Both the average (arithmetic mean) and the median of a set of 7 numbers equal 20. If the smallest number in the set is 5 less than half the largest number, what is the largest possible number in the set?
(A) 40
(B) 38
(C) 33
(D) 32
(E) 30
Answer: B
Source: Manhattan GMAT
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Both the average (arithmetic mean) and the median of a set of 7 numbers equal 20. If the smallest number in the set is 5
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Gmat_mission
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deloitte247
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Average (mean ) = 20
median = 20
Let last number be = x
First number
$$\frac{1}{2}\left(x\right)-5=\frac{x}{2}-5$$
In finding the largest (maximum) possible value of x .. The first number and median will be minimized
Sum = Average*no of items = 20*7 = 140
So, by repeating the first number ( median) as the sixth one, x will be maximised.
$$Therefore,\ \ \left(\frac{x}{2}-5\right)+\left(\frac{x}{2}-5\right)+\left(\frac{x}{2}-5\right)+26+20+20+x=140$$ $$x+\frac{3x}{2}-15+60=140$$ $$x+\frac{3x}{2}=140-45$$ $$\frac{\left(2x+3x\right)}{2}=95$$ $$\left(\frac{5x}{5}=\frac{190}{5}\right)$$ $$x=38$$ $$Answer\ is\ Option\ B$$
median = 20
Let last number be = x
First number
$$\frac{1}{2}\left(x\right)-5=\frac{x}{2}-5$$
In finding the largest (maximum) possible value of x .. The first number and median will be minimized
Sum = Average*no of items = 20*7 = 140
So, by repeating the first number ( median) as the sixth one, x will be maximised.
$$Therefore,\ \ \left(\frac{x}{2}-5\right)+\left(\frac{x}{2}-5\right)+\left(\frac{x}{2}-5\right)+26+20+20+x=140$$ $$x+\frac{3x}{2}-15+60=140$$ $$x+\frac{3x}{2}=140-45$$ $$\frac{\left(2x+3x\right)}{2}=95$$ $$\left(\frac{5x}{5}=\frac{190}{5}\right)$$ $$x=38$$ $$Answer\ is\ Option\ B$$
