Thursday, February 27, 2014

PROJECT: BEDANS

Submit your project in the comment section of this blog.  THanks

23 comments:

  1. Kim Icmat
    BAIST-2B
    MA101

    PERCENT
    Problem 1:

    The original price of a shorts was $20. It was decreased to $15 . What is the percent decrease of the price of this shirt.
    Solution to Problem 1:

    The absolute decrease is

    20 - 15 = $5

    The percent decrease is the absolute decrease divided by the the original price (part/whole).

    percent decease = 5 / 20 = 0.25

    Multiply and divide 0.25 to obtain percent.

    percent decease = 0.25 = 0.25 * 100 / 100 = 25 / 100 = 25%
    Problem 2:

    Anna has a monthly salary of $1200. She spends $280 per month on food. What percent of her monthly salary does she spend on food?
    Solution to Problem 2:

    The part of her salary that is spent on food is $280 out of her monthly salary of $1200

    percent = part / whole = 280 / 1200 = 0.23 (rounded to 2 decimal places)

    Multiply and divide 0.23 by 100 to convert in percent

    percent = 0.23 * 100 / 100 = 23 / 100 = 23%

    Problem 3:

    The price of a pair of socks was decreased by 22% to $30. What was the original price of the socks?
    Solution to Problem 3:

    Let x be the original price and y be the absolute decrease. If the price was decreased to $30, then

    x - y = 30

    y is given by

    y = 22% of x = (22 / 100) * x = 0.22 x

    Substitute y by 0.22 x in the equation x - y = 30 and solve for x which the original price.

    x - 0.22 x = 30

    0.78 x = 30

    x = $38.5

    Check the solution to this problem by reducing the origonal price found $38.5 by 22% and see if it gives $30.

    Problem 4:

    The price of an item changed from $120 to $100. Then later the price decreased again from $100 to $80. Which of the two decreases was larger in percentage term?
    Solution to Problem 4:

    First decrease in percent

    part / whole = (120 - 100) / 120 = 0.17 = 17%

    Second decrease in percent

    part / whole = (100 - 80) / 100 = 0.20 = 20%

    The second decrease was larger in percent term. The part were the same in both cases but the whole was smaller in the second decrease.
    Problem 5:

    The price of an item decreased by 20% to $200. Then later the price decreased again from $200 to $150. What is the percent of decrease from the original price to the final price of $150?
    Solution to Problem 5:

    We first need to find the original price x. The first decrease gives

    x - 20% x = 200

    0.8 x = 200

    x = 200 / 0.8 = 250

    The percentage decrease fro the original price 250 to 150 is given by

    part / whole = (250 - 150) / 250 = 0.4 = 40%

    ReplyDelete
  2. Almazan, Katrina Joyce S.
    BAIST – 2B

    Integers:
    1.) The sum of the three consecutive integers is 306. Find the numbers.
    Solution: x+(x+1) + (x+2) = 306
    3x+3 = 306
    3x= 306 – 3
    3x= 303
    3x= 303
    3
    x= 101

    2.) Twelve less than five times the smaller of two consecutive even numbers is equal to four times the larger of the numbers. Find the numbers.
    Solution: Let smaller number= n
    Let larger number = n+2
    5(sn) – 12 = 4(lg)
    5(n) – 12 = 4(n+2)
    5n – 12 = 4n+8
    5n-4n= 8+12
    n= 20

    3.) When the smaller of two even consecutive integers is added to three times the larger. The result is 230. Find the smaller integer.
    Solution: Let smaller integer= n
    Let larger integer= n+2
    si + 3(li) = 230
    n + 3(n+2) = 230
    n+ 3n + 6= 230
    n+ 3n= 230 – 6
    4n = 224
    n= 56

    4.) If two times the smaller of two consecutive integers is added to six times the larger, the result is 150. Find the smaller integer.
    Solution: Let smaller integer= n
    Let larger integer= n+1
    2(si) + 6(li)= 150
    2(n) + 6(n+1) = 150
    2n + 6n+6 = 150
    2n+ 6n= 150- 6
    8n= 144
    8n=144
    8
    n= 18
    5.) If one times the smaller of two consecutive integers is added to 3 times the larger, the result is 39. Find the smaller integer.
    Solution: Let smaller integer= n
    Let larger number= n+2
    1(si)+ 3(li)= 39
    1(n) + 3(n+1)= 39
    n+ 3n+ 3= 39
    n+ 3n= 39- 3
    4n=36
    4n= 36
    4
    n= 9

    ReplyDelete
  3. CAUBANG, Grace
    BSBA – MM – 2C

    1. The sum of Patrick’s age and Marko’s age is 58. Eight years ago, Patrick was twice as old as Marko then. How old is Marko?

    NOW
    x = Marko’s present age
    x – 58 = Patrick’s present age

    PAST ( 8 years ago )
    x – 8
    50 – x

    Equate :

    50 – x = 2 (x–8)
    50 – x = 2x – 16
    3x = 66
    x = 22 Marko’s age
    36 Patrick’s age

    2. Aries is twice as old as Rico while Jake is 24 years younger than Aries. If half of Aries’ age six years ago was three less than one half the sum of Rico’s age in four years and Jake’s present age, find the ages of each.

    x = Rico’s present age
    2x = Aries’ present age
    2x – 24 = Jake’s present age

    6 years ago
    x – 6
    2x – 6
    2x – 30

    4 years from now
    x + 4
    2x + 4
    2x – 20

    Equate:

    ½ (2x – 6) = ½ [ (x + 4) + (2x – 24) ] – 3
    2x – 6 = x + 4 + 2x – 24 – 6
    2x – 6 = 3x – 26
    x = 20 Rico’s age
    2(20) = 40 Aries’ age
    2(20) – 24 = 16 Jake’s age

    3. Mrs. Canteno is 40 years old and her eldest daughter is 12. When will the mother be twice as old as her eldest daughter?
    x = number of years that mother will be twice as old as her eldest daughter
    40 + x = mother’s age in x years
    12 + x = daughter’s age in x years

    Equate:

    40 + x = 2 (12 + x)
    40 + x = 24 + 2x
    x = 16

    Mother’s age in 16 years: 40 + 16 = 56
    Daughter’s age in 16 years: 12 + 16 = 28

    56 = 2 (28)
    56 = 56


    4. Alvin is now 21 years older than his son. In 8 years, he will be twice as old as his son’s age. What are their present ages?

    x = son’s present age
    x + 21 = Alvin’s present age

    8 years from now

    x + 8
    x + 29

    Equate:

    x + 29 = 2 (x + 8)
    x + 29 = 2x + 16
    x = 13 Son’s age
    x + 21 = 13 + 21 = 34 Alvin’s age


    5. Brenda is 4 years older than Walter, and Carol is twice as old as Brenda. Three years ago, the sum of their ages was 35. How old is each now?

    x = Walter's age now
    x + 4 = Brenda's age now
    2(x + 4) = 2x + 8 = Carol's age now

    x - 3 = Walter's age 3 years ago
    x + 1 = Brenda's age 3 years ago
    2x + 5 = Carol's age 3 years ago

    Equate:

    (x - 3) + (x + 1) + (2x + 5) = 35
    x - 3 + x + 1 + 2x + 5 = 35
    4x + 3 = 35
    4x = 32
    x = 8 = Walter’s present age
    x + 4 = 12 = Brenda’s present age
    2(x + 4) = 24 = Carol’s present age

    ReplyDelete
  4. Giselle V. Ozoa
    BAIST-2B

    MOTION PROBLEM
    1.) How long will it take a bus traveling 72 km/hr to go 36 kms?

    d = rt
    36 km = (72 km/hr) (t)
    36 = 72t
    72 72
    1 = t
    2

    Therefore, it will take one‐half hour for the bus to travel 36 km at 72 km/hr.

