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If a trader purchased a few items and sold all except 1010 items at a profit of 10%10\% on each item and recovered his total cost of purchasing all the items, how many items did the trader purchase?\newlineA) 9090\newlineB) 100100\newlineC) 110110\newlineD) 120120

Full solution

Q. If a trader purchased a few items and sold all except 1010 items at a profit of 10%10\% on each item and recovered his total cost of purchasing all the items, how many items did the trader purchase?\newlineA) 9090\newlineB) 100100\newlineC) 110110\newlineD) 120120
  1. Calculate Selling Price: He sold n10n - 10 items at a 10%10\% profit.\newlineSo each item is sold for c+0.1cc + 0.1c (which is 1.1c1.1c).
  2. Equation Setup: The total revenue from selling n10n - 10 items is (n10)×1.1c(n - 10) \times 1.1c. This revenue equals the total cost, which is n×cn\times c. So, we have the equation (n10)×1.1c=n×c(n - 10) \times 1.1c = n\times c.
  3. Simplify Equation: Let's simplify the equation:\newline1.1c(n10)=nc1.1c(n - 10) = nc\newline1.1cn11c=nc1.1cn - 11c = nc
  4. Isolate Variable: Now, we subtract 1.1cn1.1cn from both sides to isolate cc:
    11c=nc1.1cn-11c = nc - 1.1cn
    11c=c(n1.1n)-11c = c(n - 1.1n)
  5. Solve for n: Divide both sides by 'c' to solve for 'n':\newline11=n1.1n-11 = n - 1.1n\newline11=0.1n-11 = -0.1n
  6. Final Calculation: Now, divide both sides by 0.1-0.1 to find nn:n=110.1n = \frac{-11}{-0.1}n=110n = 110

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