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If the average (arithmetic mean) of p, q and 15 is equal to the averag
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26 Jan 2024, 12:44
The average of p, q, and 15 = \(\frac{{p+q+15}}{3}\)
The average of p, q, 15, and 35 = \(\frac{{p+q+15+35}}{4}\)
Since the averages are equal, \(\frac{{p+q+15}}{3} = \frac{{p+q+15+35}}{4}\)
Cross-multiply to get:
\(4(p+q+15) = 3(p+q+15+35)\)
\(4p+4q+60 = 3p+3q+150\)
\(p+q = 90\)
The average of p+q = \(\frac{90}{2} = 45\)
Also, here's a time saving property: If the average of a set of numbers remains the same after adding a single element of x, the average is equal to x. In this case, the average of {p, q, 15} remained the same after adding 35. Therefore, the average of {p, q, and 15} is 35.