    2.) How fast in miles per hour must a car travel to go 600 miles in 15 hours?


    d = rt
    600 = r(15)
    600 = r(15)
    15 15

    40 = r

    3.) John and Philip who live 14 miles apart start at noon to walk toward each other at rates of 3 mph and 4 mph respectively. In how many hours will they meet?

    Solution:
    Let x = time walked.
    r t d
    John 3 x 3x
    Philip 4 x 4x
    3x + 4x = 14
    7x = 14
    x = 2
    They will meet in 2 hours.

    4.) In still water, Peter’s boat goes 4 times as fast as the current in the river. He takes a 15-mile trip up the river and returns in 4 hours. Find the rate of the current.

    Solution:
    Let x = rate of the current.
    r t d
    down river 4x + x 15 / 5x 15
    up river 4x - x 15 / 3x 15

    5.) Andreana and Heather are driving their tractors as fast as they can in OPPOSITE directions.
    Now Andreana is driving a green, John Deere tractor at a whopping 36 miles per hour. Her friend Heather is racing away from her on a red International tractor at 24 miles per hour.
    How many hours will it be before they are 30 miles apart

    The amount of the time is unknown, so we will call it “h”
    36h + 24h = 30 The sum of the two distances must be 30 miles.
    60h = 30
    60h = 30
    60 60
    h = 5. So it took a half hour.









    ReplyDelete
  5. Raemond Zaballero
    BAIST-2B


    Mixture Problems
    Problem 1: How many liters of 20% alcohol solution should be added to 40 liters of a 50% alcohol solution to make a 30% solution?

    Solution to Problem 1:

    Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hence

    x + 40 = y

    We shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But remember the alcohol is measured in percentage term.

    20% x + 50% * 40 = 30% y

    Substitute y by x + 40 in the last equation to obtain.

    20% x + 50% * 40 = 30% (x + 40)

    Change percentages into fractions.

    20 x / 100 + 50 * 40 / 100= 30 x / 100 + 30 * 40 / 100

    Multiply all terms by 100 to simplify.

    20 x + 50 * 40 = 30 x + 30 * 40

    Solve for x.

    x = 80 liters

    80 liters of 20% alcohol is be added to 40 liters of a 50% alcohol solution to make a 30% solution.


    Problem 2: John wants to make a 100 ml of 5% alcohol solution mixing a quantity of a 2% alcohol solution with a 7% alcohol solution. What are the quantities of each of the two solutions (2% and 7%) he has to use?

    Solution to Problem 2:

    Let x and y be the quantities of the 2% and 7% alcohol solutions to be used to make 100 ml. Hence

    x + y = 100

    We now write mathematically that the quantity of alcohol in x ml plus the quantity of alcohol in y ml is equal to the quantity of alcohol in 100 ml.

    2% x + 7% y = 5% 100

    The first equation gives y = 100 - x. Substitute in the last equation to obtain

    2% x + 7% (100 - x) = 5% 100

    Multiply by 100 and simplify

    2 x + 700 - 7 x = 5 * 100

    Solve for x

    x = 40 ml

    Substitute x by 40 in the first equation to find y

    y = 100 - x = 60 ml

    Problem 3: Sterling Silver is 92.5% pure silver. How many grams of Sterling Silver must be mixed to a 90% Silver alloy to obtain a 500g of a 91% Silver alloy?

    Solution to Problem 3:

    Let x and y be the weights, in grams, of sterling silver and of the 90% alloy to make the 500 grams at 91%. Hence

    x + y =500

    The number of grams of pure silver in x plus the number of grams of pure silver in y is equal to the number of grams of pure silver in the 500 grams. The pure silver is given in percentage forms. Hence

    92.5% x + 90% y = 91% 500

    Substitute y by 500 - x in the last equation to write

    92.5% x + 90% (500 - x) = 91% 500

    Simplify and solve

    92.5 x + 45000 - 90 x = 45500

    x = 200 grams.

    200 grams of Sterling Silver is needed to make the 91% alloy.

    Problem 4: How many Kilograms of Pure water is to be added to 100 Kilograms of a 30% saline solution to make it a 10% saline solution.

    Solution to Problem 4:

    Let x be the weights, in Kilograms, of pure water to be added. Let y be the weight, in Kilograms, of the 10% solution. Hence

    x + 100 = y

    Let us now express the fact that the amount of salt in the pure water (which 0) plus the amount of salt in the 30% solution is equal to the amount of salt in the final saline solution at 10%.

    0 + 30% 100 = 10% y

    Substitute y by x + 100 in the last equation and solve.

    30% 100 = 10% (x + 100)

    Solve for x.

    x = 200 Kilograms.

    Problem 5: A 50 ml after-shave lotion at 30% alcohol is mixed with 30 ml of pure water. What is the percentage of alcohol in the new solution?

    Solution to Problem 5:

    The amount of the final mixture is given by

    50 ml + 30 ml = 80 ml

    The amount of alcohol is equal to the amount of alcohol in pure water ( which is 0) plus the amount of alcohol in the 30% solution. Let x be the percentage of alcohol in the final solution. Hence

    0 + 30% 50 ml = x (80)

    Solve for x

    x = 0.1817 = 18.75%

    ReplyDelete
  6. Balugay, Paola Denisse G. 03-03-2014
    BAIST – 2B Algebra
    Problem Solving : AGE
    Problem 1:
    Cary is 9 years older than Dan. In 7 years, the sum of their ages will equal 93.
    Find both of their ages now.

    x = Dan's age now
    x + 9 = Cary's age now {Cary is 9 yrs older than Dan}

    x + 7 = Dan's age in 7 years
    x + 16 = Cary's age in 7 years

    x + 7 + x + 16 = 93 {in seven years the sum of their ages will be 93}
    2x + 23 = 93 {combined like terms}
    2x = 70 {subtracted 23 from both sides}
    x = 35 {divided both sides by 35}
    x + 9 = 44 {substituted 35, in for x, into x + 9}

    Dan is 35
    Cary is 44
    Problem 2:
    Fred is 4 times as old as his niece, Selma. Ten years from now,
    he will be twice as old as she will be. How old is each now?

    x = Selma's age now
    4x = Fred's age now {Fred is 4 times as old as Selma}

    x + 10 = Selma's age in 10 years
    4x + 10 = Fred's age in 10 years

    4x + 10 = 2(x + 10) {his age in 10 yrs is twice her age in 10 years}
    4x + 10 = 2x + 20 {used distributive property}
    2x + 10 = 20 {subtracted 2x from both sides}
    2x = 10 {subtracted 10 from both sides}
    x = 5 {divided both sides by 2}
    4x = 20 {substituted 5, in for x, into 4x}
    Selma is 5
    Fred is 20
    Problem 3:
    An eagle is 4 times as old as a falcon. Three years ago,
    the eagle was 7 times as old as the falcon. Find the present
    age of each bird.
    x = falcon's age now
    4x = eagle's age now {the eagle is 4 times as old as falcon}

    x - 3 = falcon's age 3 years ago
    4x - 3 = eagle's age 3 years ago

    4x – 3 = 7(x – 3) {three years ago, eagle was 7 times the falcon}
    4x – 3 = 7x – 21 {used distributive property}
    4x = 7x -18 {added 3 to both sides}
    -3x = -18 {subtracted 7x from both sides}
    x = 6 {divided both sides by -3}
    4x = 24 {substituted 6, in for x, into 4x}
    falcon is 6
    eagle is 24
    Problem 4:
    Brenda is 4 years older than Walter, and Carol is twice as old
    as Brenda. Three years ago, the sum of their ages was 35.
    How old is each now?

    x = Walter's age now
    x + 4 = Brenda's age now {Brenda is 4 yrs older than Walter}
    2(x + 4) = 2x + 8 = Carol's age now {Carol is twice as old as Brenda, used distributive property}

    x - 3 = Walter's age 3 years ago {subtracted 3 from x}
    x + 1 = Brenda's age 3 years ago {subtracted 3 from x + 4}
    2x + 5 = Carol's age 3 years ago {subtracted 3 from 2x + 8}

    (x - 3) + (x + 1) + (2x + 5) = 35 {sum of ages, 3 years ago, was 35}
    x - 3 + x + 1 + 2x + 5 = 35 {took out parentheses}
    4x + 3 = 35 {combined like terms}
    4x = 32 {subtracted 3 from both sides}
    x = 8 = Walter now {divided both sides by 4}
    x + 4 = 12 = Brenda now {substituted 8, in for x, into x + 4}
    2(x + 4) = 24 = Carol now {substituted 8, in for x, into 2(x + 4)}

    Walter is 8
    Brenda is 12
    Carol is 24
    Problem 5:
    Kevin is 4 years older than Margaret.
    Next year Kevin will be 2 times as old as Margaret.
    How old is Kevin?

