## Arithmetic Chapter 5 Multiplication And Division Of A Fraction By Whole Number And By Fraction

**Question 1. Multiply:**

**1. \(3 \times \frac{6}{11}\)**

**Solution:**

**2. \(\frac{2}{3} \times 11\)**

**Solution:**

**3. \(\frac{7}{3} \times 2 \frac{2}{3}\)**

**Solution:**

**4. \(\frac{3}{8} \times \frac{6}{4}\)**

**Solution:**

**Read and Learn More WBBSE Solutions For Class 6 Maths**

**5. \(\frac{6}{49} \times \frac{7}{3}\)**

**Solution:**

**6. \(\frac{15}{28} \times 2 \frac{1}{3}\)**

**Solution:**

**7. \(4 \frac{8}{13} \times 7 \frac{4}{5}\)**

**Solution:**

**8. \(2 \frac{3}{5} \times 6\)**

**Solution:**

**Question 2. Multiply:**

**1. \(\frac{6}{7} \times \frac{14}{15} \times \frac{25}{28}\)**

**Solution:**

**2. \(2 \frac{11}{12} \times 3 \frac{1}{7} \times 4 \frac{10}{11}\)**

**Solution:**

**3. \(\frac{12}{35} \times \frac{42}{91} \times \frac{36}{55} \times \frac{13}{18}\)**

**Solution:**

**4. \(1 \frac{7}{10} \times 1 \frac{3}{17} \times 1 \frac{1}{5} \times 3 \frac{3}{4}\)**

**Solution:**

**Question 3. Multiply:**

**Solution:**

**1. \(99 \frac{98}{99} \times 99\)**

**Solution:**

**2. \(999 \frac{97}{98} \times 98\)**

**Solution:**

**3. \(999 \frac{995^{\circ}}{997} \times 997\)**

**Solution:**

**4. \(999 \frac{444}{999} \times 999\)**

**Solution:**

**Question 4 Simplify:**

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

**Solution:**

**2. \(\frac{1}{5}+\frac{1}{2}-\frac{2}{15}-\frac{1}{6}\)**

**Solution:**

**3. \(\frac{7}{12}+5 \frac{2}{9}+\frac{11}{18}-2 \frac{5}{12}\)**

**Solution:**

**4. \(3 \frac{1}{9}+\frac{7}{6} \times \frac{3}{8}-\frac{5}{24}\)**

**Solution:**

**5. \(6 \frac{2}{5}+3 \frac{1}{3}+\frac{1}{2}-\frac{7}{10}\)**

**Solution:**

**6. \(1-\left[\frac{1}{2} \div\left\{2-\frac{1}{2}\left(\frac{1}{2}-\overline{\frac{1}{3}-\frac{1}{6}}\right)\right\}\right]\)**

**Solution:**

**Question 5. Evaluate:**

**1. 1/5 part of ₹ 10**

**Solution:**

**2. 1/5 part of ₹ 25**

**Solution:**

**Question 6.**

**1. 1/3 part of how many rupees is ₹ 4?**

**Solution:**

Let the whole of the money be 1.

∴ 1/3 part of the whole of the money = ₹ 4

∴ whole of the money = ₹(4 ÷ 1/3)

= ₹ (4 x 3/1)

= ₹ 12.

**Alternative Method:**

Let the required money be ₹ x

Then 1/3 part of ₹ x = ₹ 4

∴ 1/3 x x = 4

or, x/3 = 4

or, x = 4 x 3

= 12

∴ The required money = ₹ 12.

**2. 1/6 part of how many minutes are 6 minutes?**

**Solution:**

Here 1/6 part of the total minutes = 6 minutes

∴ Total minutes = (6 ÷ 1/6)

= 6 x 6/1

= 36

∴ The required time = is 36 minutes.

