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Re M2618
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16 Sep 2014, 00:25
Official Solution:If \({\frac{1}{3}} \le {x} \le {\frac{1}{5}}\) and \({\frac{1}{2}} \le {y} \le {\frac{1}{4}}\), what is the least value of \(x^2*y\) possible? A. \(\frac{1}{100}\) B. \(\frac{1}{50}\) C. \(\frac{1}{36}\) D. \(\frac{1}{18}\) E. \(\frac{1}{6}\) To get the least value of \(x^2*y\), which obviously will be negative, try to maximize absolute value of \(x^2*y\), as the more is the absolute value of a negative number, the "more" negative it is (the smaller it is). To maximize \(x^2*y\) pick largest absolute values possible for \(x\) and \(y\): \((\frac{1}{3})^2*(\frac{1}{2})=\frac{1}{18}\). Notice that: \(\frac{1}{18} \lt \frac{1}{36} \lt \frac{1}{50} \lt \frac{1}{100}\), so \(\frac{1}{100}\) is the largest number and \(\frac{1}{18}\) is the smallest number (we cannot obtain \(\frac{1}{6}\) from \(x^2*y\) or else it would be the correct answer). Answer: D
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Re: M2618
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23 Oct 2014, 03:37
Bunuel wrote: If \({\frac{1}{3}} \le {x} \le {\frac{1}{5}}\) and \({\frac{1}{2}} \le {y} \le {\frac{1}{4}}\), what is the least value of \(x^2*y\) possible?
A. \(\frac{1}{100}\) B. \(\frac{1}{50}\) C. \(\frac{1}{36}\) D. \(\frac{1}{18}\) E. \(\frac{1}{6}\) I got this question wrong in the test, because of a silly mistake, but now when I look at my mistake, I laugh. Anyways, here's the explanation If the value has to be least, possible, the value must be negative and the value must have the highest modulus. We know that the expression \(x^2*y\) will give a negative value for negative y, now our aim is to keep the modulus maximum. so, \(x=1/3\) and \(y =1/2\). Thus the lowest value will be \(1/18\) Mistakenly, I calculated the value for the minimum modulus and got it wrong earlier. Happy to learn...
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Re M2618
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07 Nov 2015, 03:16
I think this is a highquality question and I don't agree with the explanation. Isn't 1/100 lesser in value than 1/18? And 1/100 should be the answer as it could be obtained by using x=1/5 and y=1/4



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07 Nov 2015, 04:58



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07 Jul 2016, 06:06
Bunuelplease help me understand this If −1/3≤x≤−1/5 and −1/2≤y≤−1/4 then 1/25≤x^2 ≤1/9 for least value we take x^2 = 1/25 and y= 1/2 therefore x^2* y = 1/50 why do we need to consider modulus and all?? thanks



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07 Jul 2016, 06:20



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17 Jul 2016, 09:36
I think this is a highquality question and I agree with explanation. High quality question.



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29 Jul 2016, 23:48
This is indeed easy yet tricky question testing concepts. Thanks for this question



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18 Aug 2016, 07:13
aren't we also able to get (1/6) ? > x=(1/sqrt(3)) y=1/2 > (1/sqrt(3))^2*(1/2) = (1/3)(1/2) = 1/6



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19 Aug 2016, 01:51



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20 Aug 2016, 20:42
It's a pretty straightforward question 
Since the question is to find out the least possible value of an equation wherein two variables are in multiplication form, it's quite evident that we need to take the least possible values of x and y and put them into (x^2)*y in order to get the least possible value for the equation......
From the inequality, we are able to see that least value for x and y is (1/3) and (1/2) respectively.
So the answer is  ((1/3)^2)*(1/2) = (1/9)*(1/2) = 1/18 ...So the answer is indeed "D"



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Re: M2618
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26 Apr 2017, 21:44
This is a great quality question.
Fell for the trap and selected 1/100.
Will keep that in mind.



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Joined: 25 May 2017
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Using the approach of number ranges to solve the question. Two points to be noted when using range approach.
 both multiplying ranges should either be +ve or ve  when multiplying one range with a negative number, inequality signs are reversed
Given,
1/3 < x < 1/5 > 1/9 > x2 > 1/25
1/2 < y < 1/4 > 1/2 > y > 1/4
Now since the range of both x2 and y are positive we can multiply the ranges 1/9 > x2 > 1/25 and 1/2 > y > 1/4 > 1/18 > x2y > 1/100
Multiply by 1 and change signs > 1/18 < x2y < 1/100
Hence the smallest value of x2y is 1/18
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Re: M2618
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25 Oct 2017, 05:00
I rephrased this question: 3>=x>=5 and 2>=y>=4 What is the greatest value possible?
Obviously, it is 18



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30 Dec 2017, 20:00
Fell for trap and selected 1/50



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22 Jan 2018, 18:09
Fell for trap.. Bt on second attempt quickly changed the fraction to decimal which seemed to help me, who'd think that decimal will help! −0.3≤x≤−0.2 −0.5≤y≤−0.25 x^2*y has to ve. so, smallest y is 0.5 . x^2 has to be maximum positive (for the product to be smallest ) which is 0.09 replacing fraction, (1/3) ^2*(1/2)=1/9*1/2=1/18



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Re: M2618
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10 Aug 2018, 21:10
Bunuel wrote: If \({\frac{1}{3}} \le {x} \le {\frac{1}{5}}\) and \({\frac{1}{2}} \le {y} \le {\frac{1}{4}}\), what is the least value of \(x^2*y\) possible?
A. \(\frac{1}{100}\) B. \(\frac{1}{50}\) C. \(\frac{1}{36}\) D. \(\frac{1}{18}\) E. \(\frac{1}{6}\) least value of X^2*Y i.e. max value of 1/ (X^2*Y) so 5 <= (1/x) <= 3 and 4 <= (1/y) <= 2 so max (1/ (X^2*Y)) = least negative number possible = (3)^2 * (2) = 18 so ans required = (1/18)
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Re: M2618
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29 Aug 2018, 14:01
Bunuel wrote: Official Solution:
If \({\frac{1}{3}} \le {x} \le {\frac{1}{5}}\) and \({\frac{1}{2}} \le {y} \le {\frac{1}{4}}\), what is the least value of \(x^2*y\) possible?
A. \(\frac{1}{100}\) B. \(\frac{1}{50}\) C. \(\frac{1}{36}\) D. \(\frac{1}{18}\) E. \(\frac{1}{6}\)
To get the least value of \(x^2*y\), which obviously will be negative, try to maximize absolute value of \(x^2*y\), as the more is the absolute value of a negative number, the "more" negative it is (the smaller it is). To maximize \(x^2*y\) pick largest absolute values possible for \(x\) and \(y\): \((\frac{1}{3})^2*(\frac{1}{2})=\frac{1}{18}\). Notice that: \(\frac{1}{18} \lt \frac{1}{36} \lt \frac{1}{50} \lt \frac{1}{100}\), so \(\frac{1}{100}\) is the largest number and \(\frac{1}{18}\) is the smallest number (we cannot obtain \(\frac{1}{6}\) from \(x^2*y\) or else it would be the correct answer).
Answer: D This is very much NOT a hard math question as it is a hard english question.










