## 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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### 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

by BTGmoderatorDC » Fri Sep 25, 2020 5:27 pm

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## Global Stats

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

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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

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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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