Vi Class 1.Know About Natural Numbers
We use numbers 1, 2, 3, 4 … to count and calculate. Therefore, these numbers are called Counting Numbers. However, we also use the name natural numbers for counting numbers. Thus, 1, 2, 3…, 10…, 80… 1005 … are all natural numbers.
Since we start counting with the number 1, therefore, 1 is the first or the smallest natural number. The next number is 2 which is obtained by adding 1 to the first natural number. By adding 1 to 2, we get 3, the third natural number. In fact, by adding 1 to a natural number, we get the next natural number. Thus, 30 (= 29 + 1) is the natural number next to 29.
What Are Whole Numbers?
In order to describe the number of elements in a collection with no objects, we use the symbol ‘0’ called zero. The number ‘0’ together with the natural numbers gives us the numbers 0, 1, 2, 3, 4 … which are called whole numbers.
Methods of Numeration
There are two commonly used methods of numeration:
(i). Indian System of Numeration
(ii) International System of Numeration
In Indian system, a number is split up into groups or periods starting from the right. The groups are called ones, thousands, lakhs, crores and arabs. The ones in turn are split into hundreds, tens and units.
Generally, we separate the periods by marking commas (,). In the modern notation, we leave a short space between the periods, instead of using a comma. Thus, 468572309 is written as 46,85,72,309 or 46 85 72 309. It is read as forty-six crore eighty-five lakh seventy-two thousand three hundred nine.
This number represents a collection of 46 crores, 85 lakhs, 72 thousands, 3 hundreds and 9 ones. Therefore, we can write it in expanded form as under: 46,85,72,309 = 4 × 100000000 + 6 × 10000000 + 8 × 1000000 + 5 × 100000 + 7 × 10000 + 2 × 1000 + 3 × 100 + 0 × 10 + 9 × 1.
International System of Numeration
The International System of Numeration is being followed by most of the countries of the world. In this
system also, a number is split up into groups or periods. Starting from the right, the groups are called
ones, thousands, millions and billions. The ones in turn are split into hundreds, tens and ones.
Ex 1.1
(a) 1 lakh = ten ten thousand.
(b) 1 million = ten hundred thousand.
(c) 1 crore = ten ten lakh
(d) 1 crore = ten million
(e) 1 million = ten lakh
(b) Place commas correctly and write the numerals:
(a) Seventy-three lakh seventy-five thousand three hundred seven. Ans 73,75,307
(b) Nine crore five lakh forty-one. ANS:9,05,00,041
(c) Seven crore fifty-two lakh twenty-one thousand three hundred two
ans :7,52,21,302.
(d) Fifty-eight million four hundred twenty- three thousand two hundred two.
Ans: 5,84,23,202
(e) Twenty-three lakh thirty thousand ten.
Ans ;23,30,010.
(c) 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)
(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)
Exercise 1.2
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.
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
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 first week of June worth ₹2,85,891
Books sold in second week of the month worth ₹4,00,768
Therefore, total sale of books in the two weeks together
= ₹2,85,891 + ₹4,00,768 = ₹6,86,659
In the second week of the month, the sale of books was greater.
Difference of the sale of books
= ₹4,00,768 – ₹2,85,891 = ₹1,14,877
Hence, in second week of june, the sale of books was more by ₹1,14,877.
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 are 6, 2, 7, 4, 3
Greatest number = 76432
Least number = 23467
Therefore, difference = 76432 – 23467 = 52,965
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:
Number of screws manufactured in a day = 2,825.
Number of screws manufactured in month of January = 31 x 2825 = 87,575
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 = ₹78,592
Number of radio sets = 40
Price of one radio set = ₹1200
Therefore, cost of 40 radio sets = ₹1200 x 40 = ₹48,000
Remaining money with the merchant = ₹78,592 – ₹48000 = ₹30,592
Hence, amount of ₹30,592 will remain with her after purchasing the radio sets.
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:
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?
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 = 4000215
So, the number of shirts stitched = 18 and the remaining cloth = 130 cm = 1 m 30 cm
Question 11.
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 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, total distance covered in six days = 22 km 500 m.
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:
Quantity of curd in a vessel = 4 1 500 mL = (4 x 1000 + 500) mL = 4500 mL.
Capacity of 1 glass = 25 mL
Therefore number of glasses = 450025 = 180
