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

implex · 06:57 AM

Quote:

Originally Posted by implex

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Quantitative Question # 027
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Vineet has Rs 600 with him. Each day he buys either beer for Rs 100 or vodka for Rs 200 or whisky for Rs 200. In how many ways can Vineet spend all his money?

(1) 20 (2) 24 (3) 30 (4) 32 (5) none of these

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100x+200y+200z=600
x+2y+2z=6
suppose z=0
x=6,y=0( can be arranged in 1 way)
x=4,y=1( can be arranged in 5 way)
x=2,y=2(can be arranged in 6 way)
x=0,y=3( can be arranged in 1 way)

suppose z=1
x+2y=4
x=4 y=0 ( can be arranged in 5 way)
x=2 y=1 ( can be arranged in 12 way)
x=0 y=2 ( can be arranged in 3 way)
suppose z=2
x+2y=2
x=2 y=0( can be arranged in 6 ways)
x=0 y=1( can be arranged in 3 way)
suppose z=3
x=y=0 ( 1 way)

total 43 ways ..
May be some error !!
but i will go with none of these!!

smartie

let beer=b, vodka=v and whisky=w
it can be arranged as..

b b b b b b total ways of arranging=1
b b b b v/w total ways of arranging=2*5!/4!=10
b b v/w v/w total ways of arranging=2*4!/(2!*2!) +4!/2!=12+12=24
v/w v/w v/w total ways of arranging=1+1+2*3!/2!=8

so total number of ways =43
my answer wud be none of these...
i hope i am not wrong again!!

smartie

Quote:

Originally Posted by implex

100x+200y+200z=600
x+2y+2z=6
suppose z=0
x=6,y=0( can be arranged in 1 way)
x=4,y=1( can be arranged in 5 way)
x=2,y=2(can be arranged in 6 way)
x=0,y=3( can be arranged in 1 way)

suppose z=1
x+2y=4
x=4 y=0 ( can be arranged in 5 way)
x=2 y=1 ( can be arranged in 12 way)
x=0 y=2 ( can be arranged in 6 way)
suppose z=2
x+2y=2
x=2 y=0( can be arranged in 6 ways)
x=0 y=1( can be arranged in 3 way)
suppose z=3
x=y=0 ( 1 way)

total 46 ways ..
May be some error !!
but i will go with none of these!!

i think for the case z=1 ,x=0,y=2 no. of ways of arranging would be 3...

shankey_595

Quote:

Originally Posted by implex

100x+200y+200z=600
x+2y+2z=6
suppose z=0
x=6,y=0( can be arranged in 1 way)
x=4,y=1( can be arranged in 5 way)
x=2,y=2(can be arranged in 6 way)
x=0,y=3( can be arranged in 1 way)

suppose z=1
x+2y=4
x=4 y=0 ( can be arranged in 5 way)
x=2 y=1 ( can be arranged in 12 way)
x=0 y=2 ( can be arranged in 6 way)
suppose z=2
x+2y=2
x=2 y=0( can be arranged in 6 ways)
x=0 y=1( can be arranged in 3 way)
suppose z=3
x=y=0 ( 1 way)

total 46 ways ..
May be some error !!
but i will go with none of these!!

i did d same way but got d answer as 43
for z=1 d case x=0 and y=2 will have 3 cases

implex

Quote:

Originally Posted by shankey_595

i did d same way but got d answer as 43
for z=1 d case x=0 and y=2 will have 3 cases

yeah right thats a mistake, !!
it should be 3

p.s: Thanks, I knew there must be an error, I always make error in P&C

pavanpadekal · 08:10 AM

A total of 600 Must be obtained with 100's and 200's is equivalent to obtaining a total of 6 from 1 and 2
A sum of 6 can be obtained only with the following combinations.But We need to account for the 2 varieties of 2's(or 200's) in the question(i.e.,Vodka and Whisky)
2+2+2
3P3*[(2C1)^3]/3!=8 WAYS(Dividing by 3! For symmetry)
2+2+1+1
(4P4)*(1C1)*(1C1)/2!*[(2C1)^2]/2!=4*2=24 WAYS(Dividing by 2! For symmetry)
2+1+1+1+1
5*(2C1)=10 WAYS OR 5P5*(2C1)*(1C1^4)/4!=10WAYS
1+1+1+1+1+1
1 WAY OR 6P6*[1C1^6]/6!=1 WAY

43 ways in all

My Choice (5) none of these

phoenix19

Quote:

Originally Posted by smartie

let beer=b, vodka=v and whisky=w
it can be arranged as..

b b b b b b total ways of arranging=1
b b b b v/w total ways of arranging=2*5!/4!=10
b b v/w v/w total ways of arranging=2*4!/(2!*2!) +4!/2!=12+12=24
v/w v/w v/w total ways of arranging=1+1+2*3!/2!=8

so total number of ways =43
my answer wud be none of these...
i hope i am not wrong again!!

this is exactly how i did..... u've explained very well............so even i go with 43

kamalesh123

I tried the following.

No. of beers has to be one among 0, 2, 4, 6.
So

beer = 0 ==> 4 ways of buying others
beer = 2 ==> 3 ways of buying others
beer = 4 ==> 2 ways of buying others
beer = 6 ==> 1 way of buying others

So total 10 ways. Answer 5

akhilhanda_iitd

cases:

B V W Ways
6 0 0 1
4 1 0 5C4=5
2 2 0 4C2=6
0 3 0 1
4 0 1 5C4=5
2 0 2 4C2=6
0 0 3 1
0 1 2 3C2
0 2 1 3C1
2 1 1 4C2*2C1=12
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= 43
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So I'll go with (5): none of these

nbangalorekar · 09:09 AM

let 'p' be no of days on which Vineet spends 100 n 'q' be no of days on which he spends 200.
so 100p+200q=600
p+2q=6.
further he can spend the 200 in 2 ways.
consider the cases:
q=0, p=6. only 1 such case possible.
q=1, p=4, so thru 5 days he can spend money. in these 5 days, only on one day he will spend 200. so cases possible: 5*2=10.
q=2, p=2 cases: (4C2)*4 = 24
q=3, p=0 cases: 8
so total we have 1+10+24+8 = 43 cases.
so answer is '5' ie none of these