Fill in the blanks:
(a) 1 lakh = ……….. ten thousand.
(b) 1 million = ……… hundred thousand.
1.(c) 1 crore = ……… ten lakhs.
(d) 1 crore = ……… million.
(e) 1 million = ……… lakhs.
Fill in the blanks:
(a) 1 lakh = ……….. ten thousand.
(b) 1 million = ……… hundred thousand.
1.(c) 1 crore = ……… ten lakhs.
(d) 1 crore = ……… million.
(e) 1 million = ……… lakhs.
Explanation:
(a) 1 lakh is same as 100000 and 100000 is same as 10 ten thousand. It can be also written as 1,00,000. Hence,
⇨ 1 lakh = 10 ten thousand
(b) 1 million is same as 1000000 and 1000000 is same as 10 hundred thousand. It can be also written as 10,00,000. Hence,
⇨ 1 million = 10 hundred thousand
(c) 1 crore is same as 10000000 and 10000000 is same as 10 ten lakhs. It can be also written as 1,00,00,000. Hence,
⇨ 1 crore = 10 ten lakhs
(d) 1 crore is same as 10000000 and 10000000 is same as 10 ten lakhs or 10 million. It can be also written as 1,00,00,000. Hence,
⇨ 1 crore = 10 million
(e) 1 million is same as 1000000 and 1000000 is same as 10 hundred thousand or 10 lakhs. It can be also written as 10,00,000. Hence,
⇨ 1 million = 10 lakhs
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
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
Explanation:
(a) Seventy three lakh seventy five thousand three hundred seven can be represented as the numeral 7375307. After placing all the commas it looks like- 73,75,307
(b) Nine crore five lakh forty one can be represented as the numeral 90500041. After placing all the commas it looks like- 9,05,00,041
(c) Seven crore fifty two lakh twenty one thousand three hundred two can be represented as the numeral 75221302. After placing all the commas it looks like- 7,52,21,302
(d) Fifty eight million four hundred twenty three thousand two hundred two can be represented as the numeral 58423202. After placing all the commas it looks like- 5,84,23,202.
(e) Twenty three lakh thirty thousand ten can be represented as the numeral 2330010. After placing all the commas it looks like- 23,30,010
3.Insert commas suitably and write the names according to the Indian System of Numeration:
(a) 87595762 (b) 8546283 (c) 99900046 (d) 98432701
3.Insert commas suitably and write the names according to the Indian System of Numeration:
(a) 87595762 (b) 8546283 (c) 99900046 (d) 98432701
Explanation:
(a) After inserting the commas the number 87595762 looks like- 8,75,95,762.
Now, according to Indian system this can be written as-
Eight crore seventy five lakh ninety five thousand seven hundred sixty two.
(b) After inserting the commas the number 8546283 looks like- 85,46,283.
Now, according to Indian system this can be written as-
Eighty five lakh forty six thousand two hundred eighty three.
(c) After inserting the commas the number 99900046 looks like- 9,99,00,046
Now, according to Indian system this can be written as-
Nine crore ninety nine lakh forty six.
(d) After inserting the commas the number 98432701 looks like- 9,84,32,701
Now, according to Indian system this can be written as-
Nine crore eighty four lakh thirty two thousand seven hundred one.
4.Insert commas suitably and write the names according to the International System of Numeration:
(a) 78921092 (b) 7452283 (c) 99985102 (d) 48049831
4.Insert commas suitably and write the names according to the International System of Numeration:
(a) 78921092 (b) 7452283 (c) 99985102 (d) 48049831
Explanation:
(a) After inserting the commas the number 78921092 looks like- 78,921,092
Now, according to International system this can be written as-
Seventy eight million nine hundred twenty one thousand ninety two.
(b) After inserting the commas the number 7452283 looks like- 7,452,283
Now, according to International system this can be written as-
Seven million four hundred fifty-two thousand two hundred eighty three.
