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

Which of the following number(s) is divisible by 3 ?
a) 140 b) 111 c) 2232
Solution
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.

Which of the following number(s) is divisible by 5 ?
a) 245 b) 3057 c) 24580
Solution
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

Write the numbers in standard form using digits 0, 1, .... 9 .
a) One hundred twentysix million, twentythree thousand, fortysix b) Four hundred twentyfive billion, two hundred thirtytwo thousand, fiftynine.
Solution
a)
One hundred twentysix million: 126,000,000
twentythree thousand: 23,000
fortysix: 46
One hundred twentysix million, twentythree thousand, fortysix : 126,000,000 + 23,000 + 46 = 126,023,046
b)
Four hundred twentyfive billion: 425,000,000,000
two nundred thirtytwo thousand: 232,000
fiftynine: 59
Four hundred twentyfive billion, two nundred thirtytwo thousand, fiftynine: 425,000,000,000 + 232,000 +59 = 425,000,232,059

Paolo went to work at 8:40 am. He walked to the car park for 10 minutes and drove to work for another 34 minutes. At what time did Paolo arrive at work
Solution
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.

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 ?
Solution
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.

A factory produced 7200 bicycles last year. The same factory produced 10800 bicycles this year. What is the percent increase of bicylces produced from last year to this year ?
Solution
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%.

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 ?
Solution
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

Jenny is 12 older than Mary and Mary is 2 years younger than Jill who is 23 years. What is the age of Jenny?
Solution
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.

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.
Solution
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.

2/3 of x is equal to 20. What is x ?
Solution
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 = 30
Therefore, x is equal to 30.

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 ?
Solution
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 = (100 + 150) / 4 = 62.5 km/h
So the average speed of the whole journey is 62.5 km/h.

Beverly and Joe have a total of 49 story books. If Joe has 5 more story books than Beverly, how many books does each one have?
Solution
Let x be the number of story books than Beverly has. According to the problem, Joe has 5 more story books than Beverly, so Joe has:
x + 5 story books.
We know that the total number of story books 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 story books, and Joe has x + 5 = 22 + 5 = 27 story books.
Therefore, Beverly has 22 story books, and Joe has 27 story books.

Boby has a 10ft long wire. He cuts off 1/5 of it. How long is the remaining piece, in feet and inches?
Solution
The length of the piece of the wire that was cut is given by
1/5 of 10ft = ( 1 / 5 ) × 10 = 10/5 = 2ft
The remaining piece has a length:
Total length  2 = 10  2 = 8ft

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?
Solution
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.

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 $2.6.
In a period of one week, Toby bought coffe at the high class cafetria four times, and six times in the ordinary cafeteria. How much did Toby spend on coffee during the week?
Solution
Given: The coffee at the ordinary cafeteria costs \$2.6
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 × \$2.6 = \$7.8
Therefore, a coffee at the high class cafeteria costs \$7.8.
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 × \$7.8 = \$31.2
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 × \$2.6 = \$15.6
Therefore, the total amount Toby spent on coffee during the week is:
\$31.2 + \$15.6 = \$46.8
Therefore, Toby spent $46.8 on coffee during the week.

A bag has blue and red marbles. Twofifths of the marbles are red, and the rest are blue. If there are 60 blue marbles, how many are red?
Solution
If twofifths 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.

How many seconds are in the month of August?
Solution
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
\( \) \( \)\( \) 
Write as decimals.
a) \( \displaystyle \frac{2}{4} \) b) \( \displaystyle \frac{100}{1000} \) c) \( \displaystyle \frac{1}{10000} \)
Solution
a) To write the fraction $\frac{2}{4}$ as a decimal, we divide the numerator 2 by the denominator 4:
\[ \frac{2}{4} = 0.5 \]
b) To write the fraction $\frac{100}{1000}$ as a decimal, we divide the numerator 100 by the denominator 1000:
\[ \frac{100}{1000} = 0.1 \]
c) To write the fraction $\frac{1}{10000}$ as a decimal, we divide the numerator 1 by the denominator 10000: \[ \frac{1}{10000} = 0.0001 \]