**Question 7.** **You had taken 1/3 part of the mangoes from the basket of mangoes of ****Sakuntala and then you had only 7 mangoes. How many mangoes were there in the basket of Sakuntala originally?**

** Solution**: It is clear that 1/3 part of sakuntala’s mangoes in the basket = 7

Total number of mangoes in the basket of Sakuntala

= 7 ÷ 1/3

= 7 x 3/1

= 21

∴ Sakluntala had 21 mangoes originally in her basket.

**Question 8. How much money is to be taken from 1/2 part of ₹ 150 so that there will remain only ₹ 30?**

**Solution:**

1/2 part of ₹ 150

= ₹ (1/2 x 150)

= ₹ 75

We have to find the amount of money that can be taken from ‘? 75 so that there will remain ₹ 30.

∴ The required amount of money = ₹ (75 – 30)

= ₹ 45.

∴ ₹ 45 is to be taken from 1/2 part of ₹ 150 so that there will remain ₹ 30.

**Question 9. What is to be added to 3 times of 6/7 to get 2 6/7?**

**Solution: **3 times of 6/7

= 3 x 6/7

= 18/7

2 6/7 = 20/7

We have to find the value which is to be added to 18/7 so that the sum will be 2 6/7 or 20/7.

∴ The required value = 20/7 – 18/7

= 2/7

2/7 is to be added to 3 times of 6/7 to get 2 6/7

**Question 10. 1400 visitors came in the first year in an occasion of the town. The number of visitors in the next year increased by 7/10 part of the first year. What was the total number of visitors in the next year?**

**Solution:**

Total number of visitors in the first year = 1400.

∴ 7/10 part of 1400

= 7/10 x 1400

= 980

So the number of visitors increases in the next year = 980 than the first year.

∴ Total number of visitors in the next year = 1400 + 980

= 2380.

The total number of visitors in the next year = 2380.

**Question 11. You had no stamps and Madhabi gave you 2/3 part of the stamps as many stamps had with her at first and as a result of which you had only 18 stamps. How many stamps had Madhabi with her at first?**

**Solution:**

You had only 18 stamps and Madhabi gave you – parts of her stamps.

∴ 2/3 parts of the stamps of Madhabi =18

So Madhabi’s total number of stamps = (18 ÷ 2/3)

Madhabi had 27 stamps with her.

**Question 12. Ram gave 2/5 part of his money to Shyam and 3/10 part of his ****money to Jadu. If he has ₹ 180 left with him, how much money Ram had at the beginning?**

** Solution**: Let the total money of Ram = be 1 part.

∴ Ram gave Shyam 2/5 part of his money and 3/10 part of his money to Jadu

∴ Shyam and Jadu together received (2/5 + 3/10)

= 4+3 / 10

= 7/10 part of Ram’s total money.

∴ The remaining part of ram’s money after giving to Shyam and jadu

= (1 – 7/10)

= 10-7 / 10 part

= 3/10 part

∴ 3/10 part of Ram’s total money = ₹ 180

∴ Ram’s total money what had at the beginning

= ₹ (180 ÷ 3/10)

= ₹ (180 x 10/3)

= ₹ 600

₹ 600 Ram had at the beginning.

**Question 13. Your father brought 10 liters of drinking water from the nearest tubewell and from it, your mother used 1/5 part of the water for cooking. 1/4 part of the remaining water had been used for drinking purposes. How much water still was left?**

Solution Total amount of drinking water brought = 10 liters.

Let the total amount of water = 1 part

Mother used for cooking = 1/5 part

Remaining part = 1 – 1/5

= 5-1 / 5

= 4/5

Water used for drinking = 1- 1/5

= 5-1 / 5

= 4/5

Water used for drinking = (4/5 x 1/4) part ⅕ part

Remaining part of total water = 4/5 – 4/5

= 4-1 /5

= 3/5 part

∴ Remaining water = (3/5 x 10) = 6 liters

∴ Still, 6 liters of water was left.

