9th grade math problems and answers is the page, which gives all support to  students of 9th grade.  You can get help with any of your 9th grade problems.You will get detailed step by step solution with explanation.  You can solve your homework and assignment problems using this page. 

9th grade math word problems

  • Following are few of the topic we cover in the 8th grade math.  You may enter any of the questions in 8th grade curriculum and get benefited.
    1. Algebra
    2. Pre algebra
    3. Geometry
    4. Consumer Math

Following are few word problems solved by our tutors

Example 1: -

A 10 meter ladder is leaning against a building. The bottom of the ladder is 5 meter from the building. How high is the top of the ladder from the ground?

Solution: -
Let us first draw the figure of the problem. The figure is as shown below.


we can see that the figure is a right angled triangle.  So it satisfied Pythagoras theorem.
Pythagoras theorem is given by
Hypotenuse2 = Base2 + Altitude2
Here Base = 5
Hypotenuse = 10
Therefore Altitude2 = 102 - 52
                             = 100 - 25
                             = 75
Therefore Altitude = √75 = 8.66 meters
So the ladder is 8.66 meters high from the ground.

Example 2: -

The sum of the squares of the two larger of  three consecutive even integers is 12 less than 4 times the square of the smaller one. Find the even numbers.
 Solution: -
Let the numbers be  x – 2 , x and x + 2. Given the sum of the square of the two larger of  three consecutive even integers is 12 less than 4 times the square of the smaller one. 
That is x2+ (x + 2)2 = 4(x - 2)2 - 12
x2 + x2+4x+4 = 4(x2 - 4x + 4) - 12
x2 + x2 + 4x + 4 = 4x2 - 16x + 16 - 12
2 x2 + 4x + 4 = 4x2 - 16x + 4
We subtract 2x2 on both sides
4x + 4 = 2x2 - 16x + 4
We add 4x on both side
4 = 2 x2 - 20x + 4
Subtracting 4 on both sides
2x2 - 20x = 0
That is 2x (x - 10) = 0
so x can be either 0 or 10
When x = 0, the even numbers are –2, 0 and 2.
When x = 10, the even numbers are 8, 10 and 12

Example 3: -
John sells 566 drama tickets for $\$$3797.  Tickets cost $\$$5 and admission ticket cost $\$$8.  How much of each type did he sell?
 Solution: -
 Let x be the number of tickets and y be the number of admission tickets he sold.Given that sells 566 tickets for $\$$3797. 
That is x + y = 566…(1)
Also given that tickets cost $\$$5 and admission ticket cost $8.
That is 5x + 8y = 3797 ...(2)
Solving the two equations we get the required answer.
For this first we multiply the first equation by 5, we get
5x + 5y = 2830…(3)
Subtracting equation (1) from (3) we get
-3y = -967
y ≈ 322
Substituting in equation (1), we get
x  + 322 = 566
x = 566 – 322 = 244
Therefore number of tickets he sold = 244
Number of admission tickets he sold = 322.