CAT Preparation Online - Permutation and Combination

Question

Product of the distinct digits of a natural number is 60. How many such numbers are possible?

Explanation

60 = 2^2 * 3 * 5

60 cannot be written as a product of two single digit numbers. SO, the number in question should either have 3 or more digits.

Three-digit numbers

The digits could be 345 or 265

Digits being 345 - there are 3! such numbers

There are six numbers for each of these outlines. So, there are 3! + 3! = 12 three-digit numbers

Four-digit numbers

The digits could be 1345, 1265 or 2235.

Digits being 1345 - there are 4! such numbers

Digits being 1265 - there are 4! such numbers

Digits being 2235 - this is not possible as digits have to be distinct.

So, there are totally 24 + 24 four-digit numbers possible. 48 four-digit numbers.

Total number of numbers = 12 + 48 = 60.

(this post has been modified to plug error)

CAT - Question from Pipes and Cisterns

Question

There are n pipes that fill a tank. Pipe 1 can fill the tank in 2 hours, Pipe 2 in 3 hours, Pipe 3 in 4 hours and so on. Pipe 1 is kept open for 1 hour, pipe 2 for 1 hour, then pipe 3 and so on. In how many hours will the tank get filled completely?

Explanation

Almost all questions can be approached well by asking the question "What happens in 1 hour" (or 1 day, or 1 minutes)

So, let us start with that

In 1 hour, pipe 1 fills 1/2 of the tank. So, in the first hour, the tank will not be filled
In 1 hour, pipe 2 fills 1/3 of the tank. So, in two hours we would have filled 1/2 + 1/3 of the tank, or 5/6 of the tank. So, by the end of the second hour, the tank would still not be filled.
Let us move to the third hour. In 1 hour, pipe 3 fills 1/4 of the tank. So, by the end of the third hour, we should have filled 5/6 + 1/4 = 13/12 of the tank.

Oops, one cannot fill 13/12 of a tank. What this tells us is that the tank gets filled in the 3rd hour. 

When exactly during the third hour?

At the beginning of the third hour, we still have 1/6 of the tank still to fill. Pipe 3 can fill at the rate of 1/4 of the tank per hour. Or, pipe 3 will take 2/3 hours to fill the tank.

Or, the tank will be filled in 2 hours and 40 minutes.

This is an example of a sequence that is a harmonic progression. The formulae for HP are needlessly confusing. So, simple step-by-step approach works best. 

Inequalities - Interesting Question

Question
How many positive integer values can x take that satisfy the inequality (x - 8) (x - 10) (x - 12).......(x - 100) < 0?

Answer: 30

Explanation

Let us try out a few values to see if that gives us anything.

When x = 8, 10, 12, ....100 this goes to zero. So, these cannot be counted.
When x = 101, 102 or beyond, all the terms are positive, so the product will be positive. 

So, straight-away we are down to numbers 1, 2, 3, ...7 and then odd numbers from there to 99.

Let us substitute x =1,

All the individual terms are negative. There are totally 47 terms in this list (How? Figure that out). Product of 47 negative terms will be negative. So, x = 1 works. So, will x =2, 3, 4, 5, 6, and 7.

Remember, product of an odd number of negative terms is negative; product of even number of negative terms is positive. Now, this idea sets up the rest of the question.

When x = 9, there is one positive terms and 46 negative terms. So, the product will be positive. 
When x = 11, there are two positive terms and 45 negative terms. So, the product will be negative. 
When x = 13, there are three positive terms and 44 negative terms. So, the product will be positive. 

and so on.

Essentially, alternate odd numbers need to be counted, starting from 11.

So, the numbers that will work for this inequality are 1, 2, 3, 4, 5, 6, 7...and then 11, 15, 19, 23, 27, 31,..... and so.

What will be the last term on this list? 
99, because when x = 99, there are 46 positive terms and 1 negative term. 

So, we need to figure out how many terms are there in the list 11, 15, 19,....99. These can be written as 
4 * 2 + 3, 
4 * 3 + 3, 
4 * 4 + 3
4 * 5 + 3
4 * 6 + 3
...
4 * 24 + 3

A set of 23 terms. So, total number of values = 23 + 7 = 30. 30 positive integer values of x exist satisfying the condition. 

Profit and Loss Question

Question:

A merchant can buy goods at the rate of Rs. 20 per good. The particular good is part of an overall collection and the value is linked to the number of items that are already on the market. So, the merchant sells the first good for Rs. 2, second one for Rs. 4, third for Rs. 6…and so on. If he wants to make an overall profit of at least 40%, what is the minimum number of goods he should sell?

A.  24

B.  18

C.  27

D.  32

Correct Answer: (C)

Explanation:

Let us assume he buys n goods.

Total CP = 20n

Total SP = 2 + 4 + 6 + 8 ….n terms

Total SP should be at least 40% more than total CP

 2 + 4 + 6 + 8 ….n terms > 1.4 * 20 n

2 (1 + 2 + 3 + ….n terms) > 28n

n(n + 1) > 28n

n² + n  > 28n

n² -  27n  > 0

n > 27

He should sell a minimum of 27 goods.

Answer Choice (C)