Math Problems with Solutions for Grade 4

a) The smallest area is 242,400 kilometers squared and it is that of the UK.

b) The largest area is 17,098,242 kilometers squared and corresponds to Russia.

c) The difference between the areas of Russia and China is given by \[ 17,098,242 - 9,598,094 = 7,500,148 \; \text{kilometers squared} \] d) The total area of the countries listed above is given by \( 9,629,091 + 17,098,242 + 9,598,094 + 9,984,670 \) \[ + 242,400 + 3,287,263 = 49,839,760 \; \text{kilometers squared} \] e) We first order the areas from the largest to the smallest. \[ 17,098,242 \; | \; 9,984,670 \; | \; 9,629,091 \; | \; 9,598,094 \; | \; 3,287,263 \; | \; 242,400 \] Which correspond to the following countries:

Russia, Canada, USA, China, India, UK

The number of miles to drive to finish his journey is given by \[ 1200 - 768 = 432 \; \text{miles} \]

We are comparing Sofia: \(\frac{2}{6}\) of a bottle and Max: \(\frac{1}{3}\) of a bottle. Make the denominators the same to compare

Max's fraction is \(\frac{1}{3}\), but we can write \(\frac{1}{3}\) as an equivalent fraction with a denominator of 6 by multiplying numerator and denominator by 2: \[ \frac{1}{3} = \frac{1 \times 2}{3 \times 2} = \frac{2}{6} \] a) Now both fractions are \(\frac{2}{6}\), so: They used the same amount of orange juice.

b) Since they used the same amount: \[ \frac{2}{6} - \frac{2}{6} = 0 \] No one used more. The difference is 0.

To find how many sweets, we multiply 123 by 25 \[ 123 × 25 = 3075 \; \text{sweets} \]

In 100 years, which is one century, there \[ 365 \times 100 = 36,500 \; \text{days} \]

To find the number of vans, we divide 123 by 8. \[ 123 \div 8 = 15 \; \text{with remainder} \; = 3 \] So 15 vans are needed to transport 15 × 8 = 120 girls, and 1 van is needed to transport the 3 remaining girls. A total of 16 vans are needed.

Money spent on drinks \[ 5 \times 4 = \$ 20 \] Money spent on sandwiches \[ 8 \times 6 = \$48 \] Total money spent \[ 20 + 48 = \$68 \] Money left after shopping \[ 100 - 68 = \$32 \]

To know how many cards each should take, divide 175 by 4 (Tom, Julia, Mike and Fran). \[ 175 \div 4 = 43 \; \text{with remainder} = 3 \] Each one should take 43 cards and 3 are left.

Find differences between successive numbers: \[ 6 - 2 = 4 {\displaystyle \implies } 6 = 2 + 4 \] \[12 - 6 = 6 {\displaystyle \implies } 12 = 6 + 6 \] \[ 20 - 12 = 8 {\displaystyle \implies } 20 = 12 + 8 \] \[ 30 - 20 = 10 {\displaystyle \implies } 30 = 20 + 10 \] The Pattern is: Add 4, then 6, then 8, then 10, ...

Next differences: 12 and 14 \[ 30 + 12 = 42 \] \[42 + 14 = 56 \] The 7th number is 56.

a) The whole pizza has 8 equal slices. Liam ate **3 slices** out of 8. So, the fraction he ate is: \[ \dfrac{3}{8} \] b) Ava ate 2 slices out of 8. So, the fraction she ate is: \[ \dfrac{2}{8} \] c) Total eaten = Liam's \(\dfrac{3}{8}\) + Ava's \(\dfrac{2}{8}\) \[ \dfrac{3}{8} + \dfrac{2}{8} = \dfrac{5}{8} \] So, the fraction left is: \[ 1 - \dfrac{5}{8} = \dfrac{3}{8} \] d) There are 3 slices left. If Liam and Ava share them equally, they each get: \[ \dfrac{3}{2} = 1 \dfrac{1}{2} \text{ slices} \]

The perimeter P of the rectangle is equal to: \[ P_1 = 15 + 25 + 15 + 25 = 80 \] The perimeter of the square is equal to the perimeter of the rectangle and is then equal to 80. The square has 4 sides of equal lengths and its perimeter \( P_2\) is given by: \[ P_2 = 4 \times \text{length of one side} \] \[ 80 = 4 \times 20 \] then the length of a side is equal to 20 umits

a) Total number of sweets is: \[ 123 \times 25 = 3075 \; \text{sweets} \] The number of sweets used for gift packs (1 out of every 5) is: \[ \dfrac{1}{5} \times 3075 = 3075 \div 5 = 615 \; \text{sweets} \] b) Number of sweets remaining in the boxes: \[ 3075 - 615 = 2460 \; \text{sweets} \] c) Number of boxes used for gift packs:

Each box has 25 sweets, So, \[ 615 \div 25 = 24.6 \; \text{boxes} \] Since we cannot use a part of a box without opening it, we round 24.6 up to 25 and say: About 25 boxes of sweets are used to make the gift packs.

