2IIM's blog on cracking the quant section of the CAT. Frequent posts on interesting questions, tips, shortcuts in math.

Jun 2, 2015

Online CAT Coaching: A few interesting True/False questions from Geometry

State whether the following statements are true or false

1. A parallelogram that circumscribes a circle has to be a square
2. A trapezium inscribed in a circle has to be an isosceles trapezium
3. Orthocenter of a triangle can lie outside the triangle
4. Triangle with sides a, b and c has the relationship a^2 + b^2 > c^2, the triangle has to be acute-angled.
5. Diagonals of a parallelogram are angle bisectors of the angles of a parallelogram.

Scroll down for answers and explanation
























1. A parallelogram that circumscribes a circle has to be a square: FALSE

In a parallelogram, opposite sides are equal. In a quadrilateral, the sums of pairs of opposite sides are equal. So, a parallelogram that circumscribes a circle should have all 4 of its sides equal. Or, it should be a Rhombus; it need not be a square.

2. A trapezium inscribed in a circle has to be an isosceles trapezium: TRUE

An isosceles trapezium is a symmetric diagram. The two base angles should be equal and the two top angles should be equal. So, a trapezeium where the base angles were equal would be an isosceles trapezium.

In any cyclic quadrilateral, opposite angles would be supplementary. In a trapezium, co-interior angles between the parallel lines would be supplementary. So, if we took a trapezium ABCD with AB parallel to CD inscribed in a circle. Angle A and Angle D would be supplementary (co-interior angles). And Angle A and Angle C would be supplementary (opposite angles of a cyclic quadrilateral). Or angle B would be equal to angle C. Ergo, isosceles trapezium.


3. Orthocenter of a triangle can lie outside the triangle: TRUE

For any obtuse-angled triangle, two of the altitudes would lie outside the triangle, and would intersect at a point outside the triangle. So, the orthocenter can lie outside the triangle.

4. Triangle with sides a, b and c has the relationship a^2 + b^2 > c^2, the triangle has to be acute-angled: FALSE

Let us take triangle with sides 2, 3 and 4. 4^2 + 3^2 > 2^2. But  as 2^2 + 3^3 < 4^2, the triangle is obtuse-angled. Is a^2 + b^2 > c^2, we can say angle C is acute-angled. We cannot say all three angles are acute-angled. One can use cosine rule also for having a go at this question (though it should be considered inelegant)


5. Diagonals of a parallelogram are angle bisectors of the angles of a parallelogram: FALSE

Diagonals of a parallelogram bisect each other. They need not bisect the angles of the parallelogram. Imagine this, if we took a rectangle and studied its diagonals. if the diagonals bisected each other, the angle between diagonal and a side would be 45 degrees. Or, we would end up having a square. So, any rectangle that was not a square would have diagonals that were not angle bisectors. So, diagonals of a parallelogram NEED NOT be angle bisectors of the angles of a parallelogram.

Apr 29, 2015

CAT Online Coaching - Permutation and Combination, Fixing the Errors III

This post had given a series of questions with incorrect solutions. Given below are the "debugged" solutions to questions 7, 8 and 9.

7. What is the probability of selecting 3 cards from a card pack such that all three are face cards? what is the probability that none of the three is a face card? 

Given solution
Number of cards in a card pack = 52
Numbert of face cards in a card pack = 12
Number of ways of selecting 3 cards from a card pack = 52C3
Number of ways of selecting 3 face cards from a card pack = 12C3

Probability of selecting three cards such that all three are face cards = 12C3/52C3
Probability of selecting three cards such that none of the three are face cards = 1 - 12C3/52C3

Bug in the solution:
Other possibilities exist. As in, if we select three cards from a card pack, all three could be face cards, all three could be non-face cards, one could be a face card with 2 non-face cards or we could have two face cards and one non-face card. This is why we cannot use the 1 minus idea.

As a rule we can say P(A) = 1 – P(B) – P(C) if A, B and C are mutually exclusive and collectively exhaustive events. As in among them they should account for all possible events. And there should be no overlap. Creating a group of MECE equiprobable events is the most fundamentally brilliant idea in all of probability. Go on, look it up.

Correct solution:
This is simple. Probability of selecting three cards such that none of the three are face cards = 40C3/52C3


8. A die is rolled thrice. In how many outcomes will two throws be same and the third one different?

Given solution
Let the three outcomes be ABC.

