Algebra Word Problems
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Transcript of Algebra Word Problems
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AGE PROBLEMS
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PROBLEM #1 A boy is one-third as old as his brother and 8 years younger than his sister. The sum of their ages is 38 years. How old is each of them?
Let
x = age of boy
3x=age of brother
x+8= age of sister
x+3x+(x+8)=38
5x=30
x=6 years (age of boy)
3x=18 years(age of brother)
x+8=14years(ageof sister)
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PROBLEM #2 Maria is 12 years older than her sister Josie. Six years ago, Maria was four times as old as Josie. Find their ages now.
Let
x=age of Josie now
x+12=age of Maria now
(x-6) and (x+6) are their respective ages 6 years ago.
x+6=4(x-6)
X=10 (age of Josie now)
x+12=22 years(age of Maria now)
Josie
Maria
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PROBLEM #3 Eight years ago, Manny was three times as old as Ronnie. Now he is only twice as old as Ronnie. Find their ages now.
Let
x=Ronnies age 8 years ago
3x=Mannys age 8 years ago
x+8=Ronnies age now
3x+8=Mannys age now
3x+8=2(x+8)
X=8
3x=24
Ronnies age now=x+8=16 years
Mannys age now=3x+8=32 yeas
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WORK PROBLEMS
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PROBLEM #1 In what time wold A, B, and C together do a piece of job if A alone could do it in 6 hours more, B alone in 1 hour more, and C alone in twice the time?
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PROBLEM #2 A can do a job in 8 days, and A and B can do the job together in 3 days. How long would it take B to do the job alone?
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PROBLEM #3 One pipe can fill a tank alone in 6 hours; another pipe can fill it alone in 12 hours. If the tank is empty, and all three pipes are open, how long will it take to fill the tank?
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NUMBER PROBLEMS
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Let
x=the reqired number
x-3=the excess of the number over 3.
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PROBLEM #2 The escess of the sum of the forth and fifth parts over the difference of the half and third parts of a number is 119. Find the number.
Let x= the required number
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PROBLEM #3 The difference between two numbers is 24 and their sum is 60. Find the numbers.
Let x=one number
x+24= the other nmbers
x+(x+24)=60
2x=36
x=18
x=18
x+24=(18+24)=42
The numbers are 18 and 42.
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PROBLEM #4 Find two consecutive odd numbers such that thrice the smaller number exceeds the larger by 12.
Let x=smaller odd number
x+2=larger odd number
3x-(x+2)=12
2x=14
x=7
x=7
x+2=(7+2)=9The odd numbers are 7 and 9.
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PROBLEM #5 Find two consecutive positive even numbers such that the difference of their squares is 76.
Let x=smaller positive even number
x+2=the larger positive even number
The even integers are 18 and 20.
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MIXTURE PROBLEMS
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PROBLEM #1 A goldsmith has two alloys of gold, the first being 70% pure gold, and the second 60% pure gold. How many grams of each must be used to make 100g of an alloy which will be 66% pure gold?
Let x=weight in g of the 70% pure gold
100-x=weight in g of the 60% pure gold
0.70x=actual gold in 70% alloy
0.60(100-x)=actual gold in 60% alloy
0.66(100)=actual gold in 66% alloy
100-x=100-60=40 grams
Therefore, there are 60g of 70% alloy and 40g of 60% alloy. -
PROBLEM #2 A chemist has two alcohol solutions of different strengths, 30% alcohol and 45% alcohol solutions, respectively. How many cubic cm of each must be sed so as to make a mixture of 30 cubic cm which will contain 39% alcohol?
Let x=volume of 30% alcohol solution
30-x=volume of 45% alcohol solution
x=12cubic cm (30% alcohol solution)
30-x=30-12=18cubic cm (45% alcohol solution)
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PROBLEM #3 Determine how much water should be evaporated from 50kg of a 30% salt solution to produce a 60% salt solution. All percentages are by weight.
Let x=weight of water in kg to be evaporated
Since the amount of salt remains in the solution, we havce
Weight of salt in 30% solution=weight of salt in 60% solution
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PROBLEM #4 How many liters of 45% alcohol solution must be added to 60 liters of a 15% alcohol solution to obtain a 25% alcohol solution?
Let x=number of liters of 45% alcohol solution to be added
x+60=volume in liters of the mixture
Alcohol in 45% solutiion+alcohol in 15% solution=alcohol in 25% solution
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PROBLEM #5 Rice worth P15 per kg is to be mixed with rice worth P20 per kg to make up 50kg of a mixture to sell at P18 per kg. Determine the weight of each kind of rice in the mixture.
Let x=the number of kg of rice worth P15/kg
50-x=the number of kg of rice worth P20/kg
Value of P15/kg rice + value of P20/kg rice = value of mixturex=20 kg (weight of P15/kg rice)
50-x=50-20=30kg (weight of P20/kg rice)
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MOTION PROBLEMS
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PROBLEM #1 A runs around a circular track in 60sec, and B in 50sec. Five seconds after A starts, B starts from the same point in the same direction. When will they be together for the first time, assming they run around the track continuously?
Let C= circumference of the track
t=time when A and B will be together for the first time, reckoned from the time A started
t-5=time of B
A and B will be together when the difference between the disctance run is one circumference. Hence
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PROBLEM #2 Two cars A and B, with average speeds of 40 and 50km/hr, respectively, are 220km apart. Car A starts a 8am toward B, while N starts at 9am toward A. At what time will they meet?
Let t=hours A will travel before meeting B
t-1=hours will travel before meeting A
Distance travelled by A= 40t km
Distance travelled by B=50(t-1)km
40t+50(t-1)=220
40t+50t-50=220
90t=270
t=3 hours
8am+3hours=11am
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PROBLEM #3 A man started on his bicycle for Manila, a distance of 30km intenting to arrice at a certain time. After riding 10km he was detained for half an hour, and as a result he was obliged to ride the rest of the way 2km/hr faster. What was his original speed?
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v=8km/hr (original speed)
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