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

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Permutation and Combination

TCS Question Papers: TCS Interview Experience(91) TCS Company Profile

(#M40024336) 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?
option
a. 21
b. 20
c. 19
d. 18

Asked In TCS Hiba (11 years ago)

Unsolved Read Solution (14)

Is this Puzzle helpful?   (60)   (21) Submit Your Solution Permutation and Combination

Read Solution (14) :

Shashi Kumar Gupta  (11 years ago)

answer is 21..

we can write problem in form of eqtn..

25*b+10*a=1000
now we have to find all posiible value of (a,b)..
so,
for satisfying eqtn "b" must be even & the range of "b" will varry from
0 to 40 so the total number is.... 21

Is this solution Helpfull? Yes (35) | No (28)  

akhilesh tiwari  (11 years ago)

c.19
25 oranges boxes can be packed in this fashion
(2,4,6,8,.......38) (since 0 and 40 cant be possible because in dat case type a become 0 or in later one type b become 0)
so 2 to 38 will comprises of 19 box

Is this solution Helpfull? Yes (31) | No (21)  

Pooja Udayan  (11 years ago)

ans is 21
25x+10y=1000
from x=0 to x=40 with an increment of 2 and y=100 to y=0 with a decrement of 5 satisfies the equation in 21 ways.

Is this solution Helpfull? Yes (14) | No (8)  

Shankara Prabu  (9 years ago)

Box A can contain 10 oranges and Box B can contain 25 oranges so the equation is XA + YB=1000, we know the values for A&B so Equ becomes 10X + 25B=1000 => 2X + 5B=200, so fro the equation it is clear that the basic two values of X,Y are (0,40) & (100,0). Clearly if X increases by 5 then Y should decrease by 2 in order to maintain the constant value i.e 1000 oranges. so the different combinations will be (50,20) (55,18) (60,16) (65,14) (70,12) (75,10) (80,8) (85,6) (90,4) (95,2) (100,0) (45,22) (40,24) (35,26) (30,28) (25,30) (20,32) (15,34) (10,36) (5,38) (0,40 ) so there are 21 different combination by which we can arrange the oranges in the carton.

Is this solution Helpfull? Yes (13) | No (3)  

Manohar Kumar  (11 years ago)

Here we have 10a +25b=1000
=> a=(200-5b)/2
Now a and b both should be integer so we can take all the even value for b lying between 0 and 40(excluding 0 and 40).
Thus getting total of 19 possible combinations.

Is this solution Helpfull? Yes (9) | No (8)  

PRAKHAR JOHRI  (11 years ago)

answer is c.(19)
check for different combinations of box type a and b like
95 box of type a and 2 of type b
90 box of type a and 4 of type b
............
5 box of type a and 38 of type b
avoid boundary conditions(100 box of type a and 0 of type b && 0 box of type a and 40 of type b)

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

kapde moazzam wahid  (7 years ago)

for 100 oranges
take the boxes like as follows
A:10 and B:25 (capacity)
now find possible comibanations
A=0 B=25*4
A=10*5 B=25*2
A=10*10 B=25*0

add possiblilities
(0+4)+(5+2)+(10)=21 boxes

Is this solution Helpfull? Yes (3) | No   

KC  (8 years ago)

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 total = 19+1+1=21

Is this solution Helpfull? Yes (2) | No   

Ritika M  (11 years ago)

The answer is 19.

Is this solution Helpfull? Yes (1) | No (12)  

naman jha  (9 years ago)

Answer: a

Given that 10a + 25b = 1000 ⇒ 2a + 5b = 200
One general solution for the above equation is when a = 0, then b = 40.
To get other solutions, a is increased by 5 and b is decreased by 2. So next solution is (5, 38)
Now b is reduced upto 0.
So total solutions = 40−02+1 = 21

Is this solution Helpfull? Yes (1) | No (2)  

Saibal Chakraborty  (9 years ago)

c.)19
Soln.
1000=25b+10a
The pairs possible
(2,95)
(4,90)
(6,whatever)
.
.
.
.
(38,whatever)
as odd values of box b results in no proper box of a
so count no of terms in series 2,4,6,8,10...........upto 38
hence no of terms=19

Is this solution Helpfull? Yes | No (3)  

Natarajan. R  (9 years ago)

please say the correct answer pls....

Is this solution Helpfull? Yes | No   

Sourajit Saha  (8 years ago)

ans is 19

10a+25b=1000
2a+5b=200
b=40-2/5*a
so we get when a=0 b=40
a=5 b=38
a=10 b=36
.............
a=100 b=0
here we exclude a=0 and a=100
so last term,Tn=95=5+(n-1)5
so,n=19

Is this solution Helpfull? Yes | No (1)  

Prabhu  (6 years ago)

Given that 10a + 25b = 1000 ⇒ 2a + 5b = 200
One general solution for the above equation is when a = 0, then b = 40.
To get other solutions,
a
is increased by 5 and
b
is decreased by 2. So next solution is (5, 38)
Now
b
is reduced upto 0.
So total solutions =
40
−
0
2
+
1
= 21

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.