Reasoning

Who is the Liar?

The problems based on truth and lie or Sacha and Jhuta are very common in CAT and similar exams. You will be able to understand the related concepts better with the help of the following example.

Example 1:

In an isolated island there live two tribes:

• SACHAS
• JHUTAS

The SACHAS always speak the truth, whereas the JHUTAS never do.

One day SUNIL, a stranger in the island, met four tribesmen. He asked one of them whether the second was a SACHA or a JHUTA. The reply was JHUTA. Similarly, the second tribesman said that the third was a JHUTA and the third said the same about the fourth.

When he asked the fourth as to what would the first tribesman have said about the third, the reply once again was JHUTA. Who’s who?

Solution:

Before solving this question, here are some important deductions that you can make.

1. If you asked the question to any tribesman ‘Are you a “SACHA” or a “JHUTA”?’ from that island irrespective of the tribe he belongs to, his reply would always be ‘SACHA’, because

• if he were a ‘SACHA’, he will always speak the truth about himself, that he belongs to ‘SACHA’ tribe and
• if he were a ‘JHUTA’ he will lie about his actual tribe and, say that he belongs to ‘SACHA’ tribe.

2. If two tribesmen, A & B, belong to the same tribes, either both ‘SACHAS’ or both ‘JHUTAS’, and you asked the question from A ‘Is B a “SACHA” or a “JHUTA” ?’, his reply will always be ‘SACHA’, because

• If they were belongs to ‘SACHA’ tribe then they always speak the truth, that other one belongs to ‘SACHA’ tribe and
• if both belongs to ‘JHUTA’ tribe then they always lie about themselves as well as about others, and would say that other belongs to ‘SACHA’ tribe.

3. If two tribesmen, A & B belong to different tribes and you asked the question from A ‘Is B a “SACHA” or a “JHUTA”?’, his reply will always be ‘JHUTA’, because

• if A is ‘SACHA’ then he is speaking the truth about B, and answering that B is ‘JHUTA’ or
• if A is ‘JHUTA’ then he is telling alie about B, and answering that B is ‘JHUTA’

Note: In conclusion, we can say that,if two persons belong to same tribe they will always say ‘SACHA’ about other and if two belongs to different tribe always say ‘JHUTA’ about the other.

1. Now, coming back to the problem:
2. As the first tribesman told Sunil, that the second was a ‘JHUTA’, it means that first & second tribesmen belonged to different tribes.Similarly, the second and third tribesmen were also from different tribes.
3. Moreover, the third and fourth tribesmen were also from different tribes, and
4. But as there are only two tribes, it implies that the first and third belonged to the same tribe eitherboth ‘SACHAS’ or both‘JHUTAS’, and the second and fourth also belonged to the same tribe either both ‘JHUTAS’ or both ‘SACHAS’.
5. When Sunil asked from the fourth tribesman‘what would the first tribesman have said about the third?’, we know from the second point mentioned above, that the answer given by third would always be Sacha, as they belong to the same tribe. But as you asked this question from the fourth person, he will also manipulate the answer as per HIS OWN TRIBE. The fourth said ‘Jhoota’.
6. It means the fourth tribesman, himself must be a ‘JHUTA’ because if he belongs to tribe ‘SACHA’  then his reply would be‘SACHA’, as he know that fist and third tribes belongs to same tribe and same tribe always say ‘SACHA’ about other.
7. Hence, the first and third tribesmen were both ‘SACHAS’ and the second and fourth were both ‘JHUTAS’.

Algenbra

Let X Remain UNKNOWN only

In this article, we explain the non-conventional way to solve Algebra questions without formulating equations. These alternative methods will enable you to save a lot of time as compared to the traditional and conventional approaches. Further, using these methods you will avoid making silly errors that creep in when the calculation work is lengthy.