    Solution
    Denote as x Kevin's present age.
    Then Margret's present age is x-4.

    Next year Kevin will be x + 1 years old, and Margaret will be x – 4 + 1 = x-3 years old.

    Since next year Kevin will be 2 times as old as Margaret, you can write the equation x+1=2*(x-3).

    Solve this equation by simplifying it, step by step:

    x+1=2x-6 (after brackets opening at the right side)
    1+6=2x-x (after moving variable terms to the right and constant terms to the left)
    7=x (after combining like terms)

    Thus, you got that Kevin's present age is 7 years.

    ReplyDelete
  7. Saladar, Ella
    IS-2B

    Clock problem

    1. What time after 3 o'clock will the hands of the clock are together for the first time?

    Given:
    x = minute hand position when this occurs
    Minute hand; 60 min = 360 degrees
    Hour hand; 1 hr= 360/12= 30 degrees
    Min : x/60 * 360 = 6x
    Hr: x/60 * 30 = 0.5x

    Solution:

    Since it's after 3, we have to add 3 * 30 = 90 degreea to the hr angle.
    6x=0.5x+90
    6x-0.5x=90
    5.5x=90
    X= 16.36 min or 3hr 16 min 22 sec

    2. What time after 2 o'clock will the hands of the clock are together for the first time?

    Given:
    x = minute hand position when this occurs
    Minute hand; 60 min = 360 degrees
    Hour hand; 1 hr= 360/12= 30 degrees
    Min : x/60 * 360 = 6x
    Hr: x/60 * 30 = 0.5x

    Solution:
    Since it's after 2, we have to add 2 * 30 = 60 degrees to the hr angle.
    6x=0.5x+60
    6x-0.5x=60
    5.5x=60
    X= 10.90 min.

    3. What time after 10 o'clock will the hands of the clock are together for the first time?

    Given:
    x = minute hand position when this occurs
    Minute hand; 60 min = 360 degrees
    Hour hand; 1 hr= 360/12= 30 degrees
    Min : x/60 * 360 = 6x

    Hr: x/60 * 30 = 0.5x

    Solution:
    Since it's after 10, we have to add 10* 30 = 300 degreea to the hr angle.
    6x=0.5x+300
    6x-0.5x=300
    5.5x=300
    X= 54.54 min

    4. What time after 5 o'clock will the hands of the clock are together for the first time?
    Given:
    x = minute hand position when this occurs
    Minute hand; 60 min = 360 degrees
    Hour hand; 1 hr= 360/12= 30 degrees
    Min : x/60 * 360 = 6x
    Hr: x/60 * 30 = 0.5x
    Solution:
    Since it's after 5, we have to add 5 * 30 = 150 degrees to the hr angle.
    6x=0.5x+150
    6x-0.5x=150
    5.5x=150
    X= 27.27 min

    5. What time after 7 o'clock will the hands of the clock are together for the first time?
    Given:
    x = minute hand position when this occurs
    Minute hand; 60 min = 360 degrees
    Hour hand; 1 hr= 360/12= 30 degrees
    Min : x/60 * 360 = 6x
    Hr: x/60 * 30 = 0.5x
    Solution:
    Since it's after 7, we have to add 7 * 30 = 210 degrees to the hr angle.
    6x=0.5x+210
    6x-0.5x=210
    5.5x=210
    X= 38.18 min

    ReplyDelete
  8. Rojo, Maria Lareza A.
    BAIST- 2b

    1. How fast in miles per hour must a car travel to go 600 miles in 15 hours?
    First, circle what you must find— how fast (rate). Now, using the equation d = rt, simply plug in 600 for distance and 15 for time.

    d = rt
    600 = r(15)
    600/15 = r(15)/15
    40 = r

    So, the rate is 40 miles per hour.

    2. How long will it take a bus traveling 72 km/hr to go 36 kms?

    d = rt
    36km = (72km/hr)(t)
    36/72 = 72t/72
    1/2 = t

    Therefore, it will take one‐half hour for the bus to travel 36 km at 72 km/hr.

    3. A car leaves a city with a speed of 90 mph. Three hours later and out of the same city another car in pursuit of the first leaves with a speed of 120 mph. Find:
    1The time it takes for the second car to reach the first.

    90t = 120 • (t − 3)
    90t = 120t − 360 −30t = −360 t = 12 hours

    2 The distance from the city when the second car reaches the first.
    d1 = 90 • 12 = 1,080 miles

    4. Two cities, A and B are located on the same east-west highway, 180 miles from each other. At 9 am, a car leaves each city, both travelling east. The car that leaves City A travels at 90 mph, and the car that leaves City B travels at 60 mph. Find:

    1The time it takes for Car A to reach Car B:
    90t − 60t = 180 30t = 180 t = 6 hours

    2The time at which Car A reaches Car B:
    Car A reaches Car B at 3 in the afternoon.

    3 The distance traveled by each at the time of Car A reaching Car B:
    dAB = 90 • 6 = 540 miles.
    dBC = 60 • 6 = 360 miles.

    5. Two cities, A and B are located 300 miles from each other. At 9 am, a car leaves City A with a speed of 90 mph and travels towards City B. At the same time, a car leaves City B travelling towards City A with a speed of 60 mph. Find:

    1 The time it takes for the cars to pass each other.
    90t + 60t = 300 150t = 300 t = 2 hours

    2 The time at which they passed each other.
    They were at 11 of the morning.

    3 The distance traveled by each at the time of them passing each other.
    dAB = 90 • 2 = 180 miles
    dBC = 60 • 2 = 120 miles

    ReplyDelete
  9. Villanueva, Janine Anika B.
    BAIST - 2B

    INVESTMENT PROBLEMS

    1.) After considering all the expenses for the whole year projects, the Social Action Program Fund had a spare of P300,000. The club decided to put part of the money in a savings account that returned a 7% profit at the end of the year and the rest of the money in 8.5% corporate bonds. How much is invested in each account of the total annual interest earned is P24,000?

    Solution:
    Let x = the amount invested at 7%
    and 300,000-x = the amount invested at 8.5%

    I = PRT
    7% investment:
    (x)(0.07)(1) I = 0.07x

    8.5% investment:
    (300,000 - x)(0.085) = 0.0.85(300,000 - x)

    Solve:
    1000[0.07x + 0.085(300,000 - x) = 24,000]1000
    70x + 85(300,000 - x) 24,000,000
    70x + 25,500,0000 - 85x = 24,000,000
    -15x + 25,500,000 = 24,000,000
    15x - 25,500,000 = -24,000,000
    15x = 1,500,000
    x = 100,000

    Amount invested at 7%: x= P100,000
    Amount invested at 8.5%: 300,000 - x = P200,000


    2.) Miguel invested a certain amount of money at 5% per year and an amount twice as large at 6% per year. The total annnual income from the two investments was P4,250. Find the amount invested at each rate.

