Page 12
Exercise 1.1
Q1. Fill in the blanks:
(a) 1 lakh = _______ ten thousand.
(b) 1 million = _______ hundred thousand.
(c) 1 crore = _______ ten lakh.
(d) 1 crore = _______ million.
(e) 1 million = _______ lakh.
Answer:
(a) 1 Lakh = 1,00,000
|
Lakh
|
TTh
|
Th
|
H
|
T
|
O
|
|
1
|
0
|
0
|
0
|
0
|
0
|
or 10 × 10,000
Therefore, 1 lakh = 10 ten thousand.
(b) 1 million = 10,00,000
|
TLakh
|
Lakh
|
TTh
|
Th
|
H
|
T
|
O
|
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
or 10 × 100,000
Therefore, 1 million = 10 hundred thousand.
(c) 1 crore = 1,00,00,000
|
Cr
|
TLakh
|
Lakh
|
TTh
|
Th
|
H
|
T
|
O
|
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
or 10 × 1000,000
Therefore, 1 crore = 10 ten lakh.
(d) 1 crore = 1,00,00,000
|
Cr
|
TLakh
|
Lakh
|
TTh
|
Th
|
H
|
T
|
O
|
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
One million is a thousand thousands, so 1 million = 1,000,000
|
TLakh
|
Lakh
|
TTh
|
Th
|
H
|
T
|
O
|
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
Therefore, 1 crore = 10 million
(e) 1 million = 1000000
|
TLakh
|
Lakh
|
TTh
|
Th
|
H
|
T
|
O
|
|
1
|
0
|
0
|
0
|
0
|
0
|
0
|
Therefore, 1 million = 10 Lakh
Q2. 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.
Answer:
|
|
Number Name
|
Cr
|
TLakh
|
Lakh
|
TTH
|
Th
|
H
|
T
|
O
|
Number
|
|
a.
|
Seventy three lakh seventy five thousand three hundred seven
|
-
|
7
|
3
|
7
|
5
|
3
|
0
|
7
|
73,75,307
|
|
b.
|
Nine crore five lakh forty one
|
9
|
0
|
5
|
0
|
0
|
0
|
4
|
1
|
9,05,00,041
|
|
c.
|
Seven crore fifty two lakh twenty one thousand three hundred two
|
7
|
5
|
2
|
2
|
1
|
3
|
0
|
2
|
7,52,21,302
|
|
d.
|
Fifty eight million four hundred twenty three thousand two hundred two
|
5
|
8
|
4
|
2
|
3
|
2
|
0
|
2
|
58,423,202
|
|
e.
|
Twenty three lakh thirty thousand ten
|
-
|
2
|
3
|
3
|
0
|
0
|
1
|
0
|
23,30,010
|
Q3. Insert commas suitably and write the names according to Indian System of Numeration:
(a) 87595762
(b) 8546283
(c) 99900046
(d) 98432701
Answer:
|
|
Number
|
Cr
|
TLakh
|
Lakh
|
TTH
|
Th
|
H
|
T
|
O
|
Number Name
|
|
a.
|
8,75,95,762
|
8
|
7
|
5
|
9
|
5
|
7
|
6
|
2
|
Eight crore seventy five lakh ninety five thousand seven hundred sixty two
|
|
b.
|
85,46,283
|
-
|
8
|
5
|
4
|
6
|
2
|
8
|
3
|
Eighty five lakh forty six thousand two hundred eighty three
|
|
c.
|
9,99,00,046
|
9
|
9
|
9
|
0
|
0
|
0
|
4
|
6
|
Nine crore ninety nine lakh forty six
|
|
d.
|
9,84,32,701
|
9
|
8
|
4
|
3
|
2
|
7
|
0
|
1
|
Nine crore eighty four lakh thirty two thousand seven hundred one
|
Q4. Insert commas suitably and write the names according to International System of Numeration:
(a) 78921092
(b) 7452283
(c) 99985102
(d) 48049831
Answer:
|
|
Number
|
Cr
|
TLakh
|
Lakh
|
TTH
|
Th
|
H
|
T
|
O
|
Number Name
|
|
a.
