Maths 3 Mark Question

One crore seventy-five thousand three hundred two:In order to write the smallest 8-digit number using digits 0, 1, 2, 3, 5 and 7, we put the smallest digit 1 (Except 0) at the place having the highest place value. The largest digit 7 is put at the rightmost place i.e. at unit’s place, the digit 5 is put at the ten’s place, the digit 3 is put at the hundred’s place and the digit 2 is put at the thousand’s place. All other places are filled by 0. Hence, the required largest number is 10002357.
In order to write the largest 8-digit number using digits 0, 1, 2, 3, 5 and 7, we put the largest digit 7 at the place having the highest place value. The smallest digit 5 is put at the place after the highest place value. We put the next smallest digit (i.e., 3) after the previous one. After it we place the next smallest digit (i.e., 2) and after that we put the digit 1. All other places are filled by 0. Hence, the required largest number is 75321000.

Given data : Each tablet weighs = 20mg
Therefore, The weight of 2,00,000 tablets = 2,00,000 × 20 = 40,00,000mg
Therefore, The total weight of all the tablets in the box = 40,00,000mg
We know 1g = 1,000mg
Weight of the box having all tablets = 40,00,000 ÷ 1,000 = 4000g
And, as 1kg = 1,000g
Therefore, Weight of the box having all tablets = 4,000 ÷ 1,000 = 4000g = 4kg

In order to write the largest 4-digit number using digits 0, 5, 7 and 8, we put the largest digit 8 at the place having the highest place value. The smallest digit 0 is put at the right most place i.e. at unit's place, the digit 7 is put at the hundred's place and the digit 5 is put at the ten's place. Hence, the required largest number is 8750.
In order to write the smalest 4-digit number using digits 0, 5, 7 and 8, we put the smallest digit 5 (Except 0) at the place having the highest place value. The largest digit 8 is put at the right most place i.e. at unit's place, the digit 7 is put at the ten's place and the digit 0 is put at the hundred's place. Hence, the required largest number is 5078.

In 475 the last two digits i.e., 75 is greater than 50.
Hence, 1 will be added to the 3rd last digit and the last two digits will become zero.
After rounding to nearest hundred we will get 500.
In 225 the last two digit i.e., 25 is less than 50.
Hence, the 3rd last will remain same and the last digit will become zero.
After rounding to nearest ten we will get 200.
Required product 500 × 200 = 1,00,000

As given in the question,
Number of pages in 1 copy of newspaper = 12
Therefore, Number of pages in 11,980 copies of newspaper
= 11, 980 × 12 = 1,43,760
Thus, 1,43,760 pages are printed every day.
Now, number of pages printed every day = 1,43,760
Therefore, Number of pages printed in a month = 1,43,760 × 30 = 43,12,800
Thus, 43,12,800 pages are printed in a month.

The population of Sundar Nagar in 2001 = Sum of the population of city in 1991 + Increase in population over the given time period
As given in the question, The population of Sundar Nagar in 1991 = 2,35,471
As given in the question,
Increase in population over the given time period = 72.958
Therefore, The population of Sundar Nagar in 2001
= 2,35,471 + 72,958 = 3,08,429

To write the greatest 6-digit number having three different digits, we will have to use three largest digits. Clearly 9, 8 and 7 are three largest digits. In order to write the largest 6-digit number using digits 7, 8 and 9, we put the largest digit 9 at the place having the highest place value. The smallest digit
7 is put at the right most place i.e. at unit's place and the digit 8 is put at the ten's place. All other places are filled by 9.
Hence, the required number = 999987.

The smallest digit is 0. But we cannot use 0 at the place having the highest place value in six digit numbers. So, we will use the second smallest digit i.e., 1. All other places are filled by 9.
Hence, the required number = 100000
Smallest six digit number will be 100000.
The largest digit is 9.
We can use 9 at any place. In fact, we can use 9 in all places in six digit numbers.
Hence, the required number = 999999
Largest six digit number will be 999999
Required difference = 999999 - 100000 = 899999
So, the total numbers between 999999 and 100000 will be 899998.

Ravish’s initial money = Rs.78,592
He purchased 39 radio sets at Rs.1234 each.
Therefore, Money spent by him on purchasing 39 radio sets
= 1,234 × 39 = Rs. 48,126
Therefore, Remaining money with Ravish after the purchase = Initial money-Money spent on purchasing 39 radio sets = Rs. 78,592 - Rs. 48,126 = Rs. 30,466
Thus, 230,466 are left with him after the purchase.

Margin of victory in the election for the successful candidate = Number of votes registered by the winner-Number of votes secured by nearest rival candidate
Votes registered by the winner = 5,77,570
Votes secured by the rival = 3,48,685
Therefore, Margin of victory for the successful candidate
= 5,77,570 - 3,48,685 = 2,28, 885

Therefore, Distance between the school and the house of a student
= 1,875m = 1,000 + 875 = 1,875m (1km = 1,000m)
As given in the question, Distance covered by a student in a day
= 2 × 1,875 = 3,750m
Total distance covered by her in a week = 7 × 3,750 = 26,250m = 26.25km