CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[sabsebadapaagal]

Originally Posted by Aarav

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Quantitative Question # 020
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For all integers x, y, f(x, y) is defined as f(x+2, y+1) = f(f(x+1, y),f(x, y)) and f(x+1, 0) = f(x, 1), then f(f(2, 3), f(2, 2)) =

(1) f(4, 5) (2) f(3, 3) (3) f(3, 4) (4) f(4, 3) (5) none of these

Hi Aarav

Can you pls clarify whether both f(x,y) and f(x+1,o) are equal to f(x,1) or just f(x+1,o) ?

[sabsebadapaagal]

Originally Posted by sabsebadapaagal

Hi Aarav

Can you pls clarify whether both f(x,y) and f(x+1,o) are equal to f(x,1) or just f(x+1,o) ?

Sorry , I got it .... I didnt read the quest properly

My appologies once again

[Aarav]

I'm sorry for sending today's question late -> had forgot to set the timer of the NL.
Please read last year problems also and try solving that yourself 1st before looking at the solution.

If there is someone who doesn't get the NL in next 45 minutes i.e. before 10 am, then please inform us.

[sabsebadapaagal]

Originally Posted by Aarav

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Quantitative Question # 020
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For all integers x, y, f(x, y) is defined as f(x+2, y+1) = f(f(x+1, y), f(x, y)) and f(x+1, 0) = f(x, 1), then f(f(2, 3), f(2, 2)) =

(1) f(4, 5) (2) f(3, 3) (3) f(3, 4) (4) f(4, 3) (5) none of these

On plugging x=2 , y=2

we get f(f(3,2), f(2,2)) =f(4,3)

so f(f(2, 3), f(2, 2)) comes out to be f(3,4)

i.e option 3

(the above conclusion is based on following observation

f(3,2)= f(f(2,1) + f(2,0))
f(2,3)= f(f(1,2) + f(0,2))
)

I know this is not the right way of doing it

[spidy_rules]

Hey,Please answer this:
Find the sum and no. of factors of 157! divisible by 18.

choices
a)36
b)37
c)38
d)39.

Please explain the process

[rik_12]

i am gettin option 5...i.e none of these.

here's how:

f(x+2,1)= f( f(x+1,0) , f(x,0) ) (putting y=0 in d 1st function)
= f( f(x,1) ,f(x,0) ) (using 2nd function)
= f( f(x,1) ,f(x-1,1) ) (using 2nd function).........(3)


now, f(2,2) = f( f(1,1) , f(0,1) ) (x=0 n y=1 in function 1)
= f(3,1) (using 3 where x=1)

again f(2,3) = f( f(1,2) , f(0,2) ) (x=0 n y=2 in function 1).....(4)

f(1,2)=f( f(0,1) , f(-1,1) ) (putting x=-1 n y=1 in function 1)
= f(2,1) (using 3 where x=0)

f(0,2) =f( f(-1,1) , f(-2,1) ) (putting x=-2 n y=1 in function 1)
=f(1,1) (using 3 where x=-1)

Substituting d values of f(1,2) and f(0,2) in (4),

f(2,3)=f( f(2,1), f(1,1))
=f(4,1) (using 3 where x=2)

Thus, f( f(2,3),f(2,2) )= f( f(4,1),f(3,1) )
=f(6,1)

hence option 5.

Aarav plzz correct me if i'm wrong!

[dhruvasagar]

Originally Posted by Aarav

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Quantitative Question # 020
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For all integers x, y, f(x, y) is defined as f(x+2, y+1) = f(f(x+1, y), f(x, y)) and f(x+1, 0) = f(x, 1), then f(f(2, 3), f(2, 2)) =

(1) f(4, 5) (2) f(3, 3) (3) f(3, 4) (4) f(4, 3) (5) none of these

f(x+2, y+1) = f(f(x+1, y), f(x, y)) - eq1
f(x+1, 0) = f(x, 1) - eq2

=>f(2,3) = f(f(1,2), f(0,2))

=>f(1,2)=f(f(0,1),f(-1,1))
=>f(1,2)=f(f(0,1),f(0,0)) deriving from eq2
=>f(1,2)=f(f(1,0),f(0,0)) deriving from eq2
=>f(1,2)=f(2,1) [reverse eq1.]

=>f(2,3) = f(f(2,1), f(0,2))

=>f(0,2)=f(f(-1,1),f(-2,1))
=>f(0,2)=f(f(0,0),f(-1,0))=f(1,1) reversing eq1
=>f(2,3) = f(f(2,1), f(1,1)) = f(3,2) reversing eq1

=>f(f(2,3),f(2,2))=f(f(3,2),f(2,2))=f(4,3) reversing eq1 again.

Hence (4) is right.

[vineet.nitd]

Originally Posted by Aarav

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Quantitative Question # 020
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For all integers x, y, f(x, y) is defined as f(x+2, y+1) = f(f(x+1, y), f(x, y)) and f(x+1, 0) = f(x, 1), then f(f(2, 3), f(2, 2)) =

(1) f(4, 5) (2) f(3, 3) (3) f(3, 4) (4) f(4, 3) (5) none of these

f(1,2)=f(f(0,1),f(-1,1))=f(f(0,1),f(0,0))=f(f(1,0),f(0,0))=f(2,1)
Now, f(2,3)=f(f(1,2),f(0,2))=f(f(2,1),f(1,1)=f(3,2)
Also, f(f(3,2),f(2,2))=f(4,3) =>f(f(2,3),f(2,2))=f(4,3)

[krishna13]

One box contains 8 cubes, and 6 spheres. A second box contains 6 cubes and 12 spheres. One object is transferred from the first box to the second box.
Probability that an object selected from second box to be a cube is?

Ans: 46/133

Coud anyone please help me with this question?

[dewan_iitr]

Originally Posted by Aarav

I'm sorry for sending today's question late -> had forgot to set the timer of the NL.
Please read last year problems also and try solving that yourself 1st before looking at the solution.

If there is someone who doesn't get the NL in next 45 minutes i.e. before 10 am, then please inform us.

Hi Arav,

Still not received the problem