(c) After inserting the commas the number 99985102 looks like- 99,985,102
Now, according to International system this can be written as-
Ninety-nine million nine hundred eighty five thousand one hundred two.
(d) After inserting the commas the number 48049831 looks like- 48,049,831
Now, according to International system this can be written as-
Forty-eight million forty-nine thousand eight hundred thirty-one.
5.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 four days
5.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 four days
Explanation:
It has been given that,
On the first day, number of tickets sold are = 1094
On the second day, number of tickets sold are = 1812
On the third day, number of tickets sold are = 2050
On the fourth day, number of tickets sold are = 2751
So, on the final day total tickets sold = 1812 + 2751 + 2050 + 1094
= 7707 tickets.
Total tickets sold is equal to 7707.
6.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?
6.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?
Explanation:
Score made by Shekhar = 6980 runs
Total runs he wish to make = 10000 runs
More run that needs to be made = Total runs needed – Present score
= 10000 – 6980
= 3020
So, Shekhar needs to score 3020 more suns to reach his target.
7.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?
7.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?
Explanation:
The successful candidate received = 577500 votes.
His opponent received = 348700 votes.
He won the election by a margin of = 577500 – 348700 = 228800 votes.
So, The successful candidate received 228800 votes
8.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?
8.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?
Explanation:
Books that have been sold in first week of June has a cost = Rs 285891.
In the second week of June, the price of books sold was = Rs 400768.
Total number of books that have been sold in both of the weeks = Rs 400768 + Rs 285891 (simplify)
= Rs 686659
The second week has the highest book sales.
Difference in sales between the two weeks = Rs 400768 – Rs 285891 (simplify)
= Rs 114877
The second week's sales were Rs 114877 higher than the first.
9.Find the difference between the greatest and the least 5-digit number that can be written using the digits 6, 2, 7, 4, and 3 each only once
9.Find the difference between the greatest and the least 5-digit number that can be written using the digits 6, 2, 7, 4, and 3 each only once
Explanation:
The numbers are 6, 2, 7, 4, 3
76432 is the greatest obtained 5-digit number.
23467 is the smallest obtained 5-digit number.
The difference among the two numbers is calculated as 76432 – 23467 (simplify)
= 52965.
52965 is the difference among the two numbers.
10.A machine, on average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
10.A machine, on average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Explanation:
The number of screws produced per day is 2825.
Because the month of January has 31 days, the amount of screw produced in January is =
31×2825 = 87575.
As a result, the machine generated 87575 screws in January 2006.
11.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?
11.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?
Explanation:
Total money the merchant had of Rs. 78592.
40 radio sets were included in the order she placed for them.
Each radio set cost is = Rs 1200
40 radios cost Rs 1200 each, multiplied by 40, is Rs 48000.
Money that will be left with the merchant = Rs 78592 – Rs 48000 (simplify)
= Rs 30592
As a result, after purchasing radio sets, the merchant has Rs 30592 in cash.
12.A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?
12.A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer?
Explanation:
The subtraction of 65 and 56 is (65 – 56) = 9.
The discrepancy between the right answer and the wrong one is equal to 7236 × 9 (simplify)
= 65124.
Thus, by 65124, the result was larger than the right value.
13.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?
13.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?
Explanation:
It has been given that,
The cloth is 40 metres long in total.
= 40 × 100 cm (simplify)
= 4000 cm
Cloth needed to stitch each shirt is = 2m 15cm
= 2 × 100 + 15 cm (simplify)
= 215 cm
The number of shirts that may be sewn from 4000 cm is 4000/215, which equals 18 shirts.
Thus, 18 shirts could be stitched from 40 m, and still we will be left with 1 m 30 cm of cloth.
14.It has been given that,
The cloth is 40 metres long in total.
= 40 × 100 cm (simplify)
= 4000 cm
Cloth needed to stitch each shirt is = 2m 15cm
= 2 × 100 + 15 cm (simplify)
= 215 cm
The number of shirts that may be sewn from 4000 cm is 4000/215, which equals 18 shirts.
