There were some oval and circular shapes in a box.Durai put 48 more circular shapes in the box.Then 40% of the shapes in the box were oval shapes. Later Durai put another 135 oval shapes in the box.60% of the shapes in the box were oval shapes now.How many curcular shapes were in the box at first?
Q. There were some oval and circular shapes in a box.Durai put 48 more circular shapes in the box.Then 40% of the shapes in the box were oval shapes. Later Durai put another 135 oval shapes in the box.60% of the shapes in the box were oval shapes now.How many curcular shapes were in the box at first?
Define Initial Circular Shapes: Let's call the number of circular shapes at first C. After adding 48 circular shapes, the total number of circular shapes becomes C+48.
Calculate Total Circular Shapes: If 40% of the shapes are oval after adding the 48 circular shapes, then 60% of the shapes are circular.So, 60% of the total shapes equals C+48.
Calculate Total Shapes: Let's call the total number of shapes in the box after adding the 48 circular shapes T. So, 0.60×T=C+48.
Calculate Circular Shapes After Adding Oval Shapes: After adding another 135 oval shapes, 60% of the shapes are oval, which means 40% are circular.So now, 40% of the new total (T+135) equals the number of circular shapes, which is still C+48.
Solve for Total Shapes:0.40×(T+135)=C+48.
Substitute and Simplify Equations: Now we have two equations:0.60×T=C+48 (1)0.40×(T+135)=C+48 (2)
Solve for Total Shapes: We can solve equation (1) for T: T=0.60C+48.
Solve for Circular Shapes: Substitute T from equation (1) into equation (2):0.40×(0.60C+48+135)=C+48.
Final Calculation for Circular Shapes: Simplify the equation:0.40×((C+48)/0.60+135)=C+480.40×(C+48)/0.60+0.40×135=C+48(2/3)×(C+48)+54=C+48.
Final Calculation for Circular Shapes: Now, solve for C:32⋅C+32+54=C+4832⋅C+86=C+48.
Final Calculation for Circular Shapes: Now, solve for C: (32)∗C+32+54=C+48(32)∗C+86=C+48. Subtract (32)∗C from both sides:86=(31)∗C+48.
Final Calculation for Circular Shapes: Now, solve for C: (32)⋅C+32+54=C+48(32)⋅C+86=C+48. Subtract (32)⋅C from both sides:86=(31)⋅C+48. Subtract 48 from both sides:38=(31)⋅C.
Final Calculation for Circular Shapes: Now, solve for C: (32)⋅C+32+54=C+48 (32)⋅C+86=C+48. Subtract (32)⋅C from both sides: 86=(31)⋅C+48. Subtract 48 from both sides: 38=(31)⋅C. Multiply both sides by 3 to solve for C: 3⋅38=C (32)⋅C+32+54=C+480.