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Knowing Our Numbers

Class 6th Mathematics CBSE Solution
Exercise 1.1
  1. Fill in the blanks: (a) 1 lakh = .. ten thousand. (b) 1 million = .. hundred…
  2. Place commas correctly and write the numerals: (a) Seventy three lakh seventy…
  3. Insert commas suitably and write the names according to Indian System of…
  4. Insert commas suitably and write the names according to International System of…
Exercise 1.2
  1. A book exhibition was held for four days in a school. The number of tickets sold…
  2. Shekhar is a famous cricket player. He has so far scored 6980 runs in test…
  3. In an election, the successful candidate registered 5,77,500 votes and his…
  4. Kirti bookstore sold books worth Rs. 2,85,891 in the first week of June and…
  5. Find the difference between the greatest and the least number that can be…
  6. A machine, on an average, manufactures 2,825 screws a day. How many screws did…
  7. A merchant had Rs. 78,592 with her. She placed an order for purchasing 40 radio…
  8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was…
  9. To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts…
  10. Medicine is packes in boxes, each weighing 4 kg 500 g. How many such boxes can…
  11. The distance between the school and the house of a students house is 1 km 875…
  12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml…
Exercise 1.3
  1. Estimate each of the following using general rules: (a) 730 + 998 (b) 796 - 314…
  2. Give a rough estimate (by rounding off to nearest hundreds) and also a closer…
  3. Estimate the following products using general rule: (a) 578 161 (b) 5281 3491…

Exercise 1.1
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.


Answer:

(a) 1 lakh = ….10… ten thousand

ten thousand = 10,000
1 lakh = 100000
= 10 × 10000
= 10 ten thousand

1 lakh = ….10… ten thousand

(b) 1 million = …10… hundred thousand

hundred thousand = 100000

1 million = 1,000,000

= 10 ×1,00,000

1 million = 10 hundred thousands


(c) 1 crore = ….10…. ten lakh

ten lakh = 1000000

1 crore = 10000000

= 10 × 1000000

1 crore =10 ten lakh



(d) 1 crore = …10…million.

1 million = 1,000,000

1 crore = 1,00,00,000

= 10 × 1,000,000

1 crore = …10…million.


(e) 1 million = …10…. lakh.

ten lakh = 10,00,000

1 million = 1,000,000

= 10 × 1,00,000

1 million = …10…. lakh


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.


Answer:

(a) Seventy-three lakh seventy-five thousand three hundred seven;


Let’s break the sentence first,


Seventy three lakh = 73,00,000


Seventy five thousand = 75,000


Three hundred and seven = 307


By adding the above numbers we get,


73, 00,000+75,000+307


=73, 00,000+75,307


= 73,75,307


(b) Nine crores five lakh forty-one


Nine crore = 9,00,00,000


Five lakh = 5,00,000


Forty one = 41


Now add the above numbers,


9,0000,000 + 5,00,000 + 41


= 9,0000,000 + 5,00,041


= 9,05,00,041


(c) Seven crores fifty-two lakh twenty-one thousand three hundred two,


Seven crore = 7,00,00,000


Fifty two lakh = 52,00,000


Twenty one thousand = 21,000


Three hundred and two = 302


By adding all the above numbers we get,


= 7,00,00,000 + 52,00,000 + 21,000 + 302


= 7,0000,000 + 52,00,000 + 21,302


= 7,0000,000 + 52,21,302


= 7,52,21,302


(d) Fifty-eight million four hundred twenty-three thousand two hundred two


Fifty eight million = 58,000,000


Four hundred twenty-three thousand = 4 23,000


Two hundred and two = 202


By adding above numbers we get,


= 58,000,000 + 4,23,000 + 202


= 58,000,000 + 4,23,202


= 58,423,202


(e) Twenty-three lakh thirty thousand ten,


Twenty three lakh = 23,00,000


Thirty thousand = 30,000


Ten = 10


Add the above numbers,


= 23,00,000 + 30,000 + 10


= 23,00,000 + 30,010


= 23,30,010


Question 3.

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

(a) 87595762 (b) 8546283 (c) 99900046 (d) 98432701


Answer:

According to the Indian Numeral system, each digit has different value according to their place.


For example – 225


2 is used twice but it has a different value in both the places, first 2 is in hundreds and second 2 represents the tens.


(a) 8, 75,95,762 = Eight crore seventy five lakh ninety five thousand seven hundred sixty two.


