Several Word Problems on Percentage With Detailed Solutions Are Presented

8/3/2019 Several Word Problems on Percentage With Detailed Solutions Are Presented

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Problem 1:

The original price of a shirt was $20. It was decreased to $15 . What is the percent decrease of the priceof this shirt.

0Solution 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:

Mary has a monthly salary of $1200. She spends $280 per month on food. What percent of her monthlysalary 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 trousers was decreased by 22% to $30. What was the original price of the trousers?


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 = 30x=1


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Percent decrease from 40 to 30 is given by

(40 - 30) / 40 = 0.25 = 25%

In absolute term, the percent decrease is less than the percent increase.

Problem 7:

A family had dinner in a restaurant and paid $30 for food. They also had to pay 9.5% sale tax and 10% forthe tip. How much did they pay for the dinner?


They paid for food, sales tax and tip, hence

total paid = $30 + 9.5% * 30 + 10% * 30 = $35.85

Problem 8:

A shop is offering discounts on shirts costing $20 each. If someone buys 2 shirts, he will be offered adiscount of 15% on the first shirt and another 10% discount on the reduced price for the second shirt.How much would one pay for two shirts at this shop?


The reduced price for the first shirt

20 - 15% * 20 = $17

The reduced price for the second shirt. The 10% discount will be on the already reduced price, hence theprice of the second shirt is given by

17 - 10% * 17 = $15.3 The total cost for the two shirts is

17 + 15.3 = $32.3

Problem 9:

Smith invested $5000 for two years. For the first year, the rate of interest was 7% and the second year itwas 8.5%. How much interest did he earn at the end of the two year period?


Interest at the end of the first year

7% * 5000 = $350

Interest at the end of the second year

8.5% * (5000 + 350) = $454.75

Total interest at the end of the two year period is

$350 + $454.75 = $804.75


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Problem 10:

Janette invested $2000 at 5% compounded annually for 5 years. How much interest did she earn at theend of the 5 year period?


At the of the first year, she has the principal plus the interest on the principal

P1 = 2000 + 5% * 2000 = 2000(1 + 5%)

At the of the second year, she has the principal P1 plus the interest on P1

P2 = P1 + 5% * P1 = P1(1 + 5%)

Substitute P1 by 2000(1 + 5%) found above to find

P2 = 2000 * (1 + 5%)2

Continuing with this process, it can easily be shown that a the end of the 5th year, the principal is givenby

P5 = 2000 * (1 + 5%)5

= 2000 * (1 + 0.05) = $2552.56

The interest earned at the end of 5 years is

$2552.56 - $2000 = $552.56

Problem 11:

Tom borrowed $600 at 10% per year, simple interest, for 3 years. How much did he have to repay(principal + interest) at the end of the 3 year period?


The interest to pay is given by

Interest = 600 * 10% * 3 = $180

Total to repay

600 + 180 = $780

Problem 12:

Out of a world population of approximately 6.6 billion, 1.2 billion people live in the richer countries ofEurope, North America, Japan and Oceania and is growing at the rate of 0.25% per year, while the other5.4 billion people live in the lees developed countries and is growing at the rate of 1.5%. What will be theworld population in 5 years if we assume that these rates of increase will stay constant for the next 5years. (round answer to 3 significant digits)



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Let us first calculate the population PR in 5 years in the richer countries

PR = (1.2 + 0.25% * 1.2) = 1.2(1 + 0.25%) after one year

PR = 1.2(1 + 0.25%) + 0.25% * 1.2(1 + 0.25%)

= 1.2(1 + 0.25%)2after two years

Continue with the above and after 5 years, PR will be

PR = 1.2(1 + 0.25%)5

after 5 years

Similar calculations can be used to find the population PL in less developed countries after 5 years.

PL = 5.4(1 + 1.5%)5

after 5 years

The world population P after 5 years will be

P = PR + PL = 1.2(1 + 0.25%)5

+ 5.4(1 + 1.5%)5

= 7.03 billion.

Problem 13:

Cassandra invested one part of her $10,000 at 7.5% per year and the other part at 8.5% per year. Herincome from the two investment was $820. How much did she invest at each rate?


