Solutions to Problem Set 9

Math 101-06

9-27-2017

[Systems word problems]

1. How many pounds of nuts worth $2.40 per pound must be mixed with raisins worth $1.50 per pound to obtain 36 pounds of a mixture worth $2.00 per pound?

Let x be the number of pounds of nuts and let y be the number of pounds of raisins.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & pounds & & & & price per pound & & & & total value & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & nuts & & x & & $\cdot$ & & 2.40 & & $=$ & & $2.4 x$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & raisins & & y & & & & 1.5 & & $=$ & & $1.5 y$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & mixture & & 36 & & & & 2 & & & & 72 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The first and third columns give the equations

$$x + y = 36 \quad\hbox{and}\quad 2.4 x + 1.5 y = 72.$$

Multiply the first equation by 2.4 and subtract the second equation:

$$\matrix{ 2.4 x & + & 2.4 y & = & 86.4 \cr 2.4 x & + & 1.5 y & = & 72 \cr \noalign{\vskip2pt \hrule \vskip2pt} & & 0.9 y & = & 14.4 \cr & & y & = & 16 \cr}$$

Hence, $x = 36 - 16 = 20$ . Thus, 20 pounds of nuts and 16 pounds of raisins are needed for the mixture.


2. How many liters of a $20%$ salt solution and a $36%$ salt solution must be mixed to obtain 40 liters of a $24%$ salt solution?

Let x be the number of liters of the $20%$ salt solution and let y be the number of liters of the $36%$ salt solution.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & liters of solution & & $\cdot$ & & \% salt (decimal) & & & & kilograms of salt & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $20\%$ salt solution & & x & & $\cdot$ & & 0.2 & & $=$ & & $0.2 x$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $36\%$ salt solution & & y & & $\cdot$ & & 0.36 & & $=$ & & $0.36 y$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $24\%$ salt solution & & 40 & & $\cdot$ & & 0.24 & & $=$ & & 9.6 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The first and third columns give the equations

$$x + y = 40 \quad\hbox{and}\quad 0.2 x + 0.36 y = 9.6.$$

Multiply the first equation by 0.2 and subtract it from the second:

$$\eqalign{ 0.2 x + 0.36 y & = 9.6 \cr 0.2 x + 0.2 y & = 8 \cr \noalign{\vskip2pt \hrule \vskip2pt} 0.16 y & = 1.6 \cr y & = 10 \cr}$$

Then $x + 10 = 40$ , so $x
   = 30$ .

Use 30 liters of the $20%$ salt solution and 10 liters of the $36%$ salt solution.


3. $1100 is divided among two accounts. The first account pays 3% interest, while the second account pays 4% interest. At the end of one interest period, the two accounts have returned a total of $41 in interest. How much was invested in each account?

Let x be the amount invested in the first account, and let y be the amount invested in the second account.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & amount invested & & & & interest rate & & & & interest earned & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & first account & & x & & $\cdot$ & & 0.03 & & $=$ & & $0.03 x$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & second account & & y & & $\cdot$ & & 0.04 & & $=$ & & $0.04 y$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & total & & 1100 & & & & & & & & 41 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The first and third columns give two equations:

$$x + y = 1100 \quad\hbox{and}\quad 0.03 x + 0.04 y = 41.$$

Multiply the first equation by 0.03 and subtract it from the second:

$$\matrix{ 0.03 x & + & 0.04 y & = & 41 \cr 0.03 x & + & 0.03 y & = & 33 \cr \noalign{\vskip2pt \hrule \vskip2pt} & & 0.01 y & = & 8.0 \cr & & y & = & 800 \cr}$$

Then $x = 1100 - 800 = 300$ . Thus, $300 was invested in the first account and $800 was invested in the second account.


4. $1100 is divided between two accounts. The first account pays $3%$ interest, while the other pays $2%$ interest. After one interest period, the interest earned by the $3%$ account is $13 greater than the interest earned by the $2%$ account. How much was invested in each account?

Let x be the amount invested in the $3%$ account, and let y be the amount invested in the $2%$ account.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & & & amount invested & & & & interest rate & & & & interest earned & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $3\%$ account & & x & & $\cdot$ & & 0.03 & & $=$ & & $0.03 x$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $2\%$ account & & y & & $\cdot$ & & 0.02 & & $=$ & & $0.02 y$ & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & total & & 1100 & & & & & & & & & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

The first column gives $x + y =
   1100$ .

The interest earned by the $3%$ account is $13.0 greater than the interest earned by the $2%$ account, so from the third column I get

$$0.03 x = 0.02 y + 13, \quad\hbox{or}\quad 0.03 x - 0.02 y = 13.$$

Multiply $x + y = 1100$ by 0.03 and subtract $0.03 x - 0.02 y = 13$ :

$$\matrix{ 0.03 x + 0.03 y & = & 33 \cr 0.03 x - 0.02 y & = & 13 \cr \noalign{\vskip2pt \hrule \vskip2pt} 0.05 y & = & 20 \cr y & = & 400 \cr}$$

Then $x = 1100 - 400 = 700$ . Thus, $700 was invested in the $3%$ account, and $400 was invested in the $2%$ account.


There is no security in life, only opportunity. - Mark Twain


Contact information

Bruce Ikenaga's Home Page

Copyright 2017 by Bruce Ikenaga