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

Question

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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Kurtosis · Accepted Answer

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

raarun · Answer

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

manishgmat1 · Answer

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

Abhishek009 · Answer

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 Runs.PNG 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 - 4s.PNG Though a bit lengthy ( my intentions were just to expand the wonderful explanation provided by Vyshak

RatneshS · Answer

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

chetan2u · Answer

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

vitaliyGMAT · Answer

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

anox · Answer

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)  
=> n = 17

sobby · Answer

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

rahul16singh28 · Answer

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.

KPBAN · Answer

Using modular congruences as follows:

4x + 3y = 200 implies 2x+ 3y = 100. Implies 2x congruent to 100 modulo 3 implies x congruent to 50 modulo 3 implies x congruent to 2 modulo 3.

Therefore x is of the form 3n+2 for all n integers. Putting into equation we get y is of the form 32 -2n for all n integers. Therefore in integers there are infinite solutions. But in positive numbers there are finite.

32 -2n >= 0 means n <=16. Therefore. Also n >=0 otherwise 3n+2 will be negative which is not allowed. So n can be any positive integer from 0 to 16, therefore answer is 17

Kinshook · Answer

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?

Let number of 4s and 6s be x & y respectively

4x + 6y = 200
2x + 3y = 100

(x,y) = {(50,0), (47,2), (44,4), ..... (32,2)}

Number of ways Team A can score these 200 runs = (50-2)/3 + 1 = 17

IMO D

DiveshR · Answer

chetan2u  I chose 16 instead of 17 because the question says "through 4s and 6s". I assumed that the combination of 4(50) + 6(0) isn't applicable because there has to be at least one six.

How can i avoid this error?

Krunaal · Answer

It  isn't explicitly stated that at least one six was hit , we are told that 200 hundred runs were scored and batsman only hit 4s and 6s, or in other words boundaries to score these runs.

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

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