Math Problems with Solutions for Grade 5

Total slices = 8

Slices eaten = 3 (Sarah) + 1 (Jake) = 4

Slices left = 8 - 4 = 4 \[ \text{The fraction left} = \dfrac{4}{8}\] Divide both numerator and denominator by 4: \[ \text{The fraction left} = \dfrac{4\div 4}{8\div 4}\] Simplify \[ \text{The fraction left} = \dfrac{ 1 }{2}\]

Total price before discount is: \[ 3 \times 15 = 45 \; \text{dollars } \] Total discount is: \[ 25\% \; \text{of} \; 45 = \dfrac{25}{100} \times 45 \] \[= 0.25 \times 45 = 11.25 \; \text{dollars } \] Total price after discount: \[ 45 - 11.25 = 33.75 \; \text{dollars } \] The customer will pay \$33.75.

The pattern starts with 3 and then each number is multiplied by 2 to get the next number.

1st: 3

2nd: \( 3 \times 2 = 6 \)

3rd: \( 6 \times 2 = 12 \)

4th: \( 12 \times 2 = 24 \)

5th: \( 24 \times 2 = 48 \)

6th: \( 48 \times 2 = 96 \)

7th: \( 96 \times 2 = 192 \)

The 7th number in the sequence is 192.

The time it takes John to get to work is the sum of the time to walk to the car park and the time to drive: \[ \text{Total time} = 25 + 45 = 70 \text{ minutes} \] Since \( 70 \text{ minutes} = 60 \text{ minutes} + 10 \text{ minutes} \), and knowing that \( 1 \text{ hour} = 60 \text{ minutes} \), we can write: \[ \text{Total time} = 1 \text{ hour and } 10 \text{ minutes} \] John needs to leave the house 1 hour and 10 minutes before 9:00 a.m.: \[ 9:00 - 1:10 = 9:00 - 1:00 - 0:10 = 8 - 0:10 = 7:50 \text{ a.m.} \] John should leave the house at 7:50 a.m.

If Kim walks \( 4 \) km in \( 1 \) hour, then the time to walk \( 18 \) km is: \[ \dfrac{18}{4} = 4.5 = 4 + \dfrac{1}{2} \] It takes Kim \( 4.5 \) hours (or 4 hours and 30 minutes) to walk 18 kilometers.

500 TV sets were produced in the third year more than in the second year. Therefore, the number of sets produced in the third year is: \[ 4500 + 500 = 5000 \] The total number of TV sets produced in three years is: \[ 2300 + 4500 + 5000 = 11800 \] Conclusion: 11,800 TV sets were produced over 3 years.

If 5 toys are taken out of the 49 toys and the remaining ones are distributed equally between Tom and Bob, they will both have the same number of toys. \[ 49 - 5 = 44 \quad \text{(to be shared equally)} \] If distributed equally, each one will have: \[ \dfrac{44}{2} = 22 \quad \text{toys} \] Bob has 5 more toys than Tom, so Bob has: \[ 22 + 5 = 27 \quad \text{toys} \] Tom has 22 toys and Bob has 27 toys.

Note: check that the total is 49 and that Bob has 5 more toys than Tom.

2 Possible Solutions:

1) There are 4 quarters in one whole pizza, and there are 2 quarters in half a pizza. So, in total, one pizza and a half has: \[ 4 + 2 = 6 \text{ quarters} \] If John eats one quarter in one minute, he needs: \[ 6 \times 1 = 6 \text{ minutes} \] to eat one pizza and a half.

2) The problem can also be solved using fractions:

One and a half pizzas is written as the mixed number: \[ 1\dfrac{1}{2} \] John needs one minute to eat one quarter. To find out how many quarters are in \(1\dfrac{1}{2}\), we divide: \[ 1\dfrac{1}{2} \div \dfrac{1}{4} \quad (1) \] Convert the mixed number to an improper fraction: \[ 1\dfrac{1}{2} = 1+ \dfrac{1}{2} = \dfrac{2}{2} + \dfrac{1}{2} = \dfrac{3}{2} \] Now perform the division (1) to find the number of quarters in \( 1\dfrac{1}{2} \): \[ \dfrac{3}{2} \div \dfrac{1}{4} = \dfrac{3}{2} \times \dfrac{4}{1} = \dfrac{12}{2} = 6 \] So there are 6 quarters in one and a half pizzas. Therefore, John needs: \[ 6 \times 1 = 6 \text{ minutes} \] to eat the entire pizza and a half.

