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The average (arithmetic mean) of the numbers v, w, x, y, and [#permalink]
16 May 2018, 03:53
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The average (arithmetic mean) of the numbers v, w, x, y, and z is j, and the average of the numbers x, y, and z is k. What is the average of v and w in terms of j and k ? A. \(\frac{(j  k)}{2}\) B. \(\frac{(j + k)}{2}\) C. \(\frac{(5j  3k)}{3}\) D. \(\frac{(5j  3k)}{2}\) E. \(\frac{(5j  k)}{2}\)
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Re: The average (arithmetic mean) of the numbers v, w, x, y, and [#permalink]
28 Jun 2018, 22:03
amorphous wrote: The average (arithmetic mean) of the numbers v, w, x, y, and z is j, and the average of the numbers x, y, and z is k. What is the average of v and w in terms of j and k ?
A. \(\frac{(j  k)}{2}\)
B. \(\frac{(j + k)}{2}\)
C. \(\frac{(5j  3k)}{3}\)
D. \(\frac{(5j  3k)}{2}\)
E. \(\frac{(5j  k)}{2}\) Hi, We can solve this by substituting the values or by algebra. Let's first solve by substitution. There is no restriction on the w, v, x, y, and z is given. Let's assume: w = 1, v = 2, x = 3, y = 4, z = 5. Average j = 3. Similarly, the average of x = 3, y = 4, and z = 5 is k = 4. Average of w and v is (1+2)/2 = 1.5 We have j = 3 and k = 4. By substituting the value of j and k in the options we see that only option (D) gives the correct value. Hence, the answer is (D). Thanks.



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Re: The average (arithmetic mean) of the numbers v, w, x, y, and [#permalink]
01 Aug 2018, 20:33
ganand wrote: amorphous wrote: The average (arithmetic mean) of the numbers v, w, x, y, and z is j, and the average of the numbers x, y, and z is k. What is the average of v and w in terms of j and k ?
A. \(\frac{(j  k)}{2}\)
B. \(\frac{(j + k)}{2}\)
C. \(\frac{(5j  3k)}{3}\)
D. \(\frac{(5j  3k)}{2}\)
E. \(\frac{(5j  k)}{2}\) Hi, We can solve this by substituting the values or by algebra. Let's first solve by substitution. There is no restriction on the w, v, x, y, and z is given. Let's assume: w = 1, v = 2, x = 3, y = 4, z = 5. Average j = 3. Similarly, the average of x = 3, y = 4, and z = 5 is k = 4. Average of w and v is (1+2)/2 = 1.5 We have j = 3 and k = 4. By substituting the value of j and k in the options we see that only option (D) gives the correct value. Hence, the answer is (D). Thanks. Since this poster explained on how to find the answer by substitution, I'll take the liberty on how to explain it algebra wise. The formulas given to use are \(\frac{(v+w+x+y+z)}{5} = J\) and \(\frac{(x + y + z)}{3} = k\) To get both variables (J & K) on the same equation, we manipulate the second equation. \((x+y+z) = 3k\) We can now use this equation and replace it into the first one. \(\frac{(v+w+3k)}{5} = J\) At this point, everything is just basic algebra. \(v+w+3k = 5J\) \(v+w= 5J  3k\) At this point we're almost done, but because we want to find the average of v+w, we need to divide both sides by 2. \(\frac{v+w}{2} = \frac{5J3K}{2}\) The answer is D.




Re: The average (arithmetic mean) of the numbers v, w, x, y, and
[#permalink]
01 Aug 2018, 20:33





