Class 6 Mathematics

Chapter 1 Simplification Unitary Method

Unitary Method :

1. The unitary method is a method of solving a problem by obtaining the value of one unit of material from some given value of the material.
2. Suppose you are given that the cost of 10 pens is 50. You have to obtain the cost of 16 pens.
3. Here we have to find the cost of one pen, then find the total cost of 16 pens.
4. So the cost of 10 pens = 50
5. ∴ The cost of 1 pen = ₹ \(\frac{50}{10}\) = 5
6. (Here the number of pens is less, so the cost would be less.
7. Therefore, the division process must be done.)
8. ∴ The cost of 16 pens = ₹ (5 x16) = ₹ 80.
9. (Here the number of pens is more, so the cost would be more. Therefore, the multiplication process must be done.)
10. Similarly, if the cost of 5 apples is 30, then what is the cost of 12 apples?
11. So to solve this problem, we can take the help of the above unitary method.
12. Here cost of 5 apples = ₹ 30
13. ∴ Cost of 1 apple = ₹ \(\frac{30}{5}\) =₹ 6
14. So the cost of 12 apples = 6 x 12 = ₹  72.
15. In general, two variables are so related that if one variable increases which cause, the increase of the other variable, or the decrease of one variable causes the decrease of the other variable, then the relation is said to be direct relation.
16. On the other hand, if the increase of one variable causes the decrease of the other variable or the decrease of one variable causes the increase of the other variable, then the relation between the variable is said to be indirect relation or inverse relation.
17. For example, the number of books is variable and the cost of the books is another variable.
18. There exists a direct relation between these two variables.
19. This means that the increase in the number of books causes an increase in the cost of the books or the decrease in the number of books causes a decrease in the total cost of the books.
20. The relation between the variable is direct.
21. Again let us consider that some men complete work in some days.
22. Here for the same amount of work done the number of days required is a variable and the number of men required is the other variable.
23. For the same amount of work, if the number of men is more they would take less number of days to finish the work.
24. Also if the number of men is less then they would take more days to finish the work.
25. The relation between the variables is indirect or inverse.
26. The relation between the number of books and their cost is a direct relation i.e., the increase in the number of books causes an increase in their cost and the decrease in the number of books causes a decrease in their total cost.
27. On the other hand, the relation between the number of daily working hours to complete work and the number of days required is inverse relation, i.e., the increase in the daily working hours causes the decrease in the number of days required to complete the work and the decrease of the daily working hours causes the increase of the number of days required to complete the work.

class 6 math wbbse solution

So there exist two variables in general two relations: 

1. Direct Relation
2. Indirect or Inverse Relation.
   For direct relation, the value of unit quantity would be less and for inverse relation, the value of unit quantity would be more.
   To solve this type of problem, first, you have to ascertain which type of relationship exists between the variables, then solve the problem using the unitary method otherwise the wrong results may come.
   You observe the following worked-out examples, then you will have a clear concept or idea about the unitary method.

 

Class 6 West Bengal Board Math Solution

Question 1. If 40 laborers can take 35 days to construct a part of the embankment of the Matla River, then how many laborers will be required to construct the same part of the bank in 28 days?

Solution:

Given:

40 laborers can take 35 days to construct a part of the embankment of the Matla River

A part of the embankment of the Matla River can be constructed in 35 days by 40 laborers.

∴ The same part can be constructed in 1 day by 40 x 35 laborers.

∴ In 28 days the same part can be constructed by

laborers

= (10 × 5) = 50 labourers

∴ 50 laborers can be required.

 

Question 2. Debarsi, Debalina, Debmalya, and Debdut can do 150 sums in 6 days. If each of them can do a same number of sums per day, then how many days will be required to do 250 sums by Debarsi and Debalina?

Solution :

Given:

Debarsi, Debalina, Debmalya, and Debdut can do 150 sums in 6 days. If each of them can do a same number of sums per day

Here total number of men = 4, the number of days = 6, and the number of sums = 150.

It is also given that each of them can do every day a same number of sums.

So, 4 persons can do 150 sums in 6 days

1 person can do 150 sums in 6 x 4 = 24 days

1 persons can do 150 sums in \(\frac{24}{150}\)

2 persons can do 1 sums in \(\frac{24}{150 x 2}\) day

∴ 2 persons can do 250 sums in

 

 

days = 2 x 10 = 20 days

∴ The total number of required days = 20.

Read And Learn More: WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Solved Problems

Question 3. 45 laborers can dig a well in 24 days. If the well can be dug in 18 days, then how many more laborers will be required?

Solution: 

Given:

45 laborers can dig a well in 24 days. If the well can be dug in 18 days

A well can be dug in 24 days by 45 laborers.

∴ The well can be dug in 1 day by 45 x 24 laborers.

The well can be dug in 8 days by 

 

 

= 60 laborers.

There are already 45 laborers.

∴ 60 – 45 = 15 more laborers will be appointed.

Class 6 West Bengal Board Math Solution

Question 4. : If 2 men can polish \(\frac{1}{3}\) part of a table in one day, then how many men will be required to polish \(\frac{2}{3}\) part of the table in 2 days?

Solution :

Given:

2 men can polish \(\frac{1}{3}\) part of a table in one day

\(\frac{1}{3}\) part of a table can be polished in 1 day by 2 men

∴ 1 part of the table can be polished in 1 day by 2 x \(\frac{3}{1}\) men

∴ 1 part of the table can be polished in 2 days by \(\frac{2 \times 3}{1 \times 2}\)

\(\frac{2}{3}\) part of the table can be polished in 2 days by \(\frac{2 \times 3}{1 \times 2} \times \frac{2}{3}\)

∴ The required number of men = 2.

 

Question 5. 175 kg of rice is required for a week for a mid-day meal of 500 students. After 75 kg of rice has been used, how long will the remaining rice last for 400 students?

Solution:

Given:

175 kg of rice is required for a week for a mid-day meal of 500 students. After 75 kg of rice has been used

One week 7 days.

Amount or remaining rice = (175 – 75) kg = 100 kg.

175 kg of rice will last 500 students for 7 days

1 kg of rice will last for 500 students for \(\frac{7}{175}\) days

1 kg of rice will last for 1 student for \(\frac{7 \times 500}{175}\) days

100 kg of rice will last for 1 student for \(\frac{7 \times 500}{175} \times 100\) days

100 kg of rice will last 400 students for 

 

 Days = 5 days

The remaining rice will last for 400 students for 5 days.

Class 6 West Bengal Board Math Solution

Question 6. If the price of 15 books is 1275, then how many books will be purchased for 2125?

Solution:

Given:

The price of 15 books is 1275

For₹ 1275, the number of books purchased = is 15

For ₹ 1, the number of books be purchased = \(\frac{15}{1275}\)

For ₹ 2125, the number of books be purchased = \(\frac{15 \times 2125}{1275}\)

∴ The required number of books = is 25.

 

Question 7. Sita, Gita, and Rita can complete a piece of work separately in 12 hours, 15 hours, and 18 hours respectively. If they do it together then in how many hours will they complete \(\frac{1}{2}\)of the work?

Solution:

Given:

Sita, Gita, and Rita can complete a piece of work separately in 12 hours, 15 hours, and 18 hours respectively.

Here the whole of the work = 1 part. 

Then Sita can complete the work in 12 hours. 

∴ Sita in 12 hours, can do 1 part of the work In 12 hours, Sita can do 1 part of the work

∴ In 1 hour, Sita can do \(\frac{1}{12}\) part of the work. 

Gita can do in 15 hours 1 part of the work.

In 1 hour, Gita can do \(\frac{1}{15}\) part of the work.

In 18 hours, Rita can do 1 part of the work.

∴ In 1 hour, Rita can do \(\frac{1}{18}\) part of the work.

So in 1 hour Sita, Gita, and Rita together can do (\(\frac{1}{12}\) + \(\frac{1}{15}\) + \(\frac{1}{18}\) part of the work

= \(\frac{15 + 12 + 10}{180}\) part

= \(\frac{37}{180}\) part of the work.

∴ Sita, Gita, and Rita together can do \(\frac{37}{180}\) part of the work in 1 hour.

∴ Sita, Gita, and Rita together can do 1 part of the work in \(\frac{1 \times 180}{37}\) hours

They together can do \(\frac{1}{2}\)  part of work in \(\frac{1 \times 180}{2 \times 37}\) hours

= \(\frac{90}{37}\) hours = 2 \(\frac{16}{37}\)  hours.

∴ The required time = 2 \(\frac{16}{37}\)hours.

Wbbse Class 6 Maths Solutions

Question 8. : 4 tractors are required to cultivate 360 bighas of land in 20 days. How many tractors will be required to cultivate 1800 bighas of land in 10 days? 

Solution:

Given:

4 tractors are required to cultivate 360 bighas of land in 20 days.

To cultivate 360 bighas of land in 20 days 4 tractors are required.

 

 

Question 9. There are 20 boys in a hostel and 150 kg of atta is stored for them for 30 days. But 30 kg of atta was wasted and 5 boys went home from the hostel. How long will the remaining boys be fed with the remaining amount of atta?

Solution:

Given:

There are 20 boys in a hostel and 150 kg of atta is stored for them for 30 days. But 30 kg of atta was wasted and 5 boys went home from the hostel.

The total amount of atta stored in the hostel was 150 kg and the amount of atta wasted was 30 kg.

∴ Remaining amount of atta= (150 – 30) = 120 kg

The remaining number of boys in the hostel = is 20 – 5 = 15.

In mathematical language, we have

 

 

Wb Class 6 Maths Solutions

Question 10. 15 vans can carry 75 quintals of fish in 40 minutes. How long will 20 vans carry 100 quintals of fish?

Solution:

Given:

15 vans can carry 75 quintals of fish in 40 minutes.

In mathematical language, we have,

 

Wb Class 6 Maths Solutions

Question 11. 12 farmers can cultivate land in 7 days working 6 hours a day. How many farmers will be required to cultivate that land in 9 days working 4 hours a day?

Solution:

Given:

12 farmers can cultivate land in 7 days working 6 hours a day.

In mathematical language, we have,

 

 

Question 12. A compositor can compose 11 pages in 8 hours. How many days will be required to compose a book containing 264 pages working 6 hours on average per day?

Solution:

Given:

A compositor can compose 11 pages in 8 hours

In mathematical language, we have,

Wb Class 6 Maths Solutions

 

 

 

 

Chapter 1 Simplification Miscellaneous Examples

Miscellaneous Examples

Example 1. Find the greatest number by which 10019 and 10621 will be exactly divisible.

Solution: 

Given: 10019 and 10621

The required greatest number will be the H. C. F. of 10019 and 10621.

Wbbse Class6 Math Solution

 

∴ The H.C.F. of 10019 and 10621 is 43.

So the required greatest number = is 43.

Read And Learn More: WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Solved Problems

Wbbse Class6 Math Solution

Example 2. Find the greatest number that will divide 1347, 2046, and 2568 leaving the remainder 8, 5, and 7 respectively.

Solution:

Given:

1347, 2046, and 2568 leaving the remainder 8, 5, and 7 respectively

We have to find the greatest number that will divide 1347, 2046, and 2568 leaving the remainder 8, 5, and 7 respectively.

∴ 1347 – 8 = 1339; 2046 – 5 = 2041; 2568 – 7= 2561.

Therefore, the numbers 1339, 2041, and 2561 must be divisible by the required greatest number and so the H. C. F. of these numbers will be the required greatest number.

 

 

 

∴ H.C.F. 1339, 2041, and 2561 = 13.

Wbbse Class6 Math Solution

Example 3. Find the least number which when divided by 12, 18, 24, 36, and 45 will leave a remainder of 8 in each case.

Solution:

Given:

12, 18, 24, 36, and 45 will leave a remainder of 8 in each case

The L. C. M. of 12, 18, 24 36, and 45 will be the least number, which must be divisible by these numbers.

So the required least number exceeds the L. C. M. by 8 as it leaves a remainder of 9 in each case.

Now first we shall find the L. C. M. of the given numbers.

 

 

.. L. C. M. 2 x 2 x 3 x 3 x 2 x 5 = 360.

So the required least number = 360+ 8 = 368.

 

Example 4. Find the greatest number which will divide 98, 137, and 202 so as to leave the same remainder in each case.

Solution:

Given:

98, 137, and 202 so as to leave the same remainder in each case

Since the required number when divides 98, 137, and 202 leaves the same remainder in each case, therefore (137 – 98) or 39 and (202 – 137) or 65 must be divisible by the required greatest number. 

So the required greatest number will be the H. C. F. of 39 and 65.

Now the H. C. F. of 39 and 65 = 13.

The required greatest number = is 13.

 

Example 5. Find the least number which when added to 5 will be exactly divisible by 36, 54, 66, and 72.

Solution:

Given:

36, 54, 66, and 72.

Here we shall first find the L. C. M. of 36, 54, 66, and 72.

 

 

L.C. M. = 2 x 2 x 3 x 3 x 3 x 11 x 2 = 2376.

So 2376 is the least number that must be divisible by 36, 54, 66, and 72.

But the required number will be 5 less than 2376.

∴ The required least number is 2376 – 5 = 2371.

 

Example 6. From what least number should 7 be subtracted to make it divisible by 65, 91, 104, and 195?

Solution:

Given:

65, 91, 104, and 195

Here we shall first find the L. C. M. of 65, 91, 104, and 195.

 

 

∴ So L. C. M. = 5 x 13 x 7 x 8 x 3 = 10920.

So 10920 is the least number that must be divisible by 65, 91, 104, and 195.

