Math Problems with Solutions for Grade 6

Sequence:

1st : 2

2nd : \(2 \times 3 + 2 = 8\)

3rd : \(8 \times 3 + 2 = 26\)

4th : \(26 \times 3 + 2 = 80\)

5th : \(80 \times 3 + 2 = 242\)

The 5th number is \( 242\).

Original area of the cardboard \[ \text{Area} = \text{length} \times \text{width} = 12 \times 8 = 96 \, \text{cm}^2 \] Area of one square cut from a corner \[ \text{Area} = 2 \times 2 = 4 \, \text{cm}^2 \] Total area removed from all four corners \[ 4 \times 4 = 16 \, \text{cm}^2 \] Remaining area after cutting \[ 96 - 16 = 80 \, \text{cm}^2 \]

The product of two integers is equal to the product of their LCM and GCF. Therefore, \[ 16 \times N = 48 \times 8 \] Solving for \( N \): \[ N = \dfrac{48 \times 8}{16} = 24 \]

We first write the prime factorization of each number: \[ 24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3 \] \[ 40 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5 \] \[ 60 = 2 \times 2 \times 3 \times 5 = 2^2 \times 3 \times 5 \] Now, we find the common prime factors with the smallest exponents. The only common prime factor among all three numbers is 2. The smallest power of 2 common to all three numbers is \(2^2\). \[ \text{GCF} = 2^2 = 4 \] The greatest common factor of 24, 40, and 60 is 4.

a) Ratio of girls to boys:

\[ \dfrac{260}{240} \] Reduce the fraction \[ = \dfrac{13}{12} \] the ratio of girls to boys is \[ 13:12 \] b)

Total number of pupils is: \[ 240 + 260 = 500 \] Ratio of boys to the total number of pupils: \[ \dfrac{240}{500} \] Reduce the fraction \[ = \dfrac{12}{25} \] The ratio of boys to the total number of pupils is \[ 12:25\]

The tip is 20% of what he paid for lunch. Hence: \[ \text{Tip} = 20\% \text{ of } 50.50 = \left( \dfrac{20}{100} \right) \times 50.50 = 10.10 \] Total spent: \[ 50.50 + 10.10 = 60.60 \] Tim spent \$60.60 in total.

A division is related to a mutliplication and vice versa.

The division \( \dfrac{M}{N} = P \) may be written as a multiplication \( M = P \times N \) and vice versa.

Since: \[ \dfrac{64}{k} = 4 \] the above division may be written as a multiplication as follows: \[ 64 = 4 k \] Rewrite the above multiplication as a division: \[ \dfrac{64}{4} = k \]. Carry out the division: \[ \dfrac{64}{4} = 16 \]. Hence \[ k = 16 \]

Substitute \( y = 3 \) into the equation \( x + 2y = 10 \): \[ x + 2(3) = 10 \] Simplify \[ x + 6 = 10 \] To solve for \( x \), subtract 6 from both sides: \[ x + 6 - 6 = 10 - 6\] Group like terms and simplify \[ x = 4\]

Cost \( C \) for a call of 1 minute: \[ C = 0.50 + ( 0.11 )\] Cost \( C \) for a call of 2 minutes: \[ C = 0.50 + (0.11 + 0.11) = 0.50 + 2 \times 0.11 \] Cost \( C \) for a call of 3 minutes: \[ C = 0.50 + (0.11 + 0.11 + 0.11) = 0.50 + 3 \times 0.11 \] We observe that the cost \( C \) is given by: \[ C = 0.50 + (\text{number of minutes}) \times 0.11 \] If \( N \) is the number of minutes, then the cost \( C \) is: \[ C = 0.50 + 0.11 N \]

Each 40 kilometers, 1 gallon is needed. We need to determine how many 40-kilometer segments are in 180 kilometers: \[ \dfrac{180}{40} = 4.5 \] So there \( 4.5 \) segment of \( 40 \) km in the total distance 180 kilometers. Hence the number of gallons needed is given by: \[ 4.5 \times 1\ \text{gallon} = 4.5\ \text{gallons} \] The car would need \( 4.5 \) gallons of gasoline to travel \( 180 \) kilometers.

8 minutes are needed to fill 150 bottles.

How many groups of 150 bottles are there in 675 bottles? \[ \dfrac{675}{150} = 4.5 = 4 \text{ full groups and } \dfrac{1}{2} \text{ group} \] For each group, 8 minutes are needed. So, for 4 full groups and half a group: \[ 4 \times 8 + \dfrac{1}{2} \times 8 = 32 + 4 = 36 \text{ minutes} \] Alternatively, we can calculate directly: \[ 4.5 \times 8 = 36 \text{ minutes} \] It will take \( 36 \) minutes to fill 675 bottles.

During each hour, the car travels 65 miles. For 5 hours, it will travel: \[ 65 + 65 + 65 + 65 + 65 = 5 \times 65 = 325 \text{ miles} \] We can also use the formula: \[ \text{Distance} = \text{time} \times \text{speed} = 5 \times 65 = 325 \text{ miles}\] The car will travel \( 325 \) miles in 5 hours.

