
Profit = ₹ 200

Question 4.
A man bought 400 metres of cloth for ₹ 60,000 and sold it at the rate of ₹ 400 per metre. Find his profit or loss.
Solution:
C.P of 400 metres of cloth = ₹ 60,000
S.P per metre = ₹ 400
S.P of 400 metres of cloth = 400 × 400 = ₹ 1,60,000
Here S.P > C.P
Profit = C.P – S.P = 1,60,000 – 60,000 = ₹ 1,00,000

Challenge Problems

Question 5.
A fruit seller bought 2 dozen bananas at Rs 20 a dozen and sold them at Rs 3 per banana. Find his gain or loss.
Solution:
Cost of one dozen banana = ₹ 20
Cost of 2 dozen bananas = ₹ 20 × 2 = ₹ 40
C.P = ₹ 40
S.P per banana = ₹ 3
S.P for 2 dozen banana = ₹ 3 × 24 = ₹ 72
Here S.P > C.P
Profit = S.P – C.P = 72 – 40 = 32
Profit = ₹ 32

Question 6.
A store purchased pens at ₹ 216 per dozen. He paid ₹ 58 for conveyance and sold the pens at the discount of n per pen and made an overall profit of ₹ 50. Find the M.P of each pen.
Solution:
Cost of a dozen pens = ₹ 216
Paid for conveyance = ₹ 58
Cost price of 12 pens = 216 + 58 = ₹ 274 [∵ 1 dozen = 12]
Profit of 12 pens = ₹ 50
Profit = S.P – C.P
⇒ 50 = S.P – 274
⇒ S.P = 50 + 274 = ₹ 324
Also discount allowed per pen = ₹ 2
Discount for 12 pens = 2 × 12 = ₹ 24
S.P = M.P – Discount
⇒ 324 = M.P – 24
⇒ M.P = 324 + 24 = ₹ 348
Marked price for 12 pens = ₹ 348
M.P of a pen = \(\frac { 348 }{ 12 }\) = ₹ 29
M.P per pen = ₹ 29

Question 7.
A Vegetable vendor buys 10 kg of tomatoes per day at Rs 10 per kg, for the first three days of a week. 1 kg of tomatoes got smashed every day for those 3 days. For the remaining 4 days of the week, he buys 15 kg of tomatoes daily per kg. If for the entire week he sells tomatoes at Rs 20 per kg, then find his profit or loss for the week.
Solution:
Total tomatoes bought for 3 days = 3 × 10 = 30 kg
Cost of 1 kg = ₹ 10
Cost of 30kg tomatoes = 30 × 10 = ₹ 300
Total tomatoes bought for other 4 days = 4 × 15 = 60 kg
Cost of 1 kg = ₹ 8
Cost of 60 kg tomatoes = 60 × 8 = ₹ 480
Total cost of 90 kg tomatoes = 300 + 480 = ₹ 780
C.P = ₹ 780
Tomatoes smashed = 3 kg
Total kg of Tomatoes for sale = 90 – 3 = 87 kg
S.P of 1 kg tomatoes = ₹ 20
S.P of 87 kg tomatoes = 87 × 20 = ₹ 1740
Here S.P > C.P
Profit = S.P – C.P = 1740 – 780 = ₹ 960
Profit = ₹ 960

Question 8.
An electrician buys a used T.V for ₹ 12,000 and a used Fridge for ₹ 11,000. After spending ₹ 1000 on repairing the T.V and ₹ 1500 on painting the Fridge, he fixes up the M.P of T.V as ₹ 15,000 and that of the Fridge as ₹ 15,500. If he gives each ₹ 1000 discount oh each find his profit or loss.
Solution:
(i) Cost of a T.V = ₹ 12,000
Paid for repair = ₹ 1,000
C.P of the T.V = 12,000 + 1000 = ₹ 13,000
M.P of the T.V = ₹ 15,000
Discount on a TV = ₹ 1000
S.P = M.P – Discount = 15,000 – 1000 = ₹ 14,000
Here S.P > C.P
Profit = S.P – C.P = 14,000 – 13,000 = ₹ 1,000
Profit on the T.V = ₹ 1,000
(ii) Cost of the Fridge = ₹ 11,000
Painting charge = ₹ 1500
C.P of the Fridge = 11000 + 1500 = ₹ 12,500
M.P of the Fridge = ₹ 15,500
Discount allowed = ₹ 1000
S.P = M.P – Discount = ₹ 15,500 – ₹ 1000 = ₹ 14,500
Here also S.P > C.P
Profit = ₹ 14,500 – ₹ 12,500 = ₹ 2000
Total profit = Profit on T.V + Profit on Fridge = ₹ 1000 + ₹ 2000 = ₹ 3000
Profit = ₹ 3000

Students can download Maths Chapter 2 Measurements Ex 2.3 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 2 Measurements Ex 2.3

Miscellaneous practice problems

Question 1.
Two pipes whose lengths are 7 m 25 cm and 8 m 13 cm joined by welding and then a small piece 60 cm is cut from the whole. What is the remaining length of the pipe?
Solution:
Total length = 7 m 25 cm + 8 m 13 cm = 15 m 38 cm (or) 1538 cm
length detatched = 60 cm
Remaining length = 14 m 78 cm

Question 2.
The saplings are planted at a distance of 2 m 50 cm in the road of length 5 km by Saravanan. If he has 2560 saplings, how many saplings will be planted by him? how many saplings are left?
Solution:
Distance between two saplings = 2 m 50 cm = 250 cm
Total length of the road = 5000 m = 500000 cm

Question 3.
Put ✓ a mark in the circles which adds upto the given measure.