Total number of days in one century (ignoring leap years) is: \[ 100 \times 365 = 36,500 \text{ days} \] There are 7 days in one week, so the number of weeks in one century is: \[ 36,500 \div 7 = 5,214 \text{ weeks and } 2 \text{ days} \] There are 5,214 full weeks and 2 extra days in one century (if we ignore leap years).

One week is 7 days. Billy read the first book in one week with 25 pages everyday. The total pages of the first book: \[ 25 \times 7 = 175 \text{ pages} \] Billy read the second book in 12 days with 23 pages everyday, The total pages of the second book: \[ 23 \times 12 = 276 \text{ pages} \] Total number of pages read: \[ 175 + 276 = 451 \text{ pages} \]

a) Emma used \( \frac{1}{8} \) in the morning, so the amount left after the morning is: \[ 1 - \frac{1}{8} = \frac{8}{8} - \frac{1}{8} = \frac{7}{8} \] Emma used half of what was left, and what was left was \(\frac{7}{8}\): \[ \frac{1}{2} \times \frac{7}{8} = \dfrac{1 \times 7}{2 \times 8} = \frac{7}{16} \] So in the afternoon, she used \(\frac{7}{16}\) of the bag.

a) Ema used \(\frac{1}{8}\) in the morning and \(\frac{7}{16}\) in the afternoon:

Total \( T \) used: \[ T = \frac{1}{8} + \frac{7}{16} \] We convert \( \frac{1}{8} \) to a fraction with denominator 16: \[ \frac{1}{8} = \frac{1 \times 2}{8 \times 2} = \frac{2}{16} \] Substitute and add: \[ T = \frac{2}{16} + \frac{7}{16} = \frac{9}{16} \] A total of \(\frac{9}{16}\) of the bag used used,

c) The fraction of the bag left is: \[ 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{16 - 9}{16} = \frac{7}{16} \] \(\frac{7}{16}\) of the sugar bag is left.

Money spent on drinks: \[ 5 \times 4 = \$20 \] Money spent on sandwiches: \[ 8 \times 6 = \$48 \] Total money spent: \[ 20 + 48 = \$68 \] Money left after shopping: \[ 100 - 68 = \$32 \] John had \$32 left after the shopping.

To find the number of toys produced each day, divide the total number of toys produced in one week by the number of working days 4: \[ \dfrac{5500}{4} = 1375 \text{ toys} \] 1375 toys are produced each day by 55 workers, hence each worker produces: \[ 1375 \div 55 = 25 \] Each worker produces 25 toys per day.

To find how many cards each person should take, divide 175 by 4 (since there are 4 people): \[ 175 \div 4 = 43 \text{ Remainder } 3 \] Each person should take: \[ 43 \text{ cards} \] There will be: \[ 3 \text{ cards left over} \] Each one should take 43 cards, and 3 cards are left.

The area of one square is equal to: \[ 4 × 4 = 16 \; \text{centimeters squared} \] The whole shape has 5 squares. The total area is: \[ 16 + 16 + 16 + 16 + 16 = 80 \; \text{centimeters squared} \] or \[ 16 × 5 = 80 \; \text{centimeters squared} \]

Sam = 11 shells

Sam + Carla = 24 shells

So to find how many Carla collected: \[ \text{Carla} = 24 - \text{Sam} = 24 - 11 = 13 \] Carla + Sarah = 25 shells

We just found that Carla collected 13, so: \[ \text{Sarah} = 25 - \text{Carla} = 25 - 13 = 12 \] Sam collected: 11 shells , Carla collected: 13 shells and Sarah collected: 12 shells.

From Monday to Friday, there are 5 days. He runs 6 kilometers each day: \[ 6 \times 5 = 30 \text{ kilometers} \] On Saturday and Sunday, there are 2 days. He runs 12 kilometers each day: \[ 12 \times 2 = 24 \text{ kilometers} \] Total kilometers run in a week: \[ 30 + 24 = 54 \text{ kilometers} \] Mr. Joshua runs 54 kilometers in a week.

First, find the total amount of money they had together. They bought a toy for \$22 and got \$6 back: \[ 22 + 6 = 28 \] So, together they had \$28. Since they had the same amount, divide 28 by 2: \[ 28 \div 2 = 14 \] Each of them had \$14.

Two methods are presented to solve this problem. If you have not done any algebra, the table method is good for you. If you know algebra and equation solving, then both methods are good for you.

Table method

We can create a table to try different numbers of boxes with 5 sweets and see how many boxes with 4 sweets are needed to reach a total of 22 sweets.

