Nirmala

Are you preparing for any serious tests!!! Here you can get the main idea of Conversion of Decimal into Percentage. We can hear about decimals and percentages in our daily lives, like students can see about percentages in their results or in any other competitive exams. In this worksheet on Decimal to Percentage, we can see get a detailed explanation of the Conversion of Decimal into Percentage which helps students to prepare for their examinations.

First, we will get to know about Decimal and the Percentage. Here Decimal means a number that contains a decimal point, the dot in between the number is known as a decimal point. And Percentage refers to per hundred which is used to share the amount or any other thing in terms of hundred. The percentage can be represented in the % symbol.

Converting Decimal into Percentage

To convert a Decimal into a Percentage, First, we will multiply by 100, and then we will add the percent symbol %. And we also have a shortcut to convert from a Decimal into a Percentage by moving the decimal point two places to the right and then we will add the percent symbol %. Here in the below, we can see some of the examples of converting a decimal into a percentage.

1. Convert the following Decimal into Percentage:

(i) 0.6

(ii) 2.2

(iii) 3.1

Solution:

(i) 0.6
Here we will multiply the given decimal number by 100,
= 0.6 × 100
then we will add the percent symbol, so the result will be
= 60 %

(ii) 2.2
Here we will multiply the given decimal number by 100,
= 2.2 × 100
then we will add the percent symbol, and the result will be
= 220 %

(iii) 3.1
Here we will multiply the given decimal number by 100,
= 3.1 × 100
then we will add the percent symbol, so the result will be
= 310 %

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2. Show 1.6 as a percent.

Solution:

The given decimal number is
1.6
we will multiply the given decimal number by 100,
= 1.6× 100
then we will add the percent symbol, and the result will be
= 160%.

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3. Pick up the following which expresses Decimal as Percentage:

0.325

a) 325%

b) 32.5%

c) 3.25%

d) 0.0325%

Solution:

The given decimal number is
0.325
we will multiply the given decimal number by 100,
= 0.325× 100
then we will add the percent symbol, so the result will be
= 32.5%.
So, option b is correct.

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4.  What will be 14 percent in 35?

Solution:

Given 14 percent in 35, which means
14/35 × 100
on solving we will get, so the result will be
40%.

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5. 25 is what percent of 125?

Solution:

Given 25 percent in 125, which means
25/125 × 100
on solving we will get, so the result will be
20%.

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6. Express 1.25/50%

Solution:

Given 1.25/50%
1.25/50 × 100
on solving we will get, so the result will be
2.5%.

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7. Convert 0.225 as a percentage

Solution:

The given decimal number is
0.225
we will multiply the given decimal number by 100,
= 0.225× 100
then we will add the percent symbol, so the result will be
=22.5%.

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8. What percentage of 6 rupees is 50 paise?

Solution:

Given 6 rupees,
As we know 1 rupee= 100 paise,
So 6 rupees= 6×100= 600 paise,
therefore the percent is
= 50×100/600
on solving, we will get the result as
= 8.33%.

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9. What percent will be a two dollar is a quarter?

Solution:

Given two dollar, and a quarter which equals 25
As we know 1 dollar = 100 cents and 2 dollars = 2×100
which is equal to 200 cents, then
25/200 on simplifying we will get 0.125
know, we to get percent we will multiply 100
0.125×100= 12.5%.

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Worksheet on square root using prime factorization method is useful for the students to prepare well for the exams. Here to find the prime factorization of a number we will divide the number by the first even prime number and we will continue dividing until we get a decimal or a reminder after that we will divide with an odd number until the only numbers left are prime numbers and write the number as a product of prime number.

To find the square root using prime factorization first, we will resolve the number inside the square root into prime factors and then inside the square root, for every two same numbers multiplied and one number can be taken out of the square root. And in the next step, we will combine the like square root terms using mathematical operation. Here in this finding square root by prime factorization method worksheet, we will solve various types of questions to make you familiar with the concept.

1. Find the square root of the number 196 by using the method of prime factorization?

Solution:

Given the number is 196,
the prime factorization of 196 is
196= 2×2×17×17
And the square root of 196 is 14.

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2. Find the square root of the number 361 by using the method of prime factorization?

Solution:

Given the number is 361,
the prime factorization of 361 is, as the number 361 is a composite number, so
361= 19×19
And the square root of 361 is 19.

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3. Find the square root of the number 576 by using the method of prime factorization?

Solution:

Given the number is 576,
the prime factorization of 576 is
576= 2×2×2×2×2×3×3
And the square root of 576 is 24.

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4. Find the square root of the number 1849 by using the method of prime factorization?

