Math notes for grade 8:

In this article we are going to discuss about find the Math Notes for grade 8.Math notes for grade 8 are easy to understand and solve. Math notes for grade 8 problems involve basic addition, subtraction, multiplication and division problems. Those arithmetic problems involve biggest numbers. The following topics are studied in the 8 grade,

Numbers
Measures
Algebra
Geometry
Handling data.
The math notes for grade 8 solving problems are given below.

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Example problems of math notes for grade 8:


Example 1:

Sum of 3 consecutive odd numbers is 51. Find the numbers.

Solution:

Framing the equation:

Let the first odd numbers be x.

Then the second and third odd numbers are (x+2) and (x+4)

There sum is 51.

x+(x+2)+(x+4) = 51

x+x+2+x+4 = 51

3x + 6 = 51(adding the like terms)

3x+6 = 51 is the required equation.

Solving the equation:

3x+6 = 51

3x = 51 – 6 (by rule 1)

= 45

x = 45 * `1/3 ` (by rule 3)

x = 15

The consecutive odd numbers are x, (x=2), (x=4)

15, (15+2) and (15+4)

The required numbers are 15, 17 and 19.

Example 2:

If office managers fixed in the pay scale 3200 – 85 – 4900, when will he reach his maximum?

Solution:

Pay scale: 3200 – 85 – 4900

Starting salary = $ 3200= a, Annual increment = $85 = d,

Maximum salary = $4900 = tn

tn = a + (n – 1)d = 4900 = 3200+(n-1) 85

n – 1 = 1700/85 = 20

n =20+1 = 21

The manager will reach his maximum in his 21st year of service.Having problem with math problems for 3rd grade keep reading my upcoming posts, i will try to help you.


Example 3:


The base and height of a right triangular ground are 60 m and 45 m. Find the cost of leveling the ground at Rs 150 per are ( 1 are = 100 sq.m)

Solution:

Given base b = 60m; height, h = 45m

Cost of leveling 1 are = Rs.150

Area of the triangular, A = `1/2` * 60 * 45

= 1350m2

100m2 = 1are

1350m2 = `1350/100` = 13.5are

Cost of leveling 1 are = Rs.150

Cost of leveling 13.5 Ares =Rs. 13.5 * 150 = Rs. 2025

Cost of leveling the right triangular ground = Rs. 2025

Example 4:

Find the circumference of a circle whose radius is 7cm.

Solution:

Radius r = 7cm

Circumference, C = 2`pi` r

= 2 * 22/7 * 7

= 44

Circumference of the circle = 44cm.

Example 5:

Solve 6x + 12 = 4x – 2

Solution:

Given expression 6x + 12 = 4x – 2

Subtract 12 on both sides of the equation

6x +12 – 12 = 4x – 2 -12

6x = 4x -14

Subtract 4x on both sides of the equation

6x – 4x = 4x – 4x -14

2x = -14

Divide 2 on both sides of the equation

`(2x)/2` =` -14/2`

x = -7

Solution is x = -7