MHB Setting up systems of equations

AI Thread Summary
The discussion revolves around a problem involving two individuals, Ann and Betty, and their expenditures. Ann initially has 20 less than Betty, and after spending portions of their money, the remainder of Ann's is 5/6 of Betty's remainder. The key equation derived is 1/4x = 5/6(1/5(x + 20)), leading to confusion about how the term 50/3 appears in the solution. Participants agree on the setup but question the steps leading to that term, suggesting that a correct sequence of operations might clarify the confusion. The consensus is that while the answer is correct, the process leading to it may contain errors.
TracyThomas
Messages
2
Reaction score
0
I have this problem in my book:
Ann had 20(Cash) less than Betty. Ann spent 3/4 of her money, while Betty spent 4/5 of hers. Then Ann's remainder was 5/6 of Betty's remainder. If Ann had x(Cash) originally, form an equation in x and solve it.

And this solution in the answer key:

x=Ann's (Cash) x+20=Betty's (Cash)

Remainder of Ann's money = 1/4x
Remainder of Betty's money = 1/5(x+20)

1/4x = 5/6 (1/5(x + 20))
x/4 = x/6 + 50/3

12(x/4) = 12(x/6) + 12(50/3)
3x = 2x + 40
Survey says:

x = 40

The question is: How did the 50 in 50/3 get there?
(Wondering) (Worried) Can someone please explain that step so both me and my homeschool teacher can understand it?
 
Mathematics news on Phys.org
Re: Ann & Betty got rich $$$--But where did the 50 come from?

TracyThomas said:
I have this problem in my book:
Ann had 20(Cash) less than Betty. Ann spent 3/4 of her money, while Betty spent 4/5 of hers. Then Ann's remainder was 5/6 of Betty's remainder. If Ann had x(Cash) originally, form an equation in x and solve it.

And this solution in the answer key:

x=Ann's (Cash) x+20=Betty's (Cash)

Remainder of Ann's money = 1/4x
Remainder of Betty's money = 1/5(x+20)

1/4x = 5/6 (1/5(x + 20))
x/4 = x/6 + 50/3

I agree with the setup (usually the hard part!) You are right to question this last step, though. I would do this:

x/4 = 5/6 (x/5 + 4)
x/4 = x/6 + 20/6 = x/6 + 10/3
12 (x/4 = x/6 + 10/3)
3x = 2x + 40
x = 40.

I think it's a question of incorrect processes somehow coming to the right answer.

Does this make sense? I'm not sure I understand how the 50 got in there, but perhaps a correct (I hope) sequence of steps will clear things up.

12(x/4) = 12(x/6) + 12(50/3)
3x = 2x + 40
Survey says:

x = 40

The question is: How did the 50 in 50/3 get there?
(Wondering) (Worried) Can someone please explain that step so both me and my homeschool teacher can understand it?
 
Re: Ann & Betty got rich $$$--But where did the 50 come from?

Thanks! I haven't found ANY errors in the Singapore Math textbook answer keys in many years of using the program, but this answer key is written by someone else, I think, so I suppose it could be in error.

Ackbach said:
I agree with the setup (usually the hard part!) You are right to question this last step, though. I would do this:

x/4 = 5/6 (x/5 + 4)
x/4 = x/6 + 20/6 = x/6 + 10/3
12 (x/4 = x/6 + 10/3)
3x = 2x + 40
x = 40.

I think it's a question of incorrect processes somehow coming to the right answer.

Does this make sense? I'm not sure I understand how the 50 got in there, but perhaps a correct (I hope) sequence of steps will clear things up.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Back
Top