Please help me with the following problem. I can't figure it out. Thanks

A total of 270 students are in classes a and b. 30 Students moved into class b from a and then there was twice as many students in class b. Find the original number of students in classes a and b.

Off hand I would say 180 and 90. However, I don't know if this is right. Could you please help me? Thanks

Scott

I'm taking it you mean twice as many as the previous total in class b?

So it would have been 240 in class a and 30 in class b, aafter the move it would now be 210 in class a and 60 in class b.

If that's not the correct assumption then maybe you meant twice as many students in class b as in class a after the move.

In which case you started with 120 in class a and 150 in class b and after the move you had 90 in class a and 180 in class b.

Hope this helps.

But how do you set it up? That is the problem I'm having.

Yes, the problem means twice as many once the 30 is added. 180/90 is what I thought. However, I'm having problems setting it up. Can you help? Thanks once again.

I think the way to look at this is:

a + b = 270 (Formula 1)

and since we require class b to be twice the size of a we get

b = 2a (Formula 2)

Therefore substitute Formula 2 into Formula 1 we get:

a + 2a = 270

so

3a = 270

a = 270/3

a = 90

Back to formula 1


90 + b = 270

b = 270 - 90

b = 180

I think that's how to do it.

Quote:

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Originally Posted by scott4544

Please help me with the following problem. I can't figure it out. Thanks

A total of 270 students are in classes a and b. 30 Students moved into class b from a and then there was twice as many students in class b. Find the orginal number of students in classes a and b.

Off hand I would say 180 and 90. However, I don't know if this is right. Could you please help me? Thanks

Scott

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This is how I see this problem.

A + B = 270 <- there are 270 students all together
B + 30 = 2B <- if you add 30 students to class B, there will be twice as many students in class B than before (in class B)

From that, we get that B = 30 and substituting it into the first formula, we get that A = 270 - 30 = 240

Then again, I'm not sure I understood the problem correctly... perhaps it was meant like this: If you move 30 students from class A into class B, then class B will have twice as many students as in class A (after the 30 students are moved)

That would give us a different formula:
B + 30 = 2*(A - 30)
Since we know that
A + B = 270 -> A = 270 - B
we can substitute that into first formula and get
B + 30 = 2*(270 - B - 30)
which means that
3B = 540 - 60 - 30 = 450 -> B = 150 -> A = 120

But perhaps I misunderstood the problem AGAIN! :) It's all about that sentence: "30 Students moved into class b from a and then there was twice as many students in class b". Twice as many as WHERE is the key.

Also, note that 180 and 90 CANNOT be right solutions.

Reason why? The problem asks for ORIGINAL values of A and B. If A = 90 and B = 180, that means that if we add 30 students to class B (there would then be 210 in B), there should be twice as many students in there. In whatever way you understand that phrase "twice as many students in there" it is not right. If there should be twice as many students in B after adding 30 than it was before in B, that would mean that 210 = 180*2, which is not true. If there should be twice as many students in B (after adding the 30) then there is in A (without those 30), that would mean that 210 = 2*60, which is again not true.

However if B = 150, and A = 120, that means that if 30 were added to B, there would be 180 students in B, which is twice as many as in A (without the 30 that were given to class B).

180 = 150 + 30 = B + 30 = 2*(A - 30) = 2*(120 - 30) = 2*90 = 180
See that the left and right side are correct this time.

Hi,
Let a = number of studends originally in class a
Let b = number of students originally in class b.
a + b = 270
a - 30 = 2b (30 students taken from a equals now twice those in b)

a = 2b + 30
Now substitute back in 1st equation:
(2b + 30) = b = 270
3b + 30 = 270
3b = 240
b = 80
substitute,
a + 80 = 270
a = 190.

There were originally 190 in a, and 80 in b.
check:
190 + 80 = 270; or 270=270
and,
190 - 30 = 2 x 80, or 160 = 160

Answers, a = 190; b = 80
Best wishes,
fredg

Hi,
This is an Algebra problem, and most teachers I know always start the problem by letting x = something, then filling in the second equation of letting x = something else.
There are many ways of solving this problem, and none is incorrect, if you come up with the correct answers.
It would probably be up to your teacher, as to how he/she wants it done.
I wish you the best, and good luck.

The answer :-

Originally :-

Class A contained 210
Class B contained 30

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After the move:-

Class A contained 210 - 30 = 180.

Class B contained 60 + 30 = 90.

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Both equations are satisfied:-

Originally a + b = 210+60 = 270

Subsequently a + b = 180+90 = 270 and 180 in a is twice 90 in b.