Solution:

Question 4.
Make a calendar for the month of February 2020. (Hint: January 1st, 2020 is Wednesday)
Solution:
February 2020 is a leap year

Question 5.
Observe and Collect the data for a minute

Solution:
Do your self.

Challenge Problems

Question 6.
A squirrel wants to eat the grains quickly. Help the Squirrel to find the shortest way to reach the grains. (Use your scale to measure the length of the line segments)

Solution:

The Shortest way to reach the grains is the AGFKE path.

Question 7.
A room has a door whose measures are 1 m wide and 2 m 50 cm high.
Can we make a bed of 2 m and 20 cm in length and 90 cm wide into the room?
Solution:
Measures of the door length 2 m 50 cm width and lm (100 cm)
(i) Measures of the bed 2 m 20 cm width and 90 cm
Measures of the bed < measures of the door
∴ Yes, we can take the bed into the room.

Question 8.
A post office functions from 10 a.m. to 5.45 p.m. with a lunch break of 1 hour. If the post office works for 6 days a week, find the total duration of working hours in a week.
Solution:
Working hours in a day = 6 hrs 45 min
= (6 × 60 min) + 45 min
= (360 + 45) min
= 405 min
Total duration of working hours in a week
= 6 × 405 min
= 2430 min
= \(\frac{2430}{60}\)
= \(\frac{810}{20}\)
= 40 \(\frac{1}{2}\)
= 40 hours 30 minutes

Question 9.
Seetha wakes up at 5.20 a.m. She spends 35 minutes to get ready and travels 15 minutes to reach the railway station. If the train departs exactly at 6.00 am, will Seetha catch the train?
Solution:
No, Seetha will not catch the train
Time at Seetha wakes up = 5.20 am
Time is taken for getting ready = 35 min
Travelling time to station = 15 min
Reporting time = 5.20 am + 50 min = 6.10 am
But, the train departs exactly at 6.00 am
So, Seetha will not catch the train.

Question 10.
A doctor advised Vairavan to take one tablet every 6 hours once on the 1st day and once every 8 hours on the 2nd and 3rd day. If he starts to take 9.30 am the first dose. Prepare a time chart to take the tablet in railway time.
Solution:

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.3 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 5.3

Question 1.
Find the equation of a straight line passing through the mid-point of a line segment joining the points (1, -5), (4, 2) and parallel to (i) X axis (ii) Y axis
Solution:
Mid point of the line joining to points (1, -5), (4, 2)
Mid point of the line = (\(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\))
= (\(\frac { 1+4 }{ 2 } \),\(\frac { -5+2 }{ 2 } \)) = (\(\frac { 5 }{ 2 } \),\(\frac { -3 }{ 2 } \))

(i) Any line parallel to X-axis. Slope of a line is 0.
Equation of a line is y – y₁ = m (x – x₁)
y + \(\frac { 3 }{ 2 } \) = 0 (x – \(\frac { 5 }{ 2 } \))
y + \(\frac { 3 }{ 2 } \) = 0 ⇒ \(\frac { 2y+3 }{ 2 } \) = 0
2y + 3 = 0

(ii) Equation of a line parallel to Y-axis is
x = \(\frac { 5 }{ 2 } \) ⇒ 2x = 5
2x – 5 = 0

Question 2.
The equation of a straight line is 2(x – y) + 5 = 0. Find its slope, inclination and intercept on the Y axis.
Solution:
Equation of a line is
2 (x – y) + 5 = 0
2 x – 2y + 5 = 0
-2y = -2x – 5
2y = 2x + 5
y = \(\frac { 2x }{ 2 } \) + \(\frac { 5 }{ 2 } \)
y = x + \(\frac { 5 }{ 2 } \)
Slope of line = 1
Y intercept = \(\frac { 5 }{ 2 } \)
tan θ = 1
tan θ = tan 45°
∴ angle of inclination = 45°

Question 3.
Find the equation of a line whose inclination is 30° and making an intercept -3 on the Y axis.
Solution:
Angle of inclination = 30°
Slope of a line = tan 30°
(m) = \(\frac{1}{\sqrt{3}}\)
y intercept (c) = -3
Equation of a line is y = mx + c
y = \(\frac{1}{\sqrt{3}}\) x – 3
\(\sqrt { 3 }\) y = x – 3 \(\sqrt { 3 }\)
∴ x – \(\sqrt { 3 }\) y – 3 \(\sqrt { 3 }\) = 0

Question 4.
Find the slope and y intercept of \(\sqrt { 3x }\) + (1 – \(\sqrt { 3 }\))y = 3.
Solution:
The equation of a line is \(\sqrt { 3 }\)x + (1 – \(\sqrt { 3 }\))y = 3
(1 – \(\sqrt { 3 }\))y = \(\sqrt { 3 }\) x + 3

Question 5.
Find the value of ‘a’, if the line through (-2,3) and (8,5) is perpendicular to y = ax + 2
Solution:
Given points are (-2, 3) and (8, 5)
Slope of a line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 5-3 }{ 8+2 } \) = \(\frac { 2 }{ 10 } \) = \(\frac { 1 }{ 5 } \)
Slope of a line y = ax + 2 is “a”
Since two lines are ⊥^r
m₁ × m₂ = -1
\(\frac { 1 }{ 5 } \) × a = -1 ⇒ \(\frac { a }{ 5 } \) = -1 ⇒ a = -5
∴ The value of a = -5

