MHB How many stickers does Xinyi have

  • Thread starter Thread starter Johnx1
  • Start date Start date
Johnx1
Messages
48
Reaction score
0
Siti and Xinyi had the same number of stickers. After Siti gave away 55 stickers and Xinyi threw away 15 stickers. Xinyi had three times as many stickers as Siti. How many stickers did Xinyi have in the end?

My answer:

Siti = S
Xinyi = X
Gave/threw away stickers = TWe know:

S = X
S = T - 55
X = T -15
X = 3Sso, in short, I did elimination.

S = T - 55
-(3S = T - 15)

so, S = 20

Answer = 60.

My question is, is there a better way to do this besides elimination?
 
Mathematics news on Phys.org
Johnx said:
Siti and Xinyi had the same number of stickers. After Siti gave away 55 stickers and Xinyi threw away 15 stickers. How many stickers did Xinyi have in the end?. Xinyi had three times as many stickers as Siti. How many stickers did Xinyi have in the end?

My answer:

Siti = S
Xinyi = X
Gave/threw away stickers = T
Once again, "Siti" and "Xinyi" are people, not numbers. You mean "S is the number of stickers Siti had initially" and "X is the number of stickers Xinyi had initially". I have no idea what you mean by setting "T" equal to "Gave/threw away stickers" since that phrase is used twice in reference to two different numbers.

We know:

S = X
yes, initially Siti and Xinyi had the same number of stickers.

S = T - 55
X = T -15
These make no sense at all. I don't know what number "T" refers to.

X = 3S
You are making the same mistake you did in the previous problem. "X" is the number of stickers Xinyi had initially, not after giving away stickers. It should be clear that the two equations "X= S" and "X= 3S" can't both be true (unless X= S= 0).

Initially Xinyi had X stickers and Siti had S stickers and X= S because they had the same number of stickers. After Xinyi threw away 15 stickers, Xinyi had X- 15 left. Do you understand that? After Siti gave away 55 stickers, Siti had S- 55 left. NOW Xinyi had 3 times as many stickers as Siti: X- 15= 3(S- 55).

You need to solve the two equation X= S and X- 15= 3(S- 55). Since X= S, the simplest thing to do is replace X by S in X- 15= 3(S- 55): S- 15= 3(S- 55). Solve that equation for S.

so, in short, I did elimination.

S = T - 55
-(3S = T - 15)

so, S = 20

Answer = 60.
The answer to what question? How did you go from "S= 20" to "Answer is 60"? You never say what 'S' represents but, from your first equation, apparently 'S' was the number of stickers Siti had initially. The question asked was 'How many stickers did Xinyi have in the end?' If S= 20 then you are saying that Siti had 20 stickers initially and Xinyi had the same number, 20. If Xinyi then "threw away 15 stickers" how could Xinyi have 60 at the end? Frankly it looks like you are going through all that calculation randomly, then copying the 'answer' from the back of the book.

My question is, is there a better way to do this besides elimination?
It is not a question of "a better way". Unfortunately pretty much nothing you have done here is correct. You appear not to have a clear idea of what the letters you write as variables represent and are writing down equations with no logical reason. You really need to talk to your teacher who hopefully can help you get a better idea of how to convert sentences to equations.
 
Johnx said:
Siti and Xinyi had the same number of stickers.
After Siti gave away 55 stickers and Xinyi threw away 15 stickers.
Xinyi had three times as many stickers as Siti.
How many stickers did Xinyi have in the end?
A and B have the same number of stickers. [1]
A gives away 55 stickers and B gives away 15 stickers. [2]
B now has 3 times as many stickers as A. [3]
How many stickers does B now have? [4]

a = A's stickers at start, b = B's stickers at start

[2]: A now has a-55 stickers, B now has b-15 stickers

[3]: b-15 = 3(a-55)
b-15 = 3a - 165
3a - b = 150
[1] Since a=b:
3b - b = 150
2b = 150
b = 75
[2] Since B now has b-15 stickers:
b - 15 = 75 - 15 = 60 [4]
 
Country Boy said:
You are making the same mistake you did in the previous problem. "X" is the number of stickers Xinyi had initially, not after giving away stickers. It should be clear that the two equations "X= S" and "X= 3S" can't both be true (unless X= S= 0).

Thanks for pointing that out.

Country Boy said:
The answer to what question? How did you go from "S= 20" to "Answer is 60"?

I skipped a few steps a few steps, but I times X = 3S, so that's how I got my answer.

Country Boy said:
It is not a question of "a better way". Unfortunately pretty much nothing you have done here is correct. You appear not to have a clear idea of what the letters you write as variables represent and are writing down equations with no logical reason. You really need to talk to your teacher who hopefully can help you get a better idea of how to convert sentences to equations.

I'm not trying to cause any drama, but I'm actually not a student and have no teachers. In short, I'm just currently not good at these yet, but this forum helps a lot :-)

Thank you for all your help, I really appreciate it.
 
Then keep working and posting here. You are improving!
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top