In geometry test 1, the soln for "Euclid has a triangle in mind. Its longest side has length 20 and another of its sides has length 10. Its area is 80. What is the exact length of its third side?" makes no sense to me. they have solved using Pythagoras theorem on a scalene triangle. Please look into it.

Hi Olivia,The solution is correct. If you refer to the figure given in the solution, the required triangle is AMB. AO is the height of the triangle AMB. Right angled triangle AOB has been constructed to simplify the calculation. If you did not understand the method you can also do it using Heron's Formula but it will complicate the calculations, so this is the best method to solve this question. If any specific step is still not clear, please point that out.Thanks.

How to solve passagein english

Hi Chandrakant,We have a blog post where we have comprehensively explained the various tips and techniques to solve reading comprehension passages. You can find it her https://cracku.in/blog/crack-reading-comprehension-for-cat/ . Please go through it once, it will surely help you.Thanks

https://cracku.in/cat/history/1695995. Pls provide diagrammatic solution for Q5

Hey Balu,Sorry for the inconvenience but as of now we can't post images here. But I will try to give a detailed explanation so that you understand the questionLet ABCD be a square of side 10. Mark a point L which is at 2cm from A on the line AB. Similarly, mark a point m which is 2 cm from B on line AB proceeding in a similar manner, mark n and o on BC, p and q on CD, and r and s on AD such that s is 2 cm away from A on AD.So we are removing the squares of side 2 from each of the corners. Let the 4 corners of each of these squares be u, v w and x respectively.So after removing the squares, the figure which will remain is 'sulmnowpqxr'Now when we make a cuboid out of it, the length of the cuboid will be lm, breadth will be SR and height will be ulSo the required dimensions are 6*6*2 = 72.

I always get stuck at number of existing solutions for equations. What should be approach for solving it?

Hi Nidhi,Please post a question so that we can understand what kind of questions are troubling you. Within the topic of finding solutions, there is a large variety of questions. So I would request you to post some question so that we can help you better.Thanks

please recommend me the books i have to get for cat? other than the cracku content ofc. :)

Hi Shrinivas,If you are enrolled as a premium member at Cracku, then you don't need any other book. Our material is comprehensive and is sufficient to crack CAT. However, if you are preparing on your own then the following books can be followed. They cover the CAT syllabus pretty wellQuant: Arihant- Sarvesh VermaDI/LR- Arun SharmaVerbal- Pearson - Sujit KumarHowever, you need to join a test series to decide upon your test taking strategy. So if you are preparing on your own, then you must join a test series.Thanks

who is president of Australia

Hi Satish,Australia does not have a president. They only have prime minister.

What would be the cost of building a 7 metres wide garden around a circular field with diameter equal to 280 metres if the cost per sq. metre for building the garden is Rs.21?

cost = 21 * area of path = 21 * pi*(147*147-140*140) = 132594 rs

https://cracku.in/cat/history/1689698?src=test. Pls elaborate on the solution of Q5

Question:A closed barn ABCD in the shape of a Rhombus is situated inside a vast field. Each side of the barn is equal to 10m and angle A = 60 degrees. A cow is tied to point A outside the barn with a rope of length 20m. Find the area of the field that the cow can graze. Solution:The cow can graze the whole area of a circle of radius 20m if the barn wasn't there. Because the barn is there it can graze 300 degrees angle out of 360 degrees because 60 degrees is cut by the barn at the vertex A. This means it can graze $$\frac{5}{6}$$th of the area of the circle of radius 20m.This area = $$\frac{5}{6}\times\pi\times20^2=\frac{1000\pi}{3}$$. It can also graze two sectors of circles centered at B and D of radius 10m and angle 60 degrees.This area = $$2\times(\frac{1}{6}\times\pi\times10^2)=\frac{100\pi}{3}$$ Total area = $$\frac{1000\pi}{3}+\frac{100\pi}{3}=\frac{1100\pi}{3}$$

HOW DO I SOLVE PARA JUMBLES

Hi Puneet,We have a blog post where we have described in detail the techniques to tackle para jumbles. You can find the blog post here https://cracku.in/blog/solve-parajumbles-cat/ . Please go through it once. It will surely help you.Thanks