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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) \[520\]

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Last updated date: 13th Jun 2024
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Answer
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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 $\dfrac{x}{{30}}$.

Complete step-by-step answer:
Let us see what the question provides and what it demands,
 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, $\dfrac{x}{{40}} = 12 \Rightarrow x = 480$

So, the correct answer is “Option A”.

Note: While answering this type of question we must remember that to calculate the fraction of unsalable products, we need to multiply the given ratio of unsalable products per salable product and the number of bruised products.
Students are advised to go step by step to avoid calculation mistakes.