**Question 14. A bucket holds 1/2 ****liter of water. How much water 7 such buckets** **can hold?**

**Solution: One bucket can hold = **1/2 liter of water

∴ 7 buckets can hold = (1/2 x7)

= 3.5 liters of water

∴ 7 buckets can hold 3.5 liters of water.

**Question 15. If the length of the 7/8 part of a ribbon is 56 meters, then what is the ****length of the original whole ribbon?**

**Solution: Let the length of the original whole ribbon = 1 part.**

∴ The length of 7/8 part of the ribbon = 56 meters

∴ The length of 1 part of the ribbon = (56 ÷ 7/8) metres

= (56 x 8/7 ) metres

= 64 meters.

∴ The length of the original whole ribbon = is 64 meters.

**Question 16. What is to be added to 5/7 of 1 1/2 to get the sum 4 3/5?**

**Solution:**

5/7 of 1 1/2 = 5/7 x 1 1/2

= 5/7 x 3/2

= 15/14

We have to find what is to be added to 15/14 to get the sum 4 3/5.

∴ Required value = (4 3/5 – 15/14)

= 23/5 – 15/14

= 322 – 75 / 70

= 247/70

= 3 37/70

3 37/70 is to be added to 5/7 of 1 1/2 to get the sum 4 3/5

## Arithmetic Chapter 5 Multiplication And Division Of A Fraction By Whole Number And By Fraction Reciprocal Number

**Question 1. Write the meaning of each of the following mathematical processes:**

**Solution:**

**1. 55 ÷ 11**

**Solution:**

The meaning of (55 ÷ 11) is that how many times 11 is subtracted from 55?

2. 2 ÷ 1/4

**Solution:**

By ( 2 ÷ 1/4) we mean that how many times 1/4 is subtracted from 2?

3. 8 1/2 ÷ 2

**Solution:**

By (8 1/2** ÷ **2) we mean that how many times 2 is subtracted from 8 1/2?

4. 6 1/3 ÷ 1 1/3

**Solution:**

By (6 1/3 ** ÷ **1 1/3 ) we mean that how many times 1 1/3 is subtracted from 6 1/3?

**Question 2. Determine how many times to be subtracted in each case:**

**1. 11 from 77**

**Solution:**

Here 77 ÷ 11

= 7.

∴ 11 can be subtracted 7 times from 77.

2. 1/3 from 3;

**Solution:**

Here 3 ÷ 1/3

= 3 × 3/1

= 9.

∴ can be subtracted 9 times from 3.

3. 1/2 from 8;

**Solution:**

8 ÷ 1

= 8 x 2/1

= 16.

∴ can be subtracted 16 times from 8.

4. 1 from 10 1/2

**Solution:**

10 1/2 ÷ 1 1/2

= 21/2 x 2/3

= 7.

∴ 1 1/2 can be subtracted 7 times from 10 1/2.

**Question 3. ****Determine how many integral times can be subtracted and then what is the remainder in each case:**

**1. 13 2/3 ****from 119 3/4 ****;**

**Solution:**

Here 119 3/4 ÷ 13 2/3

= 479/4 ÷ 41/3

= 479/4 x 3/41

= 1437/164

= 8 125/164.

Now the fraction part is 125/164.

∴ 125/164 x 13 2/3

= 125/25

= 10 5/12

∴ 13 2/3 can be subtracted 8 full times and then the remainder is 10 5/12.

**2. 10 1/4 from 181 1/3;**

**Solution:**

Here 181 1/3 ÷ 10 1/4

= 544/3 ÷ 4/41

= 2176/123

= 17 85/123

The fractional part = 85/123

∴ 85/123 x 10 1/4

= 85/12

= 7 1/2

∴ 10 1/4 can be subtracted 17 full times and the remainder be 7 1/12.

