Class 8

Students can Download Maths Chapter 5 Information Processing Ex 5.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Ex 5.2

Question 1.
Colour the following patterns with as few colours as possible but make sure that no two adjacent sections are of the same colour.
Solution:

Question 2.
Ramya wants to paint a pattern in her living room wall with a minimum budget. Help her to colour the pattern with 2 colours but make sure that no two adjacent boxes are the same colour. The pattern is shown in the picture.

Solution:

Question 3.
Colour the countries in the following maps with as few colours as possible but make
sure that no two adjacent countries are of the same colour.

Solution:

Students can Download Maths Chapter 4 Geometry Ex 4.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.1

Question 1.
From the figure given, prove that ∆ABC ~ ∆DEF.

Solution:
From the ∆ABC,
AB = AC
It is an isosceles triangle
Angles opposite to equal sides are equal
∴ ∠B = ∠C = 65°
∴ ∠B + ∠C = 65° + 65°
= 130°
We know that .sum of three angles is a triangle = 180°
∠A + ∠B + ∠C = 180°
∠A + 130° = 180°
∠A = 180°-130°
∠A = 50°
From ∆DEF, ∠D = 50°
∴ Sum of Remaining angles = 180° – 50° = 130°
DE = FD
∴ ∠D = ∠F
From ∆ABC and ∆DEF

∠A = ∠D = 50°
∠B = ∠E = 65°
∠C = ∠F = 65°
∴ By AAA criteria ∆DEF ~ ∆ABC

Question 2.
Prove that ∆GUM ~ ∆ BOX from the given figure.

Solution:


That is their corresponding sides are proportional.
∴ By SSS similarity ∆GUM ~ ∆BOX.

Question 3.
In the given figure YH ||TE Prove that ∆WHY ~ ∆WET and also find HE and TE.

Solution:

Question 4.
In the given figure, if ∆EAT ~ ∆BUN find the measure of all angles.

Solution:
Given ∆EAT ≡ ∆BUN
∴ Corresponding angles are equal
∴ ∠E = ∠B ..(1)
∠A = ∠U ..(2)
∠T = ∠N ..(3)
∠E = x°
∠A = 2x°
Sum of three angles of a triangle = 180°
In ∆EAT, x + 2x + ∠T = 180°
∠T = 180° – (x° + 2x° )
∠T = 180°- 3x° …(4)
Also in ∆BUN
(x + 40)° + + ∠U = 180°
x + 40° + x + ∠U = 180°
2x° + 40° + ∠U = 180°
∠U = 180° – 2x – 40°
= 140° – 2x°
Now by (2)
∠A = ∠U
2x = 140° – 2x
2x + 2x = 140°
4x = 140°

∠A = 2x° = 2 × 35° = 70°
∠N = x + 40°
= 35° + 40° = 75°
∴ ∠T = ∠N = 75°
∠E = ∠B = 35°
∠A = ∠U = 70°

Question 5.
From the given figure, UB || AT and CU ≡ CB Prove that ∆CUB ~ ∆CAT and hence ∆CAT is isosceles.

Solution:

Question 6.
In the figure, ∠CIP ≡ ∠COP and ∠HIP ≡ ∠HOP. Prove that IP ≡ OP.

Solution:

Question 7.
In the given triangle, AC ≡ AD and ∠CBD ≡ ∠DEC. Prove that ∆BCF ≡ ∆EDF.

Solution:

Question 8.
In the given figure, ∆ BCD is isosceles with base BD and ∠BAE ≡ ∠DEA. Prove that AB ≡ ED .

Solution:

Question 9.
In the given figure, D is the midpoint of OE and ∠CDE = 90°. Prove that ∆ODC ≡ ∆EDC.

Solution:

Question 10.
In the figure, if SW ≡ SE and ∠NWO ≡ ∠NEO. then, prove that NS bisects WE and ∠NOW = 90°

Proof:

Question 11.
Is ∆PRQ ≡ ∆QSP ? Why ?

Solution:
In ∠PRQ = ∠PSQ = 90° given
PR = QS = 3 cm given
PQ = PQ = 5 cm common
It satisfies RHS criteria
∴ ∆PRQ congruent to ∆QSP.

Question 12.
Fill in the blanks with the most correct term from the given list.
(in proportion, similar, corresponding, congruent shape, area, equal)
Statements Reasons

Question 1.
Corresponding sides of similar triangles are ___.
Solution:
in proportion

Question 2.
Similar triangles have the same ___ but not necessarily the same size.
Solution:
shape

Question 3.
In similar triangles, ___ sides are opposite to equal angles.
Solution:
equal

Question 4.
The symbol ~ is used to represent ___ triangles.
Solution:
congruent

Question 5.
The symbol ~ is used to represent ____ triangles.
Solution:
similar

Objective Type Questions

Question 13.
Two similar triangles will always have ___ angles
(A) acute
(B) obtuse
(C) right
(D) matching
Solution:
(D) matching

Question 14.
If in triangles PQR and XYZ, \(\frac{P Q}{X Y}=\frac{Q R}{Z X}\) then they will be similar if
Solution:
(C) Q = ∠X

Question 15.
A flag pole 15 cm high casts a shadow of 3 m at 10 a.m. The shadow cast by a building at the same time is 18.6 m. The height of the building is
(A) 90 m
(B) 91 m
(C) 92 m
(D) 93 m
Solution:
(D) 93 m

Question 16.
If ∆ABC ~ ∆PQR in which ∠A = 53° and ∠Q = 77°, then ∠R is
(A) 50°
(B) 60°
(C) 70°
(D) 80°
Solution:
(A) 50°

Question 17.
In the figure, which of the following statements is true?
(A) AB = BD
(B) BD < CD
(C) AC = CD
(D) BC = CD

Solution:
(C) AC = CD

Students can Download Maths Chapter 2 Measurements Ex 2.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.2

Question 1.
Find the perimeter and area of the combined figures given below, (π = \(\frac{22}{7}\))

Solution:

∴ Perimeter = Sum of all lengths of sides that form the closed boundary
P = 11 + 10 + 7 + 10m
Perimeter = 38 m
Area = Area of the rectangle – Area of semicircle

= 50.75 m² (approx)
Area of the figure = 50.75m² approx.

