for excercise Q1: plugging in values of x as 0,-1,-2,-3,-4,-5,-6,-7,-8,-9,-10 we get 1,3,5,7,9,11,9,7,5,3,1 as values of y.Total values of y=61
I would like to thank u a lot for the effort you are taking in each post. These posts have truly helped me in many ways for which i am grateful. I am having my CAT on Nov.1 and hope the methods taught here will come in more handy( as you have already saved us with the ant question ). 🙂
Jan, glad it is helping, and hope that on your D-day you will find many questions which you can solve confidently 🙂
sir exercise 1 ka 20 arah… 😐
i think inequality mein 61 hna chahiye ans <=5..maybe..havnt tried,equality k lie toh 20 hna chahiye.
Yes, should have been “enclosed by” only, Jaspreet. I have edited it now!
Sir after getting the range like x=2,3,4,5,6 and y=0,1,2,3,4 do we have to check after plugging each and every value or i there any quicker method in this to obtain. I mean do we need to form all the combination at hand to get the result.
Harsh, look for patterns. Here if I see 1, 3, 5, 7…. samajhdar ko ishara kaafi hai na?
Is the answer for integer co-ordinates satisfying |x| + |y| = 5, 50?
I make it 61, Harsh. You have possibly missed the case where x = 0 (which gives 11 values for y)
Yes, I missed that one.
As you have pointed out in the post that the only difference between them is their centres, so I was checking the solutions….
So, effectively, in the test, can we solve for |x| + |y| = 5, if the question is |x+5| + |y-3| = 5? ;)(Integer co-ordinates)
Yes we can.
What goes wrong if we were to plot the graphs for the 2 variables, |x-4| and |y-2|, separately, in a way similar to the previous post on modulus addition? Thank You! These topics have been of great of help.
The problem is, how would you do that? Normally if you are plotting just |x-4| it means you are plotting y = |x-4| and so here you would be trying to do two separate things, both in x and y. Confusing. Here both x and y are interacting together and we can’t separate them out that easily since both are within modulus signs. So trial and error is simplest.
Here, just draw the graph of |x|+|y|=5 and then shift the graph according to the given point and we know the graph of this given mod is simple rhombus .Using the concept of 90-45-45 find its one of the side which will be equal to the 5sqrt2=50.
5sqrt2 is the side of rhombus.So area =side square=50
Hi J , As reasoning mentioned in the question above is it safe to infer that the no of points having integral coordinates for |x|+|y|=2 will be same as |x+a|+|y+b|=2
As long as “a” and “b” are integers, yes 🙂
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