Vinay Pacha – Page 8 – Eureka Math Answers

Worksheet on Problems Involving Percentage | Percentage Word Problems Worksheet with Answers PDF

July 17, 2021

On this page, the Worksheet on Percentage Word Problems will focus on finding, working out percentages. You can find different models of questions asked on percentages such as finding the percentage of a number, changing decimals to and from percentages, finding how much percent of a number is another number, etc.

All you need to do is Solve the Worksheet on Problems Involving Percentage PDF available here on a regular basis to make the most out of it. Ace up your preparation taking the help of this quick guide on Percentage Worksheet with Solutions and resolve your queries.

Do Check: Percentage Into Decimal

Percentage Word Problems with Answers

Example 1.
In a class of 30 students, 5 students are failed. What percentage of students are passed the examination?

Solution:

No. of students in the class=30
No. of students failed in the examination=5
No. of students who passed in the exam=30-5=25
Percentage of students who passed in the examination=25/30
=5/6=0.83=83%
Hence, 83% of students passed the examination.

Example 2.
In a survey of 100 people, 8 persons didn’t know how to ride a bike. What percentage could ride?

Solution:

No. of people participated in the survey=100
No. of people didn’t know riding a bike=8
No. of people could ride a bike=100-8=92
Percentage of people who could ride a bike=92/100=0.92=92%.
Hence, 92% of people could ride a book.

Example 3.
A man spends 70% of his monthly salary. If he saves 15000 every month. Find his monthly salary?

Solution:

Let his income be Rs x.
Expenditure=70% of x.
Savings=30% of x=15000
x=15000 × 100/30
=150000/3=50000.
Hence, his monthly salary is 50,000.

Example 4.
In a class of 50 students, 30% are boys. How many more boys should be admitted to the class so that girls now become 50% of the class?

Solution:

At first, no. of boys=50×30/100
=1500/100
=15
No. of girls=50-15=35
Let the number of boys admitted in the class be X.
Now total no. of students in the class=50+x
50% of (50+x)=35
50+x=35 × 100/50
50+x=3500/50
50+x=70
x=70-50=20
Hence, 20 more boys should be admitted so that girls become 50% of the class.

Example 5.
In a sports survey, students choose their favorite sport out of basketball, cricket, and Tennis. 1/5 choose basketball, 1/3 choose cricket.3 persons choose Tennis. Find how many people are there in the group?

Solution:

Let the people participated in the survey=X
Percentage of people who choose basketball=x/5
percentage of people who choose cricket=x/3
persons choose Tennis=3
x/5+x/3+3=X
lcm=15
3x+5x+45=15 x
8x-5x=45
3x=45
x=45/3=15
The number of people who participated in the survey is 15.

Example 6.
A player won 75% of matches. In his total career, he lost 50 matches. How many matches did he play overall?

Solution:

Let no. of matches he played=x.
No. of matches he won=75%
No. of matches he lost=100-75%=25%
Given that he lost 50 matches.
25% of x=50
x=50 ×100/25
=200
Therefore, He played 200 matches overall.

Example 7.
Ajay’s scored 70 marks in the examination. His friend got a 20% less score. Find how many marks were scored by his friend?

Solution:

Ajay scored marks in the examination=70
His friend got less score=20%
Marks scored by his friend=70 × 20/100
= 14
=70-14=56
Therefore, Marks scored by his friend=56.

Example 8.
For a candidate to clear the examination, he must score 50% marks. If Anil gets 80 marks and fails by 40 marks. Find out the total marks for the examination?

Solution:

Given, Marks obtained by Anil is 80 and failed by 40 marks.
Pass marks=80+40=120
Let the total marks be X.
50/100 × x=120
x=120 ×100/50
x=240
Therefore, the total marks for the examination are 240.

Example 9.
60% of a number is more than 40% of 1200 by 180. Find the number?

Solution:

(60/100)x-(40/100)*1200=180
3/5x-2/5*1200=180
3/5x-480=180
3/5x=180+480
3/5x=660
x=660 ×5/3=1100
Therefore, the number is 1100.

Example 10.
30 is subtracted from 50% of a number, the result is 80. Find the number?

Solution:

Let the number from which 50% get subtracted is x.
50%(x)-30=80
50/100(x)=80+30
1/2x=110
x=110 ×2/1=220
Hence, if 30 is subtracted from 50% of 220 the result obtained is 80.

Example 11.
Divya took a test and got 30 correct and 10 incorrect answers. Find the percentage of correct answers?

Solution:

Given,
No. of correct answers=30
No. of incorrect answers=10
Total answers=30+10=40
Percentage of correct answers=30/40 × 100=75%
Therefore, Divya got 75% correct answers.

Example 12.
There are 150 employees in the office. On Saturday, 100 were present. Find the percentage of employees who were not attending the office?

Solution:

Total no. of employees in the office=150
No. of employees present on saturday=100
No. of employees not attending on saturday=150-100=50
The percentage of employees who were not attending the office=50/150 × 100
=33.3%
Hence, 33.3% of employees were not attending the office.

Worksheet on Division of Metric Measures | Dividing Metric Measurements Worksheet with Answers

July 16, 2021

In this article, Worksheet on the Division of Metric Measures you will find various questions on the division of metric measures. In the division metric measures, first, we will place the digits in columns and then divide as usual. This is similar to the division of ordinary numbers. You can use and download this division of metric units worksheet for free and take a printout of this page for more practicing purposes. Alongside you need to know about converting the metric measures from one unit to another unit.

Do Refer:

Division of Metric Measures Worksheet PDF

Problem 1:
Divide the value by 6kl 18hl 3dal 12l by 3

Solution:

As given in the question, the value is 6kl 18hl 3dal 12l by 3
First, we place the digits in columns and then divide as usual. The division is also known as repeated subtraction.
Now, we can perform the division operation on the given number 6kl 18hl 3dal 12l by 3.
So, after division by 3, we get the value is,
6kl 18hl 3dal 12l / 3 = 2kl 6hl 1dal 4l.
After performing the division operation, the remainder is 0 and the quotient is 2614.
Therefore, the given number’s final value is 2kl 6hl 1dal 4l.

