Solutions Grade 5 Math Practice Test Questions

Solutions to Grade 5 math practice test questions including numbers, fractions, geometry and problem solving are presented.

Solutions


  1. We shall use the divisibility rules to answer these questions.
    In order to find out if a number is divisible by 3, add its digits and check whether the sum is divisible by 3.
    a)
    The digits in the number 140 are 1,4,and 0.
    The sum of the digits is: 1+4+0 = 5 is not divisible by 3 and therefore 140 is NOT divisible by 3.
    b)
    The digits in the number 111 are 1,1,and 1.
    The sum of the digits is: 1+1+1 = 3 is divisible by 3 and therefore 111 is divisible by 3.
    c)
    The digits in the number 2232 are 2, 2, 3 and 2.
    The sum of the digits is: 2+2+3+2 = 9 is divisible by 3 and therefore 2232 is divisible by 3.


  2. We shall use the divisibility rules to answer these questions.
    To find out whether a number is divisible by 5, we simply need to check whether its units digit is 0 or 5.
    a) The unit digit of 245 is equal to 5 and therefore 245 is divisible by 5
    b) The unit digit of 3057 is equal to 7 and therefore 3057 is NOT divisible by 5
    c) The unit digit of 24580 is equal to 0 and therefore 24580 is divisible by 5


  3. a)
    One hundred twenty-six million: 126,000,000
    twenty-three thousand: 23,000
    forty-six: 46
    One hundred twenty-six million, twenty-three thousand, forty-six : 126,000,000 + 23,000 + 46 = 126,023,046
    b)
    Four hundred twenty-five billion: 425,000,000,000
    two hundred thirty-two thousand: 232,000
    fifty-nine: 59
    Four hundred twenty-five billion, two hundred thirty-two thousand, fifty-nine: 425,000,000,000 + 232,000 +59 = 425,000,232,059


  4. Paolo left home at 8:40 am and walked to the car park for 10 minutes, so he arrived at the car park at 8:40 am + 10 minutes = 8:50 am.
    Paolo left the He left the car at 8:50 then drove to work for 34 minutes, so he arrived at work at 8:50 am + 34 minutes = 9:24 am.
    So Paolo arrived at work at 9:24 am.


  5. To find the average speed, we can use the formula:
    Average Speed = Total distance / Total time
    Given that Linda drove 200 kilometers in 2 hours and a half.
    Convert 2 hours and a half to decimal form: 2 + 0.5 = 2.5 h
    Average Speed = 200/2.5 = 80 km/h
    So Linda drove at an average speed of 80 km/h.


  6. To find the percent increase of bicycles produced from last year to this year, we can use the following formula:
    Percent Increase = ( (New Quantity - Old Quantity) / Old Quantity ) x 100
    New Quantity (of this year) = 10800 , Old Quantity (of last year) = 7200
    Hence,
    Percent Increase = ( (10800 - 7200) / 7200 ) × 100
    Percent Increase = (3600/7200) × 100
    Percent Increase = 50 %
    Therefore, the percent increase of bicycles produced from last year to this year is 50%.


  7. Joe bought 2 boxes of pencils at $1.40 each and a box of pens at $1.60, so the total cost of the pencils and pens is:
    2 × $1.40 + $1.60 = $2.80 + $1.60 = $4.40
    We also know that the total cost of all items Joe bought is $10.40, so the total cost of the notebooks is:
    $10.40 - $4.40 = $6
    We know that Joe bought 3 notebooks, so we can divide the total cost of the notebooks by the number of notebooks to find the price of each notebook:
    $6 / 3 = $2
    So, the price of each notebook is $2


  8. If Mary is 2 years younger than Jill who is 23 years, then Mary is 23 - 2 = 21 years old.
    If Jenny is 12 years older than Mary, then Jenny is 21 + 12 = 33 years old.
    So the age of Jenny is 33 years old.


