Q. Bao is older than Jeriel. Their ages are consecutive integers. Find Bao's age if the sum of Bao's age and 2 times Jeriel's age is 40.Answer:
Set Up Equations: Let's denote Bao's age as B and Jeriel's age as J. Since their ages are consecutive integers, we can express Bao's age as B=J+1. The problem states that the sum of Bao's age and 2 times Jeriel's age is 40. We can write this as an equation: B+2J=40 Substituting B with J+1, we get: (J+1)+2J=40
Simplify Equation: Now, let's simplify the equation:J+1+2J=40Combine like terms:3J+1=40
Isolate Jeriel's Age: Next, we will isolate J by subtracting 1 from both sides of the equation:3J+1−1=40−13J=39
Solve for Jeriel's Age: Now, we will divide both sides by 3 to solve for J: 33J=339J=13Since J represents Jeriel's age, we have found that Jeriel is 13 years old.
Find Bao's Age: Finally, since Bao is older than Jeriel by 1 year, we can find Bao's age:B=J+1B=13+1B=14Therefore, Bao is 14 years old.