The required least number will be 7 more than the L. C. M.

Hence the required least number = 10920 + 7

= 10927.

 

Example 7. Find the greatest number of 5 digits which is exactly divisible by 24, 36, 54, and 72.

Solution:

Given:

24, 36, 54, and 72

First, we shall find the L. C. M. of 24, 36, 54, 72.

 

 

 

∴ L. C. M. = 2 x 2 x 2 x 3 x 3 x 3 = 216.

Since L. C. M. of the given numbers is the least number that is divisible by the numbers individually, the required number will also be divisible by the L. C. M.

The greatest number of 5 digits = 99999.

 

 

∴ Greatest number of 5 digits divisible by 216 = 99999 – 207

= 99792

Hence the required number = 99792.

 

Example 8. Find the least number of 6 digits which is exactly divisible by 33, 55, 66, and 88.

Solution:

Given:

33, 55, 66, and 88

First, we shall find the L. C. M. of 33, 55, 66, 88.

 

 

∴ L. C. M. = 11 x 2 x 3 x 5 x 4 = 1320.

So L. C. M. of the given numbers is the least number divisible by the numbers individually and the required number will also be divisible by 1320.

The least number of 6 digits = 100000.

wbbse class 6 math solution

 

1320 – 1000 = 320

∴ 100000 + 320 = 100320.

So the least 6-digit number divisible by the given numbers = 100320.

 

Example 9. Find the least number between 800 and 900 which when divided by 46 and 69 will leave a remainder of 5 in each case.

Solution:

Given:

The least number between 800 and 900 which when divided by 46 and 69 will leave a remainder of 5 in each case

 

 

∴ L. C. M. of 46 and 69 = 23 x 2 x 3 = 138.

wbbse class 6 math solution

 

138 – 110 = 28

∴ 800 + 28 = 828.

∴ 828 is a number lying between 800 and 900 which is divisible by the L. C. M. of 46 and 69.

So 828 is divisible by 46 and 69 individually also.

But there is a remainder 5.

Hence the required number = 828 + 5

= 833.

The least number is = 833.

 

Example 10. Find the least number which when divided by 56, 70, 84, and 140 will leave the remainder of 49, 63, 77, and 133 respectively.

Solution:

Given:

56, 70, 84, and 140 will leave the remainder of 49, 63, 77, and 133 respectively

We have, 56 497; 7063 = 7; 84 77 = 7; 140 133 7.

The required number when added to 7 becomes exactly divisible by 56, 70, 84, and 140.

wbbse class 6 math solution

 

∴ L. C. M. of 56, 70, 84, and 140 = 7 x 2 x 2 x 5 x 2 x 3 = 840.

So 840 is the least number exactly divisible by 56, 70, 84, and 140.

The required number = is 840 – 7

= 833.

The least number  is= 833.

class 6 maths solutions wbbse

Example 11. What least number of 6 digits has 233 as a factor?

Solution:

Given:

6 digits has 233 as a factor

Here the required least number of 6 digits should be divisible by 233. 

The least number of 6 digits = 100000.

 

233 – 43 = 190.

∴ 100000 + 190 = 100190

So the required least number = 100190.

 

Example 12. Find the least number which when divided by 6, 8, and 10 will leave a remainder of 3 in each case but will leave no remainder when divided by 11. 

Solution:

Given:

6, 8, and 10 will leave a remainder of 3 in each case but will leave no remainder when divided by 11

L.C. M. of 6, 8, 10 = 120.

∴ The required number will be 3 more than the least multiple of 120 and is divisible by 11.

120 x 1+3 = 123, it is not divisible by 11.

120 x 2 + 3 = 243, it is not divisible by 11.

120 x 3 + 3 = 363, which is divisible by 11.

Hence the required least number = is 363.

 

Example 13. Four bells toll together and then begin to toll at intervals of 12, 15, 18, and 20 seconds respectively. When will they toll together again?

Solution:

Given:

Four bells toll together and then begin to toll at intervals of 12, 15, 18, and 20 seconds respectively.

The time between the two consecutive simultaneous tolls of the four bells will be exactly divisible by 12, 15, 18, and 20 seconds i.e., will be exactly divisible by the L. C. M. of 12, 15, 18, and 20.

 

.

wbbse math solution class 6

∴ L. C. M. = 2 x 2 x 3 x 5 x 3 = 180.

So the required time = 180 seconds 3 min.

∴ The bells will toll together again after 3 min.

Chapter 1 Simplification Decimal Fraction Examples

Question 1. Express the following numbers in decimal fractions:

1. 2 + \(\frac{3}{10}\) 

Class 6 West Bengal Board Math Solution :

2 + \(\frac{3}{10}\) =  2 + 0.3

= 2.3

Class 6 West Bengal Board Math Solution

2. 10 + 7 + \(\frac{8}{1000}\)

Solution:

10 + 7 + \(\frac{8}{1000}\)

= 17 +.008

= 17.008

 

3. 6 one-tenths

Solution:

6 one-tenths 0.6

Wbbse Class 6 Maths Solutions

4. 9 one-hundredths

Solution:

9 one-hundreths =0.09

Read And Learn More: WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Solved Problems

5. Four one-thousandths

Class 6 West Bengal Board Math Solution:

Four one-thousandths = 0.004

 

6. Two hundred three decimal four five

Solution:

Two hundred three decimal four five = 203.45

 

7. Four thousand two units five one-thousandths 

Class 6 West Bengal Board Math Solution:

Four thousand two units five one-thousandth = 4002.005

 

8. 400 + 50 + \(\frac{9}{100}\) + \(\frac{1}{1000}\)

Solution:

400 + 50 + \(\frac{9}{100}\) + \(\frac{1}{1000}\)

= 450 +0.09 + 0·001 = 450.091

 

9. Two lac two units four 100 1000 one-thousandths

Solution:

Two lac two units four one-thousandths = 20002.004

 

10. Six hundred twenty-nine decimal zero five. 

Class 6 West Bengal Board Math Solution :

Six hundred twenty-nine decimal zero five = 629.005

 

Question 2. Put the following decimal numbers in the respective places of the place value table and then express them in words:

1. 27.9

2. 1.28

3. 65.134

4. 42.009

5. 38.205

6. 4003.08

7. 712.5

8. 45.06

Solution:

 

 

Expression in Words:

1. 27.9 = Twenty-seven decimal nine or Twenty-seven nine one-tenths.

2. 128 = One unit two one-tenths eight one-hundred.

3. 65.134 = Sixty-five one one-tenths three one-hundredths four one-thousandths.

4. 42.009 = Forty-two nine one-thousandths or forty-two decimal zero nine. 

5. 38.205 = Thirty-eight two one-tenths five one-thousandths

6. 4003.08 = Four thousand three eight one-hundredths.

7. 712.5 = Seven hundred twelve five one-tenths.

8. 45.06 = Forty-five six one-hundredths or forty-five decimal zero six.

 

Question 3. Complete the following tables:

 

 

 

Class 6 West Bengal Board Math Solution:

 

 

 

Question 4. Convert the following decimal fractions to vulgar fractions:

1. 0.3

Solution:

0.3 = \(\frac{3}{10}\)

 

2. 0.039

Solution: 

0.039 = \(\frac{39}{1000}\)

 

Question 5. Put >, = or < in the blank spaces of the following:

1. 5.0         0.5

Solution:

The left-hand number is 5.0, and its integral (or whole part) part is 5. 

The right-hand number is 0.5, its integral part is 0.

∴ 5.0        0.5.

 

2. 72.1       72.10

Solution:

The whole or integral parts of both the left-hand number and right-hand number are the same (each equal to 72) and also in the decimal part both the numbers have the same one-tenths and so both the numbers are equal.

∴ 72.1        72.10.

 

3. 68.5        68.52

Solution:

The integral part of both the numbers is the same; in the decimal part both the numbers have the same one-tenths (each equal to 5) but the one-hundredth part of the left-hand number is 0 and in the right-hand number, the one-hundredth part is 2.

So the right-hand number is greater than the left.

∴ 68.5         68.52.

 

4. 72.93        729.3

Solution:

The integral part of the left-hand number is 72 and that of the right-hand number is 729.

∴ 72.97       729.3.

 

5. 42.6         42.600

Solution:

Both numbers have same the same integral part as well as the decimal part. 

So the numbers are the same.

∴ 42.6       42.600.

 

6. 2.33        3.22

Solution:

The integral part of the left-hand number is 2 and that in the right-hand number is 3.

∴ 2.33           3.22.

 

7. 92.4       924.00

Solution:

924 = 924.0 = 924.00

∴ 924        924.00.

 

8. 10.01       10.10

Solution:

Since both the numbers have the same integral parts and the one-tenths of the left-hand number is 0 but the right-hand number is 1.

∴ 10.01        10.10.

 

Question 6. Arrange the following numbers in ascending order of magnitude:

1. 0.534, 0.52, 5.34, 0.513

Class 6 West Bengal Board Math Solution:

Given

0.534, 0.52, 5.34, 0.513

The integral parts of the given numbers are 0, 0, 5, and 0.

∴ 5.34 is the greatest number.

Again,

0.534 = 0.534

0.52 = 0.520

0.513 = 0.513

As, 513 < 520 < 534

∴ 513 < 520 < 534.

So arranging the given numbers in ascending order of magnitude, we get,

0.513, 0.52, 0.534; 5.34.

 

2. 0.536, 0.335, 0.3354,. 0.52.

Solution:

Given

0.536, 0.335, 0.3354,. 0.52

The integral parts of the given numbers are the same, we have,

0.536 = 0.5360

0.335 = 0.3350

0.3354 = 0.3354

0.52 = 0.5200

∴ As, 3350 <  3354 < 5200 < 5360

0.0335 < 0.3354 < 0·5200 < 0.536

or, 0.335 < 0.3354 <0.52 < 0.536

So arranging the given numbers in ascending order of magnitude we get,

0.335, 0.3354, 0.52, 0:536

 

Question 7. Arrange the following numbers in descending order of magnitude:

1. 13.3, 11.3, 1.33, 2.31; 

Class 6 West Bengal Board Math Solution

\(\left.\begin{array}{l}
13 \cdot 3=13 \cdot 30 \\
11 \cdot 3=11 \cdot 30 \\
1 \cdot 33=1 \cdot 33 \\
2 \cdot 31=2 \cdot 31
\end{array}\right\}\)

The integral parts of the given numbers are

13, 11, 1, 2.

∵ 13 > 11 2 > 1

∴ 13.30 11.30 > 2.31> 1-33

Arranging the numbers in descending order of magnitudes, we get,

13.3, 11.3, 2.31, 1.33.

 

2. 3.007, 3.07, 37.30, 7.13

Class 6 West Bengal Board Math Solution:

\(\left.\begin{array}{l}
3 \cdot 007=3 \cdot 007 \\
3 \cdot 07=3.070 \\
37 \cdot 30=37 \cdot 300 \\
7 \cdot 13=7 \cdot 130
\end{array}\right\}\)

3.007, 3.07, 37.30, 7.13

We get, 37.300 > 7.130

The integral parts of the given numbers are

3, 3, 37, 7

∵ 37 > 7 > 3

We get,

∴ 37.300 > 7.130 > 3.070 > 3.007  ( ∵ 3.070> 3.007)

or, 37.30 > 7.13 > 3.07 > 3.007.

So arranging the given numbers in descending order of magnitudes, we get 37.30, 7.13, 3.07, 3.007.

Wbbse Class 6 Maths Solutions

Question 8. Expand the following numbers according to the place value of the digits:

1. 101.153 

Class 6 West Bengal Board Math Solution:

101-153 = 101 + \(\frac{1}{10}\) +  \(\frac{5}{100}\) +  \(\frac{3}{1000}\)

= 100 + 1 +  \(\frac{1}{10}\) + \(\frac{5}{100}\) +  \(\frac{3}{1000}\)

 

2. 57.031.

Solution:

57.031 = 57 + \(\frac{3}{100}\) +  \(\frac{1}{1000}\)

= 50 + 7 + \(\frac{3}{100}\) +  \(\frac{1}{1000}\)

Wbbse Class 6 Maths Solutions

Question 9. Find the values of the following:

1. 0.07 + 0.09

Class 6 West Bengal Board Math Solution :

 

∴ 0.07 + 0.09 = 0.16.

 

2. 4.11 + 1.6

Solution:

4.11 = 4.11;  1.6 = 1.60

 

∴ 4.11 + 1.6 = 4.11 + 1.60

= 5.71

 

3. 312.61+ 276.72

Class 6 West Bengal Board Math Solution:

312.61+ 276.72

 

 

∴ 312.61 + 276.72 = 586.33

Wbbse Class 6 Maths Solutions

4. 5 – 0.555

Solution:

5 = 5.000;  5.000 =  0.555

 

 

∴ 5 – 0.555 = 5.000 – 0.555 = 4.445.

 

5. 27.56+14.69

Class 6 West Bengal Board Math Solution:

27.56 + 14.69 = 42.25.

 

 

6. 4.3 +36.4

Solution:

4.3 = 4.3; 3 = 3.0; 6.4 = 6.4

 

 

 

∴ 4.3 + 3 – 6.4 = 4.3 + 3.0 – 6.4 = 0.9.

 

7. 3.36 – 4.62 + 2.18

Class 6 West Bengal Board Math Solution :

3.36 4.62 + 2.18 = 3.36+ 2.184.62.

 

 

∴ 3.36 = 4.62 + 2.18

= 3.36 + 2.18 – 4.62 = 0.92.