Two rectangles \( A \) and \( B \) are similar if their lengths and widths are proportional. Let

\( L_1 = 10 \, \text{cm} \), \( W_1 = 5 \, \text{cm} \) be the length and width of rectangle \( A \)

\( L_2 = 30 \, \text{cm} \), and \( W_2 \) be the length and width of rectangle \( B \)

Since the rectangles are similar, we can write: \[ \dfrac{L_2}{L_1} = \dfrac{W_2}{W_1} \] Substituting the known values: \[ \dfrac{30}{10} = \dfrac{W_2}{5} \] Cross-multiply: \[ 30 \times 5 = 10 \times W_2 \] Divide both sides by 10: \[ W_2 = 15 \, \text{cm} \] Now find the area of rectangle \( B \): \[ \text{Area} = L_2 \times W_2 = 30 \times 15 = 450 \, \text{cm}^2 \] The area of rectangle \( B \) is \( 450 \, \text{cm}^2 \).

Let \( x \) be the number of registered students in each class.

The number of students absent in the 5 classes that were half full is: \[ 5 \times \left(\dfrac{1}{2} \times x\right) = \dfrac{5x}{2} \] The percenatge of students absent in a classes that is \(\dfrac{3}{4}\) full is: \[ 1 - \dfrac{3}{4} = \dfrac{1}{4}\] Hence the number of students absent in the 3 classes that were \(\dfrac{3}{4}\) full is given by: \[ 3 \times \left(\dfrac{1}{4} \times x\right) = \dfrac{3x}{4} \] The number of students absent in the 2 classes where \(\dfrac{1}{8}\) of the students were absent is: \[ 2\times \left(\dfrac{1}{8} \times x \right) = \dfrac{2x}{8} \] The total number of absent students is given as 70, hence: \[ \dfrac{5x}{2} + \dfrac{3x}{4} + \dfrac{2x}{8} = 70 \] Multiply all terms by 8: \[ 8 \times \dfrac{5x}{2} + 8 \times \dfrac{3x}{4} + 8 \times \dfrac{2x}{8} = 8 \times 70 \] Simplify: \[ 20 x + 6x + 2 x = 560 \] Groupe like terms: \[ 28 x = 560 \] Solve for \( x \): \[ x = \dfrac{560}{28} = 20 \] So, each class has 20 students. Total number of registered students in all 10 classes in the school is: \[ 10 \times 20 = 200 \] \( 200 \) students are registered in the school.

There are are squares of 4 different dimensions: 1 × 1, 2 × 2, 3 × 3 , 4 × 4

There are:

16 squares of dimension 1 × 1.

9 squares of dimension 2 × 2.

4 squares of dimension 3 × 3

1 square of dimension 4 × 4.

The total number of squares is: \[ 16+9+4+1 = 30 \; \text{squares} \]

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

John gave half of his stamps to Jim: \[ \text{Jim received: } \quad \dfrac{1}{2}x \] Jim gave half of his stamps to Carla: \[ \text{Carla received:} \quad \dfrac{1}{2} \cdot \dfrac{1}{2}x = \dfrac{1}{4}x \] Carla gave \(\dfrac{1}{4}\) of her stamps to Thomas, so she kept \(\dfrac{3}{4}\) of what she received: \[ \text{Carla kept: } \quad \dfrac{3}{4} \cdot \dfrac{1}{4}x = \dfrac{3}{16}x \] We are told that Carla kept 12 stamps, hence: \[ \dfrac{3}{16}x = 12 \] To solve for \( x \), multiply both sides by \( \dfrac{16}{3} \): \[ x = \dfrac{16}{3} \cdot 12 = 64 \] John started with 64 stamps.

It will take: \[ \dfrac{120}{4} = 30 \text{ seconds} \] for ball A to make one full rotation.

Ball A:   1 rot: 30 sec,   2 rot: 60 sec,   3 rot: 90 sec,   4 rot: 120 sec, ...

It will take: \[ \dfrac{60}{3} = 20 \text{ seconds} \] for ball B to make one full rotation.

Ball B:   1 rot: 20 sec,   2 rot: 40 sec,   3 rot: 60 sec,   4 rot: 80 sec, ...

The first time that they have made a whole number of rotations and therefore be at the same starting point is after 60 seconds, which is the least common multiple (LCM) of 30 and 20.

It will take 60 seconds for both balls to be at the same starting point again.

