Vinay Pacha

In our day-to-day lives, we come across many situations where we need to find averages. This article is all about finding averages of various types such as Weighted Average, Calculating Arithmetic Mean, Average Speed, etc. Check out the Formulas for finding the average in the Problems provided below and have a clear idea. Continue reading the page and find various questions on Word Problems Involving Averages with Solutions. Download the Average Word Problems Worksheet over here and begin your preparation right away and enhance your conceptual knowledge.

Do Refer:

• Mean
• Mean of the Tabulated Data

Word Problems on Averages with Solutions

Example 1.
A student scores 90,80,75,85,70 in five subjects. Find the average marks scored by a student on these five subjects?
Solution:
Marks scored by a student=90,80,75,85,70
The average marks scored by a student=90+80+75+85+70/5
=400/5=80
Hence, the average mark scored by a student in these five subjects is 80.

Example 2.
The average marks of Suresh and Naresh are 70. If Suresh’s score is 80 what is the Naresh score?
Solution:
The average marks of Suresh and Naresh=70
Suresh score=80
Let Naresh score=x
80+x/2=70
80+x=140
x=140-80=60
Hence Naresh’s score is 60.

Example 3.
Surendra is preparing for the examination. The no. of hours he studied the week for the past 5 days is 5,6,9,8,7. How much time did he spend studying on average?
Solution:
The no. of hours he studied the week=5,6,9,8,7
The time he spent studying on an average=5+6+9+8+7/5
=35/5
=7
The time Surendra spent studying on an average is 7 hours.

Example 4.
The average age of brother and sister is 40. What is the brother’s age if the sister’s age is 50?
Solution:
The average age of brother and sister =40
Sisters age=50
b+s/2=40
b+50/2=40
b+50=80
b=80-50
b=30
Hence, the Brother’s age is 30 years.

Example 5.
The average of three numbers is 10. When one number is removed from the list, the average is 8. What is the number that was removed from the list?
Solution:
Let the three numbers be x,y,z.
Given
x+y+z/3=10
x+y+z=30 —->(eq 1)
Also given
x+y/2=8
x+y=16 —->( eq 2)
(eq 1)- (eq 2)
x+y+z-(x+y)=30-16
x+y+z-x-y=14
z=14
Hence, the number removed from the list is 14.

Example 6.
Sarath collected the money of Rs. 5600  for the Drought Relief Fund in 7 days. On average, how much money did he collected each day?
Solution:
Sarath collected the money in 7 days=Rs 5600
On average, money collected by Sarath on each day=5600/7=Rs 800
Hene, money collected by Sarath each day is Rs 800.

Example 7.
Radha decided to do walking and yoga for better health. She spent 60 hours doing yoga and walking in 30 days. On average, how much money did she spend in doing yoga and walking every day?
Solution:
No. of hours Radha spent on doing yoga and walking=60 days
No. of hours Radha spent on doing yoga and walking in 30 days =60/30=2 hours
Therefore. Radha spent 2 hours doing yoga and walking.

Example 8.
The average of 5 numbers is 9 . What is the sixth number so that the average becomes 12?
Solution:
The average of 5 numbers = 9
Sum of 5 numbers/5=9
Sum of 5 numbers=45
Given if the 6^thnumber is included, the average becomes 12.
Sum of 5 numbers+6^th number/6=12
(45+6^th number)/6=12
45+6^th number=72
6^th number=72-45
6^th number=27
Hence, the sixth number is 27.

Example 9.
Anant’s average score in the last 5 tests is 60. What should be his score in the next test, So that his average score will be 65?
Solution:
Anant’s average score in 5 tests=60
The sum of scores in 5 tests/5=60
The sum of scores in 5 tests=60×5
The sum of scores in 5 tests=300
Let “x” be the average of his next test.
The average score of 6 tests is 65
(sum of scores in 5 tests+6 test)/6=65
(300+x)/6=65
300+x=390
x=390-300=90
Hence, Anant’s score in the next test should be 90.

Example 10.
The average of two numbers is 8. one of the numbers is seven times as big as the other number. Find the two numbers?
Solution:
Let x, y be the two numbers.
x+y/2=8
y=7x
x+7x/2=8
8x/2=8
8x=16
x=16/8=2
y=7(2)=14
Therefore,2,14 are two numbers.

To practice, this Metric Measures Worksheet you need to know how to measure any quantity and also how to express in their respective terms. Here, you will find Problems on Converting from Larger to Smaller Units of Length, Distance, Weight, Capacity, and Vice Versa.  You can do the exercises on metric unit conversions online here itself or simply download the Worksheet on Metric Measures as a PDF and practice whenever you wish to. Let’s begin and learn more in detail about the metric conversions worksheet.

Also, Read Some More Article:

• Units of Measurement
• Addition of Metric Measures
• Multiplication of Metric Measures
• Division of Metric Measures

Metric Measurements Worksheet PDF

Example 1: 
Convert 8 km to m.

Solution:

As given in the question, the value is 8 km.
Now, we need to convert km into m.
As we know multiply the given value with 1000 to get the value in m.
So, 8 km = 8 x 1000 = 8000 m.
Therefore, after the value of 8 km is converted into m, then the value is 8000 m.

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Example 2 :
Convert the value 15 km into meter (m), centimeter (cm), hectometer (h).

Solution:

In the given question, the value is 15 km.
Now, we will convert the given value into the meter, centimeter, hectometer.
First, Multiply the given value by 1000 to get the value in meter.
So, 15km = 15 x 1000 = 15000 m.
Next, convert the value into a centimeter (cm). Actually 1 cm = 0.00001 km.
So, therefore the value 15km is converted into cm, to get the value
15 km = 1500000 cm.
Now, we can convert the km value into a hectometer. We know 1 km = 10 Hectometer.
So, now we convert the given value in H.
15 km = 15 x 10 = 150 H (or) 150 Hectometer.
Therefore, after conversion, the value of 15 km is 15000 m, 1500000 cm, 150 H.

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Example 3 :
Convert the 564 meters into millimeters.

Solution:

As given in the question, the conversion value is 564 meters.
Now, we can convert the value of the given meter into millimeters.
Generally, 1 meter = 1000 millimeters.
So, the value is 564 m is converted to mm. To get the value is,
564 meters = 564000.0 mm.
Therefore, after conversion, the value is 564000.0 mm

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Example 4: 
Convert 30 liters into the following units.
(i) centiliters
(ii) milliliters
(iii) decalitres

Solution:

As given in the question, the value is 30 liters.
Now, we will convert the given value into the centiliters, milliliters, decalitres.
The following are,
(i) First, Multiply the given value by 100 to get the value in centiliters. Because 1liters = 100 centiliters.
So, 30 L = 30 x 100 = 3000 centiliters (cL).
(ii) Next, convert the value into milliliters. Actually 1 liter = 1000 milliliters.
So, therefore the value 30 L is converted into mL, to get the value is
30 L = 30000 mL.
(iii) Now, we can convert the liter value into decaliters. We know 1 liter = 0.1 decaliters.
So, now we convert the given value in dL.
30 liters = 30 x 0.1 = 3 decaliters.
Therefore, after conversion, the value of 30 liters is 3000 centiliters, 30000 milliliters, and 3 decaliters.