Mike read 5 hours. Sasha read twice as long as Mike. Hence, Sasha read: \[ 2 \times 5 = 10 \text{ hours} \] Tom read two-fifths of the time that Sasha read. Hence, Tom read: \[ \dfrac{2}{5} \times 10 = 4 \text{ hours} \] John read a quarter of the time that Tom read. Hence, John read: \[ \dfrac{1}{4} \times 4 = 1 \text{ hour} \] Conclusion: John read for 1 hour.

This problem can be solved using algebra or a table:

Algebra Method

Let \( x \) be Jim's age.

Carla is 5 years older than Jim, hence Carla's age is \[ x + 5 \] Tomy is 6 years older than Carla, hence Tomy's age is \[ x + 5 + 6 = x + 11 \]. The sum of all 3 ages is 31, hence: \[ x + ( x + 5 ) + (x+11) = 31 \] Group like terms \[ 3 x + 16 = 31 \] Subtract 16 from both sides of the equation and simplify: \[ 3 x = 15 \] Solve for x: \[ x = 5 \] Jim's age is: 5 , Carla's age is: 5 + 5 = 10 and Tomy's age is: 10 + 6 = 16.

Table Method

If if for reasons you cannot solve this problem using algebra, this problem can be solved using a table as shown below where Jim's age is guessed then Carla's and Tomy's ages are calculated. The calculations are stopped when the condition in the problem which is "the sum of their three ages is 31 years" is reached.

Mel spent:

\$34.00 on a pair of trousers

\(2 \times 16.00 = \$32.00\) on two shirts (\$16.00 each)

\(2 \times 24.00 = \$48.00\) on two pairs of shoes (\$24.00 each)

The total amount of money spent by Mel is: \[ 34.00 + 32.00 + 48.00 = 114.00 \] He had \$32.00 left after shopping, so the total amount of money he had before shopping is: \[ 114.00 + 32.00 = 146.00 \] This total includes the \$35.00 he originally had, plus the amount he withdrew from the bank. Therefore, the amount he withdrew is: \[ 146.00 - 35.00 = 111.00 \] Mel withdrew \$111.00 from the bank.

1 week = 7 days
1 day = 24 hours
1 hour = 60 minutes
Hence in one week, there are: \[ 7 \times 24 \times 60 = 10080 \; \text{minutes in one week} \]

The area of a rectangle is calculated using the formula: \[ \text{Area} = \text{Length} \times \text{width} \] Area of the large rectangle (pool and grass) is: \[ 50 \times 20 = 1000 \; \text{metres squared} \] Area of the pool is: \[ 30 \times 10 = 300 \; \text{metres squared} \] \[ \text{Area of grass = Area of the large rectangle - Area of the pool} \] \[ = 1000 - 300 = 700 \; \text{metres squared} \]

The total area of the cardboard before cutting is: \[ \text{Area} = \text{Length} \times \text{Width} = 15 \times 10 = 150 \text{ cm}^2 \] Area of one square cut from a corner: \[ \text{Area of 1 square} = 3 \times 3 = 9 \text{ cm}^2 \] There are 4 such squares, so the total area cut out is: \[ 4 \times 9 = 36 \text{ cm}^2 \] Therefore, the area of the cardboard after cutting the 4 corners is: \[ 150 - 36 = 114 \text{ cm}^2 \]

When we subtract the cost of materials from the total cost, we get the total cost of labor: \[ 330.00 - 225.00 = 105.00 \] So \$105.00 is the total labor cost. Since the painter charges \$35.00 per hour, the number of hours is: \[ \dfrac{105.00}{35} = 3 \text{ hours} \]

Use a table and guess the price of one toy car and one toy train then check the 2 conditions:

1) condition 1: Three toy cars and 4 toy trains should cost $18

2) condition 2: Two toy cars and 3 toy trains should cost $13

Note that as long as condition 1 is not satisfied, there no need to try to satisfy condition 2 because we need both of them to be statisfied at the same time.

The results in the table below shows that conditions 1 and 2 are satisfied when
price of 1 car is $2 and the price of a train is $3.

