Question

18. A test was conducted in three sections X, Y and Z of tenth class. The average marks per student.of sections X and Y together is 66. The average marks per student of sections Y and Z together is 63. The average marks per student of sections Z andX together is 67. The average marks per student of the three sections together is A. Find the number of integral values that A can take.​

Answer 1

Given:

• there were 3 sections X, Y, Z in which the test was conducted
• The average marks per student of section X and Y together are 66
• The average marks per student of sections Y and Z together are 63
• the average marks per student of sections of Z and X together are 67
• the average marks per student of the 3 sections together is A

To Find:  The number of integral values that A can take?

By section X and Y together = $\frac{x+y}{2} = 66$

By section Y and Z together = $\frac{y+z}{2} = 63$

By section Z and X together = $\frac{z+x}{2} = 67$

Now,

       = 2×(x+y+z)

       = 2×(66+63+67)

or

        x+y+z = 196

∴ according to the question the average marks of the three section together is A

        (x+y+z)/3 = A

              196/3 = A

              65.33 = A

Hence there is NO integral value for A