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Que: On day 1, a shopkeeper increases the price of an item by k%, and on day 2, he decreases the increased price...

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Que: On day 1, a shopkeeper increases the price of an item by k%, and on day 2, he decreases the increased price...

by Max@Math Revolution » Tue Dec 08, 2020 8:44 am

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Que: 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 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
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Re: Que: On day 1, a shopkeeper increases the price of an item by k%, and on day 2, he decreases the increased price...

by Max@Math Revolution » Fri Dec 11, 2020 1:30 am
Solution: Let us apply the IVY approach to solve the question.

As we are dealing with percentage, then let the initial price of the item be 100. Since there is a common k% of increase and decrease, then let us assume k% = 10%

Day 1: Price after price increased by 10%: (100 + 10)% of 100
=> (\(\frac{110}{100}\)) $100 => $110

Day 2: Price after the new price decreased by 10%: (100 - 10)% of 110
=> (\(\frac{90}{100}\)) $110 => $99

Day 3: Price after price increased by 10%: (100 + 10)% of 99
=> $99 * \(\frac{110}{100}\) => $108.9 ≈ $109

Day 4: Price after the new price decreased by 10%: (100 - 10)% of 109
=> $109 * \(\frac{90}{100}\) => $98.1 ≈ 98

For $ 100 = final price is $98
For $x = $398 [given in the question]

=> x = \(\frac{100\ \cdot\ 398}{98}\) = 406 ≈ 400

Therefore, D is the correct answer.

Answer D
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Re: Que: On day 1, a shopkeeper increases the price of an item by k%, and on day 2, he decreases the increased price...

by Scott@TargetTestPrep » Mon Apr 26, 2021 6:29 am
Max@Math Revolution wrote: ↑
Tue Dec 08, 2020 8:44 am
 Que: 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 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
Solution:

Let the initial price be x. At the end of day 1, the price becomes x(1 + k/100) and at the end of day 2, the price becomes x(1 + k/100)(1 - k/100). We are told that this is $1 less than x; thus:
x(1 + k/100)(1 - k/100) = x - 1

Since x represents the price of some item, it must be nonzero (which can be verified by looking at answer choices). Thus, let’s divide each side by x:

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

At the end of day 3, the price of the item is x(1 + k/100)(1 - k/100)(1 + k/100) and at the end of day 4, the price of the item is x(1 + k/100)(1 - k/100)(1 + k/100)(1 - k/100). We are told that this is equal to $398; thus:

x(1 + k/100)(1 - k/100)(1 + k/100)(1 - k/100) = 398

Let’s substitute (1 + k/100)(1 - k/100) by (x - 1)/x twice:

x * (x - 1)/x * (x - 1)/x = 398

(x - 1)^2 / x = 398

Let’s test the answer choices at this point. Substituting x = 325, (x - 1)^2 / x becomes 324^2/325; which is approximately 324. Similarly, if x = 350, 375, 400 or 450; the expression (x - 1)^2 / x becomes 349, 374, 399 and 449, approximately. We see that the value of x that is closest to the final price of $398 is x = 400. Using a calculator, we can actually verify that 399^2 / 400 equals 398,0025.

Answer: D

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