If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?

Question

If x^2+y^2=1 and ||x|-0.34|+0.66=y , than how many values can x take? A. 0 B. 1 C. 2 D. 3 E. 4 ______ Smth similar I met in my GMAT exam, thus make sure you are familiar with such questions

Philipp98 · Accepted Answer

Such tasks are easy to solve graphically: solutions to the system of equation are points of intersection of graphs of each equation.

Remember:  When a module sign is "layed" on F(x), graphical part of F(x), which lays below the X-axis, is reflected over the X-axis. This operation gives you the graph of |F(x)|.

Thus, the given task is solved with consequent graphics:

https://i.paste.pics/6ijvl.png

As already discussed, to obtain ||x|-0.34| graph, we reflect "negative" part of |x|-0.34 over X-axis. Then shift ||x|-0.34| by 0.66 upwards to obtain ||x|-0.34|+0.66. Than we draw the plot of x^2+y^2=1

Finally, one can see 3 points of intersection -> 3 solutions to the system -> x can take 3 values -> answer is D

nick1816 · Answer

1 Kudos for an excellent question.
IMO graphical approach can save lot of time in this question.

We just need to figure out whether point of  minima  of ||x|-0.34|+0.66=y lies within the circle or not, and whether  local maxima  at x=0 lies within circle or not.
 x^2+y^2=1  is a curve of a circle of radius 1.

At x=0, ||x|-0.34|+0.66= 1, which lies on the circle.

Minima of  ||x|-0.34|+0.66=y  occurs when |x|-0.34=0 or x=+0.34 and -0.34, and y=0.66, lie inside the circle.

Now we are good to draw the curve.

There are 3 points where both curves meet.

D

vanam52923 · Answer

nick1816 
 Bunuel 
What is wrong with the given solution in the image:
here x=0 or 1 0r -1

but i consider case last one in pic (from d)
here i get x=.84 and y =-.52
then also x^2+y^2=1 then y are we nopt considering such cases

nick1816 · Answer

There are couple of problems with this approach for this particular question.
1. This approach can take a lot of time. Whenever you see second order equations, go for the graphical approach.

2. You didn't find the domain for x, for each equation.
For example- 
||x|-0.34|+0.66=y
y=x-0.34+0.66, when  x≥0.34 
y=0.34-x+0.66, when   0.34 ≥x ≥0  
y=0.34+x+0.66, when  0≥x≥-0.34 
y=0.34-x+0.66, when  x≤-0.34

Now you can solve each case, and look whether the solution lies in the domain or not.

Kinshook · Answer

nick1816   VeritasKarishma   Bunuel   chondro48  Is there an algebra way to solve this question?

freedom128 · Answer

Hi Kinshook nick1816 VeritasKarishma and Bunuel , Yes, there is. Below is a framework to do so. Correct me, experts, if I make mistake. Note that using algebra, the solution is laborious and time consuming. Transform x^2=1-y^2 to |x|=√(1-y^2). Substituting |x| with √(1-y^2) into the second equation, we get: |√(1-y^2)-0.34|=y-0.66. If √(1-y^2) > 0.34, then √(1-y^2)-0.34 = y-0.66 --> √(1-y^2)= y-0.34. After squaring both sides and performing tedious calculation, we will find that there are two solution points available when √(1-y^2) > 0.34. If √(1-y^2) <= 0.34, then 0.34-√(1-y^2) = y-0.66 --> √(1-y^2)=1-y. It is obvious that y=1 satisfies the equation and correspondingly |x|=√(1-y^2)=0. Thus, there is one solution point (0,1) when √(1-y^2) <= 0.34. Finally, there are 3 values that x can take. Final answer is (D)

KarishmaB · Answer

You can solve it using the usual method of definition of absolute values (as done by others here) but if you are unable to solve it graphically, it is probably a good idea to skip the question. With algebra, it will be far too time consuming and error prone.

GMATaspirant641 · Answer

nick1816

The explanation given by you to find out the value of y at the equation's minima is crystal clear. We have arrived at y = 0.66 at minima i.e x = +/- 0.34.

Now, how do we figure if y = 0.66 falls inside the circle?

