Question
Download Solution PDFThe average of the three positive integers p, q, and r is 10. If it is given that p ≤ q ≤ r and the median of three numbers is p + 2, then what is the smallest possible value of r?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
Given: the average of the three positive integers p, q, and r is 10
So, p + q + r = 30 ....(1)
Given that p ≤ q ≤ r and the median of three numbers is p + 2
So, q = p + 2
Now, p + q + r = 30
⇒ p + p + 2 + r = 30
⇒ r = 28 - 2p
q ≤ r
⇒ p + 2 ≤ 28 - 2p
⇒ 3p ≤ 26
⇒ p ≤ 8 (since it must be an integer)
Maximum value of p = 8
Maximum value of q = p + 2 = 8 + 2 = 10
Put the value of p and q in equation (1), we get
Minimum value of r = 30 - 10 - 8 = 12
So the maximum values of p & q are 8 and 10, making the minimum value of r is 12.
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