Problem 2:
Find the quotient of the given value. The given value is 8m 1dm 3cm 4mm by 7

Solution:

As given in the question, the value is 8m 1dm 3cm 4mm by 7.
First, we place the digits in columns and then divide as usual. The division is also known as repeated subtraction.
Now, we can perform the division operation on the given number 8m 1dm 3cm 4mm by 7.
So, after division by 7, we get the value is,
8m 1dm 3cm 4mm / 7 = 1kl 1hl 6dal 2mm
After performing the division operation, the remainder is 0 and the quotient is 1162.
Therefore, the quotient of a given number is 1kl 1hl 6dal 2mm.

Problem 3:
Sony made 16 cakes out of 17kg 568g flour. How much flour did she use for every cake?

Solution:

As given in the question, Sony made 16 cakes.
The total flour used to make 16 cakes is 17kg 568g.
Now, we will find the amount of flour she used for each cake.
Using the division operation, we will find the value of the given questions. The division means repeated subtraction.
So, the weight of flour used for each cake is 17kg 568g / 16, then
17kg 568g / 16 = 1098 = 1kg 98g
After division operation, the remainder is 0 and the quotient is 1kg 98g.
Therefore, Sony used 1kg 98g flour to make each cake.

Problem 4:
A rope that is 18m long is cut into 8 equal pieces. What is the length of each piece?

Solution:

In the given question,
The total length of the rope is 18m.
The number of equal parts is 8.
Now, we will find the length of each rope piece value.
Using the division operation, we will get the value. Division means repeated subtraction.
So, the length of each rope piece is 18m / 8, how to perform division perform as shown below,
After performing division operation, the remainder is 0 and the quotient is 2.25.
Therefore, the length of each rope piece is 2.25m.

Problem 5:
Four contractors construct 16.480 km of road. How much road is made by each of them?

Solution:

In the given question,
Four contractors constructed a road is 16.480 km.
Now, we will find the one constructor constructing road value.
To perform the division operation, you will get each contractor’s value. We know, that division is nothing but repeated subtraction.
So, the one constructor that will construct the road is 16.480 / 4, then the value will be
16.480 / 4 = 4.12 km
Therefore, 4.12km of road will be constructed by each constructor.

Problem 6:
Scott bought 30m 24cm of cloth. He told the shopkeeper to cut it into 3 pieces. How much will be the length of each piece of cloth?

Solution:

As given in the question,
Scott bought cloth is 30m 24cm.
He told the shopkeeper to cut it into 3 pieces.
Now, we will find the length of each piece of cloth using the division method.
The division means repeated subtraction. Before performing the division operation, you can place all digits in the column then divide as usual.
So, the given value is 30m 24cm / 3, then
30m 24cm / 3 = 10m 8cm.
The remainder is 0 and the quotient is 10m 8cm.
Therefore, the length of each piece of cloth is 10m 8cm.

Problem 7:
15 bags of wheat were delivered at a grocery shop, the entire weight being 144kg 780g. If all the baggage contains an equal amount of wheat, find the load of every bag?

Solution:

As given in the question,
The total number of wheat bags is 15bags.
The total weight of the wheat bags is 144kg 780g.
Now, we will find the weight of each wheat bag weight using the given data.
To perform the division operation, you will get the value of each bag.
So, the amount of each bag is 144kg 780g / 15, then
144kg 780g / 15 = 9.652 kg.
Therefore, the weight of each wheat bag is 9.652 kg.

Problem 8:
Find the remainder and quotient values of the below-given figure,

Solution:

As given in the question, the value is 66dl 35ml by 5
Now, we will find the value of remainder and quotient of the given values.
Using the division operation, you will get the values. The division means repeated subtraction.
So, the given values are 66dl 35ml / 5, how to perform division operation is as shown below figure,
After performing the division operation, the remainder is 0 and the quotient is 1327.
Therefore, the given number remainder value is 0 and the quotient value is 13dl 27ml.

Problem 9:
Find the value 326.18 decimeters in a meter

Solution:

As given in the question, the value is 326.18 dm.
Now, we will find the given decimeters value in meters value.
Using the division operation, we will get the value in meters. The division means repeated subtraction.
In general, 1 dm = 1/10 m.
So, divide the given value by 1/10, then
326.18 / (1/10) = 32.618m (or) 326.18 / (1/10) = 326.18 x 0.1 = 32.618m.
Therefore, the given value of 326.18 dm in meters is 32.618m.

Problem 10:
Find the value of 4674.3 cL in liters.

Solution:

As given in the question, the value is 4674.3 cL.
Now, we will find the given cL value in liters.
Using the division operation, you will get the value easily. To perform the division operation, first, we can place all the digits in columns and then divide as usual.
We all know, 1cL = 1/100 = 0.01 liters.
So, convert the given value is 4674.3 cL into liters, then
4674.3 cL = 4674.3 / (1/100) = 46.763 liters.
Therefore, the given value of 4674.3 cL in liters is 46.763 l.

Problem 11:
Divide the value 21kl 4hl 2l by 0.2. Write the values of remainder and quotient?

Solution:

As given in the question, the value is 21kl 4hl 2l by 0.2
Now, we will find the values of remainder and quotient of the given value.
Using the division rule, we can easily find the value. In division, first, you can place all digits in columns and then divide as usual.
So, now we can perform the division operation of the given number 1kl 4hl 2l by 0.2, then
1kl 4hl 2l / 0.2 = 10kl 7hl 10l
After performing the division the remainder is 0 and the quotient value is 10710.
Therefore, after the division operation, the remainder value is 0 and the quotient value is 10kl 7hl 10l of the given value.

Worksheet on Multiplication of Metric Measures | Multiplying Measurements with Different Units Worksheets

July 16, 2021

In this Worksheet on Multiplication of Metric Measures, you will find various questions on the multiplication of metric measures. The multiplication of metric measures is similar to the multiplication of ordinary numbers. You need to know about converting the metric measures from one unit to another unit. In metric multiplication, we will arrange them in order according to their units and then multiply them.