Let’s simplify it,


= 8, 0000,000 + 7, 000,000 + 5, 00,000 + 90,000 + 5, 000 + 700 + 60 + 2


= 8, 0000,000 + 7, 000,000 + 5, 00,000 + 90,000 + 5, 000 + 700 + sixty two


= 8, 0000,000 + 7, 000,000 + 5, 00,000 + 90,000 + 5, 000 + seven hundred sixty two


= 8, 0000,000 + 7, 000,000 + 5, 00,000 + 90,000 + five thousand seven hundred sixty-two


= 8, 0000,000 + 7, 000,000 + 5, 00,000 + ninty five thousand seven hundred sixty two


= 8, 0000,000 + 7, 000,000 + five lakh ninety five thousand seven hundred sixty two


= 8, 0000,000 + seventy five lakh ninety five thousand seven hundred sixty two


= Eight crore seventy five lakh ninety five thousand seven hundred sixty two


(b) 85,46,283


By simplifying it we get,


= 8, 000,000 + 5, 00,000 + 40, 000 + 6,000 + 200 + 80 + 3


= 8, 000,000 + 5, 00,000 + 40, 000 + 6,000 + 200 + 83


= 8, 000,000 + 5, 00,000 + 40, 000 + 6,000 + 283


= 8, 000,000 + 5, 00,000 + 40, 000 + 6,283


= 8, 000,000 + 5, 00,000 + 46,283


= 8, 000,000 + 5, 46,283


= 85, 46,283
Therefore 85,46,283


= Eight- five lakh forty six thousand two hundred eighty three


(c) 9,99,00,046


By simplifying it we get,


= 9, 00,00,000 + 9, 000,000 + 9, 00,000 + 0,000 + 000 + 46


= 9, 00,00,000 + 9, 000,000 + 9, 00,046


= 9, 00,00,000 + 99,00,046


= 9,99,00,046


Therefore, 9,99,00,046 = nine crore ninty nine lakh forty six


(d) 9, 84,32,701


By simplifying it we get,


= 9, 00,00,000 + 80,00,000 + 4,00,000 + 30,000 + 2,000 + 700 + 00 + 1


= 9, 00,00,000 + 80,00,000 + 4,00,000 + 30,000 + 2,000 + 701


= 9, 00,00,000 + 80,00,000 + 4,00,000 + 30,000 + 2,701


= 9, 00,00,000 + 80,00,000 + 4,00,000 + 32,701


= 9, 00,00,000 + 80,00,000 + 4,32,701


= 9, 00,00,000 + 84,32,701


= 9, 84,32,701
Therefore, 9, 84,32,701 = nine crore eighty four lakh thirty two thousand seven hundred one.


Question 4.

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


Answer:

(a) 78921092

78,921,092 = seventy-eight million nine hundred twenty-one thousand ninty two

(b) 7452283

7,452,283 = seven million four hundred fifty-two thousand two hundred eighty-three

(c) 99985102

99,985,102 = ninety-nine million nine hundred eighty-five thousand one hundred two

(d) 48049831

48,049,831 = forty-eight million forty-nine thousand eight hundred thirty-one.



Exercise 1.2
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 and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.


Answer:

Tickets sold on the first day of exhibition = 1094


Tickets sold on the second day of the exhibition = 1812


Tickets sold on the third day of the exhibition = 2050


Tickets sold on the last day of the exhibition = 2751


Now,


To get total number of tickets we, add the all above tickets sold,


By adding we get,


= 1094 + 1812 + 2050 + 2751


= 7707


So, total tickets sold = 7,707


Question 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?


Answer:

The number of runs scored by Shekhar so far = 6980


The number of runs shekhar wants to score = 10,000


Thus, the number of runs required = 10,000 – 6980


= 3020 runs


So,


Shekhar required 3,020 more runs to complete his 10,000 runs.



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?


Answer:

Successful candidate secured = 5, 77,500 votes

Rival candidate secured = 3, 48,700 votes

The margin = 5,77,500 – 3,48,700

= 2,28,800 votes

The successful candidate has won the election by the margin of 2, 28,800 votes.


Question 4.

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?


Answer:

Value of books sold in the first week = Rs 2, 85,891

Value of books sold in the second week = Rs 4, 00,768

Total sale in the two weeks = sale in the first week + sale in the second week

Total sale for the two weeks = 2,85,891 + 4,00,768 = 6,86,659


Clearly, from the sales figures, we can say that the Sales in the second week is greater than the first week.

Difference = 4,00,768 – 2,85,891 = 1,14,877


The amount by which the Sale in the second week is higher than the sale in the first week by Rs 1, 14,877.


Question 5.

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


Answer:

*Trick: To find the greatest number arrange the given numbers in descending order and to get the smallest number, arrange the given digits in ascending order.

On using the digits: 6, 2, 7, 4 and 3 and only once, we get,

Greatest number = 76432

Smallest number = 23467

Now, the difference between the greatest and the least number = 76432 – 23467
= 52965


Question 6.

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


Answer:

Screws produced by machine in one day = 2, 825


Days in January = 31 days


Screws produced in 31 days = 2, 825 × 31


= 87575 screws


So,


Screws produced in January 2006 = 87575


Question 7.

