There will be 11 points instead of 9.So,total=41
Q3. I am getting only 25 non-negative integer points. The total number of integral solutions that I am finding is 73. Am I missing something?
P.S : Thanks for your questions and detailed approaches 🙂
Thanks guys, I’ve edited the post. Let me know if any further errors – in peak season sometimes the post gets rushed a little 🙂
Can some1 please explain the solution to the last question, how is it different from 9 mod(x)+11 mod(y)=99??
its the same,modulus fnc property : |xy| = |x| |y| , and |11| = 11. inequality holds for addition as follows : |x+y| <= |x| + |y|
please explain the method to solve q.1 and q.2
A lot of tough questions on Coordinate Geometry and Trigonometry have been reported in CAT 2013.Please would you be so kind as to cover those concepts here.I have my CAT on 30th. Or can you suggest some place where I can get a feel of the type of questions that are coming?
Co-ordinate Geometry, most questions are not too different from the type I am covering already in the current posts. If in addition to modulus you know normal line plotting and basic equation of a circle centred at origin, you should be good to go.
Trigo, I will not be able to cover for CAT 13. However, as far as I can see, it is perfectly possible to solve most of the trigo questions from 1st principles if one knows sin-cos-tan values for a few standard angles (0, 30, 45, 60, 90) and are willing to do substitution, answer choice elimination and such jugaad.
@catcracker Can you please tell me the approch for exercise 1 question. Shoul i use trial and error method or is thre any other approach. ??
Saggy, I suggest you look at the previous 3-4 posts – all the required info is there! 🙂 I am somewhat reluctant to give a solution as the purpose of the posts is to make people think about the topic and the idea of the exercises is so that they can apply what they have learnt. I could “feed” you the answers but that way you would know only how to solve that exact question and not any other…I believe there is no substitute for “the hard yards” when it comes to being able to solve CAT questions efficiently.
Sir, In ques. 3 5|x|+7|y|<=35, are we supposed to find the number of integer points in the area encl by co-ordinate axes in first quadrant? because i'm getting 73 -total integral solutions, but i'm not getting 25 non-negative integral solutions. Please give me a li'l hint on this one and the Ist question. Thank you!
Yes, but also including the points ON the coordinate axis. So you could, for example, put x = 0, 1, 2, 3, 4, 5, 6 and 7 and find the corresponding value of y which are non-negative – 6 + 5 + 4 + 3 + 3 + 2 + 1 + 1 = 25.
Hi J! For the 1st question I applied the same technique as the one required in one of your previous post’s finding domain of x problem for sum of 4 mod functions. In this problem too I tried to draw an approx graph by plotting the f(a) at the points of inflection and hence got the final ans too. But just one doubt is troubling me, regarding the slope of the |5-a|. Here the slope of (y=5-a) is -1. But we are providing the individual slopes of the |f(a)|,in the order of x8 as -4, -2 , 0 , +2 , +4 .
Actually the slope of (5-a) is -1 so shouldn’t the slope of -(5-a) (in the x5 region as -1? But if we do that the slopes of |f(a)| become -2,0,+2,0, which is obviously wrong!! Kindly help me see the mistake in this assumption.
Or is it that the slopes of |f(a)| will always take the numerical value of each individual terms?
It’s my earnest request to you to please clear this doubt as understanding this will ensure my complete comprehension of this effective approach. I even tried to resolve this issue myself by re-reading your past posts, but I found that none of the terms had a negative slope as in this! I’m sorry if the ans is too obvious, but do explain. Thank you. 🙂
* Same problem as in PG, less than signs have vanished. So let me re-type the 2 sentences.
But we are providing the individual slopes of the |f(a)|,in the order of x less than -3 to x more than 8, as -4, -2 , 0 , +2 , +4 .
Actually the slope of (5-a) is -1 so shouldn’t the slope of -(5-a) be +1 (in the x less than 5 region) and in the x more than 5 region as -1?
Remember that |x – y| is the same as |y – x|. So |x – 5| = |5 – x|. Plot both by plugging in values for x and convince yourself…
Got it. Thanks for the explanation.
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