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
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
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
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.
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
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.
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
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.
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
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%
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%
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)
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.
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
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?
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?
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
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
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
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
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
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
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.
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.
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
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.
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
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
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
Kim Icmat
ReplyDeleteBAIST-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%
Almazan, Katrina Joyce S.
ReplyDeleteBAIST – 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
CAUBANG, Grace
ReplyDeleteBSBA – 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
Giselle V. Ozoa
ReplyDeleteBAIST-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.
Raemond Zaballero
ReplyDeleteBAIST-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%
Balugay, Paola Denisse G. 03-03-2014
ReplyDeleteBAIST – 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.
Saladar, Ella
ReplyDeleteIS-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
Rojo, Maria Lareza A.
ReplyDeleteBAIST- 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
Villanueva, Janine Anika B.
ReplyDeleteBAIST - 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.
Bailon, Joshua Carlos A.
ReplyDeleteBAIST-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.
Rubrico, Jan Paula J.
ReplyDeleteBAIST-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
This comment has been removed by the author.
ReplyDeleteTanaka, Michiko Yasko M.
ReplyDeleteBAIST 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
Arreza, Patrick Joseph
ReplyDeleteBAIST 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
Vidal, Michael Allen .
ReplyDeleteIS – 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
Denice Jane L. Dela Rosa
ReplyDeleteBAIS 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
Takahashi, Sarahlyn V.
ReplyDeleteBAIST-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
Jaemee Jyn Gevana
ReplyDeleteBAIST-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
DEANNA MAE ONA MA101
ReplyDeleteBAIST-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.
DEANNA MAE ONA MA101
ReplyDeleteBAIST-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.
De Leon, Danielle Julian C.
ReplyDelete1.) 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.
Mauhay, Marc Lawrence A.
ReplyDelete2012300612
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
AWATIN, BEATRIZ V.
ReplyDeleteBAIST 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