Let x and y be the amount invested at 7.5% and 8.5% respectively

Income = $820 = 7.5% * x + 8.5% * y

The total amount invested is also known

10,000 = x + y

Solve the system of the equations to find x and y.

x = $3000 and y = $7000

As a practice check that 7.5% of $3000 and 8.5% of $7000 gives $820.

Problem 14:

The monthly salary S of a shop assistant is the sum of a fixed salary of $500 plus 5% of all monthly sales.What should the monthly sales be so that her monthly salary reaches $1500?


Let S be the total monthly salary and x be the monthly sales, hence

S = 500 + 5% * x

Find sales x so that S = 1500, hence

1500 = 500 + 5% * x = 500 + 0.05 x


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Solve for x

x = (1500 - 500) / 0.05 = $20000

Problem 15:

A chemist has a 20% and a 40% acid solutions. What amount of each solution should be used in order tomake 300 ml of a 28% acid solution?


Let x be the solution at 20% and y be the solution at 40%, hence

x + y = 300 ml

We now write an equation that expresses that the total acid in the final 300 ml is equal to the sum of theamounts of acid in x and y

28% * 300 = 20% * x + 40% * y

Solve the above system of equations to find

x = 180 and y = 120

Problem 16:

What percent of the total area of the circular disk is colored red?


Total area of disk

Ad = pi * r2

Angle t in radians of central angle of red sector

t = (360-120)* pi / 180 = (4/3) pi


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Area of red sector

As = (1/2) t * r2

Percentage of total area in red

P = [ (1/2) t * r 2 ] / [ pi * r 2 ]

= 4 / 6 = 66.7% (3 significant digits)

THINK: compare 66.7% to 240 / 360, why are they equal?

Problem 17:

What percent of the total area of the rectangle is colored red?


Total area of rectangle

Ar = L * W

Area of triangle

At = (1/2) base * height = (1/2) [ L * (1/2) W ]

Percentage of area in red

P = (1/2) [ L * (1/2) W ] / [L*W] = 1/4

= 25%

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



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Let x be the quantity of the 20% alcohol solution to be added to the 40 liters of a 50% alcohol. Lety 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 ismeasured 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

Mutliply 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% alcoholsolution with a 7% alcohol solution. What are the quantities of each of the two solutions (2% and 7%) hehas to use?


Let x and y be the quatities of the 2% and 7% aclohol 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 yml 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


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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 a90% Silver alloy to obtain a 500g of a 91% Silver alloy?


Let x and y be the weights, in grams, of sterling silver and of the 90% alloy to make the 500grams 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 solutionto make it a 10% saline solution.


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 ofsalt 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)


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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 thepercentage of alcohol in the new solution?


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 theamount 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%

Problem 6: You add x ml of a 25% alcohol solution to a 200 ml of a 10% alcohol solution to obtainanother 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 isvery large? Find x so that the final solution has a percentage of 15%.Solution to Problem 6:

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 or25% (The above function is a rational function and 0.25 is its horizontal asymptote). This meansthat if you increase the amount x of the 25% solution, this will dominate and the final solution willbe 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


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Solve the above equation for x

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

x = 300 ml

Problem 1: Two cars started from the same point, at 5 am, traveling in opposite directions at 40 and 50mph respectively. At what time will they be 450 miles apart?


After t hours the distances D1 and D2, in miles per hour, traveled by the two cars are given by

D1 = 40 t and D2 = 50 t

After t hours the distance D separating the two cars is given by

D = D1 + D2 = 40 t + 50 t = 90 t

Distance D will be equal to 450 miles when

D = 90 t = 450 miles

To find the time t for D to be 450 miles, solve the above equation for t to obtain

t = 5 hours.

5 am + 5 hours = 10 am

Problem 2: At 9 am a car (A) began a journey from a point, traveling at 40 mph. At 10 am another car (B)started traveling from the same point at 60 mph in the same direction as car (A). At what time will car Bpass car A?


After t hours the distances D1 traveled by car A is given by

D1 = 40 t

Car B starts at 10 am and will therefore have spent one hour less than car A when it passes it.After (t - 1) hours, distance D2 traveled by car B is given by

D2 = 60 (t-1)

When car B passes car A, they are at the same distance from the starting point and therefore D1= D2 which gives

40 t = 60 (t-1)

Solve the above equation for t to f ind

t = 3 hours


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Car B passes car A at

9 + 3 = 12 pm

Problem 3: Two trains, traveling towards each other, left from two stations that are 900 miles apart, at 4pm. If the rate of the first train is 72 mph and the rate of the second train is 78 mph, at wthat time will they

pass each other?


