Number System

Place value is the value of the digit depending on its place in the number.

In the number 428721,

a) Place value of rightmost ‘2’ is tens place.

Hence, its place value = 2× 10

= 20

b) Place value of leftmost ‘2’ is ten thousand

Hence, its place value = 2× 10000

= 20000

∴ Product of place values = 20× 20000

= 400000

In 3 × 10000 + 7 × 1000 + 9 × 100 + 0 ×10 + 4

Place value of 3 = 3× 10000

= 30000

Place value of 7 = 7× 1000

= 7000

Place value of 9 = 9× 100

= 900

Place value of 0 = 0× 10

= 0

Place value of 4 = 4× 1

= 4

Adding all place values = 30000+7000+900+0+4

= 37904

The greatest 7 digit number = 9999999

Hence adding 1 to it,

The number we get = 9999999+1

= 10000000

= 1× 10⁷

= 1 crore

Starting from right for the number 9578,

Place value of 8 is ones = 8× 1

Place value of 7 is tens = 7× 10

Place value of 5 is hundred = 5× 100

Place value of 9 is thousand = 9× 1000

Hence, expanding form of the number is adding all above,

i.e. 9 × 1000 + 5 × 100 + 7 × 10 + 8 × 1

Hence, (B) is the answer.

Starting from the right for the number 85642,

Place value of 2 is ones

Place value of 4 is tens

Place value of 6 is hundreds

Place value of 5 is thousands

Place value of 8 is ten thousands

Since, we need to round off to nearest thousands, all digits following 5 should be replaced by 0 and as the digit succeeding 5 is 6(greater than 5), therefore digit at ten thousand's place should be increment by 1,

∴ Answer = 86000

The largest 4-digit number is obtained by writing the digits in reverse order (i.e. descending order)

But, here 1 digit can be used twice, hence using the maximum digit (here, 9 twice) and other digits in descending order,

The answer is: 9965

The given number in Indian number system in words is:

5 crore, eighty six lakhs, ninty five thousand, three hundred and seventy six.

i.e. 5, 86, 95, 376

One million = 1, 000, 000

= 10× 1, 00, 000

= 10× 10⁵

= 10 lakhs.

To round off a number to nearest thousands,

If the number at hundred place is 0, 1, 2, 3 or 4 we round down

If the number at hundred place is 5, 6, 7, 8 or 9 we round up.

5, 001 on rounding off to nearest thousands gives 5000

(since at hundred place is 0)

5, 559 on rounding off to nearest thousands gives 6000

(since at hundred place is 5)

5, 999 on rounding off to nearest thousands gives 6000

(since at hundred place is 9)

5, 499 on rounding off to nearest thousands gives 5000

(since at hundred place is 4)

Thus the number which rounds off to 5000 and is the largest is 1 less than 5000

= 5000-1

= 5499

The number is 63, 50, and 947 in Indian number system.

The place of 6 is same i.e. ten lakhs.

Digits other than 6 are 3, 5, 0, 9, 4, and 7

To obtain the smallest number, arranging them in ascending order, i.e. 0, 3, 4, 5, 7 and 9

The required number is 6034579

A. LXXX is correct and represents 50+30 = 80

B. LXX is correct and represents 50+20 = 70

C. LX is correct and represents 50+10 = 60

D. LLX does not represent anything, because LL is not a valid roman numeral (C is used for 100 instead of LL)

Hence, D is incorrect.

To obtain the largest 5 digit number, digits must be used from 9 in the descending order.

As we have to use 3 different digits, they are 9, 8 and 7

Also since we need the largest digit, only the ones and tens place be filled by 8 and 7.

Rest by 9 which is the largest digit.

Hence, the answer is 99987

To obtain the smallest 4-digit number, we need to use digits in the ascending order.

The smallest 3-different digits are 0, 1and 2

As, 0 can’t come at thousand place,

The answer will be 1, 002

Whole numbers start from 0 and increase by 1 each

They are 0, 1, 2, 3…

Number of whole numbers between 38 and 68 are-

39, 40, 41, …, 66, 67

i.e. no. of whole numbers = 67-39+1

= 29

*Important Note:

we find the number of numbers between two numbers 'a' and 'b'

(including b and a), by the formula (b - a + 1)

Successor (number after) of 999 = 999+1 = 1000

Predecessor (number before) of 999 = 999-1 = 998

Product of successor and predecessor of 999 = 1000× 998

= 998000

Let the non-zero whole number = a

The successor (number after) = a + 1

Case 1:

If a is even,

Successor = a + 1 is odd

The product of an even and odd number is always even.

(For example: 2 × 3 = 6)

Hence, the answer is an even number.

Case 2:

If a is odd,

Successor = a+1 is even

The product of a even and odd number is always even.

(For example: 5 × 6 = 30)

Hence, the answer is an even number.

Let the whole number = a

First number = 25 + a

Second number = 25 - a

Sum of resulting numbers = 25 + a + (25 - a)

= 25 + a + 25 - a

= 25 + 25 + a - a

= 50 + 0

= 50

LHS = left hand side

RHS = Right hand side

A. (7 + 8) + 9 = 7 + (8 + 9)

Is true due to associative property of addition.

B. (7 × 8) × 9 = 7 × (8 × 9)

Is true due to associative property of multiplication.

C. LHS = 7 + 8 × 9

= 7 + 72

= 79

RHS = (7+8) × (7+9)

= 15× 16 = 240

As LHS ≠ RHS,

It is incorrect.

