Given, weight of one tablet = 20 mg
Total weight of 200000 tablets in the box = 200000 \(\times\) 20 mg
=4000000 mg
Now, weight in grams =\(\frac{4000000}{1000}g=4000g\) \(​​\left[ \because \quad 1mg=\frac { 1 }{ 1000 } g \right] \)
Now, weight in kilograms =\(\frac{4000}{1000}kg=4kg\) \(​​\left[ \because \quad 1g=\frac { 1 }{ 1000 } kg \right] \).

Using the digits 6, 2, 7, 4 and 3, we get
Greatest number = 76,432 and least number = 23,467
∴ Difference = Greatest number - Least number
=76,432 - 23,467 = 52,965.

Given, distance between the school and the house
= 1 km 875 m =1\(\times\)1000+875
= 1000 + 875 = 1875 m [∵ 1 km = 1000 m]
∴ Distance covered by the student in one day
= 2\(\times\) 1875 = 3750 m [∵ She walks both ways in one day]
∴ Distance covered in 6 days = 6 x 3750
= 22500 m = (22 \(\times\) 1000 + 500) m
= 22 km 500 m    [∵1 km = 1000 m]

Rs. 16570

We have, 439 + 334 + 4317
\(\therefore \begin{matrix} 439\rightarrow 400 \\ 334\rightarrow 300 \\ 4317\rightarrow 4300 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad hundreds]\)
∴ Rough estimate = 439 + 334 + 4317
= 400 + 300 + 4300 = 5000
\(Again\begin{matrix} 439\rightarrow 440 \\ 334\rightarrow 330 \\ 4317\rightarrow 4320 \end{matrix}\} [rounding\quad off\quad to\quad nearest\quad tens]\)
and closer estimate = 439 +334 +4317
= 440 + 330 + 4320 = 5090

We can make many situations of this kind, five situations of them are as follows:
Situation 1. Marry is a maid (servant). She charges Rs. 8 per day to clean a house but gives back Rs. 3 to the house owner for saving. What amount does she carry in her hand from 7 houses?
Situation 2. Ankit's pocket money is Rs. 8 per day. He spents Rs. 3 daily. Find his total saving in 7 days.
Situation 3. What will be the seven times the difference of eight and three?
Situation 4. A clerk works 8 h daily in an office, but he remains absent for 3 h daily. In a week, for how many hours did he work?
Situation 5. Seven children with Rs. 8 each went to market. Each of them bought pencils costing Rs. 3. What is total money left with them?

According to the Indian and International system of numeration, 2004004

Distance covered in one day = 365 m
Number of days = 120
So, total distance covered by Tony in 120 days
= 365 \(\times\) 120 m = 43800 m
∴ Estimated product = 44000 m = 44 km

Total cost of 20 flats = 348642060
Cost of each flat = 348642060 + 20 = Rs. 17432103

1325 g or 1 kg 325 g

We have, 92 = XCII  [∵ 92=100-10+1+1 i.e.100=C,10=X, 1 = I]

∵ 28,292 ⟶ 28,000[rounding off to nearest thousands]
21,496 ⟶ 21,000 [rounding off to nearest thousands]
∴ Estimated difference = 28000 - 21000 = 7000

According to the question, sum of 9 and 2 = 9 + 2
Now, mutiplying it by 4, we have 4 \(\times\) (9 + 2)
which is the required expression.

80000

8000

199

Here, given digits are 3, 5, 4, 6. For the smallest number, we write the digits in ascending order. So, the smallest four-digit number is 3456.

Expanded form of given numbers as
76496 = 7 \(\times\) 10000 + 6 \(\times\) 1000 + 4 \(\times\) 100 + 9 \(\times\) 10 + 6

The place value of 6 in 762540 = 6 \(\times\)10000
=60000

In 4942, ones digit is 2 < 5.
Then, rounded form of 4942 = 4940

Given, total money = Rs.10,00,000
Money spent on a colour TV = Rs. 16,580
Money spent on a motorcycle = Rs. 45,890
and money spent on a flat = Rs. 8,70,000
Total amount spent = 16,580 + 45,890 + 8,70,000
= Rs. 9,32,470
∴ Money left with him = 10,00,000 - 9,32,470
= Rs.67,530
Hence, Rs. 67,530 was left with him.

Given, Chinmay's total money =Rs.6,10,000
Money given to Jyoti by Chinmay = Rs. 87,500
Money given to Javed by Chinmay =Rs. 1,26,380
Money given to John by Chinmay = Rs. 3,50,000
∴ Money left with Chinmay = Total money - Distributed money
=6,10,000 -(87,500 + 1,26,380 + 3,50,000)
= 6,10,000 - 5,63,880 =Rs. 46,120
Hence, Rs. 46,120 was left with him.
The value depicted by Chinmay is that he is helpful nature towards his friends.

According to the question,
Digit at ten's place = 4
Digit at unit's place = \(\frac{1}{4}\) of ten's place digit =\(\frac{1}{4}\times4=1\)
Digit at hundred's place = 0
Digit at thousand's place = 5 x Digit of unit's place
=5\(\times\)1=5
Digit at ten thousand's place = 2 x Digit of ten's place
=2\(\times\)4=8
∴ Required number = 85041

According to the question,
A garment factory produced shirts = 216315
Produced trousers = 182736
Produced jackets = 58704
∴Total production of all the three items in that year
= Sum of all items
= 216315 + 182736 + 58704 = 457755
Hence, the total production of all the three items in that year is 457755.

m 5 cm

Total expenditure = 2725 + 2275 = 5000

Difference = 10,000 - 9999 = 1

Difference = 20975 - 15875 = 5100

1 m 25 cm = 125 cm
10 m = 1000 cm
1000 ÷ 125 = 8

100 x 89 = 8900
100 x 79 = 7900
8900 - 7900 = 1000

Number of days in
November, 2009 = 30
25 x 30 = 750