CAT 2007: Quantitative Questions a Day 101 to 133 - The Discussions

[mehtamanas]

Originally Posted by Aarav

Part B has been correctly solved by most of you The answer of Part A is also there on the thread, we will just stop short from announcing it ... please figure out yourself.

We just wanted to ask if you people are liking the new format of Part (A) and Part (B) from the same topic, though it makes life a lil tough for us to get both the answers in the same set of options. If you have some suggestions, please feel free to express

Super Concept.....GR8 work frm u 2 guys...keep it up!!!..it helps to look at two completly diff concepts at the same time......

[Bhayanak Maut]

Originally Posted by Aarav

Part B has been correctly solved by most of you The answer of Part A is also there on the thread, we will just stop short from announcing it ... please figure out yourself.

We just wanted to ask if you people are liking the new format of Part (A) and Part (B) from the same topic, though it makes life a lil tough for us to get both the answers in the same set of options. If you have some suggestions, please feel free to express

Hey Aarav ...i really liked the idea. Although the question bein from the same topic ...i have seen that the appraoches r completely different.
Now if u feel more 33 Question from 2day wont b able to cover all the things which u want ... i dont think thrs ny harm in giving 2 diff set of options.
@Aarav n vineet..
Never got this opportunity to say...
Thnx a ton for all this...v all r gainin a lot from this ...

Thnx a lot...

[pratyush_sinha]

Originally Posted by mehtamanas

Super Concept.....GR8 work frm u 2 guys...keep it up!!!..it helps to look at two completly diff concepts at the same time......

yes its a great concept. Actually disorients the mind a bit that how can two so different questions can have answers in the same set.

Ans yes question did bring a smile

[pratyush_sinha]

aarav and vinnet,

answer for the part (a)???

[arindam83]

Originally Posted by pratyush_sinha

Ans a

getting 30 as answer.

explanation--

for
step 1 ---0 ways
step 2--- 0 ways
step 3 --- 6 ways
step 4 ---0 ways
step 5--- 24 ways (12 + 6 +6)



ans b

41

hi pratyush coul u explain the approach...

[ysairam]

Part A: 10 ways.
Part B: 41 ways

[roman2005]

Part A: 36 ways
The insect can complete in 3 or 5 steps. For 3 steps the no. of ways is 3C1 * 2C1 = 6.
For 5 steps: a) no of ways to traverse without repeating will be 3C1 * 2C1 = 6.
b) no of ways to traverse with repeating (first and second on the same edge) 3C1 * 2C1 * 2C1 = 12
c) no of ways to traverse with repeating (third and fourth on the same edge) = 3C1 * 2C1 * 2C1 = 12
Sum = 36 ways.

part B: 41 ways

[warrior]

Posted it on the wrong thread so posting it again

Question(s): (Part A)
An insect starts at vertex A of a certain cube and is trying to reach at vertex B, which is opposite A, in 5 or fewer steps where a step consists travelling along an edge from one vertex to another. The insect will stop as soon as it reaches B. The number of ways in which the insect can achieve its objective is
(Part B)
A committee of 5 is to be chosen from a group of 9 people. How many ways it can be done if Vikram and kaizen must serve together or none at all, and Sunit and Pratyush refuse to serve with each other?
(a) 40 (b) 36 (c) 41 (d) 30 (e) 48


Part B

We can have 6 cases

Let event A = Vikarm and Kaizen
B= Sunit
C= Pratyush

A+B = 10
A+C=10
A alone=10

B alone=5
C alone =5

none of A B C =1

total = 41

ans (c)

Part A

here the three vertices after A are symmetrical so find out the no of ways of going from A to B via one of the vertices

there are total 16 ways
and multiply it by 3

so 48 ways

ans (e)

[Aarav]

Hi Pratyush, the answer to Part A is 48.

@All ... actually in future we may not have just one set of options, but me and Vineet wanted to give this idea a real try and see how it works. But, yes it restricts us a lot.

[Aarav]

Originally Posted by warrior

Posted it on the wrong thread so posting it again

Question(s): (Part A)An insect starts at vertex A of a certain cube and is trying to reach at vertex B, which is opposite A, in 5 or fewer steps where a step consists travelling along an edge from one vertex to another. The insect will stop as soon as it reaches B. The number of ways in which the insect can achieve its objective is
(Part B)
A committee of 5 is to be chosen from a group of 9 people. How many ways it can be done if Vikram and kaizen must serve together or none at all, and Sunit and Pratyush refuse to serve with each other?
(a) 40 (b) 36 (c) 41 (d) 30 (e) 48


Part B

We can have 6 cases

Let event A = Vikarm and Kaizen
B= Sunit
C= Pratyush

A+B = 10
A+C=10
A alone=10

B alone=5
C alone =5

none of A B C =1

total = 41

ans (c)

Part A

here the three vertices after A are symmetrical so find out the no of ways of going from A to B via one of the vertices

there are total 16 ways
and multiply it by 3

so 48 ways

ans (e)

I never wanted any of the QQAD regulars of last year to come again next year as I had wished each one of them getting an admit of his/her choice.

It simply feels great to see you posting solutions and the only motivation I see for you in doing this is - purely the love for Maths and problem solving Keep posting!