D. 7 × (8 + 9) = (7 × 8) + (7 × 9)

It is true due to distributive property of addition over multiplication.

A triangle can’t be made by 3 dots, 11 dots or 12 dots and a rectangle can’t be made by 11 dots, hence option A, C and D are incorrect.

A line, a triangle and a rectangle can be drawn by 10 dots as shown,

TRIANGLE:

LINE:

RECTANGLE:

A. It is true since addition and multiplication are associative for whole numbers.

B. If zero is multiplied by a whole number, the answer is 0. Hence, it is not the identity since on multiplying the answer is not the original number. Hence, it is not true.

C. It’s true since Addition and multiplication both are commutative for whole numbers.

D. It’s true since Multiplication is distributive over addition for whole numbers.

LHS = left hand side

RHS = Right hand side

A. LHS = 0+0 = 0

RHS = 0

∴ LHS = RHS

B. LHS = 0-0 = 0

RHS = 0

∴ LHS = RHS

C. LHS = 0× 0 = 0

RHS = 0

∴ LHS = RHS

D. LHS = 0÷ 0 is not defined, since denominator is 0

∴ it is not defined.

Predecessor is the number before the original number

Predecessor of 1 lakh = 1, 00, 000-1

= 99, 999

Successor is the number after the original number

Successor of 1 million = 1, 000, 000+1

= 1, 000, 001

Even numbers between 58 and 80 are-

60, 62, 64, 66, 68, 70, 72, 74, 76, 78

Hence, the count of numbers is 10

Prime numbers are those numbers, which don't have factors other than 0 and itself.

Prime numbers are 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61 67, 73, 79 between 16-80

The count of above list is 15

Prime numbers between 90 to 100 are 97

The count is 1

Sum of counts = 15+1 = 16

A. It is true because their does not exist any number other than 1 which is a factor of both the prime numbers. Hence, 1 is also the HCF.

B. It is true as Numbers which do not have any common factor other than 1 are co-prime. Hence, for them 1 is also the HCF

C. It’s true since for two consecutive even numbers, only 2 is the HCF.

D. It’s false because an odd number can never have 2 as its factor and hence the HCF can never be even.

Largest 4-digit number is 9999

9999 = 3²× 11× 101

So distinct prime factors are 3, 11 and 101

Hence, answer = 3

Smallest 5-digit number is 10000

10000 = 2⁴× 5⁴

Hence, it has only 2 prime factors.

Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number.

7 – 2 + 5 – 4 + * - 9 + 8 = (5 + *)

For 7254 * 98 to be divisible by 11,

(5 + *) must also be divisible by 11

Hence, 5 + * = 11

∴ * = 11 - 5

= 6

Sum of two consecutive numbers is always divisible by 4, for example

1 + 3 = 4

3 + 5 = 8

5 + 7 = 12

All are divisible by 4.

But 1 + 3 = 4 is neither divisible by 6 nor by 8, therefore 4 is the largest number which always divides the sum of any pair of consecutive odd numbers.

**Additional proof (Not for exam purpose)

Any two consecutive odd numbers will be in the form, 2n – 1 and 2n + 1, for n = 1, 2, 3, …and so on

Hence, their addition

2n – 1 + 2n + 1 = 4n is divisible by 4.

A number is divisible by 5 and 6.

Let number be a = 5× 6× b (where b is any random number)

A. a = 5× 6× b

= 5× 2× 3× b

= 10× 5× b

Hence, it is divisible by 10.

B. a = 5× 6× b

= 5× 2× 3× b

= 2×15× b

Hence, it is divisible by 15.

C. a = 5× 6× b

= 30× b

Hence, it is divisible by 30.

D. a = 5× 6× b

= 30× b

It can’t be written as 60× b and hence it may not divisible by 60.

1729 = 7× 13× 19

Prime factors of 1729 are 7, 13 and 19

Hence, sum of prime factors = 07+13+19

= 39

Let a be a odd natural number

Predecessor (number before) = a - 1 = 2× b (a even number)

Successor (number after) = a+1 = 2× c (a even number)

Since,

The product of the predecessor and successor of an odd natural n

= (2× b) × (2× c)

= 4× b× c

Hence, the largest dividing number is 4

Prime factors of 75 = 3, 5

Prime factors of 60 = 2, 3, 5

Prime factors of 105 = 3, 5, 7

Hence, common prime factors are 3 and 5

I.e. count = 2

A. Both 8 and 10 are even. Hence, 2 is also a common factor (other than 1) hence, not coprime.

B. 11, 12 only have 1 as a common factor. Hence they are coprime.

C. 1, 3 only have 1 as a common factor. Hence they are coprime.

D. 31, 33 only have 1 as a common factor. Hence they are cop rime.

Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number.

A. 1011011

Sum = 1 – 0 + 1 – 1 + 1 – 1 + 1 = 2

Since, 2 is not divisible by 11, 1011011 is not divisible by 11.

B. 1111111

Sum = 1 – 1 + 1 – 1 + 1 – 1 + 1 = 1

Since, 2 is not divisible by 11, 1011011 is not divisible by 11.

C. 22222222

Sum = 2 – 2 + 2 - 2 + 2 – 2 + 2 - 2 = 0

Since, 0 is divisible by 11, 22222222 is divisible by 11.

D. 3333333

Sum = 3 – 3 + 3 – 3 + 3 – 3 + 3 = 3

Since, 3 is not divisible by 11, 3333333 is not divisible by 11.

