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In an alloy of zinc, tin and lead, quantity of zinc is ${\left( {\dfrac{3}{4}} \right)^{th}}$ that of tin and quantity of tin is ${\left( {\dfrac{4}{5}} \right)^{th}}$ that of lead. How much of each metal will be there in 12kg of the alloy?
A) $5,4,3$
B) $3,4,5$
C) $4,5,3$
D) $5,3,4$

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Last updated date: 15th Jun 2024
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Answer
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Hint:: Here we will have to first take an arbitrary unknown value of the quantity of lead present in the alloy firstly. Then according to the question, determine the quantity of tin & zinc from it. Then form the linear equation based on – Sum of the quantity of each metal present in the alloy is $12kg$.
Quantity of lead + quantity of tin+ quantity of zinc in the alloy = total quantity of alloy

Complete step by step solution:
Given: Total quantity of alloy is $12kg$.
Quantity of tin is ${\left( {\dfrac{4}{5}} \right)^{th}}$that of lead.
Quantity of zinc is ${\left( {\dfrac{3}{4}} \right)^{th}}$ that of tin.
To find the quantity of each metal (lead, tin, zinc) present in $12kg$ of alloy.
To find this, first of all, we will have to suppose the quantity of lead is $x$ kg.
Then , Quantity of tin is ${\left( {\dfrac{4}{5}} \right)^{th}}$that of lead $ = \left( {\dfrac{4}{5} \times x} \right)$ $ = \dfrac{4}{5}x$
& Quantity of zinc is ${\left( {\dfrac{3}{4}} \right)^{th}}$ that of tin. $ = \dfrac{3}{4}of\left( {\dfrac{4}{5}x} \right)$ $ = \dfrac{3}{4} \times \dfrac{4}{5}x = \dfrac{3}{5}x$
Now we know that,
Quantity of lead + quantity of tin+ quantity of zinc in the alloy = total quantity of alloy
$ \Rightarrow x + \dfrac{4}{5}x + \dfrac{3}{5}x = 12$
$ \Rightarrow \dfrac{{5x + 4x + 3x}}{5} = 12$
$ \Rightarrow \dfrac{{12x}}{5} = 12$
$ \Rightarrow x = \dfrac{{12 \times 5}}{{12}}$
$\therefore x = 5$
Hence, quantity of lead present in the alloy is \[5kg\].
Then, Quantity of tin is ${\left( {\dfrac{4}{5}} \right)^{th}}$that of lead $ = \left( {\dfrac{4}{5} \times x} \right)$ $ = \dfrac{4}{5}x = \dfrac{4}{5} \times 5 = 4kg$
& Quantity of zinc is ${\left( {\dfrac{3}{4}} \right)^{th}}$ that of tin. $ = \dfrac{3}{4}of\left( {\dfrac{4}{5}x} \right)$ $ = \dfrac{3}{4} \times \dfrac{4}{5}x = \dfrac{3}{5}x$ $ = \dfrac{3}{5} \times 5 = 3kg$

$\therefore$ The quantity of zinc, tin & lead in the alloy is $3kg,4kg\& 5kg$ respectively. Hence, option (B) is correct.

Note:
First of all, & most importantly you should read the question very attentively because there is a high chance of making a silly mistake of not reading the question properly & that will result into having improper relations between the number of different metals in the alloy. Do the calculations very carefully to ensure you don’t get an incorrect answer even after knowing the procedure. The best & the easiest way to cross-check whether you got the correct answer in this kind of problem or not is to add the quantity of each metal in the metal whether it is equal to the total quantity of alloy (i.e. $12kg$ here). The key formula of this problem is:
Quantity of lead + quantity of tin+ quantity of zinc in the alloy = total quantity of alloy