Math Problems with Solutions for Grade 6

Follow the sequence rule step-by-step:

• 1st number: 2
• 2nd number: \(2 \times 3 + 2 = 8\)
• 3rd number: \(8 \times 3 + 2 = 26\)
• 4th number: \(26 \times 3 + 2 = 80\)
• 5th number: \(80 \times 3 + 2 = 242\)

The 5th number is 242.

First, calculate the original area of the cardboard:

\[ \text{Area} = \text{length} \times \text{width} = 12 \times 8 = 96 \text{ cm}^2 \]

Next, find the area of one square cut from a corner:

\[ \text{Area} = 2 \times 2 = 4 \text{ cm}^2 \]

Since there are four corners, the total area removed is:

\[ 4 \times 4 = 16 \text{ cm}^2 \]

Subtract the removed area from the original area to find the remaining area:

\[ 96 - 16 = \mathbf{80 \text{ cm}^2} \]

The product of two integers is always equal to the product of their LCM and GCF. Therefore:

\[ 16 \times N = 48 \times 8 \]

Solving for \( N \):

\[ N = \dfrac{48 \times 8}{16} = 3 \times 8 = \mathbf{24} \]

First, write the prime factorization of each number:

• \( 24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3 \)
• \( 40 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5 \)
• \( 60 = 2 \times 2 \times 3 \times 5 = 2^2 \times 3 \times 5 \)

Identify the common prime factors with the smallest exponents. The only common prime factor among all three numbers is 2, and the smallest power of 2 common to all three is \(2^2\).

\[ \text{GCF} = 2^2 = \mathbf{4} \]

a) Ratio of girls to boys:

\[ \dfrac{260}{240} \]

Reduce the fraction by dividing the numerator and denominator by 20: \( \dfrac{13}{12} \). The ratio is 13:12.

b) Ratio of boys to total pupils:

Total pupils = \( 240 + 260 = 500 \).

\[ \dfrac{240}{500} \]

Reduce the fraction by dividing the numerator and denominator by 20: \( \dfrac{12}{25} \). The ratio is 12:25.

The tip is 20% of what he paid for lunch. Calculate the tip:

\[ \text{Tip} = 20\% \text{ of } 50.50 = \left( \dfrac{20}{100} \right) \times 50.50 = 10.10 \]

Add the tip to the original lunch cost:

\[ \text{Total} = 50.50 + 10.10 = \mathbf{\$60.60} \]

Division is related to multiplication. The equation can be rewritten as a multiplication:

\[ 64 = 4k \]

Rewrite this multiplication back as a division to solve for \( k \):

\[ k = \dfrac{64}{4} = \mathbf{16} \]

Substitute \( y = 3 \) into the equation:

\[ x + 2(3) = 10 \]

\[ x + 6 = 10 \]

Subtract 6 from both sides to isolate \( x \):

\[ x = \mathbf{4} \]

The cost \( C \) consists of a fixed initial fee ($0.50) and a variable rate ($0.11 multiplied by the number of minutes, \( N \)).

\[ C = \mathbf{0.50 + 0.11N} \]

For every 40 kilometers, 1 gallon is needed. Divide the total distance by the distance per gallon to find the number of 40-km segments:

\[ \dfrac{180}{40} = 4.5 \]

The car would need 4.5 gallons of gasoline.

First, find how many groups of 150 bottles fit into 675:

\[ \dfrac{675}{150} = 4.5 \text{ groups} \]

Since each group requires 8 minutes, multiply the number of groups by 8:

\[ 4.5 \times 8 = \mathbf{36 \text{ minutes}} \]

Use the distance formula:

\[ \text{Distance} = \text{Time} \times \text{Speed} \]

\[ \text{Distance} = 5 \times 65 = \mathbf{325 \text{ miles}} \]

Since the rectangles are similar, the ratio of their corresponding sides is equal:

\[ \dfrac{L_2}{L_1} = \dfrac{W_2}{W_1} \]

\[ \dfrac{30}{10} = \dfrac{W_2}{5} \]

\[ 3 = \dfrac{W_2}{5} \quad \Rightarrow \quad W_2 = 15 \text{ cm} \]

Now find the area of rectangle \( B \):

\[ \text{Area}_B = 30 \times 15 = \mathbf{450 \text{ cm}^2} \]

