Question

a,b,c,d and e are five numbers such that their average is 42. the average of a,b,c is 23.d is more than e by 5. what is the value of e?​

Answer 1

Answer:

pls mark as brainliest

Step-by-step explanation:

a, b, c, d, and e are five numbers such that a≤b≤c≤d≤e and e-c=4. A is the average (arithmetic mean) of the five numbers, and M is their median. Is A>M?

(1) e+c=34

(2) c=a+10

So I actually understand the solution provided but it seemed a little un-intuitive, was hoping someone here could discover some sort of new insight about this problem. Basically, the solution suggests plugging but I wonder if there is something nicer...

Answer 2

Step-by-step explanation:

given a+b+c+d+e/2=42

a+b+c+d+e=42×2=84......(1)

now,

a+b+c/2=23×d

a+b+c=23×d×2=46×d.... (2)

now, 46×d =e +5

46d-5=e.....(3)

now putting the value of eq(2) In eq(1)

a+b+c+d+e=84

46d+d+e=84

47d+46d-5=84

d=84+5

d=89

now,

putting the value in (3)

46×89-5=e

4089=e