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Example 5: 
Convert the 5 grams into milligrams.

Solution:

As given in the question, the value is 5 grams.
Now, we can convert the given values into milligrams.
Multiply the given value by 1000 to get the value in milligrams. In general, 1 gram = 1000 milligrams.
So, 5grams = 5 x 1000 = 5000 milligrams.
Therefore, after conversion of 5 grams, the value is 5000 milligrams.

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Example 6 :
Convert the 67 hectoliters to decalitres.

Solution:

As given in the question, the value is 67 hectoliters.
Now, we will convert the given hectoliters value into decaliters.
Multiply the given value by 10 to get the value in decaliters. We all know, 1 hectoliter = 10 decaliters.
So, the given value is 67 hectoliters then 67 x 10 = 670 decaliters.
Therefore, the value of 67 hectoliters after conversion is 670 decaliters.

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Example 7: 
How to convert the kilograms into grams. The value is 5.673

Solution:

In the given question, the value is 5.673.
Now, you have to convert the value of the kilograms into grams.
Multiply the given value by 1000 to get the value in grams. We all know 1 kilogram = 1000 grams.
So, the given value is 5.673 kilogram, then
5000 kilogram + 673 grams = 5.673 kg x 1000 = 5673 grams.
Therefore, the conversion gram value is 5673 grams.

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Example 8:
Fill in the following blanks.
(i) 22 cm = _____________ mm
(ii) 654 cg = _____________ g
(iii) 600 l = _____________ kl
(iv) 54 kg = _____________ dag
(v) 95 hm = _____________ m

Solution:

In the given question,
We can fill the following blanks,
(i) 22 cm to mm
Multiply the value by 10 to get the value in mm. 1 cm = 10 mm
22 x 10 = 220 mm.
(ii) 654 Cg to g
Multiply the value by 0.01 to get the value in g. 1 Cg = 0.01g
654 x 0.01 = 6.54 g
(iii) 600L to kL
Multiply the value by 0.001 to get the value in kL. 1 L = 0.001 kL
600 x 0.001 = 0.6 kL
(iv) 54 kg to dag
Multiply the value by 100 to get the value in dag. 1 kg = 100 dag
54 x 100 = 5400 dag
(v) 95 hm to m
Multiply the value by 100 to get the value in m. 1 hm = 100 m
95 hm = 95 x 100 = 9500 m.

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Example 9 :
Ram got a piece of cloth which is 6 meters long. Find the length of cloth in centimeters?

Solution:

As given in the question,
Ram got a piece of cloth which is 6 meters long.
Now, we need to find the length of cloth in centimeters,
1 meter = 100 centimeters
6 meters = ? cm
6 x 100 = 600 centimeters
So, therefore the length of cloth in centimeters is 600 cm.

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Example 10: 
Sony placed five pieces of wood of length 20 centimeters each from one end to the other. How long are the pieces of wood altogether, in meters?

Solution:

In the given question,
5 pieces of wood of length 20 centimeters each from one end to the other.
Now, 20 centimeters x 5 = 100 centimeters.
We all know, 1 meter = 100 centimeters.
Therefore, the pieces of wood altogether are 1 meter long.

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Example 11:
Grandfather walked 520 m from his house to the dairy farm and back again. What distance did he walk in all?

Solution:

As given in the question,
Distance from house to the dairy = 520 m
Distance from dairy to house = 520 m
Now, we will find the total distance,
The total distance traveled is 640 + 640 = 1280 m. The value 1040 is divided by 1000m. We get the value is,
1 km 280 m.
Therefore, the grandfather walked or traveled a total distance is 1 km 280m.

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Example 12 :
Find the value of 2 1/2 km into m.

Solution:

As given in the question, the value is 2 1/2 km
Now, we convert the given km value into m.
We know, 1 km = 1000 m.
So, the given value gets,
2 1/2 = 2 x 1000 + 500 m = 2500 m (1/2 km = 500 m)
Therefore, the final value in meters is 2500 m.

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Worksheets on Temperature Conversion available here will assist you in learning conversion of temperature from one scale of the unit to another easily. Practice the Temperature Conversion Problems covering questions on covering from Fahrenheit to Celsius and Celsius to Fahrenheit, Kelvin to Celsius, and Vice Versa. Step by Step Solutions provided for all the Temperature Conversion Word Problems Worksheet makes it easy for kids to use them as references if in case they needed any of them.

Do Check: Converting the Temperature from Celsius to Fahrenheit

Temperature Conversion Practice Problems

Example 1.
Convert the temperature given below in the Celsius scale to Fahrenheit scale:
(i)35⁰c
(ii)49⁰c
(iii)68⁰c
(iv)17⁰c
(v)25⁰c
(vi)98⁰c

Solution:

To convert from Celsius to Fahrenheit we need to follow three steps
1. Multiply the Celsius by 9
2. Divide by 5
3. Add 32
(i) 35⁰c
By Multiplying  35 with 9 we get 315.
Divide 315 by 5 we get 63.
Add 32 to 63 we get 95⁰F.
Hence, 35⁰c is 95⁰F.
(ii) 49⁰c
By Multiplying  49 with 9 we get 441.
Divide 441 by 5 we get 88.2.
Add 32 to 88.2 we get 120⁰F.
Hence, 49⁰c is 120⁰F.
(iii)68⁰c
By Multiplying  68 with 9 we get 612.
Divide 612 by 5 we get 122.4.
Add 32 to 122.4 we get 154.4⁰F.
Hence, 68⁰c is 154.4⁰F.
(iv)17⁰c
By Multiplying  17 with 9 we get 153.
Divide 153 by 5 we get 30.6.
Add 32 to 30.6 we get 62.6⁰F.
Hence, 17⁰c is 62.6⁰F.
(v)25⁰c
By Multiplying  25 with 9 we get 225.
Divide 225 by 5 we get 45.
Add 32 to 45 we get 77⁰F.
Hence, 25⁰c is 77⁰F.
(vi)98⁰c
By Multiplying  98 with 9 we get 882.
Divide 882 by 5 we get 176.4.
Add 32 to 176.4 we get 208.4⁰F.
Hence, 25⁰c is 208.4⁰F.