Note the the total number of boxes is 5, hence once "Number of boxes with 5 sweets" is selected in the first column on the left, the "Number of boxes with 4 sweets" is equal to 5 - "Number of boxes with 5 sweets".

John has 2 boxes with 5 sweets and 3 boxes with 4 sweets.

Algebra method

Let's say John has \( x \) boxes with 5 sweets each. Then he must have \( 5 - x \) boxes with 4 sweets each (because he has 5 boxes in total). Now we find how many sweets he has: \[ \text{Sweets from boxes with 5 sweets} = 5x \] \[ \text{Sweets from boxes with 4 sweets} = 4(5 - x) \] Total sweets: \[ 5x + 4(5 - x) = 22 \] Now solve the equation: \[ 5x + 20 - 4x = 22 \] \[ x + 20 = 22 \] \[ x = 2 \] So, John has: \[ x = 2 \quad \text{boxes with 5 sweets each} \] \[ 5 - x = 3 \quad \text{boxes with 4 sweets each} \] John has 2 boxes with 5 sweets and 3 boxes with 4 sweets. ?

Let the number of chickens be denoted by \( C \) and the number of rabbits by \( R \).

The total number of animals is 16: \[ C + R = 16 \] The chickens have 2 legs each and the rabbits have 4 legs each.

The total number of legs is 50: \[ 2C + 4R = 50 \] Below is the table with different possibilities for \( C \) (chickens) and their corresponding number of legs:

Note that: \( C + R = 16 \) \[ \begin{array}{|c|c|c|} \hline \text{Chickens (C)} & \text{Rabbits (R)} & \text{Total Legs} = 2 C + 4 R\\ \hline 0 & 16 & 2 \times 0 + 4 \times 16 = 64 \\ 1 & 15 & 2 \times 1 + 4 \times 15 = 2 + 60 = 62 \\ 2 & 14 & 2 \times 2 + 4 \times 14 = 4 + 56 = 60 \\ 3 & 13 & 2 \times 3 + 4 \times 13 = 6 + 52 = 58 \\ 4 & 12 & 2 \times 4 + 4 \times 12 = 8 + 48 = 56 \\ 5 & 11 & 2 \times 5 + 4 \times 11 = 10 + 44 = 54 \\ 6 & 10 & 2 \times 6 + 4 \times 10 = 12 + 40 = 52 \\ 7 & 9 & \color{red}{ 2 \times 7 + 4 \times 9 = 14 + 36 = 50 \quad \text{(Correct Answer)}}\\ \hline \end{array} \]

From the table, when there are 7 chickens and 9 rabbits, the total number of legs is exactly 50.

Thus, there are 7 chickens and 9 rabbits in the farm.

Two methods are presented to solve this problem. If you have not done any algebra, the table method is good for you. If you know algebra and equation solving, then both methods are good for you.

Algebra method

Let the number of rabbits be denoted by \( R \), and the number of chickens be denoted by \( C \).

There are 4 more chickens than rabbits, so: \[ C = R + 4 \]

Chickens have 2 legs each, and rabbits have 4 legs each.

The total number of legs is 44, so: \[ 2C + 4R = 44 \] We can substitute \( C = R + 4 \) into the second equation: \[ 2(R + 4) + 4R = 44 \] Simplifying: \[ 2R + 8 + 4R = 44 \] \[ 6R + 8 = 44 \] \[ 6R = 44 - 8 \] \[ 6R = 36 \] \[ R = 6 \] Now substitute \( R = 6 \) into \( C = R + 4 \): \[ C = 6 + 4 = 10 \] So, there are 10 chickens and 6 rabbits.

Table method

In this table we start by the number of rabbits, calculate the number of chickens and the total number of legs. We stop calculations when the condition of the total number of legs (44) is satisfied. \[ \begin{array}{|c|c|c|} \hline \text{Rabbits (R)} & \text{Chickens (C = R + 4)} & \text{Total Legs} = 2C + 4 R \\ \hline 0 & 4 & 2 \times 4 + 4 \times 0 = 8 \\ 1 & 5 & 2 \times 5 + 4 \times 1 = 10 + 4 = 14 \\ 2 & 6 & 2 \times 6 + 4 \times 2 = 12 + 8 = 20 \\ 3 & 7 & 2 \times 7 + 4 \times 3 = 14 + 12 = 26 \\ 4 & 8 & 2 \times 8 + 4 \times 4 = 16 + 16 = 32 \\ 5 & 9 & 2 \times 9 + 4 \times 5 = 18 + 20 = 38 \\ 6 & 10 & \color{red}{2 \times 10 + 4 \times 6 = 20 + 24 = 44 \quad \text{(Correct Answer)}} \\ \hline \end{array} \] Thus there are 10 chickens and 6 rabbits.