Question

two cows were purchased for $5000 if one cow was sold at a profit of 20% and other at a loss of 12% find the purchase price of each cow if they were sold for save price
Answer Fast.. ​

Answer 1

Answer:

A man bought two cows for Rs.4800. He sold one at a loss of 15% and another at a profit of 15% and found that each of the cow was sold for the same price. Find the cost price of each cow?

A) Rs.2040, Rs.2760

B) Rs.2800, Rs.2000

C) Rs.3000, Rs.1800

D) Rs.2600, Rs.2200

Correct Answer:

A) Rs.2040, Rs.2760

Step-by-step explanation:

Let the cost price of one cow be Rs.x

The cost price of another cow = Rs.4800 -x.

C.P. of 1st be Rs.x

Profit 15%

S.P of the1st cow = 115100x

C. P of the second cow = Rs.4800-x

Loss 15%

S P of 2nd cow Rs. (4800−x)100×85

115x100−(4800−x)10085

115x=4800×85−85x

200x=4800×85

x=480×85200

C.P. of one cow = Rs.2400

S.P. of another cow = Rs.4800 - 2040 = Rs.2760

Part of solved Profit and Loss questions and answers : >> Aptitude >> Profit and Loss

Answer 2

Appropriate Question :-

• Two cows were purchased for $5000. If one cow was sold at a profit of 20% and other at a loss of 12%. Find the purchase price of each cow, if they were sold for save price.

Solution :-

• Let Cost Price of first cow is A $x.

and

• So, Cost Price of another cow is $ 5000 - x

Case :- 1

We have,

• Cost Price of Cow = $ x

• Profit % = 20 %

We know,

$\rm :\longmapsto\:Selling\:Price = \dfrac{(100 + Profit\%)\times Cost\: Price}{100}$

$\rm :\longmapsto\:Selling \: Price, \: S_1 = \dfrac{(100 + 20) \times x}{100}$

$\rm :\longmapsto\:Selling \: Price, \: S_1 = \dfrac{(120) \times x}{100}$

$\bf :\longmapsto\:Selling \: Price, \: S_1 = \dfrac{6x}{5} - - - (1)$

Case :- 2

We have,

• Cost Price of Cow = $ 5000 - x

• Loss % = 12 %

We know that,

$\rm :\longmapsto\:Selling\:Price = \dfrac{(100 - Loss\%)\times Cost\: Price}{100}$

$\rm :\longmapsto\:Selling \: Price, \: S_2 = \dfrac{(100 - 12) \times (5000 - x)}{100}$

$\rm :\longmapsto\:Selling \: Price, \: S_2 = \dfrac{88 \times (5000 - x)}{100}$

$\rm :\longmapsto\:Selling \: Price, \: S_2 = \dfrac{22 \times (5000 - x)}{25}$

According to statement,

Selling Price of both cows are same.

$\bf\implies \:S_1 = S_2$

$\rm :\longmapsto\:\dfrac{6x}{5} = \dfrac{22(5000 - x)}{25}$

$\rm :\longmapsto\:3x = \dfrac{11(5000 - x)}{5}$

$\rm :\longmapsto\:15x = 55000 - 11x$

$\rm :\longmapsto\:26x = 55000$

$\rm :\longmapsto\:x = 2115.38$

Hence,

• Cost Price of first cow = $ 2115.38

and

• Cost Price of second Cow = $ 2884.62

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