University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2023
Views:

454

Pages:

3
Author:

Shahbaz A.

Q 58 Algebra Word Problems Packet No 1 shahbaz ahmed August 2024 Q 58.The length of a rectangle is 8 inches more than its width. If the length is increased by 4 inches and the width remains the same, the area increases by 24 sq in. Find the original dimensions of the rectangle. Solution Let width=x inches. The length of a rectangle is 8 inches more than its width.=x+8 inches. Area=x(x + 8) = x2 + 8x sq in. If the length is increased by 4 inches=x+8+4=x+12 inches. Increased area= x(x+12)=x2 + 12x sq in. Difference in areas= x2 + 12x − (x2 + 8x) sq in. 1 Difference in areas= x2 + 12x − x2 − 8x sq in. Difference in areas= 4x sq in. Given difference in areas= 24 sq in. According to the condition 4x = 24 x= 24 4 x = 6 inches. Width=6 inches Putting in . Length= x+8 Length=6+8=14 inches Length= 14 inches Checking Width=6 inches. The length of a rectangle is 8 inches more than its width.=6+8=14 inches. Area=6 × 14= 84 sq in. If the length is increased by 4 inches=14+4=18 inches. Increased area=6 × 18=108 sq in. 2 Difference in areas= 108- 84=24 sq in. Given difference in areas= 24 sq in. Hence all conditions are satisfied 3