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Example 2.
Convert the temperatures below from the Fahrenheit scale to the Celsius scale.
(i)85⁰F
(ii)48⁰F
(iii)268⁰F
(iv)97⁰F
(v)35⁰F
(vi)198⁰F

Solution:

For converting the temperature from Fahrenheit to Celsius we need to follow the below steps.
1.Subtract 32 from the Fahrenheit
2.Multiply by 5.
3.Divide by 9.
(i)85⁰F
By Subtracting 32 from 85 we get 53.
By Multiplying 53 with 5 we get 265.
By dividing 265 with 9 we get 29.44 C.
Therefore, 85⁰F is converted to 29.44⁰c.
(ii)48⁰F
By Subtracting 32 from 48 we get 16.
By Multiplying 16 with 5 we get 80.
By dividing 80 with 9 we get 8.88 C.
Hence, 48 F is converted to 8.88⁰c.
.(iii)268⁰F
By Subtracting 32 from 268 we get 236.
By Multiplying 236 with 5 we get 1,180.
By dividing 1,180 with 9 we get 131.11 C.
Therefore, 268⁰F is converted to 131.11⁰c.
(iv)97⁰F
By Subtracting 32 from 97 we get 65.
By Multiplying 65 with 5 we get 325.
By dividing 325 with 9 we get 36.11 C.
Hence, 97⁰F is converted to 36.11⁰ c.
(v)35⁰F
By Subtracting 32 from 35 we get 3.
By Multiplying 3 with 5 we get 15.
By dividing 15 with 9 we get 1.66 C.
Therefore, 35⁰F is converted to 1.66⁰ c.
(vi)198⁰F
By Subtracting 32 from 198 we get 166.
By Multiplying 166 with 5 we get 830.
By dividing 830 with 9 we get 92.22⁰ C.
Hence, 198⁰F is converted to 92.22 c.

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Example 3.
On Saturday the minimum temperature in Hyderabad was 78⁰F. Find the temperature on the Celsius scale?

Solution:

The minimum temperature in Hyderabad = 78⁰F
The temperature on the Celsius scale=5/9(78-32)
=5/9(46)
=25.55⁰ c.
Hence, the temperature in Hyderabad is 25.55⁰ c.

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Example 4.
In July month the maximum temperature in Vijayawada was 39⁰c. Find out the temperature in degree celsius of Vijayawada on that day?

Solution:

The maximum temperature in Vijayawada=39⁰c
The temperature in degree celsius of Vijayawada=9/5(39+32)
=9/5(71)
=639/5
=127.8⁰F.

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Example 5.
The temperature in the Afternoon was 32c. It was dropped 3c in the evening. Find the temperature in the evening?

Solution:

The temperature in the Afternoon=32c
The temperature dropped in the evenig=3c
The temperature in the evening=32-3=29c.
Hence, the temperature in the evening is 29c.

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Example 6.
The temperature was 32c on Thursday and 36c on Friday. How much did the temperature rise on Friday?

Solution:

The temperature on Thursday=32c
The temperature on Friday=36c
The temperature rise on Friday=36c-32c
=4c
Hence, the temperature rise on Friday was 4c.

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Example 7.
Calcutta’s highest temperature was recorded as 48c and the lowest temperature is recorded as 32c. Find the difference in temperature?

Solution:

The highest temperature in Calcutta was recorded = 48c
The lowest temperature is recorded = 32c
The difference in temperature=48-32=16c
Therefore, the difference in temperature is 16c.

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Example 8.
A liquid is heated at a rate of 7 degrees Celsius per minute. If the temperature now is at 43 degrees Celsius, how long will the temperature by 92 degrees Celsius?

Solution:

The given rate is 7c.
Initial temperature t1=43c
Final temperature t2=92c
Time taken t=t2-t1/rate
=92-43/7
=49/7=7minutes.
Therefore, it will take 7 minutes for the temperature to be 92 degrees celsius.

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Simple Interest is the interest calculated on a sum borrowed over a certain period of time. Check Simple Interest Formula and Steps on How to Calculate Simple Interest by referring to the Simple Interest Problems. Try answering the questions on simple interest available here and assess your preparation standards. Below Provided Simple Interest Word Problems makes it easy for you to understand the concept and be clear with the terms. Practice the questions on a frequent basis and make the most out of them and score better grades in exams.

Do Read:

• In Simple Interest when the Time is given in Months and Days
• Worksheet on Simple Interest
• What is Simple Interest?

Solving Problems Involving Simple Interest

Example 1.
What sum would yield an interest of Rs 100 in 4 years at 5% per anum?

Solution:

Interest=Rs 100
Time=4 years
Annual rate=3%
principal=interest ×100/Rate×time
=100 × 100/3× 4
=10000/12
=833.33
Hence the principal is Rs 833.33.

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Example 2.
In what time Rs 6000 amount to 8000 if the simple interest is calculated at 15% per anum?

Solution:

Amount=8000
principal=6000
Rate=15%
Simple interest=amount-principal
=8000-6000
=2000
Time=S.I × 100/principal × rate
=2000 × 100/6000 × 15
=200000/90000
=20/9=2.22=2 11/50
Hence, in 2 11/50 years, 6000 amounts to 8000.

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Example 3.
A sum amount to Rs 5000 at 6% interest rate per annum after 4 years. Find the sum?

Solution:

Amount=5000
Annual rate=6%
Time=4 years
Sum=?
Amount=principal(100+r×t)/100
5000=principal(100+6×4)/100
5000=principal(100+24)/100
5000×100/124=principal
500000/124=principal
4032.25
Hence, the principal is 4032.25

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Example 4.
Simple interest on a sum of money at the end of 6 years is 3/4 of the sum itself. Find the annual rate per anum?

Solution:

Time=6
Letbprincipal is x.
SI=3/4 x.
We know that si=Principal ×time× Rate/100
3/4x=x × 6 × r/100
300/24=r
r=12.5
Hence, the Annual rate of interest is 12.5% per anum.

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Example 5.
At what rate of simple interest will 12000 amount to 13000 in 3 years 3 months?

Solution:

Given principal=12000
Amount=13000
Time=2 years 3 months=9/4 years
S.I=amount-principal
=13000-12000
=1000
Rate=(simple interest×100)/p × t
=1000×100/12000×9/4
=100/12 × 9/4
=900/48
=18.75
Hence, at an 18.75% annual rate, the principal 12000 will amount to 13000 in 2 years 3 months.

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Example 6.
At what rate per anum will Rs 3000 amount to 3450 in 3 years?