 

8. 2.67 – 3.727 + 4.2

Class 6 West Bengal Board Math Solution:

2.67 – 3.727 + 4.2

 

 

 

∴ 2.67 – 3.727 + 4.2

= 2.670 + 4·200 – 3.727 = 3.143

 

Question 10. For an occasion at our house, Father bought rice for 200, pulses for 125.50, and fish for what was spent by Father? 242.50. How much total amount of money

Solution :

 

Father spent 568.00 as the total amount of money for the occasion.

 

Question 11. Your exercise book is rectangular in shape and its length is 24.25 cm and its breadth is 10.75 cm. What is the perimeter of your exercise book?

Solution:

The perimeter of the exercise book = 2 (Length + Breadth)

= 2 (24.25 + 10.75) cm (2 x 35.00) cm 70 cm.

 

 

∴ The perimeter of the exercise book = is 70 cm.

 

Question 12. You had 5. You bought a pen for 3.50. How much money is left with you at present?

Solution :

₹ 5 = ₹ 5.00

 

 

∴ 1.50 is left with you at present.

 

Question 13. What must be added to 2.75 to get 3?

Solution

Now 3 = 3.00

 

 

∴ 0.25 must be added.

 

Question 14. Taniya cuts off a length of 8.5 cm of string from a string of length 12.5 cm. What is the length of the string left?

Solution:

The total length of the string = is 12.5 cm.

From it, a length of 8.5 cm is cut off.

 

 

∴ The length of the string is still left = 4 cm.

 

Question 15. What must be added to 2.172 to get 5?

Solution: 

5 = 5.000

 

 

∴ 2.828 is to be added.

 

Question 16. 2.647 is subtracted from 4.15. How much is to be added to the result of subtraction to get 10?

Solution:

4.15 = 4.150

 

∴ The result of subtraction = 1.503.

Now we have to find how much is to be added to 1-503 to get 10.

Now 10 = 10.000

 

∴ 8.497 must be added.

 

Question 17. In a long-jump competition, Susmita jumped 179.25 cm and Sagarika jumped 182.88 cm. Who and how much more length did jump? 

Solution: Susmita jumped 179.25 cm and Sagarika jumped 182.88 cm.

As 179.25 cm < 182.88 cm, therefore Sagarika jumped more length than Susmita.

 

 

∴ Sagarika jumped 3.63 cm more in length than Susmita.

 

Question 18. Simplify :

1. \(\frac{10 \cdot 573+2 \cdot 227-1 \cdot 8}{2 \cdot 347-4 \cdot 32+12 \cdot 973}\)

Solution:

The given expression = \(\frac{10 \cdot 573+2 \cdot 227-1 \cdot 8}{2 \cdot 347-4 \cdot 32+12 \cdot 973}\)

 

 

2. \(\frac{4 \cdot 5-6 \cdot 12+7 \cdot 432-1 \cdot 1}{5 \cdot 234+10 \cdot 2-2 \cdot 33-8 \cdot 392}\)

Solution:

The given expression = \(\frac{4 \cdot 5-6 \cdot 12+7 \cdot 432-1 \cdot 1}{5 \cdot 234+10 \cdot 2-2 \cdot 33-8 \cdot 392}\)

 

 

Question 19. A man gave 0-5 part of his own property to his wife, 0.2 part to his son, and 0.25 part to his daughter. If he had still the property whose value is ₹ 5000. What is the value of the whole property? What is the value of the property of his wife?

Solution:

Let the whole property = 1 unit.

∴ The total part of the property that his wife, son, and daughter got (0.502+0.25) = 0.95 part

∴ The remaining part of the property = (1 – 0.95) = 0.05 part

∴ Value of 0.05 part of the property = ₹ 5000

∴ Value of the whole property = ₹ (5000 ÷ 0.05) = ₹ 100000.

∴ The value of his wife’s property =  ₹ 100000 × 0.5 = ₹ 50000

So the value of the whole property = ₹ 100000. and the value of the wife’s property

= ₹ 50,000.

 

 

 

Chapter 1 Simplification Decimal Fraction

Decimal Fraction:

1. You have already studied integers and vulgar fractions in detail.
2. In the present article, we shall discuss decimal fractions in detail.

What is Decimal Fraction :

1. When we express a proper fraction or an improper fraction or a mixed fraction by a decimal point (.), then the fraction is called a Decimal Fraction.
2. For example, \(\frac{1}{2}\) is a proper fraction; when we express it by a decimal fraction then we write it as 0.5 i.e., \(\frac{1}{2}\) = 0.5.
3. Again, \(\frac{13}{4}\) is an improper fraction; it is expressed in decimal fraction as 3.25 i.e.,
4. \(\frac{13}{4}\) = 3.25
5. In a similar way a mixed fraction 4 \(\frac{1}{2}\) can be expressed in a decimal fraction as 4.5.

Read And Learn More: WBBSE Notes For Class 6 Maths Chapter 1 Simplification

Wbbse Class 6 Maths Solutions

Examples of Decimal Fractions :

0.32, 1.57, 11.004, 102.59, etc. are examples of Decimal Fractions.

 

 

Role of Decimal Point:

1. The role of a decimal point in any number is to make out a clear concept about the integral part and the fractional part of the number.
2. The left-hand part of the decimal point is called the integral part and the right-hand part of the decimal point including the decimal point is called the fractional part. 
3. For example, 2.04 is a decimal number and it contains 2 on the left side of the decimal point. 
4. So the integral part of the number is 2 and the number contains 04 after the decimal point in the right side of the decimal point including the decimal point and so the fractional part is .04.
5. If a decimal-number be such that there is no significant digit in the number in the left side of the decimal point then the integral part of the number is taken as 0.
   For example, the integral part of the number 0.0025 is 0 and the integral part of the number 0.9015 is zero.

Wbbse Class 6 Maths Solutions

How to write a decimal fraction?

Decimal fraction:

1. The last digit i.c., the digit in the extreme right side place of an integral part of a number is called the unit’s place digit.
   After this unit’s place digit (the just right side of the unit’s place digit) there is a point (.) written which is called the decimal point.
2. After writing this decimal point, the digits of the fractional part of the number, then only the complete decimal fraction are written.
3. For example, if for a number, the integral part is 245 and the fractional part is 356 then the decimal fraction is 245.356.

Why the name of the point (.) is the decimal point?

1. The point (.) is used for multiplication or division by 10 only, that’s why the point () is called the decimal point.
2. Suppose a decimal point exists in a number.
3. If the decimal point is shifted one place towards the right, then the value of the number increases 10 times, which means that the new number becomes 10 times the previous number, or in other words, the previous number is multiplied by 10.
4. Again if the decimal point is shifted one place towards the left, then the value of the new number is obtained by dividing the previous number by 10.
5. Suppose the given number is 2456.1251.
6. If we write 24561.251, this means that 2456.1251 is multiplied by 10.
7. If we write 245612.51, this means that 2456.1251 is multiplied by 100.
8. If we write 2456125.1, this means that 2456.1251 is multiplied by 1000.

In a reverse way:

1. If we write 245.61251, this means that 2456.1251 is divided by 10.
2. If we write 24.561251, this means that 2456.1251 is divided by 100.
3. If we write 2.4561251, this means that 2456.1251 is divided by 1000.

The Usefulness Of The Decimal System

The utility of the Decimal system is:

1. Large multiplication or division by 10 or by its multiplier is very easy in the decimal system, unlike multiplication or division by other numbers.
   In the decimal system only shifting of decimal point towards right or left can be done.
2. The units of length, mass and time, etc. can be expressed easily.

Wbbse Class 6 Maths Solutions

The Face-value and Place-value of the digits in Decimal Fractions:

The face value of a digit in any number is its own value while is the same everywhere.

In other words, the face value of a digit in any number is its absolute value.

For example, in the number 246, the face value of 2 is 2; the face value of 4 is 4 and the face value of 6 is 6.

Again in the number 24.567, the face value of 7 is 7; the face value of 2 is 2, etc.

But the place value of any digit in a number is the product of the digit and the place value of that place where the digit is placed.

For example, the place value of 5 in the number 258 is 5 x 10 = 50, because, 5 is placed in the ten’s place.

In order to determine the place value of any digit in any number, we follow the following rule which you have also learned already 

 

 

For decimal fractions, the place value of any digit after the decimal point can be determined according to the following rule:

Wbbse Class 6 Maths Solutions

In the decimal fraction 84.7325.

The place value of 7=7 x \(\frac{1}{10}\)

= \(\frac{7}{10}\)

= 0.7

The place value of 3 = 3 x \(\frac{1}{100}\)

\(\frac{3}{100}\)

= 0.002

The place value of 2 = 2 × \(\frac{1}{1000}\)

= \(\frac{2}{1000}\)

The place value of 5 = 5 x \(\frac{1}{10000}\)

= \(\frac{5}{10000}\)

= 0.0005

 

How to read Decimal Fraction:

1. Let 6.25 be a given decimal fraction. How will we read it? We see that the integer before the decimal point is 6 and the number after the decimal point is 25.
2. The given decimal fraction can be read as “six decimal two five” or “six decimal two one-tenths five one-hundredths.
3. Now let the decimal fraction 0-002 be given. This decimal fraction can be read as “zero decimal zero two” “zero decimal 2 one-thousandth” or simply “two one-thousandths”.
4. Similarly, 1247-253 = 1 thousand 2 hundred 4 tens 7 units decimal 2 one-tenths 5 one-hundredths 3 one-thousandths.
5. 42.538 4 tens 2 units decimal 5 one-tenths 3 one-hundredths 8 one-thousandths. 0.237 2 one-tenths 3 one-hundredths 7 one-thousandths.

 

Conversion Of Decimal Fraction Into Vulgar Fraction

1. In order to convert a decimal fraction into a vulgar fraction, omit the decimal point.
2. Take the number thus obtained as the numerator of the required vulgar fraction.
3. The denominator will be the number obtained by putting as many zeroes as there is a number of digits after the decimal point of the given decimal fraction towards the right of 1.
4. Then reduce this fraction to the lowest term.
5. If the fraction thus obtained is an improper fraction, then convert it into a mixed fraction.

 

Example 1. Convert 2.175 into Vulgar fraction.

Class 6 Wb Board Math Solution :

Given :

2.175

2.175 = \(\frac{2175}{1000}\)

= \(\frac{87}{40}\)

= \(2 \frac{7}{40}\)

Omitting the decimal point from the given decimal fraction 2:175, we get the number 2175. This is taken as the numerator.

There are 3 digits after the decimal point in the given decimal fraction. So the denominator will be 1000 which is obtained by putting 3 zeroes to the right side of 1.

∴ The vulgar fraction = \(\frac{2175}{1000}\)

Reducing it to the lowest term, we get, \(\frac{87}{40}\)

‍∴ But this is an improper fraction.

Converting it into a mixed fraction, we get 2 \(\frac{7}{40}\)

The required vulgar fraction = 2 \(\frac{7}{40}\)

 

Example 2. Convert 0.06235 into Vulgar fraction.

Solution:

Given: 0.06235

0.06235 = \(\frac{6235}{100000}\)

= \(\frac{1247}{20000}\)

 

Example 3. Convert the following decimal fractions into Vulgar fractions: 2.39; 0.0255; 1.3608; 0.045045.

Class 6 Wb Board Math Solution :

Given: 2.39; 0.0255; 1.3608; 0.045045

2.39 = \(\frac{239}{100}\) = 2 \(\frac{39}{100}\)

0.0255 = \(\frac{255}{10000}\)

= \(\frac{51}{2000}\)

1.3068 = \(\frac{13608}{10000}\)

= \(\frac{1701}{1250}\)

= 1 \(\frac{451}{1250}\)

0.045045 = \(\frac{45045}{1000000}\)

= \(\frac{9009}{20000}\)

 

Conversion of Vulgar Fraction Into Decimal Fraction

In order to convert a Vulgar fraction into a decimal fraction, divide the numerator of the Vulgar fraction by its denominator.

Example 1. Express as a decimal fraction.

Solution :

So the required decimal fraction is 0.4016.

 

Example 2. Express 3 \(\frac{7}{50}\) as a decimal fraction.

Class 6 Wb Board Math Solution:

3 \(\frac{7}{50}\)

3 \(\frac{7}{50}\)

= \(\frac{157}{50}\)

= 3.14

 

 

Example 3. Express \(\frac{9}{10}, \frac{73}{100}, \frac{31}{1000}, 2 \frac{3}{1000}\) as decimal fractions.

Solution:

\(\frac{9}{10}, \frac{73}{100}, \frac{31}{1000}, 2 \frac{3}{1000}\)

\(\frac{9}{10}\) = 9 ÷ 10

= 0.9

\(\frac{73}{100}\) = 73 ÷ 100

= 0.73

\(\frac{31}{1000}\) = 31 ÷ 1000

= 0.031

2 \(\frac{3}{1000}\) = \(\frac{2003}{1000}\)

= 2003 ÷ 1000

= 2.003

 

Example 4. Convert 13 \(\frac{17}{75}\) into decimal fractions up to 3 places of decimal.

Solution:

13 \(\frac{17}{75}\)

The given mixed fraction can be converted into decimal fractions up to 3 places of decimal as worked out in example 2.

The given mixed vulgar fraction can also be converted into decimal fractions up to 3 places of decimal alternatively as follows:

The integer contained in the given mixed fraction is 13, this integer will also remain in the decimal fraction.

Therefore, at first, the fraction is to be converted \(\frac{17}{75}\) into a decimal fraction and then put integer 13 to the left of the decimal point.

 

 

∴ \(\frac{17}{75}\) = 226 (upto 3 places of decimal).