You can use and download this Worksheet on Multiplication of Metric Units without paying, and take a printout of the page for practicing purposes. Exercises in Metric Measurements Multiplication Worksheet will improve subject knowledge on the concept.

Also, Read:

Multiplication of Metric Measures Worksheet PDF

Problem 1:
Find the product of the 5kl 6 hl 3dal 1l x 3

Solution:

As given in the question, the value is 5kl 6 hl 3dal 1l x 3
Now, we will find the product value of the given value.
In between the given number, we have a symbol (x). It means multiplication, to perform multiplication operation we get the value.
While multiplying, you can place all the digits in columns and then multiply as usual.
So, the given value is 5kl 6 hl 3dal 1l x 3.
After multiplication with 3, we get the value is,
5kl 6 hl 3dal 1l x 3 = 16kl 8hl 9dal 3l
Therefore, the product value of the given values is 16kl 8hl 9dal 3l.

Problem 2:
Find the value of the 6m 5dm 2cm 1mm x 10.

Solution:

As given in the question, the value is 6m 5dm 2cm 1mm x 10.
Now, we will find the value of the given value.
To perform the multiplication operation we get the final value. While multiplying, you place all the digits in columns and then multiply as usual.
So, the given value is 6m 5dm 2cm 1mm x 10.
After multiplication with 10, we get the value is,
6m 5dm 2cm 1mm x 10 = 65m 2dm 1cm 0mm.
Hence, the final value of the given values is 65m 2dm 1cm 0mm.

Problem 3:
Rosy bought four bags of wheat. The weight of one bag is 30kg 500g. Find the weight of the total four bags of wheat?

Solution:

As given in the question,
The total number of wheat bags is 4.
The weight of each bag is 30kg 500g.
Now, we will find the total four bags of wheat weight.
Using multiplication operation, we get the total four bags weight.
So, each bags’ weight is 30kg 500g x 4.
After multiplication with 4, we get the value is,
30 kg 500g x 4 = 122000 kg
Therefore, the total 4 bags of wheat weight is 122000 kg.

Problem 4:
One bottle holds 2 and a half-liter of oil. How much do 59 bottles hold?

Solution:

In the given question, the bottle holds 2 and a half-liter of oil.
Now, we will find the 59 bottles hold oil liters values.
Using multiplication operation, we get the value. While performing the multiplication operation, first you can place the digits in columns and then multiply as usual.
So, the given values is 2 liter 500 ml x 59 , then
2 litres 500 ml x 59 = 147l 500 ml.
The below figure shows, how to perform the multiplication.
Therefore, the 59 bottles of oil total liter are 147l 500ml.

Problem 5:
One round of a running track is 275m. Amir walks that track 5 times every day. How much does he walk?

Solution:

As given in the question,
Amir one round of a running track is 275m.
Total his walks that track every day is 5times.
Now, we will find the total distances he walks.
Using the multiplication operation, we get the total value.
In multiplication, first, place all the digits in columns and then multiply as usual.
So, Amir’s total walk = Length of one round x Total number of rounds.
Substitute the values above, we get
275 m x 5 = 1375m
Therefore, the total he walked is 1km 375m.

Problem 6:
Eshani purchased 10 kilograms of rice from a local shop. What amounts of grams of rice did she buy?

Solution:

As given in the question, Eshani purchased 10kg of rice in local grams.
Now, we will find the rice in grams value using the multiplication rule.
We all know that 1 km = 1000 grams.
So, the given value is 10 kilograms. Now, we find that kilograms of rice into grams.
After multiplication with 1000, we get the value is,
10 x 1000 grams = 10,000grams.
Therefore, Eshani bought 10,000 grams of rice.

Problem 7:
Multiply 46l 326ml by 21. Write the final product value.

Solution:

As given in the question, the value is 46l 326ml by 21.
Now, we will find the product value of the given value.
In multiplication, first place all the digits in columns then multiply as usual.
The below figure shows, how to place the digits and how to perform the multiplication operation.
Therefore, after multiplication, the product value is 972l 378ml.

Problem 8:
Find the value of the below figure,

Solution:

As given the question, the figure consists of value is 250kl 53dal x 20
Now, we will find the value of the given value.
Using multiplication, we get the final product value.
So, the given value multiplication operation is as shown below figure,
Therefore, the final value of the given value 250kl 53dal is 5010.60kl.

Problem 9:
Find the value of the 679.54 km in meters

Solution:

As given in the question, the value is 679.54
Now, we will find the value of the given value in meters.
To perform the multiplication operation we get the final value. While multiplying, first you can place the digits in columns and then multiply as usual. We all know that 1 km = 1000m
So, the given value is 679.54 km.
After multiplication with 1000, we get the value is,
679.54 x 1000 = 679540 m
Hence, the final value of the given values 679.54 km in meters is 679540m.

Problem 10:
If 2m 25 cm of cloth is needed for each shirt, how much cloth will be needed for 12 such shirts?

Solution:

As given in the question,
The Cloth needed for each shirt is 2m 25cm. (Also, write 2.25m)
Now, we will calculate the amount of cloth is needed for 12 shirts.
Using the multiplication rule, we get the value. place all digits in columns and then multiply.
So, the cloth required for 12 shirts is,
2.25 m x 12 = 27.00 m
Therefore, the cloth required for 12 shirts is 27m.

Worksheet on Subtraction of Metric Measures | Subtracting Metric Units Worksheets with Answers

July 16, 2021

Use this Subtraction of Measurement Worksheet to learn and practice how to subtract metric measures. Answer the questions on the subtraction of metric measures worksheet available and test your understanding. You can use and download this subtracting metric measures worksheet without paying any amount and take a printout of it for more practicing purposes.

Make the most out of the subtraction of metric measures worksheets and use them as a quick reference to understand the concept easily. You will find step-by-step solutions to all subtracting metric measures problems along with units conversion.

