Chapter-1 Number system

Solution: -

(C) 400000

Here, in the above question we have to find the product of the value 2’s according to their places.

In the above question there is two 2’s and the first 2 is at the tenth place.

So, the value of that 2 is

= 2 × 10

= 20

Now, the second 2 is at ten thousand place.

So, the value of that 2 is

= 2 × 10000

= 20000

Hence, the product of both the 2’s place values = 20 × 20000

= 400000

Solution: -

(C) 37904

Here, in the above question we have to solve a numerical problem containing addition and multiplication.

So, we know that, while solving any numerical equation first we have to solve all the multiplication terms than addition terms.

Now, by following above rule we can first solve the multiply terms than addition one.

3 × 10000 = 30000

7 × 1000 = 7000

9 × 100 = 900

0 × 10 = 0

4 = 4

Now, by adding all the multiplied terms we obtain the value of the question. 

= 30000 + 7000 + 900 + 0 + 4

= 37904

Solution: –

(D) 1 crore

In the above equation we have asked to add 1 to the greatest 7-digit number.

We know that 9999999 is the greatest 7-digit number.

So, by adding 1to 99,99,999 we get 

=99,99,999 + 1

= 1,00,00,000 = 1 crore

Solution: -

(B) 9 × 1000 + 5 × 100 + 7 × 10 + 8 × 1

Here, in the above question, we have to expand 9578.

At unit place, we have 8. So, we can write 8 as 8 × 1

At tens place, we have 7. So, we can write 7 as 7 × 10

At hundred place, we have 5. So, we can write 6 as 5 × 100

At thousand place, we have 9. So, we can write 9 as 9 × 1000

Solution: -

(D) 86000

Here, in the above question we have asked to round-off the given number to the nearest thousand. 

Here, in the above number the value given before thousand place are 5842.

So, we have to round off to nearest thousands, which is 6000.

Hence the number will become 86000.

Solution: -

(D) 9965

Here, we have to make the greatest number using the number 5, 9, 2 and 6 and we can use any one number twice.

So, using 9 twice we can create the greatest number as 9965.

Solution:-

(C) 5,86,95,376

Here, we have t write the given number in the Indian system of Numeration. So, the number 58695376 is written as five crore, eighty six lakh, ninety five thousand, three hundred and seventy six = 5,86,95,376

Solution: -

(B) 10 Lakh

Here, we have to write the one million into the Indian number of systems.

One million is equal to ten lakhs.

1,000,000 = 10,00,000

Solution: -

(D) 5499

Here, we have to round off the greatest number nto the nearest 5000.

We know that according to the rule of the round off, the number less than 5499 will return to 5000. So, the answer is 5499.

Solution: -

(C) 6034579

Here, we have to rearrange the above number by keeping the place of 6.

Now, to make the smallest number we have to rearrange the remaining number in the acceding number.

So, by doing that we obtain the value as 6034579.

Solution: -

(D) LLX

Here, we have to find the incorrect roman number.

We know that the symbol L can never be repeated.

Hence, LLX is incorrect.

Solution: -

(C) 99987

Here, we have to create the largest number of 5-digita using three different number. So from the option we can say that the number 99987 having the three digit and is the largest number in all the option.

Solution: -

Here, we have to create the smallest number using four digits in which there should be three different number. So, from the above option we can say that 1012 having three different digits and the value is the lowest.

Solution: -

(c) 29

Here, we have to find the whole number between the number 38 and 68. So, the total number of whole number between 38 and 68 is 29.

Solution: -

(B) 998000

Here, we have to find the product of the successor and the predecessors of the number 999.

So, the number come before the number 999 is known as the predecessors and the number come after the number 999 is the successor.

So, Successor of 999 = 999 + 1 = 1000

Predecessor = 999 – 1 = 998

Now, the product of successor and predecessor of 999 is = 1000 × 998= 998000

Solution: -

(A) an even number

Here, we have to find the product of the non-zero whole number and its successors. So, when we multiply the non-zero whole number and its successor, the answer will always be an even number. For example: – 4 × 5 = 20, 7 × 8 = 56

Solution: -

(C) 50

Here, we have to first add the whole number to the number 25 than we have to subtract the same whole number from the 25. 

Now, we have to add the result obtained from both the equations.

So, let’s assume that the whole number is x.

Now, adding whole number to the number 25 = x + 25 

And subtracting the same whole number from the number 25 = 25 – x

Now, adding both the results = (x + 25) + (25 – x)

= x + 25 + 25 – x

= 50 + x – x

= 50 + 0

= 50

Solution: -

(C) 7 + 8 × 9 = (7 + 8) × (7 + 9)

Here, the option C is not true.