    Solution:
    Let x = the amount invested at 5%
    and 2x = the amount invested at 6%

    I=PRT
    Some of the money:
    (x)(.05)(1) I= .05x

    The rest of the money:
    (2x)(.06)(.06)(2x)

    Solve:
    0.05x + 0.06(2x) = 4,250
    0.05x + 0.12x = 4,250
    0.17x = 4,250
    x = 25,000

    Amount invested at 5%: x= P25,000
    Amount invested at 6%: 2(25,000) = P50,000


    3.) Lorenzo invested some of her savings in real state which ended up yielding 15% at the end of 1 year. At the same time she invested P100,000 more than this amount into a friend's restaurant business, which ended up yielding a 32% profit at the end of 1 year. If her total profit was P126,000, how much did she invest in each business venture?

    Solution:
    Let x = amount invested at 15%
    and x + 100,000 = amount invested at 32%

    I = PRT
    15% investment:
    (x)(.015)(1) I = 0.15x

    32% investment:
    (x + 100,000)(0.32)(1) I = 0.32(x + 100,000)

    Solve:
    100[0.15x + 0.32(x + 100,000) = 1256,000]100
    15x + 32(x + 100,000) = 12,600,000
    15x + 32x + 3,200,000 = 12,600,000
    47x + 3,200,000 = 12,600,000
    47x = 9,400,000
    x = 200,000

    Annual investment at 15%: x = P200,000
    Amount invested at 32%: x + 100,0000 = P300,000
    Annual profit at 15%: (200,000)(0.15) = P300,000
    Annual profit at 32%: (300,000)(0.32) = P96,000
    Total annual profit: 30,000 + 96,000 = P126,000


    4.) You put $1000 into an investment yielding 6% annual interest; you left the money in for two years. How much interest do you get at the end of those two years?

    In this case, P = $1000, r = 0.06 (because I have to convert the percent to decimal form), and the time is t = 2. Substituting, I get:

    I = (1000)(0.06)(2) = 120

    I will get $120 in interest.


    5.) You invested $500 and received $650 after three years. What had been the interest rate?
    For this exercise, I first need to find the amount of the interest. Since interest is added to the principal, and since P = $500, then I = $650 – 500 = $150. The time is t = 3. Substituting all of these values into the simple-interest formula, I get:

    150 = (500)(r)(3)
    150 = 1500r
    150/1500 = r = 0.10

    Of course, I need to remember to convert this decimal to a percentage.

    I was getting 10% interest.

    ReplyDelete
  10. Bailon, Joshua Carlos A.
    BAIST-2B

    Age Problems

    1. Gabriel is 12 years older than her brother Michael. Four years from now, Michael will be two thirds as old as Gabriel. How old are Gabriel and Michael now?

    Let x + 12 = Gabriel’s current age
    x = Michael’s current age
    In 4 years,
    x + 12 + 4 = Gabriel’s age
    x + 4 = (x + 12 + 4)2/3 = Michael’s age

    Solution:
    x+4=(x+16)2/3
    3x+12=2x+32
    3x-2x=32-12
    x=20

    ANSWER:

    The present age of Gabriel is 32 and 20 for Michael. In four years, it will be 36 for Gabriel and 24 for Michael.

    2. In May of the year 2005, I was one more than eleven times as old as my cousin Brian Shaqtinmore. In May of 2014, I was seven more than three times as old as him. How old was my cousin in May of 2005?

    Solution:
    M = 11B + 1
    B + 9 = 3B + 34
    B + 9 = 3B + 34
    (11B + 1) + 9 = 3B + 34
    11B – 3B = 34 – 9 – 1
    8B = 24
    B = 3

    ANSWER:

    The age of Brian Shaqtinmore in May for the year 2005 is 3.

    3. In three more years, LeBron’s grandfather will be six times as old as LeBron was last year. When LeBron’s present age is added to his grandfather's present age, the total is 68. How old is each one now?

    LeBron + Grandfather = 68
    Grandfather + 3 = 6(L – 1)

    Solution:
    G + 3 = 6L – 6
    G + 3 = 6(68 – G) – 6
    G + 3 = 408 – 6G – 6
    G + 3 = 402 – 6G
    G + 6G = 402 – 3
    7G = 399

    G = 57 L= 11

    ANSWER:

    The current age of LeBron is 11 and his grandfather is 57.

    4. Five years ago, John Carlo’s age was half of the age he will be in 8 years. How old is he now?

    Solution:
    JC – 5 = ½ (JC + 8)
    JC – 5 = ½ JC + 4
    JC – ½ JC = 4+ 5
    ½ JC = 9
    JC = 18

    ANSWER:

    John Carlo’s age now is 18.

    5. One-half of Herbie’s age two years from now plus one-third of her age three years ago is twenty years. How old is she now?

    Solution:
    ( 1/2 )(H + 2) = H/2 + 1
    ( 1/3 )(H – 3) = H/3 – 1
    H/2 + 1 + H/3 – 1 = 20
    H/2 + H/3 = 20
    3H + 2H = 120
    5H = 120
    H = 24

    ANSWER:

    Herbie’s current age is 24 years old.


    ReplyDelete
  11. Rubrico, Jan Paula J.
    BAIST-2B

    Clock Problems

    1. What time after 6 o'clock will the hands of the clock be together for the first time?

    Let x= minus hand position when it occur
    Minute hand; 60 min = 360 degrees
    Hour hand; 1hr = 360/12= 30 degrees

    Min: x/60*360= 6x
    Hr: x/60*30= .5x

    Since it's after 3, we have to add 6*30= 180 degrees to the hr angle

    Hour hand position= .5x+180
    Minute hand position= 6x

    6x=.5x+180
    6x-.5x=180
    x=180/5.5
    x= 32.73 min or 6hr 32 min

    2.At exactly 11 o'clock. the hands are 5 minutes apart. That is 5/60 = ½
    of a full circle, and that is 30 degrees
    The minute hand rotates at a rate of 360/60 degrees / minute. Or 6 degrees / minute

    The hour hand rotates at a rate of (5/60)*360/60 degrees/minute 30/60=1/2 degrees / minute

    I want to find when the differences in the positions of the hands is 270 degrees, which is the 2nd time that they are 90 degrees apart (360-270=90 )

    Let T = time in minutes
    For the minute hand:
    position = ( degrees/min x minutes ) + initial difference
    position = 6T+30

    For the hour hand:
    position =(1/2)*T

    Let the difference = 270 degees
    (6T+30)-(1/2)*T= 270
    6T+30-(1/2)*T=270
    5.5T= 240
    T= 240/5.5
    T=43.635
    .636*60= 38.2

    The hands will be 90 degrees apart for
    the 2nd time in 43 min 38.2 sec

    3.What is the first time after 4 o'clock that the hands of the clock make an angle of 65 degrees?
    :
    We know:
    minute hand travels 360/60 = 6 deg/min
    hour hand travels 360/(60*12) = .5 deg/min
    :
    Starting out at 4 o'clock: min hand at 0 degrees, hr hand at 120 degrees
    :
    Let m = number of minute when hands are 65 degrees apart
    :
    hr hand - min hand = 65
    (120 + .5m) - 6m = 65
    .5m - 6m = 65 - 120
    -5.5m = -55
    m = -55/-5.5
    m = 10 minutes
    We can say then, that at 4:10 the hands will be 65 deg apart:

    4. What time after 3 o'clock will the hands of the clock are together for the first time?
    :
    Let x = minute hand position when this occurs
    :
    Minutes hand; 60 min = 360 degrees
    Hour hand; 1 hr = 360/12 = 30 degrees
    :
    Min: x/60 * 360 = 6x
    Hr: x/60 * 30 = .5x
    :
    Since it's after 3, we have to add 3 * 30 = 90 degrees to the hr angle
    :
    hr hand position = .5x + 90
    minute hand position = 6x
    :
    6x = .5x + 90
    6x-.5x = 90
    5.5x = 90
    x = 90/5.5
    x = 16.36 min or 3 hr 16 min 22 sec

