Inequalities Question and Solution

Question

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

Explanation

Inequalities Question and Solution

Question

|x + 2| + | x - 3| + | x - 5| < 15, find the range of x that satisfies this inequality.

Explanation

Functions Question and Solution

Question

Consider set A with 'a' elements, set B with 'b' elements, set C with 'c' elements. We can define a function that is one-one but not onto from set A to set B, a function that is onto but not one-one from set B to set C; and a function that is injective but not surjective from C to A. Arrange a, b, c in ascending order.

Explanation

Algebra Question and Solution

Question

|x| + |2y| + |3z| = 13, x * y * z is non-zero; x, y, z are all integers. How many sets of values are possible?

Answer
64 sets of values

Explanation

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)

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CAT Averages,

Question

1.       5. Scores in a classroom are broken into 5 different ranges, 51-60, 61-70, 71-80, 81-90 and 91-100. The number of students who have scored in each range is given below

51 -60 - 3 students

61 -70 - 8 students

71 -80 - 7 students

81 -90 - 4 students

91 -100 - 3 students

Furthermore, we know that at least as many students scored 76 or more as those who scored below 75. What is the minimum possible average overall of this class?

1. 72 
2. 71.2 
3. 70.6 
4. 69.2 

Answer: Choice (C)

Explanatory Answer:

Let's employ the idea of a total of 25 students (all of the same weight) sitting on a see-saw, which has numbers from 51 to 100 marked on it. At least as many students are sitting on 76 (or to its right), as there are sitting to the left of 75. Now this means that you can have only one person sitting to the left of 75 and all the rest sitting beyond 76. But you can't do that, as you have other constraints as well.

First of all, you have to seat 3 students from 51 to 60, and 8 students from 61 to 70. Secondly, you also have to make sure that the average is the least. This means that the see-saw should be tilting as much to the left as possible, which in turn means that the number of people sitting to the left of 75 should be the highest possible.

This makes it 12 students to the left of 75, and the remaining 13 students on 76 or to its right.

Next, how do you ensure that the average is least, i.e. how do you ensure that the balance tilts as much as possible to the left? Make each student score as little as possible given the constraints.

So, the first 3 students only score 51 each. The next 8 students score only 61 each. 11 Students are now fixed. The 12^th student has to be below 75, so seat him on 71. The remaining 6 students (who are in the 71 to 80 range) have to score 76. The next 4 score 81 and the next 3 score 91.

This would give you the least average.

The lowest possible average would be:

= [(3 * 51) + (8 * 61) + 71 + (6 * 76) + (4* 81) +  (3 * 91)]/25 

= 70.6 

Answer Choice (C)

CAT Functions, Onto functions

  Question

How many onto functions can be defined from the set A = {1, 2, 3, 4} to {a, b, c}?

1. 81
2. 79
3. 36
4. 45

Answer: Choice (C)

Explanatory Answer:

First let us think of the number of potential functions possible. Each element in A has three options in the co-domain. So, the number of possible functions = 3⁴ = 81.

Now, within these, let us think about functions that are not onto. These can be under two scenarios

Scenario 1: Elements in A being mapped on to exactly two of the elements in B (There will be one element in the co-domain without a pre-image).

Ø  Let us assume that elements are mapped into A and B. Number of ways in which this can be done = 2⁴ – 2 = 14

o   2⁴ because the number of options for each element is 2. Each can be mapped on to either A or B

o   -2 because these 2⁴ selections would include the possibility that all elements are mapped on to A or all elements being mapped on to B. These two need to be deducted

Ø  The elements could be mapped on B & C only or C & A only. So, total number of possible outcomes = 14 * 3 = 42.

Scenario 2: Elements in A being mapped to exactly one of the elements in B. (Two elements in B without pre-image). There are three possible functions under this scenario. All elements mapped to a, or all elements mapped to b or all elements mapped to c.

Total number of onto functions = Total number of functions – Number of functions where one element from the co-doamin remains without a pre-image - Number of functions where 2 elements from the co-doamin remain without a pre-image

ð  Total number of onto functions = 81 – 42 – 3 = 81 – 45 = 36  

Answer Choice (C)