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03 Apr 2020, 06:34
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E
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Difficulty:
65%
(hard)
Question Stats:
66% (02:59) correct
34%
(02:58)
wrong
based on 185
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While multiplying three numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. What is the minimum possible value of the sum of squares of the other two numbers ?
A. 20
B. 25
C. 30
D. 35
E. 40
Are You Up For the Challenge: 700 Level Questions
A. 20
B. 25
C. 30
D. 35
E. 40
Are You Up For the Challenge: 700 Level Questions
Archit3110
03 Apr 2020, 07:34
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Bunuel
given
73xy-37xy=720
xy=720/36 ; 20
if max value is 37 then x=y to be least = 20 each
but square of values to be min each let x=y=√20 each so sum of squares ; (√20)^2+ (√20)^2 ; 40
IMO E
03 Apr 2020, 07:50
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Bunuel
Set the other two numbers are x and y.
The two products would be \(73*x*y\) and \(37*x*y\), and their difference would be 720 so we have an equation \(73xy - 37xy = 720\).
This gives us \(xy = 720/(73-37) = 720/36 = 20\). Now given \(xy = 20\) we want to minimize \(x^2 + y^2\), to do that we can try a couple of ideas:
(1) x = 20 and y = 1, gives us \(x^2 + y^2 = 401\)
(2) Split the x and y equally, hence let x = y. With xy = 20 we get \(x = y = \sqrt{20}\), then \(x^2+y^2 = 20 + 20 = 40\).
We can try x = 5 and y = 4 as well to see the sum of squares can only get bigger compared to (2) so 40 is as small as it gets.
Ans: E
