Question

a mixture of iron, sand, glass contains the three in the ratio 4: 7: 9. calculate % of sand iron and glass in the mixture. how much and what quantity of iron, sand and glass is there in 500g of mixture

Answer 1

Given:-

• Ratio of iron, sand, glass in a mixture = 4:7:9
• Weight of mixture = 500g

To find:-

• Percentage of sand, iron and glass in the mixture.
• Quantity of iron, sand and glass in the mixture.

Assumption:-

Let the common multiple of ratio of iron, sand and glass in the mixture be x

Weight of Iron = 4x

Weight of sand = 7x

Weight of glass = 9x

Solution:-

ATQ,

Weight of (Iron + sand + glass) = Weight of mixture

= $\sf{4x + 7x + 9x = 500}$

$\sf{\implies 20x = 500}$

$\sf{\implies x =\dfrac{500}{20}}$

$\sf{\implies x = 25}$

Therefore,

$\sf{Weight \:of\: iron = 4x = 4\times25 = 100\:g}$

$\sf{Weight\:of\:sand = 7x = 7\times25 = 175\:g}$

$\sf{Weight\:of\:glass = 9x = 9\times25 = 225\:g}$

Now,

$\sf{Percentage\:of\:iron = \dfrac{100}{500}\times100 = 20\%}$

$\sf{Percentage\:of\:sand = \dfrac{175}{500}\times 100 = 35\%}$

$\sf{Percentage\:of\:glass = \dfrac{225}{500}\times100 = 45\%}$