Ratios & Proportion
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Q.

Two alloys $A$ and $B$ are composed of two basic elements. The ratios of the compositions of the two basic elements in the two alloys are $5:3$ and $1:2$, respectively. A new alloy $X$ is formed by mixing the two alloys $A$ and $B$ in the ratio $4:3$. What is the ratio of the composition of the two basic elements in alloy $X$?

 A.

$1:1$

 B.

$2:3$

 C.

$5:2$

 D.

$4:3$

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Solution:
Option(A) is correct

The new alloy $X$ is formed from the two alloys $A$ and $B$ in the ratio $4:3$. 

Hence, 7 parts of the alloy contains 4 parts of alloy $A$ and 3 parts of alloy $B$.

Let $7x$ ounces of alloy $X$ contain $4x$ ounces of alloy $A$ and $3x$ ounces of alloy $B$.

Now, alloy $A$ is formed of the two basic elements mentioned in the ratio $5:3$.

Hence, $4x$ ounces of alloy $A$ contains \(\left(\dfrac{5}{5+3}\right)4x=\dfrac{5x}{2}\) ounces of first basic element
and \(\dfrac{3}{5+3}4x=\dfrac{3x}{2}\) ounces of the second basic element.

Also, alloy $B$ is formed of the two basic elements mentioned in the ratio $1:2$.

Hence, let the $3x$ ounces of alloy A contain \(\left(\dfrac{1}{1+2}\right)3x=x\) ounces of the second basic element. ounces of the first basic element and \(\left(\dfrac{2}{1+2}\right)3x=2x\) ounces of the second basic element.

Then  the  total  compositions  of  the  two  basic  elements  in  the  $7x$  ounces  of  alloy  $X$  would  contain

 \(\dfrac{5x}{2}\) ounces (from A) + $x$ ounces (from B) = \(\dfrac{7x}{2}\) ounces of first basic element,

and basic elements in alloy X is  \(\dfrac{3x}{2}\)(from A) + \(2x\)(from B) =  \(\dfrac{7x}{2}\) ounces of the second basic element.

Hence, the composition of the two basic elements in alloy $X$ is \(\dfrac{7x}{2}:\dfrac{7x}{2}=1:1\)


(2) Comment(s)


Vineet
 ()

how to convert these ratios into a single ratio:

3:2, 4:3, 5:7, 5:3, 4:7,


Ritwik
 ()

Say,

$A:B=3:2$, $B:C=4:3$, $C:D=5:7$, $D:E=5:3$, $E:F=4:7$

Now, take the first two ratio,

$B$ is common in both the ratios, so we need to $B$ a common number.

$\Rightarrow A:B=\dfrac{A}{B}=\dfrac{3}{2} \times \dfrac{4}{4}$

$\Rightarrow B:C=\dfrac{B}{C}=\dfrac{4}{3} \times \dfrac{2}{2}$

$\Rightarrow A:B=12:8$, $B:C=8:6$

$\Rightarrow A:B:C=12:8:6=6:4:3$

Again we take the next ratio,

$C:D=5:7$

Here $C$ is common. So make it a common number.

$A:B:C=6:4:3$ ------- multiply by 5

$C:D=5:7$ ------ multiply by 3

$\Rightarrow=A:B:C=30:20:15$

and $C:D=15:21$

So,

$A:B:C:D=30:20:15:21$

You can keep going this way till the last ratio.

Hope this helps.