    5. At exactly 4 o'clock, the big hand is on the 12
    and the little hand is on the 4. They span 20
    minutes out of 60, or 120 degrees out of 360 degrees

    The minute hand will move at the rate of 360 degrees / hr
    The hour hand will move at the rate of 5/60 degrees / hr

    I want the difference between them to be 90 degrees with the difference starting out at 120 degrees
    Let a = the angle between the hands
    Let t = the time in hours for the angle between them to equal 90 degrees

    a=(120+30t)-(360t)
    90=(120+30t)-(360t)
    90=120-330t
    330t=120-90
    330t= 30

    t= 1/11 hrs
    1*60/11 = 5+5/11 minutes
    5*60/11=27.27

    The hands will be 90 degrees apart at 4:05 and 27 sec

    ReplyDelete
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    ReplyDelete
  13. Tanaka, Michiko Yasko M.
    BAIST 2B

    MONEY PROBLEM

    PROBLEM 1
    Tamar has four more quarters than dimes. If he has a total of $1.70, how many quarters and dimes does he have?
    First, circle what you must find— how many quarters and dimes. Let x stand for the number of dimes, then x + 4 is the number of quarters. Therefore, .10 x is the total value of the dimes, and .25( x + 4) is the total value of the quarters. Setting up the following chart can be helpful.
    number value amount of money
    dimes x .10 .10x
    quarters x + 4 .25 .25(x + 4)

    So, there are two dimes. Since there are four more quarters, there must be six quarters.
    PROBLEM 2
    Sid has $4.85 in coins. If he has six more nickels than dimes and twice as many quarters as dimes, how many coins of each type does he have?
    First, circle what you must find— the number of coins of each type. Let x stand for the number of dimes. Then x + 6 is the number of nickels, and 2 x is the number of quarters. Setting up the following chart can be helpful.
    number value amount of money
    dimes x .10 .10x
    nickels x + 6 .05 .05(x + 6)
    quarters 2x .25 .25(2x)
    Now, use the table and problem to set up an equation.

    So, there are seven dimes. Therefore, there are thirteen nickels and fourteen quarters.
    PROBLEM 3
    Paul has $31.15 from paper route collections. He has 5 more nickels than quarters and 7 fewer dimes than quarters. How many of each coin does Paul have?
    Solution:
    Let x be the number of quarters
    x + 5 be the number of nickels
    x – 7 be the number of dimes
    25x + 5(x + 5) + 10(x – 7) = 3,115
    25x + 5x + 25 + 10x – 70 = 3,115
    40x = 3,160
    x = 79
    Paul has 79 quarters, 84 nickels and 72 dimes.

    PROBLEM 4
    A number is doubled and then increased by seven. The result is ninety-three. What is the original number?
    Write an equation.
    The number is doubled (2x) and then increased by seven (2x+ 7). The result is ninety-three (2x+ 7 = 93)

    Solve and state the answer.
    2x+ 7 = 932x= 86x= 43
    The number is 43.

    Check.
    Is seven more than two times forty-three ninety-three?
    2(43) + 7 = 86 + 7 = 93


    PROBLEM 5

    Six less than five times a number is the same as seven times the number. What is the number?

    Understand the problem
    . We are looking for a number, let’s call itx
    .

    Write an equation.
    Six less than five times a number:5x−6
    Seven times the same number:7x
    These things are the same:
    5x−6 = 7x
    .

    Solve and state the answer.
    5x−6 = 7x
    −2x= 6x=−3
    The number is −3.

    Check.
    5(−3)−6?= 7(−3)⇒−21?=−21

    ReplyDelete
  14. Arreza, Patrick Joseph
    BAIST 2-B
    Integer Problems

    1. Three more than twice a number is eleven more than the same number. Find the number.

    Solution:
    Let x = be the number
    Three more than twice a number = 2x + 3
    Eleven more than the number = x + 11
    Solve for x
    2x + 3 = x + 11
    2x – x = 11 – 3
    x = 8
    Answer:
    The number is 8

    2. The sum of three consecutive numbers is 216. Find the numbers.

    Solution:

    Let x = first number

    Let x + 1 = second number

    Let x + 2 = third number

    Solve for x
    x + x + 1 + x + 2 = 216

    3x = 213

    x = 71

    x + 1 = 72

    x + 2 = 73
    Answer:
    The three consecutive numbers are 71, 72 and 73

    3. The sum of three positive numbers is 33. The second number is 3 greater than the first and the third is 4 times the first. Find the three numbers.

    Solution:
    Let x = first number
    Let x + 3 = second number
    Let 4x = third number
    Solve for x
    x + x + 3 + 4x = 33
    6x + 3 = 33
    6x = 33 - 3
    6x = 30
    x = 30/6
    x = 5
    x + 3
    5 + 3 = 8
    4x = 4*5 = 20
    Answer:
    The three numbers are 5, 8 and 20

    4. The sum of three positive numbers is 24. The second number is 4 greater than the first and the third is 3 times the first. Find the three numbers.

    Solution:

    Let x = first number

    Let x + 4 = second number

    Let 3x = third number
    Solve for x
    x + (x + 4) + 3x

    x + x + 4 + 3x

    5x + 4

    5x + 4 = 24

    5x = 24 - 4

    5x = 20

    x = 20/5

    x = 4

    x + 4 = 4 + 4 = 8

    3x = 3 * 4 = 12
    Answer:
    The three consecutive numbers are 4, 8 and 12

    5. The sum of two consecutive numbers is 121. Find the numbers.

    Solution:

    Let x = first number

    Let x + 1 = second number

    Solve for x
    x + (x + 1) = 121
    x + x + 1 = 121
    2x + 1 = 121
    2x = 120
    x = 60
    Answer:
    The two consecutive numbers are 60 and 61

    ReplyDelete
  15. Vidal, Michael Allen .
    IS – 2B
    Percent
    1. Our meal was P39.50, but we got a 20% discount because our food was late. What did our meal cost after the discount?
    Solution: Solve for part.
    Whole = P39.50
    Percent = 20%
    Part = x
    X = (Whole) (Percent)
    X = (39.5) (.20)
    X = 7.90
    Then subtract the whole from the part
    39.5 – 7.90 = 31.60 Answer: P31.60 is the cost of the meal

    2. Sochia got a 6% commission for selling a house. Her commission was P7,200. Find the selling price of the house.
    Solution: Solve for whole.
    Whole = x
    Percent = 6%
    Part = P7,200
    X = Part / Percent
    X = .06 / 7,200
    X = 120,000 Answer: P120,000 is the selling price of the house

    3. Stevensonsy went on a mall where everything is on sale for 50%. He bought a watch which costs P3000. How much did the watch costs after the discount?
    Solution: Solve for the part.
    Whole = P3000
    Percent = 50%
    Part = x
    X = (Whole) (Percent)
    X = (3000) (.50)
    X = 1500
    Then subtract the whole from the part
    3000-1500 = 1500 Answer: P1500 is the cost of the watch

    4. Klatskycuspo bough a car for only P50000, she then researched the original price of the car where she knew it was P85000. How much percent of discount did she get?
    Solution: Solve for percent.
    Whole: P85000
    Percent: x
    Part: P50000
    X = Part / Whole
    X = 50000 / 85000
    X = .58
    Then change it to percent by moving the decimal two place backwards and then adding the percentage sign
    .58 58% Answer: 58% is the discount Katarina got from her car.

