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If the average (arithmetic mean) of b and c is 5, and the average of c and d is 10, then b - d = A. Cannot be determine

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Re: If the average (arithmetic mean) of b and c is 5, and the average of c and d is 10, then b - d = A. Cannot be deter

by psarma » Sat Sep 26, 2020 11:41 am
Given \(\frac{b+c}{2}\) =5
So, b+c = 10

Likewise, \(\frac{c+d}{2}\) =10
So, c+d=20

Subtracting 2nd equation from 1st,
we get b-d=-10

Answer is C.
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Re: If the average (arithmetic mean) of b and c is 5, and the average of c and d is 10, then b - d = A. Cannot be deter

by AAPL » Sat Sep 26, 2020 11:47 am
BTGmoderatorDC wrote: ↑
Fri Sep 25, 2020 5:27 pm
 If the average (arithmetic mean) of b and c is 5, and the average of c and d is 10, then b - d =

A. Cannot be determined
B. -5
C. -10
D. -15
E. -20


OA C

Source: Magoosh
\begin{cases}
b + c = 10 \qquad (1) \\
c + d = 20 \qquad (2)
\end{cases}

Subtracting \((1)\) from \((2)\) we have,

\(b - d = -10\)
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