Question

Raj, Gaurav and Amit are fast friends. They started painting a wall. Raj paints \[\dfrac{2}{{15}}\] of the wall, Gaurav paints $\dfrac{1}{{24}}$ of the wall and Amit paints $\dfrac{3}{{10}}$ of the wall. Read the statements related to the situation and answer the question given.
Statement $1$: Gaurav painted the least portion of the wall.
Statement $2$: $\dfrac{1}{{18}}$ portion of the wall is unpainted.
Which of the following options holds?
A) Both Statement $1$ and Statement $2$ are true
B) Both Statement $1$ and Statement $2$ are false
C) Statement $1$ is true but Statement $2$ is false
D) Statement $1$ is false and Statement $2$ is true

Answer 1

Hint: First statement says about the least portion. So to compare the values, we find the LCM and convert to a common denominator. To find the portion unpainted, we calculate the total portion painted so that their sum constitutes the whole portion.

Useful formula:
If the parts of something are given as fractions, then the whole of it is considered as $1$.

Complete step by step solution:
Given that Raj paints \[\dfrac{2}{{15}}\] of the wall, Gaurav paints $\dfrac{1}{{24}}$ of the wall and Amit paints $\dfrac{3}{{10}}$ of the wall.
Let it be ${P_R},{P_G},{P_A}$ respectively.
To compare these three values easily let us change the fractions to a common denominator.
For, take the LCM of the three denominators $15,24,10$.
$LCM(15,24,10) = 120$
Since $120 = 15 \times 8 = 24 \times 5 = 10 \times 12$,
We have the numerators as $2 \times 8,1 \times 5,3 \times 12$
That is, $\dfrac{2}{{15}} = \dfrac{{2 \times 8}}{{15 \times 8}} = \dfrac{{16}}{{120}},\dfrac{1}{{24}} = \dfrac{{1 \times 5}}{{24 \times 5}} = \dfrac{5}{{120}},\dfrac{3}{{10}} = \dfrac{{3 \times 12}}{{10 \times 12}} = \dfrac{{36}}{{120}}$
So, the numbers become ${P_R} = \dfrac{{16}}{{120}},{P_G} = \dfrac{5}{{120}},{P_A} = \dfrac{{36}}{{120}}$
Comparing these values we get the least portion as ${P_G} = \dfrac{5}{{120}}$
This means the least portion is painted by Gaurav and so Statement $1$ is true.
Now we have to find the portion of the wall unpainted.
Since the total wall painted and those unpainted together constitute the whole wall, we need to find the total portion of the wall painted by all of them.
Total portion painted, ${P_T} = {P_R} + {P_G} + {P_A} = \dfrac{{16}}{{120}} + \dfrac{5}{{120}} + \dfrac{{36}}{{120}} = \dfrac{{57}}{{120}}$
Cancelling the common factor $3$ from the numerator and denominator we have ${P_T} = \dfrac{{19}}{{40}}$.
Therefore portion of the wall unpainted ${P_U} = 1 - {P_T} = 1 - \dfrac{{19}}{{40}} = \dfrac{{21}}{{40}}$.
From this we can say Statement $2$ is false.

So Statement $1$ is true and Statement $2$ is false, gives option B as the correct answer.

Note:We converted the fractions to a common denominator since it is easy for comparison. When parts of something are considered as fractions, its total is considered as $1$. So, we subtracted the painted portion from $1$ in order to get the unpainted portion.