Ofcourse we can substitute x = 0.34 in the circle equation to find out the value of y. And if the value of y > 0.66, we can deduce that 0.66 is inside the circle. However, this is a time consuming process that requires squaring and square-roots

Is there are short cut to find if 0.66 falls inside the circle?

VeritasKarishma   Bunuel   Philipp98  - Please help me with this.

nick1816 · Answer

0.34≃1/3
0.66≃2/3
 (\frac{1}{3})^2+(\frac{2}{3})^2 =  \frac{1}{9}+\frac{4}{9} =  \frac{5}{9} <1

KarishmaB · Answer

Check the solution given by Philipp98 above. Notice how the graph of ||x| - 0.34| + 0.66 = y is drawn. Also note that each green line has slope 1 or -1. Since the graph intersects the circle at (0, 1), the line of slope 1 will be inside the circle (45 degrees line). Hence you know that it will intersect the circle in 3 points. Check out these two posts on how to draw absolute value graphs: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... h-to-mods/ https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/1 ... solutions/

Thonereturns · Answer

We just need to figure out whether point of minima of ||x|-0.34|+0.66=y lies within the circle or not, and whether local maxima at x=0 lies within circle or not.
x2+y2=1x2+y2=1 is a curve of a circle of radius 1.

At x=0, ||x|-0.34|+0.66= 1, which lies on the circle.

Minima of ||x|−0.34|+0.66=y||x|−0.34|+0.66=y occurs when |x|-0.34=0 or x=+0.34 and -0.34, and y=0.66, lie inside the circle.

Now we are good to draw the curve.

There are 3 points where both curves meet.

D

Thonereturns · Answer

Transform x2=1−y2x2=1−y2 to |x|=√(1-y^2). Substituting |x| with √(1-y^2) into the second equation, we get:
|√(1-y^2)-0.34|=y-0.66.

If √(1-y^2) > 0.34, then √(1-y^2)-0.34 = y-0.66 --> √(1-y^2)= y-0.34. After squaring both sides and performing tedious calculation, we will find that there are two solution points available when √(1-y^2) > 0.34.

If √(1-y^2) <= 0.34, then 0.34-√(1-y^2) = y-0.66 --> √(1-y^2)=1-y. It is obvious that y=1 satisfies the equation and correspondingly |x|=√(1-y^2)=0. Thus, there is one solution point (0,1) when √(1-y^2) <= 0.34.

Finally, there are 3 values that x can take.

Akashhhhh · Answer

can anyone share an algebraic way to solve this?
 GMATNinja  - I came across this problem while using the 13 week study plan that you have posted. v grateful for any support here.
 Bunuel  - long time fan and learner from your post. any guidance you can provide here?

alternatively, can someone suggest a quick way to identify whether an inequality problem needs to be solved algebraically or graphically? atleast I'd save the time I spent solving algebraically that I did today.

arushi118 · Answer

It is better to not solve this in an algebraic way. And when you have questions like "how many solutions to ..." -> where we have to find solutions to an equation which has nested absolute values -> in those questions it is easier to draw a graph and then check.

Kinshook · Answer

If x^2+y^2=1 and ||x|-0.34|+0.66=y , than how many values can x take? Since x^2+y^2=1 -1<=x<=1 & -1<=y<=1 ||x|-0.34|+0.66=y ||x|-0.34| = y - .66>=0 y>=.66 Case 1: |x|=0 or x = 0 y = .34 + .66 = 1 Case 2: |x| >= .34 |x| - .34 = y - .66 |x| = y - .32 Combining with x^2+y^2=1 (y- .32)^2 + y^2 = 1 2y^2 - .64y + .32^2 - 1 = 0 2y^2 - .64y - .8976 = 0 y = .8487 |x| = .8487 - .32 = .5287 = .53 approx x = .53 & - .53 Case 3: |x| <.34 .34 - |x| = y - .66 |x| = 1 - y (1-y)^2 + y^ 2 = 1 2y^2 - 2y = 0 2y(y-1) = 0 y = 0 or y = 1 y = 0 is not possible since y>=.66; y = 1 is already a solution when x = 0 x = {-.53,0,.53} : 3 values IMO D

If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?

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GMAT Problem Solving (PS) Questions

If x^2+y^2=1 and ||x|-0.34|+0.66=y, how many values can x take?

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615 to 715 in 15 Days (Ria's Strategies)

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