Question

A scenery costs Rs. R1. A shopkeeper gives a discount of x% and reduces its price to R2. He gives a further discount of x% on the reduced price R2 to reduce it further to R3, which reduces it by Rs. 415. A customer bargains with him and takes an x% discount on R3 and buys the scenery for Rs. 3362.80. Find the original price R1 of the scenery.

Answer 1

Answer:

R₁ - $\frac{R_{1}x}{100}$ = R₂ ------------- ( 1 )

R₂ - $\frac{R_{2}x}{100}$ = R₃ ------------- ( 2 )

R₂ - R₃ = 415 --------------- ( 3 )

$\frac{R_{2}x}{100}$ = 415 --------------- ( 4 )

R₃ - $\frac{R_{3}x}{100}$ = 3362.80 ----------- ( 5 )

From equation 2 we get

R₂ ( 100 - x ) = 100 R₃ ------- ( 6 )
And from equation 5 we get

R₃ ( 100 - x ) = 336280 ------- ( 7 )

Now divide equation 6 by equation 7 , we get

$\frac{R_2}{R_3}=\frac{100R_3}{336280}$
From equation 3 we get R₂ = R₃ + 415 , SO
$\frac{R_3+415}{R_3}=\frac{100R_3}{336280}$

100 ( R₃)² = 336280 R₃ + ​139556200
100 ( R₃)² = 336280 R₃ + ​139556200
100 ( R₃)² - 336280 R₃ - ​139556200 = 0
Now fom quardetic equation we get

R₃ = $\frac{336280\pm \sqrt{113084238400-55822480000}}{200}$

R₃ = $\frac{336280\pm \sqrt{57261758400}}{200}$

So,
R₃ = 3736.3131
And
R₃ = -373.5131
SO value of R₃ can't be negative So,
R₃ = 3736.3131
From equation 3 we get

R₂ - 3736.3131 = 415

R₂ = ​4151.3131

And from Equation 4 we get
4151.3131 x = 41500
x = 10%

NOw from equation 1 we get

R₁ - 0.1R₁ = 4151.3131

0.9 R₁ = 4151.3131

R₁ = ​4612.570 Rs ( Ans )