  9. Let w and l be the width and the length of the rectangle respectively.
    We know that the perimeter of the rectangle is the sum of twice the length and twice the width, which is:
    P = 2l + 2w = 260
    We also know that the length is 30 meters more than the width:
    l = w + 30
    We now substitute l = w + 30 in the equation 2l + 2w = 260:
    2(w + 30) + 2w = 260
    2w + 60 + 2w = 260
    4w + 60 = 260
    4w = 200
    w = 200/4 = 50 meters
    Now that we know the width, we can use the equation l = w + 30 to find the length:
    l = 50 + 30 = 80 meters
    So the width of the rectangle is w = 50 meters, and the length is l = 80 meters.


  10. We can set up an equation to solve for x using the information given.
    Given: 2/3 of x is equal to 20
    We can write this as an equation: 2/3 × x = 20
    To find the value of x, we need to divide both sides of the equation by 2/3.
    So x = 20 / (2/3) = 20 × 3/2
    Simplify
    x = (20 × 3)/2 = 30
    Therefore, x is equal to 30.


  11. Average speed = total distance / total time
    Time driven for the first 100 km
    t1 = 100 / 50 = 2 hours
    Time driven for the remaining 150 km
    150 / 75 = 2 hours
    So the total time is 2 + 2 = 4 hours
    To find the average speed, we use the formula:
    average speed = total distance / total time
    total distance = 100 + 150 = 250 km
    Substitute total distance and total time in the formula
    average speed = 250 / 4 = 62.5 km/h
    So the average speed of the whole journey is 62.5 km/h.


  12. Let x be the number of storybooks Beverly has. According to the problem, Joe has 5 more storybooks than Beverly, so Joe has:
    x + 5 storybooks.
    We know that the total number of storybooks they have together is 49. Therefore, we can write an equation:
    x + (x + 5) = 49
    Simplifying the equation, we get:
    2x + 5 = 49
    Subtracting 5 from both sides of the equation, we get:
    2x = 44
    Dividing both sides by 2, we get:
    x = 22
    So Beverly has x = 22 storybooks and Joe has x + 5 = 22 + 5 = 27 storybooks.
    Therefore, Beverly has 22 storybooks and Joe has 27 storybooks.


  13. The length of the piece of the wire that was cut is given by
    1/5 of 10 cm = ( 1 / 5 ) × 10 = 10/5 = 2 cm
    The remaining piece has a length:
    Total length - 2 = 10 - 2 = 8 cm


  14. If six packs of biscuits cost $7.50, we can start by finding out the cost of one pack:
    Cost of one pack = $7.50 / 6 = $1.25
    Therefore, one pack of biscuits costs $1.25.
    To find out how much it would cost to buy 8 packs, we can multiply the cost of one pack by 8:
    Cost of 8 packs = $1.25 × 8
    Cost of 8 packs = $10
    Therefore, it would cost $10 to buy 8 packs of biscuits at this shop.


  15. Given: The coffee at the ordinary cafeteria costs $2.60
    If a coffee at a high class cafeteria costs three times as much as a coffee in the ordinary cafeteria, then a coffee at the high class cafeteria costs:
    3 × $2.60 = $7.80
    Therefore, a coffee at the high class cafeteria costs $7.80.
    During the week, Toby bought coffee at the high class cafeteria four times. The total cost of the coffee Toby bought at the high class cafeteria is:
    4 × $7.80 = $31.20
    During the week, Toby bought coffee six times in the ordinary cafeteria. The total cost of the coffee Toby bought at the ordinary cafeteria is:
    6 × $2.60 = $15.60
    Therefore, the total amount Toby spent on coffee during the week is:
    $31.2 + $15.60 = $46.80
    Therefore, Toby spent $46.8 on coffee during the week.


