This is another one that is taking me a long time to solve (but I think I got it?). It is from a Kumon sheet and it feels like maybe I'm missing some simpler way to get to the answer.
The reason I think I might be missing a simpler method is because the previous problems on the sheet were way way simpler. For example, [MATH]x \cdot 5= \frac{1}{2} + \frac{1}{3}[/MATH]. Usually Kumon does a good job of gradually increasing... but I'm really struggling with this jump.
Here is the original problem:
[MATH] (1999 \cdot \frac{1}{24} + x)1\frac{3}{5} - 3\frac{5}{11} \cdot 3\frac{2}{3} = 127 [/MATH]
1. I converted to improper fractions and multiplied 3 5/11 by 3 2/3:
[MATH] (1999 \cdot \frac{1}{24} + x)1\frac{3}{5} - \color{blue}3\frac{5}{11} \cdot 3\frac{2}{3}\color{black} = 127 \quad \rightarrow\quad (1999 \cdot \frac{1}{24} + x)\frac{8}{5} - \frac{38}{3} = 127 [/MATH]
2. I added 38/3 to both sides:
[MATH](1999 \cdot \frac{1}{24} + x)\frac{8}{5} - \color{red} \frac{38}{3}\color{black} = 127 \quad \rightarrow\quad (1999 \cdot \frac{1}{24} + x)\frac{8}{5} = \frac{419}{3}[/MATH]
3. I multiplied both sides by 5/8:
[MATH](1999 \cdot \frac{1}{24} + x)\color{green}\frac{8}{5}\color{black} = \frac{419}{3} \quad \rightarrow\quad (1999 \cdot \frac{1}{24} + x) = \frac{2095}{24}[/MATH]
4. I multiplied 1999 by 1/24:
[MATH](\color{blue}1999 \cdot \frac{1}{24}\color{black} + x) = \frac{2095}{24} \quad \rightarrow\quad \frac{1999}{24} + x = \frac{2095}{24}[/MATH]
5. I subtracted 1999/24 from both sides:
[MATH]\color{red}\frac{1999}{24}\color{black} + x = \frac{2095}{24} \quad \rightarrow\quad x = \frac{96}{24}[/MATH]
6. Simplified:
[MATH]x = 4[/MATH]
P.S. Mainly I think it is taking me so long because I'm trying to figure out stuff like can I reduce 1999/24? Can I reduce 419/3? Because I have always been taught that it's best to reduce first.
The reason I think I might be missing a simpler method is because the previous problems on the sheet were way way simpler. For example, [MATH]x \cdot 5= \frac{1}{2} + \frac{1}{3}[/MATH]. Usually Kumon does a good job of gradually increasing... but I'm really struggling with this jump.
Here is the original problem:
[MATH] (1999 \cdot \frac{1}{24} + x)1\frac{3}{5} - 3\frac{5}{11} \cdot 3\frac{2}{3} = 127 [/MATH]
1. I converted to improper fractions and multiplied 3 5/11 by 3 2/3:
[MATH] (1999 \cdot \frac{1}{24} + x)1\frac{3}{5} - \color{blue}3\frac{5}{11} \cdot 3\frac{2}{3}\color{black} = 127 \quad \rightarrow\quad (1999 \cdot \frac{1}{24} + x)\frac{8}{5} - \frac{38}{3} = 127 [/MATH]
2. I added 38/3 to both sides:
[MATH](1999 \cdot \frac{1}{24} + x)\frac{8}{5} - \color{red} \frac{38}{3}\color{black} = 127 \quad \rightarrow\quad (1999 \cdot \frac{1}{24} + x)\frac{8}{5} = \frac{419}{3}[/MATH]
3. I multiplied both sides by 5/8:
[MATH](1999 \cdot \frac{1}{24} + x)\color{green}\frac{8}{5}\color{black} = \frac{419}{3} \quad \rightarrow\quad (1999 \cdot \frac{1}{24} + x) = \frac{2095}{24}[/MATH]
4. I multiplied 1999 by 1/24:
[MATH](\color{blue}1999 \cdot \frac{1}{24}\color{black} + x) = \frac{2095}{24} \quad \rightarrow\quad \frac{1999}{24} + x = \frac{2095}{24}[/MATH]
5. I subtracted 1999/24 from both sides:
[MATH]\color{red}\frac{1999}{24}\color{black} + x = \frac{2095}{24} \quad \rightarrow\quad x = \frac{96}{24}[/MATH]
6. Simplified:
[MATH]x = 4[/MATH]
P.S. Mainly I think it is taking me so long because I'm trying to figure out stuff like can I reduce 1999/24? Can I reduce 419/3? Because I have always been taught that it's best to reduce first.
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