(ii) Perimeter = sum of outside lengths

Question 2.
Find the area of the shaded part of the following figures. (π = 3.14 )

Solution:
Area of the shaded part = Area of 4 quadrant circles of radius \(\frac{10}{2}\) cm
= 4 × \(\frac{1}{4}\) × πr² = 3.14 × \(\frac{10}{2} \times \frac{10}{2}\) cm²
= \(\frac{314}{4}\) cm² = 78.5cm²
Area of the shaded part = 78.5 cm²
Area of the unshaded part = Area of the square – Area of shaded part
= a² – 78.5 cm² = (10 × 10) – 78.5 cm²
= 100 – 78.5 cm² = 21.5 cm²
Area of the unshaded part = 21.5 cm² (approximately)

(ii) Area of the shaded part = Area of semicircle – Area of the triangle

∴ Area of the shaded part = 27.93 cm² (approximately)

Question 3.
Find the area of the combined figure given which is got by joining of two parallelograms.

Solution:
Area of the figure = Area of 2 parallelograms with base 8 cm and height 3 cm
= 2 × (bh) sq. units
= 2 × 8 × 3 cm² = 48 cm²
∴ Area of the given figure = 48 cm²

Question 4.
Find the area of the combined figure given, formed by joining a semicircle of diameter 6 cm with a triangle of base 6 cm and height 9 cm. (π = 3.14 )

Solution:
Area of the figure = Area of the semicircle of radius 3 cm + 2 (Area of triangle with b = 9 cm and h = 3 cm)

= 14.13 + 27 cm² = 41.13 cm²
∴ Area of the figure = 41.13 cm² (approximately)
The door mat which is in a hexagonal shape

Question 5.
The door mat which is in a hexagonal shape has the following measures as given in the figure. Find its area.

Solution:
Area of the doormat = Area of 2 trapezium
Height of the trapezium h = \(\frac{70}{2}\) cm;
a = 90 cm; b = 70 cm
∴ Area of the trapezium
= \(\frac{1}{2}\)h (a + b) sq. units
Area of the door mat
= 2 × \(\frac{1}{2}\) × 35 (90 + 70) cm²
= 35 × 160 cm² = 5600 cm²
∴ Area of the door mat = 5600 cm²

Question 6.
Find the area of an invitation card which has two semicircles attached to a rectangle as in the figure given. (π = \(\frac{22}{7}\))

Solution:
Area of the card = Area of the rectangle + area of 2 semicircles
Length of the rectangle l = 30 cm
Breadth b = 21 cm
Radius of the semicircle = \(\frac{21}{27}\)
∴ Area of the card
= (l × b) + (2 × \(\frac{1}{2}\) πr²) sq. units
= 30 × 21 + \(\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}\) cm² = 630 + 346.5
= 976.5 cm² (approximately)
∴ Area of the Invitation card = 976.5 cm²

Question 7.
Find the area of the combined figure given, which has two triangles attached to a rectangle.

Solution:
Area of the combined shape = Area of the rectangle + Area of 2 triangles
Length of the rectangle l = 10 cm
Breadth b = 8 cm
Base of the triangle base = 8 cm
Height h = 6 cm
∴ Area of the shape = (l × b) + (2 × \(\frac{1}{2}\) × base × h) cm²
= (10 × 8) + (8 × 6) cm² = 80 + 48 cm² = 128 cm²
Area of the given shape = 128 cm²

Question 8.
A rocket drawing has the measures as given in the figure. Find its area.

Solution:
Area = Area of a rectangle + Area of a triangle + Area of a trapezium
For rectangle length l = 120 – 20 – 20 cm = 80 cm
Breadth b = 30 cm
For the triangle base = 30 cm
Height = 20 cm
For the trapezium height h = 20 cm
Parallel sided a = 50 cm
b = 30 cm
∴ Area of the figure = (l × b) + (\(\frac{1}{2}\) × base × height) + \(\frac{1}{2}\) × h × (a + b) sq. units
= (80 × 30) + ( \(\frac{1}{2}\) × 30 × 20) + \(\frac{1}{2}\) × 20 × (50 + 30) cm²
= 2400 + 300 + 800 cm² = 3500 cm²
Area of the figure = 3500 cm²

Question 9.
Find the area of the glass painting which has a triangle on a square as given in the figure.

Solution:
Area of the glass painting = Area of the square + Area of the triangle
Side of the square a = 30 cm
Base of the triangle b = 30 cm
Height of the triangle h = 8 cm
∴ Area of the painting = (a)² + (\(\frac{1}{2}\) bh) sq. units
= (30 × 30) + (\(\frac{1}{2}\) × 30 × 8) cm² = 900 + 120 cm²
= 1020 cm²
Area of the glass painting = 1020 cm²

Question 10.
Find the area of the irregular polygon shaped fields given below.

Solution:
(i) Area of the irregular field = Area of ∆AHF + Area of trapezium FHIE + Area of triangle EID + Area of ∆JDC + Area of rectangle BGJC + Area of ∆AGB

(ii) Area of the field = Area of trapezium FBCH + Area of ∆DHC + Area of ∆EGD + Area of ∆EGA + Area of ∆BFA

Students can Download Maths Chapter 3 Algebra Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions

Additional Questions And Answers

Exercise 3.1

Very Short Answers [2 Marks]

Question 1.
Find the product of the following.
(i) (x, y)
(ii) (10x, 5y)
(iii) (2x², 5y²)
Solution:
(i) x × y = xy
(ii) 10x × 5y = (10 × 5) x × xy
= 50 xy
(iii) 2x² x 5y² = (2 x 5) x (x² + y²)
= 10x²y²

Short Answers [3 Marks]