Also, Refer:

Subtraction of Metric Measures Worksheet PDF | Worksheet on Subtraction of Metric Measures

Problem 1:
Subtract 661cm 32mm from 972 cm 53 mm. Write the final value?

Solution:

As given in the question, the values are 973cm 53mm, and 661cm 32mm
Now, we can perform the subtraction of given values to get the final value. Subtraction means removing the digits from one number to another number.
So, the given value is 973cm 53mm – 661cm 32mm,
973cm 53mm – 661cm 32mm = 312 cm 21 mm.
Therefore, the final value of the given values is 312 cm 21 mm.

Problem 2:
Find the value of the 65l 36 ml – 21l 14ml =?

Solution:

As given in the question, the 65l 36 ml – 21l 14ml.
Now, we will find the value of the given values.
In between the given values the ‘-‘ (minus sign) is present. So, to perform the subtraction rule we will get the final value.
Subtraction means minus or removing the digits from one number to another number.
So, the given values are 65l 36 ml – 21l 14ml,
65l 36 ml – 21l 14ml = 44 l 15 ml.
Therefore, after subtracting the final value is 44 l 15 ml.

Problem 3:
Subtract the below-given value, and write the difference value.
13 kg 6hg 8dag 9g 4dg 3 cg and 6kg 4hg 2 dag 8g 1dg 3 cg.

Solution:

As given in the question, the values are 13 kg 6hg 8dag 9g 4dg 3cg and 6kg 4hg 2dag 8g 1dg 3cg.
Now, we can perform the subtraction operation to get the value of the given numbers.
First, arrange the given numbers in columns vertically and also remember to write the highest number on the top and the least number below. Start subtracting from right to left like as shown below,
Therefore, after subtracting the difference value is 7kg 2hg 6dag 1g 3dg 0cg.

Problem 4:
Subtract 326 hm 68dam from 547hm 46dam. Write the difference value.

Solution:

As given in the question, the value is 547hm 46dam – 326 hm 68dam.
Now, we will find the difference value of the given values.
In between given values, we have a subtraction symbol. So, we can perform a subtraction operation to get the difference value.
In subtraction, first, arrange the given numbers in the columns vertically and also remember to write the highest number on the top and the least number below. Start subtracting from right to left like as shown below,
Therefore, the difference value of the given value is 220 hm 78dam.

Problem 5:
From 1L of the milk bottle, Arun gave 425ml milk to the dog and 250 ml to the cat. How much milk was left?

Solution:

As given in the question,
Milk in the bottle is 1 L. (We know that 1L = 1000 mL)
Milk is given to the dog = 325 mL
Milk is given to the cat = 250 mL
Now, we will find the left of milk in 1liter. Using subtraction, we get the value.
So, the total milk used is 325 mL + 250 mL = 575 mL.
Now, find the left of milk,
Milk left = Total milk – used milk
Substitute the values within the above expression, we get
Milk left = 1000 mL – 575 mL
Milk left = 425 mL
Therefore, 425 mL milk was left in the bottle.

Problem 6:
Rakesh has a 6m 45 cm long rope. He cuts one piece of 3m 24cm. Find the length of the left of the rope.

Solution:

In the given question,
The total length of the rope is 6m 45 cm long.
Length of rope Rakesh cut is 3m 24 cm.
Now, we will perform the subtraction operation to find the remaining length of the rope.
So, Rope left = Total length of the rope – Length of cut Rope.
Substitute the values in the above expression, we get the value.
Rope left = 6m 45 cm – 3m 24 cm = 3m 21 cm.
Therefore, the length of the remaining rope is 3m 21 cm.

Problem 7:
A can of 15 liters of oil was purchased for a function. After the function, 8 l 300ml of oil was left in the can. In function, How much oil was used?

Solution:

As given in the question,
The total no. of liters of oil can have 15 liters.
The left of the oil in the can is 8l 300ml.
Now, we will find the total no. of liters of oil used in Function.
Using the subtraction rule, we get the used oil value.
So, Oil used = Total oil – Left of oil.
Substitute the value, we get
Oil used = 15 litres – 8lit 300ml = 6 lit 700 ml
Hence, the total liters of oil used in function is 6 lit 700 ml.

Problem 8:
On Sunday, Rupal walked 6km 300 m and Gauri walked 4 km 850m. Who walked more and by how much?

Solution:

In the given question,
Rupal walked 6km 300m.
Gauri walked 4 km 840 m.
Now, first, the who walked more and then find the difference between both walked distances.
Based on the given data, Rupal walked more.
So, Rupal extra walked = Rupal walked – Gauri walked.
Substitute the given values above, we get
Rupal extra walked = 6 km 300m – 4 km 840 m = 1 km 460 m.
Therefore, Rupal walked more and she walks an extra 1 km 460 m more than Gauri.

Problem 9:
Subtract the value 240.317kl from 634.526kl. Write the final value.

Solution:

As given in the question, the values are 240.317kl from 634.526kl.
Now, we will find the final value of the given values.
Subtraction means removing the digit from one number to another number. In subtraction, first, arrange the given numbers in the columns vertically and also remember to write the highest number on the top and the least number below. Start subtracting from right to left like as shown below,
Therefore, after subtraction, the final value is 394.209kl.

Problem 10:
Find the value of the following,
375.42 hm – 304.30 hm

Solution:

As given in the question, the values are 375.42 hm – 304.30 hm.
Now, we will find the value of the given values.
In between given numbers, the ‘-‘(minus sign) is present. So, we will perform the subtraction operation then we get the value.
So, the given values are,
375.42 hm – 304.30 hm = 071.12hm
Hence, the final value of the given values is 71.12 hm.

Maths Worksheet on Average | Free Printable Finding Average Worksheets with Solutions

July 16, 2021

On this page, you will find Math Worksheets based on Average Calculations. We have covered the Various Questions on Average right from the basic ones to the most challenging ones. Our Comprehensive Collection of Finding Average Worksheets include finding the average for a given set of data, finding the missing value using a given average, calculation of new average upon modification of existing average, etc. Overcome all your shortcomings in the concept by practicing the Average Problems regularly from the Printable Average Calculation Worksheets PDF.

