Solve the following problems: 1. The speed of a boat in still water is 5 km/h. If it takes 1 hour more to go 5.25 km upstream than to return downstream to the same point, find the... Solve the following problems: 1. The speed of a boat in still water is 5 km/h. If it takes 1 hour more to go 5.25 km upstream than to return downstream to the same point, find the speed of the stream. 2. A jet plane covers 4500 km in some time. If the regular speed is decreased by 150 km/h, it takes one and a half hours more to complete the journey. Find the original speed of the jet plane. 3. Around a square pool, there is a footpath of width 2m. If the area of the footpath is 21% that of the pool, find the area of the pool. 4. A rectangular garden 40 m x 30 m is surrounded from outside by a path of equal width. If the area of the path is 456 m², find the width of the path. Read more

Steps to Solve

1. Boat Problem: Define variables and set up equations. Let $v$ be the speed of the stream. The speed upstream is $(5 - v)$ km/h, and the speed downstream is $(5 + v)$ km/h. Time is distance divided by speed. The time taken to go upstream is $\frac{5.25}{5 - v}$ hours, and the time taken to go downstream is $\frac{5.25}{5 + v}$ hours. The problem states that it takes 1 hour longer to go upstream than downstream.

2. Boat Problem: Formulate the equation and solve for $v$. We have the equation: $$ \frac{5.25}{5 - v} = \frac{5.25}{5 + v} + 1 $$ Multiplying both sides by $(5 - v)(5 + v)$ to eliminate the fractions: $$ 5.25(5 + v) = 5.25(5 - v) + (5 - v)(5 + v) $$ $$ 26.25 + 5.25v = 26.25 - 5.25v + 25 - v^2 $$ $$ 10.5v = 25 - v^2 $$ $$ v^2 + 10.5v - 25 = 0 $$ Multiply by 2 to clear the decimal: $$ 2v^2 + 21v - 50 = 0 $$ Using the quadratic formula: $$ v = \frac{-21 \pm \sqrt{21^2 - 4(2)(-50)}}{2(2)} = \frac{-21 \pm \sqrt{441 + 400}}{4} = \frac{-21 \pm \sqrt{841}}{4} = \frac{-21 \pm 29}{4} $$ Since speed cannot be negative, we take the positive root: $$ v = \frac{-21 + 29}{4} = \frac{8}{4} = 2 $$ The speed of the stream is 2 km/h.

3. Jet Plane Problem: Define variables and set up equations. Let $s$ be the original speed of the jet plane. The original time taken is $\frac{4500}{s}$ hours. If the speed is decreased by 150 km/h, the new speed is $(s - 150)$ km/h, and the new time taken is $\frac{4500}{s - 150}$ hours. This new time is 1.5 hours more than the original time.

4. Jet Plane Problem: Formulate the equation and solve for $s$. We have the equation: $$ \frac{4500}{s - 150} = \frac{4500}{s} + 1.5 $$ Multiplying both sides by $s(s - 150)$ to eliminate the fractions: $$ 4500s = 4500(s - 150) + 1.5s(s - 150) $$ $$ 4500s = 4500s - 67500 + 1.5s^2 - 225s $$ $$ 0 = 1.5s^2 - 225s - 67500 $$ Dividing by 1.5: $$ s^2 - 150s - 45000 = 0 $$ $$ (s - 300)(s + 150) = 0 $$ Since speed cannot be negative, $s = 300$. The original speed of the jet plane is 300 km/h.

5. Pool Problem: Define variables and set up equations. Let $x$ be the side length of the square pool. The area of the pool is $x^2$. The side length of the outer square (pool + path) is $x + 2(2) = x + 4$. The area of the outer square is $(x + 4)^2$. The area of the footpath is the difference between the areas of the outer square and the pool: $(x + 4)^2 - x^2$. We are given that the area of the footpath is 21% of the area of the pool.

6. Pool Problem: Formulate the equation and solve for $x$. We have the equation: $$ (x + 4)^2 - x^2 = 0.21x^2 $$ $$ x^2 + 8x + 16 - x^2 = 0.21x^2 $$ $$ 8x + 16 = 0.21x^2 $$ $$ 0.21x^2 - 8x - 16 = 0 $$ Multiply by 100 $$ 21x^2 - 800x - 1600 = 0 $$ Using the quadratic formula: $$ x = \frac{800 \pm \sqrt{(-800)^2 - 4(21)(-1600)}}{2(21)} = \frac{800 \pm \sqrt{640000 + 134400}}{42} = \frac{800 \pm \sqrt{774400}}{42} = \frac{800 \pm 880}{42} $$ Since length cannot be negative, we take the positive root: $$ x = \frac{800 + 880}{42} = \frac{1680}{42} = 40 $$ The side length of the pool is 40 m. The area of the pool is $x^2 = 40^2 = 1600$ m$^2$.

7. Garden Path Problem: Define variables and set up equations. Let $w$ be the width of the path. The length of the outer rectangle is $40 + 2w$, and the width is $30 + 2w$. The area of the outer rectangle is $(40 + 2w)(30 + 2w)$. The area of the garden is $40 \times 30 = 1200$. The area of the path is the difference between the area of the outer rectangle and the area of the garden.

8. Garden Path Problem: Formulate the equation and solve for $w$. We have the equation: $$ (40 + 2w)(30 + 2w) - 1200 = 456 $$ $$ 1200 + 80w + 60w + 4w^2 - 1200 = 456 $$ $$ 4w^2 + 140w - 456 = 0 $$ $$ w^2 + 35w - 114 = 0 $$ $$ (w - 3)(w + 38) = 0 $$ Since width cannot be negative, $w = 3$. The width of the path is 3 m.