LCM of 10, 15 and 20:-

10 = 2× 5

15 = 3× 5

20 = 2× 2× 5

2: 2 occurrences

3: 1 occurrence

5:1 occurrence

LCM = 2× 2× 3× 5 = 60

LCM is the multiple of a number while HCF is the highest factor

Hence, HCF must completely divide the LCM.

A. Since it completely divides (with integer quotient), it may be HCF.

B. Since it completely divides (with integer quotient), it may be HCF.

C. Since the answer is not an integer quotient, it can’t be the HCF.

D. Since it completely divides (with integer quotient), it may be HCF.

True.

It is a general rule for writing Roman numbers.

∴ It is true.

False.

In Roman numeration, if a symbol is repeated it means its value is added as many times as it occurs.

∴ It is false

True.

Since, 5 × 1000 + 5 × 100 + 5 × 10 + 5 × 1 = 5555.

∴ It is true.

True

Since, 3 × 10000 + 9 × 1000 + 7 × 100 + 4 × 10 + 6 = 39746.
∴ It is true.

False

Since, 8 × 1000 + 2 × 1000 + 5 × 100 + 4 × 10 + 6

= 8000 + 2000 + 500 + 40 + 6 = 10546
⇒ 10546 ≠ 82546

∴ It is false

True.

Since, 5 × 100000 + 3 × 10000 + 2 × 1000 + 2 × 100 + 3 × 10 + 5 = 532235
∴ It is true.

False.

XXIX is read as 29, i.e. XXIX = 10 + 10 + 9 = 29
⇒ 29 ≠ 31

31 is written as XXXI
∴ It is false.

True.

LXXIV is read as 74, i.e. LXXIV = 50 + 10 + 10 + 4 = 74
∴ It is true.

False.

Number LIV is read as 54, i.e. LIV = 50 + 4 = 54 and

Number LVI is read as 56, i.e. LVI = 50 + 6 = 56

⇒ 56 > 54 ⇒ LVI > LIV
LIV is smaller than LVI
∴ It is false.

False

Since, 5487 > 5478 > 4587 > 4578

On arranging the number in descending order, we get :
5487, 5478, 4587, 4578

False.

The number 85764 is nearest to 85800 than 85700
∴ on rounding off 85764 to nearest hundred we get 85800.
85800 ≠ 85700

False.

Rounding off 7826 to nearest hundred = 7800
Rounding off 12469 to nearest hundred = 12500
Sum = 7800 + 12500 = 20300
20300 ≠ 20, 000

False.

Arranging the numbers in descending order, we get :
8 > 7 > 5 > 4 > 3 > 0.
Thus the largest six digit number that can be formed using these digits is 875430.
875430 ≠ 875403

False.

81652318 is read as eight crore sixteen lakh fifty two thousand three hundred eighteen.

True.

Arranging the numbers in descending order, we get :
9 > 7 > 6 > 0.
Thus the largest four digit number that can be formed using these digits is 9760.

False.
From the metric conversion table :

1 kilo = 10000 centi and 1 kilo = 1000000 milli
Also, 1 centi = 10 milli

∴ kilo > centi > milli
∴ Milli is the smallest.

False.
If we consider a 1-digit number 9, then its successor will be
9 + 1 = 10, which is a 2-digit number.
∴ It is false.

False.
If we consider a 3-digit number 999, then its successor will be
999 + 1 = 1000, which is a 4-digit number.
∴ It is false.

False.

If we consider a 2-digit number 10, then its predecessor will be
10 - 1 = 9, which is a 1-digit number.
∴ It is false.

True
Whole number = 0, 1, 2, 3……
∴ every whole number has its successor as + 1 can be added to each number.

False
Whole number = 0, 1, 2, 3……
If we consider whole number 0, we get 0 – 1 = -1 which is an integer but not a whole number.
∴ It is false.

False.
If we take any two natural numbers say (x, y), Then their will be
x –y -1 (x > y) natural number.
eg. For natural number 2 and 6 there are 3, 4, 5 natural numbers in between.
∴ It is false

True.
Smallest 4-digit number = 1000
Largest 3- digit number = 999
∴ Successor of 999 is 999 + 1 = 1000.
Thus, it is true

True.
It is always true as it is a general rule.

True.

It is general property of natural number.
Sum of any two natural numbers is always a natural number.
∴ Natural numbers are closed under addition.

False.

Multiplication of any two natural number is always a natural number.
∴ Natural numbers are closed under multiplication.

False.

Subtraction of two natural numbers can give three results
Positive integer, Negative integer and 0.
In case of negative integer and 0, they are not natural numbers.
∴ Natural numbers are not closed under subtraction.

True.
It is general property of natural number .
If a and b are natural numbers then they follow the commutative property i.e. a + b = b + a.
∴ Addition is commutative for natural numbers.

False.

If we add 1 to any whole number we will not get the same whole number. If x is the whole number, then x + 1 ≠ x.
∴ 1 is not the identity for addition of whole numbers.

True.

If we multiply 1 to any whole number we will get the same whole number. If x is the whole number, then n × 1 = n
∴ 1 is the identity for multiplication of whole numbers.

True.

We know that 0 is the identity for addition of whole numbers i.e.
If we add 0 to any whole number we will get the same whole number. If x is the whole number, then x + 0 = x.

False.

∴ 0 is the identity for addition of numbers i.e.
If n is a natural number, then n + 0 = n.
But 0 is not a natural number.
∴ It is false.

True
It is always true.

True
It is always true as any non-zero number when divided by itself gives quotient 1.

