RD SHARMA SOLUTION CHAPTER – 4 Operations on Whole Numbers| CLASS 6TH MATHEMATICS-EDUGROWN

Exercise 4.1

Question: 1

Fill in the blanks to make each of the following a true statement:

Solution:

(i) 359 + 476 = 476 + 359 (Commutativity)

(ii) 2008 + 1952 = 1952 + 2008 (Commutativity)

(iii) 90758 + 0 = 90758 (Additive identity)

(iv) 54321 + (489 + 699) = 489 + (54321 + 699) (Associativity)

Question: 2

Add each of the following and check by reversing the order of addends:

Solution:

(i) 5628 + 39784 = 45412

And,

39784 + 5628 = 45412

(ii) 923584 + 178 = 923762

And,

178 + 923584 = 923762

(iii) 15409 + 112 = 15521

And,

112 + 15409 = 15521

(iv) 2359 + 641 = 3000

And,

641 + 2359 = 3000

Question: 3

Determine the sum by suitable rearrangements:

Solution:

(i) 953 + 407 + 647

Therefore, 53 + 47 = 100

Therefore, (953 + 647) + 407 = 1600 + 407 = 2007

(ii) 15409 + 178 + 591 + 322

409 + 91 = 500

And,

78 + 22 = 100

Therefore, (15409 + 591) + (178 + 322) = (16000) + (500)

= 16500

(iii) 2359 + 10001 + 2641 + 9999

Therefore, 59 + 41 = 100

And, 99 + 01 = 100

Therefore, (2359 + 2641) + (10001 + 9999)

= (5000) + (20000)

= 25000

(iv) 1  + 2 + 3 + 4 + 1996 + 1997 + 1998 + 1999

Therefore, 99 + 1 = 100

98 + 2 = 100

97 + 3 = 100

And

96 + 4 = 100

Therefore, (1 + 1999) + (2 + 1998) + (3 + 1997) + (4 + 1996)

= 2000 + 2000 + 2000 + 2000

= 8000

(v) 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20

10 + 20 = 30

1 + 9 = 10

2 + 8 = 10

3 + 7 = 10

And,

4 + 6 = 10

Therefore, (10 + 20) + (11 + 19) + (12 + 18) + (13 + 17) + (14 + 16)

= 30 + 30 + 30 + 30 + 30 + 15

= 150 + 15

= 165

Question: 4

Which of the following statements are true and which are false?

(i) The sum of two odd numbers is an odd number.

(ii) The sum of two odd numbers is an even number.

(iii) The sum of two even numbers is an even number.

(iv)The sum of two even numbers is an odd number.

(v) The sum of an even number and an odd number is an odd number.

(vi)The sum of an odd number and an even number is an even number.

(vii) Every whole number is a natural number.

(viii) Every natural number is a whole number.

(ix) There is a whole number which when added to a whole number, gives that number

(x) There is a natural number which when added to a natural number, gives that number.

(xi) Commmutativity and associativity are properties of whole numbers.

(xii) Commmutativity and associativity are properties of addition of whole number.

Solution:

(i) FALSE (3 + 5 = 8; 8 is an even number)

(ii) TRUE (3 + 5 = 8; 8 is an even number)

(iii) TRUE (2 + 4 = 6; 6 is an even number)

(iv) FALSE (2 + 4 = 6; 6 is an even number)

(v) TRUE (2 + 3 = 5; 5 is an odd number)

(vi) FALSE (3 + 2 = 5; 5 is not an even number)

(vii) FALSE [The whole number set is {0, 1, 2, 3, 4 …), whereas the natural number set is {1, 2, 3, 4 …)]

(viii) TRUE [The whole number set is {0, 1, 2, 3, 4 …), whereas the natural number set is {1, 2, 3, 4 …)]

(ix) TRUE [That number is zero.]

(x) FALSE

(xi) FALSE

(xii) TRUE

Exercise 4.2

Question: 1

Perform the following subtractions and check your results by performing corresponding additions:

Solution:

(i) 57839 – 2983 = 54856

Verification: 54856 + 2983 = 57839

(ii) 92507 – 10879 = 81628

Verification: 81628 + 10879 = 92507

(iii) 400000 – 98798 = 301202

Verification: 301202 + 98798 = 400000

(iv) 5050501 – 969696 = 4080805

Verification: 4080805 + 969696 = 5050501

(v) 200000 – 97531 = 102469

Verification: 102469 + 97531 = 200000

(vi) 3030301 – 868686 = 2161615

Verification: 2161615 + 868686 = 3030301

Question: 2

Replace each * by the correct digit in each of the following:

Solution:

Here, we can we see that in the units digit, 6 – * = 7, which means that the value of * is 9, as 1 gets carried from 7 at tens place to 6 at unit place and 6 at unit digit becomes 16 then 16 – 9 = 7.

Now, when 7 gives 1 to 6, it becomes 6, so 6 – 3 = 3.

Also, it can be easily deduced that in (8 – * = 6 ), the value of * is 2.

(ii) Here, it is clear that in the units place, 9 – 4 = 5;

And in the tens place,

8 – 3 = 5.

