Solutions to Problem Set 20

Math 101-06

10-25-2017

[Fraction word problems]

1. If a number is added to the top and the bottom of $\dfrac{8}{11}$ , you get $\dfrac{10}{11}$ . What is the number?

Let n be the number to be added.

If a number is added to the top and the bottom of $\dfrac{8}{11}$ , you get $\dfrac{10}{11}$ :

$$\dfrac{8 + n}{11 + n} = \dfrac{10}{11}.$$

Clear the fractions and solve:

$$\eqalign{ 11(11 + n) \cdot \dfrac{8 + n}{11 + n} & = 11(11 + n) \cdot \dfrac{10}{11} \cr 11(8 + n) & = 10(11 + n) \cr 88 + 11 n & = 110 + 10 n \cr n & = 22 \cr}$$

Check: When $n = 22$ ,

$$\dfrac{8 + n}{11 + n} = \dfrac{30}{33} = \dfrac{10}{11}.$$

The number is 22.


2. Leopold can eat 784 cupcakes in 14 hours. Molly and Leopold, eating together, can eat 720 cupcakes in 6 hours. How long will it take Molly to eat 192 cupcakes alone?

Let x be Leopold's rate and let y be Molly's rate.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & minutes & & $\cdot$ & & cupcakes per minute & & $=$ & & cupcakes & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Leopold & & 14 & & $\cdot$ & & x & & $=$ & & 784 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Leopold and Molly & & 6 & & $\cdot$ & & $x + y$ & & $=$ & & 720 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Molly & & t & & $\cdot$ & & y & & $=$ & & 192 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The first equation in the table says $14 x = 784$ , so $x = 56$ .

The second equation in the table says $6(x + y) = 720$ . Plug in $x = 56$ and solve for y:

$$\eqalign{ 6(56 + y) & = 720 \cr 56 + y & = 120 \cr y & = 64 \cr}$$

The last equation in the table says $y t = 192$ . Plug in $y = 64$ and solve for t:

$$\eqalign{ 64 t & = 192 \cr t & = 3 \cr}$$

It takes Molly 3 hours.


3. Bob can eat 266 strombolis in 7 hours.

Bob and Jose, eating together, can eat 1040 strombolis in 13 hours.

Bob, Jose, and Irving, eating together, can eat 1494 strombolis in 9 hours.

How long will it take Irving to eat 1118 strombolis eating alone?

Let x be the number of strombolis Bob can eat in 1 hour.

Let y be the number of strombolis Jose can eat in 1 hour.

Let z be the number of strombolis Irving can eat in 1 hour.

Let t be the time it takes Irving to eat 1118 strombolis eating alone.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & hours & & $\cdot$ & & strombolis per hour & & $=$ & & strombolis & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Bob & & 7 & & $\cdot$ & & x & & $=$ & & 266 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Bob and Jose & & 13 & & $\cdot$ & & $x + y$ & & $=$ & & 1040 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Bob, Jose, and Irving & & 9 & & $\cdot$ & & $x + y + z$ & & $=$ & & 1494 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & Irving & & t & & $\cdot$ & & z & & $=$ & & 1118 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The first line of the table says

$$7 x = 266, \quad\hbox{so}\quad x = 38.$$

The second line of the table says

$$13(x + y) = 1040.$$

Plug in $x = 38$ and solve for y:

$$\eqalign{ 13(38 + y) & = 1040 \cr 38 + y & = 80 \cr y & = 42 \cr}$$

The third line of the table says

$$9(x + y + z) = 1494.$$

Plug in $x = 38$ and $y
   = 42$ solve for z:

$$\eqalign{ 9(38 + 42 + z) & = 1494 \cr 38 + 42 + z & = 166 \cr z & = 86 \cr}$$

The last line of the table says

$$t z = 1118.$$

Plug in $z = 86$ and solve for t:

$$\eqalign{ t \cdot 86 & = 1118 \cr t & = 13 \cr}$$

It takes Irving 13 hours.


4. A tank has an inlet pipe and an outlet pipe. The inlet pipe, by itself, can fill an empty tank in 4 hours. The outlet pipe, by itself, can drain a filled tank in 6 hours. Starting with an empty tank, how long will it take to fill the tank if both pipes are opened?

Let x be the rate at which the inlet pipe fills, let y be the rate at which the outlet pipe drains, and let t be the amount of time it takes to fill the tank with both pipes open.