**Question 4. Divide:**

**1. 15 ÷ 5/3**

**Solution:**

15 ÷ 5/3 = 9

**2. 14 ÷ 7/2**

**Solution:**

=4

14 ÷ 7/2 =4

3. 6/13 ÷ 1 1/5

**Solution:**

= 2/13

6/13 ÷ 1 1/5 = 2/13

4. 12/19 ÷ 6

**Solution:**

= 2/19

12/19 ÷ 6 = 2/19

**5. 5 1/5 ÷ 13/2**

**Solution:**

= 4/5

5 1/5 ÷ 13/2 = 4/5

**6. 2 2/5 ÷ 1 1/5**

**Solution:**

= 12/5 ÷ 6/5

= 2

2 2/5 ÷ 1 1/5 = 2

7. 4 3/7 ÷ 3 2/7

**Solution:**

= 31/7 ÷ 23/7

= 31/7 x 7/23

= 31/23

= 1 8/23

4 3/7 ÷ 3 2/7 = 1 8/23

**Question 5: Find the reciprocal of the following number:**

**1. 7/5;**

**Solution:**

The reciprocal of 7/5 is 5/7;

2. 1/3;

**Solution:**

The reciprocal of 1/3 is 3/1

= 3;

3. 5/8;

**Solution:**

The reciprocal of 5/8 is 8/5

= 1 3/5;

4. 9/7;

**Solution:**

The reciprocal of 9/7 is 7/9;

5. 12/5;

**Solution:**

The reciprocal of 12/5 is 5/12;

6. 7/18;

**Solution:**

The reciprocal of 7/18 is 18/7

= 2 4/7;

**7. 1/8;**

**Solution:**

The reciprocal of 1/8 is 8/1

= 8;

**8. 5 6/7;**

**Solution:**

The reciprocal of 5 6/7 or the reciprocal of 41/7 is 7/41;

**9. 10 1/3;**

**Solution:**

The reciprocal of 10 1/3 or the reciprocal of 31/3 is 3/31;

**10. 14; **

**Solution:**

The reciprocal of 14 is 1/14.

**Question 6. Simplify:**

**1. 3/8 ÷ 2/3 of 1/9 of 1/16;**

**2. \(\left\{\frac{11}{16} \div\left(\frac{5}{6}+\frac{2}{3}\right)\right\}-\frac{1}{3}\)**

**3. 4 2/3 ÷ 2/3 – 3/8;**

**4. (2 3/4 + 31/2 ÷ 2 1/7) ÷ 13 1/4;**

**5. 2 – 1/10 x 1/3 ÷ 4/25 ÷ 1/8**

**6. \(1 \frac{1}{2}\left[3 \frac{1}{2} \div 2 \frac{1}{3}\left\{1 \frac{1}{4} \div\left(2+3 \frac{2}{3}\right)\right\}\right]\)**

**7. (1 1/13 x 2 3/5) ÷ (7 1/2 x 3 1/10) ÷ 28/279**

**8. \(2 \frac{1}{2}+\frac{2}{3}\left[4 \frac{1}{2}+3\left\{7 \div 4 \frac{2}{3}\right\}\left(8 \frac{1}{2}+\overline{4+7 \frac{1}{3}}\right)\right]\)**

