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Given A, B, C, D & E are integers and A B C D E. Mean of A, B,
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30 Jul 2023, 23:40
30 Jul 2023, 23:40
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Question Stats:
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Given A, B, C, D & E are integers and A ≥ B ≥ C ≥ D ≥ E. Mean of A, B, C, D & E is 120 and median is 60. What will be the minimum possible value of A?
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210
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Given A, B, C, D & E are integers and A B C D E. Mean of A, B,
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11 Aug 2023, 02:38
11 Aug 2023, 02:38
1
Arranged in ascending order : e,d,c,b,a
Median=c=60
In order to get the minimum value of a, we need to maximize e,d.
Mean=120
Sum = 120 * 5 = 600
e=d=c=60
a+b=600-(e+d+c)=600-180=420
a is minimum when a=b.
Thus, a=210
Posted from my mobile device
Median=c=60
In order to get the minimum value of a, we need to maximize e,d.
Mean=120
Sum = 120 * 5 = 600
e=d=c=60
a+b=600-(e+d+c)=600-180=420
a is minimum when a=b.
Thus, a=210
Posted from my mobile device
Re: Given A, B, C, D & E are integers and A B C D E. Mean of A, B,
[#permalink]
01 Sep 2023, 18:05
01 Sep 2023, 18:05
1
Given the conditions:
A ≥ B ≥ C ≥ D ≥ E
Mean (average) of A, B, C, D, and E is 120.
Median is 60.
To find the minimum possible value of A while keeping these conditions:
The mean of A, B, C, D, and E is 120, which means that their sum is
5
×
120
=
600
5×120=600.
The median is 60, and since the numbers are ordered in descending order, C is the middle number, so C = 60.
Now, we have:
Median = C = 60
Mean = 120
As A ≥ B ≥ C ≥ D ≥ E why can't the following be a possibility????
We want to minimize the value of A while maintaining these conditions. To do this, we need to maximize the values of B, D, and E while keeping them less than or equal to C.
So, let's set B = D = E = 60 (to maximize them) and calculate A:
A + 60 + 60 + 60 + 60 = 600
Now, solve for A:
A + 240 = 600
A = 600 - 240 = 360
So, the minimum possible value of A is 360 while maintaining the corrected given conditions.
A ≥ B ≥ C ≥ D ≥ E
Mean (average) of A, B, C, D, and E is 120.
Median is 60.
To find the minimum possible value of A while keeping these conditions:
The mean of A, B, C, D, and E is 120, which means that their sum is
5
×
120
=
600
5×120=600.
The median is 60, and since the numbers are ordered in descending order, C is the middle number, so C = 60.
Now, we have:
Median = C = 60
Mean = 120
As A ≥ B ≥ C ≥ D ≥ E why can't the following be a possibility????
We want to minimize the value of A while maintaining these conditions. To do this, we need to maximize the values of B, D, and E while keeping them less than or equal to C.
So, let's set B = D = E = 60 (to maximize them) and calculate A:
A + 60 + 60 + 60 + 60 = 600
Now, solve for A:
A + 240 = 600
A = 600 - 240 = 360
So, the minimum possible value of A is 360 while maintaining the corrected given conditions.
