Algebra Questions and Problems with Detailed Solutions for Grade 6

Detailed solutions to Algebra Questions and Problems for Grade 6 are presented along with explanations.

  1. List all like terms included in each of the expressions given below.
    A) 6 x + 5 + 12 x - 6
    B) 2 x 2 - 4 + 9 x 2 + 9
    C) x / 5 + x / 7 + 7
    D) 0.2 x + 1.2 x + x / 2 + 5
    E) 5 x - 8 + 7 x - 2 x 2 - 4 + 9 x 2 + 4 x 3
    F) 5 a + 8 - 7 b
    G) 5 a b + 8 + 6 b a - a + 3 b
    Solution
    Like terms have the same variable with equal exponents. Also all numbers are like terms.
    A) 6 x and 12 x have same variable x with exponents equal to 1 and are therefore like terms
    5 and - 6 are numbers and therefore like terms
    B) 2 x 2 and 9 x 2 have same variable x with exponents equal to 2 , they are like terms.
    - 4 and + 9 are numbers and therefore like terms.
    C) x / 5 and x / 7 have the same variable with exponents equal to 1, they are like terms
    D) 0.2 x , 1.2 x and x / 2 have the same variable x with exponents equal to 1; they are like terms
    E) 5 x and 7 x are like terms
    - 8 and - 4 are like terms
    - 2 x 2 and + 9 x 2 are like terms
    F) There are no like terms in this expression.
    G) 5 a b and 6 b a are like terms
  2. Evaluate each of the expressions for the given value(s) of the variable(s).
    A) 6 x + 5 for x = 2
    B) 12 x 2 + 5 x - 2 for x = 1
    C) 2( x + 7) + x for x = 0
    D) 2 a + 3b - 7 for a = 2 and b = 4
    Solution
    Substitute x by the given numerical value and simplify.
    A) 6 (2) + 5 = 12 + 5 = 17
    B) 12 (1) 2 + 5 (1) - 2 = 12(1) + 5 - 2 = 12 + 5 - 2 = 15
    C) 2( 0 + 7) + 0 = 2 (7) = 14
    D) 2 (2) + 3 (4) - 7 = 4 + 12 - 7 = 9
  3. Expand (if needed) and simplify each of the expressions below.
    A) 3 x + 5 x
    B) 2( x + 7) + x
    C) 2(x + 3) + 3(x + 5) + 3
    D) 2 (a + 1) + 5b + 3(a + b) + 3
    Solution
    Notes
    1) We may add or subtract like terms as follows:
    2 x + 6 x = (2 + 6) x by factoring first
    = 8 x simplifying the terms in brackets.
    2) We use distributive property to expand an expression as follows: a(b + c) = a b + a c

    A) 3 x + 5 x
    = (3 + 5) x , factor x out
    = 8 x , simplify

    B) 2( x + 7) + x
    = 2(x) + 2(7) + x = 2x + 14 + x , expand and simplify
    = (2x + x) + 14 , group like terms
    = (2 + 1) x + 14 = 3 x + 14 , simplify

    C) 2(x + 3) + 3(x + 5) + 3
    = 2(x) + 2(3) + 3(x) + 3(5) + 3 = 2x + 6 + 3x + 15 + 3 , expand and simplify
    = (2x + 3x) + (6 + 15 + 3) , group like terms
    = (2 + 3) x + 24 = 5 x + 24 , simplify

    D) 2 (a + 1) + 5 b + 3(a + b) + 3
    = 2(a) + 2(1) + 5 b + 3(a) + 3(b) + 3 = 2 a + 2 + 5 b + 3 a + 3 b + 3 , expand and simplify
    = (2 a + 3 a) + (5 b + 3 b) + (2 + 3) , group like terms
    = (2 + 3) a + (5 + 3) b + 5 = 5 a + 8 b + 5 , simplify
  4. Factor each of the expressions below.
    A) 3 x + 3
    B) 8 x + 4
    C) a x + 3 a
    D) (x +1) y + 4 (x + 1)
    E) x + 2 + b x + 2 b
    Solution
    Notes
    1) To factor an expression is to write it as a product.
    2) The distributive property may be used to expand an algebraic expression : a ( x + y) = a x + a y
    3) The same distributive property may be used as : a x + a y = a ( x + y) to factor an expression where a is called the common factor

    A) 3 x + 3
    = 3 (x) + 3 (1) , 3 is a common factor
    = 3(x + 1) , factored form

    B) 8 x + 4
    = 4 (2) (x) + 4 , write 8 as 4 (2)
    = 4 (2 x ) + 4 (1) , 4 is a common factor
    = 4 (2x + 1) , factored form

    C) a x + 3 a , a is a common factor
    = a(x + 3) , factored form

    D) (x +1) y + 4 (x + 1)
    = (x + 1) (y) + (x + 1) (4) , x + 1 is a common factor
    = (x + 1)(y + 4) , factored form

    E) x + 2 + b x + 2 b
    = (x + 2) + b(x + 2) , factor b out in b x + 2 b
    = (x + 2)(1) + (x + 2) b , x + 2 is now a common factor
    = (x + 2)(1 + b) , factored form

    Notes
    In order to check factoring, expand the factored form, simplify and make sure it is equivalent to the given expression.
  5. Solve each of the equations below and check your answer.

