**Download Class 6 Maths NCERT Solutions Chapter 1- Knowing our Numbers**

**Class 6 Maths NCERT Solutions Chapter 1 – Knowing our Numbers comprises of the 3 Exercises, **

## Class 6 Maths NCERT Solutions Chapter 1 – Exercise 1.1

**Ques. 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 1**

This question requires conversion of units from one form to another

(a) 1 lakh = **10** ten thousand – **Explanation** – (10 * 10000 = 1 lakh )

(b) 1 million = **10** hundred thousand – **Explanation** – (10 * 100,000 = 10 lakh = 1 million )

(c) 1 crore = **10** ten lakh. – **Explanation** – (10 * 10,00,000 = 100 lakh = 1 crore)

(d) 1 crore = **10** million. – **Explanation** – (1 crore = 100 lakh = **10** million – Since 1 million = 10 lakh )

(e) 1 million = **10** lakh.

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Ques. 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 2**

(a) Seventy three lakh seventy five thousand three hundred seven = 73,75,307

(b) Nine crore five lakh forty one = 9,05,00,041

(c) Seven crore fifty two lakh twenty one thousand three hundred two = 7,52,21,302

(d) Fifty eight million four hundred twenty three thousand two hundred two = 58,423,202

(e) Twenty three lakh thirty thousand ten = 23,30,010

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Ques. 3.

**Insert commas suitably and write the names according to Indian System of Numeration : – **

(a) 87595762

(b) 8546283

(c) 99900046

(d) 98432701

**Solution 3**

(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

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**Ques. 4. Insert commas suitably and write the names according to International System of Numeration :**

(a) 78921092

(b) 7452283

(c) 99985102

(d) 48049831

**Solution 4**

(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 eighty million forty nine thousand eight hundred thirty one

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** **Class 6 Maths NCERT Solutions Chapter 1 – Exercise 1.2

**Question 1 – Computing Total number of tickets sold when tickets sold at the counter on various dates are given **

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

**Solution 1**

Number of tickets sold on the first day = 1094

Number of tickets sold on the second day = 1812

Number of tickets sold on the third day = 2050

Number of tickets sold on the final day = 2751

Total number of tickets sold on all four days = 1,094 + 1,812 + 2,050 + 2,751 = 7,707

Therefore, 7,707 tickets were sold on all four days.

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**Question 2 – Runs to be scored by a batsman to reach 10,000 runs**

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

**Solution 2**

Runs Shekhar wishes to complete **or** total runs required to be scored = 10,000

Runs already scored by Shekhar = 6,980

Runs required to achieve the target = 10,000 – 6,980 = 3,020

Therefore, Shekhar needs 3,020 more runs to complete 10,000 runs.

**Check = **

Total Runs scored by Shekhar = Runs already scored by Shekhar + Runs Shekhar needs to score

= 6,980 + 3,020 = 10,000

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**Question 3 – Margin of successful candidate in the election**

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 3**

Margin by which the successful candidate won is the votes received by him , which are higher than runs scored by his rival, i.e difference between votes registered by the successful candidate and the votes registered by his nearest rival.

Number of votes registered by the successful candidate = 5,77,500

Number of votes registered by his nearest rival = 3,48,700

Margin by which the successful candidate won = 5,77,500 – 3,48,700 = 2,28,800

Therefore, the successful candidate won by a margin of 2,28,800 votes.

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**Question 4 – Sale for two weeks by bookstore when sale of each week is given and computing excess sales**

Kirti bookstore sold books worth Rs 2,85,891 in the first week of June and books worth Rs 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 4**

Sale for the two weeks together = Books sold in the first week of June + Books sold in the second week of June

Worth of books sold in the first week of June = Rs. 2,85,891

Worth of books sold in the second week of June = Rs. 4,00,768

Sale for the two weeks together = Rs.2,85,891 + Rs.4,00,768 = Rs. 6,86,659

Hence, the sale was worth Rs. 6,86,659 for the two weeks.

Since, books sold in the second week of June is Rs. 4,00,768 > books sold in the first week of June is Rs. 2,85,891, the sale was greater in the second week of June than in the first week of June .

Amount by which Sale is greater in the second week of June than first week of June = Rs.4,00,768 – Rs.2,85,891 = Rs.1,14,877

Therefore, the sale was greater in second week than the first week by Rs 1,14,877.

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**Question 5 – Difference between the greatest and the least number that can be written using 5 digits only once**

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

**Solution 5**

The **greatest five digit number** can be obtained by arranging the digits in Descending order (highest first to lowest in last) , from the highest place to one’s place.

