Question

7. A father's age is 3 times the sum of ages of his two sons. Five years later he will be
twice the sum of ages of his two sons. Find the present age of the father
8. The digits of a two digit number differ by 7. If the digits are interchanged and the
number is added to the original 121 Find the original
sulting​

Answer 1

Step-by-step explanation:

7. Let the age of father =x years

Let the age of father =x yearsThe sum of the age of 2 children =y years

Let the age of father =x yearsThe sum of the age of 2 children =y yearsAccording to the first condition

Let the age of father =x yearsThe sum of the age of 2 children =y yearsAccording to the first condition⇒x=3y.....eq1

Let the age of father =x yearsThe sum of the age of 2 children =y yearsAccording to the first condition⇒x=3y.....eq1After 5 years

Let the age of father =x yearsThe sum of the age of 2 children =y yearsAccording to the first condition⇒x=3y.....eq1After 5 years⇒ Father's age =x+5

Let the age of father =x yearsThe sum of the age of 2 children =y yearsAccording to the first condition⇒x=3y.....eq1After 5 years⇒ Father's age =x+5⇒ The sum of ages of his two children =y+10

Let the age of father =x yearsThe sum of the age of 2 children =y yearsAccording to the first condition⇒x=3y.....eq1After 5 years⇒ Father's age =x+5⇒ The sum of ages of his two children =y+10According to the second conditiona

⇒x+5=2(y+10)⇒x+5=2y+20

⇒x+5=2(y+10)⇒x+5=2y+20⇒x−2y=15....eq2

⇒x+5=2(y+10)⇒x+5=2y+20⇒x−2y=15....eq2Put the value of x from eq1

⇒x+5=2(y+10)⇒x+5=2y+20⇒x−2y=15....eq2Put the value of x from eq1⇒3y−2y=15⇒y=15

⇒x+5=2(y+10)⇒x+5=2y+20⇒x−2y=15....eq2Put the value of x from eq1⇒3y−2y=15⇒y=15Put y=15 in eq1

⇒x+5=2(y+10)⇒x+5=2y+20⇒x−2y=15....eq2Put the value of x from eq1⇒3y−2y=15⇒y=15Put y=15 in eq1⇒x=3×15⇒x=45

⇒x+5=2(y+10)⇒x+5=2y+20⇒x−2y=15....eq2Put the value of x from eq1⇒3y−2y=15⇒y=15Put y=15 in eq1⇒x=3×15⇒x=45Hence, father age =45 years

.