1. Topic: WORDED PROBLEMS
WORDED PROBLEMS
AGE PROBLEM
General Principle: The time elapsed for the people involved are equal.
Illustration:
A father is three times as old as his son. Five years ago, he was five times as old as his son was at that time. How old is his
son?
PAST PRESENT
Father 3x-5 3x
Son x-5 x
3x − 5 = 5(x − 5) “5 years ago, the father was 5 times as old as his son”
3x − 5 = 5x − 25
x = 10 years old, the age of the son
MIXTURE PROBLEM
Amount of substance = [% concentration][total amount of the mixture]
[Amount of substance in original mixture] ± [amount of substance added/subtracted] = [amount of substance in resulting mixture]
Illustration:
How much silver and copper must be added to a 20
kg alloy containing 10% silver and 25% copper to
produce an alloy containing 36% silver and 38%
copper.
Let x = amount of silver, in kg
Y = amount of copper, in kg
For silver:
(0.10)(20 kg) + (1)(x) = (0.36)(20 kg + x + y)
For copper:
(0.25)(20 kg) + (1)(y) = (0.38)(20 kg + x + y)
Solving simultaneously:
x = 16 kg
y = 14 kg
CLOCK PROBLEM
Let x = the number of minute spaces traveled by the minute hand
x
= the number of minute spaces traveled by the hour hand
12
Illustration:
At what time between 4 and 5 o’clock does the
minute hand of a clock be coincident with each
other?
x
x = 20 +
12
⎡ 12 ⎤
x = 20 ⎢ ⎥
⎣ 11 ⎦
x = 21.8181
Time is: 4:21:8181
General equation in solving clock problem:
⎡ 12 ⎤
x =t ⎢ ⎥
⎣ 11 ⎦
Where: t = time traveled by the minute hand
VARIATIONS
Direct variation: When x varies directly as y
x α y
x = ky
Inverse variation: When x varies inversely as y
1
x α
y
⎡1 ⎤
x = k ⎢ ⎥
⎣y⎦
Joint variation: When x varies directly as y and inversely as z
⎡y⎤
x = k ⎢ ⎥
⎣z⎦
DAY 2 Copyright 2010 www.e-reviewonline.com

2. Topic: WORDED PROBLEMS
Illustration:
The kinetic energy KE of an object varies directly as the square of its velocity, v. A particular object traveling at 80 ft/sec
has kinetic energy of 240 ft-lbs. What would be its kinetic energy if it’s traveling at 100 ft/sec.
KE α v 2
KE = kv 2
When the particle is traveling at 80 ft/sec, kinetic energy is 240 ft-lbs
240 ft − lbs = k(80 ft / sec)2
k = 0.0375
Hence:
KE = (0.0375)(100 ft / sec) 2
KE = 375 ft-lbs
WORK PROBLEM
Let: R = rate of working Illustration:
T = time of working An input pipe A can fill the tank in 9 hours. Another
n = number of workers input pipe can fill the same empty tank in 6 hours. A
W = number of job units drain pipe can empty a full tank in 12 hrs. If all the
For single worker: pipes are open, how long will it take them to fill the
RT= 1 tank?
For n workers working at different rates and in different time
lengths:
R 1 T1 + R 2 T2 + ... + R n Tn = 1
For n workers working together at same time rates:
(R1 + R 2 + ... + R n )T = 1
For worker 1 working alone for some period of time and
worker 2 helped at time T:
R 1 T1 + (R 1 + R 2 )T = 1
Relationship among n, T and W:
The more workers, the less time to finish the job:
1 (R A + R B − R C )T =1
nα
T ⎡1 1 1 ⎤
The more number of job units, the more workers required: ⎢ + − ⎥T = 1
⎢
⎣ TA TB TC ⎥
⎦
Wαn
The more number of job units, the more time is required: ⎡1 1 1 ⎤
⎢ + − ⎥T = 1
Wα T ⎣ 9 6 12 ⎦
n1 T1 n T T = 5.1429 hrs
Combining: = 2 2
W1 W2
NUMBER PROBLEM MOTION PROBLEM
dis tan ce
Let x = the first number speed or rate =
time
y = the second number
s
Interpret number problems in equation form. v=
t
Illustration: Illustration:
Find 3 consecutive odd integers whose sum is 39. Peter can walk from his house to his office at the rate of 5
Let x = the first odd integer mph and back at the rate of 2 mph. Find the average speed in
x + 2 = second odd integer mph.
x + 4 = third odd integer s s + sB−A s + sB−A
v ave = T = A −B = A −B
x + (x + 2) + (x + 4) = 39 tT t A −B + t B − A s A −B s
+ B− A
x = 11 v A −B v B−A
Hence; the three consecutive odd integers are 11, 13 and 15 s A −B + s B − A 10 (s A −B + s B − A )
v ave = =
s A −B sB− A 2 s A −B + 5 s B− A
+
5 mph 2 mph
Note: s A − B = sB − A = s
20 ⎡ s ⎤
v ave = ⎢ ⎥
7 ⎣s⎦
v ave = 2.8571 mph
INTEREST PROBLEM
Interest = (Principal)(Period)(Interest rate per period)
Illustration:
Ricardo loans an amount of P6,000, part at 80% annual interest and the rest at 20%. Calculate the amount of each loan if
the total annual income is at P2,000.
Let: x = money at 80% interest
P6000 – x = money at 20%
0.80(x) + 0.20(6000 – x)= 2,000
x = P1,333.3333 at 80% interest
P6,000 – x = P4,666.6667 at 20% interest
DAY 2 Copyright 2010 www.e-reviewonline.com