Grade 5 math practice test questions are presented along with their detailed step-by-step solutions and explanations covering numbers, fractions, geometry, and problem-solving.
Questions and Step-by-Step Solutions
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Question 1: Which of the following number(s) is divisible by 3?
a) 140 b) 111 c) 2232► View Step-by-Step Solution
To test divisibility by 3, add the digits of each number:
- For 140: \(1 + 4 + 0 = 5\) (Not divisible by 3)
- For 111: \(1 + 1 + 1 = 3\) (Divisible by 3)
- For 2232: \(2 + 2 + 3 + 2 = 9\) (Divisible by 3)
Therefore, 111 and 2232 are divisible by 3. -
Question 2: Which of the following number(s) is divisible by 5?
a) 245 b) 3057 c) 24580► View Step-by-Step Solution
A number is divisible by 5 if its units digit is 0 or 5.
- 245 ends in 5 (Divisible)
- 3057 ends in 7 (Not divisible)
- 24580 ends in 0 (Divisible)
Therefore, 245 and 24580 are divisible by 5. -
Question 3: Write the numbers in standard form using digits 0, 1, ..., 9.
a) One hundred twenty-six million, twenty-three thousand, forty-six
b) Four hundred twenty-five billion, two hundred thirty-two thousand, fifty-nine► View Step-by-Step Solution
a) 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, two hundred thirty-two thousand, fifty-nine: \[ 425,000,000,000 + 232,000 + 59 = 425,000,232,059 \] -
Question 4: Paolo went to work at 8:40 am. He walked to the car park for 10 minutes and then drove to work for another 34 minutes. At what time did Paolo arrive at work?
► View Step-by-Step Solution
Arrived at car park: \(8:40\text{ am} + 10\text{ minutes} = 8:50\text{ am}\).
Arrived at work: \(8:50\text{ am} + 34\text{ minutes} = 9:24\text{ am}\).
Paolo arrived at work at 9:24 am. -
Question 5: Linda drove from her place to another town a distance of 200 kilometers in 2 hours and a half. At what average speed did she drive?
► View Step-by-Step Solution
Convert 2 hours and a half to decimal form: \(2 + 0.5 = 2.5\text{ hours}\).
\[ \text{Average Speed} = \dfrac{\text{Total Distance}}{\text{Total Time}} = \dfrac{200}{2.5} = 80\text{ km/h} \] -
Question 6: A factory produced 7200 bicycles last year. The same factory produced 10800 bicycles this year. What is the percent increase of bicycles produced from last year to this year?
► View Step-by-Step Solution
\[ \text{Percent Increase} = \left(\dfrac{\text{New Quantity} - \text{Old Quantity}}{\text{Old Quantity}}\right) \times 100 \]
\[ = \left(\dfrac{10800 - 7200}{7200}\right) \times 100 = \left(\dfrac{3600}{7200}\right) \times 100 = 50\% \] -
Question 7: Joe bought 3 notebooks; two boxes of pencils at $1.40 each and a box of pens at $1.60. She paid a total of $10.40. What was the price of each notebook?
► View Step-by-Step Solution
Cost of pencils and pens: \(2 \times \$1.40 + \$1.60 = \$2.80 + \$1.60 = \$4.40\).
Total cost of notebooks: \(\$10.40 - \$4.40 = \$6.00\).
Price of each notebook: \(\dfrac{\$6.00}{3} = \$2.00\). -
Question 8: Jenny is 12 years older than Mary, and Mary is 2 years younger than Jill, who is 23 years old. What is the age of Jenny?
► View Step-by-Step Solution
Mary's age = \(23 - 2 = 21\text{ years old}\).
Jenny's age = \(21 + 12 = 33\text{ years old}\). -
Question 9: A rectangle has a length that is 30 meters more than its width. The perimeter of the rectangle is 260 meters. Find the length and the width of this rectangle.
► View Step-by-Step Solution
Let width be \(w\) and length be \(l = w + 30\).
Perimeter \(P = 2l + 2w = 260\).
Substitute \(l\): \(2(w + 30) + 2w = 260 \implies 4w + 60 = 260 \implies 4w = 200 \implies w = 50\text{ m}\).
Length \(l = 50 + 30 = 80\text{ m}\).
Width is 50 meters and length is 80 meters. -
Question 10: \(\dfrac{2}{3}\) of \(x\) is equal to 20. What is \(x\)?
► View Step-by-Step Solution
\[ \dfrac{2}{3} \times x = 20 \implies x = 20 \div \dfrac{2}{3} = 20 \times \dfrac{3}{2} = 30 \]
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Question 11: Joe drove 100 km at the speed of 50 km/h and drove 150 km at the speed of 75 km/h. What is the average speed of the whole journey?
► View Step-by-Step Solution
Time for first leg: \(100 \div 50 = 2\text{ hours}\).
