CAT preparation - Ratio and Proportion (Tough)

Question

John has chocolates of types A and B in the ratio 3 : 7, while Mike has chocolates of types B and C in the ratio 5 : 4, Ram has chocolates of types C and A in the ratio 3 : 5. If there are more chocolates of type C than of type B, and more of type B than of type A, what is the minimum possible number of chocolates overall?

Explanation

Again, big thanks to Mukund Sukumar for excellent solution.

Let john have 3x chocolates of type A and 7x of type B

Let Mike have 5y chocolates of type B and 4y of type C

Let john have 3z chocolates of type C and 5z of type A

So in total A=3x+5z ; B=7x+5y ; C=4y+3z

Since C>B we get solving y<3z-7x --->(1)

Since B>A we get solving 5y>5z-4x ---> (2)

What gets inferred from above 2 statements is z>=3. so when x=1,z=3, we get only y=1 as choice, for which second condition doesnt satisfy.

So, when x=1,z=4, we get y<5 from first condition and when y > 3.2 from second condition. so which gives choice the only y=4.

Hence x=1,y=4 and z=4 works and is the best possible answer.

For these values, we get A=23,B=27,C=28.

Minimum possible number of chocolates overall is 76. 

CAT Preparation Online - Tough one from Permutation and Combination

Question

A wonderful, but very tough question from Permutation and Combinations.

In how many ways, can we rearrange the word MONSOON such that no two adjacent positions are taken by the same letter? (Tough one. Tougher than what we will see in CAT)

Explanation

First up, lets get the facts out of the way – The three O’s need to be kept apart, then the 2 N’s. 

Let us focus on the three O’s.

We can place the three O’s in some blanks around the other letters. Or, three O’s can be placed in 3 slots out of the 5 in __M__N__S__N__ . This can be done in ⁵C₃, or 10 ways.

Or, the O’s can be in slots {1, 3, 5} or {1, 3, 6} or {1, 3, 7} or {1, 4, 6} or {1, 4, 7} or {1, 5, 7} or {2, 4, 6} or {2, 4, 7} or {2, 5, 7} or {3, 5, 7}  - Whew.

Now, for each of these arrangements, there are 4!/2! = 12 arrangements for the other 4 letters. But the one thing we need to keep in mind now is the fact that 2 N’s could be adjacent in these arrangements. We will need to eliminate these.

O’s in slots {1, 3, 5} or O__O__O__ __ - Ns could be in the last two slots. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 3, 6} or O__O__ __O__ - Ns could be two slots 4 and 5. There are 2! = 2 

words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 3, 7} or O__O__ __ __O - Ns could be in the slots {4, 5} or {5, 6}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 –  4 = 8

O’s in slots {1, 4, 6} or O__ __O__O__  - Ns could be in the slots {2, 3}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {1, 4, 7} or O__ __O__ __O - Ns could be in slots {2, 3} or {5, 6}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 –  4 = 8

O’s in slots {1, 5, 7} or O__ __ __O__ O - Ns could be in the slots {2, 3} or {3, 4}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 –  4 = 8

O’s in slots {2, 4, 6} or __O__O__O__  - Tehre are no possible slots for N. So, we count all 12 words on this list.

O’s in slots {2, 4, 7} or __O__O__ __ O - Ns could be in slots {5, 6}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

O’s in slots {2, 5, 7} or __O__ __O__ O - Ns could be in slots {3, 4}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10.

O’s in slots {3, 5, 7} or __ __ O__O__O - Ns could be in the first two slots. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10

Total number of words = 10 + 10 + 8 + 10 + 8 + 8 + 12 + 10 + 10 + 10 = 96.

Phew!.

There is a far more elegant method for accounting for the words where the 2 N’s appear together. This one came from Mukund Sukumar.