You will understand how to solve Algebra questions without ‘x’ with the help of following examples:

Example 1:

The sum to n terms of the infinite series 1. 3² + 2. 5² + 3. 7² +....is:

1.  n/6 (n+1)(6n² + 14n + 7)
2.  n/6 (n+1)(2n + 1)(3n + 1)
3.  4n³ + 4n² + n
4. None of these

Solution:

The series in the question is obtained by multiplying 1, 2, 3, 4, … and 3², 5², 7², 9², … i.e. all natural number starting with 1, and the squares of all odd natural numbers, except one. Here, you need to find the general formula that will hold true for all the values of n.

Now, for the formula to satisfy all values of n, it should be applicable for n = 1 and n = 2 also. You can see that the first term in the given series is 9, and the second term is 50. This means that the sum of first two terms will be 9 + 50 = 59. So, that option in which you get 9, for n = 1 and 59, for n = 2 will be the answer. Here, the 1^st option gives the required values, and hence is the answer.

Note: n = 1 will be used for rejection or elimination of options only (as two different series can start with same number), and n = 2 will confirm the answer.

Example 2:

The sum to n terms of the series 1/2 + 3/4 + 7/8 + 15/16 +........ is equal to

1. 2ⁿ - n - 1                    2.  1 - 2^-n
3. 2^-n + n - 1                  4.  2ⁿ - 1

Solution:

Here, the first term is 1/2 and sum of the first two terms is 1/2  + 3/4  = 5/4 . So, by taking n = 1, you should get ½ as the answer, and by taking n = 2, you should get 5/4 as the answer. Here, the third option satisfies both of these and hence it is the answer.

However, you should be careful as n = 1 may satisfy one more of the options. This condition should be used only for rejection or elimination of options and the answer should always be confirmed by taking n = 2.

Example 3:

The nth term of the series 1+ 4/5 + 7/5² + 10/5³+.... will be:

1. 3n+1/5^n-1                  2. 3n - 2/5^n-1
3. 3n - 1/5ⁿ                    4. 3n+2/5^n-1

Solution:

By taking n = 1, we should get the 1^st term i.e. 1, and by taking n = 2, we should get the second term i.e. 4/5. Here, only the second option satisfies both of those and hence it is the answer.

DI

The Answer is a NO but is the Data Sufficient?

Data sufficiency questions test your knowledge of basic math facts and skills coupled with reasoning, analytical and problem-solving abilities. Each data sufficiency item presents you with a question where you need to decide whether or not the information presented along with the question would be sufficient to answer the question.

The challenge in DS questions, as they are popularly called, is not question solving but rather establishing whether the question has a solution or not. A special array of five answer choices is provided, each of which categorizes the relationship between the question and the information provided in a different way. You must select the answer choice that describes this relationship accurately.

Let’s have a cursory look at these answer options which generally feature in this question type:

Give answer (A) If the data in Statement I alone are sufficient to answer the question, while the data in Statement II alone are not sufficient to answer the question.
Give answer (B) If the data in Statement II alone are sufficient to answer the question, while the data in Statement I alone are not sufficient to answer the question.
Give answer (C) If the data either in Statement I or in Statement II alone are sufficient to answer the question.
Give answer (D) If the data even in both Statements I and II together are not sufficient to answer the question.
Give answer (E) If the data in both Statements I and II together are necessary to answer the question.

One small tip for this question type: Students often confuse the different answer options, and end up marking the incorrect choice. Always double check whether you are marking the correct option, and do not assume that the examiner would always present the options in a default order.

Go through the answer options to check whether the order of statements is as expected.

Example 1: Is the product of two numbers greater than 100?

A. The sum of the two numbers is greater than 50.
B. Each of the numbers is greater than 10.

Solution: Statement A alone is not sufficient to answer the question and this can be proved by examples. If the two numbers are 30 and 31, their sum is greater than 50 and their product is greater than 100; but if the two numbers are 50 and 1, though their sum is greater than 50, their product is only 50, and less than 100. Statement B is sufficient. If both of the numbers are greater than 10, then their product must be greater than 10 x 10, or greater than 100.
Hence only second statement is sufficient to solve the question.

