 # CH 1 – Knowing Our Numbers

## Fill in the blanks:

1 lakh = _______ ten thousand.
Solution
1 lakh = 1,00,000
= 100 thousand

= 10 ten thousand

1 million = _______ hundred thousand.
Solution
1 million    = 1,000,000
= 1000 thousand
= 10 hundred thousand

1 crore = _______ ten lakh.
Solution
1 crore = 1,00,00,000
= 100 lakh
= 10 ten lakh

1 crore = _______ million.
Solution
1 crore = 1,00,00,000
Adding commas to the number 10000000 according to the International system, we have
10,000,000 = 10 million
1 crore = 10 million

1 million = _______ lakh.
Solution
1 million = 10,000,000
Adding commas to the number 10000000 according to the Indian system, we have
1,00,00,000 = 1 crore = 100 lakh
1 million = 100 lakh

### Place commas correctly and write the numerals:

Seventy three lakh seventy five thousand three hundred seven.
73,75,307

Nine crore five lakh forty one.
9,05,00,041

Seven crore fifty two lakh twenty one thousand three hundred two.
7,52,21,302

Fifty eight million four hundred twenty three thousand two hundred two.
58,423,202

Twenty three lakh thirty thousand ten.
23,30,010

### Insert commas suitably and write the names according to Indian System of Numeration:

87595762
Inserting commas according to the Indian system: 8,75,95,762
Number name: Eight crore seventy five lakh ninety five thousand seven hundred sixty two.

8546283
Inserting commas according to the Indian system: 85,46,283
Number name: Eighty five lakh forty six thousand two hundred eighty three.

99900046
Inserting commas according to the Indian system: 9,99,00,046
Number name: Nine crore ninety nine lakh forty six.

98432701
Inserting commas according to the Indian system: 9,84,32,701
Number name: Nine crore eighty four lakh thirty thousand seven hundred one

### Insert commas suitably and write the names according to International System of Numeration:

78921092
Inserting commas according to the Indian system: 78,921,092
Number name: Seventy eight million nine twenty one thousand ninety two

7452283
Inserting commas according to the Indian system: 7,452,283
Number name: Seven million four hundred fifty two thousand two hundred eighty three

99985102
Inserting commas according to the Indian system: 99,985,102
Number name: Ninety nine million nine hundred eighty five thousand one hundred two.

48049831
Inserting commas according to the Indian system: 48,049,831
Number name: 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:
Total number of tickets sold = Sum of the number of tickets sold on each day.
= Number of tickets sold on the first day + second day + third day + final day
= 1094 tickets + 1812 tickets + 2050 tickets + 2751 tickets
= 7707 tickets
A total of 7707 tickets were sold on all the four days.

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:
Runs scored by Shekhar = 6,980
Runs Shekhar targets to score = 10,000
Shekhar wishes to complete more number of runs than what he has scored so far.
So Shekhar needs 10,000 – 6,980 runs = 3,020 runs

Hence, to complete 10,000 runs, he needs to score 3,020 more runs.

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:
Margin = 5,77,500  – 3,48,700 = 2,28,800 votes
The successful candidate won the election by a margin of 2,28,800 votes.

4. Kirti bookstore sold books worth Rs 2,85,891 in the first week of June and books worth Rs 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:
Sale for the first week = Rs. 2,85,891
Sale for the second week =Rs. 4,00,768
Total sale for the two weeks = 2,85,891 + 4,00,768 = Rs. 6,86,659
Which of the two numbers is greater? 2,85,891 OR 4,00,768
4,00,768 > 2,85,891
The sale for the second week is greater than the sale for the first week by Rs. 1,14,877 (4,00,768 – 2,85,891)

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:
The digits to be used are 6,2,7,4,3
The greatest number can be obtained by writing the digits from the largest to smallest – 76,423
The least number is obtained by writing the digits from smallest to largest – 23,467
Difference = 76,423 – 23,467

Difference between the greatest and the least number that can be written using the given digits only once is 52,965

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 is 2,825 screws
Number of days in the month of January – 31 days
So, number of screws produced in the month of January is 2,825 × 31 = 87,575 screws

7. A merchant had Rs 78,592 with her. She placed an order for purchasing 40 radio sets at Rs 1200 each. How much money will remain with her after the purchase?

