d Word Problem 18 (Percentage part, mixture) | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

# Word Problem 18 (Percentage part, mixture)

Problem 18: One liquid contains 20% of water, another contains 35% of water. A glass is filled with 5 parts of first liquid and 10 parts of second liquid, the percentage of water in the new mixture is ;

a) 37%                        b) 30%                       c) 45%                              d) 20%

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Solution:

Given data is ;

Water in first liquid = 20%

water in second liquid = 35%

5 parts of 1st liquid +  10 parts of 2nd liquid = water %age =?

We proceed as follows;

1st liquid

Parts of water = 5

percentage of water = 20%

So we have;

$Water = \frac{20}{100}\times 5= 1$

i-e 1 part of water!

Second liquid

Parts = 10

percentage of water = 35%

So we have;

$Water = \frac{35}{100}\times10=3.5$

i-e 3,5 parts of water!

So we finally can say now ;

In the filled glass which was filled by 5 parts of first liquid — brought 1 part of water , and 10 parts of second liquid — brought 3.5 parts of water !

In other words we can say now that;

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In the glass with total 5+10 =15 parts of liquids contains 1+3.5=4.5 parts of water !

so,

$Water PercentageIn The Glass= \frac{Water In Glass}{liquid In Glass}\times 100$

$Water Percentage In Glass= \frac{4.5}{15}\times 100$

$Water Percentage In Glass=30$

I-e there is 30 % of water in the glass !

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