[WORK PROBLEM]
1. A woodworker normally makes a certain number of parts in 24 days. But he was able to increase
his productivity by 5 parts per day, and so he not only finished the job in only 22 days but also
he made 80 extra parts. How many parts does the woodworker normally makes per day and
how many pieces does he make in 24 days?
Solution:
Let x be the number of parts the woodworker normally makes daily. In 24 days he makes 24 x pieces. His
new daily production rate is x + 5 pieces and in 22 days he made 22(x + 5) parts. This is 80 more than 24
x. Therefore the equation is:
24 x + 80 = 22(x +5)
30 = 2x
x = 15
Normally he makes 15 parts a day and in 24 days he makes (15)( 24) = 360 parts.
2. In a certain electronic factory, the ratio of the number of male to female workers is 2:3. If one
hundred new female workers are hired, the number of female workers will increase to 65% of
the total number of workers. Find the original number of workers in the factory.
Solution:
Let x = male; y = female
x/y = 2/3
y + 100 = 0.65(x + y + 100)
Substitute value of x = 2y/3
y + 100 = 13y/30 + .65y + 65
y= 420
x= 2(420)/3 = 280
Answer: female = 420
Male = 280
OR
3/5x +100=13/20(x+100)
3. 150 workers are engaged to complete a job and it is known that if they all work together the job
will be completed in a certain number of days. However, after the first day of work, 4 workers
resign. After the second day, another 4 resign. This pattern continues until the job is finally
completed, 8 days over schedule.
Find the number of days in which the work was completed.
Solution:
Equating total work,
150 + 146 + 142 + ... + [150-4(n+7)] = 150n
a + (n-1) d  formula ... but the total days is 8 days overtime... thus n-1+8 = n+7
Sn=n/2 *(a1+an); an=a1+ (n-1) d
(n + 8)/2 * [300 - 4(n+7)] = 150n
n = 17
Thus the total days is 25 days
4. A farmer planned to plough a field by doing 120 hectares a day. After two days of work he
increased his daily productivity by 25% and he finished the job two days ahead of schedule.
a.) In how many days did the farmer get the job?
b.) In how many days did the farmer plan to get the job done?
Solution:
120x = y equation 1
(120 * 2) + (120*1.25)*(x-2-2) = y equation 2
120 * x = (120*2) + (120*1.25)*(x-4)
X = 12 days
A. 10 days
B. 12 days
OR
120(2) + 120(1.25) (X-2-2) =120X
X=12 days (planned)
He finished the job 2 days ahead (12-2) = 10 days
5. Steve, Ashley and Jeremy work at the same restaurant. Steve works every 4 days. Ashley works
every 6 days. Jeremy works every 3 days. if on June 1 they are all working at the restaurant,
what is the next date that three of them will work together?
Answer: June 13th
[DISTANCE PROBLEM]
6. A and B travel from Angeles to Guagua and back. A started 2 hours after B started and overtakes
B at a Point 4 km from Guagua. He then meets B 3 hours and 12 minutes thereafter. If it reaches
Angeles 4 hours and 40 minutes ahead of B, find the distance of Angeles and Guagua.
A. 6 km
B. 8 km
C. 10 km
D. 12 km
E. None of the above
Solution:
Time=distance/velocity
At distance x-4:
tb - ta =2
[(x-4)/Vb] - [(x-4)/Va] = 2 eq. 1
At distance 2x (one round trip)
tb - ta =2 +(14/3)
[2x/Vb] - [2x/Va] = 2+(14/3) eq.2
Eq.1 divided by Eq.2:
(x-4)/2x=2/[2+(14/3)]
Solve for x:
x=10km
7. It takes a boat 3 hours to travel down a river from point A to point B, and 5 hours to travel up
the river from B to A. How long would it take the same boat to go from A to B in still water?
Answer: 3.75 or 3 hrs. and 45 mins.
Let:
S= be the speed of the boat in still water,
r = be the rate of the water current and
d = the distance between A and B.
Equation 1: d = 3(S + r) : boat traveling down river (A to B)
Equation 2: d = 5(S - r) : boat traveling up river (B to A)
3(S + r) = 5(S - r): equate the two equation above.
3S + 3r = 5S – 5r
5r + 3r = 5S - 3S
8r=2S
1/8(8r=2S)1/8 : multiply both sides by 1/8 or divide bothsides by 8 to get r
r=2S/8
r=S/4
d = 3(S + r): use equation1 to get what is ask in the problem.
d = 3[S + (S/4)]: substitute the value of r in the equation
d= 3S + 3S/4
1/S(d= 3S + 3S/4)1/S: multiply both sides by 1/S or divide both sides by S
d/S = 3 + 3/4
d/S= 15/4
d / S = 3.75 hours = 3 hours and 45 minutes.