Solution:

Given the number is 1,849,
the prime factorization of 1,849 is
1,849= 43×43
And the square root of 1,849 is 14.

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5. Find the square root of the number 2401 by using the method of prime factorization?

Solution:

Given the number is 2,401,
the prime factorization of 2,401 is
2,401= 7×7×7×7
And the square root of 2,401 is 49.

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6. Find the smallest number by which 2925 must be divided to obtain a perfect square. Also, find the square root of the perfect square so obtained?

Solution:

Given number is 2,925
Prime factors of 2,925 are
2,925= 3×3×5×5×13
And here the new number is 13, so 2,925÷13
= 225.
and the square root of 225 is 15.

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7. If 2,025 students are seated in the auditorium in such a way that each row contains as many students as the number of rows. Find the number of rows and the number of students in each row.

Solution:

The number of students are 2,025
To find the number of rows, we will find the square root of 2,025
So the square root of 2,025 is 45.
And the number of rows and the number of students are 45.

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8. The people in a community arranged a picnic. Each one contributed as many rupees as the number of people in the community. If the total contribution is Rs 1,089, find the strength of the people in the community.

Solution:

Let the number of people in the community be  X
And the contribution by each person is Rs. X
X×X= 1,089
X^2= 1,089
X= 33.
So the strength of the people in the community is 33.

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Worksheet on Square Root of Numbers in Decimal and Fractions contains the questions on the square root of the numbers in decimals and fractions for your practice. To calculate the square root of the numbers in decimals and fractions, we will convert the decimal number to a rational number and then we will find the square root of the by the prime factorization or by the long division method. Here, below we can see the examples on the square root of numbers in decimal and fraction form with solutions.

1. Find the square root of the √1.44 in decimal form?

Solution:

√1.44
Here, we will convert the decimal number to a rational number and then we will find the square root.
= √144/√100
= 1.2

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2. Find the square root of √6.25  in decimal form?

Solution:

Given √6.25
To find the square root, we will convert the decimal number to a rational number and then we will find the square root.
= √625/√100
= 2.5

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3. Find the square root of √43.56 in Decimal Form?

Solution:

Given √43.56

To find the square root, we will convert the decimal number to a rational number and then we will find the square root.
= √4356/√100
= 6.6

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4. Determine √5 up to two places of decimal.

Solution:

Here, we place a dot and zeros to the given square root and then divide into pairs, after that we will choose a number and square the number. So that we will make sure the number is less than or equal to 5 and then we will subtract it. And we will follow the division method as below.

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5. Estimate √3.6 correct up to two places of decimal?

Solution:

Here, we place a dot and zeros to the given square root and then divide into pairs, after that we will choose a number and square the number. So that we will make sure the number is less than or equal to 3 and then we will subtract it. And we will follow the division method as below.

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6. Find the Square root of a number √4/25 in fraction form and evaluate the square root of the following numbers in fraction form:

Solution:

Given √4/25
Now we will apply the square root for both numeration and denominator
= √4/√25
on applying the square root, we will get
= 2/5

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7. Find the Square root of a number √9/36 in fraction form and evaluate the square root of the following numbers in fraction form?

Solution:

Given √9/36
Now we will apply the square root for both numeration and denominator
= √9/√36
on applying the square root, we will get
= 3/6

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Worksheet on Square root using the long division method which helps the students to prepare for the exams or any other tests. To determine the long division method, we will first divide the digits of the number into pairs of segments, starting with the digit in the units place. Then identify each pair and the remaining final number. And next, we will divide the digits into segments, which will be starting from the left side of the segment.

The number which is largest and whose square is equal to or less than the first segment is called as divisor which is also called as quotient. Then we will subtract the square of the divisor from the first segment and bring down the next segment to the right of the reminder to get the dividend. And in this step, the new divisor is obtained by taking two times the previous quotient. In the final step, we will repeat the above steps until all the segments have been taken and the quotient so obtained will be the required square root of the given number. Here, below we can see the examples on square root using the long division method with solutions.

 

1. Evaluate using the long division method: √961

Solution:

We will perform the long division method for the number 961 and the square root of the 961 is
961= 31×31

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2. Evaluate using the long division method: √1089

Solution:

The square root of the number 1089 is
1089= 33×33.
Here in the below, we can see the long division method

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3. Evaluate using the long division method: √10404

Solution:

The square root of the number 10404 is
10404= 102×102

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4. What will be the least number which must be subtracted from 3,606 to make it a perfect square

Solution:

As the square of 60 is 3600, and by subtracting 3606-3600= 6. And 6 is the least number that must be subtracted.

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5. What will be the least number that must be subtracted from 8,125 to obtain a perfect square.

