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01 Mar 2024, 02:30
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The mean of three numbers \(2x\), \(3x\), \(x^2\) is \(12\), where \(x<0\). What is the range of the three numbers?
A. 108
B. 144
C. 176
D. 228
E. 256
A. 108
B. 144
C. 176
D. 228
E. 256
jack5397
Joined: 13 Sep 2020
Last visit: 15 May 2025
Posts: 146
Given Kudos: 279
Location: India
Concentration: General Management, Strategy
Schools: ISB '27 IIMA '26 INSEAD '26
GMAT Focus 1: 575 Q79 V79 DI77 
GMAT Focus 2: 575 Q80 V81 DI75
GMAT Focus 3: 635 Q82 V83 DI79 
GMAT 1: 460 Q36 V18 (Online)
GPA: 3.8
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01 Mar 2024, 05:26
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IMO - A
x^2 , 2x ,3x.
\(x^2+5x=36 \)
given x < 0
x has 2 roots x = -9 and 4
we choose x = -9
\(x^2\) = 81
3x = -27
Range = Highest - lowest ;
= 81+27 = 108
x^2 , 2x ,3x.
\(x^2+5x=36 \)
given x < 0
x has 2 roots x = -9 and 4
we choose x = -9
\(x^2\) = 81
3x = -27
Range = Highest - lowest ;
= 81+27 = 108
01 Mar 2024, 22:11
Kudos
Bookmarks
Bunuel
The mean of three numbers \(2x\), \(3x\), \(x^2\) is \(12\), where \(x<0\). What is the range of the three numbers?
A. 108
B. 144
C. 176
D. 228
E. 256
A. 108
B. 144
C. 176
D. 228
E. 256
Let's set up the equation based on the given information:
The mean of three numbers 2x, 3x, and x^2 is 12.
The mean is calculated by adding up all the numbers and then dividing by the total count. In this case:
(2x+3x+x^2)/3=12
Now, solve for x:
(5x+x^2)/3=12
Multiply both sides by 3 to get rid of the fraction:
5x+x^2=36
Combine like terms:
x^2+5x−36=0
Now, factor the quadratic equation:
(x−4)(x+9)=0
So, the solutions for x are x = 4 and x = -9. Since it's given that x < 0, we discard the x = 4 solution.
Now, the three numbers are 2x, 3x, and x^2. We can find these for x = -9:
2(−9)=−18
3(−9)=−27
(−9)^2=81
So, the three numbers are -18, -27, and 81.
Now, to find the range, subtract the smallest from the largest:
81−(−27)=108
Therefore, the range of the three numbers is 108.