The **smallest five digit number** can be obtained by arranging the digits in ascending order ( lowest first , to highest in last)from the highest place to one’s place.

Hence, using the digits 6, 2, 7, 4, 3, each only once,

The greatest five digit number obtained = 76,432 – (Note that we have arranged the digits in Descending order)

The smallest five digit number obtained = 23,467 – (Note that we have arranged the digits in Ascending order)

The difference between the greatest and smallest numbers so obtained = 76,432 – 23,467 = 52,965

Hence, the difference between the greatest and the least number that can be written using the digits 6, 2, 7, 4, 3 , each only once is 52,965.

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**Question 6 – Computing screws produced by machine in a month**

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

**Solution 6**

Number of screws produced in the month of January 2006 = Number of screws manufactured in a day X Number of days in the month of January 2006

Number of screws manufactured in a day = 2,825

Number of days in the month of January 2006 = 31

Number of screws produced in the month of January 2006 = 2,825 x 31 = 87575

Therefore, the number of screws produced in January 2006 = 87,575.

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**Question 7 – Money remaining with Merchant after purchasing radio sets**

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

**Solution 7**

Amount left with the merchant after purchase = Total amount with the merchant – Total cost of purchase of radio sets

Total amount with the merchant = Rs 78,592

Cost of one radio set = Rs 1200

Number of radio sets ordered for purchase = 40

Total cost of purchase = Cost of one radio set X Number of radio sets ordered for purchase = Rs 1200 x 40 = Rs 48,000

Amount left with the merchant after purchase = Rs 78,592 – Rs 48,000 = Rs 30,592

Therefore, Rs 30,592 will remain with the merchant after the purchase of 40 radios.

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**Question 8 – Difference in answer if multiplication done by a wrong number**

A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer? (Hint: Do you need to do both the multiplications?)

**Solution 8**

Wrong answer = 7236 x 65

Right answer = 7236 x 56

Difference between the wrong and right answers = 4,70,340 – 4,05,216 = 65,124

Therefore, the wrong answer is greater than the right answer by 65,124.

Alternate Solution : –

We know that (a X b) – (a X c) = a X (b – c)

Here a = 7236, b = 65 and c = 56

Substituting in the equation, we get the answer as

7236 X ( 65 – 56 ) = 7236 X ( 9 ) = 65,124

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**Question 9 -Number of shirts that can be stitched from given cloth and remaining cloth**

To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain? (Hint: convert data in cm.)

**Solution 9**

Cloth required to stitch 1 shirt = 2 m 15 cm = (2 x 100) cm + 15 cm = (200 + 15) cm = 215 cm – **1 m = 100 cm**

Cloth available = 40 m = 40 x 100 cm = 4,000 cm

Number of shirts that can be stitched from this cloth = Total Cloth available ÷ Cloth required to stitch 1 shirt = 4000 ÷215

Therefore, 18 shirts can be stitched and 130 cm = 1m 30 cm of the cloth will remain thereafter .

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**Question 10 – Number of boxes that can be loaded in a van**

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 10**

Weight of one box = 4 kg 500 g = (4 x 1000)g + 500 g = 4,000 g + 500 g = 4,500 g – **– 1 kg = 1000g**

Maximum weight that the van can carry = 800 kg = 8,00,000 g

Number of boxes that can be loaded in the van = Maximum weight that the van can carry ÷ Weight of one box = 800000 ÷ 4500

Therefore, 177 boxes can be loaded in the van.

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**Question 11 – Total distance covered by student when walking from house to School**

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

**Solution 11**

Distance between the school and the house = 1 km 875 m = 1000m + 875 m = 1875 m **– 1 km = 1000m**

Distance covered by the girl in a day = 2 x 1875 m = 3750 m – Since she walks both ways and distance of one side is 1875 m, total distance would be 2 x 1875 m

Distance covered by the girl in 6 days =Distance covered by the girl in a day X 6 = 3750m x 6 = 22500 m = 22 km 500 m

Therefore, the girl covers 22 km 500 m distance in 6 days.

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**Question 12 – Number of glasses required to fill curd**

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

**Solution 12**

Total Quantity of curd in the vessel = 4 litres 500 ml = 4000 ml + 500 ml = 4500 ml – **– 1 litre = 1000ml**

Capacity of one glass = 25 ml

Number of glasses that can be filled from the vessel = Total Quantity of curd in the vessel ÷ Capacity of one glass

= 4500 ÷ 25

Therefore, 180 glasses can be filled with the curd in the vessel.

