## 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.
- Algebra
- Pre algebra
- Geometry
- 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

Hypotenuse^{2} = Base^{2} + Altitude^{2}

Here Base = 5

Hypotenuse = 10

Therefore Altitude^{2} = 10^{2 }- 5^{2}

= 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 x^{2}+ (x + 2)^{2} = 4(x - 2)^{2} - 12

x^{2} + x^{2}+4x+4 = 4(x^{2} - 4x + 4) - 12

x^{2 }+ x^{2} + 4x + 4 = 4x^{2} - 16x + 16 - 12

2 x^{2} + 4x + 4 = 4x^{2} - 16x + 4

We subtract 2x^{2} on both sides

4x + 4 = 2x^{2} - 16x + 4

We add 4x on both side

4 = 2 x^{2} - 20x + 4

Subtracting 4 on both sides

2x^{2} - 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.**