In a family of 4 children under 10 years old, the sum of the square of the ages of the 3 youngest is equal to the square of the age of the oldest. How old are these children

2, 3, 6 and 7.

Step-by-step explanation:

We have four natural numbers A, B, C and D.

such that all of these numbers are smaller than 10, and:

A < B < C < D.

A, B and C are the ages of the smaller ones, and D is the age of the larger kid.

Now we have the relation

A^2 + B^2 + C^2 = D^2

We can write this as:

D = √(A^2 + B^2 + C^2)

And remember that D must be smaller than 10, then now we can play with the numbers A, B and C in order to find D.

Suppose for example:

A = 1, B = 2, C  = 3

D = √( 1^2 + 2^2 + 3^2) = √(1 + 4 + 9) = √14

This is not a whole number.

Then we have that A^2 + B^2 + C^2 must be a perfect square.

The perfect squares we can aim for are:

4*4 = 16

5*5 = 25

6*6 = 36

7*7 = 49

8*8 = 64

9*9 = 81

now, looking at that comes to my mind to try to reach the 49.

because we can take the 36, add 4 to that (4 = 2*2) and then add 9 (9 = 3*3)

A^2 + B^2 + C^2 = 49

where we can use:

A = 2. B = 3 and C = 6

2^2 + 3^2 + 6^2 = 4 + 9 + 36 = 49

Then we have:

D = √(2^2 + 3^2 + 6^2 ) = √49 = 7

Then the ages of the children (for smallest to largest) are:

2, 3, 6 and 7.

24

step-by-step explanation:

the graph shows 30% of students walk. 80 students were questioned, so the number that said they walk is

30% × 80 = 0.30 × 80 = 24