With both pipes open, the net rate at which the tank is being filled is $x - y$ , because the inlet and outlet pipes are working against each other.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & hours & & $\cdot$ & & tanks per hour & & $=$ & & tanks & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & inlet alone & & 4 & & $\cdot$ & & x & & $=$ & & 1 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & outlet alone & & 6 & & $\cdot$ & & y & & $=$ & & 1 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & inlet and outlet & & t & & $\cdot$ & & $x - y$ & & $=$ & & 1 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The first row of the table says

$$4 x = 1, \quad\hbox{so}\quad x = \dfrac{1}{4}.$$

The second row of the table says

$$6 y = 1, \quad\hbox{so}\quad y = \dfrac{1}{6}.$$

The third row of the table says $t(x - y) = 1$ . Now

$$x - y = \dfrac{1}{4} - \dfrac{1}{6} = \dfrac{1}{12}.$$

Hence,

$$t \cdot \dfrac{1}{12} = 1, \quad\hbox{so}\quad t = 12.\quad\halmos$$


5. A small plane can fly at a speed of 54 miles per hour in still air. Flying with the wind, it can travel 360 miles in the same time that it can travel 288 miles against the wind.

Assume that the wind's speed adds to the plane's speed when the plane flies with the wind, and that the wind's speed subtracts from the plane's speed when the plane flies against the wind. Find the speed of the wind.

Let w be the speed of the wind.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & time & & & & speed & & & & distance & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & with wind & & t & & $\cdot$ & & $54 + w$ & & $=$ & & 360 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & against wind & & t & & $\cdot$ & & $54 - w$ & & $=$ & & 288 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The two rows give the equations

$$t (54 + w) = 360 \quad\hbox{and}\quad t (54 - w) = 288.$$

Solve the equations for t:

$$t = \dfrac{360}{54 + w} \quad\hbox{and}\quad t = \dfrac{288}{54 - w}.$$

Set the two expressions for t equal, then solve for w by clearing fractions:

$$\eqalign{ \dfrac{360}{54 + w} & = \dfrac{288}{54 - w} \cr \noalign{\vskip2pt} (54 + w)(54 - w) \dfrac{360}{54 + w} & = (54 + w)(54 - w) \dfrac{288}{54 - w} \cr \noalign{\vskip2pt} 360 (54 - w) & = 288 (54 + w) \cr 19440 - 360 w & = 15552 + 288 w \cr 3888 & = 648 w \cr 6 & = w \cr}$$

The speed of the wind is 6 miles per hour.


6. A plane can fly at a speed of 300 miles per hour in still air. It takes 2 hours longer to fly 4480 miles against the wind as it does to fly the same distance with the wind. Assume that the wind's speed adds to the plane's speed when the plane flies with the wind, and that the wind's speed subtracts from the plane's speed when the plane flies against the wind. Find the speed of the wind.

Let w be the speed of the wind, and let t be the time in hours it takes to fly 4480 miles against the wind. Then it takes $t + 2$ hours to fly the same distance with the wind.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & time & & & & speed & & & & distance & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & against the wind & & $t + 2$ & & $\cdot$ & & $300 - w$ & & $=$ & & 4480 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & with the wind & & t & & $\cdot$ & & $300 + w$ & & $=$ & & 4480 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The table gives the two equations

$$(t + 2)(300 - w) = 4480 \quad\hbox{and}\quad t (300 + w) = 4480.$$

Solving the second equation for t, I obtain

$$t = \dfrac{4480}{300 + w}.$$

Plug this into the first equation and solve for w:

$$\eqalign{ \left(\dfrac{4480}{300 + w} + 2\right) (300 - w) & = 4480 \cr \noalign{\vskip2pt} (300 + w) \left(\dfrac{4480}{300 + w} + 2\right) (300 - w) & = 4480 (300 + w) \cr \noalign{\vskip2pt} \left(4480 + 2 (300 + w)\right) (300 - w) & = 4480 (300 + w) \cr 4480 (300 - w) + 2 (300 + w) (300 - w) & = 4480 (300 + w) \cr 4480 (300 - w) + 2 (90000 - w^2) & = 4480 (300 + w) \cr 1344000 - 4480 w + 180000 - 2 w^2 & = 1344000 + 4480 w \cr -2 w^2 - 8960 w + 180000 & = 0 \cr 2 w^2 + 8960 w - 180000 & = 0 \cr w^2 + 4480 - 90000 & = 0 \cr (x - 20)(x + 4500) & = 0 \cr}$$

The roots are 20 and -4500, but the wind speed can't be negative. Hence, the speed of the wind is 20 miles per hour.


Nothing is so exhausting as indecision, and nothing is so futile. - Bertrand Russell


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