Question 1.
Find the product of the following.
(i) 3ab² c³ by 5a³b²c
(ii) 4x²yz by \(\frac{3}{2}\) x²yz²
Solution:
(i) (3ab³c³) × (5a³b²c)
= (3 × 5)(a × a³ × b² × b² × c² × c)
= 15a¹⁺³.b²⁺².c³⁺¹ = 15a⁴b⁴c⁴

(ii) 4x²yz by \(\frac{3}{2}\) x²yz²
= (4 × \(\frac{3}{2}\)) × (x² × x² × y × y × z × z²)
= -6x²⁺² y¹⁺¹ x¹⁺² = -6x⁴y²z³

Long Answers [5 Marks]

Question 1.
Simplify (3x – 2) (x – 1) (3x + 5).
Solution:
(3x – 2) (x – 1) (3x + 5)
= {(3x – 2) (x – 1)} × (3x + 5) [∴Multiplication in associative]
= {3x (x – 1) – 2 x – 1)} × (3x + 5)
= (3x² – 3x – 2x + 2) × (3x + 5)
= (3x² – 5x + 2) (3x + 5)
= 3x² × (3x + 5) – 5x (3x + 5) + 2 (3x + 5)
= (9x³ + 15x²) + (-15x² – 25x) + (6x + 10)
= 9x³ + 15x² – 15x² – 25x + 6x + 10
= 9x³ – 19x + 10

Question 2.
Simplify (5 – x) (3 – 2x) (4 – 3x).
Solution:
(5 – x) (3 – 2x) (4 – 3x)
= {(5 – x)(3 – 2x)} × (4 – 3x) [∴ Multiplication in association]
= {5 (3 – 2x) -x (3 – 2x)} × (4 – 3x)
= (15 – 10x – 3x + 2x²) × (4 – 3x)
= (2x² – 13x + 15) (4 – 3x)
= 2x² × (4 – 3x) – 13x (4 – 3x) + 15 (4 – 3x)
= 8x³ – 6³ – 52x + 39x² + 60 – 45x
= -6x³ + 47x² – 97x + 60

Exercise 3.2

Very Short Answers [2 Marks]

Question 1.
Divide.
(i) 12x³y³ by 3x²y
(ii) -15a² bc³ by 3ab
(iii) 25x³y² by – 15x²y
Solution:

Short Answers [3 Marks]

Question 1.
Divide
(i) 15m²n³ by 5m²n²
(ii) 24a³b³ by -8ab
(iii) -21 abc² by 7 abc
Solution:

Question 2.
Divide
(i) 16m³y² by 4m²y
(ii) 32m² n³p² by 4mnp
Solution:

Long Answers [5 Marks]

Question 1.
Divide.
(i) 9m⁵ + 12m⁴ – 6m² by 3m²
(ii) 24x³y + 20x²y² – 4xy by 2xy
Solution:

Exercise 3.3

Question 1.
Evaluate:
(i) (2x + 3y)²
(ii) (2x – 3y)²
Solution:
(i) (2x + 3y)²
= (2x)² + 2 × (2x) × (3y) + (3y)²
[using (a + b)² = a² + 2ab + b²]
= 4x² + 12xy + 9y²

(ii) (2x – 3y)²
= (2x)² – 2(2x) (3y) + (3y)²
[∵ using (a – b)² = a² – 2ab + b²]
= 4x² – 12xy + 9y²

Short Answers [3 Marks]

Question 1.
Evaluate the following
(i) (2x – 3) (2x + 5)
(ii) (y – 7) (y + 3)
(iii) 107 × 103
Solution:
(i) (2x – 3) (2x + 5)
= (2x)² + (-3 + 5) (2x) + (-3) (5)
[∵ (x + a) (x + b) = x² + (a + b)x + ab]
= 2²x² + 2 × 2x + (-15)
= 4x² + 4x – 15

(ii) (y – 7) (y + 3)
= y² + (-7 + 3)y + (-7) (3)
[∵ (x + a)(x + b) = x² + (a + b)x + ab]
= y² – 4y + (-21) = y² – 4y – 21

(iii) 107 × 103
= (100 + 7) × (100+ 3)
= 100² + (7 + 3) × 100 +(7 × 3)
= 10000 + 10 × 100 + 21 = 10000 + 1000 + 21 = 11021

Long Answers [5 Marks]

Question 1.
If x + y = 12 and xy = 14 find x² + y².
Solution:
(x + y)² = x² + y² + 2xy
12² = x² + y² + 2 × 14
144 = x² + y² + 28
x² + y² = 144 – 28
x² + y² = 116

Question 2.
If 3x + 2y = 12 and xy = 6 find the value of 9x² + 4y².
Solution:
(3x + 2y)² = (3x)² + (2y)² + 2 (3x) (2y)
= 9x² + 4y² + 12xy
12² = 9x² + 4y² + 12 × 6
144 = 9x² + 4y² + 72
144 – 72 = 9x² + 4y²
∴ 9x² + 4y² = 72

Exercise 3.4

Question 1.
Factorize:
(i) 7(2x + 5) + 3 (2x + 5)
(ii) 12x³y⁴ + 16x²y⁵ – 4x⁵y²
Solution:
(i) 7(2x + 5) + 3 (2x + 5)
= (2x + 5) (7 + 3)
(ii) 12x³y⁴ + 16x²y⁵ – 4x⁵y²
= 4x²y² (3xy² + 4y³ – x³)

Short Answers [3 Marks]

1. Factorize
(i) 81a² – 121b²
(ii) x² + 8x + 16
Solution:
(i) 81a² – 121b²
= (9a)² – (11b)²
[∵ using a² – b² = (a + b)²]
= (9a + 11b) (9a – 11b)

(ii) x² + 8x + 16 = x² + 2 × x × 4 + 4²
[∵ using a² + 2ab + b² = (a + b)²]
= (x + 4)² = (x + 4)(x + 4)

Long Answers [5 Marks]