Solution:

As the square of 90 is 8,100, and by subtracting 8,125-8,100= 25. And 25 is the least number that must be subtracted.

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6. What will be the least number which must be added to 61,500 to make it a perfect square.

Solution:

To find the least number, we will add 61,500 to obtain a perfect square.
So √61500= 247.99,
And the square root is 247.99. So, the perfect square is 248.
So 248×248= 61,504.
And the least number which must be added is 61,504-61,500= 4.
So the least number which should be added is 4.

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7. What will be the least number that must be added to 6,200 to obtain a perfect square.

Solution:

To find the least number, we will add 6,200 to obtain a perfect square.
we know that
78×78= 6,084
79×79= 6,241
6,241 is in between the squares of the numbers 78 and 79.
Consider,
6,241- 6,084= 157
And the least number that must be added is 157.

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The Worksheet on Squares is useful for students to prepare for their examinations. In this worksheet to calculate squares of the numbers, we will multiply the number by itself and the square root of a number is a value that can be multiplied by itself and to get the original value. If we have a negative number to calculate the square then by multiplying a negative number by itself then the number will be a positive number.

And here to calculate perfect squares by using the prime factorization method we will find the prime factorization of that number and then we will divide the prime factorization into equal pairs of factors by which we can get the perfect squares. And to find the squares of even numbers we can check the last digit of the number, if the number is odd then that number will never be the square of the even number. In the below, we can see the examples with the solutions.

1. Find which of the following numbers are perfect squares by using the prime factorization method?

(i) 343

(ii) 2048

Solution:

(i) 81
81= (3×3)×(3×3)
As it has equal pairs of factors.
So it is a perfect square.

(ii) 2048
2048=(2×2)×(2×2)×(2×2)×(2×2)×(2×2)×2
As 2048 does not have any equal pairs of factors.
So it is not a perfect square.

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2. Give reason in each case and show that none of the following is a perfect square.

(i) 189

(ii)3549

Solution:

(i) 189
189=3^2×3×7
As the number 189 does not have equal pairs of factors it is not a perfect square.

(ii)3549
3549=(13×13)×3×7
As the number 3549 does not have equal pairs of factors it is not a perfect square.

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3. Show that the following are the squares of even numbers?

(i) 784

(ii) 144

Solution:

(i) 784

784: 28^2 as 28 is an even number.

(ii) 144

144: 12^2 as 12 is an even number.

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4. Show that the following are the squares of odd numbers?

(i) 169

(ii) 289

Solution:

(i) 169

169: 13^2 as 13 is an odd number.

(ii) 289

289: 17^2 as 17 is an odd number.

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5. Calculate:

(i) (48)² – (47)²

(ii) (57)² – (56)²

Solution:

(i) (48)² – (47)²
= 48+47
= 95

(ii) (57)² – (56)²
= 57+56
= 113

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6. Find the sum without adding:

(i) (1 + 3 + 5 + 7 + 9 + 11 + 13)

(ii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17)

Solution:

(i) (1 + 3 + 5 + 7 + 9 + 11 + 13)
As the total consecutive odd numbers are 7
and n = 7
Therefore sum = n×n
= 7×7
= 49

(ii) (1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17)
As the total consecutive odd numbers are 9
and n = 9
Therefore sum = n×n
= 9×9
= 81

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7. Determine 32 as the sum of 4 odd numbers.

Solution:

x+x+2+x+4+x+6=32
4x+12=32
4x= 32-12
4x= 20
x= 5
5,7,9,11= 32

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8. Write a Pythagorean triplet whose smallest member is 8?

Solution:

The Pythagorean triplet is 2m, m^2 – 1 and m^2+1
and the smallest member is 8,
So
2m= 8
m= 8/2
= 4.
As m= 4
2m= 2×4
= 8.
m^2-1= (4)^2 – 1
= 16 – 1
= 15.
m^2+1= (4)^2 + 1
= 16+1
= 17.
So, the Pythagorean triplet is 8,15,17.

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9. Prove that 4,520 is not a perfect square.

Solution:

To get the perfect square the number should have an equal pair of factors, so it will be a perfect square.
Here the number is 4,520 and the prime factorization of the number 4,520 is
4,520= 2×2×2×5×113
Here the number 4,520 didn’t have a perfect square, so the number 4,520 is not a perfect square.

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10. Is the number 15,625 a perfect square or not?

Solution:

The prime factorization of 15,625 is
15,625= 5×5×5×5×5×5
which has an equal pair of factors. So the number 15,625 is a perfect square.

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11. Write Yes or No for each of the statements given below:

(i) For a perfect square, the number of digits should be even?

(ii) Do the square of a prime number is also prime?