'A' can take all values from 1 to 6
'B' can also take all values from 1 to 6 
'C' can take all values except A - so it has 5 possibilities

Total number of outcomes = 6 * 6 * 5 = 180.

Bug in the solution:
The given solution is actually absurd. If two throws are to be same, then if A can take values from 1 to 6 and if B were equal to A, then b can take only one value. There can be no 6 * 6 * 5

Correct solution:
'A' can take all values from 1 to 6
'B' should be equal to A --- One possibility 
'C' can take all values except A - so it has 5 possibilities

6 * 1 * 5 = 30 outcomes.

There are 30 possible outcomes when A = B but not equal to C. Likewise, we could have A = C but not equal to B and B = C but not equal to A. So, there are totally 90 different possibilities.

9. How many 7 letter words can we have in English that have two distinct vowels and 5 distinct consonants.

Given solution
Now, we know there are 5 vowels and 21 onsonants. So, we need to select 2 from these 5 and 5 from the remaining 21. Since order is important, we need to select keeping order in mind.

So, we have 5P2 * 21P5.

Bug in the solution:
In this solution we do not account for intermingling of vowels and consonants. As in, we do not count words such as BACEDFG. We account for order within vowels and order within consonants, but we do not account for order across both categories.

Correct solution:
Select without accounting for order, then arrange everything put together

So, we have 5C2 * 21C5 * 7!.

Apr 28, 2015

CAT Online Coaching - Permutation and Combination, Fixing the Errors II


This post had given a series of questions with incorrect solutions. Given below are the "debugged" solutions to questions 4, 5 and 6. The questions and the incorrect solutions here

4. In how many ways can we select 2 cards from a card pack such that both belong to the same suit?

Given solution
The first card can be selected in 52 ways, as it can be any card in the pack. Now, there are 12 different ways of selecting the second card. Depending on the suit of the first card, the choice set we have for the second card gets limited. If the first card were a club, then the second card must be from among the 12 remaining clubs and so on.

So, total number of outcomes = 52 * 12.

Bug in the solution:
We are double-counting when we do this. If we select 2H as the first card and 5D as the second card; this is same as selecting 5D as the first card and 2H as the second card. However, in our current approach, we will end up counting both of these.

Correct solution:
The correct approach would be to select one of the four suits first and then select 2 cards from 13 in that suit. Or, the number of ways of doing this = 4 * 13C2 = 52 * 6.


5. When a coin is tossed 10 times, what is the probability of getting more heads than tails?

Given solution

When a coin is tossed 'n' times, the probability of getting more heads than tails = Probability of getting more tails than heads. So, we do not need to really compute this. Probability = 1/2.

Bug in the solution:
We are not counting the scenarios where there are an equal number of heads and tails. If we threw the coin an odd number of times, the probability of getting more heads than tails will be ½.

Correct solution:
Number of heads could be 6, 7, 8, 9, or 10. Probability = ( 10C6 + 10C7 + 10C8 + 10C9 + 10C10)/ 210

The other way of thinking about this would be ½ (1 – 10C5)/ 210. If we removed the scenario where the number of heads and tails are equal, the probability of more heads than tails should be equal to probability of more tails than heads.

6. In how many ways can we arrange 3 boys and 3 girls on a straight line such that no two boys stand next to each other

Given solution
We can arrange the 6 people as BGBGBG or GBGBGB. If we arranged them as BGBGBG, we would have 3! * 3! ways of arranging the 3 boys and the 3 girls. So, total number of possibilities = 2 * 3! * 3! = 72 ways.

Bug in the solution:
We are not accounting for arrangement such as BGBGGB and BGGBGB. Note that the question says that no two boys should stand next to each other. Two girls can stand next to each other.

Correct solution:

There are 4 possible outlines – BGBGBG, GBGBGB, BGBGGB or BGGBGB. So, total number of possibilities = 3! * 3! * 4 = 144. 

Apr 25, 2015

CAT Online Coaching - Permutation and Combination, Fixing the Errors

This post had given a series of questions with incorrect solutions. Given below are the "debugged" solutions to the first three questions.

1. Five boys need to be allotted to 4 different rooms such that each boy is allotted a room and no room is empty. In how many ways can this be done?

