Subject

    Quantitative Aptitude

    Topic

    Miscellaneous

    Two vessels A and B contain alcohol and water in the ratio 7 : 5 and 17 : 7 respectively. In what ratio should the mixtures from two vessels A and B be mixed to get a new mixture containing alcohol and water in the ratio 5 : 3?

    ক)
    3 : 2
    খ)
    2 : 1
    গ)
    1 : 2
    ঘ)
    3 : 4

    Explanation

    When we form a new mixture by adding two mixtures in each other, then we will solve the question by considering the part of only one thing in the whole mixture – First method: Alcohol in vessel A=77+5=712A = \frac{7}{{7 + 5}} = \frac{7}{{12}} Alcohol in vessel B=1717+7=1724B = \frac{{17}}{{17 + 7}} = \frac{{17}}{{24}} Required alcohol in the new mixture =55+3=58= \frac{5}{{5 + 3}} = \frac{5}{8} Now, required of the two mixtures would be – ⇒first mixtureSecond mixture=58∼172458∼712=112124=21\Rightarrow \frac{{first\ mixture}}{{Second\ mixture}} = \frac{{\frac{5}{8} \sim \frac{{17}}{{24}}}}{{\frac{5}{8} \sim \frac{{7}}{{12}}}} = \frac{{\frac{1}{{12}}}}{{\frac{1}{{24}}}} = \frac{2}{1} Hence, required ratio of the two mixtures would be 2:1 Second method: Water in vessel A=57+5=512A = \frac{5}{{7 + 5}} = \frac{5}{{12}} Water in vessel B=717+7=724B = \frac{7}{{17 + 7}} = \frac{7}{{24}} Required water in new mixture =35+3=38= \frac{3}{{5 + 3}} = \frac{3}{8} Now required ratio of the two mixtures would be ⇒first mixtureSecond mixture=724∼38512∼38=112124=2:1\Rightarrow \frac{{first\ mixture}}{{Second\ mixture}} = \frac{{\frac{7}{{24}} \sim \frac{3}{8}}}{{\frac{5}{{12}} \sim \frac{3}{8}}} = \frac{{\frac{1}{{12}}}}{{\frac{1}{{24}}}} = 2:1 Hence required in the new mixture would be 2:1

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