Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.5

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 2 Algebra Ex 2.5

Miscellaneous Practice Problems

Question 1.
The sum of three numbers is 58. The second numberis three times of two-fifth of the first number and the third number is 6 less than the first number. Find the three numbers.
Solution:
Here what we know
a + b + c = 58 (sum of three numbers is 58)
Let the first number be ‘x’
b = a + 3 (the second number is three times of \(\frac{1}{2}\) of the first number)
b = 3 x \(\frac{2}{5}\)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 x 5
5x + 6x + 5x – 30 = 290
∴ 16x = 290 + 30
∴ 16x = 320
∴ x = \(\frac{320}{16}\)x = 20
Answer:
1st number =20
2nd number = 3 x \(\frac{2}{5}\) x 20 = 24
3rd number = 24 – 6 = 14

Question 2.
In triangle ABC, the measure of ∠B is two third the measure of ∠A. The measure of ∠C is 20° more than the measure of ∠A. Find the measures of the three angles.
Solution:
Let angle ∠A be a°
Given that ∠B = \(\frac{2}{3}\) x ∠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°
\(\frac{3a+2a+3a}{3}\) + 20 = 180°
\(\frac{8a}{3}\) = 180 – 20 = 160°

∠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.
Solution:
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
∴ 1st side = 5y – 2 = 5 x 11-2 = 55 – 2 = 53
2nd side = 53
3rd side = 2y + 5 = 2 x 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.

Solution:
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 = 176 ⇒ x = \(\frac{176}{8}\) ⇒ x = 22°
XOZ = 3x – 2 = 3 x 22 – 2 = 66 – 2 = 64°
YOZ = 5x + 6 = 5 x 22 + 6
= 110 + 6 = 116

Question 5.
Draw a graph for the following data:

Does the graph represent a linear relation?
Solution:
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.
Solution:
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 = 74
9x = 63 ⇒ x = \(\frac{63}{9}\) = 7
First number = 7
Second numbers = x + 1
Third numbers = x + 2
∴ 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?
Solution:
Let the number of students in each bus be ‘x’
number of students in 6 buses = 6 × x = 6x
Apart 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?
Solution:
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 x p + 20 x b
= 15p + 20b which is given to be 380.
∴ 15p + 20b = 380
Dividing by 5 throughout,
\(\frac{15p}{5}\) + \(\frac{20b}{5}\) = \(\frac{380}{5}\)
⇒ 3p + 4b = 76
Multiplying equation (1) by 3 we get
3 x 7 + 3 x b = 22 x 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?
Solution:

1. y = x
2. y = 2x
3. y = 3x
4. y = 5x

1. y = x
When x = 1, y = 1
x = 2, y = 2
x = 3, y = 2

2. y = 2x
When x = 1, y = 2
X = 2, y = 4
X = 3, y = 6

3. y = 3x
when x = 1, y = 3
x = 2, y = 6
x = 3, y = 9

4. y = 5x
When x = 1, y = 5
x = 2, y = 10
x = 3, y = 15

When we plot the alijpve 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 draw a graph illustrating the relations hip between the number of angles and the number of sides of a polygon.
Solution:
Angles: