I am not sure on what should be the quick way of approach to solve these kind of questions. As we grow, we can think of other possible ways to solve the same problem. Many genius can solve the algebra questions orally. I am not sure how can they solve. But, If I need to solve the equations then certainly I can't solve it orally. Am I correct on it? Does somebody has a more quicker way to solve the question below - using arithmetic or other concept to solve this question orally?
Problem - Given that sum of Abi and Iris age is 42 years. 11 years ago, Abi was three times as old as Iris. how old Abi will be in 2 years?
This is my approach -- Please tell me how you would have solved such questions.
Also, how much time will you take to solve this type of question.
make two equations - considering the present age always -
a + i = 42
a-11 = 3 ( i - 11)
Now, solve for them -
a = 26 and i = 16.
The answer is a +2 i.e. 28 years.
Solving problems such as these (and far more complicated than this, too) without algebra is encouraged in the Singapore system. Before I started teaching my son, I would've immediately reached for algebra, but now simple arithmetic and model drawing are my weapons of choice.
$11$ years ago, the sum of their ages would've been $42-2 \times 11 = 42 - 22 = 20$.
Since Ari was $3$ times as old as Iris, think of the sum of their ages as $3+1 = 4$ parts. $4$ parts $= 20$ so $1$ part = $5$ years.
Therefore Iris was $5$ and Ari was $3 \times 5 = 15$ then, making Ari's present age $15 + 11 = 26$ and her age in $2$ years hence $26+2 = 28$ years.