Example 2: Is x a prime number?

A. 91 <x< 97
B. x is a factor of 121

Solution: Here the first statement is sufficient to answer the question as we see that there is no prime number between 91 <x< 97. Hence ‘x’ is not a prime number. What do we learn from this question? Remember, even if a question has an answer as ‘no’, even then it is a valid answer.

In second statement, the factors of 121 are 1, 11 and 121. Here 1 and 121 are not prime numbers whereas 11 is a prime number. Hence in this case ‘x’ may or may not be a prime number.
Hence, only the first statement is sufficient to solve the question.

Example 3: Is x = - 5?

A. x² = 36
B. x is a natural number.

Solution: Here the question directly asks whether x is equal to – 5 or not. From statement A, we have x = 6 or – 6. In both the cases x is not equal to – 5. Hence first statement is sufficient to get the answer. Statement B says that x is a natural number. Since x is a natural number, it cannot be negative. Hence it is not equal to – 5. So, the second statement is also sufficient to solve the question.
Hence, both statements are independently sufficient to answer the question.

To conclude, it is very important to read the question carefully in the case of data sufficiency questions. One major mistake committed by a number of students is that when the answer has to be yes/no and normally whenever you get the answer as no, you mark the answer as insufficient.

Remember: ‘NO’ is also an answer for Data Sufficiency questions.

 Higher Degree Equations & Waving Curve Method

Higher Degree Equations:

For higher degree equations of degree n.

We can use the following formulae:

Sum of roots taken one at a time (S₁) = - (coefficient of x^n-1/ coefficient of xⁿ)

Sum of roots taken two at a time (S₂) = coefficient of x^n-2/ coefficient of xⁿ

Sum of roots taken 3 at a time (S₃) = - (coefficient of x^n-3/ coefficient of xⁿ)

And so on with alternatively positive and negative sign.

Product of roots = (+ or -) constant term / coefficient of xⁿ

The positive sign is to be used for an even numbered degree equation and negative sign for an odd numbered degree equation.

Let’s take the below an example:

Example 1:

If p, q, r are the roots of the equation x³ + 6x² + 12x + 24 = 0, find the equation with the roots 2p, 2q and 2r.

We know, S₁ = p + q + r = - 6

• S₂ = 12
• P = - 24

Now, for the equation, whose roots are 2p, 2q and 2r.

S₁ = 2p + 2q + 2r = 2(p + q + r) = 2*(-6) = -12
S₂ = 2p*2q + 2q*2r + 2p*2r = 4*12 = 48
P = 2p*2q*2r = 8*(-24) = -192

Thus, required equation is x³ + 12x² + 48x + 192 = 0

Waving Curve Method:

This method is helpful in solving the inequalities involving higher degree.

Let’s take the below an examples:

Example 2:

Solve the inequality (x-3)(x-2)(x+1) ≤ 0

Step 1: First we find the zero’s of all factors (x-3)(x-2)(x+1). Zero’s can be found by putting all factors equal to zero.

In this case, the zero’s are 3,2 and -1, x-3 =0 we get x = 3;  x-2 = 0 we get x = 2; and x+1 = 0 we get x = -1

Step 2: Now mark the above said zero’s on a number line as shown:

Step 3: Now we will draw a curve starting from the greatest number (Extreme right hand side) in this case it is 3 and the curve would start from the upper side of the number line and should cross each of the marked point.

The regions covered by the curve above the number line are marked as +ve and the regions below the number line are marked as –ve.

Step 4: In our question we have to find the region ≤ so the answer is: (-∞, -1] U [2, 3].

This method is however, only applicable if the following conditions are met:

When the equality can be factorised in the form (x-a)(x-b)(x-c) where a, b anc are integers.
On the right hand side of the inequality, there should be a zero.
Negative sign should not be there with x. If it’s there, remove it by reversing the sign of the inequality.

