Grade 7 Maths Problems

With Solutions and Explanations

Detailed solutions and full explanations to grade 7 maths problems are presented.

Problems

Problem
In a bag full of small balls, 1/4 of these balls are green, 1/8 are blue, 1/12 are yellow and the remaining 26 white. How many balls are blue?

Solution
Let us first find the fraction of green, blue and yellow balls
1/4 + 1/8 + 1/12 = 6 / 24 + 3 / 24 + 2 / 24 , common denominator
= 11 / 24 , add numerators
The fraction of white balls is given by
24 / 24 - 11 / 24 = 13 / 24
So the fraction 13 / 24 corresponds to 26 balls. If x is the total number of balls then
(13 / 24) of x = 26 balls
or (13 / 24) × x = 26
x = 26 × (24 / 13) = 48 , total number of balls
The fraction of blue balls is 1 / 8 of x. The number of blue balls is given by
(1 / 8) of 48 = 6 balls.
Problem
In a school 50% of the students are younger than 10, 1/20 are 10 years old and 1/10 are older than 10 but younger than 12, the remaining 70 students are 12 years or older. How many students are 10 years old?

Solution
Let us write the fraction for each group of students
Group A: younger than 10: 50% = 50/100 = 1/2
Group B: 10 years old: 1/20
Group C: older that 10 but younger than 12: 1/10
Group D: 12 years or older: 70 students
The fraction for group A, B and C together is given by
1 / 2 + 1 / 20 + 1 / 10 = 10 / 20 + 1 / 20 + 2 / 20 , common denominator
= 13 / 20 , add numerators
The fraction for group D is given by
20 / 20 - 13 / 20 = 7 / 20 and corresponds to 70 students
If X is the total number of students, then
7 / 20 of X = 70
or (7 / 20) × X = 70
Solve for X
X = 70 × (20 / 7) = 200
Students who are 10 years old have a fraction of 1 / 20 of the total X and their number is equal to
(1 / 20) of 200 = 10 students
Problem
If the length of the side of a square is doubled, what is the ratio of the areas of the original square to the area of the new square?

Solution
If x be the side of the original square, then its area is equal to
x²
If x is doubled to 2x, then the new area is equal to
(2x)² = 4 x²
The ratio of the areas of the original square to the area of the new square
x² / (4 x²) = 1 / 4 or 1:4
Problem
The division of a whole number N by 13 gives a quotient of 15 and a remainder of 2. Find N.

Solution
According to the division process of whole numbers, N can be written, using multiplication, as follows
N = quotient × divisor + remainder = 15 × 13 + 2 = 197
Problem
In the rectangle below, the line MN cuts the rectangle into two regions. Find x the length of segment NB so that the area of the quadrilateral MNBC is 40% of the total area of the rectangle.

Solution
We first note that the lengths of the rectangle are equal, hence
MC + 5 = 20 + x
Subtract 5 from both sides of the equation
MC + 5 - 5 = 20 + x - 5
and simplify to obtain
MC = 15 + x
The quadrilateral MNBC is a trapezoid and its area A is given by
A = (1/2) × 10 × (x + MC)
Simplify (1/2) × 10 and substitute MC by 15 + x
A = 5 ( x + x + 15)
Simplify to obtain
A = 5 (2x + 15)
40% of the area of the rectangle is equal to
40% × (20 + x) × 10 = (40 / 100) × 10 × (20 + x) = 4 (20+x)
Since the area of MNBC is equal to 40% the area of the rectangle, we can write
5(2 x + 15) = 4 (20 + x)
Expand products
10x + 75 = 80 + 4x
Solve the above equation
6x = 5
x = 5/6 meters
Problem
A person jogged 10 times along the perimeter of a rectangular field at the rate of 12 kilometers per hour for 30 minutes. If field has a length that is twice its width, find the area of the field in square meters.

Solution
Let us first find the distance d jogged
distance = rate × time = (12 km / hr) × 30 minutes
= (12 km/hr) × 0.5 hr = 6 km
The distance of 6 km corresponds to 10 perimeters and therefore 1 perimeter is equal to
6 km / 10 = 0.6 km = 0.6 × 1000 meters = 600 meters
Let L and W be the length and width of the field. The length is twice the width. Hence
L = 2 W
The perimeter is 600 meters and is given by
2 (L + W) = 600
Substitute L by 2 W
2 (2 W + W) = 600
Simplify and solve for W
4 W + 2 W = 600
6 W = 600
W = 100
Find L
L = 2 W = 200
Find the area A of the rectangle
A = L * W = 200 * 100 = 20,000 square meters
Problem
Four congruent isosceles right triangles are cut from the 4 corners of a square with a side of 20 units. The length of one leg of the triangles is equal to 4 units. What is the area of the remaining octagon?