After t hours, the two trains will have traveled distances D1 and D2 (in miles) given by

D1 = 72 t and D2 = 78 t

After t hours total distance D traveled by the two trains is given by

D = D1 + D2 = 72 t + 78 t = 150 t

When distance D is equal to 900 miles, the two trains pass each other.

150 t = 900

Solve the above equation for t

t = 6 hours.

Problem 4: John left home and drove at the rate of 45 mph for 2 hours. He stopped for lunch then drovefor another 3 hours at the rate of 55 mph to reach his destination. How many miles did John drive?


The total distance D traveled by John is given by

D = 45 * 2 + 3 * 55 = 255 miles.

Problem 5: Linda left home and drove for 2 hours. She stopped for lunch then drove for another 3 hoursat a rate that is 10 mph higher than the rate before she had lunch. If the total distance Linda travelled is230 miles, what was the rate before lunch?


If x is the rate at which Linda drove before lunch the rate after lunch is equal x + 10. The totaldistance D traveled by Linda is given by

D = 2 x + 3(x + 10)

and is equal to 230 miles. Hence

2 x + 3 (x + 10) = 230

Solve for x to obtain

x = 40 miles / hour.


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Problem 6: Two cars left, at 8 am, from the same point, one traveling east at 50 mph and the othertravelling south at 60 mph. At what time will they be 300 miles apart?


A diagram is shown below to help you understand the problem.

.

The two cars are traveling in directions that are at right angle. Let x and y be the distancestraveled by the two cars in t hours. Hence

x = 50 t and y = 60 t

Since the two directions are at right angle, Pythagora's theorem can used to find distance Dbetween the two cars as follows:

D = sqrt ( x2

+ y2

)

We now find the time at which D = 300 miles by solving

sqrt ( x2

+ y2

) = 300

Square both sides and substitute x and y by 50 t and 60 t respectively to obtain the equation

(50 t)2

+ (60 t)2

= 3002

Solve the above equations to obtain

t = 3.84 hours (rounded to two decimal places) or 3 hours and 51 minutes (to the nearest minute)

The two cars will 300 miles apart at

8 + 3 h 51' = 11:51 am.


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Problem 7: By Car, John traveled from city A to city B in 3 hours. At a rate that was 20 mph higher thanJohn's, Peter traveled the same distance in 2 hours. Find the distance bewteen the two cities.


Let x be John's rate in traveling between the two cities. The rate of Peter will be x + 10. We use

the rate-time-distance formula to write the distance D traveled by John and Peter (same distanceD)

D = 3 x and D = 2(x + 20)

The first equation can be solved for x to give

x = D / 3

Substitute x by D / 3 into the second equation

D = 2(D/3 +20)

Solve for D to obtain D = 120 miles

Problem 8: Gary started driving at 9:00 am from city A towards city B at a rate of 50 mph. At a rate that is15 mph higher than Gary's, Thomas started driving at the same time as John from city B towards city A. IfGary and Thomas crossed each other at 11 am, what is the distance beween the two cities?


Let D be the distance between the two cities. When Gary and Thomas cross each other, theyhave covered all the distance between the two cities. Hence

D1 = 2 * 50 = 100 miles , distance traveled by Gary

D1 = 2 * (50 + 15) = 130 miles , distance traveled by Gary

Distance D between the two cities is given by

D = 100 miles + 130 miles = 230 miles

Problem 9: Two cars started at the same time, from the same point, driving along the same road. Therate of the first car is 50 mph and the rate of the second car is 60 mph. How long will it take for thedistance bewteen the two cars to be 30 miles?


Let D1 and D2 be the distances traveled by the two cars in t hours

D1 = 50 t and D2 = 60 t

The second has a higher speed and therefore the distance d between the two cars is given by

d = 60 t - 50 t = 10 t


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For d to be 30 miles, we need to have

30 miles = 10 t

Solve the above equation for t to obtain

t = 3 hours.

