Question

A man sold two bicycles for ₹6000 each, gaining 20% on the one and losing 20% on the other. Find his gain or loss per cent on the whole transaction.
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Answer 1

Answer:

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Answer 2

❍  Given Info. :

➽ A man sold two bicycles for ➠ ₹ 6000.

➽ He was gaining on the first bicycle ➠ 20%

➽ He was losing on the second bicycle ➠ 20%

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❍ Exigency To Find :

• His gain or loss per cent on the whole transaction.

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❍ Method Used For Solving it :

➥ Here, S.P. and Gain% of 1st bicycle is given,then we have to find its C.P. . And in the second bicycle S.P. and Loss% is given, then we also have to find its C.P. . After finding C.P.'s of the bicycles we have to find the total C.P. of the two bicycles and total S.P. also. Then we have to compare them (i.e. S.P. and C.P.). And if there is a gain we have to subtract C.P. from S.P. or if there is loss then we have to subtract S.P. from the C.P. . And then, we have to find the Gain% or Loss%.

Let's Do It !

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❍ Formulae Used :

❒ If S.P. and Gain% is given, then :

• ➽$\;\sf C.P. : \mapsto \: \bigg \lbrace \tt \dfrac{100}{100 \: + \:Gain\% } \times \sf S.P. \bigg \rbrace$

❒ If S.P. and Loss% is given, then :

• ➽$\;\sf C.P. : \mapsto \: \bigg \lbrace \tt \dfrac{100}{100 \: + \:Loss\% } \times \sf S.P. \bigg \rbrace$

❒ To find Loss :

• ➽ Loss$\;:\mapsto\;$(C.P. - S.P.)

❒ To find Loss% :

• ➽$\;\sf Loss \% \;:\mapsto\; \tt\bigg( \dfrac{Loss}{C.P.} \: \times \: 100 \bigg)$

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❍ Required Solution :

❒ Finding the C.P. of First Bicycle :

• ➽ S.P.$\;:\mapsto\;$₹6000.
• ➽ Gain%$\;:\mapsto\;$20%

➽$\;\sf C.P. : \mapsto \: \bigg \lbrace \tt \dfrac{100}{(100 \: + \:Gain\% )} \times \sf S.P. \bigg \rbrace$

$\dag\: \: \frak{ \: \underline {Substituting \: the \:known \: values \: : }}$

➽$\;\sf C.P. : \mapsto \: Rs. \bigg \lbrace \tt \dfrac{100}{100 \: + \:20 } \times 6000 \bigg \rbrace$

➽$\;\sf C.P. : \mapsto \: Rs. \bigg \lbrace \tt \dfrac{100}{\cancel{120} } \times \cancel{6000}\;^{50} \bigg \rbrace$

➠$\;\underline{\boxed{\bf{Rs.\;5000}}}\;\;$✰

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❒ Finding the C.P. of Second Bicycle :

• ➽ S.P.$\;:\mapsto\;$₹6000.
• ➽ Loss%$\;:\mapsto\;$20%

➽$\;\sf C.P. : \mapsto \: \bigg \lbrace \tt \dfrac{100}{(100 \: - \:Loss\%) } \times \sf S.P. \bigg \rbrace$

$\dag\: \: \frak{ \: \underline {Substituting \: the \:known \: values \: : }}$

➽$\;\sf C.P. : \mapsto \: Rs. \bigg \lbrace \tt \dfrac{100}{100 \: -\:20 } \times 6000 \bigg \rbrace$

➽$\;\sf C.P. : \mapsto \: Rs. \bigg \lbrace \tt \dfrac{100}{80} \times 6000\bigg \rbrace$

➠$\;\underline{\boxed{\bf{Rs.\;7500}}}\;\;$✰

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❒ Total C.P. of two bicycles :

➽ ₹ (5000 + 7500)

➠$\;\underline{\boxed{\bf{Rs.\;12500}}}\;\;$✰

❒ Total S.P. of two bicycles :

➽ ₹ (6000 × 2)

➠$\;\underline{\boxed{\bf{Rs.\;12000}}}\;\;$✰

❒ Then, comparing S.P. and C.P. :

$\dag\: \: \frak{ \: \underline {Substituting \: the \:known \: values \: : }}$

➽ ₹ 12000 $\boxed{\:}$ ₹12500

➽ ₹ 12000 $\boxed{<}$ ₹12500

❒ Since, S.P. < C.P., so there's a loss.

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.°. Loss$\;:\mapsto\;$(C.P. - S.P.)

$\dag\: \: \frak{ \: \underline {Substituting \: the \:known \: values \: : }}$

➽ Loss$\;:\mapsto\;$(₹12500 - ₹12000)

➠$\;\underline{\boxed{\bf{Rs.\;500}}}\;\;$✰

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${\therefore\;\sf Loss \% \;:\mapsto\; \tt\bigg( \dfrac{Loss}{C.P.} \: \times \: 100 \bigg)}$

$\dag\: \: \frak{ \: \underline {Substituting \: the \:known \: values \: : }}$

➽ $\;\sf Loss \% \;:\mapsto\; \tt\bigg( \dfrac{\cancel{500} \: {}^{4} }{\cancel{12500} \: {}^{\cancel{125}} } \: \times \: \cancel{100} \bigg)$

➠$\;\underline{\boxed{\bf{4\%}}}\;\;$✰

${ \therefore{ \underline{ \boldsymbol{Hence,} \sf\:the\: man \:loses\: \bf \: 4 \% \: \: \sf on\: the\: whole\: transaction .}}}$

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