Question 1.
Factorize
(i) x² + 2xy + y² – a² + 2ab – b²
(ii) 9 – a⁶ + 2a³ – b⁶
Solution:
(i) x² + 2xy + y² – a² + 2ab – b².
= (x² + 2xy + y²) – (a² – 2ab + b²)
= (x + y)² – (a – b)²
= {(x + y) + (a – b)} {(x + y) – (a – b)}
= (x + y + a – b) (x + y – a + b)

(ii) a – a⁶ + 2a³b³ – b⁶
= 9 – (a⁶ – 2a³b³ + b⁶)
= 3² -{(a³)² – 2 × a³ × b³ + (b³)²}
= 3² – (a³ – 6³)²
= {3 + (a³ – b³)} {3 – (a³ – b³)}
= (3 + a³ – b³) (3 – a³ + b³)
= (a³ – b³ + 1){-a³ + b³ + 3)

Question 2.
Factorize
(i) 100 (x + y )² – 81 (a + b)²
(ii)(x + 1)² – (x – 2)²
Solution:
(i) 100 (x + y)² – 81 (a + b)²
= {10 (x + y)}² – {(a (a + b)}²
= {10 (x + y) + 9 (a + b)}
{10 (x + y) – 9(a + b)}
= (10x + 10y + 9a – 9b)}
(10x + 10y – 9a – 9b)

(ii) (x – 1)² – (x – 2)²
= {(x – 1 +(x – 2)}
{(x – 1) – (x – 2)}
= (2x – 3) – (x – 1 – x + 2)
= (2x – 3) × 1 = 2x – 3

Students can Download Maths Chapter 4 Geometry Ex 4.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.2

Miscellaneous Practice Problems

Question 1.
In the given figure, find PT given that l₁ || l₂

Solution:
Given that l₁ || l₂
∴ In ∆PQS and ∆PRT
∠P is common
∠Q = ∠R [∵ PR is the transversal for l₁ and l₂ corresponding angles]
∠S = ∠T [∵ corresponding angles]
∴ ∆PQS ~ ∆PRT [∵ By AAA congruency]
In similar triangles, corresponding angles are proportional.

Question 2.
From the diagram, prove that ∆SUN ~ ∆RAY

Proof:
From the ∆SUN and ∆RAY
SU = 10;
UN = 12;
SN = 14;
RA = 5,
AY = 6;
RY = 7
We have

Question 3.
The height of a tower is measured by a mirror on the ground by which the top of the tower’s reflection is seen. Find the height of the tower.
Solution:
The image and its reflection make similar shapes

Question 4.
In the figure, given that ∠1 = ∠2 and ∠3 ≡ ∠4 Prove that ∆MUG ≡ ∆ TUB.

Proof:

Question 5.
If ∆WAR ≡ ∆MOB, name the additional pair of corresponding parts. Name the criterion used by you.
Solution:

Given ∆WAR ≡ ∆MOB
∠RWA ≡ ∠BMO [∵ sum of three angles of a triangle are 180°]
∴ Criteria used here is angle sum property of triangles.

Question 6.
In the figure, ∠TMA ≡ ∠IAM and ∠TAM ≡ ∠IMA. P is the midpoint of MI and N is the midpoint of AI. Prove that ∆ PIN ~ ∆ ATM.

Solution:

Question 7.
In the figure, if ∠FEG = ∠1 then, prove that DG² = DE.DF.

Proof:

Question 8.
In the figure, ∠TEN ≡ ∠TON = 90° and TO = TE. Prove that ∠ORN ≡ ∠ERN

Solution:

Question 9.
In the figure, PQ ≡ TS, Q is the midpoint of PR, S is the midpoint TR and ∠POU ≡ ∠TSU. Prove that QU ≡ SU.

Proof:

Question 10.
In the figure ∆TOP ≡ ∆ARM . Explain why?
Solution:
In ∆TOP and ∆ARM
OP = RM given
∠TOP = ∠ARM = 90°
given ∠OTP = ∠RAM
given ∠OPT = ∠RMA Remaining angle, by angle sum property.
∴ By ASA criteria we can say that ∆TOP ≡ ∆ARM

Students can Download Maths Chapter 3 Algebra Ex 3.5 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.5

Miscellaneous Practice Problems

Question 1.
Subtract: -2(xy)² (y³ + 7x²y + 5) from 5y² (x²y³ – 2x⁴y + 10x²)
Solution:
5y² (x²y³ – 2x⁴y + 10x²) – [(-2)(xy)² (y³ + 7x²y + 5)]
= [5y² (x²y³) – 5y² (2x⁴y) + 5y² (10x²)] – [(-2)x²y² (y³ + 7x²y + 5)]
= (5y⁵5x² – 10x⁴y³ + 50x²y²)
= 5x²y⁵ – 10x⁴y³ + 50x²y² – [(-2x²y⁵) – 14x⁴y³ – 10x²y²]
= 5x²y⁵ – 10x⁴y³ + 50x²)y² + 2x² y⁵ + 14x⁴)y³ + 10x²)y²
= (5 + 2)x²y⁵ + (-10 + 14)x⁴y³ + (+50 + 10)x²y²
= 7x²y⁵ + 4x⁴y³ + 60x²y²

Question 2.
Multiply (4x² + 9) and (3x – 2).
Solution:
(4x² + 9) (3x – 2) = 4x²(3x – 2) + 9(3x – 2)
= (4x²)(3x) – (4x²)(2) + 9(3x) – 9(2) = (4 × 3 × x × x²) – (4 × 2 × x²) + (9 × 3 × x) – 18
= 12x³ – 8x² + 27x – 18 (4x³ + 9) (3x – 2) = 12x³ – 8x² + 27x – 18

Question 3.
Find the simple interest on Rs. 5a²b² for 4ab years at 7b% per annum.
Solution:

Question 4.
The cost of a note book is Rs. 10ab. If Babu has Rs. (5a²b + 20ab² + 40ab). Then how many note hooks can he buy?
Solution:
For ₹ 10 ab the number of note books can buy = 1.