Question 6.
The hill in the form of a right triangle has its foot at (19,3). The inclination of the hill to the ground is 45°. Find the equation of the hill joining the foot and top.
Solution:
Slope of AB (m) = tan 45°
= 1
Equation of the hill joining the foot and the top is
y – y₁ = m(x – x₁)
y – 3 = 1 (x – 19)

y – 3 = x – 19
– x + y – 3 + 19 = 0
– x + y + 16 = 0
x – y – 16 = 0
The required equation is x – y – 16 = 0

Question 7.
Find the equation of a line through the given pair of points.
(i) (2,\(\frac { 2 }{ 3 } \)) and (\(\frac { -1 }{ 2 } \),-2)
(ii) (2,3) and (-7,-1)
Solution:
(i) Equation of the line passing through the point (2,\(\frac { 2 }{ 3 } \)) and (\(\frac { -1 }{ 2 } \),-2)


-5(3y – 2) = -8 × 2 (x – 2)
-15y + 10= -16 (x – 2)
-15y + 10= -16x + 32
16x – 15y + 10 – 32 = 0
16x – 15y – 22 = 0
The required equation is 16x – 15y – 22 = 0

(ii) Equation of the line joining the point (2, 3) and (-7, -1) is

-9 (y – 3) = -4 (x – 2)
-9y + 27 = – 4x + 8
4x – 9y + 27 – 8 = 0
4x – 9y + 19 = 0
The required equation is 4x – 9y + 19 = 0

Question 8.
A cat is located at the point(-6, -4) in xy plane. A bottle of milk is kept at (5,11). The cat wish to consume the milk travelling through shortest possible distance. Find the equation of the path it needs to take its milk.
Solution:

Equation of the line joining the point is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
\(\frac { y+4 }{ 11+4 } \) = \(\frac { x+6 }{ 5+6 } \)
\(\frac { y+4 }{ 15 } \) = \(\frac { x+6 }{ 11 } \)
15(x + 6) = 11(y + 4)
15x + 90 = 11y + 44
15x – 11y + 90 – 44 = 0
15x – 11y + 46 = 0
The equation of the path is 15x – 11y + 46 = 0

Question 9.
Find the equation of the median and altitude of AABC through A where the vertices are A(6,2), B(-5, -1) and C(1,9).
Solution:
(i) To find median


Equation of the median AD is

2(x – 6) = -8 (y – 2)
2x – 12= -8y + 16
2x + 8y – 28 = 0
(÷ by 2) x + 4y – 14 = 0
∴ Equation of the median is x + 4y – 14 = 0
Equation of the altitude is 3x + 5y – 28 = 0

(ii) To find the equation of the altitude

Slope of BC = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac { 9+1 }{ 1+5 } \)
= \(\frac { 10 }{ 6 } \) = \(\frac { 5 }{ 3 } \)
Slope of the altitude = – \(\frac { 3 }{ 5 } \)
Equation of the altitude AD is
y – y1 = m (x – x₁)
y – 2 = – \(\frac { 3 }{ 5 } \) (x – 6)
-3 (x – 6) = 5 (y – 2)
-3x + 18 = 5y – 10
-3x – 5y + 18 + 10 = 0
-3x – 5y + 28 = 0
3x + 5y – 28 = 0

Question 10.
Find the equation of a straight line which has slope \(\frac { -5 }{ 4 } \) and passing through the point (-1,2).
Solution:
Slope of a line (m) = \(\frac { -5 }{ 4 } \)
The given point (x₁, y₁) = (-1, 2)
Equation of a line is y – y₁ = m (x – x₁)
y – 2= \(\frac { -5 }{ 4 } \) (x + 1)
5(x + 1) = -4(y – 2)
5x +5 = -4y + 8
5x + 4y + 5 – 8 = 0
5x + 4y – 3 = 0
The equation of a line is 5x + 4y – 3 = 0

Question 11.
You are downloading a song. The percenty (in decimal form) of mega bytes remaining to get downloaded in x seconds is given by y = -0.1x + 1.
(i) graph the equation.
(ii) find the total MB of the song.
(iii) after how many seconds will 75% of the song gets downloaded?
(iv) after how many seconds the song will be downloaded completely?
Solution:
(i) y = – 0. 1x + 1

(ii) y = -0.1x + 1
x → seconds
y → Mega byte of the song.
The total mega byte of the song is at the beginning that is when x = 0
y = 1 mega byte

(iii) When 75% of song gets downloaded 25% remains.
In other words y = 25%
y = 0.25
By using the equation we get
0.25 = -0.1x + 1
0.1 x = 0.75
x = \(\frac { 0.75 }{ 0.1 } \) = 7.5 seconds

(iv) When song is downloaded completely the remaining percent is zero.
i.e y = 0
0 = -0.1 x + 1
0.1 x = 1 second
x = \(\frac { 1 }{ 0.1 } \) = \(\frac { 1 }{ 1 } \) × 10
x = 10 seconds

