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705-805 (Hard)|   Arithmetic|      
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twobagels
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The harmonic mean of two quantities 'a' and 'b' is defined as 08 Feb 2021, 11:08
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75% (hard)

Question Stats:

63% (02:46) correct 38% (02:59) wrong based on 176 sessions

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The harmonic mean of two quantities 'a' and 'b' is defined as \(\frac{(2*a*b)}{(a+b)}\). If the harmonic mean of the quantities x and 12 is one less than the average of the same two quantities, which of the following can be the value of the sum of digits in x?

A. 1
B. 2
C. 3
D. 4
E. 5
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B
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nick1816
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The harmonic mean of two quantities 'a' and 'b' is defined as 11 Feb 2021, 18:23
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\(\frac{2*12*x}{(12+x) }= \frac{(12+x)}{2} - 1\)

\(\frac{2*12*x}{(12+x)} = \frac{(10+x)}{2} \)

\(48x = x^2+22x+120\)

\(x^2-26x+120 = 0\)

\(x^2- 6x-20x+120 = 0\)

x=6 or 20

sum of digits is either 2 or 6

twobagels
The harmonic mean of two quantities 'a' and 'b' is defined as \(\frac{(2*a*b)}{(a+b)}\). If the harmonic mean of the quantities x and 12 is one less than the average of the same two quantities, which of the following can be the value of the sum of digits in x?

A. 1
B. 2
C. 3
D. 4
E. 5
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sthahvi
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The harmonic mean of two quantities 'a' and 'b' is defined as 14 Feb 2021, 05:07
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nick1816
\(\frac{2*12*x}{(12+x) }= \frac{(12+x)}{2} - 1\)

\(\frac{2*12*x}{(12+x)} = \frac{(10+x)}{2} \)

\(48x = x^2+22x+120\)

\(x^2-26x+120 = 0\)

\(x^2- 6x-20x+120 = 0\)

x=6 or 20

sum of digits is either 2 or 6

twobagels
The harmonic mean of two quantities 'a' and 'b' is defined as \(\frac{(2*a*b)}{(a+b)}\). If the harmonic mean of the quantities x and 12 is one less than the average of the same two quantities, which of the following can be the value of the sum of digits in x?

A. 1
B. 2
C. 3
D. 4
E. 5
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can you explain how did you find the other ones like \frac{(10+x)}{2}
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samarpan.g28
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The harmonic mean of two quantities 'a' and 'b' is defined as 01 Feb 2025, 11:02
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The mean or average of two numbers is (num1+num2) / 2. Here we are dealing with x and 12 so their mean is (x+12)/2.
sthahvi
nick1816
\(\frac{2*12*x}{(12+x) }= \frac{(12+x)}{2} - 1\)

\(\frac{2*12*x}{(12+x)} = \frac{(10+x)}{2} \)

\(48x = x^2+22x+120\)

\(x^2-26x+120 = 0\)

\(x^2- 6x-20x+120 = 0\)

x=6 or 20

sum of digits is either 2 or 6

twobagels
The harmonic mean of two quantities 'a' and 'b' is defined as \(\frac{(2*a*b)}{(a+b)}\). If the harmonic mean of the quantities x and 12 is one less than the average of the same two quantities, which of the following can be the value of the sum of digits in x?

A. 1
B. 2
C. 3
D. 4
E. 5

can you explain how did you find the other ones like \frac{(10+x)}{2}
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Dean19
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The harmonic mean of two quantities 'a' and 'b' is defined as 04 Feb 2025, 07:17
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The wording of the question is veryyyy ambiguous, it said quantities "a" and "b" then ask sum of digits (IN) "x", the question insinuates that x is a set and not a variable as and asks for the sum of values in set x.
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vibhorrokstar19
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The harmonic mean of two quantities 'a' and 'b' is defined as 04 Feb 2025, 08:05
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I think we are discussing about the quantities X AND 12, in my opinion it represents a set of values and not the sum of the values present in the set.
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