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On the sale of a certain item a shopkeeper offered two successive discounts such that in percentage terms the second discount was only 1n\frac{1}{n}n1
Question

On the sale of a certain item a shopkeeper offered two successive discounts such that in percentage terms the second discount was only 1n\frac{1}{n}​th of the first. If the overall discount was p% of the marked price of the item, then the selling price of the item after applying only the first discount was Rs y and the marked price of the item was Rs x.

Which one of the following is correct in respect of the Question and the Statements given below?

Statement 1: The values of n, p, y and x are respectively given as 3, 15.52, 8900 and 10000.

Statement 2: The values of n, p, y and x are respectively given as 4, 21.69, 12730 and 15500.

A.

Only Statement 1 is feasible.

B.

Only Statement 2 is feasible.

C.

Neither of the two statements is feasible.

D.

Both Statements are feasible.

Correct option is C

Given:

D2=1nD1D_2 = \frac{1}{n} D_1​​

Effective discount D = p%p\%​​

Marked price ×D1=y\times D_1 = y   

x = Marked rice

y = Selling price after 1st discount

Formula Used:

For finding D1D_1​:

D1=MPSPMP×100D_1 = \frac{MP - SP}{MP} \times 100​​

For the overall effective discount:

p=(D1D2+D1×D2100)%p = \left( - D_1 - D_2 + \frac{D_1 \times D_2}{100}\right)\%​​​

Solution:

For Statement 1: The values of n, p, y and x are respectively given as 3, 15.52, 8900 and 10000.

n = 3, p = 15.52, y = 8900, x = 10000

Discount (MP - SP) = 1100, MP = 10000, and SP 8900

D1=10000890010000×100=110010000×100=11%D_1 = \frac{10000 - 8900}{10000} \times 100 = \frac{1100}{10000} \times 100 = 11\%​​

D2=1nD1=13×11=1133.67%D_2 = \frac{1}{n} D_1 = \frac{1}{3} \times 11 = \frac{11}{3} \approx 3.67\%​​

Calculating the effective discount:

p=113.67+11×3.67100=113.67+0.4037=14.26%15.52%p = - 11 - 3.67 + \frac{11 \times 3.67}{100} = -11 - 3.67 + 0.4037 = -14.26\% \ne15.52\%​​

This statement does not matches the given effective discount, so Statement 1 is incorrect.

For Statement 2: The values of n, p, y and x are respectively given as 4, 21.69, 12730 and 15500.

n = 4, p = 21.69, y = 12730, x = 15500

D1=1550012730=2770D_1 = 15500 - 12730 = 2770​ 

D1%=277015500×100=17.87096%D_1\% = \frac{2770}{15500} \times 100 = 17.87096\%​​

D2=17.870964=4.46774%D_2 = \frac{ 17.87096}{4} = 4.46774\%​​

Calculating the overall discount:

p=17.870964.46774+17.87096×4.46774100=17.870964.4677+0.79842=21.5402421.69%p = -17.87096 - 4.46774 + \frac{17.87096 \times 4.46774}{100} = -17.87096 - 4.4677 + 0.79842 = -21.54024 \ne - 21.69\%​​

This statement does not  match the given effective discount of 21.69%. Therefore, Statement 2 is incorrect.

Thus, the correct option is (c) Neither of the two statements is feasible.

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