  16. The total number of marbles can be written as a fraction equal to 5/5
    If two-fifths of the marbles are red, then the fraction of blue marbles is given by
    5/5 - 2/5 = 3/5
    Let x be the total number of marbles.
    Therefore, the number of red marbles is:
    (2/5) × x
    There are 60 blue marbles and this is 3/5 of the total number of marbles x, so we can set up an equation:
    (3/5) × x = 60
    Multiplying both sides by 5/3, we get:
    x = 100
    Therefore, there are 100 marbles in total.
    The number of red marbles is:
    (2/5) × x = (2/5) × 100 = 40
    Therefore, there are 40 red marbles.


  17. There are 24 hours in a day,
    60 minutes in an hour,
    and 60 seconds in a minute.
    Therefore, the number of seconds in a day is:
    24 hours/day × 60 minutes/hour × 60 seconds/minute = 86,400 seconds/day
    Multiplying the number of seconds in a day by the number of days in August which is 31, we get:
    86,400 seconds/day × 31 days/month = 2,678,400 seconds

    \( \) \( \)\( \)

  18. a) To write the fraction \( \frac{2}{4} \) as a decimal, we divide the numerator 2 by the denominator 4:
    \[ \frac{2}{4} = 2 \div 4 = 0,5 \]
    b) To write the fraction \( \frac{100}{1000}\) as a decimal, we divide the numerator 100 by the denominator 1000:
    \[ \frac{100}{1000} = 100 \div 1000 = 0,1 \]
    c) To write the fraction \(\frac{1}{10000}\) as a decimal, we divide the numerator 1 by the denominator 10000: \[ \frac{1}{10000} = 1 \div 10000 = 0,0001 \]


  19. a)
    To write 0.1 as a fraction, we divide 0.1 by 1 \[ 0.1 = \frac{0.1}{1} \] We then mutliply the top and bottom of the ratio \( \frac{0.1}{1} \) by 10 to eliminate the decimal point:
    \[ 0.1 = \frac{0.1 \times 10}{1 \times 10 } = \frac{1}{10} \]
    b)
    To write 2.5 as a fraction, we first write 2.5 as a sum of the whole part and a the fractional part \[ 2.5 = 2 + 0.5 \] We now write 0.5 as a ratio of 0.5 and 1 \[ 2.5 = 2 + \frac{0.5}{1} \] Multiply the top and bottom of the ratio \( \frac{0.5}{1} \) by 10 to change into a fration \[ 2.5 = 2 + \frac{0.5 \times 10}{1 \times 10} = 2 + \frac{5}{10}\] Simplify and write as a mixed number \[ 2.5 = 2 \frac{5}{10} \] c)
    To write 5.01 as a fraction, we first write 5.01 as a sum of the whole part and a the fractional part \[ 5.01 = 5 + 0.01 \] We now write 0.01 as a ratio of 0.01 and 1 \[ 5.01 = 5 + \frac{0.01}{1} \] Multiply the top and bottom of the ratio \( \frac{0.01}{1} \) by 100 to change into a fration \[ 5.01 = 5 + \frac{0.01 \times 100}{1 \times 100} = 5 + \frac{1}{100} \] Simplify and write as a mixed number \[ 5.01 = 5 \frac{1}{100} \]


  20. Write the fractions in decimal form
    a) 1.1
    b) \( \displaystyle \frac{123}{100} = 1.23 \)
    c) \( \displaystyle \frac{6}{5} = 1.2 \)
    Comparin the decimal forms, the smallest is 1.1, then 1.2 and then 1.23
    Hence the given numbers from smallest to largest : \[ 1.1 \quad , \quad \displaystyle \frac{6}{5} \quad , \quad \frac{123}{100} \]


  21. a)
    \( \displaystyle 1.191 \) to the nearest one is : \( 1 \)
    \( \displaystyle 1.191 \) to the nearest tenth is : \( 1.2 \)
    \( \displaystyle 1.191 \) to the nearest hundredth is : \( 1.19\)
    b)
    \( \displaystyle 2.578 \) to the nearest one is : \( 3 \)
    \( \displaystyle 2.578 \) to the nearest tenth is : \( 2.6\)
    \( \displaystyle 2.578 \) to the nearest hundredth is : \( 2.58\)


