# Solutions to Maths Problems for Grade 5

Solutions and explanations to grade 5 math word problems are presented.

1. A large box contains 18 small boxes and each small box contains 25 chocolate bars. How many chocolate bars are in the large box?
Solution
The number of chocolate bars is equal to
18 × 25 = 450

2. It takes John 25 minutes to walk to the car park and 45 to drive to work. At what time should he get out of the house in order to get to work at 9:00 a.m.?
Solution
The time it takes John to get to work: time to walk to car park + time to drive
time = 25 + 45 = 70 minutes = 60 minutes + 10 minutes
Knowing that 1 hour = 60 minutes,
time = 1 hour + 10 minutes
John needs to get out of the house 1 hour and 10 minutes before 9:00 am at
9:00 - 1:10 = 7:50 a.m.

3. Kim can walk 4 kilometers in one hour. How long does it take Kim to walk 18 kilometers?
Solution
The time it takes Kim to walk 18 kilometers is equal to
Time = Distance / speed
Time = 18 / 4 = 4.5 hours = (4 + 0.5) hours = 4 and 1/2 hours
Knowing that 1/2 an hour = 30 minutes, we write Time = 4 hours and 30 minutes.

4. A factory produced 2300 TV sets in its first year of production. 4500 sets were produced in its second year and 500 more sets were produced in its third year than in its second year. How many TV sets were produced in three years?
Solution
500 TV sets were produced in the third year than in the second year. The number of sets produced in the third year is equal to
4,500 + 500 = 5,000
The number of TV sets produced in three years is equal to sum of the number of TV sets produced in each year
2,300 + 4,500 + 5,000 = 11,800

5. Linda bought 3 notebooks at \$1.20 each; a box of pencils at \$1.50 and a box of pens at \$1.70. How much did Linda spend?
Solution
Linda spent
1.20 × 3 = \$3.60 on notebooks
The total amount of money that Linda spent is equal to
3.60 + 1.50 + 1.70 = \$6.80

6. Tom and Bob have a total of 49 toys. If Bob has 5 more toys than Tom, how many toys does each one have?
Solution
If 5 toys are taken out of 49 toys and the remaining ones distributed to Tom and Bob, they will both have equal numbers of toys
49 - 5 = 44 for Tom and Bob
If distributed equally, each one will have
44 ÷ 2 = 22 toys
Bob has 5 more toys than Tom, so Bob has
22 + 5 = 27 toys
Bob has 27 and Tom has 22 toys and it is easy to check that between them they have 49 and the difference is 5.

7. John can eat a quarter of a pizza in one minute. How long does it take John to eat one pizza and a half?
Solution
2 possible solutions:
1) There are 4 quarters of a pizza in one pizza and there 2 quarters of a pizza in a half a pizza. So there is a total of 6 quarters in one pizza and a half. If John eats a quarter in one minutes, he needs 6 minutes to eat all 6 quarters.
2) The above problem could also be solved as follows:
One pizza and a half is written using mixed number as 1(1/2).
John needs one minute to eat one quarter. The number of quarters in one pizza and a half is found using the division:
(1 (1/2)) ÷ 1/4
Write the mixed number 1 (1/2) as an improper fraction:
1 (1/2) = 2/2 + 1/2 = 3/2
The number of quarters in one pizza and a half is given by
(3/2) � 1/4
The division of two fractions is transformed into a mutliplication of the first fraction by the multiplicative inverse of the second.
(3/2) � 1/4 = (3/2) × (4/1) = 12 / 2 = 6 quarters
John needs 6 minutes to eat one pizza and a half.

8. John can eat a sixth of a pizza in two minutes. It takes 3 minutes for Billy to eat one quarter of the same pizza. If John and Billy start eating one pizza each, who will finish first?
Solution
In one pizza, there are 6 sixths and John will take .
2 × 6 = 12 minutes to finish one pizza
In one pizza, there are 4 quarters and Billy will take .
3 × 4 = 12 minutes to finish one pizza.
It takes each one of them 12 minutes and they will finish at the same time.

Solution
2 × 5 = 10 hours
(2 / 5) × 10 = 4 hours
(1 / 4) × 4 = 1 hour

10. Jim, Carla and Tomy are members of the same family. Carla is 5 years older than Jim. Tomy is 6 years older than Carla. The sum of their three ages is 31 years. How old is each one them?
Solution
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.

Jim's age Carla's age Tomy's age The sum of all ages
1 1 + 5 = 6 6 + 6 = 12 1 + 6 + 12 = 19
2 2 + 5 =7 7 + 6 = 13 2+ 7 + 13 = 22
3 3 + 5 = 8 8 + 6 = 14 3 + 8 + 14 = 25
4 4 + 5 = 9 9 + 6 = 15 4 + 9 + 15 = 28
5 5 + 5 = 10 10 + 6 = 16 5 + 10 + 16 = 31

The column on the right, where all ages are added, shows whether the main condition ("The sum of their three ages is 31 years") is satisfied or not: last row of the table shows: Jim 5 , Carla 10 and Tomy 16 satisfy the condition in the problem.

