The minimum number of flowers with which Roopa leaves home is : (CAT 1999)

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Solution

The correct option is B 15
Ans: (b)

Minimum value of x is available for y = 16. $x = \frac{15}{16 x y} = \frac{15}{16} \times 16 = 15$

Let Roopa have x flowers with her; then

$\begin{matrix} B a l a n c e b e f o r e o f f e r i n g & F l o w e r s o f f e r e d & B a l a n c e a f t e r o f f e r i n g I^{s t} P l a c e & 2 x & y & 2 x - y I I^{n d} P l a c e & 4 x - 2 y & y & 4 x - 3 y I I I^{r d} P l a c e & 8 x - 6 y & y & 8 x - 7 y I V^{t h} P l a c e & 16 x - 14 y & y & 16 x - 15 y \end{matrix}$

$16 x - 15 y = 0 \Rightarrow 16 x = 15 y \Rightarrow y = \frac{16 x}{15}$

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