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In the next Cricket World Cup T-20, Team A has decided to score 200

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In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 04 Apr 2016, 07:15
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In the next Cricket World Cup T-20, Team A has decided to score 200 runs only through 4s and 6s. In how many ways can the Team A score these 200 runs?

A. 13
B. 15
C. 16
D. 17
E. 18


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Re: In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 04 Apr 2016, 07:22
4x + 6y = 200...Sorry, no idea how to solve this. Very interested in the approach though. Nice questions, thanks for posting!
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Re: In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 04 Apr 2016, 07:38
3
Team A can score a maximum of 50 fours and a minimum of 2 fours with an interval or spacing of 3 units to accommodate the 6's.

So the number of fours scored forms an AP 2, 5, 8, ... 50 with a common difference of 3.

Number of ways of scoring 200 only through 4's and 6's = (50 - 2)/3 + 1 = 17

Answer: D


Alternate solution: 4x + 6y = 200 --> 2x + 3y = 100

x = (100 - 3y)/2 --> 100 - 3y should be even. This is possible when 3y is even.

There are 17 even multiples of 3 between 0 and 100.

Answer: D
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Re: In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 04 Apr 2016, 10:28
it takes max 34 balls ( 6* 33= 198 +1*4=202) And max 50 balls by hitting only fours all other combinations are in between hence calculating from 34 balls to 50 balls
ans = 17
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Re: In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 04 Apr 2016, 11:26
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chetan2u wrote:
In the next Cricket World Cup T-20, Team A has decided to score 200 runs only through 4s and 6s. In how many ways can the Team A score these 200 runs?
A. 13
B. 15
C. 16
D. 17
E. 18

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IMHO the best way has already been posted by Vyshak just expanding a bit -

The point to remember here is that the No of 4s and 6s must be positive whole numbers.

Further as vp101 has put forward the equation - 4x + 6y = 200

Here substitute x for the minimum value that will yield y as a positive real number as below

Attachment:
Runs.PNG
Runs.PNG [ 4.86 KiB | Viewed 2109 times ]


Scrutinize U will see a pattern of 4s scored increases in a definite manner { 2, 5 , 8 , 11......50 }

Our goal is to maximize 4s ( That will automatically minimize 6s )

So, count the possible number of 4s (The same can also be calculated by AP formula ) as follows -

Attachment:
4s.PNG
4s.PNG [ 5.42 KiB | Viewed 2108 times ]


Though a bit lengthy ( my intentions were just to expand the wonderful explanation provided by Vyshak
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Re: In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 05 Apr 2016, 08:39
chetan2u wrote:
In the next Cricket World Cup T-20, Team A has decided to score 200 runs only through 4s and 6s. In how many ways can the Team A score these 200 runs?
A. 13
B. 15
C. 16
D. 17
E. 18

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Kudos for BEST and SHORTEST Method..


So together we have 200 runs
1st way->200/4=50...which means this score can be attained in 50 balls (22/6 is not divisible so its not the other way)

2nd way-> (2 fours 32 sixes=200runs),(5 fours 30 sixes) and so on
we can see a pattern here for number of sixes 32 sixes then 30 sixes bla bla bla

so 32/2= 16 ways

combining 1st and 2nd we have 16+1=17 ways
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Re: In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 05 Apr 2016, 08:55
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1
For the combination of 4s and 6s take LCM of 4 and 6.
The answer is 12.
Dividing 200 by 12 we get 16 ways (With 8 as the remainder)

One cannot score 200 with 6s alone , so its 0 way

One can score 200 with 50 4s, so its 1 way

Total possibilities =16+0+1 =17 ways

Ans D
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Re: In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 05 Apr 2016, 19:59
chetan2u wrote:
In the next Cricket World Cup T-20, Team A has decided to score 200 runs only through 4s and 6s. In how many ways can the Team A score these 200 runs?
A. 13
B. 15
C. 16
D. 17
E. 18

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Kudos for BEST and SHORTEST Method..


