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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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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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[Reveal] Spoiler: OA

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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, 06:22
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, 06:38
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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, 09:28
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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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

Self Made : OA - 2 days
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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 1625 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 1624 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, 07:39
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

Self Made : OA - 2 days
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, 07:55
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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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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

Self Made : OA - 2 days
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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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, 15:41
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


Kudos for BEST and SHORTEST Method..


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 02 Feb 2018, 23:15
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, 04:52
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, 06:14
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   [#permalink] 03 Feb 2018, 06:14
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