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Question
A dealer bought two tables for Rs 3120. He sold one of them at a loss of 15% and other at a gain of 36%. Then, he found that each table was sold for the same price. Find the cost price of each table.
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Solution
\[\text { Given that the selling price is same for both the tables } . \]
\[\text { Let the C . P of 1 table be Rs . x, then the C . P of the other will be Rs } . \left( 3120 - x \right) . \]
\[\text { Loss on the first table = 15 % } \]
\[\text { Therefore, S . P = C . P }\left( \frac{100 -\text { loss % }}{100} \right)\]
\[ S . P = \frac{85x}{100} = Rs . 0 . 85x\]
\[\text { Gain on the second table = 36 % } \]
\[\text { Therefore, S . P = C . P }\left( \frac{100 + \text { gain % }}{100} \right)\]
\[\text { S . P = Rs . 1 . 36 }\left( 3120 - x \right)\]
\[\text { Since both tables have the same S . P }, \]
\[\text { So,} 0 . 85x = 1 . 36\left( 3120 - x \right)\]
\[0 . 85x = 4243 . 20 - 1 . 36x\]
\[2 . 21x = 4243 . 20\]
\[ x = \frac{4243 . 20}{2 . 21}\]
\[ x = Rs . 1920\]
\[\text { So, the cost price of the first table is Rs } . 1920 . \]
\[\text { Cost price of the second table = Rs }. ( 3120 - 1920) = Rs . 1200\]
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