SSC GD 2024 PYQ — Question 43

The correct answer is Option 2 i.e. $\frac{20}{3}$.

Let, x = $2.\bar{3}$, and y = $4.\bar{3}$

The repeating decimal $2.\bar{3}$ can be expressed as

⇒ x = 2 + $0.\bar{3}$

Similarly, the repeating decimal $4.\bar{3}$ can be expressed as

⇒ y = 4 + $0.\bar{3}$

Now, we will convert $0.\bar{3}$ into a fraction

Let, z = $0.33333....$(1)

Multiply both sides by 10

⇒ 10z = 3.33333......(2)

Now, subtraction eq (1) from (2)

⇒ 10z - z = 3.33333... - 0.33333...

⇒ 9z = 3

⇒ z = 1/3

Now,

⇒ x = 2 + 1/3 = 7/3 and,

⇒ y = 4 + 1/3 = 13/3

Adding x + y 

⇒ x + y = 7/3 + 13/ = 20/3

Thus,

The value of $2.\bar{3}$ + $4.\bar{3}$ is $\frac{20}{3}$