Shuffling Students Between Examination Rooms

• From 1 condition, A-10 = B+10 ;; A-B = 20.
• From 2 condition (A+20)=2(B-20) ;; 2B-A=60.
• On solving both the equation we get B = 80 & A = 100

To solve this problem, we may convert the statements into linear equation model.

Let initial number of student in room A be "a" and initial number of student in room B be "b"

"If 10 additional students were sent from room A to room B, then the number of students in each room will be the same."

Hence, a - 10 = b + 10 --- (i)

"If however, 20 additional students were sent from room B to room A, then the number of students in room A would be double the number of students in room B."

Hence, 2(b - 20) = a + 20 ---> 2b - 40 = a + 20 --- (ii)

(i) - (ii)

• b + 10 - (2b - 40) = a + 20 - (a - 10)
• b + 10 - 2b + 40 = a + 20 - a + 10
• -b + 50 = 30
• -b = -80
• b = 80

Substitute b=80 into (i)

• a - 10 = 80 + 10
• a = 100

Hence, there are 100 students initially in Room A.

When 10 students are sent from room A to B the equation is- A-10=B+10 and this can be further simplified as- A-B=20. When 20 students are sent from room B to A the equation becomes- A+20=2(B-20) and this can be further simplified as - B=A/2 +30. Now putting the value of B in the first equation- A-A/2-30=20 OR A/2=50 OR A=100

If A is the original number of students in Room A, and B is the original number of students in Room B, then if 10 students are transferred from Room A to Room B, the new number of students in Room A is A-10, and the new number of students in Room B is B+10, and since the new numbers of students in Rooms A and B are the same, A-10 = B+10. If 20 students are transferred from Room B to Room A, the new number of students in Room A is A+20, and the new number of students in Room B is B-20, and since the new number of students in Room A is twice the new number of students in Room B, A+20 = 2(B-20) = 2B-40. A-10 = B+10, so A = (B+10)+10 = B+20. A+20 = 2B-40, so A = (2B-40)-20 = 2B-60, and since A was found out to equal B+20, A = B+20 = 2B-60, and so B+20 = 2B-60, and so (B+20)-B = 20 = (2B-60)-B = B-60 = 20 = B-60, and so 20 = B-60, and so 20-20 = 0 = (B-60)-20 = B-80 = 0 = B-80, and so 0+80 = 80 = (B-80)+80 = B = 80, and since A was found out to equal B+20, A = B+20, and since B was found out to equal 80, A = 20+80 = 100 = A =100, and so A = 100, and since the answer is the original number of students in Room A, which by definition equals A, the answer is A, and since A was found out to be 100, the answer is 100.

A- 10 = B + 10

A + 20 = 2(B - 20)

Solving for x:

A + 20 = 2(A - 40)

A + 20 = 2A - 80

so A=100

let the number of students in class A be 'x' and number of students in class B be 'y'. by given conditions, we find 2 equations which are as follows: 1)x-y=20 2)x-2y=-60 by subtracting eqn-2 from eqn-1, we get y=80 by putting the value of 'y' in eqn-1, we get, x=100 So, the number of students in class A is 100

2 equations are formed:- Let no. of students in A=x and B=y then:- x-10=y+10 2(y-20)=x+20 Solving these we get x=100,y=80 so anwer is 100

$\left\{\begin{array}{ccc}A-B=20\\2(B-20)=A+20\end{array}\right.\\\left\{\begin{array}{ccc}A-B=20\\2B-40=A+20\end{array}\right.\\\left\{\begin{array}{ccc}A-B=20\\2B-A=20+40\end{array}\right.\\ +\left\{\begin{array}{ccc}A-B=20\\2B-A=60\end{array}\right.\\ -----------\\B=80\\A-80=20\\A=20+80\\A=100$

First instance: A-10=B+10 so, A=B+20.

Second instance: Substituting B+20 for A.........B+20 +20 = 2(B-20). Therefore B+40 = 2B - 40 and then 80 = B

So if B=80, A = 80 + 20 (as shown in the first instance)

a - 10 = b + 10 ; a + 20 = 2 (b-20) use elimination method and the answer is a =100

100/80 90/90 120/60

Let x be the no. of students in A and y be in that of B. x-10=y+10 x+20=2(y-20) Solving, x=100 & y=80

Let Class A = A and Class B = B

From the first statement,
A-10 = B+10
Therefore, B = A- 20..........[1]
From the next statement,
2(B- 20) = A+ 20...........[2]
To find the number of students in classroom A:
Substitute [1] in [2]:
2(B- 20) = A+ 20
2(A- 20- 20) = A+ 20
2(A- 40) = A+ 20
2A- 80 = A+ 20
2A- A = 80+ 20
A = 100. There there are 100 students in class A.