Let \( x \) be the number of registered students in one class. We need to express the absent students as fractions of \( x \):

• 5 classes are half full (meaning \( \frac{1}{2} \) absent): \( 5 \times \left(\dfrac{1}{2}x\right) = \dfrac{5}{2}x \)
• 3 classes are \( \frac{3}{4} \) full (meaning \( \frac{1}{4} \) absent): \( 3 \times \left(\dfrac{1}{4}x\right) = \dfrac{3}{4}x \)
• 2 classes have \( \frac{1}{8} \) absent: \( 2 \times \left(\dfrac{1}{8}x\right) = \dfrac{2}{8}x = \dfrac{1}{4}x \)

Set the sum of absences equal to 70:

\[ \dfrac{5}{2}x + \dfrac{3}{4}x + \dfrac{1}{4}x = 70 \]

Find a common denominator (4):

\[ \dfrac{10}{4}x + \dfrac{3}{4}x + \dfrac{1}{4}x = \dfrac{14}{4}x = \dfrac{7}{2}x = 70 \]

Solve for \( x \):

\[ x = 70 \times \dfrac{2}{7} = 20 \text{ students per class} \]

Since there are 10 classes, the total registered student population is \( 10 \times 20 = \mathbf{200 \text{ students}} \).

There are squares of 4 different dimensions. We count them systematically:

• 16 squares of dimension \( 1 \times 1 \).
• 9 squares of dimension \( 2 \times 2 \).
• 4 squares of dimension \( 3 \times 3 \).
• 1 square of dimension \( 4 \times 4 \).

Total number of squares = \( 16 + 9 + 4 + 1 = \mathbf{30} \)

Let \( x \) be the number of stamps John started with.

Jim received: \( \dfrac{1}{2}x \)

Carla received: \( \dfrac{1}{2} \times \left(\dfrac{1}{2}x\right) = \dfrac{1}{4}x \)

Carla gave \( \frac{1}{4} \) away, meaning she kept \( \frac{3}{4} \) of what she received:

\[ \text{Carla kept} = \dfrac{3}{4} \times \left(\dfrac{1}{4}x\right) = \dfrac{3}{16}x \]

We are told she kept 12 stamps. So:

\[ \dfrac{3}{16}x = 12 \]

\[ x = 12 \times \dfrac{16}{3} = \mathbf{64 \text{ stamps}} \]

First, find the time it takes for each ball to make one full rotation.

Ball A: \( \dfrac{120}{4} = 30 \text{ seconds per rotation} \)

Ball B: \( \dfrac{60}{3} = 20 \text{ seconds per rotation} \)

They will meet at the starting point at the Least Common Multiple (LCM) of their rotation times.

\[ \text{LCM}(30, 20) = \mathbf{60 \text{ seconds}} \]

If a 3-unit segment is divided into 9 equal parts, the length of 1 part is:

\[ 1 \text{ part} = \dfrac{3}{9} = \dfrac{1}{3} \text{ of a unit} \]

Therefore, 2 parts equal:

\[ 2 \times \dfrac{1}{3} = \mathbf{\dfrac{2}{3} \text{ of a unit}} \]

Convert kilometers to meters and hours to minutes:

\[ 1 \text{ kilometer} = 1000 \text{ meters} \]

\[ 1 \text{ hour} = 60 \text{ minutes} \]

Now, apply the conversions to the speed:

\[ 75 \text{ km/h} = \dfrac{75 \times 1000 \text{ meters}}{60 \text{ minutes}} \]

\[ = \dfrac{75000}{60} = \mathbf{1250 \text{ meters per minute}} \]

Let \( x \) be her total savings. If she spent \( \frac{3}{4} \) of her savings, she had \( \frac{1}{4} \) left.

She spent half of that remainder on a fridge, which cost $150:

\[ \dfrac{1}{2} \times \left(\dfrac{1}{4}x\right) = 150 \]

\[ \dfrac{1}{8}x = 150 \]

Multiply both sides by 8:

\[ x = 150 \times 8 = \mathbf{\$1200} \]

Let \( x \) be the side length of square A and \( y \) be the side length of square B. Their perimeters are \( 4x \) and \( 4y \).

\[ 4x = 3(4y) \quad \Rightarrow \quad 4x = 12y \quad \Rightarrow \quad x = 3y \]

The ratio of their areas is:

\[ \dfrac{\text{Area}_A}{\text{Area}_B} = \dfrac{x^2}{y^2} \]

Substitute \( x = 3y \):

\[ \dfrac{(3y)^2}{y^2} = \dfrac{9y^2}{y^2} = \mathbf{9} \] (The ratio is 9:1).