Time for second leg: \(150 \div 75 = 2\text{ hours}\).
Total time = \(2 + 2 = 4\text{ hours}\).
Total distance = \(100 + 150 = 250\text{ km}\).
\[ \text{Average Speed} = \dfrac{250}{4} = 62.5\text{ km/h} \] -
Question 12: Beverly and Joe have a total of 49 storybooks. If Joe has 5 more storybooks than Beverly, how many storybooks does each one have?
► View Step-by-Step Solution
Let Beverly have \(x\) books. Joe has \(x + 5\).
\(x + (x + 5) = 49 \implies 2x + 5 = 49 \implies 2x = 44 \implies x = 22\).
Beverly has 22 storybooks and Joe has \(22 + 5 = 27\) storybooks. -
Question 13: Boby has a 10 cm long wire. He cuts off \(\dfrac{1}{5}\) of it. How long is the remaining piece, in centimeters?
► View Step-by-Step Solution
Length cut off = \(\dfrac{1}{5} \times 10 = 2\text{ cm}\).
Remaining length = \(10 - 2 = 8\text{ cm}\). -
Question 14: A shop charges a fixed amount for each pack of biscuits. Six packs of biscuits cost $7.50 at this shop. How much would it cost to buy 8 packs?
► View Step-by-Step Solution
Cost of one pack = \(\dfrac{\$7.50}{6} = \$1.25\).
Cost of 8 packs = \(\$1.25 \times 8 = \$10.00\). -
Question 15: A coffee at a high-class cafeteria costs three times as much as a coffee in the ordinary cafeteria. The coffee at the ordinary cafeteria costs $2.60. In a period of one week, Toby bought coffee at the high-class cafeteria four times, and six times in the ordinary cafeteria. How much did Toby spend on coffee during the week?
► View Step-by-Step Solution
High-class coffee cost = \(3 \times \$2.60 = \$7.80\).
Cost for 4 high-class coffees = \(4 \times \$7.80 = \$31.20\).
Cost for 6 ordinary coffees = \(6 \times \$2.60 = \$15.60\).
Total spent = \(\$31.20 + \$15.60 = \$46.80\). -
Question 16: A bag has blue and red marbles. Two-fifths of the marbles are red, and the rest are blue. If there are 60 blue marbles, how many are red?
► View Step-by-Step Solution
Fraction of blue marbles = \(1 - \dfrac{2}{5} = \dfrac{3}{5}\).
Let total marbles be \(x\): \(\dfrac{3}{5}x = 60 \implies x = 60 \times \dfrac{5}{3} = 100\text{ total marbles}\).
Number of red marbles = \(\dfrac{2}{5} \times 100 = 40\). -
Question 17: How many seconds are in the month of August?
► View Step-by-Step Solution
August has 31 days.
Seconds in a day = \(24 \times 60 \times 60 = 86,400\text{ seconds}\).
Seconds in August = \(86,400 \times 31 = 2,678,400\text{ seconds}\). -
Question 18: Write as decimals:
a) \(\dfrac{2}{4}\) b) \(\dfrac{100}{1000}\) c) \(\dfrac{1}{10000}\)► View Step-by-Step Solution
a) \(\dfrac{2}{4} = 2 \div 4 = 0.5\)
b) \(\dfrac{100}{1000} = 100 \div 1000 = 0.1\)
c) \(\dfrac{1}{10000} = 1 \div 10000 = 0.0001\) -
Question 19: Write as fractions or mixed numbers:
a) \(0.1\) b) \(2.5\) c) \(5.01\)► View Step-by-Step Solution
a) \(0.1 = \dfrac{1}{10}\)
b) \(2.5 = 2 \dfrac{5}{10} = 2 \dfrac{1}{2}\)
c) \(5.01 = 5 \dfrac{1}{100}\) -
Question 20: Order from smallest to largest:
a) \(1.1\) b) \(\dfrac{123}{100}\) c) \(\dfrac{6}{5}\)► View Step-by-Step Solution
Convert all to decimals: \(1.1\), \(\dfrac{123}{100} = 1.23\), \(\dfrac{6}{5} = 1.2\).
