The average (arithmetic mean) of the numbers v, w, x, y, and z is j, a

Question

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}

EgmatQuantExpert · Answer

Solution

Given:  
 • Average of v, w, x, y, and z is j
• Average of x, y, and z is k

To find:  
 • The average of v and w, in terms of j and k

Approach and Working:  
 • Sum of v, w, x, y, and z = 5 * j = 5j
• Sum of x, y, and z = 3 * k = 3k
• Therefore, sum of v and w = 5j – 3k
• Hence, average of v and w =  \frac{5j – 3k}{2}

Hence, the correct answer is option D.

Answer: D

pushpitkc · Answer

Given:   
1. The average (arithmetic mean) of the numbers v, w, x, y, and z is j 
2. The average of x, y, and z is k

The mathematical expression is as follows:
 \frac{v+w+x+y+z}{5} = j  ->  v+w+x+y+z = 5j  -> (1) |  \frac{x+y+z}{3} = k  ->  x+y+z = 3k  -> (2)

Substituting (2) in (1), we get  v+w+3k = 5j  ->  v+w = 5j-3k

Therefore, the average of v and w is  \frac{v+w}{2}  or  \frac{5j-3k}{2}   (Option D)

KSBGC · Answer

Given,

\frac{(v+w+x+y+z)}{5}  =j
v+w+x+y+z = 5j ...................................................................1

again,

\frac{(x+y+z)}{3}  = k
x+y+z = 3k

Now we can substitute the value of x+y+z into the equation no. 1

v+w+x+y+z = 5j
v+w + 3k = 5j
v+w = 5j - 3k
 \frac{(v+w)}{2}  =  \frac{(5j-3k)}{2} ......................................As we are asked to find out the average of (v+w)

Thus the best answer is D.

MAMannan · Answer

Lets,
v + w = a and x + y = b;
So, a - 4 = b - 5
or, a + 1 = b;
The equation above clearly states that b (x + y) > a (v + w) by vale 1. 
that means the average of x + y will always be greater than the average of v + w.
Correct me, If I am wrong.

The average (arithmetic mean) of the numbers v, w, x, y, and z is j, a

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