Worksheet on Temperature | Measuring Temperature Worksheet with Answers

July 16, 2021

Worksheet on Temperature available here has got plenty of questions regarding conversion of temperature, reading the temperature on the thermometer, matching temperatures to real lives. Solve the Temperature Questions on your own and then verify the answers provided. Get to know the areas of improvement on the concept temperature and remediate the knowledge gap. Download the Temperature Worksheet with Answers PDF and keep them handy to prepare online too.

Do Refer:

Read the temperatures of the thermometer and write which is higher or lower than the other
Example 1.

Solution:

A is 29⁰ C and B is 32⁰C.

We can observe that A reads a lower temperature than B.

Example 2.

Solution:

A is 14⁰C and B is 0⁰C.

We can observe that A reads a higher temperature than B.

Example 3.
Convert 150⁰F to celsius

Solution:

To convert Fahrenheit to Celsius we need to follow three steps.
1. Subtract 32 from the given Fahrenheit temperature i.e. 150-32=118.
2. Multiply this number by five i.e. 118×5=590.
3. Divide with 9 i.e. 590 ÷ 9=65.55
Hence, 150⁰F is converted to 65.55⁰ c.

Example 4.
Convert 230⁰F to celsius

Solution:

To convert Fahrenheit to Celsius we need to follow three steps.
1. Subtract 32 from the 230 Fahrenheit temperature i.e. 230-32=198.
2. Multiply this number by five i.e. 198×5=990.
3. Divide with 9 i.e. 990 ÷ 9=110
Hence,230⁰F is converted to 110⁰ c.

Example 5.
Convert 90⁰ c to Fahrenheit

Solution:

To convert Celsius to Fahrenheit we need to follow three steps.
1. Divide by 5 i.e. 90 ÷ 5=18.
2. Multiply by 9 i.e. 18 × 9=162.
3. Add 32i.e. 162+32=194.
Hence, 90⁰ C is converted to 194⁰ Fahrenheit.

Example 6.
Convert 130⁰c to Fahrenheit

Solution:

To convert Celsius to Fahrenheit we need to follow three steps.
1. Divide 130 by 5 i.e. 130 ÷ 5=26.
2. Multiply 26 by 9 i.e. 26 × 9=234.
3. Add 32 to 234 i.e. 234+32=266.
Hence, 130⁰ C is converted to 266⁰ Fahrenheit.

Fill in the blanks
1. The freezing temperature of the water is _______
2. Normal human body temperature is __________
3. The temperature of the boiling water is_________

Answers:

1. 32⁰F
2. 98.6⁰F
3. 100⁰c

Write down the temperature indicated by each of the following thermometers
Example 1.

Solution:

The temperature indicated by the thermometer is 28⁰ C, 82⁰ F.

Example 2.

Solution:

The temperature indicated by the thermometer is 18⁰ C, 64⁰ F.

Example 3.

Solution:

The temperature indicated by the thermometer is -21⁰c,6⁰F.

Convert the Temperature Fahrenheit to Celsius and Celsius to Fahrenheit
Example 1.
Convert 115⁰ F into celsius

Solution:

To convert Fahrenheit to Celsius we need to follow three steps.
1. Subtract 32 from the 115 Fahrenheit temperature i.e. 115-32=83.
2. Multiply this number by five i.e. 83×5=415.
3. Divide with 9 i.e. 415 ÷ 9=46.11
Hence,115⁰F is converted to 46.11⁰c.

Example 2.
Convert 228⁰F into celsius

Solution:

1. Subtract 32 from the 228 Fahrenheit temperature i.e. 228-32=196.
2. Multiply this number by five i.e. 196×5=980.
3. Divide with 9 i.e. 980 ÷ 9=108.88
Hence,228⁰F is converted to 108.88⁰c.

Example 3.
Convert 60⁰ c into Fahrenheit

Solution:

To convert Celsius to Fahrenheit we need to follow three steps.
1. Divide 60⁰ c by 5 i.e. 60 ÷ 5=12.
2. Multiply by 9 i.e. 12 × 9=108.
3. Add 32i.e. 108+32=140.
Hence, 60 ⁰C is converted to 140⁰ F Fahrenheit.

Example 4.
Convert 125⁰ c into Fahrenheit.

Solution:

1. Divide 125⁰ c by 5 i.e. 125 ÷ 5=25.
2. Multiply by 9 i.e. 25 × 9=225.
3. Add 32i.e. 225+32=257.
Hence, 125⁰C is converted to 257⁰ F Fahrenheit.

Practice Test on Area | Questions on Area and Perimeter of Different Figures

July 16, 2021

Make use of this excellent opportunity and practice the questions in the area. You can improve your knowledge regarding the concept of finding areas for different figures. Assess your strengths and weaknesses by solving from the Area Practice Test as a part of your preparation. To make it easy for you we have compiled the step-by-step solutions for all the questions. Download the Practice Test on Area without any further delay and ace up your preparation.

Also, Refer:

Area Questions and Answers

Example 1.
Find the area of a square if the side of the square is 8 cm?

Solution:
Given,
side of the square=8 cm
Area of the square=side × side
=8 × 8=64 sq cm.
Therefore, the area of the square is 64 sq cm.

Example 2.
Find the area of the rectangle of length 13 cm and breadth 8 cm?

Solution:
Given,
The length of the rectangle=13 cm
The breadth of the rectangle=8 cm
Area of the rectangle=Length × breadth
=13 × 8=104 sq cm
Hence, the area of the rectangle is 104 sq cm.

Example 3.
Find the area of the rectangular room of length whose length is 180 m and breadth is 50 m?

Solution:
Given,
The length of the rectangular room=180 m
The breadth of the rectangular room=50 m
Area of the rectangular room=length × breadth
=180 × 50
=9000 sq m
Therefore, the area of the rectangular room is 9000 sq m.

Example 4.
The area of the square field is 121 sq cm. Find the side of the square field?

