Latest tcs Aptitude Question SOLUTION: Oranges can be packed in sets of 10 oranges in box type A or 25 oranges in box type B. A carton comprising of 1000 oranges of type a and b is packed. How many different combination

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(#M40166217) TCS Numerical Ability question Keep an EYE

Oranges can be packed in sets of 10 oranges in box type A or 25 oranges in box type B. A carton
comprising of 1000 oranges of type a and b is packed. How many different combinations are possible in the
number of type A and type B boxes while organizing the oranges?

Asked In tcs Ggopinath (4 years ago)

Unsolved Read Solution (9)

Is this Puzzle helpful?   (45)   (15) Submit Your Solution Permutation and Combination

Read Solution (9) :

Devendra Marghade  (4 years ago)

21

Let number of type A box=a and type B box=b, then 10a + 25b = 1000
Possible combinations of (a,b) are (0,40), (5,38), (10,36), ..... (100,0) which forms an A.P.
Here number of possible combinations are the number of terms of any of the A.P. series.
Consider, series 0, 5, 10, ...., 100 where a=0, d=5, number of terms=n, then Last term=a+(n-1)d
=> 100=0+(n-1)5
5n=105
n=21
Therefore, different possible combinations with type A and type B boxes are 21.

Is this solution Helpfull? Yes (13) | No (5)  

rittika  (4 years ago)

the prev sol. is wrong...guys dont do that

Is this solution Helpfull? Yes (3) | No (7)  

Aditya Vikram Singh  (4 years ago)

The expression will be 10x+25y=1000 ----------(1)
The exp (1) satisfied only with multiples of 2 up to 38.
so no. of combinations are 19.
if oranges are packed in type A boxes only= 1 way
if oranges are packed in type B boxes only= 1 way
Hence total = 19+1+1 = 21

Is this solution Helpfull? Yes (2) | No   

NABIL MAHTAB  (4 years ago)

21 combinations
A B
0 40
5 38
10 36
. .
. .
95 2

when you count the total possible number is 21

Is this solution Helpfull? Yes (1) | No   

Smriti Singh  (4 years ago)

201

Let number of type A box=a and type B box=b, then 10a + 25b = 1000
Possible combinations of (a,b) are (0,40), (5,38), (10,36), ..... (100,0) which forms an A.P.
Here number of possible combinations are the number of terms of any of the A.P. series.
Consider, series 0, 5, 10, ...., 100 where a=0, d=5, number of terms=n, then Last term=a+(n-1)d
=> 1000=0+(n-1)5
n-1=200
n=201
Therefore, different possible combinations with type A and type B boxes are 201.

Is this solution Helpfull? Yes (1) | No (9)  

Gayatri Desika  (3 years ago)

10a+25b=1000
lcm(10,25)=50
ans=(1000/50)+=21

Is this solution Helpfull? Yes (1) | No   

Swati Singh  (4 years ago)

here total number of oranges are 1000 .so the total no of oranges will be in the form of 10x+25y=1000 as type a oranges can be put 10 ina box and 25 of type b . so on making the algebra easy 10x + 25 y = 1000 can be written as x = 1000 - (25/10)y . which is also x = 1000 -(5/2)y. here y should be multiple of 2 and the product should be atmost 100.tus our combination starts from 0 ,2,4,6 ....to 20.now doing ap for total no of combinations we get 20=0+(n-1)2. n=11

Is this solution Helpfull? Yes | No (3)  

  (4 years ago)

no of A type boxes=40
No of B type boxes =1
So there will be 40 ways...select box of type A and the type B

Is this solution Helpfull? Yes | No   

Ankit Sharma  (4 years ago)

1st combination A0 B40
Last combination A100 B0
Difference in A.p of A =5
No. Of combination=21

Is this solution Helpfull? Yes | No   

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