8. A Cessna single engine has an air-speed (speed in still air) of 125 kph. A west wind of 25 kph is
blowing. The plane is to patrol due east and then return to its base. How far east can it go if the
round trip is to consume 4 hrs?
Answer:
let x distance it can go before coming back.
time of going against the wind + time going back with the wind = 4hrs
From v=d/t; t=d/v
(x/ 125-25) +(x/125+25) = 4
x=240 km
OR
Same distance
Deast = Dwest
Veast*Teast=Vwest*Twest
(125-25)*Te = (125+25)*Tw
Te + Tw = 4, Tw = 4 - Te
Te = 2.4 hrs
D = VT
D = (125-25)*2.4 = 240
9. Some hikers start on a walk at 9am and return at 2pm. One quarter of the distance is uphill, one
half is level, and one quarter is downhill. If their speed is 4miles per hour on level land, 2miles
per hour uphill, and 6miles per hour downhill, approximately how far did they walk?
Answer:
d/8 + d/8 + d/24 = 5
Then find d (yung back and forth) which 120/7
OR
Solution:
Let 4x=total distance travelled so that x~uphill, x~downhill, & 2x~level.
This takes a total time of ( x/2 + x/6 + 2x/4 ) hours and equating to 5 hours,
we find x=30/7 and the total distance 4x= 120/7~17.1miles
[TIME PROBLEM]
10. A man leaving his office one afternoon noticed the clock at past two o'clock. Between two to
three hours, he returned to his office noticing the hands of the clock interchanged. At what time
did he leave the office and the time that he returned to the office?
Answer:
Let x be the minutes after 2pm
y be the minutes after 5pm
x= 25+y/12
y=10 +x/12
Solve simultaneously
x=26.014 min
y= 12. 168 min
Leaving
2:26.014 pm
Return at
5:12.168 pm
TIME/DISTANCE PROBLEM
11. Catherine left point A at 1:00pm with a constant speed of 60kph. 20 minutes later, Anamae also
left point A with a constant speed of 75kph. Assuming they have the same destination, by what
time will Ana Mae be 71km ahead of Catherine?
Answer:
20 minutes later, Catherine is (60 km/h) × (1/3 h) = 20 km.
Now, let Xc be Catherine's distance from starting point, Xa be Anamae's distance from starting point.
Assume t be the time for Anamae. Then
Xc = X0c + Vc × t (X0c = 20 km, Vc = 60 km/h)
Xa = X0a * Va × t (X0a = 0 km, Va = 75 km/h)
Now, we want to find t such that
Xa - Xc = 71 km
Equating, we have
(Va × t) - (X0c + Vc × t) = 71 km
Solving for t, we have
t = (71 km + X0c)/(Va - Vc)
t = (71 km + 20 km)/(75 km/h - 60 km/h) = 91/15 h = 6 1/15 h, or 6 h 4 min
So Anamae is 71 km away from Catherine 6 h 4 min after she left from starting point. But remember,
Catherine is already driving 20 minutes before Anamae started, so Anamae stated 1:20 pm. So after 6 h
4 min, or 7:24 pm, she is 71 km aheas of Catherine.
So the answer is 7:24 pm.
OR
D = RT
Let x be the time (in hours) that Catherine traveled.
Catherine traveled 60(x) km, while Anamae traveled 75(x - 1/3) km. But since Anamae is 71 km ahead of
Catherine,
60(x) = 75(x - 1/3) -71
60x = 75x-25-71
-15x = -96
x = 32/5 or 6 2/5
-> 6 hours and 24 minutes. :)
So, 1:00pm + 6:24 = 7:24pm. :) That was the time that Anamae was seventy-one kilometers ahead to
Catherine.
TIME PROBLEM
12. Suppose that an airplane climbs at an angle of 30°. If its speed is maintained at 550kph, how
long will it take to reach a height of 15km?
Answer:
No need for sine law here. If you will sketch it, it will form a 30-60-90 triangle. The opposite of 30° is
15km, thus it's hypotenuse is 15*2=30km. 30/550 = 3/55 hours
NUMBER PROBLEM
13. Four children have small toys. The first child has 1/10 of the toys, the second child has 12 more
toys than the first, the third child has one more toy of what the first child has and the fourth
child has double the third child. How many toys are there?
