Hi Harsh Pandey,
We assume the numbers to be in ascending order, as given below-
a<b<c<d<e
We can definitely say that a+b is the minimum and d+e is the maximum.
Now, for the second highest weight, it is c+e or either c+d; we know that c is common in both, and e > d, then with certainty, we can say that c + e is the second highest, and we can't be sure of the third highest number, it can be c+d or b+e, as c>b but e>d.
So, we only form equations
a+b = 110 and d+e = 121
We get c = 58 by substituting in a+b+c+d+e = 289
We are certain of the second highest number, that is c+e = 120
c = 58,  Hence, e = 62
I hope this helps!

Hi Rishabh,

Thanks for pointing it out. We have updated it now.

Hi,

e>d. Adding a postive value c to both sides c+e > c+d.

Thank you.

e+d is greatest and when you compare d+c and e+c , e is greater than d and c is common so e+c will always be greater than d+c , therefore e+c is the 2nd largest term.

Hi Phinehas Prakash Jupudi,
There seems to be a formatting issue in the doubt that you posted. Could you post it again so I can understand your doubts and clarify?
Regards.

can you explain me this question step by step,i did not understand the video

Hi Phinehas Prakash Jupudi,
Let us assume the five weights are a, b, c, d, and e in increasing order. a < b < c < d < e
We are asked to find the value of 'e'.
The given 10 values represent values of a + b, a + c, a + d, a + e, b + c, b + d, b + e, c + d, c + e, d + e in some order.
We can definitely say that a+b is the minimum and d+e is the maximum.
Now, for the second highest weight, it is c+e or either c+d; we know that c is common in both and e > d. So, with certainty, we can say that c + e is the second-highest. We can't be sure of the third-highest number; it can be c+d or b+e, as c>b but e>d.
So, we only form equations a+b = 110 and d+e = 121
We get c = 58 by substituting in a+b+c+d+e = 289
We are certain of the second highest number, that is c+e = 120 => c = 58; hence, e = 62
I hope this helps.

Hi Harsh Sahu,
How did you get C+D = 118? That is equation is incorrect. Please look into that once.
I hope this helps!!

Hi Yash,

Assuming a1<a2<a3<… if you put a5= 61 in the respective equations you’ll get a4=121-61=60, and then a3=118-60=58 and a2=115-58=57, but given that a2+a5=117 in your equations, 61+57=118 shouldn’t be possible. You’ve made the equations incorrectly.

Hope this helps, feel free to ask if you have any doubts