Question 5.
Factorise : (7y² – 19y – 6)
Solution:
7y² – 19y – 6 is of the form ax² + bx + c where a = 7; b = -19; c = – 6

The product a × c = 7 × -6 = -42
sum b = – 19

The middle term – 19y can be written as – 21y + 2y
7y² – 19y – 6 = 7y² – 21y + 2y – 6
= 7y(y – 3) + 2(y – 3) = (y – 3)(7y + 2)
7y² – 19y – 6 = (y – 3)(7y + 2)

Challenging Problems

Question 6.
A contractor uses the expression 4x² + 11x + 6 to determine the amount of wire to order when wiring a house. If the expression comes from multiplying the number of rooms times the number of outlets and he knows the number of rooms to be (x + 2), find the number of outlets in terms of ‘x’ [Hint: factorise 4x² + 11x + 6]
Solution:
Given Number of rooms = x + 2
Number of rooms × Number of outlets = amount of wire.

Now factorising 4x² + 11x + 6 which is of the form ax² + bx + c with a = 4 b = 11 c = 6.
The product a × c = 4 × 6 = 24
sum b = 11


The middle term 11x can be written as 8x + 3x
∴ 4x² + 11x + 6 = 4x² + 8x + 3x + 6 = 4x (x + 2) + 3 (x + 2)
4x² + 11x + 6 = (x + 2) (4x + 3)
Now from (1) the number of outlets

∴ Number of outlets = 4x + 3

Question 7.
A mason uses the expression x² + 6x + 8 to represent the area of the floor of a room. If the decides that the length of the room will be represented by (x + 4), what will the width of the room be in terms of x ?
Solution:
Given length of the room = x + 4
Area of the room = x² + 6x + 8
Length × breadth = x² + 6x + 8

Question 8.
Find the missing term: y² + (-) x + 56 = (y + 7)(y + -)
Solution:
We have (x + a) (x + b) = x² + (a + b)x + ab
56 = 7 × 8 .
∴ y² + (7 + 8)x + 56 = (y + 7)(y + 8)

Question 9.
Factorise: 16p⁴ – 1
Solution:
16p⁴ – 1 = 2⁴p⁴ – 1 = (2²)²(p²)² – 1²
Comparing with a² – b² = (a + b)(a – b) where a = 2²p² and b = 1
∴ (2²p²)² – 1² = (2²p² + 1) (2²p² – 1) = (4p² + 1) (4p² – 1)
∴ 16p⁴ – 1 = (4p² + 1)(4p² – 1)(4p² + 1)(2²p² – 1²)
= (4p² + 1) [(2p)²– 1²] = (4p² + 1) (2p + 1)(2p – 1)
[∵ using a² – b² = (a + b) (a – b)]
∴ 16p⁴ – 1 = (4psup>2 + 1)(2p + 1)(2p – 1)

Question 10.
Factorise : x⁶ – 64y³
Solution:
x⁶ – 64y³ = (x²)³ – 4³y³ = (x²)³ – (4y)³
This is of the form a³ – b³ with a = x², b = 4y
a³ – b³ = (a – b)(a² + ab + b²)
(x²)³ – (4y)³ = (x² – 4y) [(x²)² + (x²)(4y) + (4y)²]
= (x² – 4y) [x⁴ + 4x²y + 16y²]
∴ x⁶ – 64y³ = (x² – 4y) [x⁴ + 4x²y + 16y²]

Students can Download Maths Chapter 5 Information Processing Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Intext Questions

Try these (TextBook Page No 122)

Question 1.
Find the number of all possible triangles that can be formed from the triangle given below.
Solution:

Question 2.
Use the given figure to form a 3 × 3 magic square.

Solution:

Question 3.
Convert the tree diagram into a numeric expression

Solution:
(10 × 5) + (9 × 4)

Question 4.
(i) Find out the total time taken by the bus to reach from A to E via B , C and D.
(ii) Find which is the shortest route from A to E.
Solution:

(i) Route ⇒ 2A ➝ B ➝ C ➝ D ➝ E
Time taken ⇒ (7 + 5 + 3 + 6) hrs = 21 hrs

Try this (TextBook Page No 127)

Question 1.
Kumar has four different hats. He always wears a hat. Sometimes he wears the same hat more than once in a week as in the figure.

In how many different ways might he decide to wear his hats in one week?
Solution:

H1, H2, H3, H4 be Hat 1, Hat 2, Hat 3, Hat 4 respectively.
∴ In 7 × 4 = 28 different ways kumar may decide to wear his hat in one week.

Exercise 5.1

Try these (TextBook Page No 134)

Question 1.
Find any four SETs among these set of cards.

Solution:

Question 2.
This is an example for a magic square in SETs, can you make another one?

Solution:

Try this (TextBook PageNo 137)

Question 1.
Think why this graph colouring is invalid?

Solution:
The objective of graph colouring is to assign minimum number of colours to the verties do not have the same colour.
Here adjacent edges have the same colour.
∴ It is invalid.

Students can Download Maths Chapter 1 Rational Numbers Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 1 Rational Numbers Intext Questions

Exercise 1.1
Try This (Textbook Page No. 6)
Question 1.
Is the number \(\frac{7}{-5}\) a fraction or a rational number ? Why?
Solution:
\(\frac{7}{-5}\) is a rational number
Because of rational number is a rational number which is of the form \(\frac{p}{q}\) , q ≠ 0 and p and q are integers.
But fraction is part of a whole.

Question 2.
Write any 6 rational numbers of your choice.
Solution:
0, \(\frac{-1}{2}\), \(\frac{1}{2}\), \(\frac{3}{4}\), \(\frac{6}{7}\), -5, 6

Try This (Textbook Page No. 7)

Question 1.
Explain why the following statements are true?
(i) 0.8 = \(\frac{4}{5}\)
(ii) 1.4 > \(\frac{1}{4}\)
(iii) 0.74 < \(\frac{3}{4}\) (iv) 0.4 > 0.386
(v) 0.096 < 0.24
(vi) 1.128 = 0.1280
Solution:


(vi) 0.128 = \(\frac{128}{1000}=\frac{1280}{10000}\)
0.1280 = \(\frac{1280}{10000}\)
∴ Both are equal. i.e., 0.128 = 0.1280

Try This (Textbook Page No. 9)

Question 1.
Which of the pairs are equivalent rational numbers?