  22. a)
    \( \quad \displaystyle 1 \text{ m} = 100 \text{ cm} \), hence \( \quad \displaystyle 0.2 \text{ m} = 0.2 \times 100 \text{ cm} = 20 \text{ cm} \)
    b)
    \( \quad \displaystyle 1 \text{ cm} = 0.01 \text{ m} \), hence \( \quad \displaystyle 35 \text{ cm} = 35 \times 0.01 \text{ m} = 0.35 \text{ m} \)
    c)
    \( \quad \displaystyle 1 \text{ km} = 1000 \text{ m} \), hence \( \quad \displaystyle 3.5 \text{ km} = 3.5 \times 1000 \text{ m} = 3500 \text{ m} \)
    d)
    \( \quad \displaystyle 1 \text{ in} = \frac{1}{12} \text{ ft} \), hence \( \quad \displaystyle 36 \text{ in} = 36 \times \frac{1}{12} \text{ ft} = 3 \text{ ft} \)
    e)
    \( \quad \displaystyle 1 \text{ L} = 100 \text{ cl} \), hence \( \quad \displaystyle 0.035 \text{ L} = 0.035 \times 100 \text{ cL} = 3.5 \text{ cL}\)
    f)
    \( \quad \displaystyle 1 \text{ mL} = \frac{1}{1000} \text{ L} \), hence \( \quad \displaystyle 350 \text{ mL} = 350 \times \frac{1}{1000} \text{ L} = 0.35 \text{L} \)
    g)
    \( \quad \displaystyle 1 \text{ ft} = 12 \text{ in} \), hence \( \quad \displaystyle 3.5 \text{ ft} = 3.5 \times 12 \text{ in} = 42 \text{ in} \)
    h)
    \( \quad \displaystyle 1 \text{ in} = 2.54 \text{ cm} \), hence \( \quad \displaystyle 36 \text{ in} = 36 \times 2.54 \text{ cm} = 91.44 \text{cm} \)


  23. Add/Subtract whole part together and add/subtract fractional parts together. For the fractional parts, you sometimes need to put all fractions to a common denominator.
    a)
    \( \quad \displaystyle 2\frac{1}{3} + 3\frac{2}{3} = (2 +3)+(\frac{1}{3} + \frac{2}{3}) \\ \qquad = 5 + \frac{3}{3} = 5 + 1 = 6 \)
    b)
    \( \quad \displaystyle 4 \frac{4}{5} - 3\frac{1}{2} = (4 - 3) + (\frac{4}{5} - \frac{1}{2}) \\ \qquad = 1 + (\frac{8}{10} - \frac{5}{10}) \\ \qquad = 1 + \frac{3}{10} = 1 \frac{3}{10} \)

    c) \( \quad \displaystyle 1 \frac{1}{4} + 3 \frac{3}{5} - 2 \frac{1}{2} = (1 + 3 - 2) + ( \frac{1}{4} + \frac{3}{5} - \frac{1}{2} ) \\ \qquad = 2 + ( \frac{1}{4} + \frac{3}{5} - \frac{1}{2} ) \\ \qquad = 2 + ( \frac{5}{20} + \frac{12}{20} - \frac{10}{20} ) \\ \qquad = 2 + \frac{7}{20} = 2 \frac{7}{20} \)


