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28. three - eighths of the seventh grade students were taking advanced …

Question

  1. three - eighths of the seventh grade students were taking advanced math at the beginning of the year, but seven dropped out by the end of the year. if there were 140 students taking advanced math at the end of the year, how many total 7th grade students are there?

Explanation:

Step1: Calculate initial advanced math students

$\frac{3}{8} \times 140 = 52.5$ (Wait, this is a fraction. Maybe the problem has a typo? Assuming 144 students instead of 140 for integer result: $\frac{3}{8} \times 144 = 54$)

Step2: Subtract dropped students

$54 - 7 = 47$ (If 140, 52.5 -7 is invalid. Correcting to 144 gives 47. But original problem says 140, maybe it's a mistake. Alternatively, maybe the question is about total 7th graders? Wait no, the question says "how many total 7th grade students are there?" Wait no, re-read: "Three-eighths of the seventh grade students were taking advanced math... 7 dropped out... 140 taking advanced math at end. So let total students be $x$. Initial advanced: $\frac{3}{8}x$. After dropping: $\frac{3}{8}x -7 =140$. Then $\frac{3}{8}x=147$, $x=147\times\frac{8}{3}=392$. Oh! I misread earlier. The 140 is at the end, not beginning.

Step1: Let total 7th graders be $x$

Initial advanced math: $\frac{3}{8}x$

Step2: Set up equation for end advanced math

$\frac{3}{8}x -7 =140$

Step3: Solve for $x$

$\frac{3}{8}x =147$ → $x=147\times\frac{8}{3}=392$

Step1: Define total students as $x$

Initial advanced: $\frac{3}{8}x$

Step2: Equation for end advanced

$\frac{3}{8}x -7 =140$

Step3: Solve for $x$

$\frac{3}{8}x=147$ → $x=392$

Answer:

133