Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 16 Jul 2019, 19:57 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # M26-18

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 56244

### Show Tags

2
10 00:00

Difficulty:   75% (hard)

Question Stats: 50% (01:01) correct 50% (00:56) wrong based on 313 sessions

### HideShow timer Statistics 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. $$-\frac{1}{6}$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 56244

### Show Tags

1
1
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).

_________________
Manager  Joined: 02 Jul 2012
Posts: 186
Location: India
Schools: IIMC (A)
GMAT 1: 720 Q50 V38 GPA: 2.6
WE: Information Technology (Consulting)

### Show Tags

1
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...       _________________
Give KUDOS if the post helps you... Current Student B
Joined: 29 Nov 2014
Posts: 22

### Show Tags

I think this is a high-quality 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
Math Expert V
Joined: 02 Sep 2009
Posts: 56244

### Show Tags

1
sashankmv wrote:
I think this is a high-quality 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

No. -1/100 > -1/18.
>> !!!

You do not have the required permissions to view the files attached to this post.

_________________
Intern  Joined: 27 Feb 2015
Posts: 43
Concentration: General Management, Economics
GMAT 1: 630 Q42 V34 WE: Engineering (Transportation)

### Show Tags

Bunuel
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
Math Expert V
Joined: 02 Sep 2009
Posts: 56244

### Show Tags

deepak268 wrote:
Bunuel
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

-1/18, which the correct answer is less than -1/50: -1/18<-1/50.
_________________
Manager  B
Joined: 23 Apr 2014
Posts: 57
Location: United States
GMAT 1: 680 Q50 V31 GPA: 2.75

### Show Tags

I think this is a high-quality question and I agree with explanation. High quality question.
Manager  B
Status: Bouncing back from failure
Joined: 08 Mar 2010
Posts: 66
Schools: Wharton,MIT, Tepper, Kelly,
WE 1: 7 years- Service Managament, poject Management, Business Consultant- Retail

### Show Tags

This is indeed easy yet tricky question testing concepts. Thanks for this question
Intern  Joined: 22 Dec 2015
Posts: 5

### Show Tags

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
Math Expert V
Joined: 02 Sep 2009
Posts: 56244

### Show Tags

federig wrote:
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

x cannot be $$-\frac{1}{\sqrt{3}} \approx -0.6$$ because it's not in the range ($${-\frac{1}{3}} \le {x} \le {-\frac{1}{5}}$$).
_________________
Intern  B
Joined: 26 Sep 2015
Posts: 8

### Show Tags

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"
Manager  S
Joined: 27 Aug 2014
Posts: 54
Concentration: Strategy, Technology
GMAT 1: 660 Q45 V35 GPA: 3.66
WE: Consulting (Consulting)

### Show Tags

This is a great quality question.

Fell for the trap and selected -1/100.

Will keep that in mind.
Intern  B
Joined: 26 May 2017
Posts: 25
GMAT 1: 620 Q48 V27 ### Show Tags

1
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

Manager  B
Joined: 31 Oct 2016
Posts: 106

### Show Tags

1
I rephrased this question:
-3>=x>=-5 and -2>=y>=-4
What is the greatest value possible?

Obviously, it is -18
Senior Manager  P
Joined: 17 Mar 2014
Posts: 440

### Show Tags

Fell for trap and selected -1/50
Manager  B
Joined: 16 Jan 2018
Posts: 62
Concentration: Finance, Technology
GMAT 1: 600 Q40 V33 ### Show Tags

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
Director  D
Joined: 08 Jun 2013
Posts: 560
Location: France
GMAT 1: 200 Q1 V1 GPA: 3.82
WE: Consulting (Other)

### Show Tags

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)
_________________
Everything will fall into place…

There is perfect timing for
everything and everyone.
Never doubt, But Work on
improving yourself,
Keep the faith and
It will all make sense.
Intern  B
Joined: 12 Jan 2017
Posts: 34
Location: United States (NY)
Schools: Booth '21 (D)
GMAT 1: 710 Q47 V41 GPA: 3.48

### Show Tags

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).

This is very much NOT a hard math question as it is a hard english question. Re: M26-18   [#permalink] 29 Aug 2018, 15:01
Display posts from previous: Sort by

# M26-18

Moderators: chetan2u, Bunuel  