Question 75
There are two numbers such that the first number is three times of the second number. If the average of these two numbers is 72, then what is the first number?
Solution
Let's assume the second number is 'y'.
There are two numbers such that the first number is three times of the second number.
first number = 3y
If the average of these two numbers is 72.
$$\frac{3y+y}{2}\ =\ 72$$
$$\frac{4y}{2}\ =\ 72$$
2y = 72
y = 36
First number = 3y
= $$3\times36$$
= 108
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