    5. Bardagul bought an ornament that costs P400 he then realize that the ornament was 35% off. How much did the ornament cost after the discount?
    Solution: Solve for the part.
    Whole = P400
    Percent = 35%
    Part = x
    X = (Whole) (Percent)
    X = (400) (.35)
    X = 140
    Then subtract the whole from the part
    400-140 = 260 Answer: P360 is the cost of the ornament

    ReplyDelete
  16. Denice Jane L. Dela Rosa
    BAIS 2-B

    Area and Perimeter of a Rectangle Problem with Solution
    Formula

    P = 2 (L+W) *P = Perimeter
    A= LW *A = Area

    1. A rectangle table has a length of 10ft and a width of 8ft. Find the area and perimeter of the rectangle table.

    Solution

    P = 2(10ft+8ft) A = (10ft)(8ft)
    P = 2(18ft) A = 180ft2
    P = 36ft

    2. Find the perimeter and area of a rectangle whose length is 20ft and whose width is 12ft.

    Solution

    P = 2(20ft+12ft) A = (20ft)(12ft)
    P = 2( 32ft) A = 240ft2
    P = 64ft

    3. Jason has rectangle t-shirt with a length of 15ft and a width of 22ft he wants to know the area and perimeter of his shirt. Solve the problem.

    Solution

    P = 2 (15ft+22ft) A = (15ft)(22ft)
    P = 2(37ft) A = 330ft2
    P = 74ft


    4. A carpenter made a rectangle window with the length of 5ft and a width of 7ft find its perimeter and area using the given equations.

    Solution

    P = 2(5ft+7ft) A = (5ft)(7ft)
    P = 2(12ft) A = 35ft2
    P = 24ft

    5. What is the area and perimeter of a rectangle mirror with a length of 9ft and a width of 17ft.

    Solution

    P = 2( 9ft+17ft) A = (9ft)(17ft)
    P = 2(26ft) A = 153 ft2
    P = 52ft



    ReplyDelete
  17. Takahashi, Sarahlyn V.
    BAIST-2B

    AGE PROBLEM

    Problem 1
    Jilian is 3 years older than JC. The sum of their ages is 53. How old are they now? 

    Let x = JC's age
      x+3 = Georgina's age

    so, age of JC and age of Georgina = 53

    x+(x+3)=53
        2x+3=53
            2x=53-3
            2x=50
              x=25

    Therefore JC is 25 years old
    While Georgina is 28 years old since 25+3= 28


    Problem 2
    Meisa is one year older than twice her sister's age. The difference of their ages is 18. Find their present ages.

    Let x = Meisa's sister's age
    2x+1 = Meisa's age

    so, age of Meisa - age of Meisa's sister = 18

    (2x+1)-x=18
           x+1=18
               x=18-1
               x=17

    Therefore Meisa's sister is 17 years old
    While Meisa is 35 years old since 2(17)+1=35


    Problem 3
    Allen is 3 more than twice the age of Thirdy. In 10 years, the sum of their ages will be 47. How old are they now?

    Let x = Thirdy's age
    2x+3 = Allen's age
    Age in 10 years
    x+10 = Thirdy's age in 10 years
    2x+13 = Allen's age in 10 years

    so, age of Allen in 10 years + age of Thirdy in 10 years = 47

    (x+10)+(2x+13)=47
                  3x+23=47
                        3x=47-23
                        3x=24
                          x=8

    Therefore Thirdy is 8 years old
    While Allen is 19 years old since 2(8)+3=19

    Problem 4
    Mico is three times Cody's. In 3 years, he will be twice Cody's age then. Find their present ages.

    Let x = Cody's age
        3x = Mico's age
    Age in 3 years
      x+3 = Cody's age in 3 years
    3x+3 = Mico's age in 3 years

    so, Mico's age in 3 years = Twice Cody's age in 3 years 

    3x+3=2(x+3)
    3x+3=2x+6
    3x-2x=6-3
          x=3

    Therefore Cody is 3 years old
    While Mico is 9 years old since 3(3)=9


    Problem 5
    A man has a daugther and a son. The son is three years older than the daughter. In one year the man will be six time as old as the daughter is now. In ten years the man will be fourteen years older than the combined ages of his children at that time. What is the man's present age?

    Let x = Daughter's present age
      x+3 = Son's present age

    Since in one year the man will be six time as old as the daughter is now, the man's present age is 6x-1

    In ten years the man's age will be (6x-1)+10, while the daughter's age will be x+10 and the son's age will be x+13

    so, Since in ten years the man will be fourteen yearsolder than the combined ages of his children at that time

    (6x-1)+10-14=(x+10)+(x+13)
                  6x-5=2x+23
                6x-2x=23+5
                     4x=28
                       x=7

    Therefore the daughter's age is 7 years old while the son's age is 10 years old since 7+3=10 and the man's age is 41 years old since 6(7)-1=41

    ReplyDelete
  18. Jaemee Jyn Gevana
    BAIST-2B
    MA101

    GEOMETRY

    Problem 1:

    A rectangle is 4 times as long as it is wide. If the length is increased by 4 inches and the width is decreased by 1 inch, the area will be 60 square inches. What were the dimensions of the original rectangle?

    Solution:

    Step 1: Assign variables:
    Let x = original width of rectangle
    Sketch the figure

    Step 2: Write out the formula for area of rectangle.
    A = lw
    Step 3: Plug in the values from the question and from the sketch.
    60 = (4x + 4)(x –1)
    Use distributive property to remove brackets
    60 = 4x2 – 4x + 4x – 4
    Put in Quadratic Form
    4x2 – 4 – 60 = 0
    4x2 – 64 = 0
    This quadratic can be rewritten as a difference of two squares
    (2x)2 – (8)2 = 0
    Factorize difference of two squares
    (2x)2 – (8)2 = 0
    (2x – 8)(2x + 8) = 0
    We get two values for x.

    Since x is a dimension, it would be positive. So, we take x = 4
    The question requires the dimensions of the original rectangle.
    The width of the original rectangle is 4.
    The length is 4 times the width = 4 × 4 = 16

    Answer: The dimensions of the original rectangle are 4 and 16.

    Problem 2:

    In a quadrilateral two angles are equal. The third angle is equal to the sum of the two equal angles. The fourth angle is 60° less than twice the sum of the other three angles. Find the measures of the angles in the quadrilateral.

    Solution:
    Step 1: Assign variables:
    Let x = size of one of the two equal angles
    Sketch the figure

    Step 2: Write down the sum of angles in quadrilateral.
    The sum of angles in a quadrilateral is 360°
    Step 3: Plug in the values from the question and from the sketch.
    360 = x + x + (x + x) + 2(x + x + x + x) – 60
    Combine like terms
    360 = 4x + 2(4x) – 60
    360 = 4x + 8x – 60
    360 = 12x – 60
    Isolate variable x
    12x = 420
    x = 35
    The question requires the values of all the angles.
    Substituting x for 35, you will get: 35, 35, 70, 220

    Answer: The values of the angles are 35°, 35°, 70° and 220°

    Problem 3:

    Find the dimensions of the rectangle that has a length 3 meters more that its width and a perimeter equal in value to its area?

    Solution:

    Let L be the length and W be the width of the rectangle. L = W + 3

    Perimeter = 2L + 2W = 2(W + 3) + 2W = 4W + 6

    Area = L W = (W + 3) W = W2 + 3 W

    Area and perimeter are equal in value; hence

    W2 + 3 W = 4W + 6

    Solve the above quadratic equation for W and substitute to find L

    W = 3 and L + 6

    Problem 4:

    The semicircle of area 1250 pi centimeters is inscribed inside a rectangle. The diameter of the semicircle coincides with the length of the rectangle. Find the area of the rectangle.

    Solution:

    Let r be the radius of the semicircle. Area of the semicircle is known; hence

    1250Pi = (1/2) Pi r2 (note the 1/2 because of the semicircle)

    Solve for r: r = 50

    Length of rectangle = 2r = 100 (semicircle inscribed)

    Width of rectangle = r = 50 (semicircle inscribed)

    Area = 100 * 50 = 5000

    Problem 5:

    In the figure triangle OAB has an area of 72 and triangle ODC has an area of 288. Find x and y.