Question 12.
Find the equation of a line whose intercepts on the x and y axes are given below.
(i) 4, -6
(ii) -5, – 4
Solution:
(i) x intercept (a) = 4; y intercept (b) = – 6
Equation of a line is \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
\(\frac { x }{ 4 } \) + \(\frac { y }{ -6 } \) = 1 ⇒ \(\frac { x }{ 4 } \) – \(\frac { y }{ 6 } \) = 1
(LCM of 4 and 6 is 12)
3x – 2y = 12
3x – 2y – 12 = 0
The equation of a line is 3x – 2y – 12 = 0

(ii) x intercept (a) = -5; y intercept (b) = \(\frac { 3 }{ 4 } \)
Equation of a line is \(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1 ⇒ \(\frac{x}{-5}+\frac{y}{\frac{3}{4}}=1\)
\(\frac { x }{ -5 } \) + \(\frac { 4y }{ 3 } \) = 1
(LCM of 5 and 3 is 15)
– 3x + 20y = 15
– 3x + 20y – 15 = 0
3x – 20y + 15 = 0
∴ Equation of a line is 3x – 20y + 15 = 0

Question 13.
Find the intercepts made by the following lines on the coordinate axes.
(i) 3x – 2y – 6 = 0
(ii) 4x + 3y + 12 = 0
Solution:
(i) 3x – 2y – 6 =0.
x intercept when y = 0
⇒ 3x – 6 = 0
⇒ x = 2
y intercept when x = 0
⇒ 0 – 2y – 6 = 0
⇒ y = -3
(ii) 4x + 3y + 12 = 0. x intercept when y = 0
⇒ 4x + 0 + 12 = 0
⇒ x = -3
y intercept when x = 0
⇒ 0 + 3y + 12 = 0
⇒ y = -4

Question 14.
Find the equation of a straight line.
(i) passing through (1,-4) and has intercepts which are in the ratio 2 : 5
(ii) passing through (-8, 4) and making equal intercepts on the coordinate axes.
Solution:
(i) Let the x-intercept be 2a and the y intercept 5 a
The equation of a line is
\(\frac { x }{ a } \) + \(\frac { y }{ a } \) = 1 ⇒ \(\frac { x }{ 2a } \) + \(\frac { y }{ 5a } \) = 1
The line passes through the point (1, -4)
\(\frac { 1 }{ 2a } \) + \(\frac { (-4) }{ 5a } \) = 1 ⇒ \(\frac { 1 }{ 2a } \) – \(\frac { 4 }{ 5a } \) = 1
Multiply by 10a
(L.C.M of 2a and 5a is 10a)
5 – 8 = 10a ⇒ -3 = 10a
a = \(\frac { -3 }{ 10 } \)
The equation of the line is

Multiply by 3
-5x – 2y = 3 ⇒ -5x – 2y – 3 = 0
5x + 2y + 3 = 0
The equation of a line is 5x + 2y + 3 = 0

(ii) Let the x-intercept andy intercept “a”
The equation of a line is
\(\frac { x }{ a } \) + \(\frac { y }{ a } \) = 1
The line passes through the point (-8, 4)
\(\frac { -8 }{ a } \) + \(\frac { 4 }{ a } \) = 1
\(\frac { -8+4 }{ a } \) = 1
-4 = a
The equation of a line is
\(\frac { x }{ -4 } \) + \(\frac { y }{ -4 } \) = 1
Multiply by -4
x + y = -4
x + y + 4 = 0
The equation of the line is x + y + 4 = 0

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Multiple Choice Questions

Question 1.
The value of sin2 θ + \(\frac{1}{1+\tan ^{2} \theta}\) is equal to ………………
(1) tan² θ
(2) 1
(3) cot² θ
(4) 0
Answer:
(2) 1
Hint:
sin² θ + \(\frac{1}{1+\tan ^{2} \theta}\) = sin² θ + \(\frac{1}{\sec ^{2} \theta}\) = sin² θ + cos² θ = 1

Question 2.
tan θ cosec² θ – tan θ is equal to ………………
(1) sec θ
(2) cot² θ
(3) sin θ
(4) cot θ
Answer:
(4) cot θ
Hint:
tan θ cosec² θ – tan θ = tan θ (cosec² θ – 1)
= tan θ × cot² θ = \(\frac{1}{\cot \theta}\) × cot² θ = cot θ

Question 3.
If (sin α + cosec α)² + (cos α + sec α)² = k + tan² α + cot² α, then the value of k is equal to
(1) 9
(2) 7
(3) 5
(4) 3
Solution:
(2) 7
(sin α + cos α)² + (cos α + sec α)²
= sin² α + cosec² α + 2 sin α cosec α + cos² α + sec² α + 2 cos α sec α
= 1 + cosec² α + 2 + sec² α + 2
= 1 + cot² α + 1 + 2 + tan² α + 1 + 2
= 7 + tan² α + cot² α
k = 7

Question 4.
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b (a2 – 1) is equal to ……………
(1) 2 a
(2) 3 a
(3) 0
(4) 2 ab
Answer:
(1) 2 a
Hint:
b (a² – 1) = (sec θ + cosec θ) [(sin θ + cos θ)² – 1]
= \(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\) [sin² θ + cos² θ + 2 sin θ cos θ – 1]

Question 5.
If 5x = sec θ and \(\frac { 5 }{ x } \) = tan θ, then x2 – \(\frac{1}{x^{2}}\) is equal to …………….
(1) 25
(2) \(\frac { 1 }{ 25 } \)
(3) 5
(4) 1
Answer:
(2) \(\frac { 1 }{ 25 } \)
Hint:



Question 6.
If sin θ = cos θ , then 2 tan² θ + sin² θ – 1 is equal to ………………
(1) \(\frac { -3 }{ 2 } \)
(2) \(\frac { 3 }{ 2 } \)
(3) \(\frac { 2 }{ 3 } \)
(4) \(\frac { -2 }{ 3 } \)
Answer:
(2) \(\frac { 3 }{ 2 } \)
Hint:

Question 7.
If x = a tan θ and y = b sec θ then …………..
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
(2) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)
(3) \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0\)
(4) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=0\)
Answer:
(1) \(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\)
Hint:
x = a tan θ
\(\frac { x }{ a } \) = tan θ
\(\frac{x^{2}}{a^{2}}\) = tan² θ
\(\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}\) = sec² θ – tan² θ = 1
y = b sec θ
\(\frac{y}{b}\) = sec θ
\(\frac{y^{2}}{b^{2}}\) = sec² θ

Question 8.
(1 + tan θ + sec θ) (1 + cot θ – cosec θ) is equal to ……………
(1) 0
(2) 1
(3) 2
(4) -1
Answer:
(3) 2
Hint:
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)

Question 9.
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p² – q² is equal to
(1) a² – b²
(2) b² – a²
(3) a² + b²
(4) b-a
Solution:
(2) b² – a²
(a cot θ + b cosec θ)² = p²
(b cot θ + a cosec θ )² = q²
p² – q² = a² cost²θ + a² cot²θ + 2ab cot θ cosec θ – (b²cot²θ + a² cosec²θ + 2ab cot θ cosec θ) = (a² – b²) cot²θ + (b² – a²)cosec²θ = (a² – b²) (cosec²θ – 1) + (b² – a²) (cosec²θ)
= (a² – b²)cosec²θ – (a² – b²) – (a² – b²) cosec²θ
= b² – a²

Question 10.
If the ratio of the height of a tower and the length of its shadow is \(\sqrt { 3 }\) : 1, then the angle of elevation of the sun has a measure
(1) 45°
(2) 30°
(3) 90°
(4) 60°
Answer:
(4) 60°
Hint:
Ratio of length of the tower : length of the shadow = \(\sqrt { 3 }\) : 1

Let the tower be \(\sqrt { 3 }\) x and the shadow be x
tan C = \(\frac { AB }{ BC } \) ⇒ tan C = \(\frac{\sqrt{3} x}{x}\) = \(\sqrt { 3 }\)
tan C = tan 60° ⇒ ∴ ∠C = 60°

Question 11.
The electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point ‘6’ metres above the first, the depression of the foot of the tower is 60° . The height of the tower (in metres) is equal to ……………
(1) \(\sqrt { 3 }\) b
(2) \(\frac { b }{ 3 } \)
(3) \(\frac { b }{ 2 } \)
(4) \(\frac{b}{\sqrt{3}}\)
Answer:
(3) \(\frac { b }{ 2 } \)
Hint:
Let the height of the pole BC be h
AC = b + h
Let CD be x
In the right ∆ BCD, tan 30° = \(\frac { BC }{ AB } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h }{ x } \)
x = \(\sqrt { 3 }\) h ………. (1)

In the right ∆ ACD, tan 60° = \(\frac { AC }{ CD } \)
\(\sqrt { 3 }\) = \(\frac { b+h }{ x } \)
x = \(\frac{b+h}{\sqrt{3}}\) ………(2)
From (1) and (2) we get
\(\sqrt { 3 }\) h = \(\frac{b+h}{\sqrt{3}}\) ⇒ 3 h = b + h
2 h = b ⇒ h = \(\frac { b }{ 2 } \)

Question 12.
A tower is 60 m height. Its shadow is x metres shorter when the sun’s altitude is 45° than when it has been 30° , then x is equal to
(1) 41. 92 m
(2) 43. 92 m
(3) 43 m
(4) 45. 6 m
Answer:
(2) 43. 92 m
Hint:
In the right ∆ ABC, tan 30° = \(\frac { AB }{ BC } \) = \(\frac { 60 }{ x+y } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { 60 }{ x+y } \) ⇒ x + y = 60 \(\sqrt { 3 }\)

y = 60 \(\sqrt { 3 }\) – x …….(1)
In the right ∆ ABD, tan 45° = \(\frac { AB }{ BD } \)
1 = \(\frac { 60 }{ y } \) ⇒ y = 60 ………..(2)
From (1) and (2) we get
60 = 60 \(\sqrt { 3 }\) – x
x = 60 \(\sqrt { 3 }\) – 60 = 60 (\(\sqrt { 3 }\) – 1) = 60 (1.732 – 1)
= 60 × 0.732
x = 43.92 m

Question 13.
The angle of depression of the top and bottom of 20 m tall building from the top of a multistoried building are 30° and 60° respectively. The height of the multistoried building and the distance between two buildings (in metres) is …………….
(1) 20,10\(\sqrt { 3 }\)
(2) 30, 5 \(\sqrt { 3 }\)
(3) 20, 10
(4) 30, 10\(\sqrt { 3 }\)
Answer:
(4) 30, 10\(\sqrt { 3 }\)
Hint:
Let the height of the multistoried building AB be “h”
AE = h – 20
Let BC be x
In the right ∆ ABC, tan 60° = \(\frac { AB }{ BC } \) ⇒ \(\sqrt { 3 }\) = \(\frac { h }{ x } \)
x = \(\frac{h}{\sqrt{3}}\) ………..(1)

