A barrel contains a mixture of wine and water in the ratio 3 : 1. How much fraction of the mixture must be drawn off and substituted by water so that the ratio of wine and water in the resultant mixture in the barrel becomes 1 : 1 ?
(A) $$\frac{1}{4}$$
(B) $$\frac{1}{3}$$
(C) $$\frac{2}{3}$$
(D) $$\frac{1}{2}$$
Solution:
Let the quantity of liquid drawn out = x $$\eqalign{ & \Rightarrow \frac{{3 - \frac{3}{4}x}}{{1 - \frac{1}{4}x + x}} = \frac{1}{1} \cr & \Rightarrow 12 - 3x = 4 - x + 4x \cr & \Rightarrow 8 = 6x \cr & \Rightarrow x = \frac{4}{3} \cr} $$ Hence, required part of quantity $$\eqalign{ & {\text{ = }}\frac{{\frac{4}{3}}}{4} \cr & {\text{ = }}\frac{1}{3} \cr} $$
12.
Find the ratio in which rice at Rs. 7.20 a kg be mixed with rice at Rs. 5.70 a kg to produce a mixture worth Rs. 6.30 a kg.
(A) 1 : 3
(B) 2 : 3
(C) 3 : 4
(D) 4 : 5
Solution:
By the rule of alligation: ∴ Required ratio = 60 : 90 = 2 : 3
13.
In what ratio must water be mixed with milk costing Rs. 12 per litre to obtain a mixture worth of Rs. 8 per litre?
(A) 1 : 2
(B) 2 : 1
(C) 2 : 3
(D) 3 : 2
Solution:
By the rule of alligation : Ratio of water to milk = 4 : 8 = 1 : 2
14.
A container has 20% milk and 80% water in it. It is mixed with another sample (inequal quantity) having 80% milk and 20% water. What would be the milk content in the final mixture?
From a container, full of pure milk, 20% is replaced by water and this process is repeated three times. At the end of third operation, the quantity of pure milk reduces to:
(A) 40.0%
(B) 50.0%
(C) 51.2%
(D) 58.8%
Solution:
Let pure milk was 100L. So, Water is replaced 20% in per process = 20% of 100 = 20L. Now, we use short-cut formula for it. Quantity of Milk reduced to, $$\eqalign{ & = {\text{X}} \times {\left[ {1 - {\frac{{\text{Y}}}{{\text{X}}}} } \right]^{\text{n}}} \cr & = 100 \times {\left[ {1 - {\frac{{20}}{{100}}} } \right]^3} \cr & = \frac{{100 \times 64}}{{125}} \cr & = 51.2\,{\text{%}} \cr} $$ Here, X = Initial quantity of milk. Y = Replaced water in per process. n = No. of process repeated. Note: The formula used in above problem is quite similar to depreciation formula or Compound interest formula. Alternatively, Let pure milk be 100 litres initially. After third operation, milk will be 100 == 20%↓(- 20L) ⇒ 80 == 20%↓(- 16L) ⇒ 64 == 20%↓(-12.8L) ⇒ 51.2 %
16.
A sugar solution of 3 litres contain 60% sugar. One litre of water is added to this solution. Then the percentage of sugar in the new solution is :-
Three containers whose volumes are in the ratio of 2 : 3 : 4 are full of mixture of spirit and water. In the 1st container the ratio of spirits and water is 4 : 1, in 2nd container the ratio is 11 : 4 and on the 3rd container ratio is 7 : 3. All the three mixtures are mixed in a big container. The ratio of spirit and water in the resultant mixture is ?
(A) 4 : 9
(B) 9 : 5
(C) 11 : 4
(D) 5 : 10
Solution:
In 2 litres of first container, Spirit = $$2 \times \frac{4}{5} = \frac{8}{5}$$ water = $$2 \times \frac{1}{5} = \frac{2}{5}$$ In 3 litres of first container, Spirit =$$3 \times \frac{11}{15} = \frac{11}{5}$$ water = $$3 \times \frac{4}{15} = \frac{4}{5}$$ In 4 litres of first container, Spirit =$$4 \times \frac{7}{10} = \frac{14}{5}$$ water = $$4 \times \frac{3}{10} = \frac{6}{5}$$ ∴ Required ratio $$ = \left( {\frac{8}{5} + \frac{{11}}{5} + \frac{{14}}{5}} \right) : \left( {\frac{2}{5} + \frac{4}{5} + \frac{6}{5}} \right) $$ $$ = \frac{{33}}{5} : \frac{{12}}{5} $$ = 33 : 12 = 11 : 4
18.
If X beakers of 100 ml containing 1 : 4 acid-water solution are mixed with Y beakers of 200 ml containing 3 : 17 acid-water solution then the ratio of acid to water in the resulting mixture becomes 19 : 91. Find X : Y.
(A) 5 : 3
(B) 3 : 5
(C) 7 : 13
(D) 13 : 7
Solution:
Answer & Solution Answer: Option A No explanation is given for this question Let's Discuss on Board
19.
How many kg of salt costing Rs. 28 per kg must be mixed with 39.6 kg of salt costing Rs. 16 per kg, so that selling the mixture at Rs. 29.90, there is a gain of 15%.
(A) 198
(B) 135
(C) 132
(D) 131
20.
The ratio, in which tea costing Rs. 192 per kg is to be mixed with tea costing Rs. 150 per kg so that the mixed tea when sold for Rs. 194.40 per kg, gives a profit of 20%.
(A) 2 : 5
(B) 3 : 5
(C) 5 : 3
(D) 5 : 2
Solution:
CP of first tea = Rs. 192 per kg. CP of Second tea = Rs. 150 per kg. Mixture is to be sold in Rs. 194.40 per kg, which has included 20% profit. So, SP of Mixture = Rs. 194.40 per kg. Let the CP of Mixture be Rs. X per kg. Therefore, X + 20% of X = SP $$\frac{{6{\text{X}}}}{5}$$ = 194.40 6X = 194.40 *5 X = Rs. 162 per kg. Let N kg of first tea and M kg of second tea to be added. Now, Using Alligation, We get, $$\frac{{\text{N}}}{{\text{M}}} = \frac{{12}}{{30}}$$ N : M = 2 : 5