Solution:

Amount=3450
Principal=3000
Time=3 years
S.I=amount-principal
=3450-3000
=450
R=(S.I × 100)/(principal× time)
=450 × 100/3000 × 3
=45000/9000
=45/9= 5
Hence, At 5% per anum3000 will amount to 3450 in 3 years.

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Example 7.
At what rate of simple interest will the sum of money double itself in 5 years?

Solution:

Let us consider the principal as x.
t=5 years
Amount after 5 years will be doubled=2X
Simple intrest=2x-x=x.
Rate per anum=sI ×100/p × t
=x × 100/x × 5
=100x/5x
=20
Therefore, at 20% the principal is doubled in 5 years.

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Example 8.
Sarath invested 30000 in different bank accounts which pay 7% and 5% as annual interest. After one year he received 1450 as interest. How much did he invest in each account?

Solution:

Let x be the amount invested in the 7% bank account.
The amount invested in the 5% bank account is 30000-x.
Total interest earned in both the accounts =1450
Interest in 7% account+interest in 5% account=1450
x.7/100.1+(30000-x).5/100.1=1450
0.07x+(30000-x).0.05=1450
0.07x+1500-0.05 X=1450
0.02x=50
x=2500
30000-x=30000-2500
=27500
Therefore, Sarath invested 2500 in one account and 27500 in another account.

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Example 9.
If a sum of money produces 4500 as an interest in 4 years 3 months at 15% per year simple interest, Find the principal?

Solution:

I=4500
r=15%
t=4 years 3 months
the value of ‘t’ always be in years.
So convert months into years by dividing months by 12.
Then t=4 years 3 months
=43/12 years
=41/4 years
=17/4 years
I=prt
4500=p.15/100.17/4
p=7058.823
Hence, the principal is 7058.823.

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Example 10.
In how many years 8000 will double itself in 10% of the simple interest?

Solution:

Given principal=8000
r=10%
t=?
We know that SI=prt/100
=8000.10.t÷100
=800
simple interest is 800
So 8000+8000=16000
800 ÷ 16000=20 months=1 year 8 months
Hence,8000 will be doubled in 1 year 8 months at a 10% annual rate.

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Example 11.
Simple interest on a certain sum is 15/30 of the sum. Find the rate percent and time if both are numerically equal.

Solution:

Let the sum be x.
Time=Rate=R
SI=15/30 x
We know that SI=PRT/100
15/30x=(x.R.R)/100
R²=15/30 ×100
=50
R=7.071
It means Time=7 years
Therefore, the annual rate is 7% and the time is 7 years.

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Example 12.
Find the simple interest
p=40000,T=5 years,R=10%

Solution:

We know that Simple interest=(P× R ×T)/100
=40000.×10 ×5/100
=20000

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Use the Word Problems on finding Selling Price and Cost Price to practice. Answer the Questions on Cost Price and Selling Price available and test your understanding. Make the most out of the Cost Price and Selling Price Worksheets and use them as a quick reference to understand the concept better. You will find step-by-step solutions to all the cost price and selling price problems along with respective formulas, etc. Learn the Problem Solving Approach used and apply the related knowledge to the problems you encounter as a part of your exams.

Also, Refer:

• Calculate Cost Price using Sell Price and Loss Percent
• Calculate Cost Price using Sell Price and Profit Percent
• Calculate Selling Price using Cost and Loss Percent
• Calculate Selling Price using Cost and Profit Percent

Cost Price and Selling Price Questions

Example 1.
Rachana bought 5 kg of sugar at Rs 180. She sold the sugar and gained a profit of Rs 30. Find the selling price of the sugar?

Solution:

The cost price of 5 kg of sugar=Rs 180
Profit gained by selling the sugar=Rs 30
The selling price of sugar=Cost price of sugar+Profit gained by selling the sugar
=180+30
=210
Hence, the selling price of sugar is Rs 210.

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Example 2.
A shopkeeper sold 20 toys for Rs3000 and gained a profit of Rs 500. Find the cost price of 20 toys?

Solution:

The selling price of 20 toys=3000
Profit gained by selling the toys=500
The cost price of 20 toys=selling prices of toys-profit gained by selling the toys
=3000-500
=2500
Therefore, the cost price of toys is Rs 2500.

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Example 3.
The scooter is bought at Rs 25000 and the money spent for repairs is 2000. He sold the scooter at a profit of Rs 5000. At what price did he sell the scooter?

Solution:

The scooter is bought at= Rs 25000
Money spent on repairs=Rs 2000
Total money spent on scooter=Rs 25000+Rs 2000
=27000
Hence 27000 will be the cost price of the scooter.
Profit gained by selling the scooter=5000
The selling price of the scooter=cost price of the scooter+profit gained by selling the scooter
=27000+5000
=32000
Therefore, the scooter is sold at Rs 32000.

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Example 4.
Rama bought the material for the dress at Rs 800. She spent Rs 580 on stitching. She gained a profit of Rs 1800. At what price did Rama sell the dress?

Solution:

Rama bought the dress material=Rs 800
Rama spent on stitching=Rs 580
Total money spent on dress=Rs 800+Rs 580
=1,380
Hence, the cost price of the dress is Rs 1,380.
Profit gained=1800
The selling price of dress=cost price+profit
=1380+1800
=3180
Therefore, Rama sold the dress at Rs 3180.

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Example 5.
On selling a computer for Rs 20,000 a trader gains a profit of 15%. Find the cost price of the computer?

Solution:

selling price=20,000
Profit=15%
The cost price of the computer=[100/(100+profit%)] * selling price
=100/(100+15)*20,000
=100/115*20,000
=17,391.30.
Hence, the cost price of the computer is 17,391.30

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Example 6.
A man sold two shares at Rs 2500 each. On one he made a gain of 15% and on the other, he lost 8%. Find the cost price of both the shares and the total cost price of both the shares?

Solution:

The selling price of first share=2500
Profit=15%
The cost price of the first share=[100/(100+profit%)]* selling price
100/(100+15) * selling price
=100/115 *2500
=Rs 2,173.91
Hence, the cost price of the first share is 2,173.91.
Similarly cost price of second share=100/(100- loss%) * selling price
=100/(100-8) * 2500
=100/92 *2500
=2,717.39
Hence, the cost price of the second share is 2,717.39.
Total cost price of both the shares=2,173.91+2,717.39=4,891.3.

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Example 7.
By selling a wooden table for Rs 8750 trader lost 10%. Find the cost price of the table?

Solution:

The selling price of wooden table=Rs 8750
Loss=10%
The cost price of wooden table=[100/(100-loss%)]* selling price
=100/100-10*8750
=100/90*8750
=9722.22
Hence, the cost price of the wooden table is 9722.22.

