Quantitative Aptitude – Arithmetic – Mixtures – The strength of a salt solution is p%

Quantitative Aptitude – Arithmetic - Mixtures – The strength of a salt solution is p%

Slot -2 – Quantitative Aptitude – Arithmetic – Mixtures – The strength of a salt solution is p%

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?

a) 1 : 3

b) 2 : 5

c) 1 : 4

d) 3 : 10

Answer: a) 1 :3

Solution:

let the concentration of A, B & C are a% , b% & c% respectively .

Then (a+2b+3c )/600=20/100=1/5

a+2b+3c=120——1)

And (3a+2b+c )/600=30/100=3/10

3a+2b+c=180——-2)

By eq 1)/ eq 2) , (a+2b+3c )/(3a+2b+c )=2/3
3a+6b+9c=6a+4b+2c

(2b+7c)/a=3

So correct answer : a) 1 :3

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Quantitative Aptitude – Arithmetic – Mixtures – A 20% ethanol solution is mixed

Quantitative Aptitude – Arithmetic - Mixtures – A 20% ethanol solution is mixed

Slot -2 – Quantitative Aptitude – Arithmetic – Mixtures – A 20% ethanol solution is mixed

A 20% ethanol solution is mixed with another ethanol solution, say, S of unknown concentration in the proportion 1:3 by volume. This mixture is then mixed with an equal volume of 20% ethanol solution. If the resultant mixture is a 31.25% ethanol solution, then the unknown concentration of S is?

a) 55%

b) 52%

c) 48%

d) 50%

Answer: d) 50 %

Solution:
Let the volume of 20% ethanol solution in first mixtures = x so
Volume of S = 3x

Total volume of first solution = 4x

Total volume of final solution =2×4x=8x

Volume of ethanol solution in final mixture = x + 4x = 5x ,

So volume of S in final solution = 3x

Thus (31.25-20)/(s-31.25) = 3x/5x = 3/5
56.25=3s-93.75
3s=150
s=50%

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Quantitative Aptitude – Arithmetic – Mixtures – A jar contains a mixture of 175 ml

Quantitative Aptitude – Arithmetic - Mixtures – A jar contains a mixture of 175 ml

Slot -2 – Quantitative Aptitude – Arithmetic – Mixtures – A jar contains a mixture of 175 ml

A jar contains a mixture of 175 ml water and 700 ml alcohol. Gopal takes out 10% of the mixture and substitutes it by water of the same amount. The process is repeated once again. The percentage of water in the mixture is now?

a) 20.5

b) 30.3

c) 25.4

d) 35.2

Answer: d) 35.2

Solution:
initial volume of Alcohol = 700, total volume = 700+ 175 = 875 , replaced quantity = 87.5 & n =2

Using, final volume =initial volume ( 1-(replaced quantity)/(total volume))^n = 700(1-87.5/875)^2 = 700×81/100 = 567
So water in final mixture = 875 – 567 = 308
Required percentage = 308/875×100 = 35.2

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Quantitative Aptitude – Arithmetic – Mixtures – There are two drums, each containing a mixture

Quantitative Aptitude – Arithmetic - Mixtures – There are two drums, each containing a mixture

Slot -2 – Quantitative Aptitude – Arithmetic – Mixtures – There are two drums, each containing a mixture

There are two drums, each containing a mixture of paints A and B. In drum 1, A and B are in the ratio 18 : 7. The mixtures from drums 1 and 2 are mixed in the ratio 3 : 4 and in this final mixture, A and B are in the ratio 13 : 7. In drum 2, then A and B were in the ratio?

a) 251 : 163

b) 239 : 161

c) 229 : 141

d) 220 : 149

Answer: b) 239 : 161

Solution:
Let in drum 2, A and B are in the ratio m : n.

Given, In drum 1, A and B were in the ratio 18 : 7 = 18x : 7x .

Final mixtures ratio of contribution of drum 1 and drum 2 = 3: 4 = 75x : 100x

So A in final mixture = 18x × 3+100x × m/(m+n) = ( 154m+54n)/(m+n) x
B in final mixture = 7x × 3+100x × n/(m+n) = ( 21m+121n)/(m+n) x

Given A : B in final mixture = 13:7
Or (154m+54n )/(21m+121n )=13/7

1078m+378n=273m+1573n
805m=1195n

m:n=1195∶805=239∶161

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