Download 1. The perimeter of a rectangle is 28 cm and its area is 48

1.
The perimeter of a rectangle is 28 cm and its area is 48 cm2. Find
its dimensions.
Solution:
Let the dimensions of the rectangle be x and y.
Perimeter = 28, hence,
2(x + y) = 28
x + y = 14
y = 14 – x
Area = 48, hence,
xy = 48
Solving by substitution
x(14 – x) = 48
14x – x2 – 48 = 0
x2 – 14x + 48 = 0
(x – 6)(x – 8) = 0
x = 6 or x = 8
Finding y:
2.
y = 14 – 6 = 8 or 14 – 8 = 6
Write the following in equation form and solve simultaneously to find
x and y. The sum of two numbers is 42 and their difference is 12.
What are the two numbers?
Solution:
x  y  42
 x  y  12
2 x  54
x  27
 y  42  27  15
The two numbers are 27 and 15.
IB Taskbank Mathematics
1
3.
Solve the following word problem.
The total of three consecutive numbers is 66. What are the numbers?
Solution:
x  x  1  x  2  66
3x  3  66
x  21
Then, the numbers are 21, 22 and 23.
4.
When a number is doubled and then added to 3, the new result is equal
to 15 decreased by the original number. Find the number.
Solution:
Let x = the number
2x + 3 = 15 – x
3x = 12
x=4
The number is 4.
5.
The length of a rectangle is 7 m less than twice the width. If the
perimeter is 82 m, find the area of the rectangle.
Solution:
Let x = width of rectangle
 length of rectangle = 2x – 7
Perimeter = 2x + 2(2x – 7) = 82
6x – 14 = 82
6x = 96
x = 16
 the width of the rectangle is 16 m
the length of the rectangle is 2(16) – 7 = 25 m
 the area of the rectangle is 25 × 16 = 400 m2
IB Taskbank Mathematics
2
6.
A family of six (two adults and four children) buy tickets to see
a film. The total cost of tickets is $35 and a child’s ticket is three
quarters of the cost of an adult ticket. How much does each of the
children’s tickets cost?
Solution:
Let x = cost of adult ticket
 cost of child ticket =
3
x
4
3
2 x  4  x  35
4
5 x  35
x=7
 cost of adult ticket is $7
the cost of child’s ticket is
7.
3
 7  $5.25
4
Anne is 3 years older than Michael, and Jason is twice as old as Anne.
If their average age is 19, find Jason’s age.
Solution:
Let Anne’s age = x
Michael’s age = x – 3
Jason’s age = 2x
Average =
x  x – 3  2x
 19
3
4x – 3
 19
3
4x – 3 = 57
4x = 60
x = 15
 Jason’s age is 2 × 15 = 30 years old
IB Taskbank Mathematics
3
8.
A wire is cut into two unequal pieces so that:
 the first piece forms a square
 the second piece forms a rectangle
 the width of the rectangle equals the double of the side of the
square
 the length of the rectangle equals the treble of its width.
(a) Let the side of the square be s. Find the sum of the perimeters
of the square and the rectangle in terms of s.
(b) Given that the length of the wire is 40 cm, find the value of
s.
(c) Hence, find the dimensions of the rectangle.
Solution:
(a)
The perimeter of the square = 4s.
The sides of the rectangle are: width = 2s and length = 3 × 2s = 6s.
The perimeter of the rectangle equals
2(2s + 6s)
= 2(8s)
= 16s
Hence, the sum of perimeters
= 4s + 16s
= 20s
(b)
Since the sum = 40 cm, then
20s = 40
s = 2 cm
(c)
Hence, the dimensions of the rectangle are
Width = 2 × 2 = 4 cm
Length = 6 × 2 = 12 cm
Check:
4 × 2 + 2(4 + 12)
= 8 + 2(16)
= 8 + 32
= 40 cm
IB Taskbank Mathematics
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9.
ABC is a triangle such that:

the measure of angle B is double the measure of angle A

the measure of angle C is equal to the sum of measures of angles
A and B.
Find the measure of each of the angles of triangle ABC.
Solution:
Let the measure of A  x so B  2x.
Notice that A  B  C means that C  90 and since C  3x
Then
3x = 90
x = 30
Hence, A = 30°, B  2  30  60 and C  90.
Check: they all add up to 180°.
10. A number plus its half, plus its third, plus its quarter is 25. What
is the number?
Solution:
x x x
x     25
2 3 4
25x
 25
12
x = 12
IB Taskbank Mathematics
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