Note that this problem can be solved algebraically as follows:

Let \( x \) be the cost of one toy car (in dollars)

Let \( y \) be the cost of one toy train (in dollars) The above statement is translated as:

1. Three toy cars and four toy trains cost $18: \[ 3x + 4y = 18 \] 2. **Two toy cars and three toy trains cost $13: \[ 2x + 3y = 13 \] So we end up with a system of equations: \[ \begin{cases} 3x + 4y = 18 \\ 2x + 3y = 13 \end{cases} \] Solve the system to obtain the price of one car \( x = \$2 \) and the price of one train \( y = \$3 \).

\[ \text{Volume = length } \times \text{width} \times \text{height} \] \[ = 5 \times 3 \times 4 = 60 \; \text{cubic centimeters} \] The volume of the box is 60 cm³.

Total number of balls is: \[ 6 + 4 + 10 = 20 \] Number of blue balls = 4

The probability of picking a blue ball is: \[ \dfrac{\text{ Number of blue balls }}{\text{Total number of balls}} = \dfrac{4}{20} = \dfrac{1}{5} \] The probability is \( \dfrac{1}{5} \).

Total fabric used ls: \[ 2.4 + 0.85 = 3.25 \; \text{ meters } \] Fabric left is: \[ 3.75 - 3.25 = 0.5 \; \text{ meters } \] John has 0.5 meters of fabric left.

Let \( x \) be Linda's total savings. If she spent \( \dfrac{3}{4} \) of her savings on furniture, then \[ 1 - \dfrac{3}{4} = \dfrac{4}{4} - \dfrac{3}{4} = \dfrac{1}{4} \] of her savings remain, which can be expressed as: \[ \dfrac{1}{4} x \] She then spent \( \dfrac{1}{2} \) of her remaining savings on a fridge that costs $150. Therefore, we have the equation: \[ \dfrac{1}{2} \times \left( \dfrac{1}{4} x \right) = 150 \] Simplifying the above expression: \[ \dfrac{x}{8} = 150 \] Now, multiply both sides of the equation by 8 to solve for \( x \): \[ 8 \times \left( \dfrac{x}{8} \right) = 8 \times 150 \] \[ x = 1200 \] Linda's original savings were $1200.

Let \( x \) be the length of the side of square A and \( y \) be the length of the side of square B.

The perimeters of the two squares are given by:

Perimeter of square A: \[ 4x \] Perimeter of square B: \[ 4y \] The expression "The perimeter of square A is 3 times the perimeter of square B" is written mathematically as: \[ 4x = 3(4y) = 12y \] Divide left and right sides by 4: \[ x = 3y \] Square both sides of the equation: \[ x^2 = (3y)^2 \] Simplify: \[ x^2 = 9y^2 \] The areas of the two squares are:

Area of square A: \[ x^2 \] Area of square B: \[ y^2 \] The ratio of the area of square A to the area of square B is: \[ \dfrac{x^2}{y^2} \] Using the equation \( x^2 = 9y^2 \), divide both sides by \( y^2 \): \[ \dfrac{x^2}{y^2} = \dfrac{9y^2}{y^2} \] Simplify: \[ \dfrac{x^2}{y^2} = 9 \] The ratio of the area of square A to the area of square B is \( 9:1 \).

The length of the box is given by (subtract 3 cm twice): \[ 15 - 3 - 3 = 9 \text{ cm} \] The width of the box is given by (subtract 3 cm twice): \[ 10 - 3 - 3 = 4 \text{ cm} \] The height of the box, after folding, is equal to \[ 3 \text{ cm} \] The din=mensions of the open rectangular box are: length = 9, width = 4 and height = 3.

Hence, the volume \( V \) of the open rectangular box is given by \[ V = \text{length} \times \text{width} \times \text{height} = 9 \times 4 \times 3 = 108 \text{ cm}^3 \]

Let us first find the total area \( A \) of the rectangle before the square is cut: \[ A = \text{length} \times \text{width} = 20 \times 10 = 200 \] A square of side \( 2x \) has an area \( B \) given by: \[ B = (2x) \times (2x) = 4x^2 \] The small square of area \( B \) is cut from the large rectangle of area \( A \). Hence, the area of the remaining shape is: \[ A - B = 200 - 4x^2 \]

Using the distance formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Let \( A = (x_1 - y_1) = (3, 8) \) and \( B = (x_2 - y_2) = (11, 2) \) and substitute in the formula given above. \[ d = \sqrt{ (11 - 3)^2 + (2 - 8)^2 } \] \[= \sqrt{ 8^2 + (-6)^2} \] \[ = \sqrt{64 + 36} = \sqrt{100} \] \[ d = \sqrt{100} = 10 \; \text{units} \] The distance between points A and B is 10 units.