Question

shyamlal Babu is four times old as his first son and five times as old as his son. Second son is four years younger than first son. find their ages
please solve the equation step by step
please ​

Answer 1

Answer:

The age for Shyamlal Babu is 80 years.

His First Son is 20 years old and his Second Son is 16 years old.

Step-by-step explanation:

Let us consider the age of Shyamlal Babu as x years, age for his first son as y years and his second son be z years.

Now from question, it says that Shyamlal Babu is 4 times old as his first son and 5 times old as his second son. This can be represented in the equation form as follows

x=4×y---->eqn1

x=5×z---->eqn2

And also it is given that the second son is 4 years younger than the first son. In other words we can say that the second son is 4 years smaller than the first son. The equation for the same is as follows

y=z+4---->eqn3

Now, we can see that the LHS side for both equations 1 & 2 are same. This means the RHS side should also be equal. Thus equating the RHS of both equations we get

4×y=5×z

Dividing 4 on both sides

⇒$\frac{1}{4}$×(4×y)=$\frac{1}{4}$×(5×z)

⇒y=$\frac{5}{4}$×z---->eqn4

Substituting eqn4 in eqn3, we get

$\frac{5}{4}$×z=z+4

Subtracting z on both sides of the equation

⇒ $\frac{5}{4}$×z-z=z+4-z

⇒ ($\frac{5}{4}$-1)×z=4           (taking z common in both terms on LHS side)

⇒ $\frac{5-4}{4}$×z=4            (taking LCM and solving on LHS side)

⇒$\frac{1}{4}$×z=4

Multiplying 4 on both sides of the equation

⇒4×$\frac{1}{4}$×z=4×4

⇒ z=16---->eqn5

Substituting eqn5 in eqn3

y=z+4                    (put z=16)

⇒ y=16+4

⇒ y=20---->eqn6

Substituting eqn6 in either eqn1 or eqn2

x=4×y                     (put y=20)

⇒ x=20×4              (from eqn1)

⇒ x=80

∴ x=80(Shyamlal Babu's age), y=20(first son's age), z=16(second son's age)