We need to account for the number of possibilities of N,N being together. So to subtract that part, consider 'NN' being together as one letter and place O's. In 3 out of the 4 slots in _M_NN_S_

The O’s can be selected in ⁴C₃ ways. The MNNS can be rearranged in 3! Ways if the N’s have to appear together.

Or, we get ⁴C₃ * 3! = 24 ways. So, we have 120-24 = 96 ways totally.

CAT Permutation Combination and Functions

Question

How many functions can be defined from Set A -- {1, 2, 3, 4} to Set B = {a, b, c, d} that are neither one-one nor onto? 

Explanation

To start with, if you do not know the meaning of one-one or onto, look these up. For good measure know the meanings of the terms surjective, injective etc also. CAT tests these terms.

Let us start by answering a far simpler question. How many functions are possible from Set A to Set B. This is equal to 4^4 = 256. 

Note that any function from Set A to Set B that is one-one will also be onto and vice versa. How? Why? - Think about this. Remember, tutors can only take the horse to the pond. :-)

So, we need to subtract only those functions that are one-one AND onto. Or, effectively we have to eliminate those functions where {1, 2, 3, 4} are mapped to {a, b, c, d} such that each is mapped to a different element. This is effectively same as the number of ways of rearranging 4 elements. Or, number of ways of doing this is 4! .

So, total number of functions that are neither one-one nor onto = 256 - 24 = 232. 

CAT Number Systems - Fun question based on Armstrong numbers

Question

A 3-digit Armstrong number is a three-digit number where the number is equal to the sum of the cubes of the three digits. Give a few Armstrong numbers.

Explanation

To start with, be clear that CAT does not ask questions like these. You do not need to know what Armstrong number are; nor do you need to know how to get these to crack these exam. This is just a fun question to do some trial and error with.

The 3-digit Armstrong numbers are 153, 370, 371 and 407, obtained mostly by trial and error (although a lil scientifically)

CAT Number Theory - Interesting Question from Factorial

Question

How many trailing zeroes will be present in the base 12 representation of 55!?

Explanation

The question can be restated as follows - "What is the highest power of 12 that divides 55!"

Now, 12 = 2^2 * 3. So, a number that is a multiple of 2^2 and 3 will be a multiple of 12.

So, in order to find the highest power of 12 that divides 55!, we need to look at the highest powers of 2 and 3 that divide 55!. 

Successive division by 2 will give us the highest power of 2 that divides 55!:

55/2 = 27, 27/2 = 13, 13/2 = 6, 6/2 = 3, 3/2 = 1.

Highest power of 2 that divides 55! = 27 + 13 + 6 + 3 + 1 =  50

Successive division by 3 will give us the highest power of  3 that divides 55!:

55/3 = 18, 18/3 = 6, 6/3 = 2.

Highest power of 3 that divides 55! = 18 + 6 + 2 = 26. 

So, 55! is a multiple of 2^50 and 3^26. What is the highest power of 12 that will divide 55!. Or, what is the highest value n can take such that (2^2 * 3)^n is a factor of 55!

We haev 26 threes's; but we can accommodate 2^2 only 25 times. Or, 12^25 will be a factor of 55!, while 12^26 will not, since we need 52 2's for 12^26. We have only 50 2's. 

Thus 12^25 will divide 55! - there are 25 trailing zeros in base 12 representation of 55!.

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

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)

CAT - Percents

Question: 

 

P is x% more than Q. Q is (x - 10)% less than R. If P > R, what is the range of values x can take?

A.  10% to 28%

B.  10% to 25%

C.  10% to 37%

D.  10% to 43% 

Correct Answer: (C)

Explanation:

P = Q

Q = R

R =

P > R

(100 + x) (110 – x) > 100 x 100

11,000 + 110x – 100x – x² > 10000

1000 + 10x – x² > 0

x² – 10x – 1000 < 0

x² – 10x + 25 < 1000 + 25

(x – 5)² < 1025

x – 5 < 32

x < 37

x could range from 10% to 37%

Answer Choice (C)

Difficulty level II