InterviewSolution
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1). 1 ∶ 602). 1 ∶ 503). 1 ∶ 554). 1 ∶ 45 |
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Answer» ? Cost of 150 ml tasty solution = Rs. 27 ∴ Cost of 1 ml tasty solution = Rs. $(\frac{{27}}{{150}})$ ∴ Cost of 1 litre of tasty solution = 1000 × (9/50) = Rs. 180 Cost of 1 kg of salt = Cost price of cheaper = c = Rs. 30 Cost of 1 litre of soup = Cost price of dearer = d = Rs. 183 Cost of 1 litre of tasty solution = MEAN price = m = Rs. 180 By formula, $(\BEGIN{array}{l} \frac{{{\rm{Quantity\;of\;salt}}}}{{{\rm{Quantity\;of\;soup}}}} = \frac{{{\rm{Quantity\;of\;Cheaper}}}}{{{\rm{Quantity\;of\;Dearer}}}} = \frac{{{\rm{d}} - {\rm{m}}}}{{{\rm{m}} - {\rm{c}}}}\\ \Rightarrow \frac{{{\rm{Quantity\;of\;Salt}}}}{{{\rm{Quantity\;of\;Soup}}}} = \frac{{183 - 180}}{{180 - 30}} = \frac{3}{{150}} \end{array})$ ∴ The required ratio = Salt ? Soup = 1 ? 50. |
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