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# Basic Concepts Of Whole Numbers

Whole Numbers are a part of the Number System and are an important topic for various competitive exams like SSC, Banking, and UPSC CDS. It covers 1-2 questions in the SSC tier 1 exam and covers 8-9 questions in main exams. This topic is an easily understandable and scoring topic.

Whole Numbers:-

Whole numbers are the numbers that are a set of Natural numbers along with zero. A set of whole numbers is given as {0,1,2,3,4……..}. It is denoted by W. W = {0,1,2,3…..}

a) All positive numbers are whole numbers, including 0.

b) All whole numbers are real numbers.

c) All the Natural numbers are whole numbers. Natural Numbers start with 1 except 0.

d) 0 is the smallest whole number.

e) Fractions, decimals, and negative numbers are not considered whole numbers as well as natural numbers.

0 is the number which is neither positive nor negative.

Comparison between Whole Numbers and Natural Numbers:-

From the above definition, it is clear that all-natural Numbers are whole numbers and all whole numbers are natural numbers except 0.

Set of Natural numbers:- {1,2,3,4……}

Properties of Whole Numbers:-

Addition Property:- If 0 is added to a whole number, then the result is the number itself. For Example:- 7+0 = 7.

Multiplication Property:- If 1 is multiplied by a whole number, then the result is the number itself. For example:- 7×1 = 7. If the whole number is multiplied by 0 then the result is 0. For example:- 7×0 = 0.

Division Property:- If a whole number is divided by 0 then the result is not defined. For example:- 7/0 = not defined.

Distributive Property:- This property is represented as P×(Q+R) = (P×Q)+(P×R). It is applicable for both addition and subtraction. For example:- let P=11, Q=12, R=14, 11×(12+14) = (11×12)+(11×14) = 286

Commutative Property:- This property is represented as P+Q = Q+P. This property is also applicable for multiplication, but not for subtraction and division. For example:- P=11, Q=12, 11+12 = 12+11 = 23.

Questions on Whole Numbers:

Q1. A number in which one-sixth part is increased by 30 is equal to one-eighth part is decreased by 100. Find the number.

a) 225
b) 240
c) 196
d) 216

Solution:- Let the number be x.

x/6 + 30 = 100 – x/8

x/6 + x/8 = 70

7x/24 = 70

x = 240

The number is 240. Option b) is correct.

Q2. The Product of two numbers is 140 and the sum of squares of numbers is 449. Find the sum of both numbers.

a) 24
b) 72
c) 33
d) 27

Solution:- let the two numbers be P and Q respectively.

P×Q = 140 ……..(1)

P²+Q² = 449 …..(2)

(P+Q)² = P²+Q²+2PQ

(P+Q)² = 449+2×140 = 449+280 = 729

P+Q = 27

Hence, option d) is the correct.

Q3. Which is the largest five-digit number divisible by 91?

a) 99190
b) 99008
c) 99099
d) None of these

Solution:- The Largest 5-digit number:- 99999

Largest five-digit number divisible by 91:- 91×1090 = 99190

Q4. If the number 35x47x is divisible by 6, then what will be the value of x?

a) 4
b) 5
c) 6
d) 7

Solution:- Divisibility of 6:- The number should have to be divisible by 2 & 3 both.

Divisibility of 3:- sum of numbers divisible by 3 then number divisible by 3

Possible values of x:- 1, 4,7

1 and 7 are not possible because both values can’t divisible by 2. So the value of x is 4.

Hence, option a) is correct.

Q5.  Which of the following numbers is divisible by 13?

a) 8970
b) 8465
c) 7814
d) 9765

Solution:- Multiply the unit digit of the number by 9 and subtract it from the rest of the number. If the resultant number is divisible by 13 then the number is divisible by 13.

Unit digit 0 × 9 = 0

897 – 0 = 897

8970 is divisible by 13.  Hence, option a) is correct.

Divisibility rule of 13:- Multiply unit digit with 9 and subtract from the remaining number.

Q6. 3⁵⁵ + 3⁵⁶ + 3⁵⁷ + 3⁵⁸ is divisible by which of the following.

a) 2
b) 5
c) 4
d) All of these

Solution:- 3⁵⁵(1+3+3²+3³)

3⁵⁵ × 40 which is divisible by 2, 3,4,5,10

i.e 3×40 = 120 is divisible by 2,3,4,5,6,8,10,12 etc. So the number is divisible by all the given options.

Hence, option d) is correct.

Q7. Find the value of 1³+2³+3³+…….+11³.

a) 4356
b) 4576
c) 4646
d) 4816

Solution:- Sum of cubes of natural numbers:- [n(n+1)/2]²

⇒ n = 11

⇒ [11(11+1)/2]² = 121×36

⇒ 4356

Hence, option a) is correct.

Q8. What is the value of 101, 104, 107…….131?

a) 1466
b) 1576
c) 1276
d) 1386

Solution:- All numbers are in Arithmetic Progression

Common difference d = 104-101 = 3

Final term an = a + (n-1)d

Where a is the first term, n is number of terms

131 = 101 + (n-1)3

30 = 3n-3

n = 11

Sum of numbers in A.P = Sn = n/2(a+l) where l = last term

Sn = 11/2 (101+131) = 1276

Hence, option c) is correct.

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