### Ex 1.1 Class 6 Maths Question 1.

Fill in the blanks:**(a)** 1 lakh = ………….. ten thousand.**(b)** 1 million = …………… hundred thousand.**(c)** 1 crore = ………………. ten lakh.**(d)** 1 crore = ……………. million.**(e)** 1 million = ……………. lakh.

#### Solution:

**(a)** 10**(b)** 10**(c)** 10**(d)** 10**(e)** 10

### Ex 1.1 Class 6 Maths Question 2.

Place commas correctly and write the numerals:**(a)** Seventy three lakh seventy five thousand three hundred seven.**(b)** Nine crore five lakh forty one.**(c)** Seven crore fifty two lakh twenty one thousand three hundred two.**(d)** Fifty eight million four hundred twenty three thousand two hundred two.**(e)** Twenty three lakh thirty thousand ten.

#### Solution:

**(a)** 73,75,307**(b)** 9,05,00,041**(c)** 7,52,21,302**(d)** 58,423,202**(e)** 23,30,010

### Ex 1.1 Class 6 Maths Question 3.

Insert commas suitably and write the names according to Indian System of Numeration:**(a)** 87595762**(b)** 8546283**(c)** 99900046**(d)** 98432701

#### Solution:

Inserting the commas according to Indian System of Numeration and also writing the number names: .**(a)** 8,75,95,762: Eight crore, seventy five lakh, ninety five thousand, seven hundred sixty two.**(b)** 85,46,283: Eighty five lakh, forty six thousand, two hundred eighty three.**(c)** 9,99,00,046: Nine crore, ninety nine lakh, forty six.**(d)** 9,84,32,701: Nine crore, eighty four lakh, thirty two thousand, seven hundred one.

### Ex 1.1 Class 6 Maths Question 4.

Insert commas suitably and write the names according to International System of Numeration:**(a)** 78921092**(b)** 7452283**(c)** 99985102**(d)** 48049831

#### Solution:

Inserting the commas according to International System of Numeration and also writing the number names:**(a)** 78,921,092: Seventy eight million, nine hundred twenty one thousand, ninety two.**(b)** 7,452,283: Seven million, four hundred fifty two thousand, two hundred eighty three.**(c)** 99,985,102: Ninety nine million, nine hundred eighty five thousand, one hundred two.**(d)** 48,049,831: Forty eight million, forty nine thousand, eight hundred thirty one.

## NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.2

### Ex 1.2 Class 6 Maths Question 1.

A book: exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third andfihpkday was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.

#### Solution:

Total number of tickets sold on four days = Number of tickets sold on the first day + Number of tickets sold on the second day + Number of tickets sold on the third day + Number of tickets sold on the fourth day

= 1094 +1812+2050 +2751

Now,

∴ Total number of tickets sold on four days is 7707.

### Ex 1.2 Class 6 Maths Question 2.

Shekhar is a famous cricket player. He has so far scored 6980 runs iii test matches. He wishes to complete 10,000 runs. How many more runs does he need?

#### Solution:

Shekhar wishes to complete 10000 runs.

Shekhar has scored 6980 runs so far.

Runs needed to score 10000 runs = 10000 – 6980

Now,

∴ 3020 runs are needed to complete 10000 runs.

### Ex 1.2 Class 6 Maths Question 3.

In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?

#### Solution:

Votes registered by the successful candidate = 577500

Votes registered by the nearest rival = 348700

Margin by which the successful candidate won= 577500 – 348700

Now,

∴The margin by which the successful candidate won is 228800.

### Ex 1.2 Class 6 Maths Question 4.

Kirti bookstore sold books worth ₹ 2,85,891 in the first week of June and books worth ₹ 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?

#### Solution:

Price of books sold in the first week = ₹ 285891 .

Price of books sold in the second week = ₹ 400768

Price of books sold in two weeks = ₹ 285891 + ₹ 400768

Now,

∴ Sale in two weeks is of ₹ 686659

Since 400768 > 285891

∴ Sale in the second week is greater.

Difference in the sales amount = ₹ 400768 – ₹ 285891

Now,

∴ Sale in the second week is more by ₹ 114877.

### Ex 1.2 Class 6 Maths Question 5.

Find the difference between the greatest and the least number that can be written using thedigits 6, 2, 7, 4,3 each only once.

#### Solution:

Using the digits 6, 2, 7, 4 and 3, we get

Greatest number = 76432 and, smallest number = 23467

Their difference = 76432 – 23467

Now,

∴ Difference between the greatest and least dumber is 52965.

