Learning Task 3
Solve the word problems
1. The sides of the quadrilateral are consecutive numbers.
how long is each side?
2. Inzo is 10 years older than Migz. Five years ago Mig
Inzo. What are their present ages?
3. An express train travels 100 kph from station A to
travelling at 45 kph, takes 45 minutes longer for the
is station A from station B.
4. The sum of two consecutive even integers is less tha
Find at least 3 pairs of numbers
5. The perimeter of the square is less than 152 m. Find
sides of the square.​

Answer:

1. The sides are: 38.5, 39.5, 40.5 and 41.5

2. Presently, Inzo is 15 years old and Miz is 10 years old

3. [tex]Distance = 3682km[/tex]

4. Possible pairs are: (12, 14), (16, 18) and (20, 24)

5. [tex]L < 38m[/tex]

Step-by-step explanation:

The question has lots of missing details. See the comment section for complete question

Solving (a):

Given

Shape: Quadrilateral

Sides: Consecutive

[tex]Perimeter = 160[/tex]

Let one of the sides be x. The other sides will be x + 1, x + 2 and x + 3.

So, the perimeter is:

[tex]Perimeter = x + x + 1 + x + 2 + x + 3[/tex]

This gives:

[tex]160 = x + x + 1 + x + 2 + x + 3[/tex]

Collect Like Terms

[tex]160 -1-2-3= x+x+x+x[/tex]

[tex]154= 4x[/tex]

Divide both sides by 4

[tex]38.5 = x[/tex]

[tex]x =38.5[/tex]

So, the sides are: 38.5, 39.5, 40.5 and 41.5

Solving (b):

Given

Presently: [tex]Inzo = 10 + Miz[/tex]

Five years ago: [tex]Miz - 5= \frac{1}{3} * (Inzo-5)[/tex] --- (Missing from the question)

Required: Determine their present ages

Substitute [tex]10 + Miz[/tex] for Inzo in the second equation

[tex]Miz - 5 = \frac{1}{3} * (10 + Miz - 5)[/tex]

[tex]Miz - 5 = \frac{1}{3} * (Miz + 5)[/tex]

Multiply both sides by 3

[tex]3Miz - 15 = Miz + 5[/tex]

Collect Like Terms

[tex]3Miz -Miz = 15 + 5[/tex]

[tex]2Miz= 20[/tex]

Divide both sides by 2

[tex]Miz = 10[/tex]

Substitute 10 for Miz in [tex]Inzo = 10 + Miz[/tex]

[tex]Inzo = 10 + 5[/tex]

[tex]Inzo = 15[/tex]

So presently, Inzo is 15 years old and Miz is 10 years old

Solving (3):

Given

Express Train:

[tex]Speed = 100kph[/tex]

[tex]Time = t[/tex]

Local Train:

[tex]Speed = 45kph[/tex]

[tex]Time = t + 45[/tex]

Required: Determine the distance between A and B

Distance is calculated as:

[tex]Distance = Speed * Time[/tex]

For the express train:

[tex]Distance = 100 * t[/tex]

For the local train

[tex]Distance = 45 * (45 + t)[/tex]

Distance between both stations are equal; So,

[tex]100 * t = 45 * (45 + t)[/tex]

[tex]100 t = 2025 + 45t[/tex]

Collect Like Terms

[tex]100 t -45t= 2025[/tex]

[tex]55t= 2025[/tex]

Divide by 55

[tex]t = 36.82[/tex]

Substitute 36.82 for t in [tex]Distance = 100 * t[/tex]

[tex]Distance = 100 * 36.82[/tex]

[tex]Distance = 3682km[/tex]

Solving (4):

Let the even numbers be: n and n + 2.

Sum greater than 24: [tex]n + n + 2 > 24[/tex]

Sum less than 60: [tex]n + n + 2 < 60[/tex]

So, we have:

[tex]n + n + 2 > 24[/tex]

[tex]2n + 2 > 24[/tex]

Collect Like Terms

[tex]2n > 24-2[/tex]

[tex]2n > 22[/tex]

[tex]n > 11[/tex]

[tex]n + n + 2 < 60[/tex]

[tex]2n + 2< 60[/tex]

Collect Like Terms

[tex]2n < 60-2[/tex]

[tex]2n < 58[/tex]

Divide by 2

[tex]n<29[/tex]

So, we have:

[tex]n > 11[/tex] and [tex]n < 29[/tex]

So, possible pairs are: (12, 14), (16, 18) and (20, 24)

Solving (5):

Given

[tex]Perimeter < 152m[/tex]

Required: Find the possible lengths

Let the length be L.

Perimeter is calculated as:

[tex]Perimeter =4 * L[/tex]

[tex]Perimeter =4 L[/tex]

So, we have:

[tex]4L < 152m[/tex]

Divide through by 4

[tex]L < 38m[/tex]

Hence, length is less than 38m (e.g. 37m)

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