# Knowing Our Numbers ## Ex 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 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

## Ex 1.2  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 wishes to complete  = 10,000 runs.

Shekhar has so far scored = 6980 runs
∴ Total number of runs needed by him = 10,000 – 6980 = 3020 runs

## Ex 1.2  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
∴Margin of votes to win the election = 5,77,500 – 3,48,700 = 2,28,800

## Ex 1.2  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-
Books sold in the first week of June  = ₹2,85,891
Books sold in the second week of the June  =  ₹4,00,768
∴ Total books sold
= ₹2,85,891 + ₹4,00,768 = ₹6,86,659
In the second week of the month, the sale of books was greater.
The difference between the sale of books
= ₹4,00,768 – ₹2,85,891 = ₹1,14,877
Hence, in the second week, the sale of books was more by ₹1,14,877.

## Ex 1.2  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
∴ Difference = 76432 – 23467 = 52,965

## Ex 1.2  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-
The number of screws manufactured in a day = 2,825.
Number of screws manufactured in month of January (31 days) = 31 x 2825 = 87,575

## Ex 1.2  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-
Number of radio sets = 40
Cost of one radio set = ₹1200
∴Cost of 40 radio sets = ₹1200 x 40 = ₹48,000

Amount of money with the merchant = ₹78,592

money spent by her =₹48,000
Remaining money with the merchant = ₹78,592 – ₹48000 = ₹30,592

## Ex 1.2  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-
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
Hence, the answer greater than the correct answer is 65,124.

## Ex 1.2  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-
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, the number of shirts stitched = 4000/215 So, the number of shirts stitched = 18

Remaining cloth = 130 cm = 1 m 30 cm

## Ex 1.2  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 = 4 kg 500 g = 4 x 1000 + 500 = 4500 g
Maximum load can be loaded in van = 800 kg = 800 x 1000 = 800000 g

Number of boxes =800000/4500 Therefore, 177 boxes can be loaded in the van.

## Ex 1.2  Question 11.

The distance between the school and the house of a student is 1 km 875 m. Every day 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.
Total ditance covered in one day = 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

## Ex 1.2  Question 12.

A vessel has 4 liters 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 liters 500 mL = (4 x 1000 + 500) mL = 4500 mL.
Capacity of 1 glass = 25 mL
Number of glasses can be filled =4500/25 ∴The number of glasses can be filled =180

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