there are two boxes containing certain number of packets. if 10 packets were moved from first to the second box then there will be the same number of packets in both boxes, if 20 packets are moved from second to the first box then the first box will have double the number of packets as the second box to remind the number of packets in the first box
Answers
Answer:
first box has 100packets and second box has 80 packets
Step-by-step explanation:
let first box contain x packets and box B contain y packets
in first case
x-10=y+10
x-y=20. ......eq 1
according to second case,
x+20=2(y-20)
x-2y=-60. ........eq2
now from eq 1 and 2 we have,
y=80
x=100
Answer :
No. of packets in 1st box = 100
No. of packets in 2nd box = 80
Solution :
Let the number of packets in 1st box and 2nd box be x and y respectively .
Case 1 :
When 10 packets are moved from the 1st box to the 2nd box .
Then ,
No. of packets in 1st box = x - 10
No. of packets in 2nd box = y + 10
Now ,
According to the question , if 10 packets are moved from the 1st box to the 2nd box , then there will be same number packets in both the boxes .
Thus ,
=> x - 10 = y + 10
=> x = y + 10 + 10
=> x = y + 20 ---------(1)
Case 2 :
When 20 packets are moved from the 2nd box to the 1st box .
Then ,
No. of packets in 1st box = x + 20
No. of packets in 2nd box = y - 20
Now ,
According to the question , if 20 packets are moved from the 2nd box to the 1st box , then the 1st box will have double the number of packets as the 2nd box .
Thus ,
=> x + 20 = 2(y - 20)
=> y + 20 + 20 = 2y - 2•20 { using eq-(1) }
=> y + 40 = 2y - 40
=> 2y - y = 40 + 40
=> y = 80
Now ,
Putting y = 80 in eq-(1) , we get ;
=> x = y + 20
=> x = 80 + 20
=> x = 100