### Ex 1.2 Class 6 Maths Question 6.

A machine, on an average, manufactures 2,825 Screws a day. How many screws did it produce in the month of Junuary 2006?

#### Solution:

Number of screws manufactured by the machine in a day = 2825

Number of screws manufactured by the machine in January 2006 (i.e., in

31 days) = 2825 x 31

Now,

∴Number of screws manufactured in the month of January 2006 = 87575.

### Ex 1.2 Class 6 Maths Question 7.

A merchant had ₹ 78,592 with her. She placed an order for purchasing 40 radio sets at ? 1200 each. How much money will remain with her after the purchase?

#### Solution:

Total money with the merchant = ₹ 78592

Cost of 1 radio set = ₹ 1200

Cost of 40 radio sets = ₹ (1200 x 40)

Now,

∴ Cost of 40 radio sets = ₹ 48000

∴Money left with the merchant after purchase of radio sets

Now,

∴ Money left after the purchase = ₹ 30592

### Ex 1.2 Class 6 Maths Question 8.

A student multiplied 7236 by 65 instead of multiplying by 56. By . how much was his answer greater than the correct answer?

#### Solution:

Required difference = 7236 x 65 – 7236 x 56 = 7236 x (65 – 56)

= 7236 x 9 = 65124

### Ex 1.2 Class 6 Maths Question 9.

To stitch a shirt, 2 in 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will

remain?

#### Solution:

Total cloth = 40 m = 40 x 100 cm

= 4000 cm

Cloth needed for 1 shirt = 2 m 15 cm

= 2 x 100 cm + 15 cm = 215 cm

Number of shirts stitched out of total cloth = 4000 ÷ 215

Now,

∴ 18 shirts can be stitched and cloth left over is 130 cm i.e., 1 m 30 cm.

### Ex 1.2 Class 6 Maths Question 10.

Medicine is packed in boxes, each weighing 4 kg 500g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?

#### Solution:

Van can carry a weight of 800 kg i.e., 800000 g

Weight of one packet = 4 kg 500 g = 4500 g

Number of packets that can be loaded in the van = 800000 + 4500

Now,

∴177 packets can be loaded in the van.

### Ex 1.2 Class 6 Maths Question 11.

The distance between the school and the house of a student is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.

#### Solution:

Distance of school from house = 1 km 875 m

Distance walked by the student both ways between school and home

= 1 km 875m x 2

Now,

∴ Distance walked in 1 day = 3 km 750 m

Distance walked in 6 days = 3 km 750 m x 6

Now,

∴ Distance walked in 6 days = 22km 500 m.

### Ex 1.2 Class 6 Maths Question 12.

A vessel has 4 litres and 500 ml of curd. In how-many glasses, each of 25 ml capacity, can it be filled?

#### Solution:

Capacity of vessel = 4l 500 ml

= 4 x 1000 ml + 500 ml = 4500 ml

Capacity of one glass = 25 ml

∴ Number of glasses of curd filled out of vessel = 4500 ml +25 ml

Now,

Thus, 180 glasses can be filled.

## NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Exercise 1.3

### Ex 1.3 Class 6 Maths Question 1.

Estimate each of the following using general rule :**(a)** 730 + 998**(b)** 796 – 314**(c)** 12,904 + 2,888**(d)**28,292 – 21,496

Make ten more such examples of addition, subtraction and estimation of their outcome.

#### Solution:

### Ex 1.3 Class 6 Maths Question 2.

10 more of such examples:**(i)** 840 + 897**(ii)** 697 – 213**(iii)** 744 + 785**(iv)** 975 – 687**(v)** 13805 + 2868**(vi)** 38,393 – 31,495**(vii)** 14915 + 2868**(viii)** 28,283 – 21, 527**(ix)** 25677 + 4213**(x)** 48,457 – 23,624

#### Solution:

### Ex 1.3 Class 6 Maths Question 3.

Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):**(a)** 439 + 334 + 4,317**(b)** 1,08,734 – 47,599**(c)** 8325 – 491**(d)** 4,89,348 – 48,365

Make four more such examples.

#### Solution:

### Ex 1.3 Class 6 Maths Question 4.

Four more such examples are as under:**(i)** 593 + 434 + 5317**(ii)** 109834 – 48598**(iii)** 7625 – 591**(iv)** 479548 – 47465