Ordered from smallest to largest: \(1.1, \; \dfrac{6}{5}, \; \dfrac{123}{100}\). -
Question 21: Round the numbers to the nearest one, nearest tenth, and nearest hundredth:
a) \(1.191\) b) \(2.578\)► View Step-by-Step Solution
a) For \(1.191\): Nearest one = \(1\), Nearest tenth = \(1.2\), Nearest hundredth = \(1.19\).
b) For \(2.578\): Nearest one = \(3\), Nearest tenth = \(2.6\), Nearest hundredth = \(2.58\). -
Question 22: Convert:
a) \(0.2\text{ m} = \dots\text{ cm}\) b) \(35\text{ cm} = \dots\text{ m}\)
c) \(3.5\text{ km} = \dots\text{ m}\) d) \(36\text{ in} = \dots\text{ ft}\)
e) \(0.035\text{ L} = \dots\text{ cL}\) f) \(350\text{ mL} = \dots\text{ L}\)
g) \(3.5\text{ ft} = \dots\text{ in}\) h) \(36\text{ in} = \dots\text{ cm}\)► View Step-by-Step Solution
a) \(0.2\text{ m} \times 100\text{ cm/m} = 20\text{ cm}\)
b) \(35\text{ cm} \times 0.01\text{ m/cm} = 0.35\text{ m}\)
c) \(3.5\text{ km} \times 1000\text{ m/km} = 3500\text{ m}\)
d) \(36\text{ in} \div 12\text{ in/ft} = 3\text{ ft}\)
e) \(0.035\text{ L} \times 100\text{ cL/L} = 3.5\text{ cL}\)
f) \(350\text{ mL} \div 1000\text{ mL/L} = 0.35\text{ L}\)
g) \(3.5\text{ ft} \times 12\text{ in/ft} = 42\text{ in}\)
h) \(36\text{ in} \times 2.54\text{ cm/in} = 91.44\text{ cm}\) -
Question 23: Simplify and write the result as a fraction or mixed number:
a) \(2\dfrac{1}{3} + 3\dfrac{2}{3}\) b) \(4\dfrac{4}{5} - 3\dfrac{1}{2}\) c) \(1\dfrac{1}{4} + 3\dfrac{3}{5} - 2\dfrac{1}{2}\)► View Step-by-Step Solution
a) \((2 + 3) + \left(\dfrac{1}{3} + \dfrac{2}{3}\right) = 5 + 1 = 6\)
b) \((4 - 3) + \left(\dfrac{8}{10} - \dfrac{5}{10}\right) = 1 + \dfrac{3}{10} = 1\dfrac{3}{10}\)
c) \((1 + 3 - 2) + \left(\dfrac{5}{20} + \dfrac{12}{20} - \dfrac{10}{20}\right) = 2 + \dfrac{7}{20} = 2\dfrac{7}{20}\) -
Question 24: Complete writing equivalent fractions:
a) \(\dfrac{1}{3} = \dfrac{?}{9}\) b) \(\dfrac{10}{4} = \dfrac{5}{?}\) c) \(\dfrac{?}{4} = \dfrac{15}{20}\)► View Step-by-Step Solution
a) \(\dfrac{1 \times 3}{3 \times 3} = \dfrac{3}{9}\)
b) \(\dfrac{10 \div 2}{4 \div 2} = \dfrac{5}{2}\)
c) \(\dfrac{15 \div 5}{20 \div 5} = \dfrac{3}{4}\) -
Question 25: Reduce the fractions:
a) \(\dfrac{10}{12}\) b) \(\dfrac{21}{42}\) c) \(\dfrac{15}{65}\)► View Step-by-Step Solution
a) \(\dfrac{10 \div 2}{12 \div 2} = \dfrac{5}{6}\)
b) \(\dfrac{21 \div 21}{42 \div 21} = \dfrac{1}{2}\)
c) \(\dfrac{15 \div 5}{65 \div 5} = \dfrac{3}{13}\) -
Question 26: Evaluate the following expressions:
a) \(6^3\) b) \(1000^0\) c) \(2^3 + 10^2\)► View Step-by-Step Solution
a) \(6 \times 6 \times 6 = 216\)
b) \(1000^0 = 1\)
c) \(8 + 100 = 108\) -
Question 27: Which of the following numbers are prime?
\(21, 13, 55, 41, 201\)► View Step-by-Step Solution
Prime numbers have divisors only of 1 and themselves.
- 21: Composite (divisible by 3, 7)
- 13: Prime
- 55: Composite (divisible by 5, 11)
- 41: Prime
- 201: Composite (divisible by 3, 67)
The prime numbers are 13 and 41. -
Question 28: ABCD is a rectangle of length 10 cm and width 5 cm. Find the area of the colored (orange) region.
► View Step-by-Step Solution
Area of rectangle ABCD = \(10 \times 5 = 50\text{ cm}^2\).
Area of triangle FED = \(\dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 8 \times 5 = 20\text{ cm}^2\).
Area of colored region = \(50 - 20 = 30\text{ cm}^2\). -
Question 29: Find the volume V of the composite rectangular solid shown below.
► View Step-by-Step Solution
Completing the solid into a larger rectangular prism:
Volume of large solid (\(V_1\)) = \(7 \times 12 \times 8 = 672\text{ mm}^3\).
Volume of added section (\(V_2\)) = \(4 \times 9 \times 8 = 288\text{ mm}^3\).
Total volume \(V = V_1 - V_2 = 672 - 288 = 384\text{ mm}^3\).