Now, first consider the left-hand side = 7 + 8 × 9

= 7 + (8 × 9)

= 7 + 72

= 79

Now, consider the right-hand side = (7 + 8) × (7 + 9)

= 15 × 16

= 240

Now, comparing both the side.

LHS = RHS

LHS ≠ RHS

79 ≠ 240

Solution: -

(B) 10

Here, we have to arrange the number of dots in such a way that it can form a line, triangle and rectangle.

So, by seeing the option we can say that the 10 number of the dots can be used to create a line, triangle and rectangle.

Solution: -

(B) Zero is the identity for multiplication of whole numbers.

Here, we have to find the false statement from the options.

From seeing the options, we can say that the zero is not the identity for the whole number. Example: - 1 × 0 = 0

Solution: -

(D) 0 ÷ 0 = 0

Here, from seeing the above options ew can say that in option D is not correct.

Any number divided by zero is not defined.

Solution:-

(B) 99999

Here, we have to find the predecessor of the number one lakh.

We know that the number that just come before any number is known as the predecessor of that number.

So, the predecessor of 1 lakh is = 1,00,000 – 1 = 99,999

Solution: -

(B) 1000001

Here, we have to find the successor of the number one lakh.

We know that the number that just come after any number is known as the successor of that number.

So, the successor of 1 million = 10,00,000 + 1

= 10,00,001

Solution: -

(A) 10

Here, we have to find the number of even numbers between the number 58 and 80.

So, the total number of even numbers between the number 58 and 80 is 12 and they are 58 and 80 are 60, 62, 64, 66, 68, 70, 72, 74, 76, 78.

25. Sum of the number of primes between 16 to 80 and 90 to 100 is

(A) 20 (B) 18 (C) 17 (D) 16

Solution: -

(C) 17

Here, we have to add the prime numbers between the number 16 to 80 and 90 to 100.

Now, the total Prime numbers between 16 to 80 = 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73 and 79 which is 16. Total prime number between the number 90 and 100 is 97 which is 1.

Now, addition of number of prime numbers are,

= 16 + 1

= 17

Solution: -

(D) The HCF of an even and an odd number is even.

Here, we have to find the HCF of the even and the odd number.

We know that, the HCF of an even and an odd number is odd number.

Solution: -

(B) 3

Here, we have to find the total number of distinct prime number of the largest 4-digit number.

So, the largest 4 – digit number is 9999

And the prime factors of 9999 = 3 × 3 × 11 × 101

9999 = 32 × 11 × 101

So, distinct prime factors are = 3, 11 and 101

Hence, the total number of distinct prime factors of the largest 4-digit number is 3.

Solution: -

(A) 2

Here, we have to find the total number of distinct prime number of the smallest 5-digit number.

So, the smallest 5 – digit number is 10000

And the prime factors of 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

10000 = 24 × 54

So, the number of total prime distinct prime factors are = 2 and 5

Hence, the total number of distinct prime factors of the smallest 5-digit number is  2

Solution: -

(C) 6

Here, we have to find that, whether the above number is divisible by 22 or not.

So, to find that we can simply take 11 instead of 22.

Now, to find the number is divisible by 11, we have to take the alternating sum of the digits in the number by read it from left to right. If that is divisible by 11, then the original number is also divisible by 11 which means number is divisible by 22.

7 – 2 + 5 – 4 + * – 9 + 8 = (5 + *)

Now, if (5 + *) is divisible by 11 then the whole number is also divisible by 11.

So, 5 + * = 11

* = 11 – 5

* = 6

Hence, the value of * is 6.

Solution: -

(B) 4

Here, we have to find the largest number which can divide the pair of consecutive odd numbers.

So, we know that the first odd number is 1 which means second is 3

1 + 3 = 4 = 4/4 = 1

3 + 5 = 8 = 8/4 = 2

Hence, 4 is the largest number which always divides the sum of any pair of consecutive odd numbers.

Solution: -

(D) 60

Here, we have to find that which number is not divisible by the numbers in the option is any number is divisible by 5 and 6.

So, to find that first we have to find the LCM of 6 and 5 which is 30.

Here, we know that 30 is divisible by 10, 15 and 30 in the given options.=

But, 30 is not divisible by 60.

Solution: -

(D) 39

Here, we need to find the print factor of 1729.

