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# A crate of mangoes contains one bruised mango for every $30$ mango in the crate. If $3$ out of every $4$ bruised mango are considered unsalable, and there are $12$unsalable mangoes in the crate, how may mangoes be there in the crate?(A) $480$ (B) $500$ (C) $420$ (D) $520$

Last updated date: 19th Sep 2024
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Hint: In such types of questions, the solution is obtained by simple unitary operations.
Thus the students are advised to carefully note down the given data and establish a relationship between them.
For the ease of understanding it is always better to assume the unknown quantity with a variable.
To solve this question, we must understand that if the total number of mangoes in the crate is considered to be $x$, then the number of bruised mangoes is equal to $\dfrac{x}{{30}}$.

The question provides, A crate of mangoes contains one bruised mango for every $30$ mango in the crate.
The question demands-If $3$ out of every $4$ bruised mango are considered unsalable, and there are $12$unsalable mangoes in the crate, how may mangoes be there in the crate?
Let the total number of mangoes be ‘$x$’
As $1$ out of every $30$ mango n the crate is bruised then, number of bruised mangoes $= \dfrac{x}{{30}}$
And, $3$ out of every $4$ bruised mango are considerably unsalable and there are $12$ unsalable mangoes in the crate. Total number of unsalable mangoes in the crate $= \left( {\dfrac{3}{4} \times \dfrac{x}{{30}}} \right) = \dfrac{x}{{40}}$
$\Rightarrow$ As per question, $\dfrac{x}{{40}} = 12 \Rightarrow x = 480$