[word problems part 4][section 29]

(1.)  Sam has 45 coins in dimes and quarters.   The total value

of the coins is \$7.65 .   How many coins of each kind

does he have?

d + q = 45

.10d + .25q = 7.65                here is the problem

- .10d - .10q = -4.5      multiply equation 1 thru by -.10
______________________
.15q =  3.15     subtract equations
_____  ______
.15     .15       divide each side by .15

q = 21         divide and cancel

d + 21 = 45                replace q with 21

-21  -21         subtract 21 from each side
_______________
d    = 24           subtract

result:  24 dimes and 21 quarters

(2.)   Cliff puts quarters and nickels aside for paying tolls.

He has a total of 18 coins with a total value of \$2.90 .

How many coins of each kind does he have?

n + q = 18

.05n + .25q = 2.90                 here is the problem

-.05n - .05q = -.90    multiply equation 1 thru by -.05
______________________
.2q = 2                subtract equations
___ ___
.2   .2           divide each side by .2

q = 10           divide and cancel

n + 10 = 18            replace q with 10

-10   -10        subtract 10 from each side
____________________
n       = 8           subtract

result :   8 nickels and 10 quarters

(3.)  Joe invested \$5000, part at a yearly rate of 7% and

part at a yearly rate of 9%.   The total interest

earned for one year was \$368 .   How much did he

invest at each rate?

x + y = 5000

.07x + .09y = 368                here is the problem

-.07x - .07y = -350         multiply eq 1 thru by -.07
_________________________
.02y =   18        subtract equations
_____   ____
.02     .02          divide each side by .02

y = 900           divide and cancel

x + 900 = 5000         replace y with 900

-900   -900         subtract 900 from each side
__________________
x           =  4100          subtract

result:  \$4100 at 7 % and \$900 at 9 % .

(4.)  Two loans were taken out totaling \$35,000.   The yearly

interest rates were 12% and 15% and the total yearly

interest was \$4650.   Find the amount of each loan.

x + y = 35,000

.12x + .15y = 4650             here is the problem

-.12x - .12y = -4200      multiply equation 1 thru by -.12
_____________________________
.03y =   450        subtract equations
____    _____
.03     .03           divide each side by .03

y = 15,000          divide and cancel

x + 15,000 = 35,000    replace y with 15,000

-15,000  -15,000     subtract 15,000 fr ea side
____________________________
x             = 20,000        subtract

result:  \$20,000 at 12 % and 15,000 at 15 % .

(5.)  7500 tickets were sold for a concert.   Total receipts

amounted to \$45,000 .  Tickets sold for \$5.50 and \$7.00 .

How many tickets of each kind were sold?

x + y = 7500

5.5x + 7y = 45,000              here is the problem

-5.5x - 5.5y = -41,250    multiply eq 1 thru by -5.5
__________________________
1.5y =   3750    subtract equations
____     ____
1.5      1.5     divide each side by 1.5

y = 2500        divide and cancel

x + 2500 = 7500          replace y with 2500

-2500  -2500     subtract 2500 from each side
________________________
x           = 5000      subtract

result:  5000 tickets at \$5.50 per ticket,

2500 tickets at \$7.00 per ticket

(6.)  Five full-fare bus tickets and one half-fare ticket

cost \$90.75 .  Four full-fare tickets and two half-fare

tickets cost 82.50 .   How much does each type of

ticket cost?

5f + h = 90.75

4f + 2h = 82.50              here is the problem

-2f - h = -41.25    multiply eq 2 thru by -1/2, cancel

5f + h = 90.75      put this here
____________________
___      _______
3          3        divide each side by 3

f = 16.50       divide and cancel

4(16.50) + 2h = 82.50   replace f with 16.50

66 + 2h = 82.50       multiply

- 66      - 66       subtract 66 from each side
________________________
2h = 16.50          subtract
___  ______
2      2        divide each side by 2

h = 8.25      divide and cancel

result:  full-fare tickets cost \$16.50 and half-fare

tickets cost \$8.25 .

(7.)   Bob makes a bank deposit of \$595 with 96 five- and

ten- dollar bills.   How many of each kind of bill

did she deposit?

f + t = 96

5f + 10t = 595                 here is the problem

-5f - 5t = -480     multiply eq. 1 thru by -5
_____________________
5t = 115        subtract equations
___  ____
5     5          divide each side by 5

t = 23         divide and cancel

f + 23 = 96     replace t with 23

-23  -23   subtract 23 from each side
___________________
f     = 73      subtract

result:  73 fives and 23 tens

(8.)  Ralph paid \$6.60 for some 15 cent stamps and some 20

cent stamps.  He bought 37 stamps in all.  How many of

each kind of stamp did he buy?

x + y = 37

.15x + .20y = 6.60              here is the problem

- .15x - .15y = -5.55       multiply eq 1 thru by -.15
_______________________
.05y =  1.05      subtract equations
___  _______
.05    .05          divide each side by .05

y = 21          divide and cancel

x + 21 = 37            replace y with 21

-21  -21      subtract 21 from each side
______________
x        = 16        subtract

result:  Sixteen 15 cent stamps.  Twenty-one 20 cent stamps.

(9.)   There were two models of calculators.   One model sold

for \$22.75 and the other model sold for 15.95.   In

all, there were 31 calculators.   The total sale of

these 31 calculators was \$576.06 .   How many of each

model were sold?

A + B = 31

22.75A + 15.95B = 576.05           here is the problem

-15.95A - 15.95B = -494.45      multiply eq 1 thru by -15.95
____________________________
6.80A          = 81.60       subtract equations
_____            _____
6.80            6.80      divide each side by 6.80

A = 12                 divide and cancel

12 + B = 31                 replace A with 12

-12       -12         subtract 12 from each side
_________________
B = 19          subtract

result:   12 calculators and 19 calculators

(10.)   A library charges a fixed amount for the first day that

a book is overdue and an additional charge for each day

thereafter.  Sam paid \$0.75 for one book that was 7 days

overdue and \$1.95 for a book that was 19 days overdue.

Find the fixed charge and the charge for each additional

day.

f + 6a = 0.75

f + 18a = 1.95              here is the problem

f + 6a =  0.75        put this here
______________________
12a = 1.20          subtract equations
____  ___
12     12      divide each side by 12

a = .10       divide and cancel

f + 6(.10) = 0.75     replace a with 10

f + .60 = 0.75          multiply

- .60   -.60    subtract .60 from each side
_____________________
f       =  .15         subtract

result:  The fixed charge is \$0.15 and the additional charge

is \$0.10 per day.