Solutions to Problem Set 4

Math 101-06

9-11-2017

[Linear equation word problems]

1. The sum of two numbers is 108. The second number is 3 times the first number. Find the two numbers.

Let F be the first number and let S be the second number.

The sum of the two numbers is 108, so

$$F + S = 108.$$

The second number is 3 times the first number, so

$$S = 3 F.$$

Plug $S = 3 F$ into $F + S =
   108$ and solve for F:

$$\eqalign{ F + S & = 108 \cr F + 3 F & = 108 \cr 4 F & = 108 \cr \noalign{\vskip2pt} \dfrac{4 F}{4} & = \dfrac{108}{4} \cr \noalign{\vskip2pt} F & = 27 \cr}$$

Therefore, $S = 3 F = 3 \cdot 27 =
   81$ .

The first number is 27 and the second number is 81.


2. The sum of two numbers is 154. The second number is 2 less than 5 times the first number. Find the two numbers.

Let F be the first number and let S be the second number.

The sum of the two numbers is 154, so

$$F + S = 154.$$

The second number is 2 less than 5 times the first number, so

$$S = 5 F - 2.$$

Plug $S = 5 F - 2$ into $F
   + S = 154$ and solve for F:

$$\eqalign{ F + s & = 154 \cr F + (5 F - 2) & = 154 \cr 6 F - 2 & = 154 \cr 6 F - 2 + 2 & = 154 + 2 \cr 6 F & = 156 \cr \noalign{\vskip2pt} \dfrac{6 F}{6} & = \dfrac{156}{6} \cr \noalign{\vskip2pt} F & = 26 \cr}$$

Therefore, $S = 5 F - 2 = 5 \cdot
   26 - 2 = 128$ .

The first number is 26 and the second number is 128.


3. The promoters of a concert sell regular tickets for $8 each and premium tickets for $14 each. A total of 120 tickets are sold, and the total value of the tickets is $1230. How many of each kind of ticket were sold?

Let r be the number of regular tickets sold. Since 120 tickets were sold all together, the number of premium tickets sold is $120 - r$ .

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & number of tickets & & & & cost per ticket & & & & value of tickets & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & regular tickets & & r & & $\cdot$ & & 8 & & $=$ & & $8 r$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & premium tickets & & $120 - r$ & & $\cdot$ & & 14 & & $=$ & & $14(120 - r)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & total & & & & & & & & & & 1230 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The third column gives an equation which I can solve for r:

$$\eqalign{ 8 r + 14(120 - r) & = 1230 \cr 8 r + 1680 - 14 r & = 1230 \cr -6 r + 1680 & = 1230 \cr -6 r & = -450 \cr r & = 75 \cr}$$

Then $120 - r = 120 - 75 = 45$ .

There were 75 regular tickets sold and 45 premium tickets sold.


4. $1100 is divided between two accounts. The first account pays $2%$ interest, while the second account pays $6%$ interest. After one interest period, the total interest earned by both accounts was $56. Find the amounts that were invested in the two accounts.

Let x be the amount invested in the $2%$ account. Since there was $1100 all together, the amount invested in the $6%$ account was $1100
   - x$ .

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & amount invested & & & & interest rate & & & & interest earned & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $2\%$ account & & x & & $\cdot$ & & 0.02 & & $=$ & & $0.02 x$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $6\%$ account & & $1100 - x$ & & $\cdot$ & & 0.06 & & $=$ & & $0.06 (1100 - x)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & total & & & & & & & & & & 56 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The third column gives an equation that I can solve for x:

$$\eqalign{ 0.02 x + 0.06 (1100 - x) & = 56 \cr 0.02 x + 66.0 - 0.06 x & = 56 \cr -0.04 x + 66.0 & = 56 \cr -0.04 x & = -10.0 \cr x & = 250 \cr}$$

Therefore, $1100 - x = 1100 - 250
   = 850$ . So $250 was invested in the $2%$ account and $850 was invested in the $6%$ account.


5. Two planes, which are 3840 miles apart, fly toward each other. The faster plane's speed is 40 miles per hour greater than the slower plane's speed. They pass each other after 6 hours. Find their speeds.

Let x be the speed of the slow plane. The fast plane flies 40 miles per hour faster, so the fast plane's speed is $x + 40$ .

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & time & & $\cdot$ & & speed & & $=$ & & distance & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & slow plane & & 6 & & $\cdot$ & & x & & $=$ & & $6 x$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & fast plane & & 6 & & $\cdot$ & & $x + 40$ & & $=$ & & $6(x + 40)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & & & & & & & & & 3840 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

Since the planes started 3840 miles apart, when they pass each other they must have combined to cover the 3840 miles. So the sum of their distances is equal to 3840. This equation comes from the third column in the table. Solve it for x:

$$\eqalign{ 6 x + 6(x + 40) & = 3840 \cr 6 x + 6 x + 240 & = 3840 \cr 12 x + 240 & = 3840 \cr 12 x & = 3600 \cr x & = 300 \cr}$$

One plane's speed is 300 miles per hour. The other plane's speed is $300 + 40 = 340$ miles per hour.


6. Calvin spends 3.5 hours training for an upcoming race. He runs full speed at 6 miles per hour for the race distance; then he walks back to his starting point at 1 miles per hour. How long does he spend walking? How long does he spend running?

Let x be the time he spent running. Since he spent 3.5 hours all together, he must have spent $3.5
   - x$ hours walking.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & time & & $\cdot$ & & speed & & $=$ & & distance & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & running & & x & & $\cdot$ & & 6 & & $=$ & & $6 x$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & walking & & $3.5 - x$ & & $\cdot$ & & 1 & & $=$ & & $1(3.5 - x)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

Since he ran out, then turned around and walked back, his running and walking distances must be equal. Set the distances equal and solve for x:

$$\eqalign{ 6 x & = 1(3.5 - x) \cr 6 x & = 3.5 - 1 x \cr 7 x & = 3.5 \cr x & = 0.5 \cr}$$

He spends 0.5 hours running and $3.5 - 0.5 = 3$ hours walking.


7. A slow train leaves the city at 12 a.m. A fast train leaves the city at 3 p.m. and follows the slow train. The fast train's speed is 30 miles per hour faster than the slow train's. If the fast train overtakes the slow train at 10 p.m., what are the speeds of the two trains?

Let x be the speed of the slow train. Then the fast train's speed is $x + 30$ .

The slow train started at 12 a.m., so it travels 10 hours till it's overtaken at 10 p.m..

The fast train started at 3 p.m., so it travels 7 hours till it overtakes the slow train at 10 p.m..

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & time & & $\cdot$ & & speed & & $=$ & & distance & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & slow train & & 10 & & $\cdot$ & & x & & $=$ & & $10 x$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & fast train & & 7 & & $\cdot$ & & $x + 30$ & & $=$ & & $7(x + 30)$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

At the point where the fast train overtakes the slow train, they must have travelled the same distance. Set their distances equal and solve for x:

$$\eqalign{ 10 x & = 7(x + 30) \cr 10 x & = 7 x + 210 \cr 3 x & = 210 \cr x & = 70 \cr}$$

The slow train's speed is 70 miles per hour. The fast train's speed is $70 + 30 = 100$ miles per hour.


As he thinketh in his heart, so is he. - Proverbs 23:7


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