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There are two numbers such that the first number is four times of the second number. If their average is 65, then what is the second number?
Let's assume the first and second numbers are 'y' and 'z' respectively.
There are two numbers such that the first number is four times of the second number.
y = 4z Eq.(i)
If their average is 65.
$$\frac{y+z}{2}\ =\ 65$$
y+z = 130 Eq.(ii)
Put Eq.(i) in Eq.(ii).
4z+z = 130
5z = 130
z = 26
The second number = 26
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