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16 Sep 2014, 01:25
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Difficulty:
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Question Stats:
64% (02:17) correct
36%
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If set A consists of 7 consecutive odd integers and the sum of the 5 largest integers in set A is -185, what is the sum of the 5 smallest integers in set A?
A. -165
B. -175
C. -195
D. -205
E. -215
A. -165
B. -175
C. -195
D. -205
E. -215
16 Sep 2014, 01:25
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Official Solution:
If set A consists of 7 consecutive odd integers and the sum of the 5 largest integers in set A is -185, what is the sum of the 5 smallest integers in set A?
A. -165
B. -175
C. -195
D. -205
E. -215
Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).
Given: \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185\)
Question: \(x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=\)?
Given that \(5x+40=-185\), the value of \(5x+20\) will be 20 units smaller, resulting in -205.
Answer: D
If set A consists of 7 consecutive odd integers and the sum of the 5 largest integers in set A is -185, what is the sum of the 5 smallest integers in set A?
A. -165
B. -175
C. -195
D. -205
E. -215
Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).
Given: \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185\)
Question: \(x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=\)?
Given that \(5x+40=-185\), the value of \(5x+20\) will be 20 units smaller, resulting in -205.
Answer: D
General Discussion
27 Nov 2014, 16:02
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Average of largest 5 = -185/5 =-37
So the largest numbers are -41, -39, -37, -35, -33
Now if we consider smallest 5, we have to eliminate last 2 numbers, which are -35 and -33
Notice here that the numbers are equally spaced.
So the average of smallest number will just shift 2 place to the left.
Avg= -41
Therefore, total = -41*5= -205
So the largest numbers are -41, -39, -37, -35, -33
Now if we consider smallest 5, we have to eliminate last 2 numbers, which are -35 and -33
Notice here that the numbers are equally spaced.
So the average of smallest number will just shift 2 place to the left.
Avg= -41
Therefore, total = -41*5= -205
