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6th Grade Math Problems

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

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Students who find it difficult to keep up with math lessons in school can enlist the help of live tutors who are experienced and know the subject well. Students work with tutors daily or weekly, as per their convenience and receive real-time help. These virtual sessions allow students to clear all their doubts and get explanations on topics that are not clear. Tutors also provide 6th grade math problems and answers enabling students to learn on the spot. Math word problems for 6th grade are also taught and students can ask their tutor when they have trouble with any words or phrases.

Solved Examples

Question 1: Solve (2x + y - 3z)2
Solution:
Given, (2x + y - 3z)2

 We know

(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

=> (2x + y - 3z)= (2x)2 + (y)2 + (-3z)2 + 2 * 2x * y + 2 * y * (- 3z) + 2 * (- 3z) * 2x

= 4x2 + y2 + 9z 2 + 4xy - 6yz - 6zx

=> (2x + y - 3z)2 = 4x2 + y2 + 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.

 

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

=> LCM = 72

 

Math Word Problems for 6th Grade

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Math word problems for 6th grade are available on a number of websites along with solved examples and step-by-step solutions. Worksheets with 6th grade math practice problems can be printed out and shared with friends to work out together. 6th grade math problems with answers let you test yourself to see how well you have grasped the concept. Exams do not have to be a problem anymore if you are using online math solvers as they are taken care of with exam prep and mock tests.

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, 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

 

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, First number = 7

Second number = 3