False
We know that, whole numbers are closed under multiplication i.e. Product of two whole numbers always gives a whole number.

True.
∴ It is general rule so it is true.

True
It is always true as it is a standard property.

False
As we know that numbers are infinite, they have infinite number of multiplies.

True
When any number is multiplied by 1 we get the same number.
Hence, every number is a multiple of itself.

True
It is a general property,
e.g. Let two consecutive odd numbers be 1 and 3
⇒ Sum = 1 + 3 = 4, which is divisible by 4.
∴ Sum of two consecutive odd numbers is always divisible by 4.

True
It is a general property so it is always true.
e.g. Say 3 divides 6, 9 and 12 exactly
Then, Sum = 6 + 9 + 12 = 27, which is exactly divisible by 3.

False

It is not true.
We know that, If a number divides three numbers exactly, it must divide their sum exactly but its vice versa is not necessarily to be true.

False.
Consider a number 6 it is divisible by 2 and 3, But it is not divisible by 12.

∴ It is false.

False
It is not necessarily to be true.
Consider 3-digit number 166 in this 66 formed by ones and tens digit is divisible by 6 but the number 166 is not divisible by 6.

True.
It is true as any number is divisible by 8 if its last 3-digit is divisible by 8.

False.
According to divisibility test of 9, A number is divisible by 9 if the sum of all the digits of that number is divisible by 9.

∴ It is False.

True

It is true.
e.g. consider a number 12 it is divisible by 4 but not by 8.

False.
It is false as the Highest Common Factor of two or more numbers is lower than their Lowest Common Multiple.

True

It is always true.

e.g. consider two number 4 and 6. Its LCM is 12 and HCF is 2
12 is divisible by 2.

False

It is not possible as LCM is not exactly divisible by HCF.

True.

If the two numbers are such that one is the multiple of other, then their LCM will be the greater of two numbers.

True.

If the two numbers are such that one is the multiple of other, then their HCF will be smallest of the two numbers.

False
0 is the only number which is not the successor of any whole number.

False.
In case of 0 and 1. Sum = 1 and Product = 0
⇒ Sum is greater than product
Also, In case of 2 and 2. Sum is equal to product.
∴ It is not necessary that sum of two whole numbers is always less than their product.

True
It is general property so it is always true.
e.g. consider two number 2 (even) and 9 (odd).
Sum = 2 + 9 = 11 (odd) and Difference = 9 – 2 = 7 (odd)

True
It is always true as one is odd and is even. Thus they always have HCF as 1.

True

We know that, HCF × LCM = First number × Second number

If HCF = First number
⇒ HCF × LCM = HCF × Second number

⇒ LCM = Second number

False
We know that, HCF of two numbers is either greater than or equal to smaller of the two number.
∴ It is false.

False

It is not always necessary that LCM of two numbers is greater than the larger of the number, sometimes LCM is equal to larger of the number (if one number is multiple of the other number).

True.
It is always true.
e.g. consider two co-prime numbers 7 and 13.
LCM = 91 and their product is 7 × 13 = 91.

A. We know that 1 crore = 1, 00, 00, 000 million

= 10, 000, 000

= 10 1, 000, 000

= 10 million

Therefore, 1crore = 10 million

So, 10 million = 1 crore

B. We know that 10 lakh = 10, 00, 000

= 1, 000, 000

= 1 million

Therefore, 10 lakhs = 1 million

A. We know that 1 metre = 100 centimetres.

Also 1 centimetre = 10 milimetres.

So, 1 metre = 100 1 centimetre

⇒ 1 metre = 100 10 milimetres

⇒ 1 metre = 1000 milimetres.

B. We know that 1 centimetre = 10 milimetres.

C. We know that 1 kilometre = 1000 metres.

1 metre = 100 centimetre.

1 centimetre = 10 milimetre.

1 kilometre = 1000 1 metre

1 kilometre = 1000 100 1 centimetres

1 kilometre = 1000 100 10 milimetres

1 kilometre = 1000000 milimetres.

A. We know that 1 gram = 1000 miligrams.

B. We know that 1 litre = 1000 mililitres.

C. We know that 1 kilogram = 1000 grams.

1 gram = 1000 miligrams.

1 kilogram = 1000 1 g

1 kilogram= 1000 1000 miligrams.

1 kilogram = 1000000 miligrams.

We know that 100 thousands = 100, 000

= 1, 00, 000

= 1 lakh

Height of a person is 1m 65cm.

As we know that 1 m = 100 cm……..eq(1)

1 cm = 10 mm…….eq(2)

So, height of the person is = 1m 65 cm

= 1m + 65cm

= 100cm + 65cm(from eq(1))

= 165cm

= 165 10 mm ( from eq(2))

= 1650 mm

Therefore, Height of the person in millimetre is 1650mm.

We know that 1km = 1000m

Length of river ‘Narmada’ is about 1290km.

Length of river ‘Narmada’ in metres = 1290 1000 m

= 1290000m

The distance between Srinagar and Leh is 422km.

We know that 1km = 1000m

Therefore, the distance between Srinagar and Leh is 4221000m

the distance between Srinagar and Leh is 422000m

Descending order means arrangement of writing of numbers from the greatest to the smallest.

Snallest

We know that, Ten lakhs = 1000000

Subtracting 1 from 1000000 = 1000000 – 1= 999999

Now if we add 1 to 999999 we get 10 lakhs.

So, by adding 1 to the greatest 5 digit number, we get ten lakhs.

It can be written as 5, 23, 78, 401.

