[Exercise 1.2] NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers.
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# Class 6 Maths Chapter 1 - Knowing Our Numbers.
Maths Class 6 Chapter 1 Exercise 1.2 Solutions :-
Here we have provided Exercise 1.2 class 6th Maths chapter 1 Solutions which includes NCERT Solutions for exercise 1.2 class 6 maths chapter 1. These ex 1.2 class 6 questions are related to the topics that are discussed in the chapter knowing our numbers class 6 chapter 1. We suggest you to go through these NCERT Solutions for class 6 maths chapter 1 to strengthen your understanding about knowing our numbers class 6 exercise 1.2 of class 6 mathematics.
Here at MSEducator.in you get Complete FREE Study Material for Class 6 Maths Chapter 1 : Knowing Our Numbers.
Which Includes :-
# Class 6 Maths Chapter 1 NCERT Book. (pdf)
# Class 6 Maths Chapter 1 Video Explanation.
# NCERT Solutions for Ex - 1.2
Class 6 Maths Chapter 1 Exercise 1.2 Question 1.
Q1. 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 :
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 the four days = 1094 + 1812 + 2050 + 2751 = 7,707.
Ans. Total number of tickets sold on all the four days are 7,707.
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Class 6 Maths Chapter 1 Exercise 1.2 Question 2.
Q2. 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 :
Shekhar has scored runs so far = 6980
Runs he wishes to complete = 10,000
Total number of runs needed by him = 10,000 – 6980 = 3020 runs
Ans. Total number of runs Shekhar needs are 3020
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Class 6 Maths Chapter 1 Exercise 1.2 Question 3.
Q3. 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 :
Number of Votes successful candidate registered = 5,77,500
Number of votes secured by his nearest rival = 3,48,700
Difference of votes between winner and rival = 5,77,500 – 3,48,700 = 2,28,800
Ans. Successful candidate win by 2,28,800 votes from his rival.
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Class 6 Maths Chapter 1 Exercise 1.2 Question 4.
Q4. 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 :
Part 1 -
Books sold in first week of June worth ₹2,85,891
Books sold in second week of the month worth ₹4,00,768
∴ Total sale of books in the two weeks together = ₹2,85,891 + ₹4,00,768 = ₹6,86,659
Part 2 -
Books sold in first week of June worth ₹2,85,891
Books sold in second week of the month worth ₹4,00,768
Difference in the sale of books = ₹4,00,768 – ₹2,85,891 = ₹1,14,877
∴ In second week of June, the sale of books was more by ₹1,14,877.
Ans. They sold 6,86,659 books together in the two weeks and In the second week of the month, the sale of books was greater by 1,14,877 books.
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Class 6 Maths Chapter 1 Exercise 1.2 Question 5.
Q5. Find the difference between the greatest and the least numbers that can be written using the digits 6, 2, 7, 4, 3 each only once.
Solution :
Given digits are 6, 2, 7, 4, 3
Greatest number = 76,432
Least number = 23,467
Difference = 76,432 – 23,467 = 52,965
Ans. The difference between the greatest and the least given numbers is 52,965.
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Class 6 Maths Chapter 1 Exercise 1.2 Question 6.
Q6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January, 2006?
Solution :
Number of screws manufactured in a day = 2,825.
Number of days in month of January = 31 days
Number of screws manufactured in the month of January = 31 x 2825 = 87,575
Ans. Number of screws manufactured in the month of January is 87,575.
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Class 6 Maths Chapter 1 Exercise 1.2 Question 7.
Q7. 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 :
Amount of money with the merchant = ₹78,592
Number of radio sets = 40
Price of one radio set = ₹1200
Therefore, cost of 40 radio sets = ₹1200 x 40 = ₹48,000
Amount of money remain with the merchant = ₹78,592 – ₹48000 = ₹30,592
Ans. Amount of ₹30,592 will remain with her after purchasing 40 radio sets.
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Class 6 Maths Chapter 1 Exercise 1.2 Question 8.
Q8. A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?
Solution :
Student has multiplied 7236 by 65 instead of multiplying by 56. Difference between the two multiplications is -
= (65 – 56) x 7236
= 9 x 7236
= 65124
(We don’t need to do both the multiplication)
Ans. The answer is 65,124 greater than the correct answer.
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Class 6 Maths Chapter 1 Exercise 1.2 Question 9.
Q9. 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 :
Total length of the cloth = 40 m ⇒ 40 x 100 cm = 4000 cm.
Cloth needed to stitch a shirt = 2 m 15 cm ⇒ 2 x 100 + 15 cm = 215 cm
Therefore, number of shirts stitched = 4000 ÷ 215 = 18
So, the number of shirts stitched = 18
The remaining cloth
= 4000 - (18 x 215)
= 4000 - 3870
= 130 cm
= 1 m 30 cm
Ans. The number of shirts stitched = 18 and the remaining cloth is 1 m 30 cm.
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Class 6 Maths Chapter 1 Exercise 1.2 Question 10.
Q10. Medicine is packed in boxes, each weighing 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Solution :
Weight of one box
= 4 kg 500 g
= 4 x 1000 + 500
= 4500 g
Maximum weight that van can carry
= 800 kg
= 800 x 1000
= 8,00,000 g
Therefore, number of boxes van can carry
= 8,00,000 ÷ 4500
= 177.777
Ans. The number of boxes that van can carry is 177.
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Class 6 Maths Chapter 1 Exercise 1.2 Question 11.
Q11. 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 between school and house
= 1 km 875 m
= (1000 + 875) m
= 1875 m.
Distance covered by the student both ways
= 2 x 1875
= 3750 m
Distance covered in 6 days
= 6 x 3750 m
= 22,500 m
= 22 km 500 m.
Ans. Total distance covered by student in six days is 22 km 500 m.
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Class 6 Maths Chapter 1 Exercise 1.2 Question 12.
Q12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Solution :
Quantity of curd in a vessel
= 4 litre 500 ml
= (4 x 1000 ) + 500 ml
= 4500 ml
Capacity of 1 glass = 25 ml
∴ Number of glasses that can be filled
= 4500 ÷ 25
= 180
Ans. Number of glasses that can be filled by curd is 180.
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