The height of the box will be the size of the square cut out (\( 3 \text{ cm} \)).

The new length is the original length minus the two squares cut from each side:

\[ \text{Length} = 15 - 3 - 3 = 9 \text{ cm} \]

The new width is the original width minus the two squares cut from each side:

\[ \text{Width} = 10 - 3 - 3 = 4 \text{ cm} \]

Calculate the volume:

\[ V = \text{length} \times \text{width} \times \text{height} = 9 \times 4 \times 3 = \mathbf{108 \text{ cm}^3} \]

Find the total area of the original rectangle:

\[ A_{\text{rect}} = \text{length} \times \text{width} = 20 \times 10 = 200 \]

Find the area of the small square being cut:

\[ A_{\text{square}} = (2x) \times (2x) = 4x^2 \]

The remaining area is the rectangle's area minus the square's area:

\[ \text{Remaining Area} = \mathbf{200 - 4x^2} \]

The area of a circle is given by \( \pi r^2 \):

\[ \pi r^2 = 81\pi \]

Divide both sides by \( \pi \):

\[ r^2 = 81 \quad \Rightarrow \quad r = 9 \text{ feet} \]

The circumference is \( 2\pi r \):

\[ C = 2\pi(9) = \mathbf{18\pi \text{ feet}} \]

Let Peter's current age be \( P \). This makes Jim's current age \( P - 13 \).

One year ago, their ages were:

• Peter: \( P - 1 \)
• Carla: \( 5 - 1 = 4 \)
• Jim: \( (P - 13) - 1 = P - 14 \)

Set up the equation based on the condition from one year ago:

\[ P - 1 = 2 \times [4 + (P - 14)] \]

\[ P - 1 = 2(P - 10) \]

\[ P - 1 = 2P - 20 \]

Subtract \( P \) and add 20 to both sides:

\[ P = \mathbf{19} \]

Peter is 19, Jim is 6 (\(19 - 13\)), and Carla is 5.

Add the distances she has already run:

\[ 2.4 + 1.75 = 4.15 \text{ km} \]

Subtract this from her total goal:

\[ 6 - 4.15 = \mathbf{1.85 \text{ km}} \]

Let the amount of flour be \( x \). Set up the ratio:

\[ \dfrac{x}{4} = \dfrac{3}{2} \]

Cross multiply:

\[ 2x = 12 \]

\[ x = \mathbf{6 \text{ cups}} \]

The sum of the angles in a triangle is always \( 180^\circ \). Let \( x \) be the third angle.

\[ x + 47 + 58 = 180 \]

\[ x + 105 = 180 \]

\[ x = 180 - 105 = \mathbf{75^\circ} \]

Subtract 4 from both sides:

\[ 3x = 19 - 4 \]

\[ 3x = 15 \]

Divide by 3:

\[ x = \dfrac{15}{3} = \mathbf{5} \]

Start at \( -4^\circ\text{C} \) and add the rise in temperature:

\[ -4 + 11 = 7^\circ\text{C} \]

Subtract the drop in temperature:

\[ 7 - 6 = \mathbf{1^\circ\text{C}} \]

Let the original price be \( x \). A 30% discount means she paid 70% of the original price:

\[ 100\% - 30\% = 70\% \]

\[ 0.70x = 84 \]

Divide by 0.70:

\[ x = \dfrac{84}{0.70} = \mathbf{\$120} \]

Since they are walking toward each other, their combined speed is the sum of their individual speeds:

\[ \text{Combined Speed} = 40 + 60 = 100 \text{ m/min} \]

The time it takes for them to meet is the distance divided by their combined speed:

\[ \text{Time} = \dfrac{2500}{100} = 25 \text{ minutes} \]

The dog runs continuously for those 25 minutes at 120 meters per minute. The total distance the dog travels is:

\[ \text{Distance} = 120 \text{ m/min} \times 25 \text{ min} = \mathbf{3000 \text{ meters}} \]