# Grade 6 Maths word Problems With Solutions and Explanations

Detailed solutions and full explanations to grade 6 maths word problems are presented.

1. Two numbers N and 16 have LCM = 48 and GCF = 8. Find N.
Solution
The product of two integers is equal to the product of their LCM and GCF. Hence.
16 × N = 48 × 8
N = 48 × 8 / 16 = 24

2. If the area of a circle is 81π square feet, find its circumference.
Solution
The area is given by π × r × r. Hence
π × r × r = 81π
r × r = 81 ; hence r = 81 feet
The circumference is given by
2 × π × r = 2 × pi × 9 = 18 π feet

3. Find the greatest common factor (GFC) of 24, 40 and 60.
Solution
We first write the prime factorization of each given number
24 = 2 × 2 × 2 × 3 = 23 × 3
40 = 2 × 2 × 2 × 5 = 23 × 5
60 = 2 × 2 × 3 × 5 = 22 × 3 × 5
GFC = 22 = 4

4. In a given school, there 240 boys and 260 girls.
a) What is the ratio of the number of girls to the number of boys?
b) What is the ratio of the number of boys to the total number of pupils in the school?
Solution
a) ratio of girls to boys
260:240 or 13:12
b) ratio of boys to the total number of pupils
240:(240+260) or 240:500 or 12:25

5. If Tim had lunch at $50.50 and he gave 20% tip, how much did he spend? Solution The tip is 20% of what he paid for lunch. Hence tip = 20% of 50.50 = (20/100)*50.50 = 101/100 =$10.10
Total spent
50.50 + 10.10 = $60.60 6. Find k if 64 ÷ k = 4. Solution Since 64 ÷ k = 4 and 64 ÷ 16 = 4, then k = 16 7. Little John had$8.50. He spent $1.25 on sweets and gave to his two friends$1.20 each. How much money was left?
Solution
John spent and gave to his two friends a total of
1.25 + 1.20 + 1.20 = $3.65 Money left 8.50 - 3.65 =$4.85

8. What is x if x + 2y = 10 and y = 3?
Solution
Substitute y by 3 in x + 2y = 10
x + 2(3) = 10
x + 6 = 10
If we substitute x by 4 in x + 6 = 10, we have 4 + 6 = 10. Hence
x = 4

9. A telephone company charges initially $0.50 and then$0.11 for every minute. Write an expression that gives the cost of a call that lasts N minutes.
Solution
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 × 0.11
Cost C for a call of 3 minutes
C = 0.50 + 0.11 + 0.11 + 0.11 = 0.50 + 3 × 0.11
We note that the cost C is equal to
C = 0.50 + (number of minutes) × 0.11
If N is the number of minutes, the cost C is given by
C = 0.50 + N × 0.11

10. A car gets 40 kilometers per gallon of gasoline. How many gallons of gasoline would the car need to travel 180 kilometers?
Solution
Each 40 kilometers, 1 gallon is needed. We need to know how many 40 kilometers are there in 180 kilometers?
180 ÷ 40 = 4.5 × 1 gallon = 4.5 gallons

11. A machine fills 150 bottles of water every 8 minutes. How many minutes it takes this machine to fill 675 bottles?
Solution
8 minutes are needed to fill 150 bottles. How many groups of 150 bottles are there in 675 bottles?
675 ÷ 150 = 4.5 = 4 and 1/2
For each of these groups 8 minutes are needed. For 4 groups and 1/2
8 × 4 + 4 = 32 + 4 = 36 minutes. (4 is for 1/2 a group that needs half time)
We can also find the final answer as follows
4.5 x 8 = 32 minutes

12. A car travels at a speed of 65 miles per hour. How far will it travel in 5 hours?
Solution
During each hour, the car travels 65 miles. For 5 hours it will travel
65 + 65 + 65 + 65 + 65 = 5 × 65 = 325 miles

13. A small square of side 2x is cut from the corner of a rectangle with a width of 10 centimeters and length of 20 centimeters. Write an expression in terms of x for the area of the remaining shape.
Solution
Let us first find the total area A of the rectangle before cutting the small is cut
A = length × width = 20 × 10 = 200
A square of side 2x has an area B given by
B = (2x) × (2x) = 4 × x × x = 4 x2
The small square of area B is cut from the large rectangle of area A. Hence the area of the remaining shape is given by
A - B = 200 - 4 x2