To divide a 3-unit segment into 9 equal parts, each unit must be divided into 3 parts. Hence: \[ \text{1 part} = \dfrac{1}{3} \text{ of a unit} \] Therefore: \[ \text{2 parts} = 2 \times \dfrac{1}{3} = \dfrac{2}{3} \text{ of a unit} \]

Rewrite the rate of 75 kilometers per hour by converting kilometers to meters and hours to minutes. \[ 1 \, \text{kilometer} = 1000 \, \text{meters} \] \[ 1 \, \text{hour} = 60 \, \text{minutes} \] Thus, \[ 75 \; \text{kilometers per hour} = \dfrac{75 \times \text{1 kilometers}}{ \text{1 hour}} \] Substitute 1 kilometer by 1000 meters and 1 hour by 60 minutes \[ 75 \; \text{kilometers per hour} = \dfrac{75 \times 1000 \;\text{meters }}{60 \; \text{minutes}} \] \[ = \dfrac{75000}{60} \, \dfrac{\text{meters}}{ \text{minute}} = 1250 \, \text{meters per minute} \] The car travels 1250 meters in one minute.

Let \( x \) be Linda's total savings. If she spent \( \dfrac{3}{4} \) of her savings on furniture, then \( \dfrac{1}{4} \) 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 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: \[ A - B = 200 - 4x^2 \]

The area is given by \( \pi r^2 \). Hence, \[ \pi r^2 = 81\pi \] Dividing both sides by \( \pi \): \[ r^2 = 81 \] Hence \[ r = \sqrt{81} = 9 \text{ feet} \] The circumference is given by: \[ 2\pi r = 2\pi \times 9 = 18\pi \text{ feet} \]

Let \( P \) be Peter's current age. The present ages of Carla, Jim, and Peter are:

\[ \text{Carla}: 5 \] \[ \text{Peter}: P \] \[ \text{Jim}: P - 13 \] One year ago, their ages were (subtract 1 from the current ages):

\[ \text{Carla}: 5 - 1 = 4 \] \[ \text{Peter}: P - 1 \] \[ \text{Jim}: P - 13 - 1 = P - 14 \] According to the problem, one year ago Peter's age was twice the sum of Carla's and Jim's ages. So, \[ P - 1 = 2(4 + P - 14) \] Simplify the right-hand side: \[ P - 1 = 2(P - 10) \] \[ P - 1 = 2P - 20 \] Add 20 to both sides: \[ P - 1 + 20 = 2P - 20 + 20 \] \[ P + 19 = 2P \] Subtract \( P \) from both sides: \[ P + 19 - P = 2P - P \] \[ 19 = P \] Conclusion: \[ \text{Peter's age} : P = 19 \] \[ \text{Jim's age} : P - 13 = 19 - 13 = 6 \] \[ \text{Carla's age} : 5 \]

Add the distances already run: \[ 2.4 + 1.75 = 4.15 \text{ km} \] Subtract from the total goal: \[ 6 - 4.15 = 1.85 \text{ km} \] Lisa needs to run 1.85 kilometers more.

Let the amount of flour be \( x \). The ratio is: \[ \dfrac{x}{4} = \dfrac{3}{2} \] Cross multiply: \[ 2x = 12 \] Solve for \( x \) \[ x = 6 \] You need 6 cups of flour.

Let \( x \) be the third angle. Sum of angles in a triangle is \( 180^\circ \). \[ x + 47 + 58 = 180^\circ \] Group like terms: \[ x + 105 = 180^\circ \] Solve for x: \[ x = 180 - 105 = 75^\circ \] The third angle is \( 75^\circ \).

\[ 3x + 4 = 19 \] Subtract 4 from each side and simplify: \[ 3x = 19 - 4 = 15 \] Divide by \( 3 \) and simplify the left and right sides: \[x = \dfrac{15}{3} = 5 \]

Start at \( - 4^\circ \) C

Rise by \( 11^\circ \) \[ -4 + 11 = 7^\circ C \] Drop by \( 6^\circ \) C: \[ 7 - 6 = 1^\circ C \] The temperature at 6 PM was \( 1^\circ\).

Let original price be \( x \). A 30% discount means she paid: \[ 100\% - 30\% = 70\% \] 70% of the price that Amira paid is written as: \[ 70\% \; \text{of} \; x = \dfrac{70}{100} x = 0.7 x \] It is also known that she paid \( \$84 \), hence: \[ 0.70x = 84 \] Solve for \( x \) \[ x = \dfrac{84}{0.7} = \$120 \] The original price was \$120.

Since they are walking toward each other, their combined speed is the sum of their individual speeds: \[ \text{Combined Speed} = 40 \, \text{m/min} + 60 \, \text{m/min} = 100 \, \text{m/min} \] The time it takes Harry and Kate to meet is the total distance divided by their combined speed: \[ \text{Time} = \dfrac{\text{Distance}}{\text{Combined Speed}} = \dfrac{2500 \, \text{m}}{100 \, \text{m/min}} = 25 \, \text{minutes} \] The dog is running continuously for the 25 minutes at a speed of 120 meters per minute. Therefore, the total distance traveled by the dog is: \[ \text{Distance} = \text{Dog's Speed} \times \text{Time} = 120 \, \text{m/min} \times 25 \, \text{min} = 3000 \, \text{m} \]