So, the prime factors of 1729 = 7 × 13 × 19

Hence, the addition of this prime numbers = 7 + 13 + 19 = 39

Solution:-

(B) 4

Here, we need to find out the greatest number which can divide the product of the predecessor and the successor of ab of number.

So, let’s assume that the odd natural number is 5.

Hence, the predecessor of 5 is = 5 -1 = 4

And successor of 5 is = 5 + 1 = 6

So, the product of predecessor and successor = 4 × 6

= 24

And we know that, 24 is divided by 4 = 24/4 =4

 Hence, the greatest odd natural number other than 1, is 4.

Solution:-

(A) 2

Here, we have to first find out the print factor of all the number.

Prime factors of,

75 = 3 × 5 × 5

60 = 2 × 2 × 3 × 5

105 = 3 × 5 × 7

So, as we can see that, 3 and 5 are the common prime factors for the given in three numbers.

Hence, there is 2 money of common prime factors for the numbers 75, 60, 105

Solution:-

(A) 8, 10

Here, from the above for options we need to find out which one is not the coprime.

So, for the coprime pair, both the number should not the even number and the common factor should be 1.

So, option A has both the number even and the common factor is 2.

Hence, the pair which is not coprime are 8 and 10.

Solution:-

(B) 60

Here, we need to find the LCM of 10,15 and 20.

So, to find the LCM, first we need to find the factor of all the numbers.

Hence, the LCM of 10, 15 and 30 is 2 × 2 × 3 × 5 × 1 = 60

Solution:-

(C) 75

Here, to find the HCF, first we need to find the factors of the number 180.

So, the factors of 180 are

Now, we can write 180 = 2 × 2 × 3 × 3 × 5

By any means we are not getting the value 75 grin the above number but if we multiply the correct numbers we are getting 45,60and 90.

Hence, 75 is not the HCF of the 180.

Solution:-

True.

Here, we know that it is a rule that in Roman numerals, a symbol should not be repeated thrice.

Solution:-

False.

Here, it is given that after repeating the symbols the value for multiplied which is incorrect. We know that after the repetition of the symbol the value get added.

Solution:-

True.

Here, on the about question the value given on the left hand side is 5555.

On the right hand side numerical equation is given So by solving equations we get the value of the equation.

So let’s solve the equation

Right Hand Side = 5 × 1000 + 5 × 100 + 5 × 10 + 5 × 1

= 5000 + 500 + 50 + 5

= 5555

Here, we are getting left hand side is equal to right hand.

So, the above condition is true.

Solution:-

True.

Here, in the above question, on the left hence side we are given a value, which is equal to 39746.

In the right hand side equation is given,

Right Hand Side = 3 × 10000 + 9 × 1000 + 7 × 100 + 4 × 10 + 6

Now, we need to find out the value of the above equation

= 30000 + 9000 + 700 + 40 + 6

= 39746

Here, we are getting same value on both the side

Hence, Left Hand Side = Right Hand Side

Solution:-

False.

Here, in the above question, on the left hence side we are given a value, which is equal to 82546.

In the right hand side equation is given.

Right Hand Side = 8 × 1000 + 2 × 1000 + 5 × 100 + 4 × 10 + 6

Now, we need to find out the value of the above equation

= 8000 + 2000 + 500 + 40 + 6

= 10,546

Here, we are not getting same value on both the side.

Hence, Left Hand Side ≠ Right Hand Side

Solution:-

True.

Here, in the above question, on the left hence side we are given a value, which is equal to 5322335.

In the right hand side equation is given,

Right Hand Side = 5 × 100000 + 3 × 10000 + 2 × 1000 + 2 × 100 + 3 × 10 + 5

Now, we need to find out the value of the above equation

= 5,00,000 + 30,000 + 2000 + 200 + 30 + 5

= 5,32,235

Here, we are getting same value on both the side

Hence, Left Hand Side = Right Hand Side

Solution:-

False.

Here, in the above question, we are given Roman numerals on the left hand side which is XXIX.

Now, we know that,  X = 10 and IX = 9

So, XXIX = 10 + 10 + 9

= 29

Hence, Left Hand Side ≠ Right Hand Side

Solution:-

True.

Here, in the above question, we are given Roman numerals on the left hand side which is LXXIV.

Now, we know that, L = 50, X = 10 and IV = 4

So, LXXIV = 50 + 10 + 10 + 4

= 74

Left Hand Side = Right Hand Side

Solution:-

False.

Here, in the above question, we are given two Roman numerals on the left hand side which is LIV and IV the right hand side which is LIV,

Now, we know that,  L = 50, IV = 4 AND VI = 6

So, LIV = 50 + 4 = 54

LVI = 50 + 6

Hence, 54 < 56

Therefore, LIV < LVI

Solution:-

False.

