The mean of three numbers is 40. All the three numbers are different natural
numbers. If lowest is 19, what could be highest possible number of remaining two
numbers?
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Let the highest number be y and the middle number be x.
We know that,
Mean=
number of observations
sum of observations
∴ Mean=
3
x+y+z
Given that, Mean =40 and the lowest number =19.
∴40=
3
x+y+19
⇒x+y+19=120
⇒ x+y=120−19=101
⇒ y=101−x
Let the highest number be y and the middle number be x.
We know that,
Mean=
number of observations
sum of observations
∴ Mean=
3
x+y+z
Given that, Mean =40 and the lowest number =19.
∴40=
3
x+y+19
⇒x+y+19=120
⇒ x+y=120−19=101
⇒ y=101−x
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