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## Class 6 Maths NCERT Solutions Chapter 1 – Exercise 1.3

Ques. 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

Sol. 1. To estimate the result of operation of two numbers, we first estimate the rounded off figures of each number, and then apply the given operation.

(a) Rounding off to hundreds

730 rounds off to = 700

998 rounds off to = 1000

Estimated sum = 700 + 1000 = 1700

(b) Rounding off to hundreds

796 rounds off to = 800

314 rounds off to = 300

Estimated difference = 800 – 300 = 500

(c) Rounding off to thousands

12,904 rounds off to = 13,000

2,888 rounds off to = 3,000

Estimated sum = 13,000 + 3,000 = 16,000

(d) Rounding off to thousands

28,292 rounds off to = 28,000

21,496 rounds off to = 21,000

Estimated difference = 28,000 – 21,000 = 7,000

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Ques. 2. 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

Solution 2

**(a) By rounding off to nearest hundreds**

439 rounds off to = 400

334 rounds off to = 300

4,317 rounds off to = 4,300

Estimated sum = 400 + 300 + 4,300 = 5,000

**By rounding off to nearest tens**

439 rounds off to = 440

334 rounds off to = + 330

4,317 rounds off to = +4,320

Estimated sum = 440 + 330 + 4,320 = 5,090

**(b) By rounding off to nearest hundreds**

1,08,734 rounds off to = 1,08,700

47,599 rounds off to = 47,600

Estimated difference = 1,08,700 – 47,600 = 61,100

**By rounding off to nearest tens**

1,08,734 rounds off to = 1,08,730

47,599 rounds off to = 47,600

Estimated difference = 1,08,730 – 47,600 = 61,130

**(c) By rounding off to nearest hundreds**

8,325 rounds off to = 8,300

491 rounds off to = 500

Estimated difference = 8,300 – 500 = 7,800

**By rounding off to nearest tens**

8,325 rounds off to = 8,330

491 rounds off to = 490

Estimated difference = 8,330 – 490 = 7,840

**(d) By rounding off to nearest hundreds**

4,89,348 rounds off to = 4,89,300

48,365 rounds off to = 48,400

Estimated difference = 4,89,300 – 48,400 = 4,40,900

**By rounding off to nearest tens**

4,89,348 rounds off to = 4,89,350

48,365 rounds off to = 48,370

Estimated difference = 4,89,350 – 48,370 = 4,40,980

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Ques. 3. Estimate the following products using general rule:

(a) 578 × 161

(b) 5281 × 3491

(c) 1291 × 592

(d) 9250 × 29

Solution 3

(a) By rounding off each factor to its greatest place

578 rounds off to = 600

161 rounds off to = 200

Estimated product = 600 x 200 = 1,20,000

(b) By rounding off each factor to its greatest place

5,281 rounds off to = 5,000

3,491 rounds off to = 3,500

Estimated product = 5,000 x 3,500 = 1,75,00,000

(c) By rounding off each factor to its greatest place

1,291 rounds off to = 1,300

592 rounds off to = 600

Estimated product = 1,300 x 600 = 7,80,000

(d) By rounding off each factor to its greatest place

9,250 rounds off to = 10,000

29 rounds off to = 30

Estimated product = 10,000 x 30 = 3,00,000

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**Maths – NCERT Solutions Class 6**

**NCERT Solutions Class 6**

**This Chapter contains the Exercises relating to the following topics , which are discussed in Chapter 1 – Number System Class 6 NCERT book : –**

- Exercise 1.1 Introduction
- Exercise 1.2 Comparing Numbers
- Exercise 1.2.1 How many numbers can you make?
- Exercise 1.2.2 Shifting digits
- Exercise 1.2.3 Introducing 10,000
- Exercise 1.2.4 Revisiting place value
- Exercise 1.2.5 Introducing 1,00,000
- Exercise 1.2.6 Larger numbers
- Exercise 1.2.7 An aid in reading and writing large numbers
- Exercise 1.3 Large Numbers in Practice
- Exercise 1.3.1 Estimation
- Exercise 1.3.2 Estimating to the nearest tens by rounding off
- Exercise 1.3.3 Estimating to the nearest hundreds by rounding off
- Exercise 1.3.4 Estimating to the nearest thousands by rounding off
- Exercise 1.3.5 Estimating outcomes of number situations
- Exercise 1.3.6 To estimate sum or difference
- Exercise 1.3.7 To estimate products
- Exercise 1.4 Using Brackets
- Exercise 1.4.1 Expanding brackets
- Exercise 1.5 Roman Numerals

**Download Class 6 Maths NCERT Solutions Chapter 1**

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