Solution:
Given,
The area of the square field is=121 sq cm
Area of the square=s × s=s²
121=s²
s²=121
s=\(\sqrt{121}\)
s=11
Hence, the side of the square is 11 sq cm.

Example 5.
The cost of cementing a square yard at Rs 3 per sq meter is 1200. Find the money needed for fencing it at a rate of Rs 6 per meter?
Solution:
Given,
The cost of cementing a square yard at Rs 3 per sq meter is= 1200
Let the side be X m.
Rate of cementing=Rs 3 per sq m
Total cost=1200
Area for cementing=1200/3=400
x²= 400
x=20
The perimeter of the yard=4a
=4(20)
=80 m
Rate of fencing=Rs 6 per meter
The total cost of fencing the square yard is 80 × 6=Rs 480
Therefore, the total money needed for fencing the square yard is Rs 480.

Example 6.
The perimeter of a square is 120 m. Find the area of the square?

Solution:
The perimeter of a square=120 m
4a=120
a=120/4
=30 sq m
Therefore, the area of the square is 30 sq m.

Example 7.
Find the length of the rectangular garden whose area is 450 sq cm and breadth is 50 cm?

Solution:
Area of the rectangular garden=450 sq cm
The breadth of the rectangular garden=50 cm
The length of the rectangular garden=450/50
=9 cm
Therefore, the length of the rectangular garden is 9 cm.

Example 8.
A square with a side of 6 cm and a rectangle with a width of 3 cm has the same area. what’s the length of the rectangle?
Solution:
area of the square=6²
=36 sq cm
The width of the rectangle=3 cm
l ×3=36 sq cm
l=36/3=12 sq cm
Hence, the length of the rectangle is 12 sq cm.

Example 9.
A square with a side of 5 cm and a rectangle with a width of 5 cm has the same area. What is the length of the rectangle?
Solution:
The area of the square is=5²=25
The width of the rectangle=5 cm
l ×5=25
l=25/5=5
Hence, the length of the rectangle is 5 cm.

Example 10.
Find the cost of painting a square board of side 49 cm at the rate of Rs 5 per sq cm?

Solution:
Given
side of a square=49 cm
Area of the square=side × side=(49)²
=2401 sq cm
The cost of painting the square board at a rate of Rs 5 per sq cm is
2401 × 5=Rs12005
Therefore, the cost of painting the square board is Rs. 12005.

Time Duration Between Two Dates/ Times | How to Calculate Time Duration in Hours & Minutes?

July 16, 2021

Time is a word that doesn’t stop anyone and plays the main role in everyone’s life. We depend on time for every situation and even we plan our routine based on time. Time is a word that organizes and schedules everything perfectly. Time is precious and every kid has to know and perceive the concept of time. By the end of this article, a student can have a grip on time duration and easily can calculate the time in hours and minutes. Let’s discuss more on time and duration of time with examples.

What is Time?

In mathematics, time is known as the continuous sequence of events that occur in order, from the past to the present and the future. Time is to evaluate and balance the duration of events or the intervals between them and to plan the sequence of events. Kids should learn and know more about the time that they can plan the events or can organize the day-to-day events.

In ancient times, we use the sundial and candlelight to measure the time but these days digital clocks and watches are used to represent or know the time. Time is measured in hours, minutes, and seconds.

Time Duration

Time Duration is no more than how long something lasts, from starting to end. Students should have to learn and know the duration of time because it helps manage the two or more events or activities. If a student knows the starting time and ending time of an activity or an event, they can easily calculate the time between two times. We get the duration of time in hours and minutes.

Do Refer:

How to Calculate Time Duration Manually?

We can do the calculation of time duration manually as per the below procedure. Follow them and calculate the time duration in hours & minutes quite simply. They are as such

Convert both the given times to 24 hours format i.e. in the case of pm hours simply add 12 to them.
If the start minutes are greater than the end minutes simply subtract 1 hour from the end minutes and add 60 minutes to the end minutes.
Next, Subtract the end time minutes from the start time minutes.
Then, Proceed to the hours and subtract them.
Place the Hours and Minutes Together in the final result.

Examples on Time Duration

Example 1:
How many hours have passed from Clock A to Clock B?

Solution:
Firstly, we observe both the time on clocks.
In Clock A, the time is 1 o’clock.
In Clock B, the time is 5 o’clock.
Now, count the hours from 1 o’clock to 5 o’clock.
The duration of time between Clock A and Clock B is 4Hours.

Example 2:
What will be the time after 4 hours from the following times?
(i) 8:30 AM (ii) 12:10 PM (iii) 7:00 PM
Solution:
(i) The given time is 8:30 AM.
To know the time after 4 hours from the given time we count the time from 8:30 AM.
The time after 4 Hours from 8:30 AM is 12:30 PM.
(ii) The given time is 12:10 PM.
Now, count the time after 4 hours from the time 12:10 PM.
The time after 4 Hours from 12:10 PM is 4:10 PM.
(iii) The given time is 7:00 PM.
Now, count the time after 4 hours from the given time 7:00 PM.
The time after 4 Hours from 7:00 PM is 11:00 PM.

Example 3:
Fill the missing times in the table.

Starting Time	Ending Time	Passed Time
7:30 a.m	10:45 a.m
12:45 p.m	5:30 p.m
10:45 a.m		2 hours 30 minutes
2:00 p.m		4 hours 10 minutes

Solution:
Now fill the missing places with the respected time.

Starting Time	Ending Time	Passed Time
7:30 a.m	10:45 a.m	3 hours 15 minutes
12:45 p.m	5:30 p.m	4 hours 45 minutes
10:45 a.m	1:15 p.m	2 hours 30 minutes
2:00 p.m	6:10 p.m	4 hours 10 minutes

Example 4:
Ram and Raghu play badminton every evening. Yesterday, their game started at 3:15 p.m. and finished at 6:00 p.m. How long did their game last?
Solution:
Given
Ram and Raghu play badminton every evening.
Their game starts at 3:15 PM and ends at 6:00 PM.
Now, count the duration of time in hours that start from 3:15 PM to 6:00 PM = 2 hours 45 minutes.
The duration of the game they played is 2 Hours 45 Minutes.