Write as fractions or mixed numbers.
a) \( \displaystyle 0.1 \) b) \( \displaystyle 2.5 \) c) \( \displaystyle 5.01 \)
Solution
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 \[ 5.01 = 5 \frac{1}{100} \]

Order from smallest to largest.
a) \( \displaystyle 1.1 \) b) \( \displaystyle \frac{123}{100} \) c) \( \displaystyle \frac{6}{5} \)
Solution
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 : \[ 1.1 \quad , \quad \displaystyle \frac{6}{5} \quad , \quad \frac{123}{100} \]

Round the numbers to the nearest one, nearest tenth, and nearest hundredth.
a) \( \displaystyle 1.191 \) b) \( \displaystyle 2.578 \)
Solution
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 : \( 2.58\)

Convert
This question is about unit conversion .
a) \( \quad \displaystyle 0.2 \text{ m} = .... \text{ cm} \) b) \( \quad \displaystyle 35 \text{ cm} = .... \text{m}\) c) \( \quad \displaystyle 3.5 \text{ km} = .... \text{m} \) d) \( \quad \displaystyle 36 \text{ in} = .... \text{ft} \)
e) \( \quad \displaystyle 0.035 \text{ L} = .... \text{ cL} \) f) \( \quad \displaystyle 350 \text{ mL} = .... \text{L} \) g) \( \quad \displaystyle 3.5 \text{ ft} = .... \text{ in} \) h) \( \quad \displaystyle 36 \text{ in} = .... \text{cm} \)
Solution
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 \( \quad \displaystyle 36 \text{ in} = 36 \times 2.54 \text{ cm} = 91.44 \text{cm} \)

Simplify and write the result as a fraction or mixed number .
a) \( \quad \displaystyle 2\frac{1}{3} + 3\frac{2}{3}\) b) \( \quad \displaystyle 4 \frac{4}{5}  3\frac{1}{2}\) c) \( \quad \displaystyle 1 \frac{1}{4} + 3 \frac{3}{5}  2 \frac{1}{3}\)
Solution
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) \( \quad \displaystyle 1 \frac{1}{4} + 3 \frac{3}{5}  2 \frac{1}{2} = (1 + 3  2) + ( \frac{1}{4} + \frac{3}{5}  \frac{1}{2} ) \\ \qquad = 2 + ( \frac{1}{4} + \frac{3}{5}  \frac{1}{2} ) \\ \qquad = 2 + ( \frac{5}{20} + \frac{12}{20}  \frac{10}{20} ) \\ \qquad = 2 + \frac{7}{20} = 2 \frac{7}{20} \)

Complete to write equivalent fractions .
a) \( \displaystyle \frac{1}{3} = \frac{?}{9} \) b) \( \displaystyle \frac{10}{4} = \frac{5}{?} \) c) \( \displaystyle \frac{?}{4} = \frac{15}{20} \)
Solution
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
\( \displaystyle \frac{15}{20} = \frac{15 \div 5}{20 \div 5} = \frac{3}{4} \)

Reduce the fractions.
a) \( \displaystyle \frac{10}{12} \) b) \( \displaystyle \frac{21}{42}\) c) \( \displaystyle \frac{15}{65} \)
Solution
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
\( \displaystyle \frac{15}{65} = \frac{15 \div 5}{65 \div 5 } = \frac{3}{13} \)

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

Which of the following numbers are prime?
a) \( \displaystyle 21 , 13 , 55 , 41 , 201 \)
Solution
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.

ABCD is a rectangle of length 10 cm and width 5 cm. Find the area of the colored (orange) region.
.
Solution
The area of the colored region is equal to the area of the rectangle ABCD minus the area of the triangle FED.
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}

Find the volume V of the composite rectangular solid shown below.
.
Solution
One way to find the volume of the given 3d shape is to complete it to make a larger rectangular solid as shown below. . The volume V_{1} of the large rectangular solid is given by
V_{1} = 7 × 12 × 8 = 672 mm^{ 3}
The volume V_{2} of rectangular solid (red) that was added is given by
V_{2} = 4 × 9 × 8 = 288 mm^{ 3}
The volume of the given 3d shape V is given by subtracting V_{2} from V_{1}
V = V_{1}  V_{2} = 672 mm^{ 3}  288 mm^{ 3} = 384 mm^{ 3}.
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