In Roman Numerals, X = 10

M = 1000

C = 100

L = 50

So, the symbol X can be subtracted from L, M and C as they are greater than X.

In roman numerals 66 can be written as LXVI.

(L= 50, X = 10, V= 5, I = 1)

66 = 60 + 6

= 50 + 10 + 5 + 1

= LXVI

The population of Pune rounded off to thousands in 2001 is 2538000.

The digit in the hundreds place is 4, which is less than 5. So, we replace each of the hundreds, tens and ones digits by 0 and keep the other digits as it is.

So, the number 2538473 rounded to nearest thousands is 2538000.

Whole numbers start from 0 .

So, the smallest whole number is 0.

We know that Successor of any no. is 1 greater than the number. Therefore, Successor of 106159 is 106159 + 1 = 106160.

We know that Predecessor of any no. is 1 less than the number. Therefore, Predecessor of 100000 is 99999.

We know that Predecessor of any no. is 1 less than the number. Therefore, if 400 is the predecessor then the number will be 1 greater than the predecessor. So, the number is 401.

We know that Successor of any no. is 1 greater than the number.

The largest 3 digit number is 999.

Therefore, its successor is 999 + 1 = 1000.

If 0 is subtracted from a whole number, then the result is the number itself.

Example: let a whole number 5 .

5 – 0 = 5. Therefore, after subtracting we get the number itself.

We know that the smallest 6 digit natural number is 100000.
Now, the smallest 6 digit natural number ending in 5 is

100005.

If a and b are two whole numbers,

Addition of a and b is a + b which is again a whole number, therefore whole numbers are closed under addition.

Multiplication of a and b is (a b) which is again a whole number, therefore whole numbers are closed under multiplication.

Natural numbers are closed under addition and under multiplication.

Addition of a and b is a + b which is again a natural number, therefore natural numbers are closed under addition.

Multiplication of a and b is (a b) which is again a natural number, therefore natural numbers are closed under multiplication.

Division of a whole number by 0 is not defined.

Multiplication is distributive over addition for whole numbers.

If a, b and c are whole numbers,

Then, a × (b + c)= (a × b) + (a × c)

We know that if a and b are two whole numbers then by commutative property,

a × b= b × a

So, 2395 × 6195 = 6195 × 2395

By expanding 2002 we can write as 2002 = 1001 + 1001

So, 1001 × 2002 = 1001 × ( 1001 + 1001)

We know that any number multiplied by 0 gives the result as 0.

So, 10001 × 0 = 0

We know that any number multiplied by 0 gives the result as 0.

So, 2916 × 0 = 0

We know that if we multiply any number by 1 we get the same number as result.

Therefore, 9128 × 1 = 9128

By associative property we know that,

a + (b + c) = (a + b) + c.

So, 125 + (68 + 17)= (125 + 68) + 17

We know that if we multiply any number by 1 we get the same number as result.

Therefore, 8925 × 1 = 8925.

By distributive property, we know that if a, b and c are whole numbers,

Then, a × (b + c)= (a × b) + (a × c)

19 × 12 + 19 = 19 × 12 + 19 × 1

19 × 12 + 19 = 19 × (12 + 1)

By expanding 35 as 18 + 17

24 × 35 = 24 × (18 + 17)

24 × 35 = 24 × 18 + 24 × 17

By multiplicative property, if a, b and c are whole numbers,

Then, a × (b × c)= (a × b) × c

32 × (27 × 19) = (32 × 27) × 19

786 × 3 + 786 × 7 = 786 × (3 + 7)

= 786 × 10

= 7860

24 × 25 = 24 × = 600

A number is a multiple of each of its factor.

1 is a factor of every number.

The number of factors of a prime number is 2.

Each prime number has only two factors : 1 and the number itself.

Perfect number.

Example : 6

Factors of 6 are 1, 2, 3 and 6.

Sum = 1 + 2 + 3 + 6 = 12 = 2 x 6 = twice the number.

Composite numbers are those numbers which have more than two factors.

Examples : 4, 6, 9, 36, 24, and so on….

4 has factors 1, 2 and 4.

6 has factors 1, 2, 3 and 6.

Similarly, 9, 36, 24, 65, etc are composite numbers

2 is the only prime number which is even.

Prime numbers are those numbers which have only two factors 1 and the number itself.

Co - prime numbers are those numbers having HCF 1.

Examples: (1, 5), (5, 7), (11, 13) etc.

25

Prime numbers are those numbers which have only two factors 1 and the number itself.

Prime numbers between 1 and 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 and 97.

From divisibility of 10 we know that if a number has 0 in ones place then it will be divisible by 10.

From divisibility of 5 we know that a number is divisible by 5, if it has 0 or 5 in its ones places.

According to divisibility test for 2, if the last digit of any number is 0 or 2 or 4 or 6 then the number is divisible by 2.

multiple.

According to divisibility test for 3, if the sum of the digits in a number is a multiple of 3 then it is divisible by 3.

According to divisibility test for 11, if the difference between the sum of digits at odd places (from the right) the sum of digits at even places (from the right) of a number is either 0 or divisible by 11, then the number is divisible by 11.

Multiple

LCM means the lowest common multiple of any two or more numbers.

Factor

HCF means Highest Common factor of any two or more numbers.

(I) – (D), (II) – (F), (III) – (B), (IV) – (E), (V) – (C)

(I) Since, the difference between two consecutive whole numbers is always 1.

(II) Since, the product of two non - zero consecutive whole numbers is always even.

(III) As if zero is divided by any whole number, the quotient is always zero.