Solution:
A. let x be the total number of toys
B. let x/10 be the number of toys for first child
C. let (x/10)+12 be the number of toys for second child
D. let (x/10) + 1 be the number of toys for the third child
E. let 2[(x/10) + 1] be the number of toys for the fourth child
F. x/10 + x/10 + 12 + x/10 + 1 + 2(x/10 + 1) = x ------------ add it all to get number of toys
G.5x/10 + 15 =x --------------------------------------- simplify
H.10(5x/10 + 15 =x)10----------------------------multiply both sides by 10
5x + 150 = 10x
150 = 10x - 5x
150 = 5x
1/5(150 = 5x)1/5
X=30
Therefore, there are 30 toys.
14. The sum of the digits of a four digit number is 22. If the original number is divided by its digits
reversed, the quotient is 2 and the remainder is 302. If the thousands and tens digit is reversed,
the value is increased by 1980. But if the hundreds and ones digit is reversed, the value decrease
by 594. What is the number?
Solution:
1000a+100b+10c+d
a + b +c +d=22
a=22-b-c-d
1000a+100b+10c+d/1000d+100c+10b+a=2 remainder 302
1000a+100b+10c+d=2(1000d+100c+10b+a)+302
998a+80b-190c-1999d=302 ; subs: a+ b +c +d=22
998(a+ b +c +d=22)+80b-190c-1999d=302
-918b-1188c-2997d=-21654  1st
1000c+100b+10a+d=1000a+100b+10c+d+1980
990c+(-990a)=1980; a=22-b-c-d
990c-990(22-b-c-d)=1980
1980c+990b+990d=234760  2nd
1000a+110a+10c+b=1000a+100b+10c+d-594
99d-99b=-594  3rd
MODE 5 2; b=8, c=7, d=2 ; a=22-8-7-2=5
thus; 5872 Answer.
[AGE PROBLEM]
15. Mary is 24. Mary is twice as old as Ana was when Mary was as old as Ana is now.
How old is Ana?
a. 18
b. 16
Solution:
24-x=x-12
2x=36
x=18
16. There are two boys, named as Ali and Jake. Given that Ali's age now is 15 more than the
logarithm of Jake's age 83 years from now, and the sum of their present ages is 34, how old is
Jake?
Answer:
Let:
A = Ali's age
J = Jake's age
A= 15 + log ( J + 83)
34 - J = 15 + log ( J + 83)
19 - J = log ( J + 83)
10^(19-J) = J + 83
By inspection, we can observe that the only situation that will satisfy the above equation is when ( 19 - J
)=2
Thus, J = 17 yrs
17. When John was as old as Paul is now, the sum of their ages was 51. When Paul will be as old as
John is now, the sum of their ages will be 1O3. John is older than Paul by how many years?
Solution:
Y-X=N, Y kay john. X kay paul.
Y-N=X, para bumata si john na kasing edad ni paul, pero kung babata si john babata din si paul
X-N yun. Bali sum,
Y-N+X-N=51
Tapos kung tatanda si paul para kasing edad ni john X+N=Y yun diba, pero since tumanda si paul
tumanda din si john. Y+N naman yun
Then sum
Y+N+X+N=103. Tapos 3 equations 3 unknowns na. Madali na isolve through substitutikn or calcu or elim.
Hehe
18. Peter is 36 years old. Peter is twice as old as Jun was when Peter was as old as Jun is now. How
old is Jun?
Solution:
Peter is 36 years old. Peter is twice as old as Jun was when Peter was as old as Jun is now. How old is
Jun?
------------------------Let P = present age of Peter, J = present age of Jun.
x years ago, their ages were P - x and J - x.
"...when Peter was as old as Jun is now"
---> P - x = J ---> x = P - J
"Peter is twice as old as Jun was [when Peter was as old as Jun is now]"
---> P = 2(J - x) = 2(J - (P - J)) = 2(2J - P) = 4J - 2P
---> J = 2P/3 = 2(36)/3 = [24 years old]
[AGE PROBLEM]
19. Two years ago, Theresa was 3 times as old as Carmela. Four years hence Theresa will be twice as
old as Carmela .Find their age now.
Solution:
X - 2 = 3(Y - 2), X + 4 = 2(Y + 4) SOLVING Y = 8, AND X = 20
Theresa = 20, Carmela = 8
20. The sum of the ages of the parents and three children is 9 decades over a century. The father is
twice as old as the eldest child. When the eldest child was born, the sum of the parent's ages
was 54. When the youngest was born, the sum of the parent's age was 7O. In 38 years, the sum
of parent's ages will be equal to the sum of children's ages. How old is the second child?
