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UPSC CDS 1 2020 Math Paper

Question : 9 of 100

Marks: +1, -0

\frac{{\rm{a}}}{{\rm{b}}} = \frac{1}{3},\frac{{\rm{b}}}{{\rm{c}}} = 2,\frac{{\rm{c}}}{{\rm{d}}} = \frac{1}{2},\frac{{\rm{d}}}{{\rm{e}}} = 3

\frac{{\rm{e}}}{{\rm{f}}} = \frac{1}{4}

\frac{{{\rm{abc}}}}{{{\rm{def}}}}?

Solution:

Concept:

Using chain of ratios to express all variables in terms of a single variable, then substituting into the required expression.

Explanation:

Let

a = k

(choose any convenient variable).

From

\frac{a}{b} = \frac{1}{3}

b = 3a = 3k

From

\frac{b}{c} = 2

c = \frac{b}{2} = \frac{3k}{2}

From

\frac{c}{d} = \frac{1}{2}

d = 2c = 2 \cdot \frac{3k}{2} = 3k

From

\frac{d}{e} = 3

e = \frac{d}{3} = \frac{3k}{3} = k

From

\frac{e}{f} = \frac{1}{4}

f = 4e = 4k

Now compute

\frac{abc}{def}

Numerator:

a \cdot b \cdot c = k \cdot 3k \cdot \frac{3k}{2} = \frac{9k^3}{2}

Denominator:

d \cdot e \cdot f = 3k \cdot k \cdot 4k = 12k^3

Thus,

\frac{abc}{def} = \frac{\frac{9k^3}{2}}{12k^3} = \frac{9}{2} \cdot \frac{1}{12} = \frac{9}{24} = \frac{3}{8}

Answer:

The value is

\frac{3}{8}

, which corresponds to option C.

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