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Question: Ahmed buys a plot of land for Rs. 480000. He sells \[\dfrac{2}{5}\] of it at a loss of 6 %. At what ...

Ahmed buys a plot of land for Rs. 480000. He sells 25\dfrac{2}{5} of it at a loss of 6 %. At what gain percent should he sell the remaining part of the plot to gain 10 % on the whole?

Explanation

Solution

Hint:First of all, find the SP1S{{P}_{1}} such that gain % would be 10 % on the whole. Now, find SP2S{{P}_{2}} at which he sold 25\dfrac{2}{5} of the land at 6 % loss. Now subtract SP2 from SP1S{{P}_{2}}\text{ from }S{{P}_{1}} to get SP3S{{P}_{3}}. Also, subtract CP2 of 25C{{P}_{2}}\text{ of }\dfrac{2}{5} of the land from the total CP1 to get CP3C{{P}_{1}}\text{ to get }C{{P}_{3}}. Now, find the gain % using (SP3CP3CP3)×100\left( \dfrac{S{{P}_{3}}-C{{P}_{3}}}{C{{P}_{3}}} \right)\times 100.

Complete step-by-step answer:
We are given that Ahmed buys a plot of land for Rs. 480000. He sell 25\dfrac{2}{5} of it at a loss of 6 %. We have to find the gain percent at which he should sell the remaining part of the plot to gain 10 % on the whole.
Now, let us consider our question. We are given that Ahmed buys a plot of Rs. 48000. So, we get,
Total CP1=Rs.480000.....(i)\text{Total }C{{P}_{1}}=Rs.480000.....\left( i \right)
We are given that the total gain % should be 10 %. We know that,
Gain % = (S.PC.PC.P)×100\left( \dfrac{S.P-C.P}{C.P} \right)\times 100
By substituting
CP=CP1=Rs.480000CP=C{{P}_{1}}=Rs.480000
Gain % = 10 %
SP=SP1SP=S{{P}_{1}}
We get,
10=(SP1480000)480000×10010=\dfrac{\left( S{{P}_{1}}-480000 \right)}{480000}\times 100
48000=SP148000048000=S{{P}_{1}}-480000
SP1=480000+48000S{{P}_{1}}=480000+48000
So, we get,
SP1=Rs.528000....(ii)S{{P}_{1}}=Rs.528000....\left( ii \right)
He sells 25\dfrac{2}{5} of it at a loss of 6 %. So, we get, CP of 25\dfrac{2}{5} of the plot
=CP2=25×480000=C{{P}_{2}}=\dfrac{2}{5}\times 480000
So, we get,
CP2=2×96000=Rs.192000....(iii)C{{P}_{2}}=2\times 96000=Rs.192000....\left( iii \right)
We know that,
Loss % =(CPSPCP)×100=\left( \dfrac{CP-SP}{CP} \right)\times 100
By substituting
CP=CP2=Rs.192000CP=C{{P}_{2}}=Rs.192000
Loss % = 6 %
SP=SP2SP=S{{P}_{2}}
We get,
6=(192000SP2192000)×1006=\left( \dfrac{192000-S{{P}_{2}}}{192000} \right)\times 100
6×1920=192000SP26\times 1920=192000-S{{P}_{2}}
So, we get,
SP2=19200011520S{{P}_{2}}=192000-11520
SP2=Rs.180480....(iv)S{{P}_{2}}=Rs.180480....\left( iv \right)
So, we have got the total CP that is CP1=Rs.480000 and CP2=Rs.192000C{{P}_{1}}=Rs.480000\text{ and }C{{P}_{2}}=Rs.192000
So, we get CP of the remaining that is
CP3=CP1CP2C{{P}_{3}}=C{{P}_{1}}-C{{P}_{2}}
CP3=Rs480000Rs.192000C{{P}_{3}}=Rs480000-Rs.192000
CP3=Rs288000.....(v)C{{P}_{3}}=Rs288000.....\left( v \right)
We have also got the total SP that is SP1=Rs.528000,SP2=Rs.180480S{{P}_{1}}=Rs.528000,S{{P}_{2}}=Rs.180480. So, we get, SP of the remaining part, that is,
SP3=SP1SP2S{{P}_{3}}=S{{P}_{1}}-S{{P}_{2}}
SP3=Rs.528000Rs.180480S{{P}_{3}}=Rs.528000-Rs.180480
SP3=Rs.347520...(vi)S{{P}_{3}}=Rs.347520...\left( vi \right)
We know that,
Gain % = (S.PC.PC.P)×100\left( \dfrac{S.P-C.P}{C.P} \right)\times 100
So, we get, the gain % of the remaining part,
=(SP3CP3)CP3×100=\dfrac{\left( S{{P}_{3}}-C{{P}_{3}} \right)}{C{{P}_{3}}}\times 100
=(347520288000)288000×100=\dfrac{\left( 347520-288000 \right)}{288000}\times 100
=59520288000×100=\dfrac{59520}{288000}\times 100
= 20.6667 %
So, we get the gain percent of the remaining part as 20.67 %.

Note: In these types of questions, always take care of taking the SP and CP of the same part or the same quantity. For example, if the CP is of the 25th\dfrac{2}{5}th part, that is, SP should also be of 25th\dfrac{2}{5}th part only. Also, students must understand that the physical significance of the profit and loss is that when SP > CP, profit is increased while when CP > SP, there is a loss. Students should also remember that we see loss, loss %, profit %, etc. with respect to CP.