Given solution
Let the boys be A, B, C, D and E. Let the rooms be 101, 102, 103 and 104. Now, we know that exactly one room will have two occupants. First, let us try to send 4 boys to 4 rooms, we can worry about the fifth occupant later on. Let us select 4 out of the 5 boys first. This can be done in 5C4 ways. Now, these 4 can be allotted to 4 different rooms in 4! ways. So, 4 boys in 4 rooms can be done in 5C4 * 4! = 5 * 24 = 120 ways.

Now, the fifth boy has to go into one of the rooms. He can do this in 4 ways as there are 4 different rooms available. So, total number of outcomes = 120 * 4 = 480.

Bug in the solution:
We end up double-counting here.

A, B, C and D could be allotted rooms 101, 102, 103 and 104 in that order. Post this, E could “double-up” with A. So, we would have A and E in 101, B in 102, C in 103 and D in 104.

In another scenario, E, B, C, D could be allotted 101, 102, 103 and 104 in that order. Post this, A could “double-up” with E. So, we would have A and E in 101, B in 102, C in 103 and D in 104.

The two scenarios mentioned above are identical. However, we end up counting both. This is the reason for the double count.

Note that in our method, we end up having a ‘first’ occupant for a room and a ‘second’ occupant. We say, A goes into room number 101 and then E ‘joins’ him. The moment you do that, ‘order’ creeps in. We end up factoring in order when we shouldn’t.

Awesome, isn’t it.

Correct solution:
The boys should be allotted into rooms as 2 + 1 + 1 + 1. As in, exactly one room has 2 occupants. Some two guys have to be room-mates. Let us first select these two room-mates. This can be done in 5C2 ways. After we have selected these two room-mates, we practically need to just allot these 4 ‘groups’ into 4 rooms. This can be done in 4! Ways.

Total number of options = 5C2 * 4! = 10 * 24 = 240 ways.


2. How many 4-digit numbers with 4 distinct digits are possible?

Given solution
Let the 4-digit number be 'abcd'. Now,
'd' can take 10 possible values - 0 to 9. 
'c' can take 9 possible values - 0 to 9 except d
'b' can take 8 possible values - 0 to 9 except d and c
'a' can take 6 possible values - 1 to 9 except d, b and c

So, total number of possibilities = 6 * 8 * 9 * 10 = 4320.

Bug in the solution:
The simple way of stating the bug is as follows “But one of b, c, d might have been zero”.

When we say a can take 6 possible values. We eliminate 0, b, c, and d from the list 0 to 9. If b, c or d were zero, then that in that case, a will still have 7 possible options.

In this question, the leading digit has a constraint. So, start with that. No point factoring in the constraint in the last step. Get that out of the way first.

Correct solution:
Let the 4-digit number be 'abcd'. Now,
'a' can take 9 possible values - 1 to 9. 
'b' can take 9 possible values - 0 to 9 except a
'c' can take 8 possible values - 0 to 9 except a and b
'd' can take 7 possible values - 1 to 9 except a, b and c

Total number of options = 9 * 9 * 8 * 7.


3. In how many ways can we rearrange the letters of the word TWOIIM such that the vowels appear together?

Given solution
Let us take the three vowels together and call it as X. Now, we need to rearrange the letters of the word TWMX. This can be done in 4! ways. Or, there are 24 ways of rearranging the letters of the word TWOIIM with vowels appearing together.

Bug in the solution:
We do not account for the different variations that ‘X’ can come in. Now, the three vowels have been replaced with X. Or, X is nothing but OII in some order. The key point here is that X could be OIO, IOO or OOI. X can take three different forms. The solution has overlooked this.

Correct solution:
Let us take the three vowels together and call it as X. Now, we need to rearrange the letters of the word TWMX. This can be done in 4! ways.

Now, the three vowels have been replaced with X. Or, X is nothing but OII in some order. The key point here is that X could be OIO, IOO or OOI. X can take three different forms.

So total number of possibilities = 4! * 3 = 72 ways.

Apr 12, 2015

CAT Online Preparation - Permutation and Combination

Permutation and Combination is a classcic topic where getting something wrong is often more instructive than getting it right the first time, provided we have another look at what exatly went wrong with the incorrect solution.

To reinforce this idea of "debugging" a wrong solution in order to understand this topic better, I have given a bunch of questions with incorrect solutions that need to be fixed. So, have a go at this. Before you try, let me give a few ground rules.

1. You should try to figure out what exactly is wrong with this solution. It is not enough to say 240 is wrong as 120 is the correct answer. Try to articulate which step of the given solution is wrong. 