**Solution:**

**1. 3/8 ÷ 2/3 of 1/9 of 1/16**

**Solution:**

= 3/8 ÷ 1/216

= 81

**2. \(\left\{\frac{11}{16} \div\left(\frac{5}{6}+\frac{2}{3}\right)\right\}-\frac{1}{3}\)**

**Solution:**

**3. 4 2/3 ÷ 2/3 – 3/8**

**Solution:**

= 14/3 ÷ 2/3 -3/8

= 7 – 3/8

= 56-3 / 8

= 53/8

= 6 5/8

4 2/3 ÷ 2/3 – 3/8 = 6 5/8

**4. (2 3/4 + 3 1/2 ÷ 2 1/7) ÷ 13 1/4**

**Solution:**

= (11/4 + 7/2 ÷ 15/7) ÷ 53/4

= (11/4 + 7/2 x 7/15) ÷ 53/4

= (11/4 + 49/30) ÷ 53/4

= (165 + 98 / 60) ÷ 53/4

= 263/795

(2 3/4 + 3 1/2 ÷ 2 1/7) ÷ 13 1/4 = 263/795

**5. 2 – 1/10 x 1/3 ÷ 4/25 ÷ 1/8**

**Solution:**

= 2 – 1/10 x (1/3 x 25/4) ÷ 1/8

= 2 – 1/10 x 25/12 ÷ 1/8

= 2 – 5/3

= 6-5 / 3

= 1/3.

2 – 1/10 x 1/3 ÷ 4/25 ÷ 1/8 = 1/3.

**6. \(1 \frac{1}{2}\left[3 \frac{1}{2} \div 2 \frac{1}{3}\left\{1 \frac{1}{4} \div\left(2+3 \frac{2}{3}\right)\right\}\right]\)**

**Solution:**

= 1 1/5[3 1/2 ÷ 2 1/3{1 1/4 ÷ (2 + 3 2/3)}]

= 3/2[7/2 ÷ 7/3{5/4 ÷(2 + 11/3)}]

= 3/2[7/2 ÷ 7/3{5/4 ÷ (6+11 / 3)}]

= 3/2[7/2 ÷ 7/3{5/4 ÷ 17/3}]

= 3/2[7/2 ÷ 7/3{5/4 x 3/17}]

= 3/2[7/2 ÷ 7/3 of 15/68]

= 3/2[7/2 ÷ 35/68]

= 51/5

= 10 1/5.

**7. (1 1/13 x 2 3/5) ÷ (7 1/2 x 3 1/10) ÷ 28/279**

**Solution:**

**8. \(2 \frac{1}{2}+\frac{2}{3}\left[4 \frac{1}{2}+3\left\{7 \div 4 \frac{2}{3}\right\}\left(8 \frac{1}{2}+\overline{4+7 \frac{1}{3}}\right)\right]\).**

**Question 7. How many times 1/16 are there in 3/4?**

**Solution:**

Here 3/4 ÷ 1/16

= 12.

∴ 1/16 lies 12 rimes in 3/4

**Question 8. From 16 2/3 meters long ribbons, 3/8 part of it is cut off. The cutout portion of the ribbon is further divided into 5 equal pieces, then what is the length of each piece?**

**Solution:**

The length of the cutout part of the ribbon = (3/8 of 16 2/3) metres

= (3/8 x 50/3) metres

= 25/4 metres

It is divided into 5 equal pieces.

∴ The length of each pieces = (25/4 ÷ 5) metres

= 5/4 metres

= 1 1/4 metres

The length of each piece = 1 1/4 metres

**Question 9. debarred bought 12 7/10 meters of cloth for window curtains. There were already 5 3/5 meters of cloth for curtains at home. If 4 5/6 meters of cloth is required to make curtains for each of the 3 windows. What length of cloth will remain?**

**Solution:**

Total length of cloth for curtain = (12 7/10 + 5 3/5) metres

= (127/10 + 28/5) metres

= (127 + 56 / 10) meters

= 183/10 meters

The total length of cloth which is used for curtains of 3 windows

= (3 x 4 5/6) metres

= (3 x 29/6)metres

= 29/2 metres

∴ The length of the remaining cloth = 3 4/5 meters

∴ 3 4/5 meters of cloth will remain.

**Question 10.**** Paromita prepared some pickles and 4/7 ****part of the pickle was put ****in a glass jar. The rest of the pickle was divided equally among 6 boys and girls. ****How many parts of the whole of the pickle did each get?**

**Solution:**

Let the whole of the pickle that paramita prepared = 1 part.

She put in a glass jar = 4/7 part.

∴ After putting the pickle in the glass jar, the remaining part of the pickle

= (1 – 4/7)

= 3/7 part

This part was divided among 6 boys and girls equally.

