Solutions to Algebra Questions and Problems for Grade 7

Solutions to these Algebra Questions and Problems are presented.

  1. Evaluate each of the expressions for the given value(s) of the variable(s).
    1. 12 x 3 + 5 x 2 + 4 x - 6 for x = -1
    2. 2 a 2 + 3b 3 - 10 for a = 2 and b = -2
    3. (- 2 x - 1) / (x + 3) for x = 2
    4. 2 + 2 |x - 4| for x = - 4
    Solution
    Substitute each variable by its given numerical value and simplify.
    1. 12 (-1) 3 + 5 (-1) 2 + 4 (-1) - 6 = 12 (-1) + 5 (1) - 4 - 6 = - 12 + 5 - 4 - 6 = - 17
    2. 2 (2) 2 + 3 (- 2) 3 - 10 = 2(4) + 3(-8) - 10 = 8 - 24 - 10 = - 26
    3. (- 2 (2) - 1) / ((2) + 3) = (- 4 - 1) / (5) = - 5 / 5 = - 1
    4. 2 + 2 |(- 4) - 4| = 2 + 2 |- 8 | = 2 + 2(8) = 2 + 16 = 18

  2. Expand and simplify each of the expressions below.
    1. - 2( x - 8) + 3 (x - 7)
    2. 2 (a + 1) + 5b + 3(a + b) + 3
    3. a (b + 3) + b (a - 2) + 2 a - 5 b + 8
    4. (1 / 2) (4x + 4) + (1 / 3)(6x + 12)
    5. 4 ( - x + 2 - 3(x - 2) )
    Solution
    Use the distribution rule a (b + c) = a b + a c = a (b) + a (c) to expand product terms and group like terms.
    1. - 2( x - 8) + 3 (x - 7) = - 2(x) - 2(-8) + 3(x) + 3(-7)
      = -2 x + 16 + 3 x - 21 = (- 2x + 3x) +(16 - 21) = x - 5
    2. 2 (a + 1) + 5b + 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 = (2 a + 3 a) + ( 5 b + 3 b) + (2 + 3) = 5 a + 8 b + 5
    3. a (b + 3) + b (a - 2) + 2 a - 5 b + 8 = a (b) + a (3) + b(a) + b(-2) + 2 a - 5 b + 8
      = a b + 3 a + b a - 2 b + 2 a - 5 b + 8 = ( a b + b a) +(3 a + 2 a) + (-2 b - 5 b) + 8
      = 2 a b + 5 a - 7 b + 8
    4. (1 / 2) (4x + 4) + (1 / 3)(6x + 12) = (1/2)(4 x) + (1/2)( 4 ) +(1/3)(6x) + (1/3)(12)
      = 4 x / 2 + 4 / 2 + 6 x / 3 + 12 / 3 = 2 x + 2 + 2 x + 4
      = (2 x + 2 x) + (2 + 4) = 4 x + 6
    5. 4 ( - x + 2 - 3(x - 2) ) = 4 ( - x) + 4 (2) + 4 (-3) (x - 2)
      = - 4 x + 8 - 12 (x - 2) = - 4 x + 8 - 12 x + 24 = (- 4 x - 12 x) +(8 + 24) = - 16 x + 32

  3. Simplify each of the expressions below.
    1. x / y + 4 / y
    2. (2 x / 4 ) (1 / 2)
    3. (3 x / 5 ) (x / 5)
    Solution
    We use the rules of multiplication, division and addition to simplify the expressions with numerators and denominators given below.
    1. x / y + 4 / y = (x + 4) / y
    2. (2 x / 4 ) (1 / 2) = (2 x 1) / (4 2) = 2 x / 8 = x / 4
    3. (3 x / 5 ) (x / 5) = (3 x / 5) (5 / x) = (3 x 5) / (5 x) = 1 5 x / 5 x = (15 / 5) (x / x) = 3 (1) = 3

  4. Simplify each of the expressions below.
    1. 3 x 2 5 x 3
    2. [ (2 y) 4 9 x 3] [ 4 y 4 (3 x) 2 ]
    Solution
    We use the rules of multiplication and division of expressions with exponents.
    1. 3 x 2 5 x 3 = (35) (x 2 x 3) = 1 5 x 2 + 3 = 1 5 x 5
    2. [ (2 y) 4 9 x 3] [ 4 y 4 (3 x) 2 ] = [ (2) 4 y 4 9 x 3 ] [ 4 y 4 (3 )2 x 2 ]
      = [ 16 y 4 9 x 3 ] [ 4 y 4 9 x 2 ]
      = [ (16 9 ) (49)] [ (y 4 x 3) ( y 4 x 2) ] = 4 y 4 - 4 x3 - 2 = 4 x

