# Kseeb Class 6 Mathematics Chapter 1 Knowing Our Numbers Solutions

## Kseeb Class 6 Mathematics Chapter 1 Knowing Our Numbers Solutions

Welcome to NCTB Solutions. Here with this post we are going to help 6th class students by providing Solutions for KSEEB Class 6 Mathematics chapter 1 Knowing Our Numbers. Here students can easily find all the solutions for Knowing Our Numbers Exercise 1.1, 1.2 and 1.3. Also here our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 1 solutions. Here all the solutions are based on Karnataka State Board latest syllabus.

Knowing Our Numbers Exercise 1.1 Solution

(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 ten thousand.

(b)  Ten hundred thousand.

(c) Ten ten lakh.

(d) Ten million.

(e) 10 lakh

(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, 73,307

(b) 9, 05, 00,041

(c) 7, 52, 21,302

(d) 58,423,202

(e) 23, 31,010

(3) Insert commas suitably and write the names according to Indian System of Numeration :

(a) 87595762

(b) 8546283

(c) 99900046

(d) 98432701

Solution :

(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.

(4) Insert commas suitably and write the names according to International System of Numeration :

(a) 78921092

(b) 7452283

(c) 99985102

(d) 48049831

Solution :

(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.

Knowing Our Numbers Exercise 1.2 Solution

(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 and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.

Solution :

First, second, third and final days were respectively 1094, 1812, 2050 and 2751

∴ Total tickets sold,

= (1094 + 1812 + 2050 + 2751)

= 7707

Therefore, the total number of tickets sold on all the four days is 7707

(2) 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 :

We will have to find 10000 is how many more than 6980 to find the solution.

Now, 10000-6980 = 3020

Check the answer by adding

6980 + 3020

= 1000 (The answer is right)

Thus, Shekhar need 3020 runs more.

(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 :

Clearly 5,77500 is more than 3,48,700. So we will find 577500 is how many more than 3,48,700.

For this 348700 should be subtracted from 5,77,500.

Now, 577500 – 348700 = 228800

Check the answer by adding

348700 + 228800

= 577500 (The answer is right)

Thus, the successful candidate won the election by the margin of 2, 28, 800.

(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 :

The sale for the two weeks together

= Sale in first week + sale in second week

= 2,85,891 + 4,00,768

= 6,86,659

Now, 285,891+400786 = 686677

Clearly the sale of first week i.e. 285891 is less than the sale of second week i.e. 400786.
We will find the second weeks sale is how many greater than the first week, by subtraction.

Now, 400786 – 285891 = 1,14,895

Thus, The second weeks sale is greater than the first week by 1,14,895

(5) Find the difference between the greatest and the least 5-digit number that can be written using the digits 6, 2, 7, 4, 3 each only once.

Solution :

The greatest and the least 5-digit number that can be written using

The digits 6, 2, 7, 43 each only once, are 76432 and 23,476 respectively.

The difference between the numbers 76432 and 23476 is

= 76432 – 23476

= 52956

∴ 52956 is the answer.

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

Solution :

Since we know that the number of days in the month of January in any year is 31.

The machine manufactures 2825 screws a day.

Therefore, the number of screws that going to manufacture be 31 × 2825 in January 2006.

Now, 2825 × 31

= 87575

Thus correct answer is 87575.

(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 :

The price of each radio set is RS 1200.

The total price of 40 radio is 40×1200

Now, 1200×40 = 48000

Now, she spent RS 48000 on purchasing radio sets.

Thus she will remain with the money is

= 78,582 – 48000

Now, 78592 – 48000 = 30592

∴ She will remain with Rs. 39592

(8) 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 :

After observing the both multiplication we found that student multiplied

65 and 56 by the same number i.e. 7236.

As we know that 65 is greater than 56 by 10.

Thus the answer of multiplication of 7236 × 65 is always be greater

than the multiplication of 7236 × 56 by 7236 × (65-56) I.e. 7236 × 10.

Now, 7236×10 = 72360

Thus the correct answer was greater than the correct answer by 72,360

(9) 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?

Solution :

First of all we will convert the data in CM.

i.e. 40 m = 200 cm + 15 cm = 215 cm

and 40 m = 4000 cm (1m = 100 m)

Each shirt uses 215 cm of cloth

Hence 4000 cm cloth makes 4000 ÷ 215 shirts

Now, 215 ÷ 4000

Quotient = 18

Reminder = 130

Thus, He can stitch 18 shirts and 130 cm cloth will remain i.e. 130 cm = 1m 30 cm.

(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 :

Medicine is packed in boxes, each weighing 4 kg 500g.

Boxes can be loaded in a van which cannot carry beyond 800 kg

= 800 kg / 4 kg 500g

= 177 boxes

Boxes can be loaded in a van which cannot carry beyond 800 kg is 177 boxes.

(11) The distance between the school and 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 :

The one way distance is 1 km 875 m i.e.

1 × 1000 + 875 = 1875 m

So both ways distance is 1875 × 2 –

Now, 1875×2 = 3750

She walks 3750 m every day.

Thus distance covered in six day is 3750 × 6

Now, 3750 × 6 = 22500

22500 m = 22 km 500 m

∴ Total distance covered by her in six days is 22 km 500 m.

(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 :

The glasses capacity is 250 ml.

A vessel has 4 litres and 500 ml of curd if 4000 ml + 500ml = 4500 ml.

