EXERCISE 6.A (OBJECTIVE TYPE QUESTIONS) (IN 6 10 SOLUTION)

6.       The average of the first nine prime numbers is:

a)      9

b)      11

c)       11 1/9

d)      11 2/9

7.       A student was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14, and X.. He found the mean to be 12.What should be the number in place of x?

a)      3

b)      7

c)       17

d)      31

8.       The average of 2, 7, 6, and x is 5 and the average of 18, 1, 6, X and Y is 10.What is the value of y?

a)      5

b)      10

c)       20

d)      30

9.       If the mean o0f 5 Observations x, X + 2, X + 2, X + 6 and X + 8 is 11, then the mean of the last three observations is:

a)      11

b)      13

c)       15

d)      17

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

a)      M²

b)      3M²

c)       6M²

d)      9M²

ANSWERS

6.C

7.B

8.C

9.B

10.B

SOLUTION 

6.       Average = (2 + 3 + 5 + 7 + 11 + 1 + 19 + 23/9) = 100/9 = 11 1/9.

7.       Clearly, we have (3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + X/12) =12.

Or 137 + X =144 or, X = 144 – 137 =7.

8.       We have: (2 + 7 + 6 ++ X / 4) = 5 or 15 + X 20 or X =5.

Also, (18 + 1 + 6 + X + y / 5) = 10 or 25 + 5 + y =50 or y =20.

9.       We have [X + (X +2 ) +X +4) + (X = 6) + (X + 8)/5] = 11  or 5x + 20 =55 o X = 7.

So, the numbers are 7, 9, 11, 13, 15,

Required mean = (11 + 13 + 15 / 3) = 39 / 3 = 13.

10.       We have: (a + b + c /3) = M or ( a + b + c) =3M.

Now,(a + b + c)²= (3M)² = 9M².

a² + b² + c²+ 2 ( 2 a b + b c + ca) = 9m²                    [(a b+ b c + ca) =0]

a² + b² + c²= 9M²

Required mean = (a² + b² + c² /c) = 9M² /3 = 3m²