    A) x + 5 = 8
    B) 2 x = 4
    C) x / 3 = 2
    D) 0.2 x = 1
    E) 3 x + 6 = 12
    F) 3 (x + 2) + 2 = 8
    Solution
    A) x + 5 = 8
    x + 5 - 5 = 8 - 5 , subtract 5 from both sides of the equation
    x = 3 , simplify and solve for x
    Substitute x by 3 (solution found above) in both sides of the given equation to check the answer
    Right side: (3) + 5 = 8
    Left side = 8
    x = 3 is the solution to the given equation.

    B) 2 x = 4
    2 x / 2 = 4 / 2 , divide both sides by 2
    x = 2 , simplify and solve for x
    Substitute x by 2 (solution found above) in both sides of the given equation to check the answer
    Right side: 2(2) = 4
    Left side = 4
    x = 2 is the solution to the given equation.

    C) x / 3 = 2
    3 (x / 3) = 3 (2) , multiply both sides by 3
    x = 6 , simplify and solve for x
    Substitute x by 6 (solution found above) in both sides of the given equation to check the answer
    Right side: 6 / 3 = 2
    Left side = 2
    x = 6 is the solution to the given equation.

    D) 0.2 x = 1
    0.2 x / 0.2 = 1 / 0.2 , divide both sides by 0.2
    x = 5 , simplify and solve for x
    Substitute x by 5 (solution found above) in both sides of the given equation to check the answer
    Right side: 0.2(5) = 1
    Left side = 1
    x = 5 is the solution to the given equation.

    E) 3 x + 6 = 12
    3 x + 6 - 6 = 12 - 6 , subtract 6 from both sides of the equation
    3 x = 6 , simplify
    3 x / 3 = 6 / 3 , divide both sides of the equation by 300
    x = 2 , simplify and solve for x
    Substitute x by 2 (solution found above) in both sides of the given equation to check the answer
    Right side: 3(2) + 6 = 6 + 6 = 12
    Left side = 12
    x = 2 is the solution to the given equation.

    F) 3 (x + 2) + 2 = 8
    3 (x ) + 3 (2) + 2 = 8 , expand 3 (x + 2)
    3 x + 6 + 2 = 8 , simplify
    3x + 8 = 8 , group like terms
    3 x + 8 - 8 = 8 - 8 , subtract 8 from both sides
    3x = 0 , simplify
    3x / 3 = 0 / 3 , divide both sides by 300
    x = 0 , simplify to solve
    Substitute x by 0 (solution found above) in both sides of the given equation to check the answer
    Right side: 3 (0 + 2) + 2 = 3(2) + 2 = 6 + 2 = 8
    Left side = 8
    x = 0 is the solution to the given equation.
  6. There are n boxes in a large bag and m toys in each box. What is the total number of toys in the bag?
    Solution
    We use examples with 1, 2 , 3 ... boxes and then generalize for n boxes.
    1 box contains     m = 1 m = m toys
    2 boxes contain     m + m = 2 m = 2 m toys
    3 boxes contain     m + m + m = 3 m = 3 m toys
    .
    .
    n boxes contain     m + m + m + .... + m = n m = n m toys
  7. Rewrite the expressions a a a - b b using exponential.
    Solution
    In exponential form, the expressions a a and b b are written as
    a a a = a 3 and b b = b 2
    and the given expression a a - b b is written as
    a a - b b = a 3 - b 2
  8. The length of a rectangle is given by x + 2 and its width is equal to 3. Give a simplified expression of the area of this rectangle.
    Solution
    The A area of a rectangle is given by the product width length.
    A = width length = 3 (x + 2) = 3 (x) + 3(2) , use distributive property
    = 3 x + 6 , simplify
  9. Two thirds of students in a class study math and the remaining eight students do not study math. What is the total number of students in this class?
    Solution
    Let x be the total number of students in the class. If 2 / 3 study math, then
    1 - 2 / 3 = 1 / 3 of the total number of students do not study math
    1 / 3 of the total number of students = (1 / 3) of x = x (1 / 3) = x / 3
    The number of students who do not study math is known and equal to 8. Hence
    x / 3 = 8
    Multiply both sides of the above equation by 3
    3 (x / 3) = 3 (8) Simplify and solve for x
    x = 24 students are in this class
  10. A car travels 60 kilometers in one hour. At the same rate, what distance will be covered by this car in y hour?
    Solution
    We find the distance covered in 1, 2, 3 ... hours and then generalize for y hours to find distance d.
    for 1 hour     d = 60 km = 1 60 km
    for 2 hours     d = 60 km + 60 km = 2 60 km
    for 3 hours     d = 60 km + 60 km + 60 km= 3 60 km
    .
    .
    for y hours     d = 60 km + 60 km +. . .+ 60 km = y 60 km = 60 y km
  11. Evaluate the expressions.
    A) 2 3 + 3 2
    B) 0.1 3
    C) 6 (2 / 3)
    Solution
    A) 2 3 + 3 2 = 2 2 2 + 3 3 = 8 + 9 = 17
    B) 0.1 3 = 0.1 0.1 0.1 = 0.001
    C) 6 (2 / 3) = 6 2 / 3 = 12 / 3 = 4
  12. A bag contains three red marbles; five blue marbles and seven green marbles. What is the ratio of blue marbles to the total number of marbles?
    Solution
    The total number of marbles is
    3 + 5 + 7 = 15
    Ratio of blue marbles to the total number of marbles is
    5 / 15
    Divide numerator and denominator by the common factor 5 to reduce the above fraction
    (5 5) / (15 5) = 1 / 3
    The ratio of blue marbles to the total number of marbles is
    1:3
  13. Solve the proportion : a / 3 = 5 / 18
    Solution
    Use the cross product on the given equation
    a 18 = 3 5
    Divide both sides by 18
    a 18 / 18 = 3 5 / 18
    Simplify and solve
    a = 5 / 6
  14. Which of the following ordered pairs is a solution to the equation 2 x + 3 y = 8?
    A) (0 , 0)
    B) (4 , 0)
    C) (1 , 2)
    Solution
    A)Substitute x by 0 and y by 0 in the given equation
    2 (0) + 3 (0) = 8
    Simplify the left side of the equation
    0 = 8 , false statement and therefore (0,0) is not a solution to the given equation