Solution:

Question 2.
Find the standard form of
(i) \(\frac{36}{-96}\)
(ii) \(\frac{-56}{-72}\)
(iii) \(\frac{27}{18}\)
Solution:

Question 3.
Mark the following rational numbers on a number line.
(i) \(\frac{-2}{3}\)
(ii) \(\frac{-8}{-5}\)
(iii) \(\frac{5}{-4}\)
Solution:
(i) \(\frac{-2}{3}\) lies between 0 and -1.
The unit part between 0 and -1 is divided into 3 equal parts and second part is taken.

(ii) \(\frac{-8}{-5}=1 \frac{3}{5}\)
\(1 \frac{3}{5}\) lies between 1 and 2. The unit part between 1 and 2 is divided into 5 equal parts and the third part is taken.

(iii) \(\frac{5}{-4}=-\frac{5}{4}=-1 \frac{1}{4}\)
\(-1 \frac{1}{4}\) lies between -1 and The unit part between -1 and -2 is divided into four equal parts and the first part is taken.

Try This (Textbook Page No. 15)

Question 1.
Are there any rational numbers between \(\frac{-7}{11}\) and \(\frac{6}{-11}\) ?
Solution:

Try This (Textbook Page No. 19)

Question 1.
Divide: (i) 5 by \(\frac{-7}{3}\)
(ii) \(\frac{-7}{3}\) by 5
Solution:

Exercise 1.2
Try This (Textbook Page No. 25)

Question 1.
The closure property on integers holds for subtraction and not for division. What about rational numbers? Verify.
Solution:
Let 0 and \(\frac{1}{2}\) be two rational numbers 0 – \(\frac{1}{2}\) = \(\frac{1}{2}\) is a rational number
∴ Closure property for subtraction holds for rational numbers.
But consider the two rational number \(\frac{5}{2}\) and 0.
\(\frac{5}{2} \div 0=\frac{5}{2 \times 0}=\frac{5}{0}\)
Here denominator = 0 and it is not a rational number.
∴ Closure property is not true for division of rational numbers.

Question 2.
Check whether \(\frac{3}{5}-\frac{7}{8}=\frac{7}{8}-\frac{3}{5}\)
Solution:

∴ Subtraction of rational numbers is not commutative

Question 3.
Is \(\frac{3}{5} \div \frac{7}{8}=\frac{7}{8} \div \frac{5}{3}\)? So, what do you conclude?
Solution:

∴ Commutative property not hold good for division of rational numbers.

Try This (Textbook Page No. 26)

Question 1.
Check whether the associative property holds for subtraction and division.
Solution:
Consider the rational numbers \(\frac{2}{3}\), \(\frac{1}{2}\) and \(\frac{3}{4}\)


∴ Associative property does not hold for division of rational numbers

Question 2.
Observe that

Use your reasoning skills, to find the sum of the first 7 numbers in the pattern given above.
Solution:

Students can Download Maths Chapter 5 Information Processing Ex 5.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Ex 5.1

I. Choose the best answer.

Question 1.
How many outcomes can you get when you toss three coins once?
(A) 6
(B) 8
(C) 3
(D) 2
Solution:
(B) 8

Question 2.
In how many ways can you answer 3 multiple choice questions, with the choices A, B, C and D ?
(1) 4
(2) 3
(3) 12
(4) 64
Solution:
(4) 64

Question 3.
How many 2 digit numbers contain the number 7 ?
(A) 10
(B) 18
(C) 19
(D) 20
Solution:
(B) 18

II. Answer the following.

Question 1.
You are going to have an ice cream or a cake. There are three flavours (chocolate, strawberry , vanilla) in ice creams, and two flavours (orange or red velvet) in the cake. In how many possible ways can you choose an ice cream or the cake?

Solution:
We are going to have either a ice cream or a cake.
Ice cream can be selected from 3 flavors and cake from two flavors. Both the events cannot occur simultaneously selecting ice cream and cake.
∴ Number of possible ways = 3 + 2
= 5 ways

Question 2.
In how many ways, can the teacher choose 3 students, in all one each from 10 students in VI std, 15 students in VII std and 20 students in VIII std to go to an excursion?
Solution:
The teacher is going to selected one student from class VI out of 10 students in 10 ways from
class VII out of 15 students in 15 ways and from class VIII out of 20 students in 20 ways.
∴ Number of ways 3 students can be selected
= 10 + 15 + 20 = 45 ways

Question 3.
If you have 2 school bags and 3 water bottles then, in how many different ways can you carry a school bag and a water bottle, while going to school?
Solution:
We can select one school bag from 2 and one bottle from 3 as follows.

∴ A bag and a water bottle can be selected in 2 × 3 = 6 ways.

Question 4.
Roll numbers are created with a letter followed by 3 digits in it. From the letter A, B, C, D, E and any 3 digits from 0 to 9. Then, in how many possible ways can the roll numbers be generated?
Solution:
We have a letter followed by 3 digits in the roll number.
The letter is selected from A, B, C, D, E.
For these 5 letters we have to select a 3 digit number using the digits 0 to 9.
Once place can be formed using any one of the 10 number 0 to 9 in 10 ways.
Tens place can be formed in 10 ways.
∴ A two digit number can be formed in 10 × 10 = 100 ways.
Thousands place can be formed in 10 ways
∴ A 3 digit number can be formed in 10 × 10 × 10 = 1000 ways.
∴ 5 letters can be attached in 5 × 1000 = 5000 ways.
∴ The roll number can be formed in 5000 ways.

Question 5.
A safety locker in a jewel shop requires a 4 digit unique code. The code has the digits from 0 to 9. How many unique codes are possible ?