Solution:
Cost of each radio set ordered is Rs. 1200
Cost of 40 radio sets ordered for purchase is 40 × 1200 = Rs 48,000
Amount of money the merchant has is Rs 78,592
Remaining amount after the purchase of 40 radio sets is 78,592 – 48,000

So, after the purchase of the 40 radio sets, Rs 30,592 would remain with the merchant.

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:
Difference between 65 and 56 is 9.
The student multiplied the number 7236 by 65 instead of 56.
So, difference in the answer is 7236 × 9 = 65,124
[Let’s check: 7236 × 65 = 4,70,340 and 7236 × 56 = 4,05,216.
Difference is 65,124]

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:
Measure of cloth needed to stitch a shirt is 2m 15 cm
Converting all measures to cm:
1 m is 100 cm
2 m is 200 cm and
40 m is 4,000 cm
Measure of cloth required for one shirt is 2 m 15 cm = 200 + 15 = 215 cm
Number of shirts that can be stitched out of 40 m, that is, 4000 cm cloth is 4,000 ÷ 215

Therefore, 18 shirts can be stitched and 130 cm (100 cm + 30 cm which is1 m 30 cm) cloth would remain.

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

Solution:
Weight of each box of medicine is 4 kg 500 g.
Converting to grams:
1 kg is 1,000 g
4 kg is 4,000 g
800 kg is 8,00,000 g
Weight of each box is 4 kg 500 g = 4,500 g.
The maximum weight the van can carry is 8,00,000 g
Number of boxes that can be loaded is 8,00,000 ÷ 4500

A maximum of 177 boxes can be loaded on the van.

11. The distance between the school and the house of a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.

Solution:
The distance between the school and the house of a student’s house is 1 km 875 m.
Converting to m:
1000 m is 1 km
The distance covered from student’s house to school is 1 km 875 m = 1000 m + 875 m = 1875 m.
The student walks both ways.
So, total distance covered in a day is 1875 + 1875 = 3750 m
Total distance covered by her in six days is 3750 × 6 = 22,500 m
Therefore, the total distance covered by the student in six days is 22,500 m. [22,000 + 500 m = 22 km 500 m]

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:
Total amount of curd in  the vessel is 4 l and 500 ml.
Converting to ml:
1 l is 1000 ml
4 l is 4000 ml
Total amount of curd is 4 l and 500 ml = 4000 + 500 = 4500 ml
Capacity of each glass to be filled is 25 ml
Number of glasses that can be filled is 4500 ÷ 25

Hence, 180 glasses can be filled.

## Exercise 1.3

1. Estimate each of the following using general rule. Make ten more such examples of addition, subtraction and estimation of their outcome.

Solution

1. a)  730 + 998

Round to nearest hundred: 700 + 1000 = 1700

1. b)796 – 314

Round to nearest hundred: 800 – 300 = 500

1. c)  12,904 +2,888

Round to nearest thousand: 13,000 + 3000 = 16,000

1. d)  28,292 – 21,496

Round to the nearest thousand: 28,000 – 21,000 = 7,000

1. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens). Make four more such examples :

Solution:

1. a)   439 + 334 + 4,317

Round to nearest hundred: 400 + 300 + 4,300 = 5,000

Round to nearest tens: 440 + 330 + 4,320 = 5,090

1. b)  1,08,734 – 47,599

Round to nearest hundred: 1,08,700 – 47,600 = 61,100

Round to nearest tens: 1,08,730 – 47,600 = 61,130

1. c)  8325 – 491

Round to nearest hundred: 8,300 – 500 = 7,800

Round to nearest tens: 8,330 – 490 = 7,840

1. d)4,89,348 – 48,365

Round to nearest hundred: 4,89,300 – 48,400 = 4,40,900

Round to nearest tens: 4,89,350 – 48,370 = 4,40,980

1. Estimate the following products using general rule. Make four more such examples.

Solution:

1. a)  578 × 161

Rounding off to nearest hundred: 600 × 200 = 1,20,000

1. b)  5281 × 3491

Rounding off to nearest thousands: 5000 × 3000 = 1,50,00,000

1. c)  1291 × 592

Rounding off 1291 to 1300 (rounding off to hundreds) and 592 to 600(rounding off to hundreds)

1300 × 600 = 7,80,000

1. d)9250 × 29

Rounding off 9250 to 9000 (rounding off to thousands) and 29 to 30(rounding off to tens)

9000 × 30 = 2,70,000

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