  5. Factor fully each of the expressions below.
    1. 9 x - 3
    2. 24 x + 18 y
    3. b x + d x
    Solution
    Find a common factor then use the rule of distribution a (b + c) = a b + a c from right to left a b + a c = a (b + c) to factor the given expression.
    1. 9 x - 3
      = 3 (3 x) + 3(1) 3 is a common factor
      = 3(3x + 1) use distribution from right to left to factor
    2. 24 x + 18 y
      = 6(4 x) + 6 (3 y) 6 is a common factor
      = 6(4 x + 3 y) factor
    3. b x + d x
      = x(b) + x(d) x is a common factor
      = x(b + d) factor

  6. Solve each of the equations below and check your answer.
    1. 2 x + 5 = 11
    2. 3 x = 6 / 5
    3. 3 (2 x + 2) + 2 = 20
    Solution
    Add, subtract, multiply by or divide by the same number both sides of the equation till it is solved.
    1. 2 x + 5 = 11
      2 x + 5 - 5 = 11 - 5 subtract 5 from both sides
      2x = 6 simplify
      2x / 2 = 6 / 2 divide both sides by 2
      x = 3 simplify
    2. 3 x = 6 / 5
      5(3 x) = 5(6 / 5) multiply both sides by 5
      15 x = 6 simplify
      15 x / 15 = 6 / 15 divide both sides by 15
      x = 6 / 15 simplify
      x = 2 / 5 reduce fraction
    3. 3 (2 x + 2) + 2 = 20
      3(2x) + 3(2) + 2 = 20 expand
      6 x + 6 + 2 = 20
      6 x + 8 = 20 simplify
      6x + 8 - 8 = 20 - 8 subtract 8 from both sides
      6x = 12 simplify
      6 x / 6 = 12 / 6 divide both sides by 6
      x = 2 simplify

  7. Rewrite the expressions 3 a a a - 5 b b using exponential.
    Solution
    Use exponents to rewrite
    a a a = a 3 and b b = b 2
    Substitute in the given expression
    3 a a a - 5 b b = 3 a 3 - 5 b 2
  8. A rectangle has a length given by 2 x + 3 units, where x is a variable. The width of the rectangle is given by x + 1 units. Find the value of x if the perimeter of the rectangle is equal to 32.
    Solution
    Perimeter P of a rectangle is given by
    P = 2 length + 2 width
    Substitute length by 2 x + 3 and width by x + 1
    P = 2 (2x + 3) + 2(x + 1)
    Perimeter P is given as 32, hence
    2 (2x + 3) + 2(x + 1) = 32
    Solve the above equation
    4 x + 6 + 2 x + 2 = 32 expand
    6 x + 8 = 32 group like terms
    6 x + 8 - 8 = 32 - 8 Subtract 8 from both sides
    6 x = 24 simplify
    6 x / 6 = 24 / 6 divide both sides by 6
    x = 4
  9. A rectangle has a length given by 2x - 1 units, where x is a variable. The width of the rectangle is equal to 3 units. Find the value of x if the area of the rectangle is equal to 27.
    Solution
    Area A of a rectangle is given by
    P = width length
    Substitute width by 3 and length by 2 x - 1
    P = 3 (2x - 1)
    Area A is given as 27, hence
    3 (2x - 1) = 27
    Solve the above equation
    6 x - 3 = 27 expand
    6 x - 3 + 3 = 27 + 3 add 3 to both sides
    6 x = 30 simplify
    6 x / 6 = 30 / 6 divide both sides by 6
    x = 5
  10. 45% of the students in a school are male? Find the ratio of the number of female to the total number of male students in this school.
    Solution
    If 45% of students are male then
    100% - 45% = 55% of students are female
    Ratio R of females to males is
    R = 55% / 45% = 55/45
    Reduce fraction
    R = 11/9 or 11:9
  11. A car travels at the speed x + 30 kilometers in one hour, where x is an unknown. Find x if this car covers 300 kilometers in 3 hours?
    Solution
    Distance = time speed, hence
    300 = 3(x + 30)
    Expand
    300 = 3 x + 90
    Subtract 90 from both sides of the equation
    300 - 90 = 3 x + 90 - 90
    210 = 3 x
    210/ 3 = 3 x / 3
    70 = x
  12. Solve the proportion : 4 / 5 = a / 16
    Solution
    Multiply both sides of the equation by the product of the two denominators: 516
    516(4 / 5) = 516(a / 16)
    Simplify
    164= 5a (which is also called cross product)
    Divide both sides by 5
    164 / 5 = 5 a / 5
    Simplify
    64 / 5 = a
  13. Find a if the ordered pair (2 , a + 2) is a solution to the equation 2 x + 2 y = 10?
    Solution
    Substitute the x (=2) and y (= a + 2) values of the ordered pair into the given equation
    2 (2) + 2 (a + 2) = 10
    Expand term in the above equation
    4 + 2 a + 4 = 10
    Simplify
    2 a + 8 = 10
    Solve for a
    2 a + 8 - 8 = 10 - 8
    2a = 2
    a = 1
  14. Find the greatest common factor of the numbers 25 and 45.
    Solution
    Factors of 25 and 45 are
    25: 1, 5, 25
    45: 1, 3, 5 , 9, 15, 45
    When we examine the two lists, 25 and 45 have two common factors: 1and 5 and the greatest is 5. Hence greatest common factor of the numbers 25 and 45 is
    5
  15. Write the number " one billion, two hundred thirty four million, seven hundred fifty thousand two " using digits. Solution
    1234750002
  16. Write the number 393,234,000,034 in words. Solution
    three hundred ninety three billion, two hundred thirty four million thirty four
  17. Find the lowest common multiple to the numbers 15 and 35.
    Solution
    Multiples of 15 and 35 are
    15: 15 , 30 , 45 , 60 , 75 , 90 , 105 , 120 , 135 , ....
    35: 35 , 70 , 105 , 140, ...
    When we examine the two lists and15 and 35 have the lowest common multiple equal to 105
  18. Find x if 2 / 3 of x is 30?
    Solution
    2 / 3 of x is 30 is mathematically written as
    (2 / 3) x = 30
    We now need to solve the above equation. Multiply the two sides of the equations by 3
    3 (2 / 3) x = 3 30
    Simplify
    2 x = 90
    Divide both sides of the equation by 2
    2x / 2 = 90 / 20
    x = 45
  19. What is 20% of 1 / 3?
    Solution
    20% of 1 / 3 is written as
    20% (1 / 3)
    = (20 / 100)(1 / 3) write 20% as a fraction: 20 / 100
    = (201) / (100 3) multiply fractions
    = 20 / 300 simplify
    = 1 / 15 reduce fraction

