NCERT Solutions Class 6th Maths Chapter – 1 Knowing Our Numbers 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 –

Number of tickets sold on first day = 1094.

Number of tickets sold on second day = 1812.

Number of tickets sold on third day = 2050.

Number of tickets sold on final day = 2751.

∴ Total number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 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?

Solution –

Shekhar has so far scored 6980 runs.

He wishes to complete 10,000 runs.

Therefore, total number of runs needed by him = 10,000 – 6980 = 3020 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 –

Number of votes secured by the successful candidate = 5,77,500.

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

Therefore, margin of votes to win the election = 5,77,500 – 3,48,700 = 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 –

Price of books sold in June first week = Rs 285891.

Price of books sold in June second week = Rs 400768.

No. of books sold in both weeks together = Rs 285891 + Rs 400768 = Rs 686659.

The sale of books is the highest in the second week.

Difference in the sale in both weeks = Rs 400768 – Rs 285891 = Rs 114877.

∴ Sale in second week was greater by Rs 114877 than in the first week.

Question 5. 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 – 6,2,7,4,3

Greatest number = 76432

Least number = 23467

Therefore, difference = 76432 – 23467 = 52,965

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

Solution –

The number of screws manufactured in a day = 2,825.

Number of screws manufactured in month of January = 31 x 2825 = 87,575 screws.

Question 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 –

Amount of money with the merchant = Rs 78,592.

Number of radio sets = 40.

Price of one radio set = Rs 1200.

Therefore, cost of 40 radio sets = Rs 1200 x 40 = Rs 48,000.

Remaining money with the merchant = Rs 78,592 – Rs 48000 = Rs 30,592.

Hence, the amount will remain with her after purchasing the radio sets is Rs 30,592.

Question 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 –

The student has multiplied 7236 by 65 instead of multiplying by 56.

Difference between the two multiplications = (65 – 56) x 7236 = 9 x 7236 = 65124 (We don’t need to do both the multiplied)

Hence, the answer greater than the correct answer is 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? (Hint: Convert data in cm.)

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 shirt.

So, the number of shirts stitched = 18 shirts and the remaining cloth = 130 cm = 1 m 30 cm.

Question 10. 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 = 4kg 500g = 4 x 1000 + 500 = 4500 g.
and 800 kg = 800 x 1000 = 800000 g.

Therefore, 177 boxes can only be loaded in the van.

Solution –

Distance between school and house

= 1 km 875 m = (1000 + 875) m = 1875 m.

Distance travelled by the student in both ways = 2 x 1875 = 3750 m.

Distance travelled in 6 days = 3750 m x 6 – 22500 m = 22 km 500 m.

Hence, the total distance covered in six days = 22 km 500 m.

Solution – 

Quantity of curd in a vessel = 4L 500 ml.

= (4 x 1000 + 500) ml = 4500 ml.

Capacity of 1 glass = 25 ml.

∴ Number of glasses = 4500/25 = 180 glasses.

Hence, 180 glasses can be filled with curd.