Here, in the above question we are given some number and we are asked to say weather they are in ascending or descending. 

As we know that the numbers in the in the increasing format are known as descending order.

So, the descending order of the given number is  5487, 5478, 4587, 4578.

Solution:-

False.

We know that, while doing round off any number near to hundreds, if the last two digit make less than 50 than it will become the previous hundreds but if it is making more than 50 than it will convert into the upcoming hundred.

So, the number 85764 will be rounded off to 85800.

Solution:-

True.

We know that, while doing round off any number near to hundreds, if the last two digit make less than 50 than it will become the previous hundreds but if it is making more than 50 than it will convert into the upcoming hundred.

So, The number 7826 will be rounded off to 7800 and the number 12469 will be rounded off to 12500

So, the sum of numbers will be

= 7800 + 12500 

= 20,300

hence, 20,300 will be done of to 20,000.

Solution:-

False.

Here, we are given six digit and we have to form the largest number.

So, as we know that to form largest  number we have to arrange the given figure in ascending order but here last two digits are not in ascending order.

Hence, if we arrange the following digit in ascending order it will be 875430.

Solution:-

False.

Here, we have to read the given number in the Indian format.

So, according to Indian format, the given number will be read as eight crore sixteen lakh fifty two thousand three hundred and eighteen.

Solution:-

True.

Here, we are given six digit and we have to form the largest number.

So, as we know that to form largest  number we have to arrange the given figure in ascending order but here last two digits are not in ascending order.

Hence, if we arrange the following digit in ascending order it will be 9760.

Solution: –

False.

Here we are given the units of distance.

Now we know that kilo is the largest unit among them for the distance and the smallest is mili.

Solution:-

False.

Here, in the above question, we asked about the successor of one digit number.

So, let's suppose that the number as 9, the successor of 9 is 10 which is not one digit number but 9 is one digit number.

Solution:-

False.

Here, in the above question, we asked about the successor of three digit number.

So. Let's suppose that the number is 999, the successor of 999 is 1000, which is not three digit number but 999 is three digit number.

Solution:-

false.

Here, in the above question, we asked about the predecessor of two digit number.

So, let's assume that the number is 10.

Now, the predecessor of 10 is 9, which is over digit number.

Solution:-

True.

Here, in the above question, we asked about the successor of whole number.

We now that, when we add 1 to the number, the new number obtained is known as the successor of that number.

For whole number there is no limit.

Solution:-

False.

Here, in the above question, we asked about the predecessor of whole number.

We know that, while number start from 0.

So, the predecessor for the number 0 will be -1, which is not whole number.

That's why, predecessor for all whole number is not possible.

Solution:-

False.

Here, in the above question, we asked about the natural numbers.

Natural numbers are numbers which start from 1 and increase as 1 i.e. 1,2,3,4,…..

Now, consider any try natural numbers 1 and 6.

So, there are more than 1 natural numbers between than i.e. 2,3,4 and 5.

Solution:-

True.

As we know that the number which come after the auction of 1 is known as the successor number.

So, the largest three digit number is 999 and the successor of this number is 1000, which is also the smallest four digit number. 

Solution:-

True.

As we know that the value of any number friends upon the number of digit that number have.

So, the number having more digit will have more value than that if smaller one.

Solution:-

True.

Here, in the above question, we asked about the natural numbers.

Natural numbers are those number which start from 1 and increases by 1.

So, the sum of any two natural numbers is always natural number.

Hence, natural numbers are closed under addition.

Solution:-

Here, in the above question, we asked about the natural numbers.

Natural numbers are those number which start from 1 and increases by 1.

So, the sum of any two natural numbers is always natural number.

Solution:-

False.

Here, in the above question, we asked about the natural numbers.

Natural numbers are those number which start from 1 and increases by 1.

So, the sum of any two natural numbers is always natural number but difference between than will not always result into natural numbers.

Solution:-

True.

Here, in the above question, we asked about the natural numbers.

Natural numbers are those number which start from 1 and increases by 1.

So, the sum of any two natural numbers is always natural number.

Let’s assume ‘a’ and ‘b’ are the two natural numbers.

So, the result of a + b will always be equal to b + a.

Solution:-

False.

As we know that whole numbers start from 0 and goes on increasing by 1.

So, for whole numbers, Zero (0) is the identity for addition.

 Now, consider any whole number i.e. 6.

Hence, 6 + 0 = 6

Solution:-

True.

