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

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


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 -


Though a bit lengthy ( my intentions were just to expand the wonderful explanation provided by Vyshak

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

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

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

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

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

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

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.

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

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

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?

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.