Solution:

(i) False.
As a perfect square number will also be an odd number.
For example:
169= 13×13
As 13 is an odd number.

(ii)False.
As the square of the prime number can be an even number, an odd number, or a composite number.
For example:
The square of the number 3 is 3×3= 9 which is an odd number.

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Even after the perfect preparation of Time and Work concepts, are you still worried about the exam? If yes, then stop your worry now! Here we are providing the Worksheet on Time and Work so that you can practice more and prepare well for the exam. You can use this as one last time revision material to test your caliber of answering the questions. Go through the below sections, to check Time and Work Problems, and important questions.

It is mandatory for every student to check themselves before going to the exam. That can also be possible with a lot of practice. The below-given questions will help you to check your capability of solving the questions at a specific time.

1.  The time required for X to complete the job is 9 hours each day for 7 days and the time required for Y to complete the same job is 7 hours each day for 6 days. If both X and Y work together for 42/5 hours each day, how many days would it take to complete the whole job?

A. 3 Days

B. 7 Days

C. 8 Days

D. 6 Days

Solution:

A (3 Days)

X completes the work in (7*9) = 63 hours

Y completes the work in (6*7) = 42 hours

X’s 1 hour’s work = 1/63 and Y’s 1 hour’s work = 1/42

(X+Y)’s 1 hour’s work = (1/63+1/42) = 5/126

Both X and Y will finish the work in (126/5) hours

Number of days of 42/5 hours each = (126/5 * 5/42) = 3 days

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2. X and Y complete a piece of work in 18 days, Y and Z complete the same work in 24 days, X and Z complete it in 36 days. How many days will X, Y, and Z take to finish it, working together and separately?

A. Together = 14, X = 68 Days, Y = 166/5 Days, Z = 166 Days

B. Together = 16, X = 48 Days, Y = 144/5 Days, Z = 144 Days

C. Together = 24, X = 64 Days, Y = 166/5 Days, Z = 166 Days

D. Together = 20, X = 36 Days, Y = 154/5 Days, Z = 154 Days

Solution:

B (Together = 16, X = 48 Days, Y = 144/5 Days, Z = 144 Days)

(X+Y)’s 1 day work = 1/18, (Y+Z) one day’s work = 1/24 and (X+Z)’s 1 day’s work = 1/36

Adding we get: 2 (X+Y+Z)’s 1 day’s work =(1/18+1/24+1/36) = 9/72 = 1/8

(X+Y+Z)’s 1 day’s work = 1/16

Thus, X, Y and Z together can finish the job in 16 days.

Now, X’s 1-day work = [(X+Y+Z)’s 1 day work] – [(Y+Z)’s 1 day work] = (1/16-1/24) = 1/48

X alone can finish the job in 48 days.

Similarly, Y’s 1 day work = (1/16 – 1/36) = 5/144

Y alone can finish the job in 144/5

And Z’s 1 day work = (1/16-1/18) = 1/144

Z alone can finish the job in 144 days.

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3. X is twice as good a workman as Y and together they finish a piece of job in 18 days. How many days will X alone take finish the work?

A. 40 Days

B. 27 Days

C. 15 Days

D. 30 Days

Solution:

B (27 Days)

(X’s 1 day’s work): (Y’s 1 day’s work) = 2:1

(X+Y)’s 1 day’s work = 1/18

Divide 1/18 in the ratio 2:1

X’s 1 day’s work =(1/18*2/3) = 1/27

Hence, X alone can finish the work in 27 days.

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4. X and Y can together complete a piece of job in 4 days. If X alone can complete the same job in 12 days, in how many days can Y alone complete that work?

A. 6 Days

B. 8 Days

C. 10 Days

D. 4 Days

Solution:

A (6 Days)

(X+Y)’s 1 day’s work = 1/4, X’s 1 day’s work = 1/12

Y’s 1 day’s work = (1/4-1/12) = 1/6

Hence, Y alone can complete the work in 6 days.

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5. Suppose, X can finish a particular job in 12 days. Y is 60% more efficient than X. How days will Y take to complete the work?

A. 14/3 Days

B. 15/2 Days

C. 28/3 Days

D. 15/4 Days

Solution:

B (15/2 Days)

The ratio of times taken by X : Y = 160 : 100 = 8 : 5

Suppose Y alone takes x days to do the job.

Then, 8 : 5 :: 12 : x = 8x = 5*12

x=7 1/2 days

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6. The time required for X to complete the work in 80 days. X works it for 10 days and then Y takes 42 Days to finish the work alone. How much time do X and Y take to complete the whole work when working together?

