will u solve for meeee ppppllll?

Question

2 groups of cadets are facing each other. Leader of first group tells the leader of second group, lend me one cadet then I will have double the cadets in my group compared to yours.

leader of first group says to the second group, why dont u lend me one cadet so that strength of both groups is same.

Can u tell me mathematically what was the strength of each group and total cadets.

Answer 1

the answer is what is given by sweet tooth....its 100% correct. all the best..

Answer 2

It seems to be an eight standard alegbric problem. The answer is: number of cadets in first and second groups are respectively 7 and 5.
Let x is the number of cadets in group A and y is that in group B.
(x+1)=2(y-1) and
(x-1)=(y+1)
Solve the above two equations and you get ur answer.

Answer 3

Let no of cadets in 1st group = x
Let no of cadets in 2nd group = y

>> leader of second group, lend me one cadet then I will have double the cadets in my group compared to yours

Therefore, y+1 = 2 (x-1)
=> y+1 = 2x-2
=> 2x-y = 3 ------------ [Equation 1]

>> leader of first group says to the second group, why dont u lend me one cadet so that strength of both groups is same.

Therefore, x+1 = y-1
=> x-y = -2 ------------ [Equation 2]

Subtracting Equation 2 from 1
2x-y - (x-y) = 3 - (-2)

x = 5

Substituting x in equation 2,
5-y = -2
=> y = 7

No of cadets in 1st group = x = 5
No of cadets in 2nd group = y = 7

Answer 4

Wow , very good brains. That would be
Grp 1 = 7
Grp 2 = 5
Through analysis:
If Grp 1 borrows 1 cadet from Grp 2. It will have 8 cadets which will not become twice compared to Grp 2.
On the other hand, if Grp 1 lends 1 cadet to Grp. 2, then It will have 6, thereby making the number equal.

Answer 5

Set up of the question is wrong.
Let Group [I] has--------X cadets and Group[2] has-------Y cadets.
In both cases in the question Group[1] says to Group[2]....
According to the question:-
X+1=2[Y-1] or X-2Y=-3 --------[a] and applying second condition;
X+1=Y-1 or X-Y=-2 ---------[b] ; Solving [a] and [b] we get values as X= -1 and Y=1 which is absurd.

Answer 6

seems that everyone knows the ans.
may be this can resolve u.
fine
let the leader one team be A and the other be B.
the no of cadets in A is x and in B is y.
step 1 and condition 1:
A takes one cadet from B and becomes
double of B(i.e)
2x=y+1

step 2 and condition 2:
But B says that he takes one from A and becomes equal with A
(i.e)
x-1=y
step 3:
replace y as x-1 in y+1=2x
(i.e)

Answer 7

One group has 7, the other group has 5. When the group with 7 borrows one, it creates a group of 8, and leaves a group of 4. When the group with 5 borrows one, it makes two groups of 6.

Answer 8

GR 1 GR 2
X Y

X-1 Y+1
But (Y+1) = 2(X-1)
So Y = 2X-3

X+1 Y-1
But (X+1) = (Y-1)
So X+2= Y
Now solve the 2 equations
Ans X= 5 and Y= 7.

see ya.

Answer 9

7 & 5 : 12

Answer 10

I think the answer could be 3 members in each group and totally 6 members.

Group1=3
Group2=3

When group1 borrows 1 person from group2 it becomes group1=4 members(double the amount of the group2)
When Group2 borrows again 1 person from Group1 it becomes equal=3+3(both groups have same strength)

Answer 11

ur answer may be 5 &7 as well as 3 &3 both