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?


Answer:

Total money merchant has = 78, 592 Rs

Cost of one radio set = 1200 Rs

Cost of 40 radio sets = 1200 × 40 = 48000 Rs

So,

Total money spent on radio sets = 48, 000 Rs

Money left with the merchant = 78,592 – 48,000 = 30,592 Rs

Therefore, total money left with her is 30,592 Rs after the purchase.


Question 8.

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


Answer:

The Student multiplied 7236 by 65 instead of 56;


He gets, a wrong answer = 7236 × 65 = 470340


But, the right answer he should have got = 7236 × 56 = 405216


Now the difference in the answers = 470340 – 405216 = 65124


So,


His answer was greater than the right answer by 65124.



Question 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?


Answer:

We know, 1 m = 100cm


Therefore, 2 m 15 cm = 215cm


40 m = 40×100 cm = 4000 cm


Now the cloth required for one shirt = 215 cm


Number of shirts that can be stitched out of the 4000cm = 4000÷215



Therefore, 18 shirts can be made from 4000cm long cloth. Out of 4000 cm, 130cm cloth will remain which is 1m 30cm.


Question 10.

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


Answer:

We know, 1 kg = 1000g


Thus, 4 kg 500g = 4500g


And, 800 kg = 800×1000 = 8,00,000g


Weight of one box = 4500g


The total weight that can be carried by van = 8,00,000 g


Thus, the number of boxes that can be loaded in the van = 800000÷4500



Hence, 177 boxes can be loaded in the van.


Question 11.

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


Answer:

Distance between the school and her house = 1km 875m


1 km = 1000m


1km 875m = 1875m


The distance she covered each day = 1875×2×6


= 22500 m


Therefore, distance covered in 6 days = 22,500m which is 22.5 km or we can say 22km 500m.


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?


Answer:

Capacity of vessel = 4 l 500ml


1 l = 1000 ml


4 l 500 ml = 4500 ml


Capacity of a glass = 25 ml


Number of glasses required to fill the vessel = 4500÷25



So,


180 glasses needed to fill the vessel.



Exercise 1.3
Question 1.

Estimate each of the following using general rules:

(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.


Answer:

(a) 730+998


By rounding off to hundreds,


730 rounds off to 700 and,


998 rounds off to 1000,


So we have;


= 700+1000


= 1700


(b) 796 - 314


By rounding off to hundreds,


796 rounds off to 800 and,


314 rounds off to 300


We get,


= 800 - 300


= 500


(c) 12904 + 2822


By rounding off to thousands,


12904 rounds off to 13000 and,


2822 rounds off to 3000.


We get,


= 13000+3000


= 16000


(d) 28,296 – 21, 496


By rounding off to nearest thousands,


28,296 rounds off to 28000


21,496 rounds off to 21,000


We get,


= 28,000 – 21,000


= 7,000


Question 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.


Answer:

(a) 439 + 334 + 4,317


Rounding off to nearest hundreds,


439 rounds off to 400


334 rounds off to 300


And 4317 rounds off to 4300,


We get,


= 400+300+4300


= 5000


By rounding off to nearest tens,


439 rounds off to 440


334 rounds off to 330


And 4317 rounds off to 4320,


We get,


= 440+330+4320


= 5090


(b) 1,08,734 – 47,599


Rounding off to nearest hundreds,


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


47,599 rounds off to 47,600


We get,


= 1, 08,700 – 47,600


= 61100


Rounding off to nearest tens,


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


47,599 rounds off to 47,600


We get,


= 1, 08,730 – 47,600


= 61130


(c) 8325 – 491


Rounding off to nearest hundreds,


8325 rounds off to 8300


491 rounds off to 500


We get,


= 8300 - 500


= 7800


Rounding off to nearest tens,


8325 rounds off to 8330


491 rounds off to 490


We get,


= 8330 – 490


= 7840


(d) 4,89,348 – 48,365


Rounding off to nearest hundreds,


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


48,365 rounds off to 48,400


We get,


= 489300 – 48400


= 440900


Rounding off to nearest tens,


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


48,365 rounds off to 48,370


We get,


= 489350 – 48370


= 440980


Question 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.


Answer:

(a) 578×161


Rounding off by general rule,


578 rounds off to 600


161 rounds off to 200,


So,


= 600×200 = 120000


(b) 5281×3491


Rounding off by the general rule,


5281 rounds off to 5000


3491 rounds off to 3500


So,


= 5000×3500


= 17500000


(c) 1291×592


Rounding off by general rule,


1291 rounds off to 1300


592 rounds off to 600


So,


= 1300×600


= 780000


(d) 9250×29


Rounding off by general rule,


9250 rounds off to 9000


29 rounds off to 30


So,


= 9000 × 30


= 27000