2 Word Problems

Question

1. The area of a square is 3 square inches bigger than the area of a rectangle. The width of the rectangle is 3 inches shorter than one side of the square, and the length of the rectangle is 4 inches longer than one side of the square. Find the length of the side of the square.

 

2. A boat can travel 8mph in still water. If it can travel 20 miles downstream (in the same direction as the current) in the same time that it can travel 12 miles upstream (in the opposite direction of the current) how fast is the current? (Hint: distance = rate*time)

 

Please help! Thanks!

Answer 1

problem 1:

let s be the side of the square

let l be the length of the rectangle

let w be the width of the rectangle

then,

s²=l*w+3, the area or the square is 3 inches longer than the area of the rectangle.

w=s-3, the width of the rectangle is 3 inches shorter than the side of the square.

l=s+4, the length of the rectangle is 4 inches longer than the side of the square.

 

if you substitute the s values of l and w into the area equation, 

s2=(s-3)(s+4)+3, solving this give the side of the square,

s=9.

You should be able to now calculate the length and width of the rectangle.

 

problem 2:

 

let v = the  speed of the boat

let vs = the speed of the stream

let t = time of trip

let x_d = distance covered downstream during time t

let x_u = distance covered upstream during time t

for the downstream trip, the distance covered is due to the speed of the boat and the speed of the stream, or:

x_d=(v+vs)*t

for the the trip upstream the distance covered is due to the speed of the boat less the speed of the stream, or:

x_u=(v-vs)*t

solve both equations for t, and set the equal to each other to get

20*(v-v_s)=12*(v+v_s),

solve for v_s.

 

 

 

Answer 2

Problem #1:

A_s = area of the square    x = the length of each side of the square    A_s = x²

 

A_R = area of the rectangle   length of the rectangle = x + 4    width of the rectangle = x - 3    A_R = (x+4)(x-3)

 

Since A_s = A_R+3  then the following equation is true:

 

x² = (x+4)(x-3) + 3 

 

x² = x² + 4x - 3x - 12 +3  (Now subtract x² from both sides of the equation)

 

0 = 4x - 3x - 9 (Solve for x)

0 = x - 9

9 = x

 

So, the square has sides of length 9 in each.  The rectangle has a length of 9+4 = 13 in and a width of 9-3 or 6 in.

The area of the square is 81 sq inches which is 3 sq inches bigger than the area of the rectangle (13 x 6 = 78 sq inches).

 

 

 

Problem #2:

 

Let c = the rate of the current.

 

The boat going downstream travels at its rate in still water plus the rate of the current:  8 + c

The boat going upstream travels at its rate minus the rate of the current: 8 - c

 

rate x time = distance

 

downstream distance = 20 miles = time downstream x (8 + c)  or time down = 20 / (8 + c)

upstream distance = 12 miles = time upstream x (8 - c)  or time up = 12 / (8 - c)

 

Since time downstream = time upstream, then  20 / (8 + c) = 12 / (8 - c)

 

Now cross multiply and solve for c.

 

20/(8+c) = 12/(8-c)

 

12(8+c) = 20(8-c)

96 + 12c = 160 - 20c

32c = 64

c = 2

 

The stream current is 2 mph.