As we know that whole numbers start from 0 and goes on increasing by 1.

For whole number, one is the identity number.

So, let's consider any whole number i.e. 5.

5 × 1 = 5

Solution:-

True.

As we know that whole numbers start from 0 and goes on increasing by 1.

For whole numbers, Zero (0) is the identity for addition.

So, when we add 0 to any whole number we get same number as answer.

Solution:-

False.

As we know that, natural numbers are those numbers which start from 1 and goes on increasing by one.

So, if we add the smallest natural number which is one to any natural numbers than the result will be different from the numbers we have taken.

Solution:-

True.

As we know that whole numbers start from 0 and goes on increasing by 1.

Now, as per the standard rule, if a whole number is divided by another whole number, which is greater than the first one, the quotient is not equal to zero.

Solution:-

True.

As we know that whole numbers start from 0 and goes on increasing by 1.

So, let's suppose any non-zero whole number i.e. 3.

Now, dividing this non-zero whole numbers by itself will give us one.

=3/3

=1

Solution:-

False.

As we know that whole numbers start from 0 and goes on increasing by 1.

So, when we multiply any two whole number the product will always be a whole number.

Solution:-

True.

As we know that whole numbers start from 0 and goes on increasing by 1.

Now, as per the standard rule, a whole number divided by another whole number greater than 1 never gives the quotient equal to the former.

Solution:-

True.

Here, in the above question, we asked about the natural numbers.

Natural numbers are those number which start from 1 and increases by 1.

Now, as the standard rule, every multiple of a number is greater than or equal to that number.

2 × 1 = 2

2 × 3 = 6

Solution:-

False.

Here, in the above question, we asked about the multiplication of number.

We know that numbers are infinite.

So, the multiple of any number goes on increasing infinitely.

Solution:-

True.

Here, in the above question, we asked about the multiplication of any number.

As we know that, 1 is the identity for multiplication of whole numbers

So, by multiplying any number by one we get that number itself.

Hence, we can say that every number is a multiple of itself.

Solution:-

True.

Here, in the above question, we asked about the divisibility of any number by four.

We know that, sum of any consecutive odd number will be always divisible by four.

1 + 3 = 4 = 4/4 = 1

11 + 13 = 24 = 24/4 = 6

Solution:-

True.

Here, in the above question, we asked about the divisibility for any number. 

Now, as per the standard rule, if a number divides three numbers exactly, it must divide their sum exactly.

So, let’s assume that one number i.e. 2, it divides 4, 6 and 8.

Now, their sum is = 4 + 6 + 8 = 18

We know that, 18 is divisible by 2.

Solution:-

False.

Here, in the above question, we asked about the divisibility rule for any number.

If any three number is divisible by any number, than it is not compulsory that all the three number seriously divisible by that number.

Solution:-

False.

Here, in the above question it is said that if a number is divisible by 2 and 3 than it is divisible by 12 also but we know that 6 is divisible by both 2 and 3 but it is not divisible by 12.

Solution:-

False.

Here, in the above question it is given that a number is divisible by 6 if the last two digit is divisible by 6, but we know according to the rule of divisibility for 6, a number is divisible by 6 only if it is divisible by 2 and 3 both.

Solution:-

True.

Here, in the above question, it is given that a number is divisible by 8 if the numbers formed by its last three digit is divisible by 8.

So, we know that, as per the rule of divisibility test, a number with 4 or more digits is divisible by 8, if the number formed by the last three digits is divisible by 8.

Solution:-

False.

Here, in the above question it is given that if a the sum of a number is divisible by 3 then the number is itself is divisible by 9 which is not true.

We know that, as per the rule of divisibility test, the sum of the digits of a number is divisible by 9, then the number itself is divisible by 9.

Solution:-

True.

It is not necessary that if a number is divisible by 4 must be divisible by 8.

Now, let’s consider the number 28, it is divisible by 4 but not divisible by 8.

Solution:-

False.

Here, in the above question it is given that HCF of two or more number is greater than LCM which is not true.

We know that, the HCF of two or more numbers is lower than their LCM

Solution:-

True.

Here, in the above question it is given that LCM of two or more numbers is divisible by their HCF, which is true.

We know that, as per the rule, LCM of two or more numbers is divisible by their HCF.

Solution:-

False.

Here, in the above question LCM and HCF of two number is given which is 28 and 8 respectively. 

We know that, from the divisibility rule that LCM of two or more numbers must buy divisible by their HCF.

But here, LCM of two numbers is 28 and their HCF is 8 and 28 is not exactly divided by 8.

Solution:-

True.

