Class – 6 CH-1 KNOWING OUR NUMBERS NCERT SOLUTIONS

      

Class – 6 CH-1 KNOWING OUR NUMBERS 

MATHS NCERT SOLUTIONS

 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 

SOLUTION 1: 

(a) 10  

(b) 10  

(c) 10  

(d) 10 

(e) 10   

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.  

SOLUTION 2:  

(a) 73,75,307  

(b) 9,05,00,041 

(c) 7,52,21,302   

(d) 58,423,202 

(e) 23,30,010 

 

Question 3:   

Insert commas suitable 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   

Eight-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. 

 Question 4:   

Insert commas suitable 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) 7,452,483   

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 

 

 

 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. 

  SOLUTION 1:  

Number of tickets sold on first day    =    1,094 

Number of tickets sold on second day  =    1,812 

Number of tickets sold on third day    =    2,050 

Number of tickets sold on fourth day  = + 2,751 

Total tickets sold  =    7,707 

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

 

 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? 

 SOLUTION 2:  

Runs to achieve  =   10,000 

Runs scored  =  – 6,980 

Runs required    =     3,020 

 Therefore, he needs 3,020 more 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?   

SOLUTION 3:  

Number of votes secured by successful candidates    =    5,77,500 

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

Margin between them  =    2,28,800 

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

 Question 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 4:  

Books sold in first week =    2,85,891 

Books sold in second week = + 4,00,768 

Total books sold =    6,86,659 

Since, 4,00,768,> 2,85,891 

Therefore sale of second week is greater than that of first week. 

Books sold in second week =    4,00,768 

Books sold in first week = – 2,85,891 

More books sold in second week =    1,14,877 

Therefore, 1,14,877 more books were sold in second week. 

  Question 5:   

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:  

Greatest five-digit number using digits 6,2,7,4,3 =    76432 

Smallest five-digit number using digits 6,2,7,4,3 = – 23467 

Difference =    52965 

Therefore the difference is 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? 

  SOLUTION 6:  

Number of screws manufactured in one day  = 2,825 

Number of days in the month of January (31 days)  = 2,825 x 31 = 87,575 

Therefore, the machine produced 87,575 screws in the month of January. 

 Question 7:   

A merchant had ₹78,592 with her. She placed an order for purchasing 40 radio sets at ₹1,200 each. How much money will remain with her after the purchase? 

  SOLUTION 7:  

Cost of one radio  = ₹ 1200 

Cost of 40 radios = 1200 x 40    

Now, = ₹ 48,000 

Total money with merchant    =    ₹ 78,592 

 Money spent by her = – ₹ 48,000 

 Money left with her =    ₹ 30,592 

 Therefore, ₹ 30,592 will remain with her 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? 

SOLUTION 8:  

 Wrong answer = 7236 x 65 = 470340  

Correct answer = 7236 x 56 =405216

Difference in answers  = 470340 – 405216 = 65,124 

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? 

  SOLUTION 9:  

Cloth required to stitch one shirt  = 2 m 15 cm 

= 2 x 100 cm + 15 cm 

= 215 cm 

Length of cloth = 40 m = 40 x 100 cm = 4000 cm 







Number of shirts can be stitched  = 4000 ÷ 215  

Therefore, 18 shirts can be stitched and 130 cm (1 m 30 cm) cloth will remain. 

 Question 10:   

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

 SOLUTION 10:  

The weight of one box = 4 kg 500 g = 4 x 1000 g + 500 g = 4500 g 

Maximum load can be loaded in van = 800 kg = 800 x 1000 g = 800000 g 

Number of boxes = 800000 ÷ 4500 








Therefore, 177 boxes can be loaded. 

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.  

SOLUTION 11:  

Distance between school and home    =    1.875 km 

Distance between home and school    = + 1.875 km 

Total distance covered in one day     =    3.750 km 

Distance covered in six days    = 3.750 x 6 = 22.500 km 

 Therefore, 22 km 500 m distance covered in six days. 

 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? 

  SOLUTION 12:  

Capacity of curd in a vessel = 4 litres 500 ml = 4 x 1000 ml + 500 ml = 4500 ml 

Capacity of one glass = 25 ml 

Number of glasses can be filled = 4500 ÷ 25 








Therefore, 180 glasses can be filled by curd.  

 

 Exercise 1.3  

Question 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   

SOLUTION 1:  

(a) 730 round off to  700  

998 round off to  1000      

Estimated sum =  1700  

(b) 796 round off to 800

 314 round off to 300 

Estimated sum = 500

(c) 12904 round off to  13000 

  2888 round off to     3000          

Estimated sum = 16000       

  (d) 28292 round off to   28000 

21496 round off to   21000 

Estimated difference = 7000 

 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 + 4317 

(b) 1,08,737 – 47,599 

(c) 8325 – 491 

(d) 4,89,348 – 48,365   

SOLUTION 2:  

(a) 439 round off to  400  

334 round off to 300  

4317 round off to  4300  

Estimated sum =    5000 

(b) 108734 round off to  108700

 47599 round off to  47600 

Estimated difference =    61100 

 (c) 8325 round off to  8300 

491 round off to  500 

Estimated difference = 7800 

(d) 489348 round off to  489300 

 48365 round off to 48400 

Estimated difference =  440900 

Question 3:   

Estimate the following products using general rule: 

(a) 578 x 161 

(b) 5281 x 3491 

(c) 1291 x 592 

(d) 9250 x 29   

SOLUTION 3:  

(a) 578 x 161 

578 round off to 600 

161 round off to 200 

The estimated product = 600 x 200 = 1,20,000 

 (b) 5281 x 3491 

5281 round of to 5,000 

3491 round off to 3,500 

The estimated product = 5,000 x 3,500 = 1,75,00,000 

 (c) 1291 x 592 

1291 round off to 1300 

592 round off to 600 

The estimated product = 1300 x 600 = 7,80,000 

 (d) 9250 x 29 

9250 round off to 10,000 

229 round off to 30 

The estimated product = 10,000 x 30 = 3,00,000 


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