If the mean of a, b, c is M and ab + bc + ca = 0, then the mean of a2, b2, c2 is:

Options:
A.3 M x M
B.3 M
C.9 M
D.9 M x M
E.None of these
Answer: Option A
 -   We have :   a + b + c 3   = M or (a + b + c) = 3M.       Now, (a + b + c)M2 = (3M)M2 = 9M2       a2 + b2 + c2 + 2 (ab + bc + ca) = 9M2           a2 + b2 + c2 = 9M2         Required mean =   a2 + b2 + c2 3   = 9M2 3 = 3M2

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More Questions Related to

Quantitative Aptitude > Average :
Question 1.

The average of a non-zero number and its square is 5 times the number. The number is:

Options:
  1.    0 , 9
  2.    0 , 8
  3.    5 , 8
  4.    6 , 0
  5.    None of these
Answer: Option A
 -  Let the number be x.
Then,
x + x2 2 = 5x       x2 - 9x = 0   x (x - 9) = 0   x = 0 or x = 9.
Question 2.

The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is:

Options:
  1.    51.4
  2.    53.6
  3.    54.4
  4.    55.6
  5.    None of these
Answer: Option A
 -  Let the required mean score be a

Then, 20 x 80 + 25 x 31 + 55 x a = 52 x 100

1600 + 775 + 55a = 5200

55a = 2825

a = 51.4
Question 3.

The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one them weighing 65 kg. What might be the weight of the new person?

Options:
  1.    65 kg
  2.    75 kg
  3.    85 kg
  4.    90 kg
  5.    None of these
Answer: Option C
 -  Total weight increased = (8 x 2.5) kg = 20 kg

Weight of new person = (65 + 20) kg = 85 kg
Question 4.

3 year ago, the average of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby is:

Options:
  1.    1 year
  2.    2 years
  3.    3 years
  4.    4 years
  5.    None of these
Answer: Option B
 -  Total age of 5 members, 3 years ago = (17 x 5) years = 85 years

Total age of 5 members now = (85 + 3 x 5) years = 100 years

Total age of 6 members now = (17 x 6) years = 102 years

Age of the baby = (102 – 100) years = 2 years
Question 5.

Of the four numbers, the first is twice the second, the second is one-third of the third and the third is 5 times the fourth. The average of the numbers is 24.75. The largest of these numbers is:

Options:
  1.    9
  2.    15
  3.    30
  4.    45
  5.    None of these
Answer: Option D
 -  Let the fourth number be a
Then, third number = 5a, second number = 5a 3 And first number = 10a 3 x + 5x + 5x 3 + 10x 3 = (24.75 x 4) or 11x = 99 or x = 9.                 So, the numbers are 9, 45, 15 and 30

Largest number = 45
Question 6.

The average of six numbers is 3.95. The average of two of them is 3.4, while the average of the other two is 3.85. What is the average of the remaining two numbers?

Options:
  1.    4
  2.    4.6
  3.    5
  4.    5.6
  5.    None of these
Answer: Option B
 -  Sum of the remaining two numbers = (3.95 x 6) – [(3.4 x 2) + (3.85 x 2)]
                                                         
                                                         = 23.70 - (6.8 + 7.7) = 23.70 - 14.5 = 9.20

Average = 9.2/2 = 4.6