It is possible, that the LCM of two or more numbers may be one of the numbers.

Let's consider the two numbers as 2 and 4.

Now, LCM of 2 and 4 is 4.

Solution:-

True.

It is possible, that the HCF of two or more numbers may be one of the numbers.

Let’s consider the two numbers as 2 and 4.

Now, HCF of 2 and 4 is 2.

Solution:-

False.

Here, in the above question it is given that every whole number is the successor of another whole number which is not true for 0.

We know that, whole numbers start from 0.

So, there is no number before 0 that's why 0 is no the successor of another whole number.

Solution:-

False.

Here, in the above question it is given that the sum of two numbers will always be greater than their product, which is not true.

Let's take an example:-

2 + 3 = 5

2 × 3 = 6

Here, in the above example sum is 5 and product is 6.

Solution:-

True.

Here, in the above question it is given that if the sum of two different whole number is odd than their difference is also odd.

So, let's suppose that the two odd numbers 2 and 5.

Now, the sum of the two whole numbers is 

= 2 + 5 = 7 

7 is also an odd number.

Hence, the difference is

= 2 – 5 = 3 

3 is also an odd number.

Solution:-

True.

Here, from the rule, we know that co-prime number is a set of numbers or integers which have only 1 as their common factor.

Now, co-prime numbers are also known as relatively prime or mutually prime numbers. It is important that there should be two numbers in order to form co-primes.

Solution:-

True.

Here, in the above question it is given that the LCM of two number is one is the number than their LCM is the other number.

Solution:-

False.

We know that the HCF of two numbers will always be greater of equal to the smallest number.

Solution:-

False.

We know that the LCM of two numbers will always be greater or equal to the larger number.

Solution:-

True.

Here, from the rule, we know that co-prime number is a set of numbers or integers which have only 1 as their common factor.

Let's take an example

The two numbers are 2 and 3.

Now, LCM of this two numbers is 6 and the product is also 6.

Solution:-

a)Here, we have to convert the numbers into the Indian format.

So, we can write 10 million as 1 crore

And also, 1 million = 10 lakh

So, 10 million = 10 × 10 = 100 lakh = 1,00,00,000

Hence, 10 million = 1 crore.

b)Here, we have to convert the number from Indian format to US numbering system.

So, we know that 10 lakh = 1 million.

Solution:-

Here, we have to convert the unit of distance from one format to another format.

We know that, 1 meter = 100 centimeter

1 centimeter = 10 millimeter

So, 100 cm = 10 × 100 = 1000 millimeters

Hence, we can write 1 meter = 1000 millimeters

b) Here, we have to convert the unit of distance from one format to another format

Solution:

1. 1000

Here, we need to convert a value from gram to milligram.

We know that,

1 gram =1000 milligrams 

2. 1000,

Here, we need to convert a value from litre to millilitres.

We know that,

1 litre =1000 millilitres

© 10,00,000

Here, we need to convert a value from               kilogram to milligram.

We know that,

1 kilogram = 1000 grams and

1 gram = 1000 milligrams

So, 1 kilogram = 1000×1000 milligrams

1 milligram = 10,00,000 milligrams

Solution:

Here, we need to convert the value in thousand into lakh.

So, we know that

1 lakh = 100 thousand

So we can write,

100 thousand = 1 lakh

Solution:

Here, we miss to convert the unit of distance.

We know that, 1 centimeter = 10 millimetre 

And 1 metre = 100 centimetres.

So, 1 m 65 cm = 100 + 65 = 165 centimetres.

Now, 165 cm = 164 × 10 mm = 1650 mm

Solution:

Here, we need to convert the value in km into metres.

We know that, 1 km = 1000 m

So, 1290 km= 1290 × 1000 m = 12,90,000 m

Solution:

Here, we need to convert the value in km into metres.

We know that, 1 km = 1000 m

So,  422 km= (422 × 1000) m = 422000 m

Solution:

Smallest*

Here, we know that we obtain the largest number by arranging the numbers from largest to smallest.

So, when we reverse the order, the new number is in the form of smallest to largest. Which is the smallest number.

Solution

Here, we need to find such number to which by adding one we get the ten lakh.

Now, we know that ten lakh is 7 digit smallest number.

So, we have to add one to the 6 digit largest number.

Hence, by adding 1 to 999999, we get 1000000.

Solution:

Here, we need to write the number into Roman numerals format.

We know that, 66 can be written as LXVI into the Roman numerals. 

66 = LXVI

Solution:

Here, we need to find off the population into the nearest value.