Solution:
The unique code has 4 digits.
Each digit is formed using any of the 10 numbers from 0 to 9.
∴ Single digit number can be formed in 10 ways.
A double digit number can be formed in 10 × 10 ways.
A three digit number can be formed in 10 × 10 × 10 ways.
A four digit number can be formed in 10 × 10 × 10 × 10 ways. = 10,000 ways

Students can Download Maths Chapter 1 Life Mathematics Ex 1.1 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 1 Life Mathematics Ex 1.1

Question 1.
Fill in the blanks:

Question (i)
If 30% of x is 150, then x is ………….
Answer:
500
Hint:
Given 30% of x is 150

Question (ii)
2 minutes is ………… % to an hour.
Answer:
3\(\frac{1}{3}\)%
Hint:
Let 2 min be x% of an hour
and 1 hr = 60 min
x% = \(\frac{2}{60}\) x 100 = \(\frac{200}{3}\) = \(\frac{10}{3}\) = 3\(\frac{1}{3}\)
x = 3\(\frac{1}{3}\)

Question (iii)
If x % of x = 25, then x = …………
Answer:
50
Hint:
Given that x % of x is 25
∴ \(\frac{x}{100}\) = 25
x² = 25 x 100 = 2500
∴ x = \(\sqrt{2500}\) = 50

Question (iv)
In a school of 1400 students, there are 420 girls. The percentage of boys in the school is ………….
Answer:
70
Hint:
Given total number of students in school = 1400
Number of girls in school = 420
Number of boys in school = 1400 – 420 = 980

% of boys = 70%

Question (v)
0.5252 is ………….. %.
Answer:
52.52%
Hint:
Given a number, and to express as a percentage, we need to multiply by 100
∴ to express 0.5252 as percentage, we should multiply by 100
∴ 0.5252 x 100 = 52.52%

The Percentage Difference Calculator (% difference calculator) will find the percent difference between two positive numbers greater than 0.

Question 2.
Rewrite each underlined portion using percentage language.

Question (i)
One half of the cake is distributed to the children.
Answer:
Answer:
50% of the cake is distributed to the children
Hint:
One half is nothing but \(\frac{1}{2}\) as percentage, we need to multiply by 100
\(\frac{1}{2}\) x 100 = 50%

Question (ii)
Aparna scored 7.5 points out of 10 in a competition.
Answer:
Aparna scored 75% in a competition
Hint:
7.5 points out of 10 is \(\frac{7.5}{10}\) = 0.75
For percentage, we need to multiply by 100
∴ We get 0.75 x 100 = 75%

Question (iii)
The statue was made of pure silver.
Answer:
96% students participated in sports
Hint:
Pure silver means there are no other metals
so, 100 out of 100 parts is made of silver = \(\frac{100}{100}\)
∴ to express as percentage, \(\frac{100}{100}\) x 100% = 100%

Question (iv)
48 out of 50 students participated in sports
Answer:
96% students participated in sports
Hint:
48 out of 50 students in fraction form is \(\frac{48}{50}\).
As a percentage, we need to multiply by 100

Question (v)
Only 2 persons out of 3 will be selected in the interview
Answer:
Only 66\(\frac{2}{3}\)% persons will be selected in the interview.
Hint:
2 out of 3 in fraction form is \(\frac{2}{3}\) to express as percentage, we need to multiply by 100
\(\frac{2}{3}\) x 100 = \(\frac{200}{3}\) = 66\(\frac{2}{3}\)%

Online percentage off calculator tool makes the calculation faster, and it displays the percentage off of the product in a fraction of seconds.

Question 3.
48 is 32% of what number?
Solution:
Let the number required to be found be ‘x’
Given that 32% of x is 48

Question 4.
A bank pays ₹ 240 as interest for 2 years for a sum of ? 3000 deposited as savings. Find the rate of interest given by the bank.
Solution:
The formula for simple interest is given by
Interest (I) = \(\frac{PxRxT}{100}\) 100
Where P is principal amount; R is rate of interest in percentage; t is time period Substituting given values in formula, we get
P = 3000, t = 2 yrs, I = Rs 240

Question 5.
A Welfare Association has a sports club where 30% of the members play cricket, 28% play volleyball, 22% play badminton and the rest play indoor games. If 30 member play indoor games.

1. How many members are there in the sports club?
2. How many play cricket, volleyball and badminton?

Solution:
Given data
Members who play Cricket = 30% …..(1)
Members who play Vollyball = 28% …..(2)
Members who play Batminton = 22% ……(3)
Members who play outdoor games = (1) + (2) + (3) = 30
Members who play indoor games = 100 – 80 = 20%
Also given 30 members play indoor games
Let total no of members be ‘x’
∴ \(\frac{30}{x}\) = 20
∴ x = \(\frac{30×100}{20}\) = 15

1. Total number of members is 150

2. Members who play Cricket = 30% of total = \(\frac{30}{100}\) x 150

Question 6.
What is 25% of 30% of 400?
Solution:
Required to find 25% of 30% of 400

Question 7.
If the difference between 75% of a number and 60% of the same number is 82.5, then find 20% of the number.
Solution:
Given that 75% of number less 60% of number is 82.5. Let the number be ‘x’
∴ \(\frac{75}{100}\) × x = 82.5
∴ 0.75 – 0.6.60x = 82.5
∴ 0.15x = 82.5
∴ x = \(\frac{82.5}{0.15}\) = \(\frac{8250}{15}\) = 550
Required to find 20% of number ie 20% of x.

Question 8.
If a car is sold for ₹ 2,00,000 from its original price of? 3,00,000 find the percentage for decrease in the value of the car.

