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23 May 2020, 06:35
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B
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What is the average (arithmetic mean) of a, b, c, d, e, and f with a < b < c < d < e < f ?
(1) The average of d, e, and f is 11.
(2) The value of a is 5 and b + c + d + e + f = 51.
DS21260
(1) The average of d, e, and f is 11.
(2) The value of a is 5 and b + c + d + e + f = 51.
DS21260
yashikaaggarwal
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23 May 2020, 06:40
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Answer should be B
Average=total/no. of terms.
Statement 1: (d+e+f)/3=11
What about a b and c
We need their values as well to determine a average (not sufficient)
Statement 2:
A=5
B+C+D+E+F=51
Total = A+B+C+D+E+F/6
=> 5+51/6
=> 56/6
=> 9.33
(Sufficient)
Posted from my mobile device
Average=total/no. of terms.
Statement 1: (d+e+f)/3=11
What about a b and c
We need their values as well to determine a average (not sufficient)
Statement 2:
A=5
B+C+D+E+F=51
Total = A+B+C+D+E+F/6
=> 5+51/6
=> 56/6
=> 9.33
(Sufficient)
Posted from my mobile device
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23 May 2020, 06:43
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Bunuel
What is the average (arithmetic mean) of a, b, c, d, e, and f with a < b < c < d < e < f ?
(1) The average of d, e, and f is 11.
(2) The value of a is 5 and b + c + d + e + f = 51.
PS21260
(1) The average of d, e, and f is 11.
(2) The value of a is 5 and b + c + d + e + f = 51.
PS21260
statment 1: The average of d, e, and f is 11.
no information about a,b,c
all we know is a<b<c
not sufficient
let a=-1, b=0, c=1; mean = [(a+b+c+d+e+f)/6] = 11/6
let a= 1, b=2, c=3; mean = [(a+b+c+d+e+f)/6] = 17/6
let a=-3, b=-2, c=0; mean = [(a+b+c+d+e+f)/6]= 6/6
different means for different values of a<b<c , so not sufficient
statement 2: a=5, (b+c+d+e+f)= 51
mean = [(a+b+c+d+e+f)/6] = (5+51)/6 = 56/6
therefore, sufficient
Answer B