    B)Substitute x by 4 and y by 0 in the given equation
    2 (4) + 3 (0) = 8
    Simplify the left side of the equation
    8 = 8 , true statement and therefore (4 , 0) is a solution to the given equation

    C)Substitute x by 1 and y by 2 in the given equation
    2 (1) + 3 (2) = 8
    Simplify the left side of the equation
    8 = 8 , true statement and therefore (1 , 2) is a solution to the given equation
  15. List all factors of the following numbers.
    A) 4
    B) 12
    C) 50
    Solution
    A) Factors of 4 are: 1 , 2 , 4
    B) Factors of 12 are: 1 , 2 , 3 , 4 , 6 , 12
    C) Factors of 50 are: 1 , 2 , 5 , 10 , 25 , 50
  16. Find the greatest common factor to each pair of the given numbers.
    A) 6 and 3
    B) 18 and 24
    C) 50 and 60
    Solution
    Make a list of all factors for each number in the given pair and select the common factor to the two numbers that is the largest
    A) Factors of 6 are: 1 , 2 , 3 , 6             Factors of 3 are: 1 , 3
    the greatest common factor of 6 and 3 is 3
    B) Factors of 18 are: 1 , 2 , 3 , 6 , 9 , 18             Factors of 24 are: 1 , 2 , 3 , 4 , 6 , 8 , 12 , 24
    the greatest common factor of 18 and 24 is 6
    C) Factors of 50 are: 1 , 2 , 5 , 10 , 25 , 50             Factors of 60 are: 1 , 2 , 3 , 4 , 5 , 6 , 10 , 12 , 15 , 20 , 30 , 60
    the greatest common factor of 50 and 60 is 10
  17. Write the number " Six hundred seventy-two million two hundred fifty-nine" using digits.
    Solution
    672,000,259
  18. Which of the following expressions are equivalent?
    A) 2(x + 3) - 2
    B) 2x + 4
    C) 3(x + 3) - x - 5
    Solution
    Expand and simplify the given expressions if necessary
    A) 2(x + 3) - 2 = 2(x) + 2(3) - 2 = 2 x + 6 - 2 = 2 x + 4
    B) 2x + 4
    C) 3(x + 3) - x - 5 = 3(x) + 3(3) - x - 5 = 3x + 9 - x - 5 = (3x - x) + (9 - 5) = 2 x + 4 All three expressions are equivalent
  19. Find the lowest common multiple to each pair of the given numbers.
    A) 2 and 3
    B) 7 and 14
    C) 25 and 15
    Solution
    Make a list of the first few multiples for each number in the given pair and select the common multiple to the two numbers that is the lowest. Multiples of a number are obtained by multiplying the number by 1 ,2 , 3 ...
    A) First few multiples of 2 are: 2 , 4 , 6 , 8 ...            First few multiples of 3 are: 3 , 6 , 9 , ...
    lowest common multiple of 2 and 3 is 6
    B) First few multiples of 7 are: 7 , 14 , 21 ...            First few multiples of 14 are: 14 , 28 , 42 ...
    lowest common multiple of 7 and 14 is 14
    C) First few multiples of 25 are: 25 , 50 , 75 ...             First few multiples of 15 are: 15 , 30 , 45 , 60 , 75 ...
    lowest common multiple of 25 and 15 is 75
  20. Add and / or subtract and simplify
    A) 1 / 3 + 2 / 3
    B) 2/5 - 1 / 7
    Solution

