asked 79.9k views
0 votes
Use the IDEAL model of problem-solving to solve the task below showing clearly all necessary actions taken in each of the stages: A teenager's age is increased by 2 gives a perfect square. His age decreased by 10 gives the square root of that perfect square. He is 5 years older than the sister. How old is his sister?​

asked
User Nathan A
by
8.3k points

1 Answer

7 votes

Answer:

I - Identify the problem:

The problem is to find the age of the teenager's sister.

D - Define the problem:

Let's define the age of the teenager as "x" and the age of his sister as "y". The problem states that:

- x + 2 = a perfect square

- √(x + 2 - 10) = a perfect square

- x = y + 5

We need to find the value of "y".

E - Explore possible solutions:

First, let's solve the equation x + 2 = a perfect square. We can try different values of perfect squares until we find one that is 2 more than a multiple of 10:

- 4 + 2 = 6 (not a multiple of 10)

- 9 + 2 = 11 (not a multiple of 10)

- 16 + 2 = 18 (not a multiple of 10)

- 25 + 2 = 27 (not a multiple of 10)

- 36 + 2 = 38 (not a multiple of 10)

- 49 + 2 = 51 (not a multiple of 10)

- 64 + 2 = 66 (not a multiple of 10)

- 81 + 2 = 83 (not a multiple of 10)

- 100 + 2 = 102 (a multiple of 10)

So, x = 100.

Now, we can solve the equation √(x + 2 - 10) = a perfect square by substituting x = 100:

√(100 + 2 - 10) = √92 = 2√23

So, the perfect square is 2√23.

Now, we can use the third equation to find y:

x = y + 5

100 = y + 5

y = 95

A - Act on the chosen solution:

The age of the teenager's sister is 95.

L - Look back and evaluate:

We can check if the solution is correct by verifying that:

- x + 2 = 102 (a perfect square)

- √(x + 2 - 10) = 2√23 (a perfect square)

- x = y + 5 = 100 (the teenager's age)

- y = 95 (the sister's age)

Therefore, the solution is correct.

answered
User Neeraj Sharma
by
7.9k points

No related questions found

Welcome to Qamnty — a place to ask, share, and grow together. Join our community and get real answers from real people.