Question

A mixture contain 88% of sulphur,another mixture contain 70% of sulphur.In order to get 84% of sulphur,in what ratio these two must be mixed.

Answer 1

Answer:

7:2 will be the ratio.

Explanation:

If we take that teh ratio of the two must be in the ratio say that  k : 1.

Therefore, we can write from the question that the equation will be

(k× 88) + (1 × 70) = 84 (k+1) .

88k + 70 = 84k + 84 .

4k = 14 .

2k = 7  .

k = 7/2.

So,on solving we will get the values of the k to be :-

k : 1 = 7/2 : 1 or 7 : 2.

Hence the amount of mixture must be in the ratio 7:2 to get the percentage f sulphur to be 84 percent.

Answer 2

answer :- 7 : 2

explanation : Let two mixtures are mixed in r : 1 ratio.

a/c to question,

first mixture contains 88% of sulphur

and 2nd mixture contains 70% of sulphur.

so, % amount of sulphur of first mixture in final mixture = 88r/(r + 1)

% amount of sulphur of 2nd mixture in final mixture = 70 × 1/(r + 1)

now, % amount of sulphur of first mixture + % amount of sulphur of 2nd mixture = % amount of sulphur of final mixture

or, 88r/(r + 1) + 70/(r + 1) = 84

or, 88r + 70 = 84(r + 1)

or, 88r + 70 = 84r + 84

or, 4r = 14

or, r = 7/2

hence, ratio is r : 1 = 7 : 2.

shortcut :

it can be easily solved with help of allegations.

here, first Mixture = 88, 2nd mixture = 70

and final = 84 ,

so, ratio is (84 - 70) : (88 - 84) = 14 : 4 or 7 : 2