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GMAT 1: 760 Q51 V42
Posts: 10,057
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Difficulty:
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64% (01:52) correct
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[GMAT math practice question]
The average of \(a, b, c,\) and \(d\) is \(5\) and their standard deviation is \(4.\) What is the sum of the average and the standard deviation of \(2a-5, 2b-5, 2c-5\) and \(2d-5\)?
A. \(7\)
B. \(9\)
C. \(11\)
D. \(13\)
E. \(15\)
The average of \(a, b, c,\) and \(d\) is \(5\) and their standard deviation is \(4.\) What is the sum of the average and the standard deviation of \(2a-5, 2b-5, 2c-5\) and \(2d-5\)?
A. \(7\)
B. \(9\)
C. \(11\)
D. \(13\)
E. \(15\)
firas92
Current Student
Joined: 16 Jan 2019
Last visit: 24 Jul 2026
Posts: 616
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
Products:
MathRevolution
This is based on the properties of Standard Deviation
Quote:
Now if the Set {\(a,b,c,d\)} has a Standard Deviation of \(4\), then the set {\(2a,2b,2c,2d\)} must have a standard deviation of \(8\) based on the first property.
If the Set {\(2a,2b,2c,2d\)} has a Standard Deviation of \(8\), then the set {\(2a-5,2b-5,2c-5,2d-5\)} must have a standard deviation of \(8\) based on the second property.
Average of {\(a,b,c,d,e\)} => \(\frac{a+b+c+d+e}{4} = 5\) => \(a+b+c+d+e=20\)
Average of {\(2a-5,2b-5,2c-5,2d-5\)}\(=\)\(\frac{2a-5+2b-5+2c-5+2d-5}{4} = \frac{2(a+b+c+d)-20}{4} = \frac{2(20)-20}{4} = 5\)
So, the sum of the average and the standard deviation of {\(2a−5,2b−5,2c−5,2d−5\)} \(= 8+5 = 13\)
Answer is (D)
_________________
General Discussion
GMAT 1: 760 Q51 V42
Posts: 10,057
=>
Remember the property \(E(aX+b) = aE(X) + b\), where \(E(X)\) is the average of a set \(X.\)
Remember the property \(S(aX+b) = |a|S(X),\) where \(S(X)\) is the standard deviation of a set \(X\).
\(E(2X-5) = 2E(X) – 5 = 2*5 – 5 = 10 – 5 = 5.\)
\(S(2X-5) = 2*S(X) = 2*4 = 8.\)
Then, the sum of their average and standard deviation is \(5 + 8 = 13.\)
Therefore, the answer is D.
Answer: D
Remember the property \(E(aX+b) = aE(X) + b\), where \(E(X)\) is the average of a set \(X.\)
Remember the property \(S(aX+b) = |a|S(X),\) where \(S(X)\) is the standard deviation of a set \(X\).
\(E(2X-5) = 2E(X) – 5 = 2*5 – 5 = 10 – 5 = 5.\)
\(S(2X-5) = 2*S(X) = 2*4 = 8.\)
Then, the sum of their average and standard deviation is \(5 + 8 = 13.\)
Therefore, the answer is D.
Answer: D
GMAT 1: 760 Q51 V40
Posts: 623
Quote:
Solution:
- Average of \(a, b, c,\) and \(d\) is \(5\) means \(a+b+c+d=5\times 4=20\)
- SD of \(a, b, c,\) and \(d\) is \(4\)
- So, SD of \(2a, 2b, 2c,\) and \(2d\) will be \(4\times 2=8\)
- And SD of \(2a-5, 2b-5, 2c-5,\) and \(2d-5\) will be \(8\) itself
- Average of \(2a-5, 2b-5, 2c-5,\) and \(2d-5=\frac{2a-5+2b-5+2c-5+2d-5}{4}=\frac{2(a+b+c+d)-20}{4}=\frac{2\times 20-20}{4}=5\)
- Sum of the average and the standard deviation of \(2a-5, 2b-5, 2c-5\) and \(2d-5=5+8=13\)
Hence the right answer is Option D
_________________
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