A. 50 Days

B. 40 Days

C. 30 Days

D. 15 Days

Solution:

C (30 Days)

Work done by X in 10 days = (1/80 * 10) = 1/8

Remaining work = (1-1/8) = 7/8

Now, 7/8 work is done by Y in 42 days

The whole work will be done by Y in (42*8/7) = 48 days

X’s 1 day’s work = 1/80 and Y’s 1day’s work = 1/48

(X+Y)’s 1 day’s work = (1/80+1/48) = 8/240 = 1/30

Hence X and Y both will finish the work in 30 days

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7. X takes 10 days less than the time taken by Y to finish a piece of work. If X and Y can do it in 12 days, then how much time will Y alone take to finish the job?

A. 30

B. 27

C. 20

D. 25

Solution:

A (30 Days)

As mentioned in the question, X and Y can complete the work in 12 days.

We Suppose that X takes 24 days and Y takes 24 days to complete the work.

But it is given that X takes less no of days when compared to Y

Now, look at the option as it is mentioned that Y takes more no. of days than X, the answer should be more than 24.

So, we can cancel option C and option D is also approximately equal to 25 so we can cancel it.

Now we go for the trial and error method with the remaining 2 options.

Let’s take option A(30), so according to the question if Y takes 30 Days, X takes 10 days less i.e., 20 Days

That is X takes 2 portions and Y takes 3 portions. If we calculate = 60/5 = 12 days.

So 30 days is the answer.

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8.  If 3 men or 4 women can do a piece of work in 43 days, how long will 7 men and 5 women take to do the same work?

A. 10 days

B. 11 days

C. 9 days

D. 12 days

Solution: 

As mentioned in the question, 3 men and 4 women can do the work in 43 days

Here we can assume that 3 men can do the work in 43 days or 4 women also can do the work in 43 days

Here we need to find how many days 7 men and 5 women would need to complete the same work.

Here, we need to observe that the question is in the form of “or” followed by “and”, so here is an easy technique to follow the solution.

If there is a condition like “or”, the number of days would be the same and they will ask us to find something in a type of relation “and”.

To find the number of days, Multiply the numbers which you have in question 4*3*43. This will be our numerator value.

For the denominator, multiply the value of men and women and add them to the value of women and men (men*women)+(women*men)

Divide the numerator to the denominator and you will get the number days value as 12 days which will be your final answer.

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In this Worksheet on Simple Interest, we can see the questions on calculating simple interest, rate of interest, and the amount, which is useful to the students so that they can practice for their examinations or any competitive tests. Given below are different types of questions with solutions that help you to understand the chapter better. Assess your preparation level by solving the Simple Interest Worksheet Questions on your own.

Here to solve the questions on simple interest we will use the formula,
Simple interest (S.I.) = (Principal × Rate× Time)/100 and to calculate the Amount
Amount (A) = Principal +Interest

1. Find the simple interest and amount in each of the following:

(a) P = $30,000 R = 10% T = 5 years

Solution: 

Given:
Principal= $30,000
Rate= 10%
Time= 5 years.
To find Simple interest we will use the formula,
Simple interest (S.I.) = (Principal × Rate× Time)/100
= (30,000×10×5)/100
By solving we will get
Simple interest = $15,000.
(b) P= $17000  R= 20%  T= 30 days
Solution:
Given:
Principal= $17,000
Rate= 20%
Time= 30 days
Here we will convert days to years by dividing with 365 i.e 30/365, then substitute the given values in the formula,
= (17000×20×30)/100×365
by solving we will get
Simple interest = $279.45
(c) P= $6000  R= 15%  T= 13 months
Solution:
Given:
Principal= $6000
Rate= 15%
Time= 13 months
Here we have to convert months to years, so we will divide with 12
13÷12= 1.08 years
Simple interest (S.I.) = (Principal × Rate× Time)/100
= $975.
(d) P= $500  R= 3%  T= 1 1/2 year
Solution:
Given:
Principal= $500
Rate= 3%
Time= 1 1/2 year= 1.5 years.
Simple interest (S.I.) = (Principal × Rate× Time)/100
= $1,350.

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2. What sum would yield an interest of $80 in 4 years at 4% p.a.?

Solution:

Given:
Interest= $80
Time= 4 years
Rate= 4%
Principal = Interest×100/ Rate×Time
= 80×100/ 4×4
= $500.

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3. At what rate percent per annum will $550 amount to $650 in 4 years?

Solution:

Given:
Amount= $650
Principal= $550
Time = 4 years
Simple interest= Amount – Principal
= $650 – $550
=$100
So Rate= Simple interest ×100 / Principal×Time
= 100×100 / 550× 4
= 4.5%

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4. In what time will $800 amount to $1050 if the simple interest is calculated at 15% p.a.?

