Question

metal base 20 times heavier than iron and metal b is 13 times heavier than iron in what proportion must these two metals be mixed to obtain an alloy which is 17 times heavier than iron​

Answer 1

QUESTION

A metal base A 20 times heavier than iron and metal B is 13 times heavier than iron in what proportion must these two metals be mixed to obtain an alloy which is 17 times heavier than iron

CALCULATION

Let the weight of Iron = w

Since metal base A 20 times heavier than iron

So weight of A is = 20w

Again metal B is 13 times heavier than iron

So weight of metal B = 13w

So their wight ratio is = 20w : 13w = 20 : 13

Now the weight of the alloy will be 17w

So the part of A in the alloy =

$\displaystyle \: = 17w \times \frac{20}{20 + 13} = \frac{340w}{33}$

Now the part of B in the alloy

$\displaystyle \: = 17w \times \frac{13}{20 + 13} = \frac{221w}{33}$

So the required ratio is

$\displaystyle \: = \frac{340w}{33} : \frac{221w}{33}$

$= 340 : \: 221$

$= 20 : 13$