## 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.