Question

(P.U., M.A. Econ. 1974; B.Z.
M.A., ECE
1991
The following table shows the age distribution of 1,143 horses.
Age (years) Number of horses (9) Average age (;)
1-4
12.
2.7
5-9
223
7.6
10 - 14
435
12.0
15 - 19
272
16.3
20 - 24
161
20.8
25 - 29
34
25.8
30 and over
6
31.8
Compute the average age of these horses (a) from the first two columns of the table by the
usual short method, (b) from the last two columns by weighting the group averages by the
number of horses in the groups. Compare the two results. Which one is more nearly the real
average age?​

Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.