In the right ∆ ABC, tan 30° = \(\frac { AE }{ ED } \) = \(\frac { h-20 }{ x } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h-20 }{ x } \)
\(\frac{1}{\sqrt{3}}\) = \(\frac { h-20 }{ x } \)
x = (h – 20) \(\sqrt { 3 }\) ………(2)
From (1) and (2) we get,
\(\frac{h}{\sqrt{3}}\) = (h – 20) \(\sqrt { 3 }\)
h = 3h – 60 ⇒ 60 = 2 h
h = \(\frac { 60 }{ 2 } \) = 30
Distance between the building (x) = \(\frac{h}{\sqrt{3}}=\frac{30}{\sqrt{3}}=\frac{30 \sqrt{3}}{3}=10 \sqrt{3}\)

Question 14.
Two persons are standing ‘x’ metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is ……………….
(1) \(\sqrt { 2 }\)x
(2) \(\frac{x}{2 \sqrt{2}}\)
(3) \(\frac{x}{\sqrt{2}}\)
(4) 2 x
Answer:
(2) \(\frac{x}{2 \sqrt{2}}\)
Hint:
Consider the height of the 2^nd person ED be “h”
Height of the second person is 2 h
C is the mid point of BD
In the right ∆ ABC, tan θ = \(\frac { AB }{ BC } \)

Question 15.
The angle of elevation of a cloud from a point h metres above a lake is β . The angle of depression of its reflection in the lake is 45° . The height of the location of the cloud from the lake is ………….

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Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.4

Question 1.
Solve the following systems of linear equations by Cramer’s rule:
(i) 5x – 2y + 16 = 0, x + 3y – 7 = 0
Solution:

(ii) \(\frac{3}{x}\) + 2y =12, \(\frac{2}{x}\) + 3y = 13
Solution:
Let \(\frac{1}{x}\) = a
3a + 2b = 12
2a + 3b = 13

(iii) 3x + 3y – z = 1 1, 2x – y + 2z = 9, 4x + 3y + 2z = 25
Solution:
Δ = \(\left| \begin{matrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{matrix} \right| \)
= 3(-2 – 6) -3 (4 – 8) -1(6 + 4)
= 3(-8) -3(-4) -1(10)
= -24 + 12 – 10 = -22 ≠ 0
Δ_x = \(\left| \begin{matrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{matrix} \right| \)
= 11 (-2 – 6) – 3(18 – 50) – 1(27 + 25)
= 11(-8) -3(32) -1(52)
= -88 + 96 – 52 = -44
Δ_y = \(\left| \begin{matrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{matrix} \right| \)
= 3(18 – 50) – 11(4 – 8) – 1(50 – 36)
= 3(32) -11(4) -1(14)
= -96 + 44 – 14 = -66
Δ_x = \(\left| \begin{matrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{matrix} \right| \)
= 3(-25 – 27) – 3(50 – 36) + 11(6 + 4)
= 3(-52) -3(14) + 11(10)
= -156 – 42 + 110
= -88

Solution:

3a – 4b – 2c = 1 ……….. (1)
a + 2b + c = 2 …………… (2)
2a – 5b – 4c = -1 ………….. (3)
Δ = \(\left| \begin{matrix} 3 & -4 & -2 \\ 1 & 2 & 1 \\ 2 & -5 & -4 \end{matrix} \right| \)
= 3(-8 + 5) + 4 (-4 – 2) – 2(-5 – 4)
= 3(-3) +4(-6) -2(-9)
= -9 – 24 + 18
= -15 ≠ 0
Δ_a = \(\left| \begin{matrix} 1 & -4 & -2 \\ 2 & 2 & 1 \\ -1 & -5 & -4 \end{matrix} \right| \)
= 1(-8 + 5) + 4(-8 + 1) – 2(-10 + 2)
= 1(-3) + 4(-7) – 2(-8)
= -3 – 28 + 16
= -15
Δ_b = \(\left| \begin{matrix} 3 & 1 & -2 \\ 1 & 2 & 1 \\ 2 & -1 & -4 \end{matrix} \right| \)
= 3(-8 + 1) – 1(-4 – 2) – 2(-1 – 4)
= 3(-7) -1(-6) – 2(-5)
= – 21 + 6 + 10 = -5
Δ_c = \(\left| \begin{matrix} 3 & -4 & 1 \\ 1 & 2 & 2 \\ 2 & -5 & -1 \end{matrix} \right| \)
= 3(-2 + 10) + 4(-1 – 4) + 1 (-5 – 4)
= 24 – 20 – 9 = -5

Question 2.
In a competitive examination, one mark is awarded for every correct answer while \(\frac{1}{4}\) mark is deducted for every wrong answer. A student answered 100 questions and got 80 marks. How many questions did he answer correctly? (Use Cramer’s rule to solve the problem).
solution:
No. of Questions answered = 100
Let the No. of questions answered correctly be x
and the No. of questions answered wrongly be y
Here, x + y = 100 and x – \(\frac { 1 }{ 4 }\) y = 80
(i.e) x + y = 100 and 4x – y = 320

correct questions = 84
wrong questions = 16.