|
78,921,092
|
7
|
8
|
9
|
2
|
1
|
0
|
9
|
2
|
Seventy eight million nine hundred twenty one thousand ninety two
|
|
b.
|
7,452,283
|
-
|
7
|
4
|
5
|
2
|
2
|
8
|
3
|
Seven million four hundred fifty two thousand two hundred eighty three
|
|
c.
|
99,985,102
|
9
|
9
|
9
|
8
|
5
|
1
|
0
|
2
|
Ninety nine million nine hundred eighty five thousand one hundred two
|
|
d.
|
48,049,831
|
4
|
8
|
0
|
4
|
9
|
8
|
3
|
1
|
Forty eight million forty nine thousand eight hundred thirty one
|
Page 16
Exercise – 1.2
Q1. 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.
Answer:
Number of tickets sold on 1st day = 1,094
Number of tickets sold on 2nd day = 1,812
Number of tickets sold on 3rd day = 2,050
Number of tickets sold on 4th day = 2,751
Total number of tickets sold = <MathML>
Therefore, the total number of tickets sold on all the four days was 7,707.
Q2. 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?
Answer:
Runs to complete = 10,000
Runs already scored = 6,980
Runs required = <MathML>
Therefore, he needs 3,020 more runs.
Q3. 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?
Answer:
Number of votes registered by successful candidate = 5,77,500
Number of votes secured by his opponent = 3,48,700
Margin between them = <MathML>
Therefore, the margin by which the successful candidate won the election was 2,28,800 votes.
Q4. 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?
Answer:
Books sold by Kirti in 1st week of June = Rs. 2,85,891
Books sold by Kirti in 2nd week of June = Rs. 4,00,768
Total sale for the two weeks = <MathML>
Therefore, the sale for the two weeks together was Rs. 6,86,659
Since, 4,00,768 > 2,85,891
Therefore sale of 2nd week was greater than that of 1st week.
Books sold in 2nd week – Books sold in first week = <MathML>
Therefore, the books sold in 2nd week were 1,14,877 more than week 1.
Q5. 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.
Answer:
By arranging the given digits in descending order we get the largest number = 76,432
By arranging the given digits in ascending order we get the smallest number = 23,467
Difference between the greatest and the least number written using digits 6, 2, 7, 4, 3 each only once will be
<MathaML>
Q6. A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Answer:
Number of screws manufactured in one day = 2,825
Number of days in the month of January is 31.
Total number of screws produced in January = <MathML>
Therefore the machine produced 87,575 screws in the month of January 2006.
Q7. 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?
Answer:
The cost of one radio set = Rs. 1200
Total cost of 40 radio sets = <MathML>
Now,
Total money with the merchant = Rs. 78,592
Money spent by her = Rs. 48,000
Money remaining with her = <MathML>
Therefore, Rs. 30,592 will remain with her after the purchase.
Q8. 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?)
Answer:
Student multiplied = 7236 × 65
Multiplication to be done = 7236 × 56
Difference between the students answer and the correct answer will be
= 7236 × (65 – 56)
= 7236 × 9
= 65,124
Therefore, the student’s answer was 65,124 greater than the correct answer.
Q9. 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.)
Answer:
Cloth required to stitch one shirt = 2 m 15 cm
= (2 x 100) cm + 15 cm
= 215 cm
Length of cloth = 40 m
= 40 x 100 cm
= 4000 cm
Number of shirts that can be stitched = 4000/215
= 18
For 18 shirts cloth required will be = 215 × 15
= 3870 cm
Total cloth was 4000, so the cloth that will be left is
= 4000 – 3870
= 130 cm
Therefore, 18 shirts can be stitched and the cloth remaining will be 130 cm.
Q10. 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?
