So plotting them will give a triangle of base -3 to +3 and height 0 to 3 and a rectangle of sides -3 to 3 and 0 to -3. By this iam getting area as 42 i.e 7 * 4 + 0.5 * 7 * 4. What am i doing wrong here?
for x = -2, we can also get y= -4, as:
-2 -y -(2+y) = 4,
which gives -2y = 8,
y= -4
because when value from modulus is considered, it will be negative? this way for one value, many values of y can come.. whats wrong in it?
Among other things, you have taken modulus of -2 as -2.
regards
J
ohh.. ok, actually i intentionally took it because i thought negatives will be taken as it is, like we do in modulus , if |x| <= 5, then we take -5<x<5, like that.. anyways,, thanks
sir, i am having hard time for getting the range for values here…like for x=2, i am getting 2y=0 so y=0 correct…then what should be the next step to get y=-2…if i open the mod as -2-y-y-2=4 then i am getting y=-4…what step i am doing wrong here…?
Thanks again for the wonderful post.
I have one query with regards to finding the range for x and y.
I have understood that the basic concept you are using for the same is making sure mod of x or y should not be equal to a negative number, that’s where you stop.
Having said that for the equation lxl+lyl+lx+yl=4 when I sub x=3 I get the values of y=-1,-2.
Now just looking at the values of y I am not able to say if it as acceptable value or not.
I did substitute back to the main equation and find out that it was not satisfying the equation hence I could discard this value.
Is this the way you are following too?
Could you please help me on this.
Not exactly. In lxl+lyl+lx+yl=4 when I put x=3 I get l3l+lyl+l3+yl=4 which is |y| + |y+3| = 1 which gives no values (the sum of the distances of y from 0 and 3 should add up to 1, which is not even possible!). But yeah, it is always safer to check your answer by plugging back in, if time permits.
regards
J
sir like for x=2 first i am doing putting |x| as x and getting y=0 as result then putting |x| as -x …for this i am getting y=-4 but the range is from -2 to 0 …i know i am making a very silly mistake but please can you point me to it…
The RHS is a number here as well? So I have no idea what you want to ask I’m afraid. (If you mean what if it is 20 instead of 4, same logic, once you know how it works for 4 you should easily be able to extrapolate to 20)
CAT-holics 
Dear J,
Is the no: of integer points = 37? i.e X varies from -3 to 3 right? And the is the figure in the form of a triangle seated on top of a rectangle?
Thanks,
Jan
Jan, the number of points does come 37 if I remember correctly. 4 + 5 + 6 + 7 + 6 + 5 + 4.
regards
J
Dear J,
So plotting them will give a triangle of base -3 to +3 and height 0 to 3 and a rectangle of sides -3 to 3 and 0 to -3. By this iam getting area as 42 i.e 7 * 4 + 0.5 * 7 * 4. What am i doing wrong here?
Thanks,
Jan
The line segment from -3 to 3 has length 6, not 7….think about it. 7 points, but length 6.
regards
J
Thanks 🙂 got it now 🙂
YES TOTAL 37 POINTS
if we solve |x| + |y| + |x+y| = 4, and put x=23, we will get y=-21, how can x and y be only -2 to 2
No, you will get |y| = the negative value. Not y. Which is absurd, right?…y can be -21, but |y| cannot.
regards
J
can you please explain how did you got +/- 2 for x and y for |x| + |y| + |x+y| = 4?
The same way. Put x = 0, find limits of y, put x = 1, find limits of y, etc. Think!
regards
J
for x = -2, we can also get y= -4, as:
-2 -y -(2+y) = 4,
which gives -2y = 8,
y= -4
because when value from modulus is considered, it will be negative? this way for one value, many values of y can come.. whats wrong in it?
one correction here* for x= -2, we have -2 -y -(-2 +y)= 4, which will give, y= -2, but you only got 0 to 2, how come?
Among other things, you have taken modulus of -2 as -2.
regards
J
ohh.. ok, actually i intentionally took it because i thought negatives will be taken as it is, like we do in modulus , if |x| <= 5, then we take -5<x<5, like that.. anyways,, thanks
sir, i am having hard time for getting the range for values here…like for x=2, i am getting 2y=0 so y=0 correct…then what should be the next step to get y=-2…if i open the mod as -2-y-y-2=4 then i am getting y=-4…what step i am doing wrong here…?
Where did you get all the negative signs? Mod of a positive number is not negative 😛 Please revisit basics…
regards
J
Hi Sir,
Thanks again for the wonderful post.
I have one query with regards to finding the range for x and y.
I have understood that the basic concept you are using for the same is making sure mod of x or y should not be equal to a negative number, that’s where you stop.
Having said that for the equation lxl+lyl+lx+yl=4 when I sub x=3 I get the values of y=-1,-2.
Now just looking at the values of y I am not able to say if it as acceptable value or not.
I did substitute back to the main equation and find out that it was not satisfying the equation hence I could discard this value.
Is this the way you are following too?
Could you please help me on this.
Not exactly. In lxl+lyl+lx+yl=4 when I put x=3 I get l3l+lyl+l3+yl=4 which is |y| + |y+3| = 1 which gives no values (the sum of the distances of y from 0 and 3 should add up to 1, which is not even possible!). But yeah, it is always safer to check your answer by plugging back in, if time permits.
regards
J
sir like for x=2 first i am doing putting |x| as x and getting y=0 as result then putting |x| as -x …for this i am getting y=-4 but the range is from -2 to 0 …i know i am making a very silly mistake but please can you point me to it…
You can’t put |x| as -x unless you know that x is negative, as I said above. You are taking |2| = -2 😦
regards
J
oh…got it now…..thanks a lot for your insight on this….
Can we expect this kind of ques in CAT17 or it is more of a concept builder question?
Well, it is closely based on an actual CAT question. So yes, you can definitely expect something of this level.
regards
J
Sir,
what should be done if RHS is a number (say) 10 or 20 ?
The RHS is a number here as well? So I have no idea what you want to ask I’m afraid. (If you mean what if it is 20 instead of 4, same logic, once you know how it works for 4 you should easily be able to extrapolate to 20)
regards
J