Solution
Let us first find area A of the square
A = 20 × 20 = 400 square units
The area B of a small triangle is
B = (1/2) × 4 × 4 = 8 square units
The area of the octagon obtained by subtracting the ares of the 4 triangles from the area of the large square
A - 4 B = 200 - 4 × 8 = 168 square units
Problem
A car is traveling 75 kilometers per hour. How many meters does the car travel in one minute?

Solution
Convert hour into minutes ( 1 hour = 60 minutes) and kilometers into meters (1 km = 1000 m) and simplify
75 kilometers per hour = 75 km/hr
= (75 × 1000 meters) / (60 minutes) = 1,250 meters / minute
Problem
Linda spent 3/4 of her savings on furniture and the rest on a TV. If the TV cost her $200, what were her original savings?

Solution
If Linda spent 3/4 of her savings on furnitute, the rest
4 / 4 - 3 / 4 = 1 / 4 on a TV

But the TV cost her $200. So 1 / 4 of her savings is $200. So her original savings are 4 times $200 = $800
Problem
Stuart bought a sweater on sale for 30% off the original price and another 25% off the discounted price. If the original price of the sweater was $30, what was the final price of the sweater?

Solution
The price with 30% off
30 - 30% of 30 = 30 - (30 / 100) × 30 = 30 - 9 = 21
The price with another 25% off
21 - 25% of 21 = 21 - (25/100) × 21
= 21 - (525 / 100) = 21 - 5.25 = $15.75
Problem
15 cm is the height of water in a cylindrical container of radius r. What is the height of this quantity of water if it is poured into a cylindrical container of radius 2r?

Solution
The volume V of water in the container of radius r
V1 = 15*(?r²)
The volume V of water in the container of radius 2r
V2 = H * (2? (2r)²) (H is the height to be found)
Since is it is the same quantity of water, V1 = V2
15*(?r²) = H * (2? (2r)²)
Solve the above for H to obtain
H = 15/4 = 3.75 cm
Problem
John bought a shirt on sale for 25% off the original price and another 25 % off the discounted price. If the final price was $16, what was the price before the first discount?

Solution
let x be the price before the first discount. The price after the first discount is
x - 25%x (price after first discount)
A second discount of 25% of the discounted price after which the final price is 16
(x - 25%x) - 25%(x - 25%x) = 16
Solve for x
x = $28.44
Problem
How many inches are in 2000 millimeters? (round your answer to the nearest hundredth of of an inch).

Solution
One inch is the same as 25.4 mm. Let x inches be the same as 1000 mm
x = 1 inch * 2000 mm / 25.4 mm = 78.74 inches
Problem
The rectangular playground in Tim's school is three times as long as it is wide. The area of the playground is 75 square meters. What is the perimeter of the playground?

Solution
Let L be the length and W be the width of the playground. "The rectangular playground in Tim's school is three times as long as it is wide means":
L = 3 W
The area A = L * W. Hence
75 = L * W = (3W) * W = 3 W²
Solve for W
3 W² = 75 give W = 75/3 = 25 gives W = &sqrt;(25) = 5 m
L = 3 W = 3 * 5 = 15 m
Perimeter = 2L + 2W = 2(15) + 2(5) = 40 m
Problem
John had a stock of 1200 books in his bookshop. He sold 75 on Monday, 50 on Tuesday, 64 on Wednesday, 78 on Thursday and 135 on Friday. What percentage of the books were not sold?

Solution
Let N be the total number of books sold. Hence
N = 75 + 50 + 64 + 78 + 135 = 402
Let M be the books NOT sold
M = 1200 - N = 1200 - 402 = 798
Percentage
Books not sold / total number of books = 798/1200 = 0.665 = 66.5%
Problem
N is one of the numbers below. N is such that when multiplied by 0.75 gives 1. Which number is equal to N?

Solution
N is such that when multiplied by 0.75 gives 1" is written mathematically as
N * 0.75 = 1
Solve for N
N = 1/0.75 = 100/75 = (75 + 25) / 75 = 75/75 + 25/75 = 1+1/3 Answer: B
A) 1 1/2
B) 1 1/3
C) 5/3
D) 3/2
Problem
In 2008, the world population was about 6,760,000,000. Write the 2008 world population in scientific notation.

Solution
a number in scientific notation is written as
m * 10ⁿ , such that |m| is greater than or equal to 1 and smaller than 10.
6,760,000,000 = 6.76 * 10⁹
Problem
Calculate the circumference of a circular field whose radius is 5 centimeters.

Solution
Circumference C is given by
C = 2?r = 2?*5 = 10 ? cm

Answers to the Above Problems

6 balls are blue
10 students are 10 years old
1:4
N = 197
x = 5/6 meter
20,000 square meters
368 square units
1250 meters per minute
$800
$15.75
3.75 cm
$28.44
78.74 inches
40 meters
66.5%
B
6.76 10⁹
10? centimeters

Grade 7 Maths Problems With Solutions and Explanations

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Grade 7 Maths Problems

With Solutions and Explanations