Maths 2 Mark Question

There are 10 digits i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
We cannot use ‘0’ at the place having the highest place value in 8 digit numbers.
So, we can use only 9 digits at the place having the highest place value in 8 digit numbers.
Also, we can use 10 digits at the remaining places in 8 digit numbers So, total numbers of 8-digit
numbers = 9 × 10 × 10 × 10 × 10 × 10 × 10 × 10 = 90000000

109 × 107 = (100 + 9) × 107
= 100 × 107 + 9 × 107
= 100 × (100 + 7) + 9 × (100 + 7)
= 10000 + 700 + 900 + 63
= 11663

12 × 105 = (10 + 2) × 105
= 10 × 105 + 2 × 105
= 10 × (100 + 5) + 2 × (100 + 5)
= 1000 + 50 + 200 + 10
= 1260

Four smallest digits are 0, 1,2 and 3. In order to generate the smallest 6-digit number using digits 0, 1, 2 and 3, we write the smallest non-zero digit at the place having the highest place value and the largest digit at the place having least place value. Thus, we put 1 in the left-most place and 3 in the right-most place. Digit 2 is put at ten's place and at all other places we write zero.
Hence, the required number = 100023.

Runs scored by cricket player in test matches = 6,978
Therefore, Remaining runs required to complete 10,000 runs
= 10,000 - 6,978 = 3,022
Thus, the player needs to score 3,022 more runs to complete 10,000 runs.

17 × 109 = (10 + 7) × 109
= 10 × 109 + 7 × 109
= 10 × (100 + 9) + 7 × (100 + 9)
= 1000 + 90 + 700 + 63
= 1853

The three-digit numbers formed using the digits 0, 2 and 5 (without repeating any digit in the number) are 250, 205, 502 and 520.
Therefore, four such numbers can be formed.

The place value of a digit depends on the place where it occurs, while the face value is the value of the digit itself.
In a number, the digits that have same face value and place value are the ones digit and all the zeroes of the number.
Therefore, in 9,20,78,634,4 (the ones digit) and 0 (the lakhs digit) have the same face value and place value.

16 × 108 = (10 + 6) × 108
= 10 × 108 + 6 × 108
= 10 × (100 + 8) + 6 × (100 + 8)
= 1000 + 80 + 600 + 48
= 1728

To write the natural numbers between 500 and 600 which do not change if the digits are written in the reverse order we must have the same digit at the hundred's place and unit's place.
Hence, the required numbers are 505, 515, 525, 535, 545. 555, 565, 575, 585, 595.

9 × 105 = 9 × (100 + 5)
= 9 × 100 + 9 × 5
= 900 + 45
= 945

6 × 112 = 6 × (100 + 12)
= 6 × 100 + 6 × 12
= 600 + 6 × (10 + 2)
= 600 + 6 × 10 + 6 × 2
= 600 + 60 + 12
= 672

101 × 105 = (100 + 1) × 105
= 100 × 105 + 1 × 105
= 100 × (100 + 5) + 105
= 10000 + 500 +105
= 10605

Place value of first 7 = 7 × 10 = 700
Place value of second 7 = 7 × 10,000 = 70,00,000
Required difference = 70,00,000 - 700 = 69,99,300

102 × 103 = (100 + 2) × 103
= 100 × 103 + 2 × 103
= 100 × (100 + 3) + 2 × (100 + 3)
= 10000 + 300 + 200 + 6
= 10506

To write the greatest 6-digit number having four different digits, we will have to use four largest digits. Clearly 9, 8, 7 and 6 are four largest digits. In order to write the largest 6-digit number using digits 6, 7, 8 and 9, we put the largest digit 9 at the place having the highest place value. The smallest digit 8 is put at the hundred's place, the smallest digit 6 is put at the right most place i.e. at unit's place and the digit 7 is put at the ten's place. All other places are filled by 9.
Hence, the required number = 999876.

The number = 7,86,54,321
The place value of 5 = 5 ten thousands = 50,000
The face value of 5 = 5
Therefore, the difference = 50,000 - 5 = 49,995

As given in the question,
Number of screws produced by a machine in a day = 2,825
Therefore, Number of screws produced by the same machine in the month of January 2006 = 2,825 × 31 = 87,575
Thus, machine-produced 87,575 screws in the month of January 2006.

Place value of first 5 = 5 × 10 = 50
Place value of second 5 = 5 × 10,000 = 50,000
Required product = 50 × 50,000 = 25,00,000

There are 10 digits i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
We cannot use ‘0’ at thousand’s place.
So, we can use only 9 digits at thousand’s place.
Also, we can use 10 digits at hundred’s, 10 digits at ten’s and 10 digits at unit’s place.
So, total numbers of four-digit numbers = 9 × 10 × 10 × 10 = 9000

Total number of tickets sold on all four days is the sum of the tickets sold on the first, second, third and final days.
Therefore, total number of tickets sold on all four days is given by :
= 1094 + 1812 + 2050 + 2751 = 7707