You could use multiple methods, one would be G/C (guess and check) with logic. First, you have to figure out that Room A has 20 more students than Room B. And, so you could try two numbers with a difference of 20. I tried with 40 and 20. I took 10 students from 40 and sent them to room B. Now both of the rooms had an equal amount of students. The I sent all the students (20) from Room B to Room A. And there were 60 students in Room A and the difference was 60, and wasn't double room B, making the statement false. Then I realized that the only way the statement would be true was if room B had 80 students since 80-20=60, and 80+20=100+20=120. Causing the answer to be 100 students.

let,x is student of A and y is student of B. So we can write A=x and B=y. In first condition 10 student is transferred to B from A. So A=x-10 and B=y+10. And now A=B according to question. If A=B then x-10=y+10. we can write it x-y=20 (let it be eqn 1). Now the second condition if 20 students are transferred from B to A then A=2B. So B=y-20 and A=x+20. Since A=2B in this situation so we can writ A=2B will be like x-2y=-60(let it be eqn..2). Now if we solve eqn..1 and eqn...2 then we will find y=80 and x=100. . .. :)

A-10 = B +10, 2(B-20) = A+20, Implies A = 100

First Equation : a=b+20 second equation: 2(b-20)=a+20

solving both

2b-40=b+40 b=80 hence A= 100

The question states there are 2 exam rooms - A&B.

If ten students are sent from A to B, A is the same as B. Thus, A-10=B+10, A=B+20.

If twenty additional students are sent from B to A, A would be double B. Thus, 2(B-20)=A+20, 2B-40=A+20.

If you substitute in A from the first part: 2B-40=B+20+20, 2B=B+80, B=80.

However, the question asks for A, so you again substitute in. This time B though. A=B+20, A=80+20, A=100.

Thus, there should be 100 people in room A - and that is the answer.

let students in classroom A=x

,and in classroom B=y

according to 1st condition x-10 = y+10 i.e. x-y=20 .....1

according to 2nd condition x+20=2(y-20) i.e. x-2y=-60 ......2

solving 1&2 we get x=100 and y=80

Let students in room A & B are x and y respectively Hence, x-10=y+10 ; x=y+20 .......(1) and 2(y-20);=x+20 2y-40=y+20+20 [Substituting value of x for (1)] y=80 x=80+20=100 Hence, Students in A & B are 100 and 80 students respectively

x-10 = y+10 ----> (1) x+20 = 2(y-20) ---> (2) where x is number of students in room A , and y is number of students in room B solve to get y = 80 and x = 100

....at present, number of students in room A = A ....and in room B = B ....As per statement: (if 10 students shifted from A to B), means A-10 and B+10, they would be equal: ....A - 10 = B + 10 ....As per statement 2: (if 29 students shifted from B to A room), means B-20 and A+20, students of A would be double than ....that in B: ....A + 20 = 2(B - 20).... now by these two equations we can get : ....A - B = 20 ------------ (1) ....A - 2B = -60 ------------ (2) ....subtracting (2) from (1), we get ....B=80 ....and than ....A = 100

assume that the number of students in room A is X and in room B is Y: X-10=Y+10 .................(1) X+20=2(Y-20).............(2) BY SOLVING 1&2 then X=100 Y=80

From the problem we come up with 2 equations:
1. A - 10 = B + 10 . If we substract 10 from room A and give that 10 to room B then both are equal.
2. (A + 20)/(B - 20) = 2 . Here we have a ratio of 2:1 if we add 20 students to room A and at the same time substract 20 from room B then room A has double students than room B.

Using equation 2 we solve for B: (A + 20) = 2(B - 20) ==> A + 20 = 2B - 40 ==> 2B = A +20 + 40 ==> 2B = A + 60 ==> B = (A + 60)/2

We then substitude avove B into equation 1: A - 10 = ((A + 60)/2) + 10 == > A - 10 = ((A + 60)/2) + (20/2) ==> A - 10 = (A + 60 +20)/2 ==> A -10 = (A + 80)/2 ==> 2A - 20 = A + 80 ==> 2A - A = 80 + 20 ==> A = 100 there you go...

Proof: using A = 100 into either equations we sovle B to be equal to 80 so A = 100 B = 80 1. 100 - 10 = 80 + 10 ==> 90 = 90 !! 2. (100 + 20)/(80 - 20) = 2 ==> 120/60 = 2 !!

room A=x,, room B=y......by the equations............. x-10=y+10.......and........ x+20=2(y-20) ......there for by simultaneous equation the answer is x=100 and y=80
at x=room A ..... and....... y=room B

Students in room A =a ;Students in room B = b ; after first shuffling a-10 =b +10; a -b =10 + 10= 20 ;.....................(1) If we shuffle the other way , a+20 = 2 *(b - 20) 2 b- 40 =a +20 ' 2 b -a =20 +40 =60 ; ...................(2) adding (1) & (2) b== 80 ' from (1) a =20 +80= 100

Let the number of students in Room A be x+10, and that in Room B be x-10.