Example 5:
Raju took the train from Jaipur at 12:15 AM and it arrives at 2:45 PM in Bangalore. How long was the train journey?
Solution:
Step 1: First, we observe both the train times when they started and when they reached.
Step 2: Now, count the number of hours from 12:15 AM to 2:15 PM. The time it took in hours is 14 Hours.
Step 3: Next, count the remaining minutes from 2:15 to 2:45. It is for 30 minutes.
Step 4: The total train time for the journey is 14 Hours 30 Minutes.

Example 6:
If a party starts at 7:00 PM and ends at 10:00 PM. Find the time duration of a party?
Solution:
Step 1: We observe and calculate both the times
Step 2: Count the hours from 10:00 PM – 7:00 PM = 3 Hours.
Step 3: Thus, the duration of a party is 3 Hours.

Worksheet on Addition of Metric Measures | Addition of Measurement Worksheets with Answers

July 15, 2021

In this Adding Units of Measurement worksheet, you will find various questions on the addition of metric measures. The addition of metric measures is similar to the addition of ordinary numbers. In metric addition, we will arrange them in columns according to their units and we will add them.

Alongside you need to know about converting the metric measures from one unit to another unit. You can use and download this Worksheet on Addition of Metric Measures for free, and take a printout of the page for the sake of practicing purposes. Free Interactive Exercises in the Metric Measurements Addition Worksheet will improve subject knowledge on the concept.

Addition of Metric Measures Worksheet PDF

Problem 1:
Find the value of the 132 g 24 mg + 64 g 50 mg =?

Solution:

As given in the question, the value is 132 g 24 mg + 64 g 50 mg
Now, we can perform the addition operation in between the given values to get the final value. The addition means adding given two numbers and get the resultant sum.
So, the given values are 132 g 24 mg + 64 g 50 mg,
After performing the addition the value is,
132 g 24 mg + 64 g 50 mg = 196 g 74 mg
Therefore, the value of the 132 g 24 mg + 64 g 50 mg = 196g 74mg.

Problem 2:
Find the value of the following:
8 kg 6 hg 3 dag 1 g + 5 kg 4 hg 7 dag 6 g

Solution:

As given in the question, the number is 8 kg 6 hg 3dag 1 g + 5 kg 4 hg 7dag 6 g
Now, we will find the value of the given number. In between the given number ‘+’ (plus sign) is there. So, we perform the addition operation to get the sum value.
The addition means adding of given numbers and the resultant value is a sum.
So, the value is 8 kg 6 hg 3dag 1 g + 5 kg 4 hg 7dag 6 g, first arrange the given numbers in the column vertically and add them from right to left as shown below
Therefore, the final value of the given numbers is 14kg 1hg 0dag 7g.

Problem 3:
Find the value of 6kl 4 hl 5 dal 2 l + 4kl 7 hl 3 dal 6 l + 2kl 9 hl 1 dal 0 l

Solution:

As given in the question, the number is 6kl 4 hl 5 dal 2 l + 4kl 7 hl 3 dal 6 l + 2kl 9 hl 1 dal 0 l
Now, we will find the value of the given number. In between the given number ‘+’ (plus sign) is there. So, we perform the addition operation to get the sum value.
The addition means adding of given numbers and the resultant value is the sum.
So, the value is 6kl 4 hl 5 dal 2 l + 4kl 7 hl 3 dal 6 l + 2kl 9 hl 1 dal 0 l first arrange the given numbers in the column vertically and add them from right to left as shown below,
Now, we can perform the addition perform as shown below,
Therefore, the sum of 6kl 4 hl 5 dal 2 l + 4kl 7 hl 3 dal 6 l + 2kl 9 hl 1 dal 0 l is 14kl 0 hl 9 dal 8l.

Problem 4:
Find the sum of the below-given value,
242 km 200 m + 186 km 136 m + 112 km 326 m

Solution:

In the given question, the value is 242 km 200 m + 186 km 136 m + 112 km 326 m
Now, we can find the sum of the given value.
In the given value, ‘+’ (plus sign) is present. So, we can perform the addition operation of the given value. In addition, we will add given two numbers and the resultant value is the sum.
So, the given value is 242 km 200 m + 186 km 136 m + 112 km 326 m.
After performing the addition operation, the resultant sum value is,
242 km 200 m + 186 km 136 m + 112 km 326 m = 540 km 662 m.
Therefore, the sum of the given value is 540 km 662 m.

Problem 5:
Raji has 8 kg 250 g of rice in a bag and she added more than 6 hg 450 g to it. How much amount of rice is in the bag?

Solution:

As given in the question,
Weight of rice in bag = 8 kg 250 g
Weight of rice she added = 6 hg 450 g
Now, we find the total weight of the rice bag using the addition law.
So, the total weight of the rice in bag = weight of rice in bag + weight of rice she added.
Therefore, the total amount of rice in the bag is 8kg 6 hg 700 g.

Problem 6:
Add 48m 43 cm and 32m 8cm. Express the sum in meters.

Solution:

As given in the question, the value is 48m 43cm and 32m 8cm.
Now, we need to find the value in meters. Using addition, we add the given two numbers and obtain the resultant value i.e. sum.
Convert the cm value into meter value at first and then we can add the given values easily.
100 cm = 1 meter
So, the given numbers in meters are,
48 m 43 cm = 48.43 m
32 m 8 cm = 32.08 m
Now, perform the addition perform,
48.43 m + 32.08 m = 80.51 m.
Hence, the sum of given values in meters is 80.51 m.

Problem 7:
John walked 5.326 km on Monday, 9.052 km on Tuesday, and some distance 18.9 km walked on Wednesday. How much did he walk on those three days?

Solution:

As given in the question,
He walked 5. 326 km on Monday.
He walked 9.052 km on Tuesday.
He walked 18.9 km on Wednesday.
Now, we will find the total km he walked on those three days.
So, the given values are,
5.326 km + 9.052 km + 18. 009 km
We have to perform the addition operation, you will get the total value.
5.326 km + 9.052 km + 18. 009 km = 32.387 km
Therefore, the total distance he walked on those three days is 32. 387 km.