∴ Each got = (3/7 ÷ 6) part

= 1/14 part.

So each of the boys and girls got 1/14 part of the whole of the pickle.

**Question 11. Rahim and his group have decided that they will construct 24 11/****15 ****km of the road in 33 days. They have constructed 11/15 km of road each day for 25 days. If they are to finish the work in due time, at what rate will they work for the remaining work?**

**Solution:** Rahim and his group have constructed 11/15 km of road each day.

.. In 25 days they already have constructed The rest of the road to be still constructed

11/15 x 25 = 55/3 km of the road.

= (24 11/15 – 55/3)km

= (371/15 – 55/3)km

= 371 – 275 / 15 km

= 96/15 km.

This 96/15 km of the road is to be constructed in the remaining (33 – 25)

= 8 days.

∴ They are to construct each day at the rate of (96/15 ÷ 8)

= 4/5 km of the road

∴ To finish the construction work of the road in due time the rate of work done by them for the remaining days = 4/5 km.

**Question 12.** **5 is added to 3/7 ****and the sum is multiplied by 4 2/3. ****Now the** **product is divided by 4 4/9 and the quotient is subtracted from 8 2/5. Find the ****result of the subtraction (write this in mathematical language and then find the number after subtraction).**

**Solution:**

Writing in Mathematical Language, we get, the result of subtraction

∴ Simplification

∴ The required number after subtraction = 2 7/10.

**Question 13. After retirement, Debkumarbabu donated 1/4 ****part of his property** **to the local library. He gave 1/6 ****part of the remaining property to his wife and the rest of the property was divided equally between his two sons. What part of the whole of the property was given to his wife and to each of his two sons? **

**Solution: **Let the whole of the property of Debkumarbabu = 1 part.

He donated 1/4 part of his property to the local library.

.. The remaining part of the property after donation to the local library

= (1 – 1/4) part

= 3/4 part

.. His wife received = (1/6 of 3/4 )part

= 1/8 part of the property.

Remaining property = 3/4 – 1/8

= 6-1 / 8

= 5/8 part

This part was divided between the two sons equally.5

∴ Each son received = (5/8 ÷ 2) part

= (5/8 x 1/2) part

= 5/16 part.

∴ Debkumarbabu gave 1/8 part to his wife and 5/16 part of his property to each son.

**Question 14. ****Ajoy has bought (5/7 of 14/25) ****part of a property and he has sold 1/2 ****part of his property for ₹10000. What is the value of the whole property?**

**Solution: **Let the whole property = 1 part.

∴ The property of Ajoy = (5/7 of 14/25) part

= 2/5 part of the whole property.

Again Ajoy sold 1/2 part of his property.

∴ Ajoy sold (1/2 of 2/5) part

= 1/5 part of the whole property.

∴ The value of 1/5 part of the whole property = ₹ 10,000

∴ The value of the whole property = ₹(10000 ÷ 1/5)

= ₹(10000 x 5/1)

= ₹ 50000

∴ The value of the whole property = ₹ 50000.

**Question 15. The distance of Surbabu’s house from the station is 14 km. He ****traveled ****1/8 ****part of the distance on foot and 11/16 part of the distance by bus. The ****rest of the distance was traveled by him by Autorickshaw. What distance did he travel by autorickshaw?**

Solution: Let the total distance of Surbabu’s house to the station = be 1 part.

∴ The part of the total distance traveled by bus and on foot = (1/8 + 11/16) part

= 13/16 part

∴ Remaining part of the distance = 1 – 13/16

= 16-13/16 part

= 3/16 part

This part of the distance was traveled by autorickshaw. But total distance = 14 2/3 km.

∴ Surbabu traveled by autorickshaw = (14 2/3 x 3/16)km.

= 11/4 km

= 2 3/4 km.

2 3/4 km distance that he travel by autorickshaw.