60
Chapter 2 | Review of Basic Algebra
2.5 |
Formulas and Applications
Formulas are similar to equations. In formulas, the relationship among many variables is written as a rule
for performing calculations. To solve for one of the variables in a formula, rearrange the formula to isolate
the required variable, simplify, and then solve for the required variable. Rearrangement can be performed
using the rules that you have learned in the previous sections of this chapter.
To solve for a required variable using a formula, it is important to know what each symbol in the formula
represents. For example, consider the formula for simple interest: I = Prt. In this simple interest formula,
'I' represents the amount of simple interest, 'P' represents the amount of investment or loan, also known
as the principal, 'r' represents the interest rate per annum, and 't' represents the time period in years.
To solve for any one of the variables, 'I', 'P', 'r', or 't' in this simple interest formula, we can rearrange the
variables, as shown below:
Solving for 'P' :
I = Prt Prt = I rt
rt
P= I
rt
is the same as Prt = I
Dividing both sides by 'rt' , we obtain,
Solving for 'r' :
Prt = I Pt
Pt
r= I
Pt
Dividing both sides by 'Pt' , we obtain,
Prt = I Pr
Pr
I
t = Pr
Dividing both sides by 'Pr' , we obtain,
Solving for 't' :
Example 2.5(a)
Rearranging to Isolate Variables
Rearrange and isolate the variables:
(i) 'M' in S = C + M
(ii) 'P' in C + E + P = S
(iii) 'R' and 'B' in P = RB
(iv) 'b' and 'm' in y = mx + b
(v) 'P' in S = P(1 + rt)
Solution
If a = b + c,
then b + c = a
(i)
S=C+M
M + C = S
Subtracting 'C' from both sides, we obtain,
M + C - C = S - C M=S-C
(ii) C+E+P=S
Subtracting 'C' and 'E' from both sides, we obtain,
C + E + P - C - E = S - C - E
P=S-C-E
(iii) P = RB
Isolating 'R'
RB = P
Isolating 'B'
RB = P
Chapter 2 | Review of Basic Algebra
Solution
continued
Dividing both sides by 'B', we obtain, RB = P B
B
Dividing both sides by 'R', we obtain,
RB = P
R
R
P
R= B
P
B= R
(iv) y = mx + b
Isolating 'b'
mx + b = y
Isolating 'm'
Subtracting 'mx' from both sides, we obtain,
mx - mx + b = y - mx b = y - mx mx + b = y
Subtracting 'b' from both sides, we obtain,
mx + b - b = y - b
mx = y - b
Dividing both sides by 'x ', we obtain,
y- b
mx
= x
(v) S = P(1 + rt) x
m= y- b
x
P(1+rt) = S
P^1 + rth
S
=
^1 + rth
^1 + rth
S
P=
^1 + rth
Example 2.5(b)
Isolating 'P' by dividing both sides by '(1 + rt)', we obtain,
Solving for Variables Using the Rearranged Simple Interest Formula
In the simple interest formula I = Prt, find:
(i) 'I', when P = $1000, r = 5% = 0.05, t = 3 years
(ii) 'P', when I = $150, r = 3% = 0.03, t = 1 year
(iii) 'r', when I = $500, P = $8000, t = 2 years
(iv) 't', when I = $40, P = $800, r = 5% = 0.05
Round your answers to two decimal places wherever applicable.
Solution
(i) Substitute the value for 'P', 'r', and 't' in the formula: I = Prt
I = 1000 # 0.05 # 3 = $150.00
I
(ii) Substitute the value for 'I', 'r', and, 't' in the rearranged formula: P = rt
150
P=
= $5000.00
^0.03 # 1h
(iii) Substitute the value for 'I', 'P', and 't' in the rearranged formula: r = I
Pt
500
= 0.03125 = 3.125% = 3.13% (Rounded to two decimal places.)
r=
8000 # 2
I
(iv) Substitute the value for 'I', 'P', and 'r' in the rearranged formula: t = Pr
40
= 1 year
t=
800 # 0.05
61
62
Chapter 2 | Review of Basic Algebra
­­­ 2.5V |= Exercises
r2h = π (62)(15) = 1696.460033... = 1696.46 cm3
Answers to the odd-numbered problems are available at the end of the textbook
For Problems 1 to 14, rearrange and isolate the indicated variables.
1. 'C' in S = C + M2. 'M' in S = C + M
3.
'E' in S = C + E + P
4. 'C' in S = C + E + P
5.
'B' in P = R × B6. 'R' in P = R × B
7.
'r' in S = P(1 + rt) 8. 't' in S = P(1 + rt)
9.
'L' in N = L(1 - d) 10. 'd' in N = L(1 - d)
11. 'a' and 'c' in b =
ac 1- a
13. 'a' and 'b' in c =
a- c
b 12 'a' and 'c' in b = c + ac
a- 2
ab - b
14. 'a' and 'b' in c = 4 a
+
For Problems 15 to 40, express the answers rounded to two decimal places, wherever applicable.
In the simple interest formula I = Prt,
15. a. find 'I', when P = $4500, r = 0.05, t = 10
12
16. a. find 'I', when P = $1200, r = 0.0325, t = 8 12
17. a. find 'r', when P = $4850, I = $162.65, t = 7 12
18. a. find 'r', when P = $5775, I = $296.75, t = 9 12
In the simple interest formula S = P(1 + rt),
b. find 'P', when I = $20.75, r = 0.0475, t = 91
365
b. find 'P', when I = $65.50, r = 0.0525, t = 180
365
b. find 't', when P = $850, r = 0.035, I = $19.50
19. find 'S', when P = $6500, r = 0.065, t = 9 12
20. find 'S', when P = $1500, r = 0.075, t = 4
12
21. find 'P', when S = $8260.80, r = 0.0425, t = 280 365
In the trade discount formula N = L(1 - d),
22. find 'P', when S = $10,135.62, r = 0.0375, t = 132
365
23. find 'N', when L = $1200, d = 0.08
24. find 'N', when L = $1800, d = 0.07
25. find 'L', when N = $3000, d = 0.40 26. find 'L', when N = $10,000, d = 0.20
27. find 'd', when L = $900, N = $675
28. find 'd', when L = $1280, N = $1126.40
b. find 't', when P = $9250, r = 0.075, I = $635.94
n
In the compound interest formula FV = PV(1 + i) ,
29. find 'FV', when PV = $5000, i = 0.0375, n = 4
30. find 'FV', when PV = $40,000, i = 0.0425, n = 12
31. find 'FV', when PV = $2500, i = 0.0004, n = 17 12
33. find 'PV', when FV = $7769.72, i = 0.00375, n = 24
32. find 'FV', when PV = $7750, i = 0.0035, n = 25
6
34. find 'PV', when FV = $4364.12, i = 0.00475, n = 39
m
In the effective interest rate formula f = (1 + i) - 1,
35. find 'f', when i = 0.03, m = 4
In the periodic interest rate formula i = (
36. find 'f', when i = 0.02, m = 2
1
n
FV
) - 1,
PV
37. find 'i', when PV = 3000, FV = 3280.85, n = 8 38. find 'i', when PV = $8600, FV = $11,587.22, n = 24
m1
In the equivalent interest rate formula i2 = (1 + i1) m - 1,
2
39. find 'i2', when i1 =
0.078
, m1 = 12, m2 = 4
12
40. find 'i2', when i1 = 0.096 , m1 = 4, m2 = 12
4