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Updated on: 13 Aug 2018, 00:48
Originally posted by EgmatQuantExpert on 11 Jul 2018, 08:00.
Last edited by EgmatQuantExpert on 13 Aug 2018, 00:48, edited 4 times in total.
Last edited by EgmatQuantExpert on 13 Aug 2018, 00:48, edited 4 times in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
53% (02:46) correct
47%
(02:50)
wrong
based on 457
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Solve any Median and Range question in a minute- Exercise Question #1
1- Jack bought five mobiles at an average price of $150. The median of all the prices is $200. What is the minimum possible price of the most expensive mobile that Jack has bought, if the price of the most expensive mobile is at least thrice that of the least expensive mobile.
Options
To solve question 2: Question 2
To read the article: Solve any Median and Range question in a minute
1- Jack bought five mobiles at an average price of $150. The median of all the prices is $200. What is the minimum possible price of the most expensive mobile that Jack has bought, if the price of the most expensive mobile is at least thrice that of the least expensive mobile.
Options
- a) $150
b) $200
c) $250
d) $300
e) $350
To solve question 2: Question 2
To read the article: Solve any Median and Range question in a minute
11 Jul 2018, 08:51
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The answer is B.
My explanation:
a) cannot be the answer as the median itself is $200.
b) correct
c) Let's assume the 5th phone (in terms of ascending prices) is of $250. Then the two extreme cases are possible for the value of x4 (x1, x2, $200, x4, $250) - it can either be $200 or $250. If it is $250, then in the eq. 152 = (x1+x2+200+250+250)/5, x1+x2 = 60 but according to the question, 3*x1 = 250 => x1 ~ 89. NOT POSSIBLE. If x4 is $200, then in the eq. 152 = (x1+x2+200+200+250)/5, x1+x2 = 110 and ideally x1 should be greater than 89, but if it is the case then x2<x1. NOT POSSIBLE.
d) Solving similarly as option c. No solution possible.
e) Solving similarly as option c. No solution possible.
Hence the correct answer is B.
My explanation:
a) cannot be the answer as the median itself is $200.
b) correct
c) Let's assume the 5th phone (in terms of ascending prices) is of $250. Then the two extreme cases are possible for the value of x4 (x1, x2, $200, x4, $250) - it can either be $200 or $250. If it is $250, then in the eq. 152 = (x1+x2+200+250+250)/5, x1+x2 = 60 but according to the question, 3*x1 = 250 => x1 ~ 89. NOT POSSIBLE. If x4 is $200, then in the eq. 152 = (x1+x2+200+200+250)/5, x1+x2 = 110 and ideally x1 should be greater than 89, but if it is the case then x2<x1. NOT POSSIBLE.
d) Solving similarly as option c. No solution possible.
e) Solving similarly as option c. No solution possible.
Hence the correct answer is B.
13 Sep 2018, 07:23
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EgmatQuantExpert
Jack bought five mobiles at an average price of $150.
So, (sum of all 5 mobiles)/5 = $150
Multiply both sides by 5 to get: sum of all 5 mobiles = $750
The median of all the prices is $200.
Let a = smallest value
Let b = 2nd smallest value
Let d = largest value
Let c = 2nd largest value
So, when we arrange the values in ASCENDING order we get: a, b, $200, c, d
From here, a quick approach is to test each answer choice, starting from the smallest value
A) $150
This answer choice suggests that d = $150, which is impossible, since the greatest value cannot be less than the median ($200)
ELIMINATE A
B) $200
This answer choice suggests that d = $200
Let's add this to our list to get: a, b, $200, c, $200
This means c must also equal $200. So we have: a, b, $200, $200, $200
Is it possible to assign values to a and b so that all of the conditions are met?
YES!
We must satisfy the condition that sum of all 5 mobiles = $750, and it must be the case that the price of the most expensive mobile is at least thrice that of the least expensive mobile
Well, if we let a = $50, then the price of the most expensive mobile ($200) is at least thrice that of the least expensive mobile ($50)
Finally, if we let b = $100, we get: $50, $100, $200, $200, $200, which meets the condition that sum of all 5 mobiles = $750
PERFECT!!
Answer: B
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