  20. Order 12 / 5 , 250% , 21 / 10 and 2.3 from the smallest to the largest.
    Solution
    Convert all number in decimal form
    12 / 5 = 2.4
    250% = 2.5
    21 / 10 = 2.1
    2.3 = 2.3
    Order from smallest to largest
    21/10 , 2.3 , 12 / 5 , 250%
  21. The sum of 3 positive consecutive integers is equal to 96. Find the largest of these numbers.
    Solution
    Let x be the smallest of these numbers. Since the three integers are consecutive, we can write them as: x , x + 1 and x + 2. Their sum is 96, hence
    x + (x + 1) + (x + 2) = 96
    x + x + 1 + x + 2 = 96 Expand terms in parentheses
    (x + x + x) + (1 + 2) = 96 group like terms
    3 x + 3 = 96 simplify
    3 x + 3 - 3 = 96 - 3 subtract 3 from both sides of the equation
    3 x = 93 simplify
    3 x / 3 = 93 / 3 Divide by 3
    x = 31 Solve
    The largest of these numbers is x + 2, hence
    x + 2 = 31 + 2 = 33
  22. Dany scored 93 in physics, 88 in mathematics, and a score in chemistry that is double his score in geography. The average score of all 4 courses is 79. What were his scores in chemistry and geography?
    Solution
    Let x be Dany's score in geography. His score in chemistry is double his score in geography and it is equal to
    2 x
    The average of all four scores is 79. Hence
    (93 + 88 + x + 2 x) / 4 = 79
    Multiply both sides of the equation by 4
    4(93 + 88 + x + 2 x) / 4 = 479
    Simplify
    93 + 88 + x + 2 x = 316
    Group like terms
    3 x + 181 = 316
    Solve for x
    3 x = 135
    3 x / 3 = 135 / 3
    x = 45
    score in geography = x = 45
    score in chemistry = 2 x = 2 45 = 90
  23. Linda scored a total of 265 points in mathematics, physics and English. She scored 7 more marks in mathematics than in English and she scored 5 more marks in physics than in mathematics. Find her scores in all three subjects.
    Solution
    Let x be Linda's score in English. Her score in mathematics is 7 more her score in English, hence her score in mathematics is
    x + 7
    Linda scored 5 more marks in physics than in mathematics, hence her score in physics is
    (x +7) + 5 = x + 12
    The total of all three scores is 265, hence
    x + (x + 7) + (x + 12) = 265
    group like terms
    3 x + 19 = 265
    Solve for x
    3 x = 246
    3 x / 3 = 246 / 3
    x = 82
    score in English = x = 82
    score in mathematics = x + 7 = 82 + 7 = 89
    score in physics = x + 12 = 82 + 12 = 94
  24. There are bicycles and cars in a parking lot. There is a total of 300 wheels including 100 small wheels for bicycles. How many cars and how many bicycles are there?
    Solution
    100 small wheels for bicycles gives
    100 / 2 = 50 bicycles
    A total of 300 wheels and 100 wheels for bicycles gives
    300 - 100 = 200 wheels for cars
    200 wheels for cars gives
    200 / 4 = 50 cars.
  25. The difference between two numbers is 17 and their sum is 69. Find the largest of these two numbers.
    Solution
    Let x be the smallest number, then the largest number is
    x + 17 since their difference is 17
    The sum of the two numbers is 69, hence
    x + (x + 17) = 69
    group like terms and solve
    2 x + 17 = 69
    2 x = 52
    2 x / 2= 52 / 2
    x = 26
    the largest of these two numbers is
    x + 17 = 43

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