∴ The required decimal fraction = 13.226 (up to 3 places of the decimal).

 

Addition And Subtraction Of Decimal Fractions

The decimal fractions which are to be added or subtracted are to be written one below the other in such a way that their decimal points must be one below the other.

The digits in the units, tens, thousands, tenths, hundredths, thousandths’ place, etc. of one decimal fraction should be written below the digits in the respective places of the other decimal fractions.

Then using the usual procedure of addition and subtraction of integers, the addition and subtraction of the given decimal fractions are done, and put the decimal point in the result is just below the decimal column.

If there are one, two, or three digits after the decimal point in a decimal fraction, that is, if there are tenths, hundredths, or thousandths’ place digits in a decimal fraction after the decimal point, one can place the zeroes according to the requirements after the last digit and then the addition and subtraction can be done.

 

Example 1. Add: 289.7, 25.379, 93.25, 7.5278

Class 6 Wb Board Math Solution :

Given: 289.7, 25.379, 93.25, 7.5278

∴ The required sum = 415.8568.

 

Example 2. Subtract 87.5923 from 205.31.

Solution:

Given: 87.5923 And 205.31

 

 

∴ The required result = 117.7177

Chapter 1 Simplification Fraction In Real Mathematical Problem And The Application Of The Rules Of Fraction

In our daily real life, we always face different types of mathematical problems related to fractions.

We solve these problems by applying the rules of fractions. 

We discuss these in the following examples

Question 1. How much money is to be taken from \(\frac{3}{5}\) part of ₹ 175 so that still there remains ₹ 45?

Solution :

Given:

Since there remains Rs 45 if some money is taken from 105, the amount of money taken = ₹ (105 – 45) = ₹ 60.

∴ The required sum of money to be taken out from ₹ 105 is ₹ 60.

Wbbse Class 6 Maths Solutions

Question 2. If ₹ 35 is added to \(\frac{5}{7}\) part of some money, then the sum should be ₹ 65. Find the amount of money.

Solution:

Given:

₹ 35 is added to \(\frac{5}{7}\) part of some money, then the sum should be ₹ 65.

Since ₹ 35 be added to \(\frac{5}{7}\)  part of the required money, then the sum should be ₹ 65

∴ \(\frac{5}{7}\) part of the money ₹ (65 – 35) = ₹ 30

∴ The required amount of money = \(₹\left(30 \div \frac{5}{7}\right)\)

 

 

= ₹ 42

∴ The required amount of money is = ₹ 42.

Wbbse Class 6 Maths Solutions

Question 3. How much will be added to \(\frac{7}{25}\) part of 4 so that the sum will be \(2 \frac{3}{5}\)

Solution:

Given:

 

Question 4. What do you mean by half of a piece of bread?

Solution:

By half of a piece of bread, are mean that \(\frac{1}{2}\) part of the piece of the bread.

Wbbse Class 6 Maths Solutions

Question 5. How much will be added to \(\frac{2}{3}\) so that the sum will be 2?

Solution: The required number = \(\left(2-\frac{2}{3}\right)\)

∴ If \(1 \frac{1}{3}\) be added to \(\frac{2}{3}\) , then the sum will be 2.

 

Question 6. What part of the total number of prime numbers between 1 and 10 is the total number from 1 to 10?

Solution: Total number of numbers from 1 to 10 = 10.

The prime numbers between them are 2, 3, 5, and 7.

So a total number of prime numbers between 1 and 10 = 4.

The required part = \(\frac{4}{10}\)

= \(\frac{2}{5}\)

So the total number of prime numbers between 1 and 10 to the total numbers from 1 to 10 = \(\frac{2}{5}\) part.

Wbbse Class 6 Maths Solutions

Question 7. \(\frac{5}{7}\) part of school gate is painted. How many parts of the gate is still to be painted?

Solution:

Given:

\(\frac{5}{7}\) part of school gate is painted.

Let the total part of the complete gate of the school = 1.

Its \(\frac{5}{7}\) part is painted.

∴ The required part of the gate of the school is still to be printed = \(\left(1-\frac{5}{7}\right)\) part

= \(\frac{7-5}{7}\)part 

= \(\frac{2}{7}\)part.

 

Question 8. I have large chocolate which is divided into 8 equal parts. Among these pieces, 3 pieces are given to my sister and 2 pieces are given to my brother. The remaining pieces are eaten by me. What part of the chocolate is distributed to each of us?

Solution:

Given:

I have large chocolate which is divided into 8 equal parts. Among these pieces, 3 pieces are given to my sister and 2 pieces are given to my brother. The remaining pieces are eaten by me.

Since the chocolate is divided into 8 equal parts,

∴ Each piece = \(\frac{1}{8}\) part of the chocolate

My sister is given 3 pieces.

So she has received \(\left(3 \times \frac{1}{8}\right)\) part = \(\frac{3}{8}\) part of the chocolate.

My brother is given 2 pieces. 

So he has received \(\left(2 \times \frac{1}{8}\right)\) part = \(\frac{1}{4}\) part of the chocolate.

I have taken {8 – (3 + 2)) pieces = 3 pieces. 

So I have received \(\left(3 \times \frac{1}{8}\right)\) = \(\frac{3}{8}\) part of the chocolate.

∴ The distribution of the chocolate is as follows:

sister = \(\frac{3}{8}\) part,

brother = \(\frac{1}{4}\) part,

Myself = \(\frac{3}{8}\) part of the chocolate.

Wbbse Class 6 Maths Solutions

Question 9. There are some oranges in the basket. Half of the oranges are given to my grandfather and then there are still 2 oranges left in the basket. How many oranges are there in the basket at first before distributing them to the grandfather?

Solution:

Given:

There are some oranges in the basket. Half of the oranges are given to my grandfather and then there are still 2 oranges left in the basket.

Let the total part of the oranges at first in the basket be 1 part.

\(\frac{1}{2}\) part of the organges be given to my grandfather.

∴ Still left \(\left(1-\frac{1}{2}\right)\) part = \(\frac{1}{2}\) part of the oranges in the basket.

By the given condition, \(\frac{1}{2}\) part of the oranges = 2.

∴ Original number of oranges in the basket = \(\left(2 \div \frac{1}{2}\right)\)

= \(2 \times \frac{2}{1}\)

= 4.

So there are 4 oranges in the basket at first, before distributing them to my grandfather.

 

Question 10. There are two glasses of the same measurement. A mixture of sweet drinks is prepared in both glasses. The first glass contains sugar \(\frac{1}{5}\)th part of it while the second glass contains sugar \(\frac{2}{7}\)th part of it. Without drinking the mixtures, determine which glass contains more sugar.

West Bengal Board Class 6 Math Solution :

Given:

There are two glasses of the same measurement. A mixture of sweet drinks is prepared in both glasses. The first glass contains sugar \(\frac{1}{5}\)th part of it while the second glass contains sugar \(\frac{2}{7}\)th part of it.

The glasses contain sugar \(\frac{1}{5}\) part and \(\frac{2}{7}\) part respectively.

The denominators of the fractions are 5 and 7.

 

Question 11. I had 20 and I spent 5. What part of my money did I spend and what part of my money had I still left?

West Bengal Board Class 6 Math Solution : I had ₹ 20 and I spent ₹ 5.

I spent \(\frac{5}{20}\) part of my money or, \(\frac{5}{20}\) part of my money.

Now money left to me ₹ (20 – 5) = ₹ 15.

It is  \(\frac{15}{20}\) part of my money or, \(\frac{3}{4}\) part of my money.

∴  \(\frac{3}{4}\) part of my money still left to me.

Alternative method:

Since I spent  \(\frac{1}{4}\)  part of my money, so I had left 1 –  \(\frac{1}{4}\) 

part or  \(\frac{3}{4}\) part of my money with me.

Question 12. Taniya have 36 oranges. She will give me \(\frac{2}{3}\) part of her oranges. How many oranges will she give me?

West Bengal Board Class 6 Math

Solution :

Given:

Taniya have 36 oranges. She will give me \(\frac{2}{3}\) part of her oranges.

Taniya has 36 oranges.

∴ Taniya will give me 24 oranges.

Question 13. Natasha has 15 metres long orange coloured tape. She has cut \(\frac{1}{3}\) part of the tape. What part of the tape still she has left and what is its length?

Class 6 Wb Board Math

Solution :

Given:

Natasha has 15 metres long orange coloured tape. She has cut \(\frac{1}{3}\) part of the tape.

Total length of the tape = 15 metres. 

She has cut \(\frac{1}{3}\) part of the tape.

∴ \(\left(1-\frac{1}{3}\right)\) = \(\frac{2}{3}\) part.

∴ Natasha has left  \(\frac{2}{3}\) part of the tape with her. Its length = 15 x \(\frac{2}{3}\)

= 10 metres.

∴ Natasha has left  \(\frac{2}{3}\) part of the tape and its length is 10 metres.

 

Question 14. At the beginning of the school, the water tank was in full; it was seen that at the time of tiffin \(\frac{1}{4}\) part of the tank was spent and at the time of closing the school \(\frac{1}{3}\) part of the tank was spent. What part of the tank water was left after closing the school?

West Bengal Board Class 6 Math

Solution :

Given:

At the beginning of the school, the water tank was in full; it was seen that at the time of tiffin \(\frac{1}{4}\) part of the tank was spent and at the time of closing the school \(\frac{1}{3}\) part of the tank was spent.

A total part of the tank water was spent at the time of closing the school

∴ After closing the school \(\frac{5}{12}\) part of the tank water was left.

 

Question 15. Dibakarbabu has 25 bighas of land and Ushadevi has 15 bighas of land. They have cultivated paddy 16 bighas and 8 bighas of land respectively. What part of their lands have been used to cultivate paddy respectively and who has used more lands to cultivate paddy?

Class 6 Wb Board Math

Solution :

Given :

Dibakarbabu has 25 bighas of land and Ushadevi has 15 bighas of land. They have cultivated paddy 16 bighas and 8 bighas of land respectively. What part of their lands have been used to cultivate paddy respectively

Dibakarbabu has 25 bighas of land and he has cultivated paddy in 16 bighas of land.

∴ Dibakarbabu \(\frac{16}{25}\) part of his land has been used to cultivate paddy.

Ushadevi has 15 bighas of land and she has cultivated paddy in 8 bighas of land.

∴ Ushadevi \(\frac{8}{15}\) part of her land has been used to cultivate paddy.

Now, the denominators of \(\frac{16}{25}\) and \(\frac{8}{15}\).

 

 

∴ L.C. M. of 25 and 15 = 5 x 5 x 3 = 75.

75 ÷ 25 = 3

75 ÷ 15 = 5

∴ \(\frac{16}{25}=\frac{16 \times 3}{25 \times 3}=\frac{48}{75} ; \quad \frac{8}{15}=\frac{8 \times 5}{15 \times 5}=\frac{40}{75}\)

As, 49 > 40, 

∴ \(\frac{48}{75}>\frac{40}{75} \text { or, } \frac{16}{25}>\frac{8}{15}\)

∴ Dibakarbabu has cultivated paddy on more land than that of Ushadevi.

 

Question 16. \(\frac{5}{12}\) part of a property is sold. If the value of the remaining part of the property is ₹ 70,000, then find the value of \(\left(\frac{7}{9} \text { of } \frac{8}{21} \div \frac{64}{27}\right)\) part of the whole property.

West Bengal Board Class 6 Math

Solution:

Given:

\(\frac{5}{12}\) part of a property is sold. If the value of the remaining part of the property is ₹ 70,000

Let the whole property = 1.

∴ The remaining part of the property = \(\left(1-\frac{5}{12}\right)\)

= \(\frac{7}{12}\)

So the value of \(\frac{7}{12}\) part of the whole property = ₹ 70,000

The value of the whole property = ₹ (70,000 ÷ \(\frac{7}{12}\))

=₹ ( 70,000 x \(\frac{12}{7}\))

= ₹ 120000

Again, 

 

 

∴ The value of \(\frac{1}{8}\) part of the property = ₹ (120000 x \(\frac{1}{8}\)) = 15000.

So the required value of ₹ is 15000.

Question 17. \(\frac{3}{5}\) of the soldiers in a regiment were killed in a battle, \(\frac{7}{27}\) of the soldiers in the regiment were captured by the enemy and the remaining 1900 soldiers fled away. How many soldiers were in the regiment? 

Class 6 Wb Board Math

Solution :

Given:

\(\frac{3}{5}\) of the soldiers in a regiment were killed in a battle, \(\frac{7}{27}\) of the soldiers in the regiment were captured by the enemy and the remaining 1900 soldiers fled away.

Let the total number of soldiers in Regiment 1.

∴ \(\frac{3}{5}\) + \(\frac{7}{27}\) =  \(\frac{81+35}{1355}\)

=  \(\frac{116}{135}\)

So \(\frac{116}{135}\) part of the soldiers in the regiment were killed and captured. 

∴ The remaining part of the soldiers in the regiment = (1 – \(\frac{116}{135}\)) = \(\frac{19}{135}\)

So,  \(\frac{19}{135}\) part of the soldiers of the regiment fled away.

∴ \(\frac{19}{135}\) part of the soldiers in the regiment = 1900

So total number of soldiers in the regiment = 1900 ÷ \(\frac{19}{135}\)

= 1900 x \(\frac{135}{19}\)

=13500

∴ There were 13500 soldiers in the regiment.

 

Question 18. A person distributed \(\frac{5}{8}\) part of his savings to his son, \(\frac{1}{6}\) part to his daughter and the rest to his wife. If his wife got savings?

West Bengal Board Class 6 Math

Solution:

Given:

A person distributed \(\frac{5}{8}\) part of his savings to his son, \(\frac{1}{6}\) part to his daughter and the rest to his wife.

Let the total of the person’s savings = 1

∴ \(\frac{5}{8}\) + \(\frac{1}{6}\) = \(\frac{15+4}{24}\)

= \(\frac{19}{24}\) part

So \(\frac{19}{24}\) part of the savings were distributed to the son and daughter.

∴ The remaining part of the savings = (1- \(\frac{19}{24}\))

= \(\frac{5}{24}\)

So \(\frac{5}{24}\) part of the savings = 15000

∴ Total savings = ₹ (15000 ++ \(\frac{5}{24}\))

= ₹ ( 15000 x \(\frac{24}{4}\))

=₹ 72000

So the total savings of the person = ₹ 72000.

 

Question 19. A person on his death bed divided his property in such a manner that his wife got part of his property and his sons got \(\frac{1}{3}\) the rest of the property equally each. If the wife’s portion was 3 times that of a son, find the number of sons.

West Bengal Board Class 6 Math

Solution:

Given:

A person on his death bed divided his property in such a manner that his wife got part of his property and his sons got \(\frac{1}{3}\) the rest of the property equally each. If the wife’s portion was 3 times that of a son,

Let the person’s total property = 1.

∴ Wife got = \(\frac{1}{3}\) part.

So the remaining part = (1 – \(\frac{1}{3}\)) part

= \(\frac{2}{3}\) part.

∴ Sons got \(\frac{2}{3}\) part of the property.

Since the wife’s portion was 3 times that of a son.

∴ Each son got = (\(\frac{1}{3}\) ÷ 3) part = (\(\frac{1}{3}\) x \(\frac{1}{3}\)) part

= \(\frac{1}{3}\) part.

∴ Number of sons = (\(\frac{2}{3}\) ÷ \(\frac{1}{9}\))

= 6

The number of sons = 6

 

Question 20. From a sum, \(\frac{1}{4}\) part of it is spent. If 45 is spent from part \(\frac{1}{3}\) of the rest, an amount of Rs. 30 still remains. What is the original sum?

Class 6 Wb Board Math

Solution :

Given:

From a sum, \(\frac{1}{4}\) part of it is spent. If 45 is spent from part \(\frac{1}{3}\) of the rest, an amount of Rs. 30 still remains.

After spending \(\frac{1}{4}\) part of the original sum there remains (1 – \(\frac{1}{4}\)) part

= \(\frac{3}{4}\) part of the total sum.

\(\frac{1}{3}\) part of the rest = \(\frac{1}{3}\) x \(\frac{3}{4}\) = \(\frac{1}{4}\) part of the original sum.

By the given condition, if ₹ 45 is spent from \(\frac{1}{4}\) part of the original sum and there still remains ₹ 30. This means that = ₹ 75.

part of the original sum = ₹ (45+30)

∴ Original sum = ₹ = (75 ÷ \(\frac{1}{4}\)) ₹ (75 x 4)

= ₹ 300.

The original sum = ₹ 300.

 

 

Chapter 1 Simplification Vulgar Fraction Examples

Question 1. From the following table identify the proper fraction, improper fraction, and mixed fraction

Wbbse Class 6 Maths Solutions:

Identifying the proper fraction, improper fraction, and mixed fraction:

Proper Fraction: \(\frac{1}{5}, \frac{2}{7}, \frac{3}{8}, \frac{6}{13}, \frac{1}{9}, \frac{2}{5}, \frac{5}{9}, \frac{4}{17}, \frac{11}{12}, \frac{3}{7}\)

Improper Fraction: \(\frac{15}{13}, \frac{29}{19}, \frac{23}{17}\)

Mixed Fraction: \(9 \frac{14}{15}, 1 \frac{22}{25}, 11 \frac{1}{9}, 2 \frac{3}{4}, 3 \frac{5}{11}\)

 

Question 2. Reduce \(6 \frac{7}{19}\) in improper fraction.

Class 6 Wb Board Math Solution :

Given:

\(6 \frac{7}{19}\) \(6 \frac{7}{19}=\frac{6 \times 19+7}{19}=\frac{114+7}{19}=\frac{121}{19}\)

Question 3. Express in mixed fraction:

Wbbse Class 6 Maths Solutions :

 

 

[Division of 123 by 7]

∴ \(\frac{123}{7}\) = \(17 \frac{4}{7}\)

∴ the required mixed fraction = \(17 \frac{4}{7}\)

 

Read And Learn More: WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Solved Problems

 

Question 4. Express with a lowest common denominator: 

\(\frac{5}{8}\), \(\frac{7}{12}\).

Class 6 Wb Board Math Solution:

Given:

\(\frac{5}{8}\), \(\frac{7}{12}\).

The denominators of the given fractions are 8 and 12.

 

 

Now 24 ÷ 8 = 3 and 24 ÷ 12 = 2

The common denominator for the given fractions will be 24.

For this, it is required to multiply both the numerator and denominator of the first fraction by 3 and to multiply both the numerator and denominator of the second fraction by 2.

∴ \(\frac{5}{8}=\frac{5 \times 3}{8 \times 3}=\frac{15}{24}\) and \(\frac{7}{12}=\frac{7 \times 2}{12 \times 2}=\frac{14}{24}\)

∴ The given two fractions with the lowest common denominator are \(\frac{15}{24}\) and \(\frac{14}{24}\)

 

Question 5. Express the following fractions with the lowest common numerator:

\(\frac{12}{13}\), \(\frac{18}{23}\)

Wbbse Class 6 Maths Solutions :

Given:

\(\frac{12}{13}\), \(\frac{18}{23}\)

The numerators of the given fractions are 12 and 18.

 

 

∴ The L. C. M. of 12, 18 = 2 x 3 x 2 x 3.36.

The lowest common numerator of the given two fractions will be 36.

Now 36 12 3 and 36 18 = 2

Multiplying both the numerator and denominator of the first fraction by 3 and multiplying both the numerator and denominator of the second fraction by 2, we get,

∴ \(\frac{12}{13}=\frac{12 \times 3}{13 \times 3}=\frac{36}{39}\) and \(\frac{7}{12}=\frac{7 \times 2}{12 \times 2}=\frac{14}{24}\)

∴ The required two fractions with the lowest common numerator are \(\frac{36}{39}\) and \(\frac{36}{46}\)

 

Question 6. Write 3 equivalent fractions of each of the following fractions:

1. \(\frac{1}{5}\)

Wbbse Class 6 Maths Solutions:

Given:

\(\frac{1}{5}\)

 We know that, if we multiply both the numerator and denominator of any fraction, then the value of the fraction is not changed.

∴ \(\frac{1}{5}=\frac{1 \times 2}{5 \times 2}=\frac{1 \times 3}{5 \times 3}=\frac{1 \times 4}{5 \times 4}=\cdots \cdots \text { etc. }\)

or, \(\frac{1}{5}=\frac{2}{10}=\frac{3}{15}=\frac{4}{20}=\cdots \cdots \text { etc. }\)

∴ The required three equivalent fractions are \(\frac{2}{10}, \frac{3}{15}, \frac{4}{20}\)

 

2. \(1 \frac{1}{3}\)

Solution:

Given:

\(1 \frac{1}{3}\) \(1 \frac{1}{3}=\frac{4}{3}=\frac{4 \times 2}{3 \times 2}=\frac{4 \times 3}{3 \times 3}=\frac{4 \times 4}{3 \times 4}=\cdots \cdots \text { etc. }\)

or, \(1 \frac{1}{3}=\frac{8}{6}=\frac{12}{9}=\frac{16}{12}=\cdots \cdots \text { etc. }\)

∴ The required 3 equivalent fractions are [late]\frac{8}{6}, \frac{12}{9}, \frac{16}{12}[/latex]

 

Question 7. Reduce the following fractions into the lowest terms:

1. \({72}{108}\)

Class 6 Wb Board Math Solution :

\(\frac{72}{108}\)

 

 

 

2. \(\frac{243}{405}\)

Solution:

\(\frac{243}{405}\)

 

 

Example 8. Arrange the following fractions in ascending order of magnitude:

1. \(\frac{7}{2}, \frac{7}{4}, \frac{7}{5}\)

Solution:

\(\frac{7}{2}, \frac{7}{4}, \frac{7}{5}\)

The denominators of the fractions are 2, 4, and 5.

 

 

 

∴  L. C. M. of 2, 4, 5 = 2 x 1 x 2 x 5 = 20

20 ÷ 2 = 10 ; 20 ÷ 4 = 5; 20 ÷ 5 = 4.

Multiplying both numerator and denominator of the 1st fraction by 10; the 2nd fraction by 5; the 3rd fraction by 4, we get,

\(\frac{7}{2}=\frac{7 \times 10}{2 \times 10}=\frac{70}{20} ; \quad \frac{7}{4}=\frac{7 \times 5}{4 \times 5}=\frac{35}{20} ; \quad \frac{7}{5}=\frac{7 \times 4}{5 \times 4}=\frac{28}{20}\)

Now the denominators of all the given fractions are 20.

∵ 28 < 35 < 70.

∴ \(\frac{28}{20}<\frac{35}{20}<\frac{70}{20} \text { or, } \frac{7}{5}<\frac{7}{4}<\frac{7}{2}\)

∴ Arranging the fraction in ascending order, we get \(\frac{7}{5}, \frac{7}{4}, \frac{7}{2}\)

 

2. \(5 \frac{3}{4}, 5 \frac{5}{9}, 5 \frac{8}{12}\)

Solution:

\(5 \frac{3}{4}, 5 \frac{5}{9}, 5 \frac{8}{12}\)

Now,

\(5 \frac{3}{4}=\frac{23}{4} ; \quad 5 \frac{5}{9}=\frac{50}{9} ; \quad 5 \frac{8}{12}=\frac{68}{12}\)

The denominators of the given mixed fractions are 4,9,12.

 

 

∴ L. C. M. of 4, 9, 12 = 2 x 2 x 3 x 3 = 36.

36 ÷ 4 = 9; 36 ÷ 9 = 4; 36 ÷ 12 = 3.

then

 

 

As 200 < 204 < 207, we get, \(\frac{200}{36}<\frac{204}{36}<\frac{207}{36} \text { or, } 5 \frac{5}{9}<5 \frac{8}{12}<5 \frac{3}{4}\)

∴ Arranging the fractions in ascending order, we get, \(5 \frac{5}{9}, 5 \frac{8}{12}, 5 \frac{3}{4}\)

 

3. \(1 \frac{1}{5}, 1 \frac{1}{7}, 1 \frac{1}{8}\)

Class 6 Wb Board Math Solution :

\(1 \frac{1}{5}, 1 \frac{1}{7}, 1 \frac{1}{8}\) Since each of the fractions has the integral part 1, we take the fractions \(\)

The denominator is 5, 7, and 8.

These numbers are prime to each other.

Their L. C. M. = 5 x 7 x 8 = 280.

280 ÷ 5 = 56

280 ÷ 7 = 40

280 ÷ 8 = 35.

∴ \(\frac{1}{5}=\frac{1 \times 56}{5 \times 56}=\frac{56}{280} ; \frac{1}{7}=\frac{1 \times 40}{7 \times 40}=\frac{40}{280} ; \frac{1}{8}=\frac{1 \times 35}{8 \times 35}=\frac{35}{280}\)

As 35 < 40 < 56, we, get \(\frac{35}{280}<\frac{40}{280}<\frac{56}{280} \text { or, } \frac{1}{8}<\frac{1}{7}<\frac{1}{5}\)

∴ \(1 \frac{1}{8}<1 \frac{1}{7}<1 \frac{1}{5}\)

∴ Arranging the fractions in ascending order, we get, \(1 \frac{1}{8}<1 \frac{1}{7}<1 \frac{1}{5}\)

 

4. \(\frac{1}{3}, \frac{4}{5}, \frac{7}{15}\)

Class 6 Wb Board Math Solution :

\(\frac{1}{3}, \frac{4}{5}, \frac{7}{15}\) The denominators of the given fractions are 3, 5, 15.

 

 

∴ L.C.M. of 3, 5, 15= 3 x 5

=15

15 ÷ 3 = 5

15 ÷ 5 = 3

15 ÷ 15 = 1.

 

∵ 5 < 7 < 12,

∴ \(\frac{5}{15}<\frac{7}{15}<\frac{12}{15} \text { or, } \frac{1}{3}<\frac{7}{15}<\frac{4}{5}\)

∴ Arranging the fractions in ascending order, we get, \(\frac{1}{3}, \frac{7}{15}, \frac{4}{5}\)

 

5. \(\frac{5}{7}, \frac{3}{4}, \frac{1}{4}\)

Solution:

\(\frac{5}{7}, \frac{3}{4}, \frac{1}{4}\)

∴ The denominators of the given fractions are 7, 4, and 4.

 

 

∴ L. C. M. of 7, 4, 4 = 4 x 7

= 28.

28 ÷ 2 = 45

90 ÷ 9 = 10

90 ÷ 5 = 18.

 

 

∵ 7 < 20 < 21,

∴ \(\frac{7}{28}<\frac{20}{28}<\frac{21}{28} \text { or, } \frac{1}{4}<\frac{5}{7}<\frac{3}{4}\)

∴ Arranging the fractions in ascending order, we get, \([latex]\frac{1}{4}, \frac{5}{7}, \frac{3}{4}\)[/latex]

 

6. \(3 \frac{1}{2}, 7 \frac{5}{9}, 7 \frac{1}{5}\)

Solution:

\(3 \frac{1}{2}, 7 \frac{5}{9}, 7 \frac{1}{5}\)

Now,

\(3 \frac{1}{2}\) = \(\frac{7}{2}\)

\(7 \frac{5}{9}\) = \(\frac{68}{9}\)

\(7 \frac{1}{5}\) = \(\frac{36}{5}\)

The denominators of the given fractions are 2, 9, and 5.