Since the number of students in Room A would be double the number of students in Room B if 20 additional students were sent from Room B to Room A, we can conclude that the equation below stands valid:

x+30 = 2(x-30)

Therefore, x+30 = 2x - 60 x-2x = -60 - 30 -x = -90 x = 90

Therefore, the number of students in Room A is x+10 = 90 + 10 = 100.

eqn fm 1st cmnt : x-100=y+10 (taking room A be x n Room B be y ) eqn fm 2nd cmnt : 2(y-20) = x+20
solve bth u'll get x =100 n y =80 !

A - 10 = B + 10 A = B + 20 .... (1)

A + 20 = 2 (B - 20) A + 20 = 2B - 40 A = 2B - 60 ... (2)

2B - 60 = B + 20 B = 80

so we get A = B + 20 = 80 + 20 = 100

let students in room A be x and that in room B be y. By the first statement we derive the relation x-10=y+10...........(1) By the second statement we derive the relation y-20= (x+20)/2..........(2) Substitute value of y from first relation ie. y= x-20 in the second relation x-20-20=(x+20)/2; x-40= (x+20)/2; 2x-80=x+20; 2x-x=20+80; x=100

A - 10 = B + 10,

therefore A = B + 20

2(B - 20) = A + 20

2B - 40 = A + 20

2B - 40 = (B + 20) + 20

2B - 40 = B + 40

B = 80

A = B + 20

A = (80) + 20

A = 100

Let a be the number of students in room A and b be the number of students in room B. The outcome of the shifting of 10 students from A to B would be a=b, which means a-10=b+10, a=b+20.

The out come of the shifting of 20 students from B to A would be a=2b, which means a+20=2(b-20), a=2b-60.

a=b+20 and a=2b-60 so b+20=2b-60 //+60, -b 80=b

a=20+b=20+80=100.

That's all.

room A had 100 students

students in room A = X and students in room B= Y
If 10 additional students were sent from room A to room B , then the number of students in each room will be the same so

X - 10 = Y + 10
X = Y + 20

If however, 20 additional students were sent from room B to room A, then the number of students in room A would be double the number of students in room B. so

X + 20 = 2 ( Y - 20 )

X + 20 = 2 Y - 40

then replacing X by Y + 20

   Y + 20 + 20 = 2 Y - 40

    40 + 40 = 2 Y - Y

    Y = 80

then replacing Y by 80 in X = Y + 20

                                            X = 80 + 20

                                      then X = 100

so students in room A = 100

Based on the problem we can say that

$A - 10 = B + 10$

$A = B + 20$

Then on the second statement, $A + 20 = 2B - 40$ .

Therefore,

$B + 40 = 2B - 40$

$B = \boxed{80}$

$A = 80 + 20 = \boxed{100}$

1. A-10 = B+10 => A=B+20
2. A+20 = 2*(B-20) Solving B=80, A=100

Let students in Room A and Room B be x and y respectively.

As per first condition, x-10=y+10 or y==x-20..............I

As per second condition, x+20=2(y-20) transposing, x+60=2y substituting value of y from equation 1 above, x+60=2(x-20) or x+60=2x-40 transposing x==100

A-10=B+10 or A=20+B

&

20+A=2(B-20) or 2B=A+60

Substituting B we get

A=100

Let $x$ be the number of students in classroom A and let $y$ be the number of students in classroom B.

Now, from the first condition,

$x - 10 = y + 10$ Thus, $x = y + 20$

From the second condition, $2 \left( y- 20 \right) = x + 20$

$2y - 40 = y + 40$ substituting the value of $x$ from the previous equation.

$y = 80$

Substituting the value of $y$ in the equation $x = y + 20$ , we get $x = 80 + 20 = \boxed{100}$

A-10=B+10

A=B+20

2A=2B+40 (1)

A+20= 2*(B-20)

A=2B-60

A=2B-60 (2)

SUBSTRACT EQ (2) FROM EQ (1)

A=100 ANSWERS

The difference in numbers is 20,that is A - B = 20 and 2 ( B - 20 ) =A + 20. On solving the two equations, we get B = 80 & A = 100

A=B+20 ---- (eq.1) A=2B-60 ----(eq.2) solving both equations: B=80 A=100

Let the number of students in room A be 'x' and room B be 'y' According to the data given, at first 10 students were sent from room A to room B. So the no of students in room A will be 'x-10' students. Whereas in room B there will be 10 extra students. So, total students in room B= y+10 By data, x-10 = y+10

              x= y+10+10

Therefore x= y+20 If 20 students were sent to room A from room B, no of students in room B will be y-20 At the same time 20 students were added to room A. So, room A will have x+20 students. By data, no of students in room A will be double the no of students in room B

            2[y- 20] = x+20
            2y-40  =  x+20 
            2y-40 = y+20+20                     [ x=y+20 see above ]
  2y-y=40+40

         y= 80

Substitution of y=80 in x=y+20

         x=80+20

         x=100

Therefore no of students in room A = 100

From the given data Let no.of students in room A is x, room B is y. x-10=y+10 x+20=2(y-20)

On solving these two you'll get x i.e number of students in Room A