Question 3.
A chemist has one solution which is 50% acid and another solution which is 25% acid. How much each should be mixed to make 10 litres of a 40% acid solution? (Use Cramer’s rule to solve the problem).
solution:
Let two solutions x and y
x + y = 10 …….. (1)
0.25 x + (0.50)y = (0.40) ……….. (2)
(2) × 100 ⇒ 25x + 50y = 400
(2) ÷ 5 ⇒ 5x + 10y = 80 …………. (3)

Question 4.
A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself? (Use Cramer’s rule to solve the problems).
solution:
pump A fills \((\frac {1}{x})^{th}\) of the tank in 1 hour.
pump B fills \((\frac {1}{y})^{th}\) of the tank in 1 hour.
Both can filled \((\frac {1}{10})^{th}\) of the tank in 1 hour.

using Cramer’s rule

Pump A takes 15 minutes
Pump B takes 30 minutes

Question 5.
A family of 3 people went out for dinner in a restaurant. The cost of two dosai, three idlies and two vadais is Rs 150. The cost of the two dosai, two idlies and four vadais is Rs 200. The cost of five dosai, four idlies and two vadais is T 250. The family has Rs 350 in hand and they ate 3 dosai and six idlies and six vadais. Will they be able to manage to pay the bill within the amount they had?
solution:
Let the Cost of dosai, Idlies and vadais be x, y, z
2x + 3y + 2z = 150
2x + 2y + 4z = 200
5x + 4y + 2z = 250
Δ = \(\left| \begin{matrix} 2 & 3 & 2 \\ 2 & 2 & 4 \\ 5 & 4 & 2 \end{matrix} \right| \)
= 2(4 – 16) – 3(4 – 20) + 2(8 – 10)
= 2(-12) – 3(-16) + 2(-2)
= -24 + 48 – 4
= 20 ≠ 0
Δ_x = \(\left| \begin{matrix} 150 & 3 & 2 \\ 200 & 2 & 4 \\ 250 & 4 & 2 \end{matrix} \right| \)
= 150(4 – 16) – 3(400 – 1000) + 2(800 – 500)
= 150(-12) – 3(-600) + 2(300)
= -1800 + 1800 + 600
= 600
Δ_y = \(\left| \begin{matrix} 2 & 150 & 2 \\ 2 & 200 & 4 \\ 5 & 250 & 2 \end{matrix} \right| \)
= 2(400 – 1000) – 150(4 – 20) + 2(500 – 1000)
= 2(-600) – 150(-16) + 2(-500)
= -1200 + 2400 – 1000
= 200
Δ_z = \(\left| \begin{matrix} 2 & 3 & 150 \\ 2 & 2 & 200 \\ 5 & 4 & 250 \end{matrix} \right| \)
= 2(500 – 800) – 3(500 – 1000) + 150(8 – 10)
= 2(-300) – 3(-500) + 150(-2)
= -600 + 1500 – 300
= 600

x = Rs 30, y = Rs 10, z = Rs 30
There are 3 dosai, 6 idlies and 6 vadais
= 3x + 6y + 6z
= 3(30) + 6(10) + 6 (30)
= 90 + 60 + 180
= Rs. 330
They can eat within the amount.

Students can download Maths Chapter 5 Information Processing Ex 5.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 5 Information Processing Ex 5.1

Question 1.
Study and complete the following pattern.
(i) 1 × 1 = 1
11 × 11 = 121
111 × 111 = 12321
1111 × 1111 = ?
11111 × 11111 = ?

Solution:
(i) 1234321, 123454321
(ii) 144, 60, 84, 36, 48, 15, 27

Question 2.
Find next three numbers in the following number patterns.
(i) 50, 51, 53, 56, 60……
(ii) 77, 69, 61, 53, ……
(iii) 10, 20, 40, 80,…
(iv) \(\frac{21}{33}\), \(\frac{321}{444}\), \(\frac{4321}{555}\)
Solution:

i) The pattern generating these numbers is
50, 50 + 1, 51 + 2, 53 + 3, 56 + 4, 60 + 5, 65 + 6, 71 + 7,
∴ 50, 51, 53, 56, 60, 65, 71, 78, ……
∴ The next three numbers will be 65, 71, 78

ii) The pattern generating these numbers is
77, 77 – 8, 69 – 8, 61 – 8, 53 – 8, 45 – 8, 37 – 8, 29
77, 69, 61, 53, 45, 37, 29, 21,
∴ The next three numbers will be 45, 37, 29.

iii) The pattern generating these numbers is
10, 10 + 10, 20 + 20, 40 + 40, 80 + 80, 160 + 160, 320 + 320,….
10, 20, 40, 80, 160, 320, 640,….
∴ The next three numbers will be 160, 320, 640.

(iv) \(\frac{54321}{66666}\), \(\frac{654321}{777777}\), \(\frac{7654321}{8888888}\)

Question 3.
Consider the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,…. Observe and complete the following table by understanding the number patterns? followed. After filling the table discuss the pattern followed in addition and subtraction, of the numbers of the sequence?

Solution:
(i) 12, 13 – 1 = 12
(ii) 33, 34 – 1 = 33
(iii) 1 + 3 + 8 + 21 + 55 = 88, 89 – 1 = 88

Question 4.
Complete the following patterns.