Solution:

Given:
Amount= $1050
Principal= $800
Rate = 15%
Simple interest= Amount – Principal
= $1050-$800
= $250
Time= Simple interest ×100 / Principal×Rate
= 250×100 / 800×15
= 2 1/12 years.

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5. A sum amount to $1500 at 5% simple interest per annum after 3 years. Find the sum.

Solution:

Given:
Amount payable = $1500
Rate= 5%
Time= 3 years
So amount payable= Principal(100+r×t)/100
1500= Principal(100+5×3)/100
By solving we will get
Principal= $1304.34

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6. Mr.Mike borrowed $5500 from Sam at 5% per annum. After 5 years he cleared the amount by giving $8,500 cash and a bag. Find the cost of the bag.

Solution: 

Given:
Principal= $5,500
Rate= 5%
Time= 5 years
Simple interest (S.I.) = (Principal × Rate× Time)/100
= $1,375.
The total amount to be cleared after 5 years= Principal+Interest
= $1,375+$5,500
= $6,875
Mike cleared by paying $8,500 and a bag,
So the cost of the bag is
= $8,500-$6,875
= $1,625

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8. In how many years will $7000 yield an interest of $4620 at 22% simple interest?

Solution:

Given:
Principal= $7,000
Simple interest= $4,620
Rate= 22%
Time=
$4,620= (7,000×22×Time)/100
By solving
Time= 3.04 years

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9. At what rate of simple interest will $2800 amount to $3500 in 2 years, 3 months?

Solution:

Given:
Principal= $2,800
Amount= $3,500
Time= 2 years 3 months = 9/4 years
Simple interest= amount – principal
= $3,500 – $2,800
= $700
Rate= (Simple interest × 100)/ Principal× time
By substituting the values we will get rate= 11.1%

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10. Find the simple interest at the rate of 8% p.a. for 4 years on that principal which in 6 years, 6 months at the rate of 4% p.a. gives $1600 as simple interest.

Solution: 

Given:
Rate= 4%
Time= 6 years 6 months which is 13/2 years
Simple interest= $1600
As Principal = (Simple interest × 100)/ Rate× time
= 1600×100×2 / 13×4
= $6153.84
Now Simple interest for 4 years is
= (6153.84×4×8)/100
= $1969.22

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11. Simple interest on a certain sum is 36/72 of the sum. Find the rate percent and time if both are numerically equal. [Hint: (T = R), P = x, S.I. = 36/72x]

Solution:

Let Sum be X
Time= Rate= R
Simple interest= 36/72X
Simple interest (S.I.) = (Principal × Rate× Time)/100
36/72X =(X × R × R)/100
on solving we will get 7% approx.

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12. Simple interest on a sum of money at the end of 5 years is 2/5 of the sum itself. Find the rate percent p.a.

Solution:

Let Principal be X
Time= 5 years
Simple interest= 2/5X
Simple interest (S.I.) = (Principal × Rate× Time)/100
2/5X= ( X × R × 5)/100
on solving we will get
Rate= 8%

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This Worksheet on Profit and Loss Percent helps the students in preparing for their examinations. In this worksheet, we can see questions on Profit and Loss percentage. Here we can see about profit and loss percent in which profit means the amount which we gained after selling a product and profit percent means the percent which is calculated with the Cost price in the base.

And loss means when we sold an item or a product less than it’s the actual price, then the Loss occurs. Loss percent means is the percent which is expressed as the percentage of the cost price. Here below we can see some examples of Profit and Loss percent with hints so that scholars can practice easily.

1. Find Profit or Loss percent when:

(a) Cost Price = 200 and Selling Price = 350

Solution:

To find profit and profit percent the formula is
Profit= SP-CP

= 350 – 200

= 150

Profit%= (Profit/CP) × 100

= (150/200)*100

= 75

Therefore, for the given data we have a 75% Profit Percent.

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2. Find the cost price when:

(a) Selling Price = 500 and Gain % = 5%

Solution:

To find the CP the formula is

CP = Selling Price × 100/100 + gain%
= 500*100/(100+5)

= 50000/105

= 476.19

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3. Find selling price when

(a) Cost Price = 900 and Gain % = 2%

Solution:

To find the SP the formula is

Selling Price = Cost Price(100 + Profit%)/100

= 900(100+2)/100

= (900*102)/100

= 918

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4. The cost price of tin is $20 and the selling price of the tin is $26. Find the profit of the tin?

Solution:

To find the profit of the tin we will use
Profit(P)= Selling Price(SP) – Cost Price (CP)

= $26-$20

= $6

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5. John sold two bicycles for $15,000 each. By selling one bicycle he gained 10% and on the other, he lost 15%.
Find his total gain or loss.