Answer:
Weight of one box = 4 kg 500 g
= (4 x 1000) g + 500 g
= 4500 g
Maximum load van can carry = 800 kg
= (800 x 1000) g
= 800000 g
Number of boxes = 800000/4500
= 177
Therefore, 177 boxes can be loaded in a van which cannot carry beyond 800 kg.
Q11. The distance between the school and the house of 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.
Answer:
Distance between school and home = 1.875 km
Total distance covered in 1 day = 1.875 × 2
= 3.750 km
Distance covered in 6 days = 3.750 × 6
= 22.500 km
Therefore, the total distance covered by the student in six days is 22 km 500 m.
Q12. A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Answer:
Quantity of curd = 4 litres 500 ml
= (4 x 1000) ml + 500 ml
= 4500 ml
Capacity of one glass = 25 ml
Number of glasses that can be filled = 4500 / 25
= 180
Therefore, 180 glasses can be filled.
Page 23
Exercise – 1.3
Q1. Estimate each of the following using general rule:
(a) 730 + 998
(b) 796 – 314
(c) 12,904 + 2,888
(d) 28,292 – 21,496
Answer:
(a) 730 + 998
Round off to hundreds
730 rounding off to 700
998 rounding off to 1000
Estimated sum = 1700
(b) 796 – 314
Round off to hundreds
796 rounding off to 800
314 rounding off to 300
Estimated difference = 500
(c) 12,904 + 2,888
Round off to thousands
12904 rounding off to 13000
2888 rounding off to 3000
Estimated sum = 16000
(d) 28,292 – 21,496
Round off to thousands
28292 rounding off to 28000
21496 rounding off to 21000
Estimated difference = 7000
Q2. Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens) :
(a) 439 + 334 + 4,317
(b) 1,08,734 – 47,599
(c) 8325 – 491
(d) 4,89,348 – 48,365
Answer:
(a) 439 + 334 + 4,317
Rough estimate (by rounding off to nearest hundreds)
439 rounding off to 400
334 rounding off to 300
4317 rounding off to 4300
Estimated sum = 5000
Closer estimate (by rounding off to nearest tens)
439 rounding off to 440
334 rounding off to 330
4317 rounding off to 4320
Estimated sum = 5090
(b) 1,08,734 – 47,599
Rough estimate (by rounding off to nearest hundreds)
108734 rounding off to 108700
47599 rounding off to 47600
Estimated difference = 61100
Closer estimate (by rounding off to nearest tens)
108734 rounding off to 108730
47599 rounding off to 47600
Estimated difference = 61130
(c) 8325 – 491
Rough estimate (by rounding off to nearest hundreds)
8325 rounding off to 8300
491 rounding off to 500
Estimated difference = 7800
Closer estimate (by rounding off to nearest tens)
8325 rounding off to 8330
491 rounding off to 490
Estimated difference = 7840
(d) 4,89,348 – 48,365
Rough estimate (by rounding off to nearest hundreds)
489348 rounding off to 489300
48365 rounding off to 48400
Estimated difference = 440000
Closer estimate (by rounding off to nearest tens)
489348 rounding off to 489350
48365 rounding off to 48370
Estimated difference = 440000
Q3. Estimate the following products using general rule:
(a) 578 × 161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Answer:
The general rule is, Round off each factor to its greatest place, then multiply the rounded off factors.
(a) 578 × 161
Round off to hundreds
578 rounds off to 600
161 rounds off to 200
The estimated product = 600 x 150
= 1,20,000
(b) 5281 × 3491
Round off to thousands
5281 rounding off to 5000
3491 rounding off to 3000
The estimated product = 5000 x 3500
= 1,50,00,000
(c) 1291 × 592
1291 rounding off to 1000
592 rounding off to 600
The estimated product = 1000 x 600
= 6,00,000
(d) 9250 × 29
9250 rounding off to 9000
29 rounding off to 30
The estimated product = 9000 x 30
= 2,70,000