6th grade math word problems can be pretty challenging, especially when you are trying to get your way around fractions, decimals, percentages and proportions. However, there is an easy way to crack those math problems and it is as simple as going online. Online 6th grade math problems give students plenty of opportunities to practice math problems for 6th grade from the comfort of their homes. Online math solvers are growing in popularity as more and more students turn to this instantly available source of help for their math needs. Free 6th grade math problems cover all the topics that students learn in class and give students the opportunity to practice every day. ‘Practice makes perfect’ holds perfectly true for math, so online 6th grade math problems are the perfect way for students to keep improving at the subject.

## 6th Grade Math Practice Problems

## Solved Examples

**Question 1:**Solve (2x + y - 3z)

^{2}

**Solution:**

Given, (2x + y - 3z)

(a + b + c)

=> (2x + y - 3z)

= 4x

=> (2x + y - 3z)

^{2}We know(a + b + c)

^{2}= a^{2}+ b^{2}+ c^{2}+ 2ab + 2bc + 2ca=> (2x + y - 3z)

^{2 }= (2x)^{2}+ (y)^{2}+ (-3z)^{2}+ 2 * 2x * y + 2 * y * (- 3z) + 2 * (- 3z) * 2x= 4x

^{2}+ y^{2}+ 9z^{ 2}+ 4xy - 6yz - 6zx=> (2x + y - 3z)

^{2}= 4x^{2}+ y^{2}+ 9z^{ 2}+ 4xy - 6yz - 6zx**Question 2:**Evaluate the expression when x = 1.

12x + 3y = 16

**Solution:**

Given, 12x + 3y = 16

Step 1:

Subtract 12x from both sides

=> 12x + 3y -12x = 16 - 12x

=> 3y = 16 - 12x

Step 2:

Divide each side by 3

=> $\frac{3y}{3} = \frac{16 - 12x}{3}$

=> y = $ \frac{16 - 12x}{3}$

Step 3:

Put x = 1

=> y = $ \frac{16 - 12 * 1}{3}$

=> y = $ \frac{16 - 12}{3}$

=> y = $ \frac{4}{3}$, is the answer.

Step 1:

Subtract 12x from both sides

=> 12x + 3y -12x = 16 - 12x

=> 3y = 16 - 12x

Step 2:

Divide each side by 3

=> $\frac{3y}{3} = \frac{16 - 12x}{3}$

=> y = $ \frac{16 - 12x}{3}$

Step 3:

Put x = 1

=> y = $ \frac{16 - 12 * 1}{3}$

=> y = $ \frac{16 - 12}{3}$

=> y = $ \frac{4}{3}$, is the answer.

**Question 3:**Find the Lowest common multiple of 36 and 24 having GCD 12.

**Solution:**

Two numbers are 36 and 24

GCD(24, 36) = 12

LCM = ?

We know that

Product of two numbers = Product of their LCM and GCD

=> 24 * 36 = LCM * 12

=> LCM = $\frac{24 * 36}{12}$

=>

GCD(24, 36) = 12

LCM = ?

We know that

Product of two numbers = Product of their LCM and GCD

=> 24 * 36 = LCM * 12

=> LCM = $\frac{24 * 36}{12}$

=>

**LCM = 72**

## Math Word Problems for 6th Grade

## Solved Examples

**Question 1:**Find the perimeter of a rectangular garden with a length of 25 feet and width of 20 feet.

**Solution:**

Length of a rectangular garden = 25

width of a rectangular garden = 20

Perimeter of rectangle = 2(Length + width)

=> Perimeter of rectangular garden = 2(25 + 20)

= 2(45)

= 90

Hence,

width of a rectangular garden = 20

Perimeter of rectangle = 2(Length + width)

=> Perimeter of rectangular garden = 2(25 + 20)

= 2(45)

= 90

Hence,

**Perimeter of rectangular garden is 90 feet****Question 2:**Find the two consecutive integers such that total of ten times the second integer and thrice the first integer is thirty-six.

**Solution:**

Let first integer = x

Second integer = x + 1

According to statement:

=> 10(x + 1) + 3x = 36

Solve for x,

=> 10x + 10 + 3x = 36

=> 13x + 10 = 36

=> 13x = 36 - 10

=> 13x = 26

=> x = $\frac{26}{13}$

=> x = 2

The first integer = 2

Second integer = 2 + 1 = 3

Second integer = x + 1

According to statement:

=> 10(x + 1) + 3x = 36

Solve for x,

=> 10x + 10 + 3x = 36

=> 13x + 10 = 36

=> 13x = 36 - 10

=> 13x = 26

=> x = $\frac{26}{13}$

=> x = 2

The first integer = 2

Second integer = 2 + 1 = 3

**Question 3:**The sum of two numbers is 10 and the difference is 4. What are the numbers?

**Solution:**

Let first number = x

and second number = y

The problem states:

x + y = 10 ..........................(1)

x - y = 4 ............................(2)

Adding equation(1) and (2)

=> 2x = 14

=> x = 7, put in (1)

=> 7 + y = 10

Subtract 7 from both sides

=> 7 + y - 7 = 10 - 7

=> y = 3

Hence,

and second number = y

The problem states:

x + y = 10 ..........................(1)

x - y = 4 ............................(2)

Adding equation(1) and (2)

=> 2x = 14

=> x = 7, put in (1)

=> 7 + y = 10

Subtract 7 from both sides

=> 7 + y - 7 = 10 - 7

=> y = 3

Hence,

**First number = 7**

Second number = 3Second number = 3