  24. a)
    \( \displaystyle \frac{1}{3} = \frac{?}{9} \)
    We need to multiply the denominator of the fracction on the left which is 3 by 3 in order to obtain the denominator 9 of the fraction on the right. Hence
    \( \displaystyle \frac{1}{3} = \frac{1 \times 3}{3 \times 3} = \frac{3}{9} \)
    b)
    \( \displaystyle \frac{10}{4} = \frac{5}{?} \)
    We need to divide the numerator 10 of the fraction on the left in order to obtain the numerator 5 of the fraction on the right. Hence
    \( \displaystyle \frac{10}{4} = \frac{10 \div 2}{4 \div 2} = \frac{5}{2} \)
    c)
    \( \displaystyle \frac{?}{4} = \frac{15}{20} \)
    We need to divide the denominator 20 of the fraction on the right in order to obtain the denominator 4 of the fraction on the left. Hence
    \( \displaystyle \frac{15}{20} = \frac{15 \div 5}{20 \div 5} = \frac{3}{4} \)


  25. This question is about reduce fractions
    a)
    2 is a common factor to both the numerator 10 and the denominator 12, hence we divide both of them by 2
    \( \displaystyle \frac{10}{12} = \frac{10 \div 2}{12 \div 2 } = \frac{5}{6} \)
    b)
    3 is a common factor to both the numerator 21 and the denominator 42, hence we divide both of them by 3
    \( \displaystyle \frac{21}{42} = \frac{21 \div 3}{42 \div 3 } = \frac{7}{14} \)
    7 is a common factor to both the numerator 7 and the denominator 14, hence
    \( \displaystyle \frac{21}{42} = \frac{7 \div 7}{14 \div 7} = \frac{1}{2} \)
    c)
    5 is a common factor to both the numerator 15 and the denominator 65, hence we divide both of them by 5
    \( \displaystyle \frac{15}{65} = \frac{15 \div 5}{65 \div 5 } = \frac{3}{13} \)


  26. This question is about exponents .
    a)
    \( \displaystyle 6^3 = 6 \times 6 \times 6 = 216\)
    b)
    \( 1000^0 = 1 \)
    c)
    \( 2^3 + 10^2 = 2 \times 2 \times 2 + 10 \times 10 = 8 + 100 = 108 \)


  27. This question is about prime and composite numbers .
    Definition: A prime number is a whole number that is divisible by 1 and itself only.
    21 is divisible by : 1 , 3 , 7 and 21 and therefore is NOT a prime number.
    13 is divisible by 1 and 13 only and therefore is a prime number.
    55 is divisible by : 1 , 5 , 11 and 55 and therefore is NOT a prime number.
    41 is divisible by 1 and 41 only and therefore is a prime number.
    201 is divisible by 1, 3, 67, 201 and therefore is NOT a prime number.


  28. The area of the colored region is equal to the area of the rectangle ABCD minus the area of the triangle FED.
    cm 2 is an abbreviation for square centimeter
    Area of the rectangle ABCD = length × width = 10 5 = 50 cm 2
    Area of the triangle FED = (1/2) × height × base
    The height of the triangle is equal to the width of the rectangle and it is equal to 5 cm. The base of the triangle is equal to 8 cm.
    The area of the triangle = (1/2) × base × height = (1/2) × 8 × 5 = 20 cm 2
    The area of the colored (orange) region = area of rectangle - area of trioangle = 50 - 20 = 30 cm 2


  29. One way to find the volume of the given 3d shape is to complete it to make a larger rectangular solid as shown below. Split Composite Rectangualr Solid.
    mm3 is an abbreviation for cubic millimeter
    The volume V1 of the large rectangular solid is given by
    V1 = 7 × 12 × 8 = 672 mm 3
    The volume V2 of rectangular solid (red) that was added is given by
    V2 = 4 × 9 × 8 = 288 mm 3
    The volume of the given 3d shape V is given by subtracting V2 from V1
    V = V1 - V2 = 672 mm 3 - 288 mm 3 = 384 mm 3.

More References and links

  1. Divisibility Rules Examples and Questions
  2. Percent Maths Problems
  3. Primary Maths (grades 4 and 5) with Free Questions and Problems With Answers
  4. 3D Shapes Volume Problems
  5. Reduce Fractions Step by Step
  6. Exponents
  7. Prime and Composite Numbers
  8. Home Page