On day 1, a shopkeeper increases the price of an item by k%, and on da

Question

On day 1, a shopkeeper increases the price of an item by k%, and on day 2, he decreases the increased price by k%. By the end of the day 2, the price of the item drops by $1. On day 3, he again increases the decreased price by k%, and on day 4, he again decreases the increased price by k%. If, at the end of day 4, the price of the item comes to $398, what was the approximate initial price of the item?

(A) $325
(B) $350
(C) $375
(D) $400
(E) $450

Show Answer

Official Answer

D

(A) $325
(B) $350
(C) $375
(D) $400
(E) $450

ashmhjn · Answer

Lets assume initial price = 100$ (Day 0)

Day 1 = 100 + k
Day 2 = 100 - k^2/100

Also it's given that:
100 - (Day 2) = 1$
on solving : k=10

plugging in k=10 in all equations- 
Day 0= 100$
Day 1= 110$
Day 2= 99$

Now Day 3 = 99 + 10% = 108.9$
Day 4: 108.9 - 10% = 98.01$

Now going back to the question -  
Given that Day 4 = 398$

Unitary method - 
98.01$ ---> 398$
1$ ---> (398/98.01)$
100$ ---> 405.7$ which is approx 400$ (option D)

rvgmat12 · Answer

P - P(1+k/100)(1-k/100) = 1

P - P(1-k^2/10^4) = 1

1-1/P = (1-k^2/10^4)

P(1-k^2/100)(1-k^2/100) = 398

P*(1-1/P)^2 = 398

(P-1)^2/P = 398

P approx 400

D

PriChats · Answer

I feel the question is easy but lengthy!!
Day wise cost trajectory looks like this with an important caveat at Day2 end which says the price drop was 1 dollar that means (initial price - Day 2 final price) = 1$

C -> C(1+k/100) -> C(1-k/100)*(1+k/100)

as per our caveat => C - C(1-(k^2/100^2)) = 1 => K^2/100^2 = 1/C -------- (1) important observation

C -> C(1+k/100) -> C(1-k/100)*(1+k/100)   ->  (1+k/100) ->  (1+k/100)(1-K/100) = C(1-(K^2/100^2))^2

using observation 1, we have :
C(C-1)^2 / C^2 => (C-1)^2 = 398C
and on expansion it gives:
C^2 - 400C + 1 = 0

we can simply use approximation in the root formula : sqr(b)-root(D) / 2a
and root(400^2 - 4) can be approximated as 400 itself

Thus, answers are either 0 or 400. 0 isnt' possible so we can approx solution to 400. Final answer!

On day 1, a shopkeeper increases the price of an item by k%, and on da

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