Now, if the there is power of x in the numerator and the denominator has some factor(s), we will use the below example:

Example 3:

[(x-3)(x+2)(x+1)²/(x-2)] ≤ 0

First we mark the zero’s on the number line. Put a circle at a point that is derived from the factor which is in the denominator. Mark X on a point that is derived from a factor having even power. In this case it is -1.

Here, the curve will retrace its path- where X is marked. If it’s on the upper side, it’ll remain on upper side and similarly for lower side.

Now, the region is ≤ 0. We will not be including the points where circle is marked as it’s derived from the denominator. Thus the required region is: (-∞,-1] U (2,3]. 

Quant

There is 50% Probability that Your CAT Registration Number will come before Your Friend!

The chance of a particular event happening is called its probability. In general, the probability of an event is the number of favorable outcomes divided by the total number of possible outcomes.

In terms of formula it can be written as:

Probability = no. of Favorable outcomes / total no. of Possible outcomes.

The questions on probability can be on coins, dices, cards and arrangement of people.

It is important to understand a few terms before seeing some examples:

Tossing a Coin:

When a coin is tossed, there are two possible outcomes: Heads (H) or Tails (T). It can be concluded that the probability of the coin landing H is half or ½ and the probability of the coin landing T is also ½.

Throwing Dice:

When a single die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, 6.

The probability of getting an even numberon a single throw of dice is ½ as there are equal number of cases for an even number and an equal number of cases for an odd number.

Arranging People:

Let us say there are 3 people A, B and C. They are to be arranged in a row. There will be a total ofsix arrangements possible viz. ABC, ACB, BCA, BAC, CBA, CAB.

Now, if we have to compare only among A and B then out of the total arrangements half the number will have A coming before B i.e. ABC, ACB, CAB and in remaining half of the cases i.e. BAC, BCA and CBA B will be before A.

Hence, the probability of A coming before B is 50% i.e. ½ and that of B before A is also 50%.

Choosing a Card from a Deck:

There are 52 cards in a deck (not including Jokers). The number of black cards is 26 and the number of red cards is also 26. If a card is selected at random from the pack, then the probability of getting a red card is ½, as there is equal number of red and black cards.

Let us take a few examples to understand the concepts:

Example 1:

There are six marbles in a bag. Out of that, three are blue and three are red.

What is the probability that a blue marble is picked?

Solution:

Since there are equal number of blue and red marbles, so the chances of getting a red marble = ½ and also, the chances of getting a blue marble = ½.

Example 2:

Delegates of six different countries appear in a UNO meet for a conference with the UNO External Affairs Chief. Among them are one delegate each, from India and Pakistan. If each delegate meets the UNO chief one at a time.

What is the probability that India’s delegate meets the UNO’s Chief before Pakistan’s?

Solution:

As we have discussed earlier, out of the total 6! arrangements, in half the cases India’s delegate will be before Pakistan’s and in remaining half Pakistan’s will be before India’s. Hence, the required probability = ½.

Example 3:

There are three mothers with each mother having a child. They appear before the school selection committee for admission of their respective child. The committee decides to interview all six of them one by one.

What is the probability that every mother is interviewed before her child?

Solution:

Let us assume the names of the three mothers as M1, M2, M3, and their respective children as C1, C2 and C3. There will be total 6! arrangements possible.

Now, if we consider only M1 and C1 then in exactly half the number of cases, M1 is interviewed before C1 i.e. 720/2 = 360 cases, and in the rest 360 cases C1 will be interviewed before M1.

We reject this second case, hence, we are left with only 360 cases with M1 before C1. Again, among these 360 cases, half the number will have M2 before C2 i.e. 360/2 = 180 whereas the remaining 180 cases will have C2 before M2.

We reject this second case, hence, we are left with 180 cases having both the conditions satisfied i.e. M1 before C1 and M2 before C2. Again, among these remaining cases, half of them will have M3 before C3, and the other half will have C3 before M3.

So, we reject the second case again, and finally we are left with 180/2 = 90 cases, where M1 is before C1, M2 is before C2 and M3 is before C3 also.