Solution:

Given Selling price and Profit% of one cycle. First, we will find the CP of the first cycle and then we will find the Loss of another cycle using the formula

C.P of one cycle when profit % is given = (Selling Price × 100)/100 + gain%

= 15, 000*100/100+10

= 1500000/110

= $13636.36

C.P of the other cycle when loss % is given = (Selling Price × 100)/100 – Loss%

= (15, 000*100)/100-15

= $17647.05

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6. A lounge set was bought for $4800 and $130 was spent on transportation and $500 on the repair. And it was sold at a loss of 10%. Find the S.P. of the lounge set?

Solution:

Here we must add the transportation and repair charges and then we must find the S.P of the lounge set using the formula

Total Cost Price = $4800+$130+$500

= $5430

Formula to find S.P when C.P and Loss% are given

S.P = (C.P(100-Loss%))/100

= $5430(100-10)/100

= $488700/100

= $4887

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7. Mr.Mike sold his car for $50,000 thereby gaining 15%. Find the C.P. of the car?

Solution:

In this S.P and profit, % is given and we should find the C.P of the car.
Let the C.P be 100,
As profit% is 15, S.P= 100+15
Let the C.P of the car be X,
Now make a proportion
100/X= 115/50,000
then solve X i.e. on performing basic math we get the C.P of the Car as $43478.26.

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8. Oliver bought a bike for $15,200 and sold it at a profit of 10 ¹/₂ %. Find the selling price of the bike?

Solution:

Given CP and the profit percent, we should find the selling price using the formula
SP= CP(100+P%)  / 100.

= ($15,200(100+21/2))/100

= ($15200*221/2)/100

= $16796

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9. Leo sells a table for $540 and loses 6%. At what price must he sell it to gain 15%?

Solution:

As SP and Loss Percent are given we must find the CP of the table.

C.P = (Selling Price × 100)/100 – Loss%

= ($540*100)/(100-6)

= $574.46

Selling Price he needs to sell to obtain a gain of 15% is

Selling Price = Cost Price(100 + Profit%)/100

= $574.46(100+15))/100

= $660.63

Leo needs to sell the table for a Cost of $660.63 to gain a Profit of 15%.

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10. By selling a laptop for $32,000, Jack loses 20%. How much percent would he gain or lose by selling it for $35,000?

Solution:

First, we must find the CP of the laptop, and then we must see that he will get profit or loss if he sells for $35,000.

S.P = $32, 000

Loss % = 20

C.P = (Selling Price × 100)/100 – Loss%

= ($32, 000*100)/100-20

= $32, 00000/80

= $40, 000

S.P given is $35,000

C.P = $40, 000

Loss = C.P – S.P

= $40, 000 – $35, 000

= $5000

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11. The selling price of 15 pens is equal to the cost price of 20 pens. Find the gain percent?

Solution:

20 C.P = 15 S.P

C.P/S.P = 15/20

C.P/S.P= 3/4

Gain % = ((S.P -C.P)/C.P)*100

= ((4-3)/3)*100

= 1/3*100

= 33.33%

Profit Percent is 33.33%

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12. The cost price of 20 apples is equal to the S.P. of 30 apples. Find the loss percent.

Solution:

20 C.P = 30 S.P

C.P/S.P = 30/20

= 3/2

Loss% = ((Cost Price – Selling Price)/Cost Price) *100

= ((3 -2)/3)*100

= (1/3)*100

= 33.33%

Loss Percent is 33.33%

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Worksheet on Discounts assists students to solve the questions on the Marked price, Selling price, and on the Discounts from the concept of Profit and Loss. If the students are preparing for any kind of test, they can prepare from this worksheet which they will find extremely helpful.

In this Worksheet, every topic was covered by some examples with hints. By this students can easily solve the problems and can prepare for their exams. Given below are some examples and answers are also provided.

1. The marked price of a study table is $ 3650. The shopkeeper offers an off-season discount of 20% on it. Find its selling price.

2. The marked price of a sweater was changed from $ 860 to $ 716 by a shopkeeper in the winter season. Find the rate of discount given by him.

3. Find the rate of discount is given on a coat whose selling price is $ 446 after deducting a discount of $ 100 on its marked price.

4. After allowing a discount of 10% on a bag, it is sold for $ 300. Find the marked price of the bag.

5. A dinner set was bought for $ 628 after getting a discount of 15% on its marked price. Find the marked price of the dinner set.

6. A dealer marks his goods at 25% above the cost price and allows a discount of 10% on the marked price. Find his gain or loss percent.