    Solution:

    area of OAB = 72 = (1/2) sin (AOB) * OA * OB

    solve the above for sin(AOB) to find sin(AOB) = 1/2

    area of ODC = 288 = (1/2) sin (DOC) * OD * OD

    Note that sin(DOC) = sin(AOB) = 1/2, OD = 18 + y and OC = 16 + x and substitute in the above to obtain the first equation in x and y

    1152 = (18 + y)(16 + x)

    We now use the theorem of the intersecting lines outside a circle to write a second equation in x and y

    16 * (16 + x) = 14 * (14 + y)

    Solve the two equations simultaneously to obtain

    Answer: x = 20 and y = 14

    ReplyDelete
  19. DEANNA MAE ONA MA101
    BAIST-2B MR. ORATA
    MIXTURE
    EXAMPLE # 1:
    Suppose a car can run on ethanol and gas and you have a 15 gallons tank to fill. You can buy fuel that is either 30 percent ethanol or 80 percent ethanol. How much of each type of fuel should you mix so that the mixture is 40 percent ethanol?

    SOLUTION:

    Le x represent number of gallons of gas that contain 30 percent ethanol

    Let 15 - x be number of gallons of gas that contain 80 percent ethanol

    Since the mixture contains 40 percent ethanol, only 40% of the 15 gallons will be ethanol

    40% of 15 = (40/100) times 15 = (40/100) times 15/1 = (40 × 15) / (100 × 1) = 600 / 100 = 6

    In order for x gallons of gas to contain 30% ethanol, we must take 30% of x or 30% times x

    In order for 15 - x gallons of gas to contain 80% ethanol, we must take 80% of 15 - x or 80% times 15 - x

    0.30 × x + 0.80 × (15 - x) = 6

    0.30x + 0.80 × 15 - 0.80 × x = 6

    0.30x + 12 - 0.80x = 6

    0.30x - 0.80x + 12 = 6

    -0.50x + 12 = 6

    -0.50x = -6

    x = 12

    So 12 gallons of gass contain 30 percent ethanol and 15 - 12 = 3 gallons contain 80 percent ethanol

    Therefore, mix 12 gallons of a 30% ethanol with 3 gallons of an 80% ethanol

    Indeed 30% of 12 = 0.30 × 12= 3.6 and 80% of 3 = 0.80 × 3 = 2.4

    3.6 + 2.4 = 6

    EXAMPLE # 2:

    You have 6 liters of water that have 20 percent strawberry juice. How many liters of a 80 percent strawberry juice should be added to the mixture to make 75 percent strawberry juice?
    SOLUTION:

    Le x be number of liters of water that contain 20 percent strawberry juice

    Let y represent number of liters of water that contain 80 percent strawberry juice

    x + y = 6

    0.20x + 0.80y = 0.75 × 6

    Solve by substitution

    x = 6 - y

    0.20 × (6 - y) + 0.80y = 4.5

    0.20 × 6 - 0.20 × y + 0.80y = 4.5

    1.2 + 0.80y - 0.20y = 4.5

    1.2 + 0.60y = 4.5

    1.2 - 1.2 + 0.60y = 4.5 - 1.2

    0.60y = 3.3

    0.60y / 0.60 = 3.3 / 0.60

    y = 5.5

    x = 6 - 5.5 = 0.5

    Therefore, if you want your juice to contain 75% strawberry juice, do the following:

    Mix 5.5 liters of water that has 80% strawberry juice with 0.5 liter of water that has 20% strawberry juice.

    EXAMPLE # 3:

    How many liters of 20% alcohol solution should be added to 40 liters of a 50% alcohol solution to make a 30% solution?
    SOLUTUION:
    Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Let y be the quantity of the final 30% solution. Hence

    x + 40 = y
    We shall now express mathematically that the quantity of alcohol in x liters plus the quantity of alcohol in the 40 liters is equal to the quantity of alcohol in y liters. But remember the alcohol is measured in percentage term.

    20% x + 50% * 40 = 30% y
    Substitute y by x + 40 in the last equation to obtain.

    20% x + 50% * 40 = 30% (x + 40)
    Change percentages into fractions.

    20 x / 100 + 50 * 40 / 100= 30 x / 100 + 30 * 40 / 100
    Multiply all terms by 100 to simplify.

    20 x + 50 * 40 = 30 x + 30 * 40
    Solve for x.

    x = 80 liters
    80 liters of 20% alcohol is be added to 40 liters of a 50% alcohol solution to make a 30% solution.

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  20. DEANNA MAE ONA MA101
    BAIST-2B MR. ORATA
    EXAMPLE # 4:

    John wants to make a 100 ml of 5% alcohol solution mixing a quantity of a 2% alcohol solution with a 7% alcohol solution. What are the quantities of each of the two solutions (2% and 7%) he has to use?
    SOLUTION:

    Let x and y be the quantities of the 2% and 7% alcohol solutions to be used to make 100 ml. Hence

    x + y = 100
    We now write mathematically that the quantity of alcohol in x ml plus the quantity of alcohol in y ml is equal to the quantity of alcohol in 100 ml.

    2% x + 7% y = 5% 100
    The first equation gives y = 100 - x. Substitute in the last equation to obtain

    2% x + 7% (100 - x) = 5% 100
    Multiply by 100 and simplify

    2 x + 700 - 7 x = 5 * 100
    Solve for x

    x = 40 ml
    Substitute x by 40 in the first equation to find y

    y = 100 - x = 60 ml


    EXAMPLE # 5:

    Sterling Silver is 92.5% pure silver. How many grams of Sterling Silver must be mixed to a 90% Silver alloy to obtain a 500g of a 91% Silver alloy?

    SOLUTION:
    Let x and y be the weights, in grams, of sterling silver and of the 90% alloy to make the 500 grams at 91%. Hence

    x + y =500
    The number of grams of pure silver in x plus the number of grams of pure silver in y is equal to the number of grams of pure silver in the 500 grams. The pure silver is given in percentage forms. Hence

    92.5% x + 90% y = 91% 500
    Substitute y by 500 - x in the last equation to write

    92.5% x + 90% (500 - x) = 91% 500
    Simplify and solve

    92.5 x + 45000 - 90 x = 45500

    x = 200 grams.
    200 grams of Sterling Silver is needed to make the 91% alloy.

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  21. De Leon, Danielle Julian C.
    1.) How many P100 and P50 bills is there in P18,000 if the amount of 50 peso bills are 4 times more than the amount of 100 peso bills.

    Solution:
    Let 4x be the number of 50 peso bills and x be the number of 100 peso bills

    4x + x = 18,000
    5x = 18,000
    X = 3,600

    Answer:
    36 pcs of 100 peso bills
    4x= 14,400
    288 pcs of 50 peso bills

    2.) Sue has $1.15 in nickels and dimes, totally 16 coins. How many nickels and how many dimes does Sue have?

    Solution:
    Let us denote as n the number of nickels Sue has.
    Then the number of dimes is equal to 16-n.
    So, Sue has 5 x n cents in nickels and 10 (16-n) cents in dimes.
    Since the total amount Sue has is equal to $1.15, or 115 cents, this leads to the equation

    5n+10 (16-n) =115

    5n + 160 -10n = 115
    -5n = 115 -160
    -5n = -45
    n = 9

    Answer:
    Sue has 9 nickels and 7 dimes.

    3.) Michael has $1.95 totally in his collection, consisting of quarters and nickels. The number of nickels is in three more than the number of quarters. How many nickels and how many quarters does Michael have?
    Solution:
    Denote as q the number of quarters Michael has. Then the number of nickels is equal to q. So, Michael has 25 x q cents in quarters and 5 (q + 3) cents in nickels. Since the total amount Michael has is equal to $1.95, or 195 cents, this leads to the equation.
    25q + 5q + 15 = 195
    30q = 195 – 15
    30q = 180
    q = 6
    Answer:
    Michael has 6 quarters and 9 nickels.
    4.) James needs interest income of $5,000. How much money must he invest for one year at 7%?

