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A = 11.0, 13.9, 15.6, 12.3, 19.7, 21.4, 9.9, x
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28 Aug 2021, 01:34
28 Aug 2021, 01:34
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Question Stats:
71% (01:55) correct
28% (01:53) wrong based on 7 sessions
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Given a set \(A = 11.0, 13.9, 15.6, 12.3, 19.7, 21.4, 9.9, x\), the average of the elements of \(A\) is equal to range of the set where \(x\) is a positive number.
A. Quantity A is greater
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.[/quote]
Quantity A |
Quantity B |
Range of \(x\) |
19 |
A. Quantity A is greater
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.[/quote]
Re: A = 11.0, 13.9, 15.6, 12.3, 19.7, 21.4, 9.9, x
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29 Aug 2021, 09:48
29 Aug 2021, 09:48
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himanshu13 wrote:
Is it option D?
for solving this we have to consider two cases, where x is largest element in set and another is smallest element.
case 1-where x is largest element
103.801+x=(x-9.9)*8
x=26.143 and
case 2-where is smallest element.
103.801+x=(21.4-x)*8
x=7.4887
range of x=26.143-7.488=18.65
less than 19.
so ans is b.
