Question

A father who is 42 years old has a son and a daughter. The daughter is three times as old as the son. In 10 years the sum of all their ages will be 100 years. How old are the two siblings at present?

Answer 1

Answer: Son’s age:7 Daughter’s age:21

Step-by-step explanation:

1. The daughter is 3 times older than the son. Son will be x. Then the daughter will be 3x

2. In ten years the sum of their ages will be 100. Father+Son+Daughter=(42+10)+(x+10)+(3x+10)=100

3. Solve the equation:

72+4x=100

x=7

4. You have found the age of the son, 7. Now find the daughter. 3*7=21. The daughter is 21.

Success.!

Answer 2

The ages at present are:

• Age of the son: 7 years old
• Age of the daughter: 21 years old

Step-by-step explanation:

Translate the word language to algebraic expressions.

1) A father who is 42 years old has a son and a daughter.

• Age of the father: 42

2) The daughter is three times as old as the son.

• Age of the son: x (this is the variable chosen, x = present age of the son)

• Age of the dagther: 3x (three times as old as the son, x)

3) In 10 years the sum of all their ages will be 100 years

• (42 + 10) + (x + 10) + (3x + 10) = 100

↑ ↑ ↑ ↑

age of the father age of the son age of the daughter sum

4) How old are the two siblings at present:

Solve the equation

• Delete the parenthesis: 42 + 10 + x + 10 + 3x + 10 = 100

• Combine like terms: 72 + 4x = 100

• Subtraction property of equalities (subtract 72 from each side)

4x = 100 - 72

4x = 28

• Division property of equalities (dive both sides by 4)

x = 28 / 4

x = 7

5) Answers:

• Age of the son: x = 7
• Age of the daughter: 3x = 3(7) = 21

6) Verification:

• In ten years:

age of the son: 7 + 10 = 17

age of the daughter: 21 + 10 = 31

age of the father: 42 + 10 = 52

sum of the ages: 17 + 31 + 52 = 100 ⇒ correct.