Math, asked by shivanikiran, 2 days ago

A man bought 2 bicycles for ₹1250. He sold 1 of them. So as to gain 6% and the other at the loss of 4%. On the whole, he neither gained nor lost. How much did each cycle cost ?

Answers

Answered by mathdude500
5

\large\underline{\sf{Solution-}}

Let assume that

  • Cost Price of first bicycle is ₹ x

So,

  • Cost Price of second bicycle is ₹ (1250 - x)

Case :- 1

  • Cost Price = ₹ x

  • Gain % = 6 %

So, Selling Price of bicycle is evaluated by using formula

\boxed{ \rm{ \:Selling \: Price \:  =  \:  \frac{(100 + Gain\%) \times Cost \: Price}{100} \: }} \\

So, on substituting the values, we get

\rm \: Selling \: Price_1 = \dfrac{(100 + 6) \times x}{100}  \\

\rm\implies \:\boxed{ \rm{ \:\rm \: Selling \: Price_1 = \dfrac{106x}{100} \:  }} -  - (1) \\

Case :- 2

  • Cost Price = ₹ (1250 - x)

  • Loss % = 4 %

So, Selling Price of bicycle is evaluated by using formula

\boxed{ \rm{ \:Selling \: Price \:  =  \:  \frac{(100 - Loss\%) \times Cost \: Price}{100} \: }} \\

So, on substituting the values, we get

\rm \: Selling \: Price_2 = \dfrac{(100 -  4) \times (1250 - x)}{100}  \\

\rm\implies \:\boxed{ \rm{ \:\rm \: Selling \: Price_2 = \dfrac{96(1250 - x)}{100} \:  }} -  - (2) \\

Now, According to statement, in the whole transaction, he neither gained nor loss.

\rm\implies \:Selling \: Price_1 + Selling \: Price_2 = 1250 \\

\rm \: \dfrac{106x}{100}  + \dfrac{96(1250 - x)}{100}  = 1250

\rm \: 106x + 96(1250 - x) = 1250 \times 100 \\

\rm \: 106x + 96 \times 1250 - 96x = 1250 \times 100 \\

\rm \: 106x  - 96x = 1250 \times 100 \times 1250 \times 96 \\

\rm \: 10x = 1250 \times (100  -  96) \\

\rm \: x = 125 \times 4 \\

\rm\implies \:x = 500 \\

Hence,

Cost Price of first bicycle = ₹ 500

Cost Price of second bicycle = 1250 - 500 = ₹ 750

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{Gain = \sf S.P. \: – \: C.P.} \\ \\ \bigstar \:\bf{Loss = \sf C.P. \: – \: S.P.} \\ \\ \bigstar \: \bf{Gain \: \% = \sf \Bigg( \dfrac{Gain}{C.P.} \times 100 \Bigg)\%} \\ \\ \bigstar \: \bf{Loss \: \% = \sf \Bigg( \dfrac{Loss}{C.P.} \times 100 \Bigg )\%} \\ \\ \\ \bigstar \: \bf{S.P. = \sf\dfrac{(100+Gain\%) or(100-Loss\%)}{100} \times C.P.} \\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}

Answered by krohit68654321
0

Step-by-step explanation:

\large\underline{\sf{Solution-}}

Solution−

Let assume that

Cost Price of first bicycle is ₹ x

So,

Cost Price of second bicycle is ₹ (1250 - x)

Case :- 1

Cost Price = ₹ x

Gain % = 6 %

So, Selling Price of bicycle is evaluated by using formula

\begin{gathered}\boxed{ \rm{ \:Selling \: Price \: = \: \frac{(100 + Gain\%) \times Cost \: Price}{100} \: }} \\ \end{gathered}

SellingPrice=

100

(100+Gain%)×CostPrice

So, on substituting the values, we get

\begin{gathered}\rm \: Selling \: Price_1 = \dfrac{(100 + 6) \times x}{100} \\ \end{gathered}

SellingPrice

1

=

100

(100+6)×x

\begin{gathered}\rm\implies \:\boxed{ \rm{ \:\rm \: Selling \: Price_1 = \dfrac{106x}{100} \: }} - - (1) \\ \end{gathered}

SellingPrice

1

=

100

106x

−−(1)

Case :- 2

Cost Price = ₹ (1250 - x)

Loss % = 4 %

So, Selling Price of bicycle is evaluated by using formula

\begin{gathered}\boxed{ \rm{ \:Selling \: Price \: = \: \frac{(100 - Loss\%) \times Cost \: Price}{100} \: }} \\ \end{gathered}

SellingPrice=

100

(100−Loss%)×CostPrice

So, on substituting the values, we get

\begin{gathered}\rm \: Selling \: Price_2 = \dfrac{(100 - 4) \times (1250 - x)}{100} \\ \end{gathered}

SellingPrice

2

=

100

(100−4)×(1250−x)

\begin{gathered}\rm\implies \:\boxed{ \rm{ \:\rm \: Selling \: Price_2 = \dfrac{96(1250 - x)}{100} \: }} - - (2) \\ \end{gathered}

SellingPrice

2

=

100

96(1250−x)

−−(2)

Now, According to statement, in the whole transaction, he neither gained nor loss.

\begin{gathered}\rm\implies \:Selling \: Price_1 + Selling \: Price_2 = 1250 \\ \end{gathered}

⟹SellingPrice

1

+SellingPrice

2

=1250

\rm \: \dfrac{106x}{100} + \dfrac{96(1250 - x)}{100} = 1250

100

106x

+

100

96(1250−x)

=1250

\begin{gathered}\rm \: 106x + 96(1250 - x) = 1250 \times 100 \\ \end{gathered}

106x+96(1250−x)=1250×100

\begin{gathered}\rm \: 106x + 96 \times 1250 - 96x = 1250 \times 100 \\ \end{gathered}

106x+96×1250−96x=1250×100

\begin{gathered}\rm \: 106x - 96x = 1250 \times 100 \times 1250 \times 96 \\ \end{gathered}

106x−96x=1250×100×1250×96

\begin{gathered}\rm \: 10x = 1250 \times (100 - 96) \\ \end{gathered}

10x=1250×(100−96)

\begin{gathered}\rm \: x = 125 \times 4 \\ \end{gathered}

x=125×4

\begin{gathered}\rm\implies \:x = 500 \\ \end{gathered}

⟹x=500

Hence,

Cost Price of first bicycle = ₹ 500

Cost Price of second bicycle = 1250 - 500 = ₹ 750

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{Gain = \sf S.P. \: – \: C.P.} \\ \\ \bigstar \:\bf{Loss = \sf C.P. \: – \: S.P.} \\ \\ \bigstar \: \bf{Gain \: \% = \sf \Bigg( \dfrac{Gain}{C.P.} \times 100 \Bigg)\%} \\ \\ \bigstar \: \bf{Loss \: \% = \sf \Bigg( \dfrac{Loss}{C.P.} \times 100 \Bigg )\%} \\ \\ \\ \bigstar \: \bf{S.P. = \sf\dfrac{(100+Gain\%) or(100-Loss\%)}{100} \times C.P.} \\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}

MoreFormulae

MoreFormulae

★Gain=S.P.–C.P.

★Loss=C.P.–S.P.

★Gain%=(

C.P.

Gain

×100)%

★Loss%=(

C.P.

Loss

×100)%

★S.P.=

100

(100+Gain%)or(100−Loss%)

×C.P.

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