We know that, of the last three digit of less than 500 than the number before them remains same but if it exceeds 500 than the number will increase by 1.

Now, here the last three digit is making less than 500, which is 473.

So, the number before them will remains same and that is 8.2,538,000

Solution

Here, we know asked to find out the smallest whole number.

We know that, whole numbers starts from 0 and increases by 1.

So, the smallest whole number is 0.

Solution

Here, we know asked to find out the smallest whole number.

We know that, whole numbers starts from 0 and increases by 1.

So, the smallest whole number is 0.

Solution:

Here, we know that, after subtracting one to any numbers the new number obtained in known as the predecessor of that numbers.

So, the predecessor of 100000 will be

100000-1, i.e., 99999

Solution:

Here, we know that, after subtracting one to any numbers the new number obtained in known as the predecessor of that numbers.

So, 400 is the predecessor of 400 + 1, i.e., 401

Solution:

Here, we know that, after adding one to the any numbers the new number obtained is known as the successor of that number.

So, the largest three digit number is 999 and the

Successor of 999 is 999 + 1, i.e., 1000

Solution:

We know that, after subtracting 0 grin any numbers their will be no effect in the value of that number.

Which means the numbers mentioned same.

Number 

Solution:

Here, we know that the smallest 6 digit number is 100000.

Now, we need to find out the smallest six digit number ending with five.

So, the numbers is 

100005

Solution:

We know that, after adding and multiplying two whole numbers we get only whole numbers.

Addition, multiplication

Solution:

We know that, after adding and multiplying two natural numbers we get only natural numbers.

Addition, multiplication

Solution:

We know that, dividing any number not only whole by zero is not defined. 0

Solution:

We know that, for whole numbers, multiplication is distributive over addition for whole numbers. Addition

Solution:

We know that, for whole numbers, multiplication is a cumulative.

So, right hand side will always equal to left hand side.

6195 

Solution:

We know that, for whole numbers, multiplication is cumulative.

So, grin solving above equation, we can obtain the value for the blank space as 1001.

Solution:

We know that, multiplication of any number with zero will always gives zero

Solution:

We know that, multiplication of any numbers with zero will always gives zero.

0

Solution:

Here, we know that, 1 is the multiplicative identity for the whole numbers.

So, multiplying any numbers with 1 will give the same numbers.

1 

Solution:

 Here, from solving the above equation we get 68 for the blank space.

Since, addition is associative for whole numbers.

68

Solution: 

Here, we know that, 1 is the multiplicative identity for the whole numbers.

So, multiplying any numbers with 1 will give the same numbers.

8925

Solution:

Since, multiplication is distributive over addition for whole numbers.

So, we get the answer as 1.

Solution:

Here, by solving the above given equation we get the value for the blank space as 17.

Solution:

Here, by solving the above given equation we get the value for the blank space as 27.

Solution:

Here, by solving the above given equation we get the value for the above equation as 7860.

7860 : 786 × 3 + 786 × 7 = 786 × (3 + 7)

= 786 × 10 = 7860

Solution:

Multiple

Here, we know that factors can be obtained by rectification if a number and by multiplying that factors we obtain that number again.

Solution:

1

Here, we know that, any number have one common factor and that is one.

Solution:

2

Here, we know that, any number have one common factor and that is one.

We know that, prime numbers are those numbers which doesn't have any factor either than one and that number itself.

Solution:

Perfect

We know that every number have factor even prime numbers also have factor as one and the number itself.

When the sun of the factors of equal to the twice of that number than the number is called perfect number.

Solution:

Composite

We know that every number have factor.

So, when these numbers have more than two factors they known as composite numbers.

Solution:

Co-prime

When the factors of two number are different, only one factor is same and that is 1, the numbers are known as co-prime numbers.

Solution:

Here, we need to write the prime numbers between 1 to 100.

We know that prime numbers are those numbers which do not have factor except one and the number itself.

So, prime numbers between 1 to 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Here, between 1 to 100 there are 25 prime numbers

Solution:

0

We know that, if there is zero at once place than the number is divisible by 10.

Solution:

0, 5

Here, we need give the number which is divisible by 5.

We know that a number is divisible by 5 if it has 0 and 5 in its once place.

Solution:

2

Here, we need to give the numbers which is divisible by 2.

We know that, if we multiply 2 than we get 0,2,4,6, and 8.

So, the number which have these numbers in its once place is divisible by 2.

Solution:

Multiple

Here, we need to apply the divisibility rule of 3.

According to rule, a number is divisible by 3 is it's sum is multiple of 3.