(IV) 2 added three times, to the smallest whole number i.e., (2x3) + 0 = 6

(V) As the smallest odd prime number is 3.

(look for the greater digit from left)

2 5 8 4 3 >1 3 5 8 4 >0 8 4 3 5 >0 5 3 4 8 >0 4 8 3 5

for example: 5^th digit(leftmost) of 25843 is more than anyone’s; so it’s the greatest number

(compare the digits from left side, more the digit, more the no.)]

(look for the greater digit from right)

38, 051, 425

30, 040, 700(smallest)

67, 205, 602(greatest)

The greatest: 67205602

The smallest: 30040700

A. 74836

= 7 × 10⁴ + 4 × 10³ + 8 × 10² + 3 × 10 + 6

B. 574021

= 5 × 10⁵ + 7 × 10⁴ + 4 × 10³ + 0 × 10² + 2 × 10 + 1

C. 8907010

= 8 × 10⁶ + 9 × 10⁵ + 0 × 10⁴ + 7 × 10³ + 0 × 10² + 1 × 10 + 0

[Use the fact that, Any number, say, ”a_na_{n - 1}a_{n - 2….}a₁a₀ can be expanded as :

= a_n × 10ⁿ + a_{n - 1} × 10^{n - 1} + a_{n - 2} × 10^{n - 2} + a_{n - 3} × 10^{n - 3} + … + a₁ × 10¹ + a₀ (note, n is the no. of digits of the number),

[this tells you why to look for greater digits from left side, while comparing two numbers. Ask yourself, how! ]

(i)ascending: Andhra Pradesh(76210007)<Bihar(82998509)<Maharashtra(96878627)<Uttar Pradesh(166197921)

(ii)descending: Uttar Pradesh(166197921)>Maharashtra(96878627)>Bihar(82998509)>Andhra Pradesh(76210007)

[Use the process mentioned in Q.152 and Q.153 ;to compare]

142, 800, 000 → One hundred forty two million eight hundred thousand .

Population in 1961 = 439 millions

= 439,000,000

Population in 2001 = 1028 millions

= 1028,000,000

Increase in population = 1028,000,000 – 439,000,000

= 589,000,000

In Indian Numeration,

5,890,000,00

Earth’s radius (6400Km, i.e., 6, 400, 000m) is greater than that of earth, by 6, 400, 000 - 4, 300, 000

= 2, 100, 000m

= 2100Km.

(Using the fact that, 1Km = 1000m).

Tripura: Three million one hundred ninety nine thousand two hundred three.

Meghalaya: Two million three hundred eighteen thousand eight hundred twenty two.

The difference is (2, 16, 813 - 2, 12, 583)

= 4, 230

The money person had = 1000000

The money he spent = Money for color TV + money for motor cycle + Money for flat

= 6580 + 45890 + 870000

= 922470

The money he left = Money he had – Money he spend

= 1000000 – 922470

He left with 77530 rupees,

Remaining vitamin tablet tablets

= 180, 000 - 18, 734

= 161, 266

Money he is left with

= Rs. [610, 000 - (126, 380 + 350, 000)]

= Rs. [610, 000 - 476, 380]

= Rs.133, 620.

The largest number of seven digits = 9, 999, 999

The smallest number of eight digits = 10, 000, 000

So, the difference is = 1

As the remaining digits lie on the middle. it can start with 0.

total remaining digits are = 10 - (4 + 3)[it’s a 10 - digit number, first four and last three digits are given]

= 3

So, these remaining digits, may start with 0 (not required, as it would not make it max.), are such that, they make the maximum number of 3 digits, so that the actual number is maximum,

= 999

The remaining digits are 9, 9, 9.

The first four digits are fixed (9, 9, 7, 9).

The remaining six digits should be such that it forms the least number possible of six digits, using only one digit twice.

Let’s use the digits in this way: the smaller digits will be used in left side [As the left side digits contributes more towards the value of the number: Any number, say, ”a_na_{n - 1}a_{n - 2….}a₁a₀ can be expanded as :

= a_n × 10ⁿ + a_{n - 1} × 10^{n - 1} + a_{n - 2} × 10^{n - 2} + a_{n - 3} × 10^{n - 3} + … + a₁ × 10¹ + a₀], and, 0 (the unique repeatable digit in this case) should be used twice as it is the smallest digit∷003568

[note, here the six digit number, unexpectedly starts with 0. Observe properly, this 0 actually lies in the middle of the actual ten digit number, which we are considering. So, no issue here.]

the smallest number is: 9979003568.

The ten’s digit = 4

The unit digit

=

The thousand’s place = 5 × 1 = 5

The ten thousand’s digit = 2 × 1 = 2

The digit in hundred’s place = 0

The number = 25041.

Let’s use the digits in this way:

To minimize the number, the smaller digits will be used in left side;

And, to maximize the number, larger digits are used in left side. [As the left side digits contributes more towards the value of the number: Any number, say, ”a_na_{n - 1}a_{n - 2….}a₁a₀ can be expanded as :

= a_n × 10ⁿ + a_{n - 1} × 10^{n - 1} + a_{n - 2} × 10^{n - 2} + a_{n - 3} × 10^{n - 3} + … + a₁ × 10¹ + a₀]

The greatest = 765420;

and the smallest = 204567

[Note, 0 can not be used in starting the number, so second smallest digit, available, 2 is used as starting digit].

