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Discussion in 'Jukwaa la Elimu (Education Forum)' started by sulphadoxine, Dec 29, 2011.

#1
Dec 29, 2011
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There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is?

2. E=mcsquaredJF-Expert Member

#2
Dec 29, 2011
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Kwa walioko F1 muhula wa kwanza tu
Solve simultaneously:
A-10=B+10
2(A+20)=B-20

3. E=mcsquaredJF-Expert Member

#3
Dec 29, 2011
Joined: Apr 1, 2009
Messages: 218
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Kwa walioko F1 muhula wa kwanza tu
Solve simultaneously:
A-10=B+10
2(A+20)=B-20

4. SanjaJF-Expert Member

#4
Dec 29, 2011
Joined: Nov 7, 2010
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Hili swali hata ukilipata cwezi toa hata zawadi ya big g. Ukitumia hata 1 min sio fair. Ok in A kuna 60stdnts. Nawewe tafuta wa B.

5. LoreenSenior Member

#5
Dec 29, 2011
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uongo kabisa embu proove!

6. MKOBA2011Senior Member

#6
Dec 30, 2011
Joined: Jul 12, 2011
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A-10=B+10
2(A+20)=B-20.

Not true it was suppose to be
A-10=B+10
2(B-20)=B-20.Then when you solve it you will get

A - B=20 ....... (i)
2B-40=B-20...... (ii)

When we solve simultaneously we get

From (ii) 2B-B=-20+40
B=20
Substitute the value of B in (i) to get the value of A which will be
A-20=20
A=20+20
A=40

But when we solve your simultaneously we are going to get the value of A as negative which contradicts for the number of people to be negative

7. S

Song'itoJF-Expert Member

#7
Dec 30, 2011
Joined: Oct 4, 2011
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A-10=B+10------------------------(i)
2(B-20)=A+20---------------------(ii)

from equation (i) making A the subject

A= B+20---------------------------(iii)

putting (iii) in (ii)

2(B-20)= (B+20) +20
2B-40=B+40
2B-B=40+40
B=80------------------------------ (iv)

putting (iv) in (iii)

A=80+20=100-------------------(v)

hence value of A=100 and B= 80

That is to say kuna wanafunzi 100 room A na wanafunzi 80 room B