Solution:

11

Here, we need to apply the divisibility rule of 11.

We know that, according to divisibility rule of 11,

If the difference between the sum of digits at odd places (from the right) and the sum of digits at even places (from the right) of a number is either 0 or divisible by 11.

Solution:

Multiple

Here, we know that, to find the LCM of two or more number.

First of all we find the factors of all the numbers, than the LCM of those numbers are the common factor among them.

Solution:

Factors

Here, we know that to find the HCF, first we find the factors if all the given numbers and then we take the highest of their common factors.

Solution:

1. ➝ (d) ; (ii) ➝ (f) ; (iii) ➝ (b) ; (iv) ➝ € ; (v) ➝ (c)

   
   

Solution:

Here, we have to find the smallest and the greatest number from the given numbers.

So, the greatest number is 67205602 and the smallest number is 30040700.

Solution:

(a)

Here, we need to write the given number in the expanded form.

So, the above number can be written as

 74836 = 7 × 10000 + 4 × 1000 + 8 × 100 + 3 × 10 + 6 × 1

(b)

Here, we need to write the given number in the expanded form.

So, the above number can be written as

 574021 = 5 × 100000 + 7 × 10000 + 4 x 1000 + 0 × 100 + 2 × 10 + 1 × 1

© 

Here, we need to write the given number in the expanded form.

So, the above number can be written as

 + 9 × 100000 + 0 × 10000 + 7 × 1000 + 0 × 100 + 1 × 10 + 0 × 1

Solution:

We know that, when the numbers are arrange in the increasing form they are said to be arranged in ascending order.

Ascending order ➝ (b), (c), (a), (d)

We know that, when the numbers are arrange in the decreasing form they are said to be arranged in descending order.

Descending order ➝ (d), (a), (c), (b)

Solution:

Here, we need to rewrite the above given number by inserting comma’s at proper place.

So, we can write as

The diameter of Jupiter is 142,800,000 metres.

Solution:

Here, we need to calculate the increase in the population of India.

So, total increase in population from 1961 to 2001= 1028 millions – 439 millions = 589 millions

Now, we need to rewrite this numbers into the Indian System of Numeration.

So, the increase in population = 58,90,00,000

Solution:

Here, we have to decide wide radius of greater and hour much.

So, the radius of the Earth = 6400 km = 6400 × 1000 m = 6400000 m

And radius of the Mars = 4300000 m

Note, to find the difference, we have to subtract the radius of the Mars from the radius of the Earth.

(6400000 – 4300000) m = 2100000 m.

Solution:

Here, the population of tripura and meghalaya of given in numbers and we need to rewrite this numbers into the word format.

So, the population of Tripura = 3,199,203 

Which can be written in word as, Three million one hundred ninety-nine thousand two hundred three.

Now, the population of Meghalaya = 2,318,822, 

Which can be written in word as, Two million three hundred eighteen thousand eight hundred twenty two.

Solution:

Here, we have to find the difference between the number if children getting polio drops in two different months.

So, the number of children getting polio drops in March 2008 = 2,12,583

And the number of children getting polio drops in April 2008 = 2,16,813

Now, you find the difference, we have to subtract the smallest numbers of children getting polio grin the larger numbers of children getting pilot drops.

Hene, the required difference of the number of children getting polio drops in the two months = 2,16,813 – 2,12,583 = 4,230

Solution:

Here, we need to find out how much amount person spend and how much is left.

Total amount a person had = ₹ 1000000

The amount he spent on a colour T.V. = ₹ 16580

The amount he spent on a motorcycle = ₹ 45890

The amount he spent on a flat = ₹ 870000

Now, the total amount he spent = ₹ (16580 + 45890 + 870000) = ₹ 932470

Hence, the amount left with him = ₹ 1000000 – ₹ 932470 = ₹ 67530

Solution:

Here, we have to find out how much tablets is left.

So, the total number of tablets of Vitamin A = 180000

And total number of tablets distributed among the students in a district = 18734

Hence, the number of remaining vitamin tablets = 180000 – 18734 = 161266

Solution:

Here, we have to find how much money does Chinmay have after hi give money to Jyoti, Javed and John.

Total amount Chinmay had = ₹ 610000

The amount he gave to Jyoti = ₹ 87500

The amount he gave to Javed = ₹ 126380

The amount he gave to John = ₹ 350009

Now, total amount given by Chinmay = ₹ (87500 + 126380 + 350000) = ₹ 563880

Hence, the total amount left along the Chinmay

= ₹ 610000 – ₹ 563880 = ₹ 46120