Their sum

= 969, 987

200 ml can be kept in 1 bottle

1ml can be kept in bottle

35874lit, i . e, 35874 × 1000ml can be kept in

= bottles

= 179, 370 bottles

[1lit = 1000ml]

14 persons contain 1 illiterates

ó1 person contain illiterate

ó450, 772 persons contain illiterates

L.C.M. = 5 × 2 × 2 × 5 × 2 × 2 × 3 × 5 × 2 = 12, 000.

Here, the ten’s place is fixed( = 5).

For the rest of the digits; Let’s use the digits in this way:

To minimize the number, the smaller digits will be used in left side;

And, to maximize the number, larger digits are used in left side. [As the left side digits contributes more towards the value of the number: Any number, say, ”a_na_{n - 1}a_{n - 2….}a₁a₀ can be expanded as :

= a_n × 10ⁿ + a_{n - 1} × 10^{n - 1} + a_{n - 2} × 10^{n - 2} + a_{n - 3} × 10^{n - 3} + … + a₁ × 10¹ + a₀]

The greatest number = 99959

The smallest number = 10050

[note, as we can not start a number with 0, we are taking the 2^nd smallest digit available (here, 1) to start with while forming the least number.]

Weight to be added to 2kg 300g to make it 5kg 68g =

5kg 68 g - 2kg 300g =

5068g - 2300g = 2768 g.

Hence, 2768 grams should be added to 2kg 300g to make it 5kg 68g.

Weight of 1 packet = 120g

Number of packets in a box = 50

∴ Weight of one box = 120g × 50 = 6000g = 6kg

Capacity of van = 900 kg

Hence, number of boxes can be loaded = 900 ÷ 6 = 150 boxes

1 billion = 1000000000

5 billion = 5000000000

1 lakh = 100000

Hence, number of lakhs in 5 billion = 5000000000÷ 100000 = 50, 000

3 crores = 30000000

1 million = 1000000

Hence, number of millions in 3 crores = 30000000÷ 1000000 = 30

(a) Rounding off 874 to nearest hundred = 900

Rounding of 478 to nearest 100 = 500

900 + 500 = 1400

(b) Rounding off 793 to nearest hundred = 800

Rounding of 397 to nearest hundred = 400

800 + 400 = 1200

(c) Rounding off 11244 to nearest hundred = 11200

Rounding of 3507 to nearest hundred = 3500

11200 + 3500 = 14700

(d) Rounding off 17677 to nearest hundred = 17700

Rounding of 3507 to nearest hundred = 13600

17700 + 13600 = 31300

(a) Rounding off 11963 to nearest tens = 11960

Rounding of 9369 to nearest tens = 9370

11960 - 9370 = 2590

(b) Rounding off 76877 to nearest tens = 76880

Rounding of 7783 to nearest tens = 7780

76880 - 7780 = 69100

(c) Rounding off 10732 to nearest tens = 10730

Rounding of 4354 to nearest tens = 4350

10730 - 4350 = 6380

(d) Rounding off 78203 to nearest tens = 78200

Rounding of 16470 to nearest tens = 16410

78200 - 16410 = 61790

(a) Rounding off 87to nearest tens = 90

Rounding of 32 to nearest tens = 30

90 × 30 = 2700

(b) Rounding off 311 to nearest tens = 310

Rounding of 113 to nearest tens = 110

310 × 110 = 34100

(c) Rounding off 3239 to nearest tens = 3240

Rounding of 28 to nearest tens = 30

3240 × 30 = 97200

(d) Rounding off 1385 to nearest tens = 1390

Rounding of 789 to nearest tens = 790

1390 × 790 = 1098100

Population of town in 1991 = 78787

Rounding off to nearest hundred = 78800

Population of town in 2001 = 95833

Rounding off to nearest 100 = 95800

Hence, increase in population by rounding off each population to nearest hundreds = 95800 - 78800 = 17000

According to general rule, Round off each factor to its greatest place, then multiply the rounded off factors.

Rounding off 758 using the general rule, as its greatest place is 100’s, therefore we will round off it to nearest hundreds = 800

Rounding off 6784 using the general rule, as its greatest place is 1000’s, therefore we will round off it to nearest thousands = 7000

800 × 7000 = 560000

In a year,

Number of shirts produced = 216315

Number of trousers = 182736

Number of jackets = 58704

Hence, total number of items produced = 216315 + 182736 + 58704 = 457755

160 = 2 × 2 × 2 × 2 × 2 × 5

170 = 2 × 5 × 17

90 = 2 × 3 × 3 × 5

We know, LCM is calculated as product of unique divisors of no’s

Here,

Lcm = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 17

= 12240

As, 1 L = 1000 mL

Capacity of the vessel = 13 L 200 mL = 13 × 1000 + 200 = 13200 mL

Capacity of each glass = 60 mL

Hence, number of glasses required to fill the vessel = 13200÷60 = 220

Successor of 32 = 32 + 1 = 33

Predecessor of 49 = 49 - 1 = 48

Predecessor of predecessor of 56 = predecessor of 56 - 1 = 56 - 1 - 1 = 54

Successor of successor of 67 = successor of 67 + 1 = 67 + 1 + 1 = 69

Hence, their sum,

= 33 + 48 + 54 + 69

= 204

Weight of each box = 15 kg

Capacity of loading tempo = 482 boxes

Capacity of a van = 518 boxes

∴ Total number of boxes of 15 kg each that can be carried by both the vehicles = 482 + 518 = 1000

Hence, total weight that can be carried by both the vehicles = 15 × 1000 = 15000 kg