Answer: 30&24
[NUMBER PROBLEM]
21. A number is expressed by 3 digits of which the first and the last are alike. By interchanging the
digits in the unit's and ten's places, the number is increased by 54, but if the digits in the ten's
and hundred's places are interchanged, 9 must be added to four times the resulting number to
make it equal to the original A number is expressed by 3 digits of which the first and the last are
alike. By interchanging the digits in the unit's and ten's places, the number is increased by 54,
but if the digits in the ten's and hundred's places are interchanged, 9 must be added to four
times the resulting number to make it equal to the original number.
1. Find the hundred's digit.
2. Find the sum of ten's digit and units digit.
3. Find the product of the cube root of the ten's digit and the cube of the hundred's digit.
Answer:
717
1.) 1
2.) 8
3.) ³√7
[MIXTURE PROBLEM]
22. In what ratio must a peanut costing P 240.00 per kg be mixed with a peanut costing P340.00 per
kg so that a profit of 20% is made by selling the mixture at P 360.00 per kg?
Solution:
Let x=quantity of P240/kg, and y=quantity of P340/kg..
Capital=240x+340x
Total sales=360(x+y)
Profit=0.20(240x+340y)
Profit =Total sales-Capital
0.20(240x+340y)=360(x+y)-240x+340y
72x=48y
Therefore the ratio is
2:3
23. How many galloons of a mixture containing 80% alcohol should be added to 5 galloons of 20%
solution to give a 30% solution?
Solution:
0.8x+0.2(5)=0.3(x+5)
x = 1 gallon
[AGE PROBLEM]
The ages of Jace, Jane, and Jade forms an arithmetic sequence. When Jace's age is decreased by 4, Jane's
age is increased by 5, and Jade' age is doubled, the "new numbers" obtained also forms an arithmetic
sequence. How old is Jade☺️?
Solution:
Let Jace's age = a-d
Jane's age = a
Jade's age = a+d
(a-d-4, a+5, 2(a+d))
a+5-a+d+4=2(a+d)-a-5
d+9=a+2d-5
a+d=14 - Jade's age
[MIXTURE PROBLEM]
A given alloy contains 20% copper and 5% tin. How many pounds of copper and of tin must be melted
with 100 lb of the given alloy to produce another alloy analyzing 30% copper and 10% tin? All
percentages are by weight.
A. 20 lb copper and 15 lb tin
B. 15 lb copper and 20 lb tin
C. 17.5 lb copper and 7.5 lb tin
D. 7.5 lb copper and 17.5 lb tin.
Solution:
100(.2)+x=(100+x+y)(.30) (equation 1)
100 (.05)+y= (100+x+y) (.10) (equation 2)
Then simplify the two equations. Then apply mode 5 1 in the calculator.
You will get x= 17.5 and y= 7.5
[MATH]
Peter is crossing the dessert with his friend, John. He starts off with a full waterbottle and drinks 1/3 of
the contents during the first day. He then lets John drink half of what is left. The next day, Peter drinks
1/4 of what has been saved from the previous day.
What fraction of the original amount did he save for John?
Solution:
(2/3)(1/2)=1/3 (1/3)(1/4)=1/12 1/3-1/12= ¼
[ANALYSIS]
Hish is a game played by two players each of them having some number of marbles with her. One of the
two players has to determine whether the number of marbles with other player is even or odd. A
particular game of Hish has seven players and starts with players P1 and P2 on field and the other
players P3, P4, P5, P6, P7 waiting in a queue for their turn in order. After each game is played, the loser
goes to the end of the queue; the winner adds 1 point to her score and stays on the field; and the player
at the head of the queue comes on to contest the next point. Game continues until someone has scored
11 points. At that moment, it was found out that a total of 43 points have been scored by all seven
players together. Who is the winner?
[AGE PROBLEM]
The ages of Alysa, Allysa, and Alyssa forms an arithmetic sequence, where in Alysa is the youngest and
Alyssa is the oldest. When Alyssa's age is increased by 8, Alysa's age is decreased by 7, and Allysa's age is
decreased by 3, the "new numbers" obtained forms a geometric progression, and the order Alysa-AllysaAlyssa is still attained. Given that Alysa is a teenager, how old is Alyssa?
Solution:
The earlist age that a teenager is thirteen so when alysa's age is decreased by 7 then Alyssa 's age would
be 13 + 7 = 20
[WORK PROBLEM]
Elizabeth can get a certain job done in 15 days, and Tony can finish only 75% of that job within the same
time. Tony worked alone for several days and then Elizabeth joined him, so they finished the rest of the
job in 6 days, working together.
For how many days have each of them worked and what percentage of the job have each of them
completed?
Answer:
Tony: 12 days, 60%; Elizabeth: 6 days, 40%