a mixture of iron, sand, glass contains the three in the ratio 4: 7: 9

1.	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» 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, $\sf{Percentage\:of\:iron = \dfrac{100}{500}\times100 = 20\%}$ $\sf{Percentage\:of\:sand = \dfrac{175}{500}\times 100 = 35\%}$ $\sf{Percentage\:of\:glass = \dfrac{225}{500}\times100 = 45\%}$