# Find value of x^2 + 3xyz + 2y^ + 3z^2 where x = Mean, y = Median and z = Mode of: 83 87 93 97 110 48 114 119 87 118 122

+1 vote
63 views

Find the value of x^2 + 3xyz + 2y^ + 3z^2 where x = Mean , y = Median and z = Mode of following numbers:
83 87 93 97 110 48 114 119 87 118 122

posted Sep 7, 2018

2532195

Assumed it reads x^2 + 3xyz + 2y^2 + 3z^2 (<<2y^2>>)
122, 119, 118, 114, 110, 97, 93, 87, 87, 83, 48
Mean = sum/count= 98
Median= middle term= 97
Mode= number given maximum number of times= 87 (appears 2 times)
x^2 + 3xyz + 2y^2 + 3z^2=98^2+3*98*97*87+2*97^2+3*87^2=2532195

answer Sep 7, 2018

Similar Puzzles

Find the value of 2x^2 + 4zy + 3z^2.
Where
x = mean, y= median and z= mode of following number
77 104 91 97
131 48 114
119 91 117 122

Find the value of x*y*2*z/5
Where x = mean, y = median and z= mode of following number
57 59
56 81 69
55 64
63 81

Find the product of half of mean, twice of median and three times of mode (i.e. you need to find (1/2)*mean* 2*median * 3*mode) for the following list of values?
67 17 37
27 27 87
17 57
47 17