14. A rectangle A with length 10 centimeters and width 5 centimeters is similar to another rectangle B whose length is 30 centimeters. Find the area of rectangle B.
Solution
Two rectangles A and B are similar if their lengths and widths are proportinal.
Let L1 = 10 cm and W1 = 5 cm be the length and width of rectangle A. Let L2 = 30 cm and W2 be the length and width of rectangle B.
Proportionality of the dimensions of the two rectangles is written as:
L2 / L1 = W2 / W1
Substitute by the known quantities and find W2
30 / 10 = W2 / 5
For the above ratios to be equal, W2 must be equal to 15. Hence the area of rectamgle B is given by
L2 × W2 = 30 cm × 15 cm = 450 cm 2

15. A school has 10 classes with the same number of students in each class. One day, the weather was bad and many students were absent. 5 classes were half full, 3 classes were 3/4 full and 1/8 of the students in the last two classes were absent. A total of 70 students were absent. How many students are registered in this school?
Solution
The given information is related to different classes and since we have 10 classes with the same number of students we need to find the number of registered students in each class. Let x be the number of registered students in each class.
Hence the number of students absent in the 5 classes that were half full is given by: (1/2 of the students were absent in each class)
5 × ( 1/2 × x) = 5 x / 2
The number of students absent in the 3 classes that were 3/4 full is given by: ( a class that 3/4 full has 1/4 of the students absent)
3 × ( 1/4 × x) = 3 x / 4
The number of students absent in the 2 classes that were 1/8 full is given by:
1/8 × (2 × x) = 2 x / 8
The total number of students who were absent is equal to 70 (given). Hence
5 x / 2 + 3 x / 4 + 2 x / 8 = 70
rewrite fractions with same denominator
(4 / 4) × 5 x / 2 + (2/ 2) × 3 x / 4 + 2 x / 8 = 70
Simplify
20 x / 8 + 6 x / 8 + 2 x / 8 = 70
28 x / 8 = 70
Multiply both sides of the above equation by 8 / 28
(8 / 28) × 28 x / 8 = (8 / 28) × 70
Simplify to solve for x.
x = 20 students per class
total number of students for all 10 classes in the school: 10 × 20 = 200 students

16. A large square is made of 16 congruent squares. What is the total number of squares of different sizes are there?

.

Solution
There are are squares of 4 different dimensions: 1 × 1, 2 × 2, 3 × 3 , 4 × 4
There are 16 squares of dimension 1 × 1. There are 9 squares of dimension 2 × 2. There are 4 squares of dimension 3 × 3 and 1 square of dimension 3 × 3.
There is a total of
16+9+4+1 = 30 squares

17. The perimeter of square A is 3 times the perimeter of square B. What is the ratio of the area of square A to the area of square B.
Solution
Let x be the size of the side of square A and y be the size of the side of square B. The perimeters of the two squares are given by:
Perimeter of A: 4 x and Perimeter of B: 4 y
The expression "The perimeter of square A is 3 times the perimeter of square B" is written mathematically as:
4 x = 3(4 y) = 12 y
Divide both sides of the above equation by 4
x = 3 y
Square both sides of the above equation.
(x) 2 = (3 y) 2
Simplify to obtain.
x 2 = 9 y 2
The area of the two squares are given by
Area of square A := x 2 and Area of square B = y 2
The ratio of the area of square A to the area of square B is given by
x 2 / y 2
Divide both sides of the equation x 2 = 9 y 2 obtained above by y 2
x 2 / y 2 = 9 y 2 / y 2
Simplify to obtain the ratio of the two areas
x 2 / y 2 = 9

18. John gave half of his stamps to Jim. Jim gave gave half of his stamps to Carla. Carla gave 1/4 of the stamps given to her to Thomas and kept the remaining 12. How many stamps did John start with?
Solution
Let x be the number of stamps that John started with.
John gave half of his stamps to Jim: Jim got (1/2) x
Jim gave gave half of his stamps to Carla: Carla got (1/2) ((1/2) x)
Carle gave 1/4 and therefore kept 3/4 of the stamps given to her. Carla kept 3/4 of (1/2) ((1/2) x ) = (3/4) ((1/2) ((1/2) ) x)
Simplify the expression (3/4) ((1/2) ((1/2) ) ) x .
(3/4) ((1/2) ((1/2) ) x) = (3 × 1 × 1) / (4 × 2 × 2) x = 3 x / 16
The number of stamps kept by Carla is equal to 12. Hence the equation to solve
3 x / 16 = 12
Mutliply both side of the above equation by 16 / 3
(16/3) × (3 x / 16) = (16 / 3) ×12
Simplify and solve for x
x = (16 / 3) ×12 = 64