Thus, 18 shirts could be stitched from 40 m, and still we will be left with 1 m 30 cm of cloth.
14.It has been given that,
The cloth is 40 metres long in total.
= 40 × 100 cm (simplify)
= 4000 cm
Cloth needed to stitch each shirt is = 2m 15cm
= 2 × 100 + 15 cm (simplify)
= 215 cm
The number of shirts that may be sewn from 4000 cm is 4000/215, which equals 18 shirts.
Thus, 18 shirts could be stitched from 40 m, and still we will be left with 1 m 30 cm of cloth.
Explanation:
One box weighs 4 kg 500 g, or = 4 × 1000 + 500 g
= 4500 g
Maximum weight that can be carried by van = 800 kg
= 800 × 1000 g (because 1kg = 1000g)
= 800000 g
As a result, 177 is the number of boxes that can be loaded into the van using the formula 800000/4500.
15.The distance between the school and 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.
15.The distance between the school and 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.
Explanation:
Distance travelled between both her house and her school = 1 km 875 m (change into metres)
= 1000 + 875 (add)
= 1875m
As the student is walking both directions again,
The length covered by the student per day = 2 × 1875 (simplify)
= 3750 m
Length covered by the student in six days = 3750 m × 6
= 22500 m (rewrite)
= 22km 500m
So, the total distance travelled by the student in the span of six days is 22 km and 500 m.
16.A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
16.A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Explanation:
Amount of curd inside the vessel = 41500 ml
= 1000×4 + 500 (simplify)
= 4500 ml
Volume of 1 glass is = 25 ml
So, Quantity of glasses which can be packed by curd = 4500/25 (divide)
= 180 glasses
Thus, 180 glasses can be packed by curd.
17.Estimate each of the following using the general rule:
(a) 730 + 998 (b) 796 – 314 (c) 12904 + 2888 (d) 28292 – 21496
Make ten more such examples of addition, subtraction and estimation of their outcome.
17.Estimate each of the following using the general rule:
(a) 730 + 998 (b) 796 – 314 (c) 12904 + 2888 (d) 28292 – 21496
Make ten more such examples of addition, subtraction and estimation of their outcome.
Explanation:
A) 730 + 998
Round to the nearest hundred
730 is rounded to 700
998 is rounded to 1000
So, 730 + 998
= 700 + 1000 (simplify)
= 1,700
(b) 314 – 796
Round to the nearest hundred
796 is rounded to 800
314 is rounded to 300
So, 796 – 314
= 800 – 300 (simplify)
= 500
(c) 12904 + 2888
Round to the nearest thousand
12904 is rounded to 13000
2888 is rounded to 3000
So, 12904 + 2888
= 13000 + 3000 (simplify)
= 16,000
(d) 28292 – 21496
Round to the nearest thousand
28292 is rounded to 28000
21496 is rounded to 21000
So, 28292 – 21496
= 28000 – 21000 (simplify)
= 7,000
Ten more examples similar to above are
(1) 430 + 380 = 400 + 400 = 800
(2) 4937 + 6990 = 5000 + 7000 = 12000
(3) 5392 – 2772 = 5000 – 3000 = 2000
(4) 4440 – 3972 = 4000 – 4000 = 0
(5) 3175 + 2206 = 3000 + 2000 = 5000
(6) 3110 – 2292 = 3000 – 2000 = 1000
(7) 610 + 375 = 600 + 400 = 1000
(8) 7400 – 5900 = 7000 – 6000 = 1000
(9) 4731 + 3300 = 5000 + 3000 = 8000
(10) 8485 – 2319 = 8000 – 2000 = 6000