Solution:

Question 5.
Find the HCF of the following pair of numbers by Euclid’s game
(i) 25 and 35
(ii) 36 and 12
(iii) 15 and 29
Solution:
(i) HCF of (25, 35 – 25)
25 = 5 × 5
10 = 2 × 5
HCF of (25, 10) = 5

(ii) HCF of (36, 36 – 12)
36 = 2 × 2 × 3 × 3
24 = 2 × 2 × 2 × 3
HCF of (36, 24) = 2 × 2 × 3 = 12

(iii) HCF of (15, 29 -15)
15 = 3 × 5 × 1
14 = 2 × 7 × 1
HCF of (15, 14) = 1

Question 6.
Find HCF of 48 and 28. Also find the HCF of 48 and the number obtained by finding their difference.
Solution:
HCF of 48 and 28
48 = 2 × 2 × 2 × 2 × 3
28 = 2 × 2 × 7
HCF of (48, 28) = 2 × 2 = 4
HCF of (48, 48 – 28)
48 = 2 × 2 × 2 × 2 × 3
20 = 2 × 2 × 5
HCF of (48, 20) = 4

Question 7.
Give instructions to fill in a bank withdrawal form issued in a bank.
Solution:

• The name should be written in capital letters from left to right.
• Write the date of withdrawal on the right top comer of the form.
• Write the amount (in words) to be withdrawn in the space provided.
• Write the amount (in figures) to be withdrawn in the box provided.
• Put your signature at the right bottom above the ‘signature of the depositor’.

Question 8.
Arrange the name of your classmates alphabetically.
Solution:

Question 9.
Follow and execute the instructions given below.

(i) Write the number 10 in the place common to the three figures
(ii) Write the number 5 in the place common for square and circle only.
(iii) Write the number 7 in the place common for triangle and circle only.
(iv) Write the number 2 in the place common for triangle and square only.
(v) Write the numbers 12, 14, and 8 only in square, circle, and triangle respectively.
Solution:

Question 10.
Fill in the following information

Solution:

Objective Type Questions

Question 11.
The next term in the sequence 15, 17, 20, 22, 25, … is
(a) 28
(b) 29
(c) 27
Hint:
Add 2 and 3 alternatively
Solution:
(c) 27

Question 12.
What will be the 25th letter in the pattern? ABCAABBCCAAABBBCCC,…
(a) B
(b) C
(c) D
(d) A
Solution:
(a) B

Question 13.
The difference between 6th term add 5th term in the Fibonacci sequence is ___.
(a) 6
(b) 8
(c) 5
(d) 3
Solution:
(d) 3

Question 14.
The 11th term in the Lucas sequence 1, 3, 4, 7, is
(a) 199
(b) 76
(c) 123
(d) 47
Solution:
(a) 199

Question 15.
If the Highest Common Factor of 26 and 54 is 2, then HCF of 54 and 28 is .
(a) 26
(b) 2
(c) 54
(d) 1
Hint: HCF (54, 28) = HCF (28, 26) = 2
Solution:
(b) 2

Students can download Maths Chapter 4 Symmetry 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 3 Chapter 4 Symmetry Ex 4.2

Miscellaneous Practice Problems

Question 1.
Draw and answer the following.
(i) A triangle which has no line of symmetry.
(ii) A triangle which has only one line of symmetry
(iii) A triangle which has three lines of symmetry.
Solution:
(i) Scalene triangle
(ii) Isosceles triangle
(iii) Equilateral triangle

Question 2.
Find the alphabets in the box which have

(i) No line of symmetry
(ii) Rotational symmetry
(iii) Reflection symmetry
(iv) Reflection and rotational symmetry.
Solution:
i) The alphabets which have no line of symmetry are P, N, S, Z
ii) The alphabets which have Rotational symmetry are I, O, N, X, S, H, Z
iii) The alphabets which have reflection symmetry are A, M, E, D, I, K, O, X, H, U, V, W.
iv) The alphabets which has reflection and rotational symmetry are I, O, X, H.

Question 3.
For the following pictures, find the number of lines of symmetry and also find the order of rotation.

Solution:
(i) 0, 2
(ii) 1, 0
(iii) 2, 2
(iv) 8, 8
(v) 1, 0

Question 4.
The three-digit number 101 has rotational and reflection symmetry. Give five more examples of three-digit numbers that have both rotational and reflection symmetry
Solution:
The digits 0, 1, 8 have rotational and reflection symmetry.
∴ The three digits numbers 181, 111, 808, 818, 888 have both rotational and reflection symmetry.

Question 5.
Translate the given pattern and complete the design in a rectangular strips?

Solution:

Challenge Problems

Question 6.
Shade one square so that it possesses

(i) One line of symmetry
(ii) Rotational symmetry of order 2
Solution:

Question 7.
Join six identical squares so that atleast one side of a square fits exactly with any other side of the square and has reflection symmetry (any three ways).
Solution:

Question 8.
Draw the following
(i) A figure which has reflection symmetry but no rotational symmetry.
(ii) A figure which has rotational symmetry but no reflection symmetry.
(iii) A figure which has both reflection and rotational symmetry.
Solution:

Question 9.
Find the line of symmetry and the order of rotational symmetry’ of the given regular polygons and complete the following table and answer the questions given below.

i) A regular polygon of 10 sides will have ______ lines of symmetry.
ii) If a regular polygon has 10 lines of symmetry then its order of rotational symmetry is ______
iii) A regular polygon of ‘n’ sides has lines of symmetry and the order of rotational symmetry is ______
Solution:
(i) 10
(ii) 10
(iii) n, n

Question 10.
Colour the boxes in such a way that it possesses translation symmetry.

Solution: