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

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Oranges can be packed in sets of 10 oranges in a box type A or 25 oranges in box type B.A carton comprising of 1000 oranges of type A and type B is packed. How many different combinations possible in the number of type A and type B boxes while organizing the oranges?
a)21 b)20 c)19 d)18

Asked In TCS sarika nayak (11 years ago)

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vemavarapu moulika  (11 years ago)

for given data, the expression will be 10x+25y=1000
the above exp will be 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
so tatal = 19+1+1=21

Is this solution Helpfull? Yes (26) | No (8)  

Kundan Kumar  (11 years ago)

ans is 21
AS for Type B it can be 0 50, 100,150,200,,(similarly(by adding 50)) upto 1000.
There are 21 possible ways for packing the 1000 oranges.

Is this solution Helpfull? Yes (5) | No (2)  

pankaj choudhary  (11 years ago)

pairs will be solution of 10a+25b=1000
that could be (5,38),(10,36),(15,34).......(95,2)
so possible nos are 19 if a and b type is there and if it may be possible that only one will be their then possible pairs will be 19+2=21....

Is this solution Helpfull? Yes (4) | No (3)  

GAURAV JAISWAL  (11 years ago)

I think its 19 bcoz there are both type A and type B oranges packed in carton

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Gaurav Kumar  (11 years ago)

.A carton comprising of 1000 oranges of type A and type B
this statement says about both type so I think ans should be 19
by removing only A and only B sets.

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Tanuma Patra  (11 years ago)

10x+25y=1000
=>2x+5y=200
now,x can take values 0,5,10,15..100
0 being the case of all type b and 100 the case of all type a.
using A.P. 0+5*n=100
=>n=20
so according to me the ans should be 20..

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S.Rajashree  (11 years ago)

is it 21 or 19? anyone pls do explain me

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Divya  (11 years ago)

answer is 21

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VJ  (9 years ago)

as per the given question the equation formed is
10a+25b=1000
=> 2a+5b=200, put a=0
we get b=40
only integer values possible so, put a=5 we get b=38 , rest combinations are
(a,b)=(0,40), (5,38),(10,36),(15,34),----------------(100,0) = 21 combinations

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