The mean of 8 numbers is 42. If 1 number is excluded from their mean it becomes 39.what will be the excluded number?
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Number is63
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8 numbers say 'a,b,c,d,e,f,g,h'
their
1st case :
mean = (a+b+c+d+e+f+g+h)/8 = 42
Transpose 8 to RHS.
then it becomes
a+b+c+d+e+f+g = 42*8 = 336 - [eq1]
2nd case :
(a+b+c+d+e+f+g)/7 =39
(because h is excluded)
Transpose 7 to RHS.
then it becomes
a+b+c+d+e+f+g = 39*7 = 273 - [eq2]
Solve [eq1 and 2], we get:
Subtract [eq2 from 1];
a,b,c,d,e,f,g will be cancelled as in the above picture...
h then comes = 63
Therefore the excluded number is 63.
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Hope you mark it as brainliest !
their
1st case :
mean = (a+b+c+d+e+f+g+h)/8 = 42
Transpose 8 to RHS.
then it becomes
a+b+c+d+e+f+g = 42*8 = 336 - [eq1]
2nd case :
(a+b+c+d+e+f+g)/7 =39
(because h is excluded)
Transpose 7 to RHS.
then it becomes
a+b+c+d+e+f+g = 39*7 = 273 - [eq2]
Solve [eq1 and 2], we get:
Subtract [eq2 from 1];
a,b,c,d,e,f,g will be cancelled as in the above picture...
h then comes = 63
Therefore the excluded number is 63.
Hope you find my answer useful
Hope you mark it as brainliest !
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GeniusYH:
this is very detailed answer
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