11. Mel had \$35.00 and withdraw some more money from his bank account. He bought a pair of trousers at \$34.00, two shirts at \$16.00 each and 2 pairs of shoes at \$24.00 each. After the shopping, he had \$32.00 left. How much money did Mel withdraw from the bank?
Solution
Mel spent :
\$34.00 on a pair of trousers
2 × 16.00 = \$32.00 on two shirts (\$16.00 each shirt)
2 × 24.00 = \$48.00 on 2 pairs of shoes (\$24.00 each pair)
The total amount of money spent by Mel is:
\$34.00 + \$32.00 + \$48.00 = \$114.00
He had \$32.00 left after shopping, therefore the amount of money he had before he started shopping is the sum of what he spent and what is left and is equal to:
\$114.00 + \$32.00 = \$146.00
The total money he had which was \$146.00 includes what he had \$35.00 and what he withdraw from the bank. Therefore he withdraw from the bank:
\$146.00 - \$35.00 = \$111.00

12. How many minutes are in one week?
Solution
1 week = 7 days
1 day = 24 hours
1 hour = 60 minutes
Hence in one week, there are:
7 × 24 × 60 = 10080 minutes in one week.

13. In Tim's house, a rectangular swimming pool (blue) whose length 30 meters and width 10 meters is surrounded by grass (green). The pool with the grassy area make a large rectangle whose length is 50 meters and width 20 meters. What area is occupied by the grass? .

Solution
The area of a rectangle is calculated using the formula:
Area = Length × width
Area of the large rectangle (pool and grass) = 50 × 20 = 1000 metres squared
Area of the pool = 30 × 10 = 300 metres squared
Area of grass = Area of the large rectangle - Area of the pool = 1000 - 300 = 700 metres squared

14. Mary wants to make a box. She starts with a piece of cardboard whose length is 15 centimeters and with is 10 centimeters. Then she cuts congruent squares with side of 3 centimeters at the four corners. What is the area of the cardboard after she cuts the 4 corners? .

Solution
The total area of the cardboard before cutting is:
Area = Length × width = 15 × 10 = 150 centimeters squared
Area of 1 square at one corner = 3 × 3 = 9 centimeters squared
There are 4 square cut at the 4 corners, therefore the total area cut from the main sheet of cardboard (4 squares) is:
4 × 9 = 36 centimeters squared
The area of the cardboard after cutting the 4 corners is:
150 - 36 = 114 centimeters squared.

15. A painter charges \$225.00 for materials and \$35.00 per hour for labor. The total cost of painting an office is \$330.00. How many hours did it take the painter to paint the office?

Solution
When we subtract the cost for material from the total cost, we will obtain the total cost of labour which is:
\$330.00 - \$225.00 = \$105.00
So \$105.00 is the cost for labor at \$35.00 per hour, therefore the number of hour is:
105.00 � 35 = 3 hours.

16. Three toy cars and 4 toy trains cost \$18. Two toy cars and 3 toy trains cost \$13. What is the price of one toy car and the price of one toy train if both prices are whole numbers of Dollars? (Hint: Use a table)
Solution
Use a table and guess the price of one toy car and one toy train then calculate the price of and check the 2 conditions:
1) condition 1: Three toy cars and 4 toy trains should cost \$18
Once the price of 3 toy cars and 4 toy trains is correct, calculate the price of
2) condition 2: Two toy cars and 3 toy trains should cost \$13
Steps to follow:
Start by gessing price of 1 car as \$1 (column 1), then start the price of 1 train at \$1 (column 2) and then increase by one unit.
Once the price of Three toy cars and 4 toy trains is above \$18, increase the price of 1 car to \$2 dollars and start again from \$1 as the price of a train and increasing. Calculate the cost of 3 cars and 4 trains (condition 1) in column 3.
Once condition 1 is satisfied which is the price of Three toy cars and 4 toy trains is \$18, check condition 2: the price of 2 cars and 3 trains and it should be \$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.
Guess
price of 1 car
Guess
price of 1 train
Calculate Price
3 cars + 4 trains
Calculate Price
2 cars + 3 trains
1 1 3×1+4×1 = 7 No need for calculations
1 2 3×1+4×2 = 11 No need for calculations
1 3 3×1+4×3 = 15 No need for calculations
1 4 3×1+4×4 = 19 No need for calculations
2 1 3×2+4×1 = 10 No need for calculations
2 2 3×2+4×2 = 14 No need for calculations
2 3 3×2+4×3 = 18 2×2+3×3 = 13