Amount spent on

Food and decoration = Rs 216766

Jewellary = Rs 122322

Furniture = Rs 88234

Kitchen = Rs 26780

Hence, total Amount spent = 216766 + 122322 + 88234 + 26780 =

454102

Number of Capsules in each strip = 12

Weight of medicine in each capsule = 500 mg = 0.5g

∴ weight of medicine in each strip = 12 × 0.5 = 6g

Number of strips in each box = 5

∴ weight of each box = 6 × 5 = 30g

Hence, the total weight in grams of medicine in 32 such boxes

= 30 × 32 = 960 g

Least number which when divided by 3, 4 and 5 leaves remainder 0 in each case = LCM of 3, 4 and 5

3 = 3 × 1

4 = 2 × 2

5 = 5 × 1

LCM = 3 × 2 × 2 × 5 = 60

Hence, the least number which when divided by 3, 4 and 5 leaves remainder 2 in each case = 60 + 2 = 62

The capacity of tins should be such that it is a factor of 120, 180, and 240 so that they can be filled in tins of equal capacity.

Hence, the greatest capacity of such a tin = HCF 0f 120, 180 and 240

120 = 2 × 2 × 2 × 3 × 5

180 = 2 × 2 × 3 × 3 × 5

240 = 2 × 2 × 2 × 2 × 3 × 5

Hence, LCM = product of unique divisors = 2 × 2 × 2 × 3 × 5 × 3 × 2 = 720

A number is divisible by 4 if the number formed by its two last digits is divisible by 4

∴ the possible last two digits of the number formed by interchanging first and last digit of the original number are 12, 24, 52

Choose any (say 52)

Now, original number is odd.

Therefore its last digit = 1

And first digit = 2

Second digit = 4(only 4 is remaining)

Hence, the original number = 2451

For a number to be divisible by 4, the number formed by its last two digits should be divisible by 4.

∴ possible last two digits here are 12, 24, 32

By taking 12, possible numbers = 3412 and 4312

By taking 24, possible numbers = 1324 and 3124

By taking 32, possible numbers = 4132 and 1432

Smallest number among above = 1324

Hence, using each of the digits 1, 2, 3 and 4 only once, the smallest 4 - digit number divisible by 4 = 1324

Since, Fatima wants to buy stamp of only one denomination

It should be a factor of 20, 28, 36

Largest such number = HCF of 20, 28, 36 = 4

Hence, the greatest denomination of stamps she must buy to mail the three parcels = Rs. 4

Number of biscuits of brand A that can be bought will be a multiple of 12

Number of biscuits of brand B that can be bought will be a multiple of 15

Number of biscuits of brand C that can be bought will be a multiple of 21

Since, shopkeeper wants to buy an equal number of biscuits of each brand

That number should be divisible by 12, 15, 21

Least such number = LCM of 12, 15, 20 = 420

Hence, minimum number of biscuits of each brand shopkeeper has to buy = 420

8m 96cm = 8 × 100 + 96 = 896 cm

6m 72cm = 6 × 100 + 72 = 672 cm

Side of largest square tile that can be used = HCF of 896 and 672 = 224 cm

Area of the tile = side × side = 224 × 224 cm²

Area of the floor = length × breadth = 896 × 672 cm²

Hence, Minimum number of tiles required = Area of floor/area of tile =

= 32 tiles

Number of English books = 780

Number of Science books = 364

Minimum no of books in each shelf is the LCM of no’s of books.

LCM of 780 and 364:

780 = 2 × 2 × 3 × 5 × 13

364 = 2 × 2 × 7 × 13

LCM = Product of common divisors = 2 × 2 × 3 × 5 × 7 × 13 = 5460

First stop at which both van and bus stops = LCM of 6 and 10 = 30

Hence, the stops at which both of them stop = 30, 30 + 30 = 60 and 30 + 30 + 30 = 90

(a) A number divisible by 11 if difference sum of its digits at odd places and sum of its digits at even places is either 0 or a multiple of 11.

Sum of digits at odd places = 5 + 3 = 8

Sum of digits at even places = 3 + 5 = 8

8 - 8 = 0

Difference is 0

Hence, 5335 is divisible by 11.

(b) A number divisible by 11 if difference sum of its digits at odd places and sum of its digits at even places is either 0 or a multiple of 11.

Sum of digits at odd places = 4 + 8 + 2 + 9 = 23

Sum of digits at even places = 1 + 0 + 0 = 1

23 - 1 = 22

Difference is multiple of 11.

Hence, 9020814 is divisible by 11.

(a) A number is divisible by 4 if the number formed by its two last digits is divisible by 4

Number formed by last two digits of 4096 = 96

96 is divisible by 4

Hence, 4096 is divisible by 4.

(b) A number is divisible by 4 if the number formed by its two last digits is divisible by 4

Number formed by last two digits of 21084 = 84

84 is divisible by 4

Hence, 21084 is divisible by 4.

(c) A number is divisible by 4 if the number formed by its two last digits is divisible by 4

Number formed by last two digits of 31795012 = 12

12 is divisible by 4

Hence, 31795012 is divisible by 4.

(a) A number is divisible by 9 if sum of its digits is divisible by 9.

Sum of digits 0f 672 = 6 + 7 + 2 = 15

15 is not divisible by 9

Hence 672 is not divisible by 9.

(b) A number is divisible by 9 if sum of its digits is divisible by 9.

Sum of digits 0f 5652 = 5 + 6 + 5 + 2 = 18

18 is divisible by 9

Hence, 5652 is divisible by 9.