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17 Jul 2019, 04:49
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61% (02:53) correct
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Mr. Kay, a shopkeeper, found the average sale for 5 consecutive days to be $160. The highest sale was of $200. If there were no two days when the sale amount was same, what is the minimum possible value of the \(3^{rd}\) highest sale? (All amounts were integral values)
A. $132
B. $133
C. $134
D. $135
E. $136
A. $132
B. $133
C. $134
D. $135
E. $136
18 Jul 2019, 21:54
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Answer D
a,b,c,d,e are sale for 5 days.
Given
All are integer and no two days has same sale.
Assume
a<b<c<d<e given e=200.
We need to find out minimum value of C.
So we have to maximize D and keep value of a and b as close as possible to c.
E=200 so d can be max 199.
Total sale was 800. So balance amount 401 need to be divide in a,b,c such as a<b<c
Each 134 will give value of 402.
So value of c has to be 135, so a and b can be 132,134..
Answer D.
Posted from my mobile device
a,b,c,d,e are sale for 5 days.
Given
All are integer and no two days has same sale.
Assume
a<b<c<d<e given e=200.
We need to find out minimum value of C.
So we have to maximize D and keep value of a and b as close as possible to c.
E=200 so d can be max 199.
Total sale was 800. So balance amount 401 need to be divide in a,b,c such as a<b<c
Each 134 will give value of 402.
So value of c has to be 135, so a and b can be 132,134..
Answer D.
Posted from my mobile device
15 Apr 2020, 05:09
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Dillesh4096
Solution:
- • Average sale of 5 days = $ 160
- o Total sale of 5 days = \(160*5 = $ 800\)
- o Sales of remaining 4 days =\(800-200=$ 600\)
- o The maximum possible value of 2nd highest sale = \($199\)
• Let the minimum possible value of 3rd highest sale be x
- o The maximum possible value of 4th highest sale = \(x-1\)
o The maximum possible value of 5th highest sale = \(x – 2\)
- o \(3x =404\)
o \(x =134.66\)
- x must be an integer, the minimum value of 3rd highest sales is $ 135
