Given that Plasticity Index (PI) of local soil = 15 and PI of sand = zero, for a desired PI of 6, the percentage of sand in the mix should be

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Solution

The correct option is B 60
(b)
Let percentage of sand is x. So plasticity of mix is
$x \times 0 + (1 - x) \times 15 = 6$

$x = \frac{9}{15} = 0.6 o r 60 %$

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