Question

4) Ahmad is twice as old as Bobby. John is 7 years younger than Ahmad. If the sum of their age is 38, how old are the three boys?

Answer 1

Ответ:Let A = Ahmad's age B = Bobby's age J = John's age

1) A = 2*B or B = A/2

2) J = A - 7

3) A + B + J = 38

You have 3 equations and 3 unknowns. Start by substituting equation 1) and equation 2) into equation 3) and solve for A:

A + A/2 + (A-7) = 38 (combine like terms)

2.5*A = 38 + 7 = 45

A = 45/2.5 = 18 yrs old (Ahmad's age)

B = A/2 = 18/2 = 9 yrs old (Bobby's age)

J = A - 7 = 18 - 7 = 11 yrs old (John's age)

Пошаговое объяснение:

Answer 2

Let's assign variables to represent the ages of the three boys.

Let's say Ahmad's age is A, Bobby's age is B, and John's age is J.

From the given information, we can form the following equations:

1. Ahmad is twice as old as Bobby: A = 2B

2. John is 7 years younger than Ahmad: J = A - 7

3. The sum of their ages is 38: A + B + J = 38

Now, let's solve these equations to find the ages of the three boys.

Substituting the value of A from equation 1 into equations 2 and 3:

J = (2B) - 7 ---(4)

A + B + (2B - 7) = 38 3B - 7 = 38 3B = 38 + 7 3B = 45 B = 45/3 B = 15

Now, substitute the value of B into equation 1 to find the value of A:

A = 2B A = 2(15) A = 30

Substitute the values of A and B into equation 4 to find the value of J:

J = (2B) - 7 J = (2(15)) - 7 J = 30 - 7 J = 23

Therefore, Ahmad is 30 years old, Bobby is 15 years old, and John is 23 years old.