The sum of the lengths of two rectangles P and Q is $60 ext{ cm}$, and their breadths are $16 ext{ cm}$ and $12 ext{ cm}$ respectively. If the ratio of the area of rectangle P to that of rectangle Q is $8 : 3$, what is the difference between the perimeters of these two rectangles?

Question

The sum of the lengths of two rectangles P and Q is $60 	ext{ cm}$, and their breadths are $16 	ext{ cm}$ and $12 	ext{ cm}$ respectively. If the ratio of the area of rectangle P to that of rectangle Q is $8 : 3$, what is the difference between the perimeters of these two rectangles?

Accepted Answer

**Solution:**

1. Let the lengths of rectangles P and Q be $x$ cm and $60 - x$ cm respectively.

2. Given breadths:
   - Rectangle P: $16 	ext{ cm}$
   - Rectangle Q: $12 	ext{ cm}$

3. Areas:
   - Area of P = $16 	imes x = 16x$
   - Area of Q = $12 	imes (60 - x) = 720 - 12x$

4. Given ratio of areas:
   $$\frac{16x}{720 - 12x} = \frac{8}{3}$$

5. Cross-multiply:
   $$3 	imes 16x = 8 	imes (720 - 12x)$$
   $$48x = 5760 - 96x$$

6. Bring terms involving $x$ to one side:
   $$48x + 96x = 5760$$
   $$144x = 5760$$

7. Solve for $x$:
   $$x = \frac{5760}{144} = 40$$

8. Lengths:
   - $x = 40 	ext{ cm}$ (P)
   - $60 - x = 20 	ext{ cm}$ (Q)

9. Calculate perimeters:
   - Perimeter of P = $2 	imes (40 + 16) = 2 	imes 56 = 112 	ext{ cm}$
   - Perimeter of Q = $2 	imes (20 + 12) = 2 	imes 32 = 64 	ext{ cm}$

10. Difference between perimeters:
    $$112 - 64 = 48 	ext{ cm}$$

**Answer:** $48 	ext{ cm}$

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