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Example 8.
Anil buys an led tv for Rs 15,580 and sells it at a loss of 13%. Find the selling price of the TV?

Solution:

The cost price of the Tv=15,580
loss=13%
Selling price=[(100-loss%)/100]* c.p
=[100-13/100] * 15,580
=87/100 *15,580
=13,554
Therefore, Anil sold the led tv for Rs 13,554.

---

Example 9.
The cost of the watch is 1050 and the handbag is 1350. The shopkeeper sold both and gained a profit of 20% on each. Find the selling price of both and how much money he got?

Solution:

The cost of the watch is =1050
The cost of the handbag is =1350
Profit=20%
The selling price of watch=[(100+profit%)/100]* cost price
=[(100+20)/100]* 1050
=(120/100)*1050
=1,260
The selling price of hand bag=[(100+profit%)/100]* cost price
=[(100+20)/100]* 1350
=(120/100)*1350
=1,620
The selling price of both=1260+1620=2880.
Hence, the shopkeeper sold both the watch and handbag and got the money of Rs 2880.

---

Example 10.
Ram bought a painting for Rs 5870 and sold it for the loss of Rs 1200. At what price did Ram sell the painting?

Solution:

The cost price of the painting=Rs 5870
Loss=Rs 1200
The selling price of the painting=cost price-loss
=5870-1200
=Rs 4670
Hence, the selling price of the painting is Rs 4670.

---

Example 11.
Anand bought a house for Rs 3557890 and spent 525500 on interior decoration. He wants to sell the house and make a profit of 20%. At what price he should sell the house?

Solution:

Anand bought a house=3557890
Money spent on interior decoration=525500
Total money spent on the house=3557890+525500
=4083390
Anand wants a profit=20%
Anand should sell the house=[(100+profit%)/100]* cost price
=[(100+20)/100]* 4083390
=[(120)/100]* 4083390
=4900068
Therefore, Anand should sell the house For Rs 4900068.

---

Example 12.
A bookseller sold two books each at Rs 500 on Saturday. He got a 5% profit on one book and a 10% profit on another book. Find the cost price of the books?

Solution:

The cost price of the books=Rs 500
Bookseller got profit on first book=5%
The cost price of first book=[100/(100+profit%)]* selling price
=100/(100+5)*500
=100/105*500
=475
The cost price of second book=[100/(100+profit%)]* selling price
=100/(100+10)*500
=100/110*500
=450
Therefore, the cost price of the books is 475,450.

---

 

Download Free Printable Worksheets on Profit and Loss and practice the different models of questions asked on the topic of profit and loss. Answer the Profit and Loss Percentage Questions available and test your understanding of the concept. Word Problems on Profit and Loss provided here have questions on finding gain, loss percent, profit, profit percent, etc. Ace your preparation using the exclusive profit and loss worksheets and understand the concept in a better way and score better grades in exams.

Do Refer:

• Concept of Profit and Loss
• Word Problems on Profit and Loss
• Practice Test on Profit and Loss

Worksheets Based on Profit and Loss

Example 1.
If the cost price of biscuit packet is Rs 35 and selling price is Rs 50. Find the profit?

Solution:

The cost price of biscuit packet=Rs 35
The selling price of biscuit packet= Rs 50
As selling price is greater than cost price, profit is gained.
Profit=selling price-cost price
=50-35
=15
Therefore, the profit is Rs 15.

---

Example 2.
Anil bought an old grinder for Rs 3780. He spent Rs 258 for repairs. Then he sold it for Rs 4680. Find the profit he gained?

Solution:

Anil bought an old grinder=Rs 3780
Money spent for repairs=Rs 258
Total money spent by anil for grinder=3780+258=4038
Anil sold the old grinder=Rs 4680
Since the selling price is greater than the cost price, there is profit.
Profit=selling price-cost price
=4680-4038
=642
Hence, Anil gained a profit of Rs 642.

---

Example 3.
Find the selling price if the cost price is Rs 3545 and the loss percent is 20?

Solution:

Cost price=Rs 3545
Loss percent=20
Selling price=\(\frac {100-loss }{ 100 } \) × c.p
=\(\frac {100-20 }{ 100 } \) × 3545
=\(\frac {80}{ 100 } \) × 3545
=\(\frac {4 }{ 5} \)× 3545
=709 × 4
=2836
Hence, the selling price is Rs 2836.

---

Example 4.
Jaya sold her diwan for 5000 at a profit of 1000. Find the cost price of diwan and profit percentage?

Solution:

The selling price of diwan=5000
profit=1000
cost price=selling price-profit
=5000-1000
=4000
Therefore, the cost price of diwan is 4000.
profit percentage=profit/cost price × 100
=1200/4000 × 100
=30
Hence, the profit percentage is 30%.

---

Example 5.
If the cost of 6 sarees is equal to the selling price of 3 sarees. Find the profit percent?

Solution:

Let the cost price of sarees be x.
Then CP of 6 sarees = 6x
CP of 3 sarees = 3x
Given: SP of 3 sarees = 6x
Using the profit formulas,
profit= SP- CP
profit = 6x – 3x = 3x
Profit % = (profit /CP) × 100
= (3x/3x)× 100 = 100%
Hence, the profit is 100 %.

---

Example 6.
On selling a toy car for Rs 560, Ram lost 5 percent. Find out how much did he purchase it for?

Solution:

Loss percent=5%
s.p=Rs 560
Let us take the cost price as 100.
If the loss percent is 5% then the loss is Rs 5.
S.p when the cost price is 100=c.p-loss=100-5=95.
c.p when s.p is Rs 95=Rs 100
c.p when s.p is Rs 560 is 100/95 × 560=589.473.
Hence, the cost price is 589.473.

---

Example 7.
Jaya Ram bought bananas for Rs 60 a dozen. He sold them for Rs7 each. What is the profit percent gained by Jaya Ram?

Solution:

Jaya Ram bought bananas a dozen=Rs 60
He sold each banana for rs=7
The selling price of bananas a dozen=Rs 84
Profit gained by Jaya ram=84-60=24
Profit percent gained by Jaya Ram=24/60 × 100
=2/5 × 100
=40%
Hence, Jaya Ram gained 40 % profit.

---

Example 8.
Suresh bought a TV for Rs 5752. He spent money Rs 325 on repairs. Find how much money he should sell to gain a profit of Rs 1000?

Solution:

Suresh bought a tv=Rs 5752
Money spent on tv for repairs=Rs 325
Total money spent on TV=5752+325=6077
The cost price of the TV will become 6077.
To gain a profit of Rs 1000 he should sell= cost price+profit
=6077+1000=7077
Hence, Suresh should sell the tv at Rs 7077 to gain a profit of Rs 1000.

