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04 Dec 2024, 01:30
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Question Stats:
65% (02:33) correct
35%
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Three numbers, p, q and r, are such that the HCF of any two numbers out of the three numbers is 5. If p + q + r = 60, what is the maximum possible value of the product p x q x r?
A) 4,800
B) 6,000
C) 7,500
D) 9,000
E) 12,000
A) 4,800
B) 6,000
C) 7,500
D) 9,000
E) 12,000
04 Dec 2024, 07:28
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HCF of \(p, q, r\) is \(5\). We need to maximize the value of \(p \times q\times r\). The sum of \(p + q + r = 60\).
As HCF of \(p, q, r\) is \(5\), the numbers are multiple of \(5\). Therefore, \(5\times (p +q+r) = 60 \Longrightarrow (p+q+r) = 12\)
Hence, the possible combinations of \(p, q, r\) are as follows:
\(4, 4, 4\quad P_{\max} = (4\times 4\times 4) \times (5^3) = 64\times 125 = 8000 \), whereas \(P_{\max} =\) Product max
\(3, 4, 5 \quad P_{\max} = (3\times 4\times 5) \times (5^3) = 60 \times 125 = 7500 \)
\(3, 3, 6 \quad P_{\max} = (3\times 3\times 6) \times (5^3) = 54\times 125 = 6750\)
\(2, 5, 5 \quad P_{\max} = (2\times 5\times 5) \times (5^3) = 50\times 125 = 6250\)
\(\vdots\)
\(1, 1, 12 \quad P_{\max} = (1\times 1\times 12) \times (5^3) = 12\times 125 = 1500\)
ANS C
As HCF of \(p, q, r\) is \(5\), the numbers are multiple of \(5\). Therefore, \(5\times (p +q+r) = 60 \Longrightarrow (p+q+r) = 12\)
Hence, the possible combinations of \(p, q, r\) are as follows:
\(4, 4, 4\quad P_{\max} = (4\times 4\times 4) \times (5^3) = 64\times 125 = 8000 \), whereas \(P_{\max} =\) Product max
\(3, 4, 5 \quad P_{\max} = (3\times 4\times 5) \times (5^3) = 60 \times 125 = 7500 \)
\(3, 3, 6 \quad P_{\max} = (3\times 3\times 6) \times (5^3) = 54\times 125 = 6750\)
\(2, 5, 5 \quad P_{\max} = (2\times 5\times 5) \times (5^3) = 50\times 125 = 6250\)
\(\vdots\)
\(1, 1, 12 \quad P_{\max} = (1\times 1\times 12) \times (5^3) = 12\times 125 = 1500\)
ANS C
04 Dec 2024, 09:34
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HCF of p, q, r is 5 & (p+q+r=60)
P, q, r are all multiples of 5. Hence, 5(p+q+r)=60
p+q+r=12
To maximise the value of p*q*r we take maximum values of p, q, r possible (4,4,4) or (3,4,5)
P*q*r= (5*4)*(5*4)*(5*4)= (20)^3 = 8,000 Not given in options
P*q*r= (5*3)*(5*4)*(5*5)= (15)*(20)*(25)= 7,500
Answer: C
P, q, r are all multiples of 5. Hence, 5(p+q+r)=60
p+q+r=12
To maximise the value of p*q*r we take maximum values of p, q, r possible (4,4,4) or (3,4,5)
P*q*r= (5*4)*(5*4)*(5*4)= (20)^3 = 8,000 Not given in options
P*q*r= (5*3)*(5*4)*(5*5)= (15)*(20)*(25)= 7,500
Answer: C
