Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 3 Algebra Ex 3.10 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 3 Algebra Ex 3.10

Question 1.

The sum of three numbers is 58. The second number is three times of two-fifth of the first number and the third number is 6 less than the first number. Find the three numbers.

Answer:

Here what we know

a + b + c = 58 (sum of three numbers is 58)

Let the first number be b ‘x’

b = a + 3 (the second number is three times of of the first \(\frac{2}{5}\) number)

b = 3 × \(\frac{2}{5}\)x \(\frac{6}{5}\)x

Third number = x – 6

Sum of the numbers is given as 58.

∴ x + \(\frac{6}{5}\)x + (x – 6) = 58

Multiplying by 5 throughout, we get

5 × x + 6x + 5 × (x – 6) = 58 × 5

5x + 6x + 5x – 30 = 290

∴ 16x = 290 + 30

∴ 16x = 320

∴ x = \(\frac{320}{16}\)

x = 20

1^{st} number = 20

3^{rd} number = 24 – 16 = 14

Question 2.

In triangle ABC, the measure of ∠B is two-third of the measure of ∠A. The measure of ∠C is 200 more than the measure of ∠A. Find the measures of the three angles.

Answer:

Let angle ∠A be a°

Given that ∠B = \(\frac{2}{3}\) × ∠A = \(\frac{2}{3}\)a

& given ∠C = ∠A + 20 = a + 20

Since A, B & C are angles of a triangle, they add up to 180° (∆ property)

∴∠A + ∠B + ∠C = 180°

⇒a + \(\frac{2}{3}\)a + a + 20 = 180°

\(\frac{3 a+2 a+3 a}{3}\) + 20 = 180°

\(\frac{8 a}{3}\) = 180 – 20 = 160

∴ a = \(\frac{160 \times 3}{8}\) = 60°

∠C = 80°

Question 3.

Two equal sides of an isosceles triangle are 5y – 2 and 4y + 9 units. The third side is 2y + 5 units. Find ‘y’ and the perimeter of the triangle.

Answer:

Given that 5y – 2 & 4y + 9 are the equal sides of an isosceles triangle.

∴ The 2 sides are equal

∴5y – 4y = 9 + 2 (by transposing)

∴ y = 11

∴ 1^{st} side = 5y – 2 = 5 × 11 – 2 = 55 – 2 = 53

2^{nd}side = 53 .

3^{rd}side = 2y + 5 = 2 × 11 + 5 = 22 + 5 = 27

Perimeter is the sum of all 3 sides

∴ P = 53 + 53 + 27 = 133 units

Question 4.

In the given figure, angle XOZ and angle ZOY form a linear pair. Find the value of x.

Answer:

Since ∠XOZ & ∠ZOY form a linear pair, by property, we have their sum to be 180°

∴ ∠XOZ + ∠ZOY 180°

∴ 3x – 2 + 5x + 6 = 180°

8x + 4 = 180 = 8x = 180 – 4

∴ 8x = 76 ⇒ x = \(\frac{176}{8}\) ⇒ x = 22°

XOZ = 3x – 2 = 3 × 22 – 2 = 66 – 2 = 64°

YOZ = 5x + 6 = 5 × 22 + 6

= 110 + 6 = 116

Question 5.

Draw a graph for the following data:

Answer:

Graph between side of square & area

When we plot the graph, we observe that it is not a linear relation.

Challenging problems

Question 6.

Three consecutive integers, when taken in increasing order and multiplied by 2, 3 and 4 respectIvely, total up to 74. Find the three numbers.

Answer:

Let the 3 consecutive integers be ‘x’, ‘x + 1’ & ‘x + 2’

Given that when multiplied by 2, 3 & 4 respectively & added up, we get 74

Simplifying the equation, we get

2x + 3x + 3 + 4x + 8 = 74

9x + 11 = 63

9x = 63 ⇒ x = \(\frac{63}{9}\) = 7

First number = 7

Second numbers = x + 1 ⇒ 7 + 1 = 8

Third numbers = x + 2 ⇒ 7 + 2 = 9

∴ The numbers are 7, 8 & 9

Question 7.

331 students went on a field trip. Six buses were filled to capacity and 7 students had to travel in a van. How many students were there in each bus?

Answer:

Let the number of students in each bus be ‘x’

∴ number of students in 6 buses = 6 × x = 6x

A part from 6 buses, 7 students went in van

A total number of students is 331

∴ 6x + 7 = 331

∴ 6x = 331 – 7 = 324

∴ x = \(\frac{324}{6}\) = 54

∴ There are 54 students in each bus.

Question 8.

A mobile vendor has 22 items, some which are pencils and others are ball pens. On a particular day, he is able to sell the pencils and ball pens. Pencils are sold for ₹ 15 each and ball pens are sold at ₹ 20 each. If the total sale amount with the vendor is ₹ 380,

how many pencils did he sell?

Answer:

Let vendor have ‘p’ number of pencils & ‘b’ number of ball pens

Given that total number of items is 22

∴ p + b = 22

Pencils are sold for ₹ 15 each & ball pens for ₹ 20 each

total sale amount = 15 × p + 20 × b

= 15p + 20b which is given to be 380.

∴ 15p + 20b = 380

Dividing by 5 throughout,

\(\frac{15 p}{5}+\frac{20 b}{5}\) = \(\frac{380}{5}\) ⇒ – 3p + 4b = 76

Multiplying equation (1) by 3 we get

3 × p + 3 × b = 22 × 3

⇒ 3p + 3b = 66

Equation (2) – (3) gives

∴ b = 10

∴ p = 12

He sold 12 pencils

Question 9.

Draw the graph of the lines y = x, y = 2x, y = 3x and y = 5x on the same graph sheet. Is there anything special that you find in these graphs?

Answer:

(i) y = x

(ii) y = 2x,

(iii) y = 3x

(iv) y = 5x

(i) y = x

When x = 1, y = 1

x = 2, y = 2

x = 3, y = 2

(ii) y = 2x

When x = 1, y = 2

x = 2, y = 4

x = 3, y = 6

(iii) y = 3x

When x = 1, y = 3

x = 2, y = 6

x = 3, y = 9

(i) y = 5x

When x = 1, y = 5

x = 2, y = 10

x = 3, y = 15

When we plot the above points & join the points to form line, we notice that the lines become progressively steeper. In other words, the slope keeps increasing.

Question 10.

Consider the number of angles of a convex polygon and the number of sides of that polygon. Tabulate as follows:

Use this to draw a graph illustrating the relationship between the number of angles and the number of sides of a polygon.

Answer:

Angles

Name of Polygon | No of angles | No. of Sides |

Triangle | 3 | 3 |

Rectangle | 4 | 4 |

Pentagon | 5 | 5 |

Hexagon | 6 | 6 |

Deptagon | 7 | 7 |

Octagon | 8 | 8 |