July 22nd, 2020 by Ravi Handa

In this post, we will learn about logical reasoning concepts on selection and group formation that is frequently asked in CAT exam. This topic generally deals with the selection of a team of say â€˜râ€™ members from â€˜nâ€™ (n>r) available for selection or it can be the selection of committee of certain number of members. Certain number of constraints drives this selection. In order to understand these constrains and the implicit details related to them, let us start the discussion with an example.
Question: Among five students of group I â€“ A, B, C, D, E and six students of group I

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Posted in CAT, IIFT, LR DI Funda, MBA, SNAP, Solved Examples, XAT, XAT Funda

March 24th, 2015 by Ravi Handa

A common type of question that gets asked in CAT is when you are given a maximum function and you are supposed to find out the minimum value of the function. Actually, the concept would remain the same even if you are given a minimum function and you are supposed to find out the maximum value of the function.
To solve such kind of questions, all you need to do is to find out the point of intersection of the individual values. More often than not, that would lead to the answer. Let us look at a question that has appeared in CAT before.
Let g(x) = max (5 âˆ’ x, x + 2). The smallest possi

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Posted in CAT, Quant Funda, Solved Examples

January 22nd, 2014 by Ravi Handa

A team ofÂ minerÂ planned to mine 1800 tonnes in a certain number of days.Due to some difficulties in one third of the planned days, the team was able to achieve an output of 20 tons of ore less than the planned output.To make up for this, the team overachieved for the rest of the days by 20 tons.The end result for this that they completed the one day ahead of time.How many tone of ore did the team initially plan to ore per day?
a. 50
b. 100
c. Â 150
d. Â 200
e. Â 250
Answer :Â
Let us assume the no. of days as '3d' and the output per day as 'x'
T

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Posted in Solved Examples

January 22nd, 2014 by Ravi Handa

Question : a, b and c are the sides of a triangle. Equations ax^2 + bx + c = 0 and 3x^2 + 4x + 5 = 0 have a common root. Then angle C is equal to??
Answer : Roots of 3x^2 + 4x + 5 = 0 are complex i.e. they are of the form (p + iq) & (p - iq) where i is iota = sqrt(-1)
For a quadratic equation, complex roots occur in conjugate pairs if the coefficients are real.
In the equation, ax^2 + bx + c = 0, the coefficients are sides of a triangle and hence real. So, if one of roots is common with the other equation say (p+ iq) then the other root will

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Posted in Solved Examples

January 22nd, 2014 by Ravi Handa

Question: Six persons try to swim across a wide river.Its known that on an average only three persons out of 10 are successful in crossing the river.Whats the probability that at most four of the 6 persons will cross the river safely?
Answer :Â
The probability distribution of the random variable X is called aÂ binomial distribution, and is given by the formula:
P(X)= nCr * p^r * q^(n-r)
where
n = the number of trials
r = 0, 1, 2, â€¦ n
p = the probability of success in a single trial
q = the probability of failure in a single trial
(i.e. q = 1 Ã¢Ë†â€™ p)
Keeping th

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Posted in Solved Examples

January 22nd, 2014 by Ravi Handa

Question : A person lent out some money for 1year at 6% per annum simple interest and after 18 months he again lent out the same money at a simple interest of 24% per annum. In both cases he got Rs.4704. What was the amount that he lent out if interest is paid half yearly?
a. Rs 4000
b. Rs 4200
c. Rs 4400
d. Rs 3600
Answer :
In the first case, the rate of interest is 3% per half-year. The money will become
1.03x after 6 months from t = 0
1.06x after 12 months from t = 0
1.09x after 18 months from t = 0
1.12x after 24 months from t = 0Â and so on ....

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Posted in Solved Examples

January 22nd, 2014 by Ravi Handa

Question : The L.C.M of 5/2, 8/9, 11/14 ?
Answer :
LCM = LCM of Numerator / HCF of Denominator
HCF = HCF of Numerator / LCM of Denominator
Â
So,in this case;
LCM (5/2, 8/9, 11/14) is :
LCM (5,8,11) / HCF (2,9,14) =Â 440 / 1 = 440

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Posted in Solved Examples

February 19th, 2013 by Ravi Handa

Question : Find the No of zeroes at the end of 25! +26! + 27! + 28! + 30!
Â
Answer :Â
To understand this, let us understand the basic idea first
What will be the number of 0s at the end of a + b + c would depend upon the least number of 0s that any one of a or b or c has.
For eg: 300 + 120000 + 17272730 will end in 1 zero
But, if they have the same number of zeroes, we will also have to consider the last non-zero digit.
For eg: 12000 + 161237000 + 1212331000 will not end in 3 zeroes but in 4 zeroes because the last non-zero digits 2, 7 and 1 will add up to generate an extra zero.

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Posted in Solved Examples

February 11th, 2013 by Ravi Handa

Question : 100 million bacteria can completely decompose a garbage dump in 15 days and 60 million can do so in 30 days. If same quantity of garbage is added to the initial quantity of the dump everyday , how many bacteria will be required to completely decompose the dump in 10 days?
Answer :
Let us say 1 Million bacteria can clear 1 unit of garbage in 1 day.
Initially there were 'x' units of garbage and everyday 'y' units of garbage gets added.
Case 1Â : 100 M bacteria can decompose in 15 days.
Total garbage present = Total garbage decompo

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Posted in Solved Examples

February 11th, 2013 by Ravi Handa

Question : A student was given 8 two-digit numbers to add,by a teacher.If the student reversed each number and added the results,the sum of the numbers would be 36 more than the actual sum.Find the excess of the sum of the units digits over that of the ten digits.
Answer :Â
Let us say that the 8 numbers are a1 b1, a2 b2, a3 b3 ... a8 b8 ...{eg: 38}
They actually represent a1*10 + b1, a2*10 + b2 ... a8*10 + b8 ...{eg: 3*10 + 8}
Sum of the numbers = (a1 + a2 + a3 .. a8)*10 + (b1 + b2 + b3 .. b8)
Reversed numbers will be b1 a1, b2 a2, ... b8 a8
Sum

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Posted in Solved Examples