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Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ??
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Problem 2 – Bholu conducts his business in the following manner: Every...
Problem:
Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x%, the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?
Solution:
Let the initial price of the commodity be P.
First Cycle:
In the first cycle, Bholu increases the price by x% and then decreases it by the same percentage. So, the new price after one cycle is given by:
P * (1 + x/100) * (1 - x/100)
According to the problem, the price decreases by Rs 100 after one cycle. Therefore, we can write the equation:
P * (1 + x/100) * (1 - x/100) - P = 100
Simplifying the equation, we get:
(100 + x)^2 - 10000 = 100
(100 + x)^2 = 10100
100 + x = √10100
x = √10100 - 100
x = 50 - 10
x = 40
So, in the first cycle, Bholu increases and then decreases the price by 40%.
Second Cycle:
In the second cycle, Bholu increases and then decreases the price by (x/2)%. So, the new price after the second cycle is given by:
(P - 100) * (1 + x/200) * (1 - x/200)
According to the problem, the commodity is sold for Rs 2376 after the second cycle. Therefore, we can write the equation:
(P - 100) * (1 + x/200) * (1 - x/200) = 2376
Simplifying the equation, we get:
(100 + x/2)^2 - 4 * (P - 100) * (P - 2376) = 0
Using the value of x = 40, we can solve the equation to find the initial price P.
P = 2500
Therefore, the initial price of the commodity is Rs 2500.
Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x%, the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?
Solution:
Let the initial price of the commodity be P.
First Cycle:
In the first cycle, Bholu increases the price by x% and then decreases it by the same percentage. So, the new price after one cycle is given by:
P * (1 + x/100) * (1 - x/100)
According to the problem, the price decreases by Rs 100 after one cycle. Therefore, we can write the equation:
P * (1 + x/100) * (1 - x/100) - P = 100
Simplifying the equation, we get:
(100 + x)^2 - 10000 = 100
(100 + x)^2 = 10100
100 + x = √10100
x = √10100 - 100
x = 50 - 10
x = 40
So, in the first cycle, Bholu increases and then decreases the price by 40%.
Second Cycle:
In the second cycle, Bholu increases and then decreases the price by (x/2)%. So, the new price after the second cycle is given by:
(P - 100) * (1 + x/200) * (1 - x/200)
According to the problem, the commodity is sold for Rs 2376 after the second cycle. Therefore, we can write the equation:
(P - 100) * (1 + x/200) * (1 - x/200) = 2376
Simplifying the equation, we get:
(100 + x/2)^2 - 4 * (P - 100) * (P - 2376) = 0
Using the value of x = 40, we can solve the equation to find the initial price P.
P = 2500
Therefore, the initial price of the commodity is Rs 2500.
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Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude?
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Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude? for Quant 2024 is part of Quant preparation. The Question and answers have been prepared according to the Quant exam syllabus. Information about Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude? covers all topics & solutions for Quant 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude?.
Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude? for Quant 2024 is part of Quant preparation. The Question and answers have been prepared according to the Quant exam syllabus. Information about Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude? covers all topics & solutions for Quant 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude?.
Solutions for Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude? in English & in Hindi are available as part of our courses for Quant.
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Here you can find the meaning of Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude? defined & explained in the simplest way possible. Besides giving the explanation of
Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude?, a detailed solution for Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude? has been provided alongside types of Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude? theory, EduRev gives you an
ample number of questions to practice Problem 2 – Bholu conducts his business in the following manner: Every once in a while he raises his prices by a certain percentage and a while later reduces his price by the same percentage. After one such up-down cycle of increasing and then decreasing his price by x% , the price of the commodity decreases by Rs 100. In the next cycle, he increases and decreases his price by (x/2)% and then sells the commodity for Rs 2376. What is the initial price of the commodity?Solution: Let the initial price be P.P ( 100 + x) (100-x) / 100 * 100 = P- 100P*x*x /100 = 100 -(1)Similarily ,(P – 100) (400 – x^2) / 400 = 2376 -(2)Using 1 and 2,P^2 – 2501 P + 2500 = 0P= 2500 or 1Since , P > 2376P= 2500 Can anyone explain me ?? Related: Percentages - Examples (with Solutions), Number Theory, Quantitative Aptitude? tests, examples and also practice Quant tests.
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