KarunMendiratta wrote:
Explanation:
a1,a2,.........,a7,..........,a12,a13
Given:
a1,a2,.........,10,..........,a12,a13, and
a1+a2+.........+10+..........+a12+a1313=15
i.e.a1+a2+.........+10+..........+a12+a13=195
Since, we need to find the minimum value of a13, we must keep all other scores maximum
10+10+10+10+10+10+10+a8+..........+a12+a13=195
70+a8+a9+a10+a11+a12+a13=195
a8+a9+a10+a11+a12+a13=125
a13=125−(a8+a9+a10+a11+a12)
Now, if a8=a9=a10=a11=a12=20
a13=125−5(20)=25
if a8=a9=a10=a11=a12=21
a13=125−5(21)=20, which cannot be the case as a13 has to be the maximum value of all
What if we take a8=a9=a10=a11=21 and a12=20
a13=125−4(21)−20=21, which again is not possible
Let us take a8=a9=a10=21 and a11=a12=20
a13=125−3(21)−2(20)=22, POSSIBLE
Let us take a8=a9=21 and a10=a11=a12=20
a13=125−2(21)−3(20)=23, the value starts increasing
Col. A: 22
Col. B: 23
Hence, option B
Alternatively, we can also divide 125/6 to get 20.833 as this will be the minimum value. Plug and check can be time consuming