∴ L. C. M. of 2, 9, 5 = 2 x 9 x 5

= 90. (Here 2, 9, and 5 are prime to each other)

90 ÷ 2 = 45

90 ÷ 9 = 10

90 ÷ 5 = 18.

 

 

As 315 648 < 680, we get, \(\frac{315}{90}<\frac{648}{90}<\frac{680}{90} \text { or, } 3 \frac{1}{2}<7 \frac{1}{5}<7 \frac{5}{9}\)

∴ Arranging the fractions in ascending order, we get, \(3 \frac{1}{2}, 7 \cdot \frac{1}{5}, 7 \frac{5}{9}\)

 

7. \(\frac{1}{8}, \frac{7}{10}, \frac{3}{5}\)

Solution:

\(\frac{1}{8}, \frac{7}{10}, \frac{3}{5}\) The denominators of the given fractions are 8, 10, and 5.

 

 

∴ L. C. M. of 8, 10, 5 = 2 x 5 x 4 = 40.

40 ÷ 8 = 5

40 ÷ 10 = 4

40 ÷ 5 = 8.

 

∵ 5 < 24, < 28,

∴ \(\frac{5}{40}<\frac{24}{40}<\frac{28}{40} \quad \text { or, } \frac{1}{8}<\frac{3}{5}<\frac{7}{10}\)

∴ Arranging the fractions in ascending order, we get, \(\frac{1}{8}, \frac{3}{5}, \frac{7}{10}\) 

 

3. \(3 \frac{1}{2}, 3 \frac{5}{9}, 3 \frac{1}{5}\)

Solution:

\(3 \frac{1}{2}, 3 \frac{5}{9}, 3 \frac{1}{5}\)

Since such of the given fractions has the integer 3.

So we have to test the fractions \(\frac{1}{2}, \frac{5}{9}, \frac{1}{5}\)

Now the denominators of the fractions \(\frac{1}{2}, \frac{5}{9}, \frac{1}{5}\) are 2, 9, 5.

L. C M of 2, 9, 5 = 2 x 9 x 5

= 90

∴ 90 ÷ 2 = 45

90 ÷ 9 = 10

90 ÷ 5 = 18

 

 

As, 18< 45 < 50;

∴ \(\frac{18}{90}<\frac{45}{90}<\frac{50}{90} \quad \text { or, } \quad \frac{1}{5}<\frac{1}{2}<\frac{5}{9}\)

or, \(3+\frac{1}{5}<3+\frac{1}{2}<3+\frac{5}{9} \quad \text { or, } \quad 3 \frac{1}{5}<3 \frac{1}{2}<3 \frac{5}{9}\)

∴ Arranging the fractions in ascending order, we get, \(3 \frac{1}{5}, 3 \frac{1}{2}, 3 \frac{5}{9}\)

 

Example 9. Add: \(\frac{7}{2}+\frac{2}{3}+1 \frac{1}{2}\)

Solution:

Given:

\(\frac{7}{2}+\frac{2}{3}+1 \frac{1}{2}\)

 

 

 

Example 10. Subtract: \(10 \frac{2}{3}-7 \frac{2}{5}\)

Solution:

Given:

\(10 \frac{2}{3}And7 \frac{2}{5}\)

 

 

Example 11. Simplify: \(4 \frac{2}{15}+8 \frac{3}{5}-9 \frac{4}{25}\)

Solution:

The given quality = \(4 \frac{2}{15}+8 \frac{3}{5}-9 \frac{4}{25}\)

 

 

 

The required answer = \(3 \frac{43}{75}\)

 

Example 12. Find the value of the following:

1. \(\frac{2}{7}-\frac{2}{3}+1 \frac{1}{2}\)

Solution:

\(\frac{2}{7}-\frac{2}{3}+1 \frac{1}{2}\)

 

 

2. \(1 \frac{2}{5}-\frac{3}{8}+\frac{1}{4}\)

Solution:

\(1 \frac{2}{5}-\frac{3}{8}+\frac{1}{4}\)

 

 

[ L.C.M of 5, 8, 4 :

 

 

∴ L.C.M. of 5, 8, 4 = 2 x 2 x 5 x 2 = 40]

 

3. \(\frac{2}{5}+\frac{3}{8}-\frac{1}{4}\)

Solution:

\(\frac{2}{5}+\frac{3}{8}-\frac{1}{4}\)

 

 

4. \(7-3 \frac{1}{8}-2 \frac{1}{3}\)

Solution:

\(7-3 \frac{1}{8}-2 \frac{1}{3}\)

 

5. \(\frac{4}{5}+\frac{5}{8}-1 \frac{1}{3}\)

Solution:

\(\frac{4}{5}+\frac{5}{8}-1 \frac{1}{3}\)

 

6. \(1 \frac{3}{10}+1 \frac{4}{5}-1 \frac{1}{4}\)

Solution:

\(1 \frac{3}{10}+1 \frac{4}{5}-1 \frac{1}{4}\)

 

 

 

∴ L.C.M of 10, 5, 4 = 2 x 5 x 2

= 20.

 

7. \(2 \frac{5}{6}-1 \frac{8}{9}+1 \frac{3}{4}\)

Solution:

\(2 \frac{5}{6}-1 \frac{8}{9}+1 \frac{3}{4}\)

 

 

∴ L.C.M of 6, 9, 4 = 2 x 3 x 3 x 2

= 36.

 

8. \(4 \frac{1}{7}+2 \frac{2}{5}-5\)

Solution:

Given

\(4 \frac{1}{7}+2 \frac{2}{5}-5\)

 

Example 13. Simplify: \(8 \frac{1}{4} \div 1 \frac{4}{17} \div 5 \frac{7}{8} \text { of } 1 \frac{2}{15}\)

Solution:

Given expression =  \(8 \frac{1}{4} \div 1 \frac{4}{17} \div 5 \frac{7}{8} \text { of } 1 \frac{2}{15}\)

 

 

∴ The required answer = \(1 \frac{1}{329}\)

 

Example 14. Simplify: \(1 \frac{1}{4}+\frac{1}{3}\left[2 \frac{1}{4}+1 \frac{1}{2}\left\{3 \frac{1}{2} \div 2 \frac{1}{3}\left(4 \frac{1}{4} \div \overline{2+3 \frac{2}{3}}\right)\right\}\right]\)

Solution:

The given expression = \(1 \frac{1}{4}+\frac{1}{3}\left[2 \frac{1}{4}+1 \frac{1}{2}\left\{3 \frac{1}{2} \div 2 \frac{1}{3}\left(4 \frac{1}{4} \div \overline{2+3 \frac{2}{3}}\right)\right\}\right]\)

 

 

∴ The required answer = 3

 

Example 15. Simplify: \(\frac{4 \frac{1}{2}}{32} \times \frac{2 \frac{2}{3} \div \frac{5}{8}}{1 \frac{1}{5} \text { of } \frac{5}{6} \div 8 \frac{1}{3}} \times \frac{2}{5}\)

Solution:

\(\frac{4 \frac{1}{2}}{32} \times \frac{2 \frac{2}{3} \div \frac{5}{8}}{1 \frac{1}{5} \text { of } \frac{5}{6} \div 8 \frac{1}{3}} \times \frac{2}{5}\)

 

 

 

∴ The required answer = 2.

 

Example 16. Simplify: \(\frac{1 \frac{1}{4}-\frac{5}{12}}{1 \frac{1}{4}+\frac{5}{12}}-4 \div \frac{6 \frac{1}{2}}{2+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}}+4 \frac{5}{7 \frac{2}{3}}\)

Solution:

The given expression = \(\frac{1 \frac{1}{4}-\frac{5}{12}}{1 \frac{1}{4}+\frac{5}{12}}-4 \div \frac{6 \frac{1}{2}}{2+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}}+4 \frac{5}{7 \frac{2}{3}}\)

 

 

 

 

 

Chapter 1 Simplification Vulgar Fraction

Introduction :

When a material is divided into some parts, then each part is called a Fraction. If an apple is divided into two equal parts, then each part is said to be half of the whole apple and if the whole apple is divided into 3 equal parts then each part is called one-third of the whole apple. These two parts are written \(\frac{1}{2}\). \(\frac{1}{3}\) respectively.

Simplification Maths Class 6

In this way, these written parts are called Vulgar Fraction.

If a bar is divided into 5 equal parts, then the length of each part is \(\frac{1}{5}\)th of the length of the whole bar.

The length of such 3 parts together is the \(\frac{3}{5}\)th part of the length of the whole bar or in other words, it is said that the length of 3 parts is among 5 parts of the original bar.

A fraction can be expressed in the following way:

First, you draw a small line segment, then write an integer in each of the above and below this small line segment.

This small line segment is called the fraction line.

The integer which is written above the fraction line is called the Numerator and the integer which is written below the fraction line is called the Denominator.

For example, in the fraction, the numerator is 3 and the denominator is 5.

Similarly \(\frac{3}{4}\)

\(\frac{5}{13}\)

\(\frac{7}{38}\)

etc are examples of fractions.

Simplification Maths Class 6

Read And Learn More: WBBSE Notes For Class 6 Maths Chapter 1 Simplification

Wbbse Class 6 Maths Solutions

The denominator of any fraction can be any integer except zero but the numerator may be any integer or zero.

From the above discussions, it is clear that the numerator and the denominator of any fraction can be considered as the dividend and divisor respectively and the fraction itself can be considered as the quotient.

Definition:

A vulgar fraction is a rational number that can be expressed by drawing a small line segment and by writing an integer above this line segment and an integer below the line segment. The integer above the line segment is called the numerator and the integer below the line segment is called the denominator.

The small line segment is called the fraction line.

Wbbse Class 6 Maths Solutions

Classification Of Fractions

Vulgar Fractions can be classified into 4 classes:

1. Proper Fraction

2. Improper Fraction

3. Mixed Fraction

4. Complex Fraction.

Now we shall discuss these.

1. Proper Fraction:

The fraction in which the numerator is less than the

denominator is called the Proper Fraction. The fractions

\(\frac{5}{7}\), \(\frac{7}{11}\), \(\frac{13}{19}\) etc. are

proper fractions. The proper fraction is always less than 1.

2. Improper Fraction:

The fraction in which the numerator is greater than

\(\frac{5}{3}\), \(\frac{9}{7}\), \(\frac{23}{12}\), etc. are

the denominator is called the Improper Fraction. The fractions are improper fractions. The improper fraction is always greater than 1.

3. Mixed Fraction:

The fraction which consists of an integral part along with a fractional part is called the Mixed Fraction.

The fractions \(2 \frac{3}{7}, 5 \frac{9}{13}, 8 \frac{4}{17}\) etc. are. mixed fractions.

So a mixed fraction has two parts: One is an integral part and the other is a proper fraction.

The mixed fraction \(3\frac{2}{5}\) has the integral part 3 and the proper fraction part is \(\frac{2}{5}\).

4. Complex Fraction:

The fraction in which either the numerator or the denominator is a fraction or both the numerator and denominator are fractions, is called the Complex Fraction. For example, fractions \(\frac{\frac{3}{7}}{5}, \frac{2}{\frac{9}{11}}, \frac{\frac{2}{3}}{\frac{5}{7}}\), etc., are complex fractions.

There are other types of fractions that are also used, other than the above 4 types of fractions.

Simple fraction:

The fractions in which both the numerators and denominators are integers are called Simple Fractions.

For example: The fractions \(\frac{5}{9}, \frac{11}{7}, \frac{9}{17}\), etc. are simple fractions.

Compound fraction:

The fraction of any fraction is called the compound fraction.

For example: The fractions of \(\frac{5}{11} \text { of } \frac{7}{9}, \frac{9}{10} \text { of } \frac{13}{17}\), etc. are compound fractions.

Reciprocal fraction:

If two fractions are such that the numerator and denominator of one are respectively the denominator and numerator of the other, then the fractions are said to be reciprocal to one another.

For example: The fractions \(\frac{3}{4}\),  and \(\frac{4}{3}\) are reciprocal to one another.

Wbbse Class 6 Maths Solutions

Conversion of the improper fraction to mixed fraction and mixed fraction to improper fraction:

Question 1. Reduce \(7\frac{9}{17}\) to improper fraction.

Solution:  Rule \(Integer \frac{Numerator}{Denominator}=\frac{Integer \times Denominator+Numerator}{Denominator}\)

\(7 \frac{9}{17}=\frac{7 \times 17+9}{17}=\frac{128}{17}\)

Question 2. Express \(\frac{49}{9}\) as a mixed fraction.

Class 6 Wb Board Math Solution : Rule: Improper fraction = \(\frac{Numerator}{Denominator}\)

(Here Numerator is greater than the Denominator)

= \(Quotient\frac{Remainder}{Denominator}\)

 

 

∴ \(\frac{49}{9}\) = \(5\frac{4}{9}\)

 

Reduction of Fractions Into Lowest Terms

Question 1. Reduce \(\frac{210}{315}\) into lowest terms

Simplification Questions For Class 6 : \(\frac{210}{315}\)

Given: \(\frac{210}{315}\)

 

= \(\frac{2}{3}\)

Wbbse Class 6 Maths Solutions

Question 2. Reduce \(\frac{54}{81}\), \(\frac{78}{130}\), \(\frac{111}{148}\) into lowest terms.

