Showing posts with label XAT. Show all posts
Showing posts with label XAT. Show all posts

Thursday, September 29, 2011

Problem of the Day 29 Sept 2011

Let a, b, c be three distinct odd natural numbers. Which of the following can be the sum of the squares of a, b and c?

(a) 3333                            (b) 5555                                (c) 9999                        (d) 7777




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Wednesday, September 28, 2011

Problem of the Day 28 Sept 2011


A good approximation of π is 3.14. Find the least positive integer d such that if the area of a circle with
diameter d is calculated using the approximation 3.14, the error will exceed 1.


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Tuesday, September 27, 2011

Problem of the Day 27 Sept 2011


The diagram below shows some small squares each with area 3 enclosed inside a larger square. Squares that touch each other do so with the corner of one square coinciding with the midpoint of a side of the other square. Find integer n such that the area of the shaded region inside the larger square but outside the smaller squares is n



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Monday, September 26, 2011

Problem of the day 26 Sept 2011


Let a1= 2 and an+1 = 2an + 1. Find the least value of an which is not prime.
(a) 47                  (b) 4                    (c) 5                 (d) 95





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Sunday, September 25, 2011

Problem of the Day 25 Sept 2011

Let P be the set of all the vertices of a regular polygon of 25 sides with its center at C. How many triangles have vertices in P and contain the point C in the interior of the triangles?


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Saturday, September 24, 2011

Problem of the Day 24 Sept 2011

In triangles ABC and DEF, DE=4AB, EF=4BC, FD=4CA The area of triangle DEF is 360 units more than the area of triangle ABC. Compute the area of triangle ABC



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Friday, September 23, 2011

Problem of the Day 23 Sept 2011


Darryl has a six-sided die with faces 1; 2; 3; 4; 5; 6. He knows the die is weighted so that one face
comes up with probability 1/2 and the other fi ve faces have equal probability of coming up. He
unfortunately does not know which side is weighted, but he knows each face is equally likely
to be the weighted one.

He rolls the die 5 times and gets a 1; 2; 3; 4 and 5 in some unspecifi ed order. Compute the probability that his next roll is a 6.



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Thursday, September 22, 2011

How to use the options to maximize scores in QA/DI/LR in IIM CAT

The good thing about CAT is, that it is not a mathematics test. So, it helps that you know the process and the steps but you should be smart and try to minimize the work. This is why there are options to help you. A smart use of options will save a lot of time and help you increase your score.

Some examples

Two different two-digit natural numbers are written beside each other such that the larger number is written on the left. When the absolute difference of the two numbers is subtracted from the four-digit number so formed, the number obtained is 5481. What is the sum of the two two-digit numbers?


(a) 70         (b) 71                (c) 72                         (d) 73


Now it is pretty much obvious the number is of the form 55xy
To find xy we can just subtract 55 from the options which gives xy= 15, 16, 17, 18

If you are lucky you will start with 73 and you will know it is right, else even if you start with 15, you will soon reach 18 and get your answer.


The direct method will be a bit long.

 Let n be the total number of different 5-digit numbers with all distinct digits, formed using 2, 3, 4, 5 and 6 and divisible by 4. What is the value of n?

1] 44 2] 32 3] 36 4] 38 5] 40

Permutations & Combinations is probably the most ‘hated’ topic. However, if you understand the basics and use logic, then it is the most fun-filled topic of all. Let us get to this question. As mentioned, we need to find out the 5 digit numbers divisible by 4 formed by the digits given in the question. To be divisible by 4, the last 2 digits should be divisible by 4. So to arrive at the answer, the first step is to find out combinations of the 2 digits from 2,3,4,5 & 6 that are divisible by 4 – eg: 24. Then, for each such combination, the last 2 digits are fixed. The remaining 3 digits can be arranged in 3! Ways = 6 ways. So the answer would be - 6 multiplied by the total no. of combinations of 2 digits divisible by 4. The answer necessarily should be a multiple of 6 and therefore the answer is 36 – option [3]. We just got lucky here, by the way, since there is just one option that is divisible by 6!



Problem of the Day 22 Sept 2011


Susan plays a game in which she rolls two fair standard six-sided dice with sides labeled one through six. She wins if the number on one of the dice is three times the number on the other die. If Susan plays this game three times, compute the probability that she wins at least once.


Wednesday, September 21, 2011

Problem of the Week 21 Sept 2011

The leftmost digit of an integer of length 2000 digits is 3. In this integer, any two consecutive digits must be divisible by 17 or 23. The 2000th digit may be either a or b. What is the value of a+b?



Monday, September 19, 2011

Problem of the Day 19 September 2011

Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between 9 a.m. and 10 a.m., and stay for exactly m mintues. The probability that either one arrives while the other is in the cafeteria is 40 \%, and m=a-b\sqrt{c},where a, b, and c are positive integers, and c is not divisible by the square of any prime. Find a+b+c.