7. A mobile phone was marked at 30% above the cost price and a discount of 15% was given at its marked price. Find the gain or loss percent made by the shopkeeper.

8. A dealer has purchased a cooler for $ 2080. After allowing a discount of 15% on its marked price, he gains 20%. Find the marked price of the cooler.

9. A dealer bought a freezer for $ 10500. After allowing a discount of 15% on its marked price, he gains 10%. Find the marked price of the freezer.

10. A carpenter allows a discount of 15% to his customers and still gains 20%. Find the marked price of a study table which costs the carpenter $ 2190.

11. After allowing a discount of 20% on the marked price, a shopkeeper still makes a gain of 15%. By what percent is the marked price above the cost price?

12. How much percent above the cost price should a dealer mark his goods so that after allowing a discount of 20% on the marked price, he gains 12%?

13. The marked price of a refrigerator is $ 29000. A dealer allows two successive discounts of 30% and 15%. For how much is the refrigerator available?

14. Find a single discount which is equivalent to two successive discounts of 30% and 15%.

Answers:

1) $2,920

2)16.74%

3)Hint. MP = (SP) + (discount).
18.31%

4) $333.33

5) $738.82

6) Gain percent is 12.5%

7) Loss percent is 10.5%

8) $2936.47

9) $13,588.23

10)$3,091

11) The shopkeeper must mark his good’s price 43.75 % more than the cost price.

12) 40%

13)$17,255

14)40.5%

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This Worksheet to Find Profit and Loss is very helpful for the students to prepare for their examinations or any other competitive tests. Here this worksheet contains various types of questions on Profit and Loss which are helpful for the students in preparation. In this worksheet, we can see problems on Profit and Loss which students can practice more.

This concept of Profit and Loss is used in our daily life, which is like we will buy some essentials goods from the shopkeeper and the shopkeeper will buy them either from the manufacturer or from wholesalers. After that, the shopkeeper will sell those goods for a higher price than they bought so that he can earn some profit. In the below, we can see some examples of Profit and Loss below.

1. Find the Profit if:
a) SP= $100 and CP= $60
b) SP= $125 and CP= $100
c) SP= $90 and CP= $75.
d) SP= $500 and CP= $450.
e) SP= $2500 and CP= $1550
f) SP= $720 and CP= $420
g) SP= $870 and CP= $560.

2. The cost price of chocolate is Rs.10 and the selling price is Rs. 15. Find the profit.

3. Mark bought 4 dozens of apples at $15 a dozen and sold at $20 a dozen. Find the profit.

4. Mr.Singh bought a table for Rs 15,000 and spent Rs 500 on transportation. He sold the table for Rs 17,000. Find his profit.

5. Find the Selling price if
a) P= $20 and CP= $100
b) P= $32 and CP= $150
c) P= $5 and CP= $22
d) P= $50 and CP= $270
e) P= $80 and CP= $500
f) P= $26 and CP= $370
g) P= $18 and CP= $660

6. The cost price of a dining set is Rs 8000 and the shopkeeper got a profit of Rs 2000. Find the selling price of the dining set.

7. Find the Cost price if
a) SP= $200 and P= $20.
b) SP= $250 and P= $50.
c) SP= $125 and P= $25.
d) SP= $1225 and P= $150.
e) SP= $555 and P= $75.
f) SP= $375 and P= $55.
g) SP= $890 and P= $80.

8. A shopkeeper sold a chair for Rs 2000 and he got a profit of Rs 500. What is the actual price of the chair?

9. Find Profit% if,
a) SP= $140 and CP= $100
b) SP= $220 and CP= $200
c) SP= $70 and CP= $50.
d) SP= $505 and CP= $425.
e) SP= $420 and CP= $300.
f) SP= $125 and CP= $100.
g) SP= $350 and CP= $325.

10. A man bought 50 bulbs for Rs 125 each. And sells them at Rs 150 each. Find the profit and profit percent.

Answers:

1)
a)  40.
b) 25.
c) 15.
d) 50.
e) 950.
f) 300.
g) 310.

2) 5.

3) $5.

4) Rs 1500.

5)
a) S.P= $120.
b) S.P= $162.
c) S.P= $27.
d) S.P= $320.
e) S.P= $580.
f) S.P= $396.
g) S.P= $678.

6) Rs. 10,000.

7)
a) C.P= 180.
b) C.P= 50.
c) C.P= 100.
d) C.P= $1,075.
e) C.P= $4,80
f) C.P= $320.
g) C.P= $810.

8) Rs. 1500.

9)
a) 40%.
b) 10%.
c) 20%.
d) 18.82%
e) 40%
f) 25%
g) 7.69%.

10) 20%.

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