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Xat Quant

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Information about Xat Quant

Published on January 4, 2008

Author: kapil1312

Source: slideshare.net

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XAT - Quant

Objective Over the last 2 years, XAT has tended to be more on the lines of CAT with very few topics of higher maths thrown in. Regular CAT Quant is good enough to help clear the XAT Quant too. This presentation will look at some of the questions from older XAT patterns where the question types were difficult to get used to. So run thro’ this presentation to get used to the ‘language’ of some of those questions.

Over the last 2 years, XAT has tended to be more on the lines of CAT with very few topics of higher maths thrown in.

Regular CAT Quant is good enough to help clear the XAT Quant too.

This presentation will look at some of the questions from older XAT patterns where the question types were difficult to get used to.

So run thro’ this presentation to get used to the ‘language’ of some of those questions.

 

 

 

 

 

 

 

 

 

 

 

 

There were also some questions similar to the ones in previous IIFT papers.. A few examples follow…..

Rule 1 : If a word ends with A, then a new word can be made by adding B at the end Rule 2 : If ‘x’ is any sequence of letters and Cx is a word, then so is Cxx Rule 3 : If AAA occurs in any word, then we may drop it and replace it by B Rule 4 : if BB occurs in any word, then we may drop it Q1 : Given CA is a word. Construct CBABBAB CAA ( rule 2 ) CAAAA ( rule 2 ) CBA ( rule 3 ) CBAB ( rule 1 ) CBABBAB ( rule 2 ) Number of steps : 5

Rule 1 : If a word ends with A, then a new word can be made by adding B at the end

Rule 2 : If ‘x’ is any sequence of letters and Cx is a word, then so is Cxx

Rule 3 : If AAA occurs in any word, then we may drop it and replace it by B

Rule 4 : if BB occurs in any word, then we may drop it

Q1 : Given CA is a word. Construct CBABBAB

CAA ( rule 2 )

CAAAA ( rule 2 )

CBA ( rule 3 )

CBAB ( rule 1 )

CBABBAB ( rule 2 )

Number of steps : 5

Sunil goes to small city in Europe on vacation, where he enjoys walking along the streets in the afternoon. He observes that there are 6 parallel roads running East - West and 5 parallel roads running North-South in the city. In order to observe the landmarks in the city, he takes different routes every time he goes out. He also observed that the distance between every consecutive pair of roads is equal. Given this, mark all the correct options. a. The number of shortest possible routes that Sunil can take to travel from one corner of the city to the other diagonal end is 126. b. The total number of possible routes that Sunil can take to travel from one corner of the city to the other diagonal end is 196. c. The number of rectangles that can be formed with their sides along the roads is 150. d. If the number of parallel roads running East-West and North - South increase by one each, the number of shortest possible routes that Sunil can take travel from one corner of the city to the other diagonal end would go up by 336.

Sunil goes to small city in Europe on vacation, where he enjoys walking along the streets in the afternoon. He observes that there are 6 parallel roads running East - West and 5 parallel roads running North-South in the city. In order to observe the landmarks in the city, he takes different routes every time he goes out. He also observed that the distance between every consecutive pair of roads is equal. Given this, mark all the correct options. a. The number of shortest possible routes that Sunil can take to travel from one corner of the city to the other diagonal end is 126. For each shortest path, you need to Cover 5 vertical steps and 4 horizontal steps 4 x’s and 5 y’s Each arrangement of these 9 items linearly would give a shortest path Thus number of routes = 9! 5! x 4! = 126

Sunil goes to small city in Europe on vacation, where he enjoys walking along the streets in the afternoon. He observes that there are 6 parallel roads running East - West and 5 parallel roads running North-South in the city. In order to observe the landmarks in the city, he takes different routes every time he goes out. He also observed that the distance between every consecutive pair of roads is equal. Given this, mark all the correct options. b. The total number of possible routes that Sunil can take to travel from one corner of the city to the other diagonal end is 196 . This has to be Infinite as he can go back n forth c. The number of rectangles that can be formed with their sides along the roads is 150 . Any 2 vertical lines and 2 horizontal lines would generate a rectangle Out of any 5 vertical, we need to choose 2 And any 6 horizontal, we need to choose 2 Number of such combinations = 5 C 2 x 6 C 2 = 150

d. If the number of parallel roads running East-West and North - South increase by one each, the number of shortest possible routes that Sunil can take travel from one corner of the city to the other diagonal end would go up by 336 . Sunil goes to small city in Europe on vacation, where he enjoys walking along the streets in the afternoon. He observes that there are 6 parallel roads running East - West and 5 parallel roads running North-South in the city. In order to observe the landmarks in the city, he takes different routes every time he goes out. He also observed that the distance between every consecutive pair of roads is equal. Given this, mark all the correct options. Similar to the first question New number of shortest paths = 11! 6! x 5! = 462 Original number was 126. Therefore it has gone up by 462-126 = 336 Therefore options A,C,D are correct

Ankit is appearing in a entrance examination for a profession course. In the General Knowledge section, the students are asked to match certain years in which the soccer world cup was held with the name of the champion team in that particular year. The information given was as follows: Champion Year WestGermany 1966 Italy 1982 France 1990 England 1998 Now, Ankit it not being a football fan, does the matching randomly. If X denotes the number of correct answer his random matching generates. mark all the correct probabilities. a. P(X >= 1) = 5/8 b. P (X = 1) = 1/4 c. P (x = 3) = 0 d. P (X = 4) = 1/24 Options C and D are easy to verify… d. Total number of ways you could arrange (possibilities) = 24 Out of which one is right. So P (X = 4) = 1/24 c. P(X=3) is probability of getting exactly 3 right which means there is exactly one wrong. This is not possible since there needs to be at least one other wrong to exchange values. i.e. if 3 countries are correct, the fourth has to be correct as well. So P (x = 3) = 0

1 + ∑ n Every new line will cut the previous lines and create one extra Polygon with that many sides Two straight lines can divide a circular disk into a maximum of 4 parts. Likewise, into how many maximum parts can four straight lines divide a circular disk? a. 8 b. 9 c.10 d. 11 11 7 4 2 Regions 1+10 1+6 1+3 1+1 4 3 2 1 Lines

All the best! If you still need answers, mail to – chari.r@gmail.com

If you still need answers, mail to – chari.r@gmail.com

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