Problem 8:
Akhila bought 2 pieces of cloth. one of them was 34 cm long, the second one was 53 cm long. How much cloth did she buy altogether?

Solution:

As given in the question, Akhila bought 2 pieces of cloth.
One of them was 34 cm long.
The second one was 53 cm long.
Now, let us find the total cm of cloth did she bought altogether.
So, the total of cloth is 34 cm + 53 cm,
34 cm + 53 cm = 87 cm.
Therefore, the total cloth she bought is 87 cm.

Problem 9:
Find the value of the below figure,

Solution:

As given in the question, the figure consists of some value.
Now, we can find the value of the given values using the addition law.
So, the sum of the given values is as shown in the below figure,
Hence, the resultant sum of the given metric values is 118.602 dal.

Problem 10 :
The length of a rectangular garden is 12.75 m and its breadth is 8.25 m. How much fencing will be required for the 4 sides of this rectangular garden?

Solution:

In the given question, rectangular length and breadth are given.
Now, we will find how much fencing is required.
The rectangular garden length is 12.75 m.
The rectangular garden breadth is 8.25 m.
The rectangular has 4 sides, so the length and breadth of both sides’ values are the same.
So, the fencing required is,
12.75 m + 8.25 m+ 12. 75m + 8.25 m = 42.00 m
Therefore, the fencing required for the 4 sides of the garden is 42 m.

Worksheet on Word Problems on Average | Free Average Math Word Problems Worksheet PDF

July 14, 2021

Finding the Average can be a quite challenging task for young learners. To aid them we have several free-to-use Average Word Problems Worksheets. After understanding the concept of average you will learn how to do average problems without any hassle. All you need to do is practice the different problems on average over here consistently to get a good hold of them. We are sure by the end of this Worksheet on Word Problems on Average you will no longer feel difficulty in solving the problems related to Average.

Also, Check:

Word Problems Involving Average Worksheet

Example 1.
The average age of 20 members in the yoga class is 9 years. A new member joined the class. If his age is included, then the average age becomes 10 years. Find the new member’s age?

Solution:

The average age of 20 members in the class=9 years
Total age of 20 members/20=9
Multiply both sides by 20
Total age of 20 members=180
(Total age of 20 members+new member age)/21=10
(180+new member age)/21=10
(180+new member age)/21=10
Multiply 21 on both sides
180+new member age=10×21
180+new member age=210
new member age=210-180
new member age=30
Hence, the new member age is 30.

Example 2.
The average of eight numbers is 40. If one number is removed, the average becomes 30. What is the number removed?

Solution:

Let the remaining seven numbers are x and the removed number be y.
x+y/8=40
x+y=320
When the number is removed
x/7=30
x=210
210+y=320
y=320-210=110
Hence, the removed number is 110.

Example 3.
The marks scored by the four students in the examination are 75,84,95,70.
1.Find the average marks?
2.How many students got higher marks than average?
3. How many students got fewer marks than average?

Solution:

The marks scored by the four students in the examination = 75,84,95,70
1. Average mark=sum of marks/no.of students
=75+84+95+70/4
=324/4
=81
Therefore, the average mark is 81.
2. No. of students got higher marks than average=2
3. No. of students got fewer marks than average=2

Example 4.
Anish collected 30 different books from his friends during the past 10 days. On average, how many books he collected every day?

Solution:

No. of books collected by Anish from his friends=30
No. of books collected by Anish everyday=30/10=3
Therefore, no. of books collected by Anish every day is 3.

Example 5.
The average daily sales generated by the store for the past 30 days is 1800. Find the total amount of sales generated by the store for the past 30 days?

Solution:

The average daily sales generated by the store for the past 30 days =1800
The total amount of sales generated by the store for the past 30 days=1800 × 30=54000
Hence, the total amount of sales generated by the store is 54000.

Example 6.
Anish and Girish both are friends. The goal scored by the Anish team in 5 matches is 3,2,1,5,4. The goal scored by the Girish team in 5 matches is 5,0,1,2,2. Find the average score of both the teams and which team scored higher average?

Solution:

The goal scored by the Anish team in 5 matches =3,2,1,5,4
The average score of the Anish team=3+2+1+5+4/5=15/5=3
The goal scored by the Girish team in 5 matches = 5,0,1,2,1
The average score of the Girish team=5+0+1+2+2/5=10/5=2
we know 3>2.
Therefore, the Anish team scored a higher average.

Example 7.
The student scored 50,60,70,80 on the four tests he took. After he took his fifth test, the average is now 70. What did he score on the fifth test?

Solution:

Let the fifth test mark be x.
50+60+70+80+x/5=70
260+x=350
x=350-260
x=90
Therefore, the student scored 90 marks on the fifth test.

Example 8.
The average age of the mother and her two sons is 30 years and that of two sons is 25 years. Find the mother’s age?

Solution:

The average age of the mother and her two sons = 30 years
Sum of their ages/3=30
Sum of their ages=90 years
The average age of two sons =25 years
Sum of the ages of two sons/2=25
Sum of their ages=50 years
Mothers age=90-50=40 years
Therefore, the Mothers age is 40 years.

Example 9.
The average of x,y,z is 65. x is as much more than the average as y is less than the average. Find the value of z?

Solution:

Given an average of x,y,z=65
i.e. x+y+z/3=65
x+y+z=195
Also given x-65=65-y
x+y=130
130+z=195
z=195-130
z=65
Hence, the value of z is 65.

Example 10.
The average mark of ten papers is 60. The average mark of the first 5 papers is 62 and that of the last 5 papers is 55. Find the marks obtained in the tenth paper?

Solution:

The average mark of ten papers=60
so the Sum will be 10 × 60=600
The average mark of the first 5 papers = 62
so the sum=5 × 62=310
The average mark of the last 5 papers = 58
So the sum=5 ×55=275
The marks obtained in the tenth paper= 600-585=15.