# IIM CAT Preparation Tips

IIM CAT Preparation Tips: May 2013

## May 31, 2013

### CAT Progressions

Question
Second term of a GP is 1000 and the common ratio is   where n is a natural number. Pn is the product of n terms of this GP. P6 > P5 and P6 > P7, what is the sum of all possible values of n?
A.            5
B.            9
C.            8
D.            11

Choice B

Common ratio is positive, and one of the terms is positive => All terms are positive
P6 = P5 * t6 => If P6 > P5, t6 > 1
P7 = P6 * t7 => If P6 > P7, t7 < 1
T6 = t2 * r4 = 1000r4;
T7 = t2 * r5 = 1000r5
1000r4 > 1 and 1000r5 < 1
1/r^4 < 1000 and 1/r^5 > 1000
1/r = n
n4 < 1000 and n5 > 1000, where n is a natural number
n4 < 1000 => n < 6
n5 > 1000 => n > 4
n could be 4 or 5. Sum of possible values = 9

### CAT Counting: Letters and Words

Question
If all words with 2 distinct vowels and 3 distinct consonants were listed alphabetically, what would be the rank of “ACDEF’?
A.      4716
B.      4720
C.      4718
D.      4717

Choice C

The first word would be ABCDE. With 2 distinct vowels, 3 distinct consonants, this is the first word we can come up with.

Starting with AB, we can have a number of words

AB __ __ __. The next three slots should have 2 consonants and one vowel. This can be selected in 20C2 and 4C1 ways. Then the three distinct letters can be rearranged in 3! Ways.

Or, number of words starting with AB = 20C2 * 4C1 * 3! = 190 * 4 * 6 = 4560

Next, we move on to words starting with ACB
ACB __ __. The last two slots have to be filled with one vowel and one consonant. = 19C1 * 4C1. This can be rearranged in 2! Ways.

Or, number of words starting with ACB =  19C1 * 4C1 * 2 = 19 * 4 * 2 = 152

Next we move on words starting with ACDB: There are 4 different words on this list – ACDBE, ACDBI, ACDBO, ACDBU

So, far number of words gone = 4560 + 152 + 4 = 4716

 Starting with AB 4560 Starting with ACB 152 Starting with ACDB 4 Total words gone 4716

After this we move to words starting with ACDE, the first possible word is ACDEB. After this we have ACDEF.

So, rank of ACDEF = 4718

## May 23, 2013

### CAT Counting: Question with digits

This is an interesting question from Counting. Simple framework, but one needs to be very careful with the enumeration. One can get wrong answers in a number of ways.

Question
If we listed all numbers from 100 to 10,000, how many times would the digit 3 be printed?
A.      3980
B.      3700
C.      3840
D.      3780

Choice A

We need to consider all three digit and all 4-digit numbers.

Three-digit numbers: A B C. 3 can be printed in the 100’s place or10’s place or units place.

Ø  100’s place:  3 B C. B can take values 0 to 9, C can take values 0 to 9. So, 3 gets printed in the 100’s place 100 times

Ø  10’s place:  A 3 C. A can take values 1 to 9, C can take values 0 to 9. So, 3 gets printed in the 10’s place 90 times

Ø  Unit’s place:  A B 3. A can take values 1 to 9, B can take values 0 to 9. So, 3 gets printed in the unit’s place 90 times

So, 3 gets printed 280 times in 3-digit numbers
Four-digit numbers: A B C D. 3 can be printed in the 1000’s place, 100’s place or10’s place or units place.

Ø  1000’s place:  3 B C D. B can take values 0 to 9, C can take values 0 to 9, D can take values 0 to 9. So, 3 gets printed in the 100’s place 1000 times.

Ø  100’s place:  A 3 C D. A can take values 1 to 9, C & D can take values 0 to 9. So, 3 gets printed in the 100’s place 900 times.

Ø  10’s place:  A B 3 D. A can take values 1 to 9, B & D can take values 0 to 9. So, 3 gets printed in the 10’s place 900 times.

Ø  Unit’s place:  A B C 3. A can take values 1 to 9, B & C can take values 0 to 9. So, 3 gets printed in the unit’s place 900 times.

3 gets printed 3700 times in 4-digit numbers.
So, there are totally 3700 + 280 = 3980 numbers