Thursday, October 23, 2014

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.





Wednesday, October 15, 2014

CAT Preparation Online - A simple question from Quadratic Equations

Questions

x^2 - 17x + |p| = 0 has integer solutions. How many values can p take?

Explanation

To begin with, if we assume roots to be a and b, sum of the roots is 17 and product of the roots is |p|. Product of the roots is positive and so is the sum  of the two roots. 

So, both roots need to be positive.

So, we are effectively solving for 

Number of positive integer solutions for a + b = 17.

We could have (1,16), (2, 15), (3, 14)......(8, 9) - There are 8 sets of pairs of roots. Each of these will yield a different product.

So, |p| can take 8 different values. Or, p can take 16 different values.

is that it? Or, are we missing something? Can p be zero? What are the roots of x^2 - 17x = 0. This equation also has integer solutions. 

So, p can also be zero.

So, number of possible values of p = 16 + 1 = 17.

Wonderful question - chiefly because there are two really good wrong answers one can get. 8 and 16. So, pay attention to detail. No point telling yourself "Just missed, I just didnt think of zero. I deserve this mark". Being just wrong, will earn us a -1 instead of the honourable 0.


Tuesday, October 14, 2014

CAT Preparation Online - Simple one from Quadratic Equations

Question

How many integer solutions exist for the equation x2 - 8|x| - 48 =0?

Explanation


One approach is to solve when x > 0 and then solve for x < 0. However, there is a slightly simpler method.

Note that x2 is the same as |x|2, so we can treat the equation as a quadratic in |x|.

Or, |x|2 – 8|x| - 48 = 0
(|x| - 12|) (|x| + 4) = 0
|x| could be 12. |x| cannot be -4.

If |x| could be 12, x can be -12 and 12.


Two possible solutions exist.