2. There might be 1-2 questions where the answer + solution are indeed correct. Otherwise, there is no thrill in this.

3. The best counter-solutions will get published on the blog with due credit to whoever contributed this. There might be some thrill in getting an admit to join IIM A or B or some such, but I am sure it will pale in comparison to getting a badge of honour from this blog. So, go for it.

Now, on to the questions and solutions.

1. Five boys need to be allotted to 4 different rooms such that each boy is allotted a room and no room is empty. In how many ways can this be done?

Solution: Let the boys be A, B, C, D and E. Let the rooms be 101, 102, 103 and 104. Now, we know that exactly one room will have two occupants. First, let us try to send 4 boys to 4 rooms, we can worry about the fifth occupant later on. Let us select 4 out of the 5 boys first. This can be done in 5C4 ways. Now, these 4 can be allotted to 4 different rooms in 4! ways. So, 4 boys in 4 rooms can be done in 5C4 * 4! = 5 * 24 = 120 ways.

Now, the fifth boy has to go into one of the rooms. He can do this in 4 ways as there are 4 different rooms available. So, total number of outcomes = 120 * 4 = 480.

This solution has been debugged here

2. How many 4-digit numbers with 4 distinct digits are possible?

Let the 4-digit number be 'abcd'. Now,
'd' can take 10 possible values - 0 to 9. 
'c' can take 9 possible values - 0 to 9 except d
'b' can take 8 possible values - 0 to 9 except d and c
'a' can take 6 possible values - 1 to 9 except d, b and c

So, total number of possibilities = 6 * 8 * 9 * 10 = 4320.

This solution has been debugged here

3. In how many ways can we rearrange the letters of the word TWOIIM such that the vowels appear together.

Let us take the three vowels together and call it as X. Now, we need to rearrange the letters of the word TWMX. This can be done in 4! ways. Or, there are 24 ways of rearranging the letters of the word TWOIIM with vowels appearing together.

This solution has been debugged here

4. In how many ways can we select 2 cards from a card pack such that both belong to the same suit?

The first card can be selected in 52 ways, as it can be any card in the pack. Now, there are 12 different ways of selecting the second card. Depending on the suit of the first card, the choice set we have for the second card gets limited. If the first card were a club, then the second card must be from among the 12 remaining clubs and so on.

So, total number of outcomes = 52 * 12.

This solution has been debugged here

5. When a coin is tossed 10 times, what is the probability of getting more heads than tails?

When a coin is tossed 'n' times, the probability of getting more heads than tails = Probability of getting more tails than heads. So, we do not need to really compute this. Probability = 1/2.

This solution has been debugged here

6. In how many ways can we arrange 3 boys and 3 girls on a straight line such that no two boys stand next to each other

We can arrange the 6 people as BGBGBG or GBGBGB. If we arranged them as BGBGBG, we would have 3! * 3! ways of arranging the 3 boys and the 3 girls. So, total number of possibilities = 2 * 3! * 3! = 72 ways.

This solution has been debugged here

7. What is the probability of selecting 3 cards from a card pack such that all three are face cards? what is the probability that none of the three is a face card? 

Number of cards in a card pack = 52
Numbert of face cards in a card pack = 12
Number of ways of selecting 3 cards from a card pack = 52C3
Number of ways of selecting 3 face cards from a card [ack = 12C3

Probability of selecting three cards such that all three are face cards = 12C3/52C3
Probability of selecting three cards such that none of the three are face cards = 1 - 12C3/52C3

This solution has been debugged here

8. A die is rolled thrice. In how many outcomes will two throws be same and the third one different?

Let the three outcomes be ABC.

'A' can take all values from 1 to 6
'B' can also take all values from 1 to 6 
'C' can take all values except A - so it has 5 possibilities

Total number of outcomes = 6 * 6 * 5 = 180.

This solution has been debugged here

9. How many 7 letter words can we have in English that have two distinct vowels and 5 distinct consonants.

Now, we know there are 5 vowels and 21 onsonants. So, we need to select 2 from these 5 and 5 from the remaining 21. Since order is important, we need to select keeping order in mind.

So, we have 5P2 * 21P5.

This solution has been debugged here

10. This is an interesting question - Of the 10 questions given here, how many answers are wrong? 

Solution: Can the 10th question have two possible answers?