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Question 33 : The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is

- 3 : 10
- 1 : 3
- 2 : 5
- 1 : 4

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Given that the strength of the salt solution is p% if 100 ml of the solution contains p grams of salt

It is also given that three salt solutions A , B , C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%.

So \\frac{A + 2B + 3C}{6}) = 0.2 ⟹ A + 2B + 3C = 1.2 ------(1)

If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%.

So, \\frac{3A + 2B + C}{6}) = 0.3 ⟹ 3A +2B + C = 1.8 -------(2)

It is given that 4^{th} solution D is produced by mixing B and C in the ratio 2 : 7

So D = \\frac{2B + 7C}{9})

We have to find the ratio of the strength of D : A

Subtracting equations 1 and 2, we get

2A – 2C = 0.6 or A – C = 0.3

Since we could not find anything from the above methods, we can eliminate the number part and get the ratio going

A + 2B + 3C = 1.2

3A + 2B + C = 1.8

So let us multiply eqn 1 and 2 with 3 and 2,

3A + 6B + 9C = 6A + 4B + 2C

2B + 7C = 3A

It is given that D = \\frac{2B + 7C}{9})

\\frac{2B + 7C}{9}) = \\frac{3A}{9}) ⟹ \\frac{2B + 7C}{9}) = \\frac{A}{3})

Hence D = \\frac{A}{3})

Therefore the ratio D : A = 1 : 3

The question is **"The strength of a salt solution is p% if 100 ml of the solution contains p grams of salt. If three salt solutions A, B, C are mixed in the proportion 1 : 2 : 3, then the resulting solution has strength 20%. If instead the proportion is 3 : 2 : 1, then the resulting solution has strength 30%. A fourth solution, D, is produced by mixing B and C in the ratio 2 : 7. The ratio of the strength of D to that of A is"**

Choice B is the correct answer

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