Problem 10: Two trains started at 10 pm, from the same point. The first train traveled North at the rate of80 mph and the second train traveled South at the rate of 100 mph. At what time were they 450 milesapart?


Let D1 and D2 be the distances traveled by the two trains in t hours.

D1 = 80 t and D2 = 100 t

Since the two trains are traveling in opposite directions, then total distance D between the two

trains is given by

D = D1 + D2 = 180 t

For this distance to be 450 miles, we need to have

180 t = 450

Solve for t to obtain

t = 2 hours 30 minutes.

10 pm + 2:30 = 12:30 am

Problem 11: Two trains started from the same point. At 8:00 am the first train traveled East at the rate of80 mph. At 9:00 am, the second train traveled West at the rate of 100 mph. At what time were they 530miles apart?


When the first train has traveled for t hours the second train will have traveled (t - 1) hours since itstrated 1 hour late. Hence if D1 and D2 are the distances traveled by the two trains, then

D1 = 80 t and D2 = 100 (t - 1)

Since the trains are traveling in opposite directios, the total distance D between the two trains isgiven by

D = D1 + D2 = 180 t - 100

For D to be 530 miles, we need to have

180 t - 100 = 530


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Solve for t

t = 3 hours 30 minutes.

8 am + 3:30 = 11:30 am

Problem 1:

It takes 1.5 hours for Tim to mow the lawn. Linda can mow the same lawn in 2 hours. How long will it takeJohn and Linda, work together, to mow the lawn?


We first calculate the rate of work of Jonh and Linda

John: 1 / 1.5 and Linda 1 / 2

Let t be the time for John and Linda to mow the Lawn. The work done by John alone is given by

t * (1 / 1.5)

The work done by Linda alone is given by

t * (1 / 2)

When the two work together, their work will be added. Hence

t * (1 / 1.5) + t * (1 / 2) = 1

Multiply all terms by 6

6 (t * (1 / 1.5) + t * (1 / 2) ) = 6

and simplify

4 t + 3 t = 6

Solve for t

t = 6 / 7 hours = 51.5 minutes.

Problem 2: It takes 6 hours for pump A, used alone, to fill a tank of water. Pump B used alone takes 8hours to fill the same tank. We want to use three pumps: A, B and another pump C to fill the tank in 2hours. What should be the rate of pump C? How long would it take pump C, used alone, to fill the tank?


The rates of pumps A and B can be calculated as follows:

A: 1 / 6 and B: 1 / 8


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Let R be the rate of pump C. When working together for 2 hours, we have

2 ( 1 / 6 + 1 / 8 + R ) = 1

Solve for R

R = 1 / 4.8 , rate of pump C.

Let t be the time it takes pump C, used alone, to fill the tank. Hence

t * (1 / 4.8) = 1

Solve for t

t = 4.8 hours , the time it takes pump C to fill the tank.

Problem 3: A tank can be filled by pipe A in 5 hours and by pipe B in 8 hours, each pump working on itsown. When the tank is full and a drainage hole is open, the water is drained in 20 hours. If initially the tankwas empty and somenone started the two pumps together but left the drainage hole open, how long does

it take for the tank to be filled?


Let's first find the rates of the pumps and the drainage hole

pump A: 1 / 5 , pump B: 1 / 8 , drainage hole: 1 / 20

Let t be the time for the pumps to fill the tank. The pumps ,add water into the tank however thedrainage hole drains water out of the tank, hence

t ( 1 / 5 + 1 / 8 - 1 / 20) = 1

Solve for t

t = 3.6 hours.

Problem 4: A swimming pool can be filled by pipe A in 3 hours and by pipe B in 6 hours, each pumpworking on its own. At 9 am pump A is started. At what time will the swimming pool be filled if pump B isstarted at 10 am?Solution to Problem 4:

the rates of the two pumps are

pump A: 1 / 3 , pump B: 1 / 6

Working together, If pump A works for t hours then pump B works t - 1 hours since it started 1hour late. Hence

t * (1 / 3) + (t - 1) * (1 / 6) = 1


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Solve for t

t = 7 / 3 hours = 2.3 hours = 2 hours 20 minutes.

The swimming pool will be filled at

9 + 2:20 = 11:20

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Several Word Problems on Percentage With Detailed Solutions Are Presented