---

Example 9.
Naresh bought toys for Rs 2000. If one-fifth of toys are sold at a profit of 5% and the remaining four-fifth of the toys at a profit of 10%. Find the net profit?

Solution:

The cost price of one-fifth of toys =1/5 × 2000
=400
The selling price of one-fifth of toys at 5% profit is
=(100+5)% × 400
=105% × 400
=420
The cost price of the remaining four-fifth of toys is
=4/5 × 2000
=1600
The selling price of the remaining four-fifth of toys at 10% profit is
=(100+10)% × 1600
=110% × 1600
=1760
The selling price of total toys=420+1760
=2180
Net profit=selling price of the total toys-cost price of total toys
=2180-2000
=180
Hence, the profit is Rs180.

---

Example 10.
A trader bought a dress for rs 2000. If he marks the dress 20% above the cost price and gives a discount of 10%. Find the profit percentage?

Solution:

The cost price of the dress=2000
A trader marks 20% above cost price=(100+20)% ×2000
=120% × 2000
=120 × 20
=2,400
selling price=(100-10)% × m.p
=90%×2,400
=2160
Profit=s.p-c.p
=2160-2000
=160
Hence, the profit is Rs 160.

---

 

Printable Area of Regular Figures Worksheets contain questions for finding areas of different 2D Shapes such as Triangles, Squares, Rectangles, Pentagons, Circles, Quadrilaterals, etc. These Sample Area of Regular Shapes Worksheets are exclusively designed for students to aid their preparation. You can check out the step-by-step explanation provided for finding areas of different regular figures along with their formulas. Try to solve the Questions on the Area of Regular Polygons on your own and then verify with the Answers Provided here to understand your subject knowledge.

Also, Read:

• Regular and Irregular Polygon
• Areas of Irregular Figures
• Find the Area of the Shaded Region

Area of Regular Shapes Worksheets

Example 1.
Find the area of the below figure
1.
2.
3.

Solution:

1. Area of the rectangle=length × breadth
The length of the rectangle=12
The breadth of the rectangle=10
Area of the given rectangle=12 × 10=120 sq cm
2. Area of the square=side × side
Given side of the square=7 m
=7 × 7=49 sq m
3. Area of the rectangle=length × breadth
The length of the given rectangle=14 cm
The breadth of the given rectangle=15 cm
=14 × 15=210 sq cm

---

Example 2.
Find the area of the figures
1.2. 3.

Solution:

1. Length of the rectangle=8 cm
The breadth of the rectangle=9 cm
Area of the rectangle=length × breadth
=8 cm × 9 cm=72 sq cm
2.Length of the rectangle=16 cm
The breadth of the rectangle=11 cm
Area of the rectangle=16 cm × 11 cm=176 sq cm
Hence, the area of the given rectangle=176 sq cm
3. Length of the rectangle=16 cm
The breadth of the rectangle=21 cm
Area of the rectangle=16 cm × 21 cm=336 sq cm
Hence, the area of the given rectangle=336 sq cm

---

Example 3.
Find the area of the following figures
1.2. 3.

Solution:

1. Length of the given rectangle=15 m
The breadth of the given rectangle=5 m
Area of the given rectangle=length × breadth
=5 m × 15 m=75 sq m
Therefore, the Area of the rectangle=75 sq cm
2. Length of the given rectangle=16 cm
The breadth of the given rectangle=5 cm
Area of the given rectangle=length × breadth
=16 m × 5 m=80 sq m
Therefore, the Area of the rectangle=80 sq m
3. Length of the given rectangle=23 m
The breadth of the given rectangle=7 m
Area of the given rectangle=length × breadth
=23 m × 7 m=161 sq m
Therefore, the Area of the rectangle=161 sq m.

---

Example 4.
Find the area of an equilateral triangle whose perimeter is 18 cm?

Solution:

Given perimeter = 18 cm
We know that the perimeter of an equilateral triangle = 3a
Therefore, 3a=18
a=18/3=6 cm
Hence, the length of the side of an equilateral triangle is 6 cm.
Area= \(\sqrt{ 3 }\)a²/4 sq cm
=\(\sqrt{ 3 }\)6²/4 sq cm
=\(\sqrt{ 3 }\)36/4 sqcm
=9\(\sqrt{ 3 }\)sq cm
Therefore, the area of an equilateral triangle is 9\(\sqrt{ 3 }\) sq cm.

---

Example 5.
Find the area of an equilateral triangle whose side is 4 cm?

Solution:

Area of an equilateral triangle=\(\sqrt{ 3 }\)a²/4
=\(\sqrt{ 3 }\) 4²/4 cm
=\(\sqrt{ 3 }\) 16/4 cm
=4\(\sqrt{ 3 }\) sq cm
Hence, the area of an equilateral triangle is 4\(\sqrt{ 3 }\) sq cm.

---

Example 6.
What is the area of the circle if the radius is 14 cm?

Solution:

Given radius=14 cm
Area of the circle= π × r²
=22/7 × 14 × 14
=22 × 28
=616
Hence, the area of the circle is 616 sq cm.

---

Example 7.
If the diameter of the circle is 14 cm. Find the area of the circle?

Solution:

Given diameter=14 cm
So radius=14/2=7 cm
Hence area of the circle=π × r²
=22/7 × 7 ×7
=22 ×7
=154 sq cm

---

Example 8.
Find the area of the pentagon of the side 14 cm and apothem length 8 cm?

Solution:

Given pentagon side=14 cm,apothem length=8 cm
Area of the pentagon=5/2 ×s ×a
=5/2×14×8
=5×7×8
=280 sq cm
Hence, the area of the pentagon is 280 sq cm

---

Example 9.
Find the area of the regular hexagon whose side is 2 cm?

Solution:

Given side of the hexagon=2 cm
Area of the hexagon=3\(\sqrt{ 3}\)/2 s²
=3\(\sqrt{ 3 }\)/2 2²
=3\(\sqrt{ 3 }\) × 2
=6\(\sqrt{ 3 }\) sq cm
Therefore, Area of the regular hexagon is 6\(\sqrt{ 3 }\)  sq cm.

---

Example 10.
Find the area of the pentagon of the side 12 cm and apothem length 4 cm?

Solution:

Given pentagon side=12 cm,apothem length=4 cm
Area of the pentagon=5/2 ×s ×a
=5/2×12×4
=5×6×4
=120 sq cm
Hence, the area of the pentagon is 120 sq cm

---

 

 

To practice math skills there is no effective way better than practicing using Worksheets. Try solving using the free printable worksheets on area of a square and rectangle and improve your analytical and problem-solving skills. Area of a Square and Rectangle Word Problems has questions on finding areas of grid and normal images, missing sides, etc. Assess your strengths and weaknesses using the Area of a Square and Rectangle Questions and Answers provided and improvise on the areas of need accordingly.

