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In a 40 litre mixture of alcohol & water, the ratio of alcohol and water is 5 : 3. If 20% of this mixture is taken out and the same amount of water is added then what will be the ratio of alcohol and water in final mixture?

Correct Answer: 1:1

Quantity of alohol in the mixture = 40 x 5/8 = 25 lit

Quantity of water = 40 - 25 = 15 lit

According to question,

Required ratio = 20 - 40 x 20100 x 5815 - 40 x 20100 x 38 + 40 x 20100 = 2015 - 3 + 8 = 1 : 1

Consider the 5 × 5 matrix \[{\text{A}} = \left[ {\begin{array}{*{20}{c}} 1&2&3&4&5 \\ 5&1&2&3&4 \\ 4&5&1&2&3 \\ 3&4&5&1&2 \\ 2&3&4&5&1 \end{array}} \right

Let A be an m × n matrix and Ban n × m matrix. It is given that determinant ($$I$$m + AB) = determinant ($$I$$n + BA), where $$I$$k is the k × k identity matrix. Using the above property, the determinant of the matrix given below is
\[\left[ {\begin{array}{*{20}{c}} 2&1&1&1 \\ 1&2&1&1 \\ 1&1&2&1 \\ 1&1&1&2 \end{array}} \right

Eigen values of the matrix \[\left[ {\begin{array}{*{20}{c}} 0&1&0&0 \\ 1&0&0&0 \\ 0&0&0&{ - 2i} \\ 0&0&{2i}&0 \end{array}} \right

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Rate constant for a first order reaction does not depend upon reaction time, extent of reaction and the initial concentration of reactants ; but it is a function of reaction temperature. In a chemical reaction, the time required to reduce the concentration of reactant from 100 gm moles/litre to 50 gm moles/litre is same as that required to reduce it from 2 gm moles/litre to 1 gm mole/litre in the same volume. Then the order of this reaction is

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There are two containers, the first one contains 1 litre pure water and the second one contains 1 litre pure milk.Now 5 cups of water from the first container is taken out is mixed well in the second container. Then, 5 cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of milk in the first container and B denote the proportion of water in the second container then:

In a first order reaction, the time required to reduce the concentration of reactant from 1 mole/litre to 0.5 mole/litre will be __________ that required to reduce it from 10 moles/litre to 5 moles/litre in the same volume.