Ncert Class 6 Mathematics Solutions Chapter 1 Knowing our Numbers
Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of NCERT Class 6 Mathematics Book, Chapter 1, Knowing our Numbers. Here students can easily find step by step solutions of all the problems for Knowing our Numbers, Exercise 1.1 and 1.2 Also here our mathematics teacher’s are 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 NCERT 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) Ten ten thousand
(b) Ten hundred thousand
(c) Ten Ten Lakh
(d) Ten million
(e) Ten 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 :
∴ Total tickets sold,
= (1094 + 1812 + 2050 + 2751)
= 7707
Hence, 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 for the above question.
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)
Hence, the successful candidate won the election by the margin of 2, 28, 800.
(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?
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
Therefore, 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
Therefore, 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, the machine will produce 87575 screws.
(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?
Solution :
As per the question,
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
Therefore, 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?
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
= Rs 72360
Thus the correct answer was greater than the correct answer by Rs 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 (1 m = 100 m)
Each shirt uses 215 cm of cloth
Hence 4000 cm cloth makes 4000 ÷ 215 shirts
Now, = 215 ÷ 4000
Quotient = 18
Reminder = 130
Hence, 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 :
First of all we will convert all the data in gram.
4 kg 500 g = 4×1000 g + 500 g
= 4000 g + 500 g = 4500 g
A van’s capacity to carry a load is 800 kg.
i.e. 800 kg
= 800 × 1000 g
= 800000 gram.
Each box weighing 4500 g
Hence the number of boxes can be loaded in the van which
has the capacity to carry a load of 800000 gram is following.
Now, 4500 ÷ 800000
Quotient = 177
Reminder = 3500
Thus a van can carry 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
Therefore, 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
Therefore, 160 glasses to be filled with curd.
Next Chapter Solution :
| Chapter 2 | Chapter 3 | Chapter 4 | Chapter 5 |
| Chapter 6 | Chapter 7 | Chapter 8 | Chapter 9 |
| Chapter 10 | Chapter 11 | Chapter 12 |