Do Refer:

• Word Problems on Area of a Rectangle   
• Word Problems on Area of a Square
• Area of a Square

Area of a Square and Rectangle Questions

Example 1.
Find the area of the square length of whose sides are given below:
1. 17
2. 43
3. 55
4. 8
5. 30

Solution:

1. The area of the square=17²
=289
2.The area of the square=43²
=1849
3.The area of the square=55²
=3025
4.The area of the square=8²
=64
5.The area of the square=30²
=900

---

Example 2.
Fill in the blanks
Length             Breadth                Area
1. 6 cm                   5 cm                  ______
2.  7 cm                  3 cm                 _______
3.  9 cm                   4 cm                 ________
4.  5 cm                    _____                 10 sq cm
5.  15 m                   ______               150 sq m

Solution:

1. Area=length × breadth=6 × 5=30 sq cm
2. Area=7×3=21 sq cm
3. Area=9 × 4=36 sq cm
4. Length=5 cm
Area=10 sq cm
Breadth=10/5=2
Hence, the breadth of the rectangle=2.
5. Length=15 m
Area=150 sq m
breadth=150/15=10 sq m
Hence, the breadth of the rectangle=10.

---

Example 3.
Find the area of the following
1. The side of the square is 12 cm. Find the area?
2. The side of a square is 19 cm. Find the area?
3. Rectangle of length 13 cm and breadth 3 cm. Find the Area?
4. Rectangle of length 17 cm and breadth 4 cm. Find the Area?

Solution:

1. The side of the square =12 cm
Area of the square=12²=144
2. The side of the square=19 cm
Area of the square=19²=361
3. Length of a rectangle=13 cm
The breadth of a rectangle=3 cm
Area of the rectangle=13 × 3=39
4. Length of the rectangle=17 cm
The breadth of the rectangle=4 cm
Area of the rectangle=17×4=68

---

Example 4.
The bed is 6 m long and the width is 5 m. Find the area occupied by the bed?

Solution:

The length of the bed=6 m
The breadth of the bed=5 m
Area of the bed=6×5=30 sq m
Hence, the Area occupied by the bed=30 sq m.

---

Example 5.
Suresh wants a new carpet for the rectangular function hall. Hall is in a rectangular shape of length 19 cm and width 10 cm. Find the area of the carpet that covers the entire hall?

Solution:

Length of the hall=19 cm
width of the hall=10 cm
Area of the carpet that covers the entire hall=19 × 10=190 sq cm.
Hence, the area of the carpet is 190 sq cm.

---

Example 6.
The area of the square-shaped artificial grass is 4900. Find the length of the side of the grass?

Solution:

The area of the square-shaped artificial grass=4900
Length of side of grass=\(\sqrt{ 4900 }\)
=70
Hence, the length of the square-shaped artificial grass is 70.

---

Example 7.
Find the area of the carpet whose perimeter is 60?

Solution:

The perimeter of the square=60
We know that the perimeter of the square=4 × side
Therefore, 4 × side=60
side=60/4=15
Hence, the area of the carpet is 15.

---

Example 8.
The perimeter of the square garden is 164 m. Find the cost of fencing it at the rate of 10 per m²?

Solution:

The perimeter of the square garden=164
Therefore, the side of the square-shaped garden=164/4=41
Therefore, the area of the square-shaped garden=41× 41 m²=1681 m²
For 1 m², cost of fencing=10
For 1681 m², cost of fencing=1681 × 10=Rs16810 m²
Hence, the cost of fencing for 1681 m is Rs16810.

---

Example 9.
Find the breadth of the rectangular plot of land whose area is 740 m and whose length is 15. Find its perimeter?

Solution:

The area of the rectangular plot of land=740 m
The length of the rectangular plot of land=15 m
The breadth of the rectangular plot of land=740/15=49
The perimeter of the rectangular plot is=2(length+breadth)
=2(15+49)
=2(64)=128 sq m
Hence, the perimeter of the rectangular plot is 128 sq m.

---

Example 10.
The length and breadth of the rectangular courtyard are 60 m and 22 m. Find the cost of leveling it at the rate of Rs 10 per m². Also, find the distance covered by the Ram to take 3 rounds of the courtyard?

Solution:

The Length of the courtyard=60 m
The breadth of the courtyard=22m
The perimeter of the courtyard=2(60+22)
=2(82)
=164 m
Distance covered by the Ram to take 3 rounds=3 × 164=492 m
Area of the courtyard=60 × 22=1320 sq m
For 1 m², the cost of levelling= Rs 10
For 1320, the cost of leveling =1320 × 10=Rs 13200.
Hence, the cost of leveling is Rs 13200 and the distance covered by Ram to take 3 rounds is 492 m.

---

 

Are you looking for help on the concept of finding the square area? If so, you have arrived at the right place that provides you complete knowledge on how to find the area of a square and its formula. Area of a Square Word Problems provided here include questions on finding areas when sides are given.

One simple logic that you need to keep in mind while solving the area of square problems is that you need to have sides of the same unit of length. If not, convert them into the same units of length and apply the formula. Start solving the area of square problems over here and cross-check your solutions with ours.

Read More:

• Area of a Square
• Areas of Irregular Figures
• Worksheet on Area of a Polygon

Area of a Square Questions and Answers

Example 1.
The area of the square is 9 sq cm. Find the side of the square?

Solution:
Area of the square=9
side of the square=\(\sqrt{9 }\)
=3
Hence, side of the square=3.

Example 2.
The perimeter of the square is 32 cm. Find the area of the square?

Solution:
The perimeter of the square=32
4 a=32
a=32/4=8
Therefore, the area of the square is 8.

Example 3.
The side of the square is 7 cm. Find the area of the square?

Solution:
The side of the square=7
Area of the square=7²=49
Hence, the area of the square=49.

Example 4.
A square and a rectangle have the same area. The rectangle has a length of 9 cm and a width of 4 cm. Find the side of the square and its area?
Solution:
Length of the rectangle=9 cm
width of the rectangle=4 cm
Area of rectangle=9 × 4=36
Since the area of the rectangle and square are equal, the area of the square=36
Side of the square=\(\sqrt{36 }\)
=6
Hence the side of a square is 6 and the area of the square is 36.

Example 5.
The side of the square is 3 cm. If its side is doubled, how many times is the area of the new square bigger than the area of the older square?
Solution:
The side of the square=3 cm
The area of the old square=3²=9
The area of the new square=6²=36
The area is 4 times bigger.

Example 6.
In a square, if the length of each diagonal is 2 cm. Find its area?