18.Give a rough estimate (by rounding off to the nearest hundreds) and also a closer estimate (by rounding off to the nearest tens):
(a) 439 + 334 + 4317 (b) 108734 – 47599 (c) 8325 – 491 (d) 489348 – 48365
Make four more such examples
18.Give a rough estimate (by rounding off to the nearest hundreds) and also a closer estimate (by rounding off to the nearest tens):
(a) 439 + 334 + 4317 (b) 108734 – 47599 (c) 8325 – 491 (d) 489348 – 48365
Make four more such examples
Explanation:
(a) 439 + 334 + 4317
Round to the nearest hundred
439 + 334 + 4317 (rewrite)
= 400 + 300 + 4300 (simplify)
= 5,000
Round to the nearest ten
439 + 334 + 4317 (rewrite)
= 440 + 330 + 4320 (simplify)
= 5,090
19.(b) 108734 – 47599
19.(b) 108734 – 47599
Explanation:
Round to the nearest hundred
108734 – 47599 (rewrite)
= 108700 – 47600 (simplify)
= 61,100
Round to the nearest ten
108734 – 47599 (rewrite)
= 108730 – 47600 (simplify)
= 61,130
20(c) 8325 – 491
20(c) 8325 – 491
Explanation:
Round to the nearest hundred
491 – 8325 (rewrite)
= 500 – 8300 (simplify)
= 7,800
Round to the nearest ten
491 – 8325 (rewrite)
= 490 – 8330 (simplify)
= 7,840
21.(d) 489348 – 48365
21.(d) 489348 – 48365
Explanation:
Round to the nearest hundred
489348 – 48365 (rewrite)
= 489300 – 48400 (simplify)
= 4,40,900
Round to the nearest ten
489348 – 48365 (rewrite)
= 489350 – 48370 (simplify)
= 4,40,980
Four more examples similar to above are
(i) 5853 + 762
Round to the nearest hundred
5853 + 762 = 5800 + 800
= 6,600
Round to the nearest ten
5853 + 762 = 5850 + 760
= 6,610
22.(ii) 875 – 490
22.(ii) 875 – 490
Explanation:
Round to the nearest hundred
875 – 490 = 900 – 500
= 400
Round to the nearest ten
875 – 490 = 880 – 500
= 380
23.(iii) 8375 – 1875
23.(iii) 8375 – 1875
Explanation:
Round to the nearest hundred
8375 – 1875 = 8400 – 1900
= 6,500
Round to the nearest ten
8375 – 1875 = 8380 – 1880
= 6,500
25.(iv) 9246 – 3312
25.(iv) 9246 – 3312
Explanation:
Round to the nearest hundred
9246 – 3312 = 9200 – 3300
= 5,900
Round to the nearest ten
9246 – 3312 = 9240 – 3310
= 5,930
26.Estimate the following products using the general rule:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Make four more such examples.
26.Estimate the following products using the general rule:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Make four more such examples.
Explanation:
(a) 578 × 161
Using the general rule to round off
578 and 161 can be round off to 600 and 200
= 600 × 200 (simplify)
= 1,20,000
(b) 5281 × 3491
Using the general rule to round off
5281 and 3491 can be round off to 5000 and 3500
= 5000 × 3500 (simplify)
= 1,75,00,000
(c) 1291 × 592
Using the general rule to round off
1291 and 592 can be round off to 1300 and 600
= 1300 × 600 (simplify)
= 7,80,000
(d) 9250 × 29
Using the general rule to round off
9250 and 29 can be round off to 9000 and 30
= 9000 × 30 (simplify)
= 2,70,000
Four more examples similar to above are-
(i) 430 × 380 = 400 × 400 = 1,60,000
(ii) 4937 × 6990 = 5000 × 7000 = 35,000,000
(iii) 5392 × 2772 = 5000 × 3000 = 15,000,000
(iv) 4440 × 3972 = 4000 × 4000 = 16,000,000
Also Read: Knowing Our Numbers Class 6 Extra Questions
Also Read: Chapter 2: Whole Numbers
