There are 6 integers and a

Question

There are 6 integers and a<=b<=c<=d<=e<=f. If the median of the integers is 30 and the average (arithmetic mean) of the integers is 40, what is the smallest possible value of f?
A) 55
B) 60
C) 65
D) 70
E) 75

amanvermagmat · Accepted Answer

In case of 6 integers arranged like this: a <= b <= c <= d <= e <= f
Median = average of middle two integers = (c+d)/2 .. this is given as 30 , so c+d = 60

Now, mean of all 6 integers = 40, so their sum = 40*6 = 240
We have now: a+b+c+d+e+f = 240, out of which c+d = 60

We have to give lowest possible value to f, so for that lets give highest possible value to others.
a, b, c, d can each take highest possible value as 30 each, which makes their sum a+b+c+d = 120
Now sum of e/f becomes: e+f = 240-120 = 120
Here f will have smallest value when both e/f are equal.. so both can be 60 each.

Smallest possible value of f = 60. Hence  B answer

rohit8865 · Answer

for getting f (min) we should consider c=d---(1)
also a+b+c+d+e+f= 240
where c+d=60

from (1)
we get d=30
for e+f to be minimum , a=b=c=d
a+b+c+d=120
leaving e+f=120---(2)
minimum f ..when e=f
putting in (2)
thus f=60

Ans B

roadrunner · Answer

From question stem, a +b + e+ f =180. To minimize f, maximize a,b, and e.

Since a + b <= 60 (c+d).
Max(a) and Max(b) = 30

thus, e+f = 120. e<=f ---> max (e) = f

Here, f is minimum when e=f ==> f =120/2 =60

gps5441 · Answer

Please note the question stem says each integer can be equal to the other.

Average is 40 hence sum =240
Median is the middle no since the set has even no of integers ie 6 the middle no =30(30+30/2)
therefore C=D=30.
Now plugin in the nos

a<=b<=30<=30<=e<=f

For f to be minimum following are the requirements
1) a and b have to be maximum possible value but cannot be more than the median which is 30 hence a and b=30
2) e has to be maximum ie greater than the median but less than f 
3) Out of 240 we have already allocated 120(a=b=c=d=30)
4) we are left with 120 and as per point no 2 e=60

Hence f is=60

AdarshSambare · Answer

for f to be minimum, set c = d.
also: a + b + c + d + e + f = 240, with c + d = 60.
from c = d → d = 30.

to minimize e + f we should make a = b = c = d, so
a + b + c + d = 120 → leaving e + f = 120.

for f minimum, set e = f, so f = 120/2 = 60.
Ans. B

There are 6 integers and a<=b<=c<=d<=e<=f. If the median of the

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