Solution:
Length of each diagonal=2
Area of the square=1/2 d²
=1/2(2)²
=1/2(4)
=4/2=2
Hence, the area of the square=2 cm.

Example 7.
If the length of diagonals of two squares is in the ratio 2:7. Find the ratio of their areas?
Solution:
let the diagonals of two squares be 2x, 7x.
Area of first square=1/2(2x)²
=1/24x²
=4x²/2
Area of the second square=1/2(7x)²
=1/249 x²
=49 x²/2
The ratio of areas=4x²/2: 49 x²/2
Multiply each term of a ratio by 2.
4x²:49 x²
Divide each term by x²
=4:49
Hence, the ratio of the squares=4:49

Example 8.
Find the area of the floor of the square-shaped room which is made up of 150 square tiles of side 20 inches?
Solution:
Area of the square tile=20 ×20=400 sq inches
As the floor is occupied by 150 square tiles, An Area of 150 square tiles=150 × 400
=60,000 sq inches
Therefore, the area of the floor is 60,000 sq inches.

Example 9.
How many squares with the side of 3 cm cover the surface of a rectangle with a length of 27 cm and a width of 10 cm?
Solution:
The area of the square=3²=9 sq cm
The length of the rectangle=27 cm
the width of the rectangle=10 cm
The area of the rectangle=27 × 10=270 sq cm
The no. of squares cover the surface of rectangle=270/9=30
Hence, 30 squares are required to cover the surface of the rectangle.

Example 10.
Find the no. of square tiles with the side of 5 cm required for covering the floor of rectangular shape which is of length 25 cm and a width 10 cm?
Solution:
The area of the square tiles=5²=25 sq cm
The length of the floor=25 cm
the width of the floor=10 cm
The area of the floor=25 × 10=250 sq cm
The no. of squares cover the surface of rectangle=250/25=10
Hence, 10 squares are required to cover the surface of the rectangle.

Word Problems on Area of Rectangle provided develops mathematical interest among kids by relating the concept to real-life situations. You will find a variety of problems in finding the area of rectangular shapes in this article. Try solving the Area of a Rectangle Problems provided on a regular basis and get a good hold of the concept. Learn how to solve the problems by finding the rectangle area using the formula and try to apply the knowledge to solve related problems.

Do check:

• Worksheet on Area
• Find the Area of the Shaded Region
• Area of a Square

Example 1.
Find the area of the rectangle of length 45 cm and width 20 cm?

Solution:
The length of the rectangle=45 cm
The width of the rectangle=20 cm
The area of the rectangle=45×20=900
Hence, the area of the rectangle is 900 cm².

Example 2.
Find the area of the rectangular room of length 11 mm and width 18 mm?

Solution:
The length of the rectangular room=11 mm
The width of the rectangular room=18 mm
The area of the rectangular room=11×18=198.
Therefore, the area of the rectangular room=198 sq mm.

Example 3.
Find the area of the rectangle of length 73 cm and width 28 cm?

Solution:
The length of the rectangle=73 cm
The width of the rectangle=28 cm
The area of the rectangle=73 × 28=2044.
Therefore, the area of the rectangle=2044 sq cm.

Example 4.
The area of the rectangle is 96 cm². The length of the rectangle is 8 cm. Find the breadth of the rectangle?

Solution:
The area of the rectangle=96 cm²
The length of the rectangle=8 cm
The breadth of the rectangle=area of the rectangle/length of the rectangle
=96/8=12
Therefore, the breadth of the rectangle is 12 cm.

Example 5.
The length of the rectangular cardboard is 6 cm, and the rectangular cardboard’s breadth is 9 cm. Find the area of the rectangular cardboard?

Solution:
The length of the rectangular cardboard =6 cm
The breadth of the rectangular cardboard=9 cm
The area of the rectangular cardboard=6 ×9=54 sq cm
Hence, the area of the rectangular cardboard is 54 sq cm.

Example 6.
The length of the rectangle is 5 cm and the breadth of the rectangle is 7 cm. If the length is greater by 2 cm, what should be the width so that the new rectangle has the same area as that of the first one?
Solution:
The area of the first rectangle=5×7=35sq cm
The length of the new rectangle=5+2=7cm
The breadth of the new rectangle=w × 7=35
w=35/7=5
Hence, the width of the new rectangle is 5 cm.

Example 7.
How many squares with the side of 4 cm cover the surface of a rectangle with a length of 16 cm and a width of 9 cm?

Solution:
The side of a square=4
The area of a square=4²=16
The length of the rectangle=16sq cm
The width of a rectangle=9 cm
The area of the rectangle=16 × 9=144 sq cm
The no. of squares is 144/16 =9
Therefore, 9 squares are required to cover the surface of the rectangle.

Example 8.
A square with a side of 8 cm and a rectangle with a width of 2 cm has the same area. What’s the length of the rectangle?
Solution:
The area of the square is 8²=64 sq cm
The width of the rectangle=6
l × 2=64
l=64/2=32
Hence, the length of the rectangle is 32 cm.

Example 9.
The length of the rectangle is 5 cm and the breadth of the rectangle is 6 cm. If the length is smaller by 2 cm, what should be the width so that the new rectangle has the same area as that of the first one?
Solution:
The area of the first rectangle=5×6=30 sq cm
The length of the new rectangle=5-2=3cm
The breadth of the new rectangle=w × 3=30
w=30/3=10
Hence, the width of the new rectangle is 10 cm.

Example 10.
How many paper squares with the side of 4 cm cover the surface of the rectangular cardboard with a length of 6 cm and width of 8 cm?

Solution:
The side of a square=4
The area of a square=4²=16
The length of the rectangle=6sq cm
The width of a rectangle=8 cm
The area of the rectangle=6 × 8=48 sq cm
The no. of squares is 48/16 =3
Therefore, 3 squares are required to cover the surface of the rectangle.

Example 11.
Srinu decided to build the house in a square shape whose side is 8 m in his rectangular field which is of the length 45 m and width 65 m. Find the area of the field?
Solution:
The length of the rectangular field=45 m
The width of the rectangular field=65 m
The area of the rectangular field=45 × 65=2,925 sq m
The field covered by the house is a square with a side of 8 m.
The area=8²=64
The area of the rectangular field=2,925-64
=2,861 sq m.
Hence, the area of the rectangular field=2,861 sq m.

Example 12.
A rectangle has a length of 14 cm and its width is 4 cm smaller than length. Find the area of the rectangle?
Solution:
Length of the rectangle=14 cm
width of the rectangle=14-4=10 cm
Area of the rectangle=14 ×10=140 sq cm
Hence, the area of the rectangle=140 sq cm.