19. Two balls A and B rotate along a circular track. Ball A makes 4 full rotations in 120 seconds. Ball B makes 3 full rotation in 60 seconds. If they start rotating now from the same, how long will take them to be at the same starting point again?
Solution
It will take
120 / 4 = 30 seconds for ball A to make one full rotation
It will take
60 / 3 = 20 seconds for ball B to make one full rotation
Let us calculate the time for whole rotations of ball A and B
Ball A : 1 rot: 30 sec , 2 rot: 60 sec , 3 rot 90 sec, 4 rot 120 sec, .... 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 lowest common (LCM) multiple of 30 and 20.

20. A segment is 3 units long. It is divided into 9 parts. What fraction of a unit are 2 parts of the segment?
Solution
To divide a 3 unit segment to make 9 parts, you need to divide each unit by 3. Hence
1 part = 1/3 of a unit
and therefore
2 parts = 2/3 of a unit

21. Mary wants to make a box. She starts with a piece of cardboard whose length is 15 centimeters and width is 10 centimeters. Then she cuts 4 congruent squares with sides of 3 centimeters at the four corners and folded at the broken lines to make the box. What is the volume of the box?

.

Solution
The length of the box is given by
15 - 3 - 3 = 9 cm The width of the box is given by
10 - 3 - 3 = 4 cm
The height of the box is equal to
3 cm
The volume V of the box is given by
V = length × width × height = 9 × 4 × 3 = 108 cm 3

22. A car is traveling 75 kilometers per hour. How many meters does the car travel in one minute?
Solution
Rewrite the rate 75 kilometers per hour converting kilometers in meters and hours in minutes
1 kilometer = 1000 meters
1 hour = 60 minutes
Hence
75 kilometers per hour = 75 × 1000 meters per 60 minutes = 75 000 / 60 meters/minute = 1250 meters/minute

23. Carla is 5 years old and Jim is 13 years younger than Peter. One year ago, Peter's age was twice the sum of Carla's and Jim's age. Find the present age of each one of them.
Solution
Let x be Peter's age now. Hence the present age of Carla, Jim and Peter are given by
Carla : 5
Peter : x
Jim : x - 13
One year ago their ages were
Carla : 5 - 1 = 4
Peter : x - 1
Jim : x - 13 - 1 = x - 14
One year ago, Peter's age was twice the sum of Carla's and Jim's age which is written mathematically as
x - 1 = 2 (4 + x - 14)
Simplify and expand the right side of the above equation
x - 1 = 2 (x - 10)
x - 1 = 2x - 20
Add + 20 to both sides and simplify
x - 1 + 20 = 2x - 20 + 20
x + 19 = 2x
Subtract x to both sides and simplify
x + 19 - x = 2x - x
19 = x , Peter's age now
Jim's age: x - 13 = 19 - 13 = 6
Carla's age: 5

24. Linda spent 3/4 of her savings on furniture. She then spent 1/2 of her remaining savings on a fridge. If the fridge cost her $150, what were her original savings? Solution Let x be Linda's savings. If she spent 3/4 of her savings on furniture then 1/4 of her savings are remaining and are written as (1/4) x She spent 1/2 of her remaining saving on a fridge that costs 150. Hence (1/2) × ((1/4) x ) = 150 Simplify the above and rewrite as x / 8= 150 Multiply both sides of the above equation by 8 and solve for x 8 × (x / 8) = 8 × 150 x =$1200

25. The distance between Harry and Kate is 2500 meters. Kate and Harry start walking toward one another and Kate' dog start running back and forth between Harry and Kate at a speed of 120 meters per minute. Harry walks at the speed of 40 meters per minute while Kate walks at the speed of 60 meters per minute. What distance will the dog have travelled when Harry and Kate meet each other?
Solution
The dog runs during all the period of time while Kate and Harry are walking. This period of time may be calculated as follows:
2500 meters / (40 + 60) meters / minutes = 25 minutes
The dog runs during 25 minutes at the rate of 120 meters per minutes. Hence the total distance covered by the dog is given by
120 meters/minute × 25 minutes = 3000 meters

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