Thus the number of glasses to be needed to fill the curd in a vessel is 4500÷25

Now, 25 ÷ 4500

Quotient = 160

Reminder = 0

∴ 160 glasses to be filled with curd.

Knowing Our Numbers Exercise 1.3 Solution

(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 :

(a) 730+998

= To get a closer estimate, let us try rounding each number to hundreds.

730 rounds off to

+ 998 rounds off to

Estimated addition = 700+1000 = 1700

(b) 796 – 314

= Round off to hundreds

796 is rounds off to

– 314 is rounds off to

Estimated subtraction = 800 – 300 = 500

(c) 12904 + 2888

= Rounds off to thousand

12, 904 is rounds off to

+ 2888 is rounds off to

Estimated sum = 13000 + 3000 = 16000

(d) 28, 292 – 21, 496

= Rounds off to thousand

28, 292 is rounds off to

– 21, 496 is rounds off to

Estimated difference = 28000-21000 = 7000

Ten more such examples of addition, subtraction and estimation of their outcome :

(a) 630 + 752

Round off to hundreds

630 is rounds off to

+ 752 is rounds off to

Estimated sum = 600+800 = 1400

(b) 896 – 225

= 896 is rounds off to

– 225 is rounds off to

Estimated difference = 900 – 200 = 700

(c) 587 + 162

= Round off to hundreds

587 is rounds off to

+ 162 is rounds off to

Estimated sum = 600+200 = 800

(d) 732 – 591

= Rounds off to nearest hundreds

732 is rounds off to

– 591 is rounds off to

Estimated difference = 700 – 600 = 100

(e) 23, 623 + 2932

= Round off to nearest towards

23, 623 is rounds off to

+ 2932 is rounds off to

Estimated addition = 24000 + 3000 = – 27000

(f) 16, 285 – 6923

= Rounds off to nearest thousands

16, 235 is rounds off to

– 6923 is rounds off to

Estimated difference = 16000 – 7000 = 9000

(g) 3,290 + 4,692

Rounds off to nearest thousand

3,290 is rounds off to

+ 4692 is rounds off to

Estimated sum = 3000 + 5000 = 8000

(h) 9999 -6662

Round off to nearest thousand

9999 is rounds off to

– 6662 is rounds off to

Estimated difference = 10000 – 7000 = 3000

(i) 12,302 + 13032

Round of to nearest thousand

12,302 is round off to

+ 13,032 is round off to

Estimated addition = 12000 + 13002 = 15002

(j) 56,023 – 25,011

Round off to nearest thousand

56,023 is round off to

– 25011 is round off to

Estimated difference

= 56000 – 25000

= 31000

(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

Make four more such examples.

Solution :

(a) 439 + 334 + 4,317

First we rounding off to closer estimate.

439

+ 334

+ 4,317
——————–
5090 (closer estimate)

5090 rounding off to nearest hundreds is 5000.

(b) 1, 08,734 – 47,599

First we rounding off to closer estimate

1, 08,734

– 47,599
——————–
61135(closer estimate)

61135 rounding off to nearest hundreds is 60000.

(c) 8325 – 491

First we rounding off to closer estimate

8325

– 491
——————–
7834(closer estimate)

7834 rounding off to nearest hundreds is 8000.

(d) 4, 89,348 – 48,365

First we rounding off to closer estimate

4, 89,348

– 48,365
——————–
4, 40,993(closer estimate)

4, 40,993 rounding off to nearest hundreds is 4, 40,980

(3) Estimate the following products using general rule:

(a) 578 × 161

(b) 5281 × 3491

(c) 1291 × 592

(d) 9250 × 29

Make four more such examples.

Solution :

(a) 578 × 161

= 578 is rounds off to nearest hundred is 600.

161 is rounds off to nearest hundreds is 200.

The estimated product

= 600 × 200

= 12000.

(b) 5281 × 3491

= 5281 is rounded off to 5000 (rounding off to thousands)

3491 is rounded off to 3500 (rounding off to hundreds)

The estimated product

= 5000 × 3500

= 1750000.

(c) 1291 × 592

= 1291 is rounded off to 1300 (rounding off to hundreds)

592 is rounded off to 600 (rounding off to hundreds)

The estimated product

= 1300 × 600

= 780000

(d) 9250 × 29

= 9250 is rounds off to 9300 (rounding off to hundreds)

29 is rounds off to 30 (rounding off to nearest ten)

The estimated product

= 9300 × 30

= 2,79,000.

Four more such examples are :

(a) 978 × 232

= 978 is rounds off to nearest thousand is 1000.

232 is rounds off to 200 (rounding off to nearest hundred)

The estimated product

= 1000 × 200

= 200000.

(b) 9350 × 39

= 9350 is rounds off to 9400 (rounding off to thousands)

39 is rounds off to 40 (rounding off to nearest tens)

The estimated product

= 9400 × 40

= 376000

(c) 4192 × 782

= 4192 is rounds off to nearest thousands i.e. 4000

782 is rounds off to nearest hundred i.e. 800

The estimated product = 4000 × 800 = 3200000

(d) 6383 × 3677

= 6383 is rounds off to 6400 (rounds off to hundreds)

The estimated products,

= 6400 × 3700

= 23680000

Next Chapter Solution :

Updated: July 27, 2023 — 4:39 am