Solution:
Original price of car = ₹ 3,00,000
actual selling price of car = ₹ 2,00,000
Decrease in amount from original = 3,00,000 – 2,00,000 = 1,00,000

Question 9.
A number when increased by 18% gives 236. Find the number.
Solution:
Let the number be x. Given that when it is increased by 18%, we get 236.
x + \(\frac{18}{100}\) = 236
\(\frac{100x+18x}{100}\) = 236
\(\frac{118}{100}\)x = 236
The number = x = \(\frac{236×100}{118}\) = 200

Question 10.
A number when decreased by 20% gives 80. Find the number.
Solution:
Let the number be x. Given that when it is increased by 20% we get 80.

Question 11.
A number is increased by 25% and then decreased by 20%. Find the change in that number.
Solution:
Method 1:
Let the number be x. First it is increased by 25%

Method 2:
[to understand, let us assume that number is 100]
So, first when we increase by 25%, we get

We get back 100 ⇒ No change

Question 12.
If the numerator of a fraction is increased by 25% and the denominator is increased by 10%, it becomes \(\frac{2}{5}\). Find the original fraction.
Solution:
Let the fraction be \(\frac{N}{D}\) where N is numerator & D is denominator
It is given that numerator is include by 25%

Question 13.
A fruit vendor bought some mangoes of which 10% were rotten. He sold 33 \(\frac{1}{3}\)% of the rest. Find the total number of mangoes bought by him initially, if he still has 240 mangoes with him.

Solution:
Let the number of mangoes bought by fruit seller initially be x.
Given that 10% of mangoes were rotten
∴ Number of rotten mangoes = \(\frac{10}{100}\) × x
Number of good mangoes = x – no. of rotten mangoes
= x – \(\frac{10}{100}\)x = \(\frac{100x-10x}{100}\) = \(\frac{90}{100}\)x …(1)
Number of mangoes sold = 33\(\frac{1}{3}\)% of good mangoes = \(\frac{100}{3}\)%
∴ Mangoes sold = \(\frac{100}{3}\) x \(\frac{90}{100}\)x × \(\frac{1}{100}\) = \(\frac{30}{100}\)x
Number of mangoes remaining = No. of good mangoes – No. of mangoes sold
From (1) and (2)

Question 14.
A student gets 31% marks in an examination but fails by 12 marks. If the pass percentage is 35%, find the maximum marks of the examination.
Solution:
Let the maximum marks in the exam be ‘x’. Pass percentage is given as 35%
∴ Pass mark = \(\frac{35}{100}\) × x = \(\frac{35}{100}\)x
Student gets 31 % marks = \(\frac{31}{100}\) × x = \(\frac{31}{100}\)x
But student fails by 12 marks → meaning his mark is 12 less than pass mark.
∴ \(\frac{31}{100}\)x = \(\frac{35}{100}\)x – 12
∴ \(\frac{35}{100}\)x – \(\frac{31}{100}\)x = 12
∴ \(\frac{35x-31x}{100}\) = 12 ⇒ \(\frac{4x}{100}\) = 12
∴ \(\frac{12×100}{4}\) = 300

Question 15.
The ratio of boys and girls in a class is 5:3. If 16% of boys and 8% of girls failed in an examination, then find the percentage of passed students.
Solution:
Let number of boys be ‘6’ and number of girls be ‘g’
Ratio of boys and girls is given as 5:3
b:g = 5:3 ⇒ \(\frac{b}{g}\) = \(\frac{5}{3}\) …..(A)
Failure in boys = 16% = \(\frac{16}{100}\) x b = \(\frac{16b}{100}\)
Failure in girls = 8% = \(\frac{8}{100}\) x g = \(\frac{8g}{100}\)
Pass in boys = 100 – 16% = 84% \(\frac{84}{100}\)b …..(1)
Pass in girls = 100 – 8% = 92% = \(\frac{92}{100}\)g
From A, we have \(\frac{b}{g}\) = \(\frac{5}{3}\), = adding 1 on both sides, we get
\(\frac{b}{g}\) + 1 = \(\frac{5}{3}\) + 1
\(\frac{b+g}{g}\) = \(\frac{5+3}{3}\) = \(\frac{8}{3}\)
∴ g = \(\frac{3}{8}\)(b + g)
Similarly b = \(\frac{5}{8}\)(b + g)
Total pass = Pass in girls + Pass in boys
= (1) + (2) = \(\frac{84}{100}\)b + \(\frac{92}{100}\)g

Objective Type Questions

Question 16.
12% of 250 litres is the same as ………. of 150 liters
(a) 10%
(b) 15%
(c) 20%
(d) 30%
Answer:
(c) 20%
Hint:
12% of 250 = \(\frac{12}{100}\) x 250 = 30 lit
Percentage : \(\frac{30}{150}\) x 100 = 20%

Question 17.
If three candidates A, B and C is 3 in a school election got 153, 245 and 102 votes respectively, the percentage of votes for the winner is ……..
(a) 48%
(b) 49%
(c) 50%
(d) 45%
Answer:
(b) 49%
Hint:
Candidate 1: 153
Candidate 2: 245 – winner [as maximum votes]
Candidate 3: 102
Total votes = 1 + 2 + 3 = 153 + 245 + 102 = 500

Question 18.
15% of 25% of 10000 = ………..
(a) 375
(b) 400
(c) 425
(d) 475
Answer:
(a) 375
Hint:
15% of 25% of 10000 is
First let us do 25% of 10,000, which is

Next 15% of the above is \(\frac{15}{100}\) x 2500 = 375

Question 19.
When 60 is subtracted from 60% of a number to give 60, the number is –
(a) 60
(b) 100
(c) 150
(d) 200
Answer:
(d) 200
Hint:
Let the number be ‘x’.
60% of the number is \(\frac{60}{100}\) × x = \(\frac{60x}{100}\)
Given that when 60 is subtracted from 60%, we get 60
i.e \(\frac{60}{100}\)x – 60 = 60
∴ \(\frac{60}{100}\)x = 60 + 60 = 120
∴ x = \(\frac{120×100}{60}\) = 200

Question 20.
If 48% of 48 = 64% of x, then x =
(a) 64
(b) 56
(c) 42
(d) 36
Answer:
(d) 36
Hint:
Given that 48% of 48 48 = 64% of x