Solution:

Given: \(\frac{54}{81}\), \(\frac{78}{130}\), \(\frac{111}{148}\)

1. \(\frac{54}{81}\)

 

 

= \(\frac{2}{3}\)

 

2. \(\frac{78}{130}\)

 

 

= \(\frac{3}{5}\)

Wbbse Class 6 Maths Solutions

3. \(\frac{111}{148}\)

 

=\(\frac{3}{4}\)

 

Expression of Fractions With Lowest Common Denominator or Numerator:

Question 1. Express \(\frac{3}{8}\) and \(\frac{5}{12}\) with lowest common denominator.

Class 6 Wb Board Math Solution : 

Given: \(\frac{3}{8}\) and \(\frac{5}{12}\)

The L.C.M. of the denominators 8 and 12 of the given fractions is 24.

The common denominator of both fractions will be 24.

24 ÷ 8 = 3, 24 ÷ 12 = 2

\(\frac{3}{8}\) = \(\frac{3 \times 3}{8 \times 3}\)

= \(\frac{9}{24}\)

\(\frac{5}{12}\) = \(\frac{5 \times 2}{12 \times 2}\)

= \(\frac{10}{24}\)

∴ The required fractions with the lowest common denominators are respectively.

Wbbse Class 6 Maths Solutions

Question 2. Express \(\frac{16}{27}\) and \(\frac{20}{41}\) with lowest common numerator. 41

Class 6 Wb Board Math Solution:

Given: \(\frac{16}{27}\) and \(\frac{20}{41}\)

The L.C.M. of the numerators 16 and 20 of the given fractions is 80.

80 ÷ 16 = 5, 80 ÷ 20 = 4

\(\frac{16}{27}=\frac{16 \times 5}{27 \times 5}=\frac{80}{135}\)

\(\frac{20}{41}=\frac{20 \times 4}{41 \times 4}=\frac{80}{164}\)

∴ The required fraction with the lowest common numerators are \(\frac{80}{135}\) and \(\frac{80}{164}\) respectively.

Question 3. Arrange the following fractions in ascending order of magnitude:

\(\frac{7}{8},\frac{9}{10},\frac{11}{16},\frac{13}{24},\frac{23}{30}\)

Class 6 Wb Board Math Solution :

Given: \(\frac{7}{8},\frac{9}{10},\frac{11}{16},\frac{13}{24},\frac{23}{30}\)

Here first we have to express the given fraction with the lowest common denominators.

 

 

∴ L.C.M. of 8, 10, 16, 24, and 30 = 2 × 2 × 2 × 3 × 5 × 2=240.

240 ÷ 8 = 30

240 ÷ 10 = 24

240 ÷ 16 = 15

240 ÷ 24 = 10

240 ÷ 30 = 8

∴ \(\frac{7}{8}=\frac{7 \times 30}{8 \times 30}=\frac{210}{240}\)

\(\frac{9}{10}=\frac{9 \times 24}{10 \times 24}=\frac{165}{240}\) \(\frac{11}{16}=\frac{11 \times 15}{16 \times 15}=\frac{130}{240}\) \(\frac{23}{30}=\frac{23 \times 30}{30 \times 8}=\frac{184}{240}\)

∴ Arranging in ascending order of magnitude, we get

\(\frac{13}{24},\frac{11}{16},\frac{23}{30},\frac{7}{8},\frac{9}{10}\)

Question 4. Arrange the following fractions in descending order of magnitude

\(\frac{5}{12},\frac{17}{20},\frac{7}{16},\frac{3}{8},\frac{13}{15}\)

Class 6 Wb Board Math Solution:

Given: \(\frac{5}{12},\frac{17}{20},\frac{7}{16},\frac{3}{8},\frac{13}{15}\)

First, we have to express the given fractions with common denominators.

 

∴ L.C.M. of 12, 20, 16, 8, 15 = 2 X 2 X 2 X 3 X 5 X 2 = 240.

240 ÷ 12 = 20

240 ÷ 20 = 12

240 ÷ 16 = 15

240 ÷ 8 = 30

240 ÷ 15 = 16

∴ \(\frac{5}{12}=\frac{5 \times 20}{12 \times 20}=\frac{100}{240}\)

\(\frac{17}{20}=\frac{17 \times 12}{20 \times 12}=\frac{204}{240}\) \(\frac{7}{16}=\frac{7 \times 15}{16 \times 15}=\frac{105}{24,0}\) \(\frac{3}{8}=\frac{3 \times 30}{8 \times 30}=\frac{90}{240}\) \(\frac{13}{15}=\frac{13 \times 16}{15 \times 16}=\frac{208}{240}\)

∴ Arranging in descending order of magnitudes, we get,

\(\frac{13}{15}, \frac{17}{20}, \frac{7}{16}, \frac{5}{12}, \frac{3}{8}\)

Question 5. How much money will have to be taken from \(\frac{3}{5}\)th part of RR 175 so that still there will remain RR 45?

Class 6 Maths West Bengal Board Solution :

Given: \(\frac{3}{5}\)th

\(\frac{3}{5}\)th part of RR 175 = \(\)

Since there will remain still RRl 45, the amount of money that will have to be taken from RR 105 is equal to RR (105 – 45)=60.

∴ The required money that will have to be taken = is RR 60.

 

Question 6. If 35 is added to \(\frac{5}{7}\)th of a number, then the sum is 65. Find the number.

Class 6 Maths West Bengal Board Solution:

Given:

35 is added to \(\frac{5}{7}\)th of a number, then the sum is 65.

Since, after adding 35 to \(\frac{5}{7}\)th part of a number, the sum is 65, we have,

\(\frac{5}{7}\)th part of the required number = 65 – 35

= 30.

∴ The required number = 30 ÷ \(\frac{5}{7}\)

= 30 x \(\frac{7}{5}\)

= 42

The required number = 42.

 

Question 7. How much is to be added to \(\frac{7}{25}\) of 4 so that the sum becomes \(2\frac{3}{5}\)?

Class 6 Maths West Bengal Board Solution:

Given:

\(\frac{7}{25}\) of 4 so that the sum becomes \(2\frac{3}{5}\)

\(\frac{7}{25}\) of 4 = \(\frac{28}{25}\)

∴ The required sum

 

Addition And Subtraction Of Fractions:

Question 1. Add: \(\frac{7}{15}+\frac{8}{25}+\frac{11}{45}+\frac{13}{75}\)

Solution:

Given:

\(\frac{7}{15}+\frac{8}{25}+\frac{11}{45}+\frac{13}{75}\)

 

 

∴ L. C. M. of 15, 25, 45 and 75 = 5 x 3 x 5 x 3 = 225

Class 6 Math Solution WBBSE

 

∴ The required sum = \(1\frac{46}{225}\)

 

Question 2. Add: \(7 \frac{3}{8}+5 \frac{7}{12}+2 \frac{11}{18}+3 \frac{13}{45}\)

Class 6 Maths West Bengal Board Solution:

Given:

\(7 \frac{3}{8}+5 \frac{7}{12}+2 \frac{11}{18}+3 \frac{13}{45}\)

 

\(7 \frac{3}{8}+5 \frac{7}{12}+2 \frac{11}{18}+3 \frac{13}{45}\)

 

= \(\frac{59}{8}+\frac{67}{12}+\frac{47}{18}+\frac{148}{45}\)

= \(\frac{2655+2010+940+1184}{360}\)

= \(\frac{6789}{360}\)

= \(18 \frac{309}{360}\)

= \(18 \frac{103}{120}\).

∴ The required sum = \(18 \frac{103}{120}\).

 

 

Question 3. Subtract: \(23 \frac{13}{25}\) – \(12 \frac{19}{35}\)

Solution:

Given:

\(23 \frac{13}{25}\) And \(12 \frac{19}{35}\)

\(23 \frac{13}{25}\) – \(12 \frac{19}{35}\)

= \(\frac{588}{25}\) – \(\frac{439}{35}\)

= \(\frac{4116 – 2195}{175}\)

= \(10 \frac{171}{175}\)

∴ The required result = \(10 \frac{171}{175}\)

 

 

∴ L.C.M. of 25 and 35

= 5 x 5 x 7

= 175.

 

Multiplication And Division Of Fractions:

Question 1. Multiply: 

1. \(\frac{65}{143}\) x 77

Class 6 Maths West Bengal Board Solution:

Given:

\(\frac{65}{143}\) And 77

 

 

∴ The required product = 35.

 

2. \(5 \frac{9}{125}\) x \(\frac{25}{32}\)

Solution:

Given:

\(5 \frac{9}{125}\) And \(\frac{25}{32}\)

\(5 \frac{9}{125}\) x \(\frac{25}{32}\)

 

 

= \(\frac{317}{80}\)

= \(3 \frac{77}{80}\)

∴ The required product = \(3 \frac{77}{80}\)

 

Question 2. Divide:

1. \(7 \frac{9}{13}\) ÷ \(3 \frac{7}{15}\)

Solution:

Given:

\(7 \frac{9}{13}\) And \(3 \frac{7}{15}\)

\(7 \frac{9}{13}\) ÷ \(3 \frac{7}{15}\)

= \(\frac{100}{13}\) ÷ \(\frac{52}{15}\)

 

 

= \(\frac{375}{169}\)

= \(2 \frac{37}{169}\)

∴ The required quotient = \(2 \frac{37}{169}\)

 

2. \(12 \frac{2}{9}\) ÷ \(27 \frac{1}{2}\)

Solution:

Given:

\(12 \frac{2}{9}\) And \(27 \frac{1}{2}\)

\(12 \frac{2}{9}\) ÷ \(27 \frac{1}{2}\)

= \(\frac{110}{9}\) ÷ \(\frac{55}{2}\)

 

 

= \(\frac{4}{9}\)

∴ The required quotient = \(\frac{4}{9}\)

 

 

 

Chapter 1 Simplification Highest Common Factor And Lowest Common Multiple

Chapter 1 Highest Common Factor

Highest Common Factor:

A composite number has two or more factors.

Simplification Questions For Class 6

Definition:

1. The H. C. F. or Highest Common Factor of two integral numbers is the highest (or greatest) among all the possible common factors of the two numbers.
2. We consider two numbers 15 and 20.
3. 15 = 1 x 3 x 5 i.e., the factors of 15 are 1, 3, 5, 15.
4. 20 = 1 x 2 x 2 x 5, i.e., the factors of 20 are 1, 2, 4, 5, 10, 20.
5. ∴ 15 and 20 have common factors 1, and 5.
6. Among them, the highest common factor is 5. 
7. So the H. C. F. of 15 and 20 5. 
8. Again, 36 1 x 2 x 2 x 3 x 3 and 48 = 1 x 2 x 2 x 2 x 2 x 3.
9. ∴ The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36 and the factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
10. ∴ The common factors of 36 and 48 are 1, 2, 3, 4, 6, and 12.
11. Among these factors, the highest common factor is 12.
12. ∴ The H. C. F. of 36 and 48 = 12.
13. There are two processes in which the H. C. F. of two numbers can be determined. These are:
    1. Resolution into prime factors and
    2. Division process.
14. In the illustrative examples below, both processes are described to determine the H. C. F. of two numbers.

Read And Learn More: WBBSE Notes For Class 6 Maths Chapter 1 Simplification

Wbbse Class 6 Maths

Chapter 1 Lowest Common Multiple

A number has an infinite number of multiples.

Two numbers may have infinite numbers of common multiples.

Lowest Common Multiple Definition

1. The lowest of all the common multiples of two numbers is called the L. C. M. or the Lowest Common Multiple of the numbers.
2. For example, we consider the numbers 15 and 25.
3. The multiples of 15 are 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, ………
4. The multiples of 25 are 25, 50, 75, 100, 125, 150,……..
5. Among these multiples, the common multiples are 75, 150,…………..
6. ∴ The lowest common multiple is 75.
7. So the required L. C. M. of 15 and 25 = 75.
8. The L. C. M. of two given numbers can be determined in two processes as follows:
   1. Resolution into prime factors
   2. Division method.
9. In the illustrative examples discussed below, both processes are described to determine the L. C. M. of two numbers.

Formula:

The product of two numbers H. C. F. x L. C. M. of two number

Chapter 1 Simplification Problems On Highest Common Factor And Lowest Common Multiple Of Integers

Question 1. The L. C. M. and H. C. F. of the two numbers are 252 and 6 respectively. What is the product of these two numbers?

Solution:

Given:

The L. C. M. and H. C. F. of the two numbers are 252 and 6 respectively.

 We know that, the product of two numbers = L. C. M. x H. C. F.

 = 252 × 6 

= 1512.

∴ The required product = 1512.

Simplification Questions For Class 6

Question 2. The H. C. F. and L. C. M. of the two numbers are 8 and 280 respectively; if one of them is 56, then what is the other number?

Solution:

Given:

The H. C. F. and L. C. M. of the two numbers are 8 and 280 respectively; if one of them is 56

We know that, the product of two numbers = H. C. F. x L. C. M.

Here the product of two numbers = 8 × 280.

One number = 56. 

∴ The other number = \(\frac{8 x 280}{56}

=40

∴ The required other number = is 40.

Simplification Questions For Class 6

Read And Learn More: WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Solved Problems

Class 6 Math Solution WBBSE

Question 3. The circumference of the front wheel of a rail engine is 140 cm and that of the rear wheel is 350 cm. Find the least distance to be covered so that both wheels complete their full rotation simultaneously.

Solution:

Given:

The circumference of the front wheel of a rail engine is 140 cm and that of the rear wheel is 350 cm.

