Question

19. Satwinder bought two fans for 3605. He sold one at a profit of 15% and the other at a loss
of 9%. If Satwinder obtained the same amount for each fan, find the cost price of each fan.Please do it using linear question.​

Answer 1

Step-by-step explanation:

:

Cost price of first fan = 1592.5 rupees,

Second fan =2012.5 rupees.

Step-by-step explanation:

Let x be the cost of first fan = x rupees,

∵ Total cost price = 3605 rupees

So, the cost of second fan = (3605 - x) rupees,

Now, profit in first fan = 15%,

Thus, the selling price of first fan = x + 15% of x = x + 0.15x = 1.15x

Also, the loss in second fan = 9%

Thus, the selling price of second fan = 3605 - x - 9% of (3605 - x)

= 3605 - x - 0.09(3605 - x)

= 3605 - x - 324.45 + 0.09x

= 3280.55 -0.91x

According to the question,

1.15x = 3280.55 -0.91x

1.15x + 0.91x = 3280.55

2.06x = 3280.55

\implies x = \frac{3280.55}{2.06}=1592.5

Thus, the cost price of first fan = 1592.5 rupees,

Second fan = 3605 - 1592.5 = 2012.5 rupees.

Answer 2

AnswEr :

$\bf{\green{\underline{\underline{\bf{Given\::}}}}}$

Satwinder bought two fans for Rs.3605. He sold one at a profit of 15% and the other at a loss of 9%. If Satwinder obtained the same amount for each fan.

$\bf{\green{\underline{\underline{\bf{To\:find\::}}}}}$

The cost price of each fan.

$\bf{\green{\underline{\underline{\bf{Explanation\::}}}}}$

Let the 1st fan of cost price (C.P.) be Rs.R

Let the 2nd fan of cost price (C.P.) be Rs.M

So,

$\leadsto\tt{R+M=3605}\\\\\leadsto\tt{\pink{M=3605-R.......................(1)}}$

$\dag\bf{\underline{\underline{\bf{Selling\:price\:(S.P.)\:of\:1st\:fan\::}}}}}}$

$\mapsto\sf{S.P.=\dfrac{100+profit\:(\%)}{100} \times C.P.}\\\\\\\mapsto\sf{S.P.=\dfrac{100+15}{100} \times R}\\\\\\\mapsto\sf{S.P.=\cancel{\dfrac{115}{100}} R}\\\\\\\mapsto\sf{\red{S.P.=\dfrac{23R}{20} }}$

$\dag\bf{\underline{\underline{\bf{Selling\:price\:(S.P.)\:of\:2nd\:fan\::}}}}}}$

$\mapsto\sf{S.P.=\dfrac{100-loss\:(\%)}{100} \times C.P.}\\\\\\\mapsto\sf{S.P.=\dfrac{100-9}{100} \times M}\\\\\\\mapsto\sf{S.P.=\dfrac{91}{100}M}\\\\\\\mapsto\sf{\red{S.P.=\dfrac{91M}{100} }}$

$\bf{\orange{\underline{\underline{\tt{A.T.Q\::}}}}}$

$\mapsto\tt{\dfrac{23R}{\cancel{20}} =\dfrac{91M}{\cancel{100}} }\\\\\\\mapsto\tt{\dfrac{23R}{1} =\dfrac{91M}{5} }\\\\\\\mapsto\tt{5(23R)=91M}\\\\\\\mapsto\tt{115R=91(3605-R)}\\\\\\\mapsto\tt{115R=328055-91R}\\\\\\\mapsto\tt{115R+91R=328055}\\\\\\\mapsto\tt{206R=328055}\\\\\\\mapsto\tt{R=\cancel{\dfrac{328055}{206} }}\\\\\\\mapsto\tt{\red{R=Rs.1592.5}}$

Putting the value of R in equation (1),we get;

$\mapsto\sf{M=Rs.(3605-1592.5)}\\\\\\\mapsto\sf{\red{M=Rs.2012.5}}$

Thus;

$\underbrace{\sf{The\:1st\:cost\:price\:of\:fan\:=R=Rs.1592.5}}}}}\\\ \underbrace{\sf{The\:2nd\:cost\:price\:of\:fan\:=M=Rs.2012.5}}}}}$