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(#M40152383) TCS Numerical Ability question permutation and combination Keep an EYE

how many ways a batsman can score 200 with only 4's and 6's?

Asked In TCS Souradeep Guhathakurta (9 years ago)

Unsolved Read Solution (13)

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Arunima   (9 years ago)

The solution is very simple.
just form the equation as said.
i.e. 4x+6y=200
or
2x+3y=100
or 2x=100-3y
100-3y has to be an even number.
y can be 0, 2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32
which makes total of 17 cases,
hence 17 is the answer.

Is this solution Helpfull? Yes (11) | No   

Dhruv  (9 years ago)

First Case:
LCM of 4 and 6 is 12.
So, total runs made by 4's and 6's =200/12= 16.
Second Case:
Only 4's= 50*4.

Total ways =16+1= 17.

Is this solution Helpfull? Yes (7) | No   

Ananya Das  (9 years ago)

batsman can score 200 in 16 ways....

1) 32*6+2*4=200
2) 30*6+5*4=200
3) 28*6+8*4=200
4) 26*6+11*4=200
5) 24*6+14*4=200
6) 22*6+17*4=200
7) 20*6+20*4=200
8)18*6+23*4=200
9) 16*6+26*4=200
10) 14*6+29*4=200
11) 12*6+32*4=200
12) 10*6+35*4=200
13) 8*6+38*4=200
14) 6*6+41*4=200
15) 4*6+44*4=200
16) 2*6+47*4=200

As maximum no. of 6's possible within 200 is....33 that is 33*6=198...but for this case 200-198=2 ,which is not possible.....so max no of 6's can be 33-1=32.....that is 6*32=192 for this min no. of 4's required is 2 as.....200-196=8,that is 2*4.

and again we know, 6*2=4*3 =12 .

so ,the max no. 6's will be 32 and it will be decreased by two that is (30,32,30......till it comes 2) and accordingly 4 will increase by 3 that is(2,5,8,11,......till no of sixes comes 2 ).

as, the batsman have to score 200 with 4's and 6's .....we can't take ....no. of 4's or 6's as 0.

Is this solution Helpfull? Yes (3) | No (2)  

purnima digra  (9 years ago)

17
ananya u missed 50*4+6*0=200

Is this solution Helpfull? Yes (1) | No   

Pratima  (9 years ago)

batsman can score 200 in 3 ways
1)50*4=200
2)4*20+6*20=200
3)4*5+6*30=200

Is this solution Helpfull? Yes | No (6)  

MARSHAL KUMAR  (9 years ago)

let x=no.of 4,y=no. of 6
4x+6y= 200
2x+3y= 100
2x= (100-3y)/2
x= 50 - 3y/2
here x and should be whole number
therefore y will be multiple of 2
therefore maximum value of y will be 32
and minimum value of x will be 2
but all y is multiple of 2 or 0
therefore total number of ways=32/2+1 =17

Is this solution Helpfull? Yes | No (1)  

Alakananda De  (9 years ago)

Ananya...i like your solution. but you did not consider the case of 50*4=200. then the no. of ways become 17. should we consider this case because then we are taking only 4's?

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Souvik Pal  (9 years ago)

200 score can be achieved by hitting 4's only can be done in 1 way that is
50*4 +0*6 =200
now 3 fours = 2 sixes
so we can take 3 fours from that 50 and convert into 2 sixes
so the combinations will be like this
(50 - 3)*4 +(0+2)*6=200
(50-3-3) * 4 +(0+2+2)*6=200
so there are 50/3= 16 ways of scoring 200 with sixes and 1 without any 6 (i.e the 50*4 +0*6=200)
so there are 17 combinations

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milan jena  (9 years ago)

For this one we can write 4X+6y=200

If we put 2 for x and 32 for y then it will satisfy.
again for x=5 and y=30 it will be satisfy,
if we continue this series we get that value of x increases 3 whereas value of y increases only 2.
so possible cases are (2,32)(5,30)(8,28)(11,26)(14,24)(17,22)(20,20)(23,18)(26,16)(29,14)(32,12)(35,10)(38,8)(41,6)(44,4)(47,2)

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Deepak Kumar  (9 years ago)

there are lot of way to score 200 with only 4's & 6's,which are following
1) 50*4=200
2) 10*6+35*4=200
3) 20*6+20*4=200
4) 30*6+5*4=200
5) 32*6+2*4=200
.
.
.
.
.
so on ......

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Pushan Sen  (9 years ago)

17 ways.
50*4+0*6
47*4+2*6
44*4+4*6
.
.
.
.
.
2*4+32*6
for every 2 sixes he hits, he has to hit 3 less fours.
min. no of fours needed is 2, 200 is not divisible by 6,

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Kashish Bhasin  (9 years ago)

In such Cases, Take the higher Number and find out the it's maximum Multiple , Here it is 32
Check with 6*1,6*2, 6*3 and so on, We see that 1,3,5,7... do not fit in .
So Divide 32 by 2 and add 1 (1 is added for the special Case i.e. 4*50) = 17

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Shyam Agrawal  (9 years ago)

For this one we can write 4X+6y=200

If we put 2 for x and 32 for y then it will satisfy.
again for x=5 and y=30 it will be satisfy,
if we continue this series we get that value of x increases 3 whereas value of y increases only 2.
so possible cases are (2,32)(5,30)(8,28)(11,26)(14,24)(17,22)(20,20)(23,18)(26,16)(29,14)(32,12)(35,10)(38,8)(41,6)(44,4)(47,2)and (25,0)

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