    A) The two fractions 1 / 3 and 2 / 3 have same (common) denominators 3 and can therefore be added by adding the numerators and keep the common denominator.
    1 / 3 + 2 / 3 = (1 + 2) / 3 = 3 / 3 = 1
    B) The two fractions 2 / 5 and 1 / 7 do not have same (common) denominators and we therefore need to first find a common denominator by finding the lowest common multiple of the denominators 5 and 7.
    First few multiples of 5 are: 5 , 10 , 15 , 20 , 25 , 30 , 35 ...            First few multiples of 7 are: 7 , 14 , 21, 18 , 35 , ...
    lowest common multiple of 5 and 7 is 35
    We now convert the fractions 2 / 5 and 1 / 7 to equivalent fractions with common denominator equal to 35.
    To covert 2 / 5 to an equivalent fraction with a denominator equal to 35, we need to multiply its numerator and denominator by 7. Hence
    2 / 5 = (7 2) / (7 5) = 14 / 35
    To covert 1 / 7 to an equivalent fraction with a denominator equal to 35, we need to multiply numerator and denominator by 5. Hence
    1 / 7= (5 1) / (5 7) = 5 / 35
    We now substitute the fractions in the expressions to simplify by their equivalent fractions with common denominator.
    2/5 - 1 / 7 = 14 / 35 - 5 / 35 = (14 - 5) / 35 = 9 / 35
  21. What is 2 / 3 of 21?
    Solution
    2 / 3 of 21 is mathematically written as
    (2 / 3) 21
    Simplify the above
    = (2 / 3) (21 / 1) = (2 21) / (3 1) = 42 / 3 = 14
  22. What is 40% of 1 / 4?
    Solution
    40% of 1 / 4 is mathematically written as
    40% (1 / 4)
    Write percentage as a fraction and simplify
    = (40 / 100) 1 / 4 = (40 1)/ (100 4) = 40 / 400 = 1 / 10
  23. What is 20% of 50%?
    Solution
    20% of 50% is mathematically written as
    20% 50%
    Write percentages as fractions and simplify
    = (20 / 100) (50 / 100) = (20 50) / (100 100) = 1000 / 10000 = 1 / 10
  24. How many seconds are there in one hour?
    Solution
    There are 60 minutes in one hour
    1 hour = 60 minutes
    There 60 seconds in 1 minute; hence
    1 hour = 60 minutes = 60 60 seconds = 3600 seconds
  25. How many minutes are there in the month of January?
    Solution
    There are 31 days in the month of January
    January = 31 days
    There 24 hours in one day; hence
    January = 31 days = 31 24 hours
    In one hour there are 60 minutes; hence
    January = 31 24 60 minutes = 44640 minutes
  26. What is the location of each of the points: (-2 , 0) , (0 , 3) and (-2 , - 3) in a coordinate plane?
    Solution
    (-2 , 0) is on the x axis 2 units to the left of the origin (0 , 0)
    (0 , 3) is on the y axis 3 units above the origin (0 , 0)
    (-2 , - 3) is in quadrant 3
  27. Order   7 / 5 , 12 / 10 , 21 / 20 and 111%   from the smallest to the largest.
    Solution
    We need to convert all the given fractions to equivalent fractions the same denominator 100
    7 / 5 = (20 7) / (20 5) = 140 / 100
    12 / 10 = (10 12) / (10 10) = 120 / 100
    21 / 20 = (5 21) / (5 20) = 110 / 100
    111% = 111 / 100
    We now have all fractions with common denominator and we can compare them by comparing the numerators.
    smallest to largest: 110 / 100 ; 111 / 100 ; 120 / 100 , 140 / 100
    which are the equivalent of the original fractions
    21 / 20 ; 111% ; 12 / 10 ; 7 / 5     order from smallest to largest.

More Middle School Math (Grades 6, 7, 8, 9) - Free Questions and Problems With Answers
More High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers
More Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers
Home Page