A mathematical problem that will sharpen your mind Bashny.Net

To pump logic, solving small problems, is a great training for the brain and the ability to independently find the right solution. Today we have selected tasks related to age determination. How old is my brother if I... But these are details. More about them in the article.



When I was 4 years old, my brother was half my age. I'm 18 now. How old is your brother then?



It seems to be nothing complicated, but it is embarrassing that age is variable and changes all the time. At the same time, it is easy to notice that the indicator “how many times” changes with age. Some people ignore this fact and give the wrong answer “nine”.

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We won’t make such an elementary mistake and start by determining the age of the older brother when he was 4. Twice younger, we divide by 2. 4:2 = 2 years.



Let's see how many years have passed. 18 - 4 = 14 years. If the youngest was 2 years old and 14 years old, now he is 16. Voila, problem solved. My older brother is 18, so my younger brother is 16.

How long will my granddaughter be twice as old? To strengthen the skill, we will try to solve another similar problem. But it's gonna be a little more complicated.



Okay, here we go. Kate is 14 and her grandfather Kola is 56. How long will my granddaughter be twice as old? How old will Grandpa and Granddaughter be?

Hmm. I don't know what to do. But you have to start somewhere. We can find out how old the granddaughter is. 56 - 14 = 42 years. What does that give us? Something gives, but what exactly is not yet known.



Okay, let's go the other way. Let's try to make an equation. Let’s say your granddaughter will be twice as old in X years. Then the following expression will be true.

(56 + x): (14 + x) = 2;
56 + x = 28 + 2x;
x = 28;

That means it will take another 28 years. Checking. Kate in 28 years will be 42 years old, and grandfather Kole - 84. 84:2 = 2. That's what it took to prove.

We hope your answer matches ours. But that doesn’t necessarily mean that the solution algorithm has to be the same. If you can find a simpler solution, share it in the comments. We'll develop logic together.