The L. C. M. of 140 cm and 350 cm will be the required distance.

 

 

∴ 140 = 2 x 2 x 5 x 7

 

∴ 350 = 2 x 5 x 5 x 7

∴ The required L. C. M. of 140 and 350 = 2 x 2 x 5 x 7 x 5

= 700

So the required distance = 700 cm

= 70 dm.

Class 6 Math Solution WBBSE

Question 4.

1. With the help of H. C. F. of 45 and 60, find their L. C. M.

Solution:

Given:

H. C. F. of 45 and 60

 

 

∴ The H.C.F. of 45, 60 = 15

Now, the product of numbers = H.C.F x L.C.M

∴ 45 x 60 = 15 x L.C.M

 

 

 2. With the help of L. C. M. of 105 and 225, find their H. C. F.

Solution:

 

 

∴ 105 = 3 x 5 x 7

 

Class 6 Math Solution WBBSE In English

∴ 225 = 3 x 3 x 5 x 5

∴ L.C. M. of 105 and 225 = 3 x 5 x 7 x 3 x 5 = 1575 

Now, the product of the numbers = H. C. F. x L. C. M.

or, 105 x 225 = H. C. F. x 1575

 

 

∴ The required H.C.F of 150 and 225 = 15.

Question 5. Find the H. C. F. of 24, 33, and 130.

Solution:

Given:

The H. C. F. of 24, 33, and 130

 

∴ 24 = 1 x 2 x 2 x 2 x 3

 

 

∴ 33 = 3 x 11

 

 

∴ 130 = 1 x 2 x 5 x 13

Here 33 and 130 are prime to each other.

The given numbers have no common factor except 1.

Hence the required H. C. F. is 1.

Simplification Questions For Class 6

Question 6. Examine whether 278 and 365 are prime to each other.

Solution:

Given:

278 and 365

Here we shall find the H. C. F. of the given numbers by the method of division

 

 

∴ The H. C. F. of 278 and 365 is 1. Hence 278 and 365 are prime to each other.

Question 7. Find the H. C. F of 906, 1057, and 1510 by the division method.

Solution:

Given:

906, 1057, and 1510

First, we find the H. C. F. of 906 and 1057.

 

 

∴ The H. C. F. of 906 and 1057 is 151.

Now we shall find H. C. F. of 151 and 1510.

 

 

∴ The H. C. F. of 151 and 1510 = 151.

Hence the required H. C. F. of 906, 1057, and 1510 = 151.

Question 8. Find the H. C. F. of 84, 112, 140.

Solution:

Given:

84, 112, 140.

Simplification For Class 6

 

Here we divide all the given numbers by their common prime factor 2 in the 1st row.

The quotients obtained are written in the 2nd row and divided into all the numbers in the 2nd row by their common prime factor 2 and the quotients are written in the 3rd row.

They are also divided by their common factor 7 and the quotients are written in the 4th row.

But the numbers in the 4th row have no common factor except 1 and the process is now completed.

∴ The required H. C. F. = product of the common prime factors

= 2 × 2 × 7 = 28.

The required H. C. F = 28.

Question 9. Find the L. C. M. of 28, 35, 63, 84, 96.

Solution:

Given:

28, 35, 63, 84, 96

 

 

∴ The required L.C.M = 2 x 2 x 3 x 7 x 5 x 3 x 8

= 10080.

The required L.C.M = 10080.

Question 10

1. Find the least number which is exactly divisible by 18, 24, and 42.

Solution:

Given:

18, 24, and 42

Find the least number which is exactly divisible by 18, 24, and 42.

 

 

∴ The L. C. M. of 18, 24, 42 = 2 x 3 x 3 x 4 x7 = 504.

∴ The required least number = 504.

2. Find the least number which is exactly divisible by 18, 39, 56, and 64.

Solution:

Given:

18, 39, 56, and 64

The required least number will be the L. C. M. of 18, 39, 56, and 64. 

Now,

 

Simplification For Class 6

L.C.M. of 18, 39, 56 and 64 = 2 x 2 x 2 x 3 x 3 x 13 x 7 x 8 = 52416.

∴ The required least number = 52416.

Question 11.

The honorable Education Minister of the West Bengal Government has sent some books on Mathematics, Physical Science, and Life Science for class VI of Raghunath Vidyamandir. These books can be arranged in the school library in 20, 24, and 30 rows so that each row contains an equal number of books. How many least number of books has the minister sent?

Solution:

Given:

The honorable Education Minister of the West Bengal Government has sent some books on Mathematics, Physical Science, and Life Science for class VI of Raghunath Vidyamandir. These books can be arranged in the school library in 20, 24, and 30 rows so that each row contains an equal number of books.

Since each row contains an equal number of books the required number of books will be the L. C. M. of 20, 24, 30.

Now,

 

Question 12.

Find the H. C. F. and L. C. M. of 35, 45, and 50. Examine whether the product of the numbers and the product of their H. C. F. and L. C. M. are equal or not.

Solution:

Given:

35, 45, and 50

 

 

∴ H. C. E. of 35, 45, 50 = 5 and their L. C. M. = 5 x 7 x 9 x 10 = 3150. Product of H. C. F. and L. C. M. = 5 x 3150 = 15750.

Again the product of the numbers 35 x 45 x 50 = 78750

∴ Product of the numbers ≠ product of H.C.F. and L.C.M.

Question 13.

The sum of the two numbers is 150 and their H. C. F. is 15. Find the numbers.

Solution:

Since the H. C. F. of two numbers is 15, the numbers must be multiples of 15.

Let the numbers be 15x and 15 y where x and y are primes to each other.

∴ Their sum = 15x + 15y= 15 (x + y)

∴ By the given condition, we get, 15 (x + y) = 150

Dividing both sides by 15, we get, x + y = [latex]\frac{150}{15}\) = 10.

Now,

10 = 1+9=2+8=3+7=4+6=5+5.

Among them, 1, and 9 are prime to each other, and 3, and 7 are also prime to each other. 

So the numbers are either 15 x 1 = 15; 15 x 9 = 135

or, 15 x 3 = 45; 15 x 7 = 105 . 

∴ The required numbers are either 15, 135, or 45, 105.

Question 14.

If the H. C. F. and L. C. M. of two numbers be 5 and 75 respectively, then find the numbers.

Solution:

Since the H. C. F. of two numbers is 5, let the numbers be 5x and 5y, where x and y are primes to each other.

Then L. C. M. of 5x and 5y = 5xy. But the L. C. M. is 75.

∴ 5xy = 75.

Now dividing both sides by 5, we get, xy= 15.

But 151 x 15 = 3 x 5. Both the sets of numbers 1, 15, and 3, 5 are prime to each other.

∴ The numbers are either 5 x 1, 5 x 15 = 5, 75.

or, 5 x 3, 5 x 5 = 15, 25.

∴ The required numbers are either 5, 75

or 15, 25.

Chapter 1 Simplification Highest Common Factor And Lowest Common Multiple Of Integers

Question 1. 

1. Write down 2 numbers of two digits which are multiples of 4

Solution:

Two multiples of 4 of two digits are 12 and 16, because 4 x 3 = 12 and 4 x 4 = 16.

 

2. Write 6 multiples of 5 except 0;

Solution:

6 multiples of 5 except 0 are 5 x 15, 5 x 2 = 10, 5 x 3 = 15, 5 x 4 = 20, 5 x 5 25 and 5 x 6 = 30.

 

3. Write two numbers whose L. C. M. is 12 and whose sum is 10;

Solution:

Two numbers are 4 and 6; because of the L. C. M. of 4 and 6 = 12 and whose sum 4+6= 10.

(Here 4 2 x 2 and 6 = 2 x 3. .. L. C. M.2 x 2 x 3 = 12).

Read And Learn More: WBBSE Solutions For Class 6 Maths Chapter 1 Simplification Solved Problems

4. Write three numbers each of which has a factor of 4

Solution:

The required 3 numbers are 8, 12, and 16 because each of these numbers has a factor of 4.

 

5. Write three multiples of 7 greater than 50.

Solution:

Three multiples of 7 greater than 50 are 56, 63, and 70 because 56 > 50; 63 50; 70 > 50

56 = 7 x 8,

63 = 7 x 9, 

70 = 7 x 10.

 

Question 2.

1. Which is the smallest even prime number?

Solution:

The smallest even prime number is 2.

 

2. Which is the smallest odd prime number?

Solution:

The smallest odd prime number is 3.

 

3. What are the prime factors of 14?

Solution:

We know that 14= 1 x 2 x 7.

The factors of 14 are 1, 2, 7, and 14.

The prime factors of 14 are 2 and 7.

 

Question 3. 42 is the multiple of which of the following numbers? 

1. 5

2. 6

3. 7

4. 13.

Solution:

 42 is divisible by both the numbers 6 and 7 but 42 is not divisible by 5 or 13.

So, 42 is the multiple of 6 and 7.

42 is not a multiple of 5. and 13.

 

Question 4. 11 is a factor of which of the following numbers?

1. 101

2. 111

3. 121

4. 112.

Solution: We know that 121 = 11 x 11.

∴ 11 is a factor of 121. 

Again the numbers 101, 111, and 112 are not divisible by 11.

So 11 is not a factor of 101, 111, or 112.

 

Question 5.

Find the H.C.F. by the resolution into prime factors of the numbers in each of the following cases:

1. 22, 44

Solution:

22 = 1 x 2 x 11

 

44 = 1 x 2 x 2 x 11

 

 

The factors of 22 are 1, 2, 11, 22 and that of 44 are 1, 2, 4, 11, 22, 44

 

 

∴ The common factors of 22 and 44 are 1, 2, 11, and 22.

Among these factors, the highest factor is 22

∴ The required H. C. F. = 22.

 

2. 54, 72.

 

∴ So the factors of 54 are 1, 2, 3, 6,9, 18, 27, and 54.

 

 

 

 

∴ So the factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

∴ The common factors of 54 and 72 are 1, 2, 3, 6, 9, and 18.

Among these factors, the highest factor is 18.

∴ The required H. C. F. = 18.

 

Question 6. Find the H. C. F. of 36 and 48 by resolution into factors.

Solution:

 

∴ 36 = 2 x 2 x 3 x 3

∴ 48 = 2 x 2 x 2 x 2 x 3

 

Now

 

 

 

∴ The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.

The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48.

The common factors of 36 and 48 are 1, 2, 3, 4, 6, and 12.

Among these factors, the highest factor is 12.

∴ The required H. C. F. = 12.

 

Alternative Method:

 

 

 

Here

markings are common to both 36 and 48.

∴ The required H.C.F = 1 x 2 x 2x 3 =12.

 

Question 7. Find the H. C. F. of 75 and 105 by factorization.

Solution:

 

 

 

 

 

Here

marking is common to both 75 and 105.

∴ the required H.C.F = 1 x 3 x 5 = 15.

 

Question 8. Write two numbers whose H. C. F. is 7.

 Solution : 7 x 2 = 14; 7 x 3 = 21

2 and 3 are prime to each other, and the H. C. F. of 14 and 21 is 7.

The required two numbers are 14 and 21.

N.B. We can get an infinite number of such two numbers.

 

Question 9. By division method find the H. C. F. in the following cases:

1. 28, 35

Solution: 

 

 

 

∴ the required H.C.F. = 7

 

2. 54, 72

 

 

∴ The required H.C.f. = 18.

 

Question 10. By division method, find the H. C. F. of 90 and 144.

Solution:

 

 

∴ The required H.C.F. = 18.

 

Question 11. What will be the greatest number by which 45 and 60 will be exactly divisible so that there will be no remainder in each case?

Solution:

The required greatest number will be the H. C. F. of 45 and 60.

Now,

 

 

∴ The H. C. F. of 45 and 60 = 15. So the required greatest number is 15.

 

Question 12.

1. The H. C. F. of two numbers is 1. How many such two numbers may exist? Find one pair of such numbers. What conclusion can you draw about such two numbers?

Solution:

There are infinite numbers of two numbers that have the H. C. F. 1. For example 2, 3; 5, 7; 9, 11, etc.

The required numbers are 9 and 11.

The numbers which have the H. C. F. 1 are prime to each other.

 

2. How many greatest number of persons can be distributed equally two kinds of sweets: 48 sweets of one kind and 64 sweets of another kind, without breaking the sweets?

Solution:

The required number of persons will be the H. C. F. of 48 and 64.

Now,

 

 

 

H.C. F. of 48 and 64 = 16

So the required greatest number of people is 16.

 

Question 13. Find the L. C. M. by factorization in each of the following cases:

 1. 25, 80

Solution:

 

 

 

∴ 25 = 1 x 5 x 5

 

∴ The required L.C.M = 1 x 5 x 5 x 2 x 2 x 2 x 2

= 400.

 

2. 36, 39

 

 

∴ 36 = 2 x 2 x 3 x 3

 

 

∴ 39 = 3 x 13

∴ The required L. C. M. = 2 x 2 x 3 x 3 x 13

= 468

 

Question 14. Find the L. C. M. by prime factors in each of the following cases:

1. 33, 132

Solution:

 

 

∴ 33 = 3 x 11

 

 

∴ 132 = 2 x 2 x 3 x 11

∴ The required L. C. M. = 3 x 11 x 2 x 2 = 132.

 

2. 90, 144

 

 

∴ 90 = 2 x 3 x 3 x 5

 

 

∴ 144 = 2 x 2 x 2 x 2 x 3

∴ The required L. C. M. = 2 x 3 x 3 x 5 x 2 x 2 x 2 = 720.