The answer is 30.
First off, the numbers are consecutive even numbers. So, the difference between every consecutive number is 2 (numbers go from even to odd to even to odd...). Since the sum of their numbers is 96, I divided that by 3 to get 32. This gives me the median of the three numbers. To find the smallest number, I simply subtracted 2 from the 32.
This may help:
96/3=32
32 + 32 + 32 = 96
(32-2)+(32+0)+(32+2)=96
30+32+34=93
Call (F) the age of the father and (J) the age of Julio
The F & J are related in this way: F=4J
Now you have a restriction in the form of inequality: The sum of both ages has to be greater or equal than 55.
Algebraically that is: F + J ≥ 55
You can substitute F with 4J to find the solution for J:
4J + J ≥ 55
5J ≥ 55
Now divide both sides by 5
5J/5 ≥ 55/5
J ≥ 11
That Imposes a lower boundary for the value of J of 11, meaning that the youngest age of Julio can be 11
4^4 + 1/2
= (4×4×4×4) + 1/2
= 256 + 1/2
= 256 + 0.5
= 256.5
The answer for A is 2,026. I set a porportion up which is 77/x = 3.8/100
The answer to B is 26