Hi,

The equation made is 4x + 6y =200 OR 2x+3y=`100....
FROM 2x + 3y = 100 we can make out y has to be even...
so 3y will be a multiple of 6...

now 100 - multiple of 6 will be div by 2..
How many such numbers are there?
100/6 = 16 numbers..

One more case is when y is 0,
So total 16+1=17 ways


ans D
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Re: In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 11 Nov 2016, 06:37
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1
4x+6y=200

Divide common factor 2

2x+3y=100
3y=100-2x
3y=2(50-x)

y=2n; 50-x=3n => x=50-3n

There is no indication that both 4s and 6s must definitely be used, that means we can use either 4s, or 6s, or 4s and 6s together. So x,y ≥0
From that \(n\) can take following values:

0, 1, 2 , 3 ….. up to 16 (from 17 \(x\) becomes negative).

Total number of elements in sequence is 17
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Re: In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 12 Nov 2016, 16:41
chetan2u wrote:
In the next Cricket World Cup T-20, Team A has decided to score 200 runs only through 4s and 6s. In how many ways can the Team A score these 200 runs?
A. 13
B. 15
C. 16
D. 17
E. 18


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Not sure why this question has been categorized as "Combinations". All it is asking you is to find out # of pairs of x and y for which the equation \(4x + 6y = 200\) holds true.

Solution:

We know that x and y have to be non-negative integers.

However, x can not be 0 or 1 as for those values y does not have integer solution (200/6, 196/6 are not integers).

So the minimum value of x for which y has an integer solution is 2 (and corresponding y = 192/6 = 32)

Also note that LCM for (4, 6) is 12, hence every 2 6s can be replaced by 3 4s.

So we increase x by 3 and decrease y by 2 to get the next pair and so on.

1st pair -> (2, 32)
2nd pair -> (5, 30)
3rd pair -> (8, 28)
...
nth pair -> (50, 0)

x => {2, 5, 8, ..., 50}
y => {32, 30, 28, ..., 0}

All we have to do is to use formula for AP on either values of x or values of y to get the value of n.

=> 50 = 2 + 3(n-1) [Last term = 50, First Term = 2; d = 3]
=> n = 17
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Re: In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 03 Feb 2018, 00:15
raarun wrote:
For the combination of 4s and 6s take LCM of 4 and 6.
The answer is 12.
Dividing 200 by 12 we get 16 ways (With 8 as the remainder)

One cannot score 200 with 6s alone , so its 0 way

One can score 200 with 50 4s, so its 1 way

Total possibilities =16+0+1 =17 ways

Ans D


Very interesting approach will be trying this on other linear equations problem to see if it works universally,
would you be any chance have further details on this approach?
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Re: In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 03 Feb 2018, 05:52
solved in this way :
reduce the Eq to : 2x + 3y = 100
now we can have a solution when Y = 0 (only 4 hits ) = 1 , but when x = 0 no solution = 0 ( only 6's )
we can have a integer solution for 2x + 3y = 100 when 3y = even(as 2x will always be even) , that is 3y = 6,12,18.....altogether 16 ways ...
total 16 + 1 = 17
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Re: In the next Cricket World Cup T-20, Team A has decided to score 200  [#permalink]

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New post 03 Feb 2018, 07:14
chetan2u wrote:
In the next Cricket World Cup T-20, Team A has decided to score 200 runs only through 4s and 6s. In how many ways can the Team A score these 200 runs?

A. 13
B. 15
C. 16
D. 17
E. 18


Kudos for BEST and SHORTEST Method..


Let a be the number of 4's and b be the number of 6's. Here a and b has to non-negative integers.

\(a = 50 - \frac{3b}{2}\). For a to be positive \(\frac{3b}{2} < 50\) and b should be even integer.

So, b value will from 0 - 33 and even integer... Total, 17.
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Re: In the next Cricket World Cup T-20, Team A has decided to score 200 &nbs [#permalink] 03 Feb 2018, 07:14
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