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There were some oval and circular shapes in a box.Durai put 4848 more circular shapes in the box.Then 40%40\% of the shapes in the box were oval shapes. Later Durai put another 135135 oval shapes in the box.60%60\% of the shapes in the box were oval shapes now.How many curcular shapes were in the box at first?

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Q. There were some oval and circular shapes in a box.Durai put 4848 more circular shapes in the box.Then 40%40\% of the shapes in the box were oval shapes. Later Durai put another 135135 oval shapes in the box.60%60\% of the shapes in the box were oval shapes now.How many curcular shapes were in the box at first?
  1. Define Initial Circular Shapes: Let's call the number of circular shapes at first CC. After adding 4848 circular shapes, the total number of circular shapes becomes C+48C + 48.
  2. Calculate Total Circular Shapes: If 40%40\% of the shapes are oval after adding the 4848 circular shapes, then 60%60\% of the shapes are circular.\newlineSo, 60%60\% of the total shapes equals C+48C + 48.
  3. Calculate Total Shapes: Let's call the total number of shapes in the box after adding the 4848 circular shapes TT. So, 0.60×T=C+480.60 \times T = C + 48.
  4. Calculate Circular Shapes After Adding Oval Shapes: After adding another 135135 oval shapes, 60%60\% of the shapes are oval, which means 40%40\% are circular.\newlineSo now, 40%40\% of the new total (T+135)(T + 135) equals the number of circular shapes, which is still C+48C + 48.
  5. Solve for Total Shapes: 0.40×(T+135)=C+480.40 \times (T + 135) = C + 48.
  6. Substitute and Simplify Equations: Now we have two equations:\newline0.60×T=C+480.60 \times T = C + 48 (11)\newline0.40×(T+135)=C+480.40 \times (T + 135) = C + 48 (22)
  7. Solve for Total Shapes: We can solve equation (11) for TT: T=C+480.60T = \frac{C + 48}{0.60}.
  8. Solve for Circular Shapes: Substitute TT from equation (1)(1) into equation (2)(2):\newline0.40×(C+480.60+135)=C+48.0.40 \times \left(\frac{C + 48}{0.60} + 135\right) = C + 48.
  9. Final Calculation for Circular Shapes: Simplify the equation:\newline0.40×((C+48)/0.60+135)=C+480.40 \times ((C + 48) / 0.60 + 135) = C + 48\newline0.40×(C+48)/0.60+0.40×135=C+480.40 \times (C + 48) / 0.60 + 0.40 \times 135 = C + 48\newline(2/3)×(C+48)+54=C+48.(2/3) \times (C + 48) + 54 = C + 48.
  10. Final Calculation for Circular Shapes: Now, solve for CC:23C+32+54=C+48\frac{2}{3} \cdot C + 32 + 54 = C + 4823C+86=C+48\frac{2}{3} \cdot C + 86 = C + 48.
  11. Final Calculation for Circular Shapes: Now, solve for CC: \newline(23)C+32+54=C+48\left(\frac{2}{3}\right) * C + 32 + 54 = C + 48\newline(23)C+86=C+48\left(\frac{2}{3}\right) * C + 86 = C + 48. Subtract (23)C\left(\frac{2}{3}\right) * C from both sides:\newline86=(13)C+4886 = \left(\frac{1}{3}\right) * C + 48.
  12. Final Calculation for Circular Shapes: Now, solve for CC: \newline(23)C+32+54=C+48(\frac{2}{3}) \cdot C + 32 + 54 = C + 48\newline(23)C+86=C+48(\frac{2}{3}) \cdot C + 86 = C + 48. Subtract (23)C(\frac{2}{3}) \cdot C from both sides:\newline86=(13)C+4886 = (\frac{1}{3}) \cdot C + 48. Subtract 4848 from both sides:\newline38=(13)C38 = (\frac{1}{3}) \cdot C.
  13. Final Calculation for Circular Shapes: Now, solve for CC:
    (23)C+32+54=C+48(\frac{2}{3}) \cdot C + 32 + 54 = C + 48
    (23)C+86=C+48(\frac{2}{3}) \cdot C + 86 = C + 48. Subtract (23)C(\frac{2}{3}) \cdot C from both sides:
    86=(13)C+4886 = (\frac{1}{3}) \cdot C + 48. Subtract 4848 from both sides:
    38=(13)C38 = (\frac{1}{3}) \cdot C. Multiply both sides by 33 to solve for CC:
    338=C3 \cdot 38 = C
    (23)C+32+54=C+48(\frac{2}{3}) \cdot C + 32 + 54 = C + 4800.

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