    Solution:

    5,000 = p (0.07)1

    p = 71,428.57

    Answer:
    $71,429

    5.) Terri has $13.45 in dimes and quarters. If there are 70 coins in all, how many of each coin does she have?

    Solution:
    Let x represent the number of dimes. Because the number of dimes and quarters is 70, 70 − x represents the number of quarters. Terri has x dimes, so she has $0.10 x in dimes. She has 70 − x quarters, so she has $0.25(70 − x ) in quarters. These two amounts must sum to $13.45.

    0.10x + 0.25(70-x) = 13.45

    0.10x + 0.25(70-x) = 13.45
    0.10x + 17.5 – 0.25x = 13.45
    -0.15x + 17.5 = 13.45
    -17.5 = -17.50
    -.015x = -4.05
    -4.05
    x= -4.05/-0.15
    x = 27
    Answer:
    Terri has 27 dimes and 70 − x = 70 – 27 = 43 quarters.

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  22. Mauhay, Marc Lawrence A.
    2012300612

    1. You have 6 liters of water that have 20 percent strawberry juice. How many liters of a 80 percent strawberry juice should be added to the mixture to make 75 percent strawberry juice?

    Let x be number of liters of water that contain 20 percent strawberry juice

    Let y represent number of liters of water that contain 80 percent strawberry juice

    x + y = 6

    0.20x + 0.80y = 0.75 × 6

    Solve by substitution

    x = 6 - y

    0.20(6 - y) + 0.80y = 4.5

    1.2 + 0.80y - 0.20y = 4.5

    1.2 + 0.60y = 4.5

    0.60y = 4.5 - 1.2

    0.60y = 3.3

    0.60y / 0.60 = 3.3 / 0.60

    y = 5.5

    x = 6 - 5.5 = 0.5

    Therefore, if you want your juice to contain 75% strawberry juice, do the following:

    Mix 5.5 liters of water that has 80% strawberry juice with 0.5 liter of water that has 20% strawberry juice

    Your juice shoud taste better now!
    ____________________________________________________________________________________________________________
    2. You add x ml of a 25% alcohol solution to a 200 ml of a 10% alcohol solution to obtain another solution. Find the amount of alcohol in the final solution in terms of x. Find the ratio, in terms of x, of the alcohol in the final solution to the total amount of the solution. What do you think will happen if x is very large? Find x so that the final solution has a percentage of 15%.

    Let us first find the amount of alcohol in the 10% solution of 200 ml.

    200 * 10% = 20 ml

    The amount of alcohol in the x ml of 25% solution is given by

    25% x = 0.25 x

    The total amount of alcohol in the final solution is given by

    20 + 0.25 x

    The ratio of alcohol in the final solution to the total amount of the solution is given by

    [ ( 20 + 0.25 x ) / (x + 200)]

    If x becomes very large in the above formula for the ratio, then the ratio becomes close to 0.25 or 25% (The above function is a rational function and 0.25 is its horizontal asymptote). This means that if you increase the amount x of the 25% solution, this will dominate and the final solution will be very close to a 25% solution.

    To have a percentage of 15%, we need to have

    [ ( 20 + 0.25 x ) / (x + 200)] = 15% = 0.15

    Solve the above equation for x

    20 + 0.25 x = 0.15 * (x + 200)

    x = 100 ml
    ____________________________________________________________________________________________________________
    3. Marie has 25 ounces of a 20% boric acid solution which she wishes to dilute to a 10% solution. How much water does she have to add to obtain a 10% solution?

    Let X = Number of Liters of 100 percent solution

    1X + 0.6*10 = 0.9(X + 10)

    100X + 60*10 = 90(X + 10)

    100X + 600 = 90X + 900

    10X = 300

    X = 30 Liters
    ____________________________________________________________________________________________________________
    4. 7 L of an acid solution was mixed with 3 L of a 15% acid solution to make a 29% acid solution. Find the percent concentration of the first solution.

    7 Liters: Percent concentration X to be determined = Ingredient 1

    3 Liters of 15 percent ACID = Ingredient 2

    Let X = resulting percent of ACID

    7X + 3(0.15) = (7+3)*0.29

    7X + 0.45 = 2.9

    7X = 2.45X

    X = 0.35

    X = 35 Percent
    ____________________________________________________________________________________________________________
    5. A given alloy contains 20% copper and 5% tin. How many pounds of copper and of tin must be melted with 100 pounds of the given alloy to produce another alloy analyzing 30% copper and 10% tin.

    Let C = Number of Pounds of Copper
    Let T = Number of Pounds of Tin



    0.2*100 + C = 0.3(100 + C + T) // Copper
    0.05*100 + T = 0.1(100 + C + T) // Tin

    20 + C = 30 + 0.3C + 0.3T // Copper
    5 + T = 10 + 0.1C + 0.1T // Tin

    0.7C = 10 + 0.3T // Copper
    0.9T = 5 + 0.1C // Tin

    C = 14.2857 + 0.428571T

    0.9T = 5 + 1.42857 + 0.0428571T

    0.857143T = 6.42857

    T = 7.5

    C = 17.5

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  23. AWATIN, BEATRIZ V.
    BAIST 2B
    MR. ORATA
    2012-300-281
    09278287105

    1.) The tens digit of a certain two-digit number exceeds the units digit by 4 and is 1 less than twice the units digit. Find the two-digit numbers.

    Let x= units digit
    x + 4 = tens digit
    Since the tens digit =2(unit digit) - 1
    x+4 = 2 (x) – 1
    - x=5
    x=5 units digit
    5 + 4 = 9
    95 = the two-digit number

    2.) Find the three consecutive integers such that the sum of their squares is 149.

    Let x = 1st integer
    x + 1 = 2nd integer
    x + 2= 3rd Integer

    x2 + (x + 1)2 + (x + 2)2 =149
    x2 + x2 + 2 x + 1 + x2 + 4X + 4 = 149
    x2 + 2x -48 = 0
    (x + 8) (x – 6) = 0
    x + 8 = 0 x – 6=0
    x= -8 x= 6 x= -8, 6


    Checking:
    If x = 6 = 1st integer
    x+ 1 = 7 = 2nd integer
    x + 2 = 8 = 3rd Integer
    If x + -8 = 1st Integer
    x + 1= -7 = 2nd Integer
    x + 2 = -6 = 3rd Integer

    3.) Twice the sum of a number and 4 is the same as four times the number decreased by 12. Find the number.
    Let : x= unknown number
    2 (x + 4) = 4x – 12
    2x + 8 = 4x -12 (Apply Distributive property)
    2x + 8 – 4x = 4x -12 (Transpose 4x to the other side)
    -2x + 8 = - 12
    -2x + 8 – 8 = -12 – 8
    -2x = -20
    -2
    X= 10

    Check: Check this solution in the problem as it was originally stated. To do so, replace “number "with ten. Twice the sum of “10” and 4 is 28, which is the same as 4 times “10” decreased by 12.
    State: The number is 10



    4.) The Sum of two consecutive integers is 35. Find the integers.

    Let x = smaller integer
    X + 1 = Bigger Integer
    X + (x+ 1) = 35
    2x + 1 = 35
    2x= 34
    2
    x= 17 smaller integer
    17 + 1 = 35 Bigger Integer

    Check:
    X= smaller number (17)
    X + 1 = Bigger Number ( 18)
    17 + 18 = 95
    35= 35

    5.) The sum of two consecutive integers is 11. What are they?

    Let x = smaller integer
    x + 1 = Bigger Integer
    x + (x+ 1) = 11
    2x + 1 = 11
    2x= 11-1

    2x=10
    2
    x= 5
    x= 5 smaller integer
    5 + 1 = 6 Bigger Integer
    6+ 5 = 11
    11=11



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