We can now easily find out the other missing blanks by subtracting 3455 from 8989. Addend (difference) = 3455

Thus, the correct answer is:

(iii)

Here, in the units digit, 17 – 8 = 9; in the tens digit, 9 – 7 = 2;

in the hundreds place, 10 – 9 = 1;

and in the thousands place, 9 – 8 = 1.

Addend difference = 5061129.

So, in order to get the addend, we will subtract 5061129 from 6000107.

Thus, the correct answer is:

(iv)

In the units place, 10 -1 = 9;

Also, in the lakhs place, 9 -0 = 9;

Addend difference = 970429.

So, in order to get the addend, we will subtract 970429 from 1000000.

Thus, the correct answer is:

(v) 

Here, in the units digit, 13 – 7 = 6;

in the tens digit, 9 – 8 = 1;

in the hundreds place, 9 – 9 = 0;

and in the thousands place, 10 – 6 = 4.

Addend difference = 4844016.

So, in order to get the addend, we will subtract 4844016 from 5001003.

(vi)

It is clear from the units place that 11 – 9 = 2.

Addend difference = 54322.

To get the other addend, we will subtract 54322 from 111111.

Thus, the other addend is 56789.

The correct answer is:

Question: 3

What is the difference between the largest number of five digits and smallest number of six digits?

Solution:

The largest five – digit number is 99999.

The smallest six – digit number is 100000.

Therefore, difference between them = 100000 – 99999 = 1

Question: 4

Find the difference between the largest number of 4 digits and the smallest number of 6 digits.

Solution:

The largest four – digit number is 9999.

The smallest seven – digit number is 1000000.

Therefore, difference between them = 1000000 – 9999 = 990001

Question: 5

Rohit deposited Rs 125000 in his savings bank account. Later he withdrew Rs 35425 from it. How much money was left in his account?

Solution:

Money deposited by Rohit = Rs 125000

Money withdrawn by Rohit = Rs 35425

Therefore, money left in the account = Rs (125000 – 35425) = Rs 89575

Question: 6

The population of a town is 96209. If the number of men is 29642 and that of women is 29167, determine the number of children.

Solution:

Total population of the town = 96209

Number of men = 29642

Number of women = 29167

Sum of men and women = (29642 + 29167) = 58809

Therefore, Number of children in the town = ( Total population ) – ( Sum of men and women )

= 96209 – 58809 = 37400

Question: 7

The digits of 6 and 9 of the number 36490 are interchanged. Find the difference between the original number and the new number.

Solution:

Original number = 39460

New number = – 36490

Difference = 39460 – 36490 = 2970

Question: 8

The population of a town was 59000. In one year it was increased by 4563 due to new births. However, 9218 persons died or left the town during the year. What was the population at the end of the year?

Solution:

Population of the town = 59000

Increase in the population = 4536

Decrease in the population = 9218

New population = 59000 + 4536 – 9218 = 54318

Exercise 4.3

Question: 1

Fill in the blanks to make each of the following a true statement:

Solution:

(i) 785 × 0 = 0

(ii) 4567 × 1 = 4567 (Multiplicative identity)

(iii) 475 × 129 = 129 × 475 (Commutativity)

(iv) 1243 × 8975 = 8975 × 1243 (Commutativity)

(v) 10 × 100 × 10 = 10000

(vi) 27 × 18 = 27 × 9 + 27 × 4 + 27 × 5

(vii) 12 × 45 = 12 × 50 – 12 × 5

(viii) 78 × 89 = 78 × 100 – 78 × 16 + 78 × 5

(ix) 66 × 85 = 66 × 90 – 66 × 4 – 66

(x) 49 × 66 + 49 × 34 = 49 × (66 + 34)

Question: 2

Determine each of the following products by suitable rearrangements:

Solution:

(i) 2 × 1497 × 50

= (2 × 50) × 1497 = 100 × 1497 = 149700

(ii) 4 × 358 × 25

= (4 × 25) × 358 = 100 × 358 = 35800

(iii) 495 × 625 × 16

= (625 × 16) × 495 = 10000 × 495 = 4950000

(iv) 625 × 20 × 8 × 50

= (625 × 8) × (20 × 50) = 5000 × 1000 = 5000000

Question: 3

Using distributivity of multiplication over addition of whole numbers, find each of the following products:

Solution:

(i) 736 × 103 = 736 × (100 + 3)

{Using distributivity of multiplication over addition of whole numbers}

= (736 × 100) + (736 × 3)

= 73600 + 2208 = 75808

(ii) 258 × 1008 = 258 × (1000 + 8)

{Using distributivity of multiplication over addition of whole numbers}

= (258 × 1000) + (258 × 8)

= 258000 + 2064 = 260064

(iii) 258 × 1008 = 258 × (1000 + 8)

{Using distributivity of multiplication over addition of whole numbers}

= (258 × 1000) + (258 × 8)

= 258000 + 2064 = 260064

Question: 4

Find each of the following products:

Solution:

(i) 736 × 93

Since, 93 = (100 – 7)

Therefore, 736 × (100 – 7)

= (736 × 100) – (736 × 7)

(Using distributivity of multiplication over subtraction of whole numbers)

= 73600 – 5152 = 68448

(ii) 816 × 745

Since, 745 = (750 – 5)

Therefore, 816 × (750 – 5)

= (816 × 750) – (816 × 5)

(Using distributivity of multiplication over subtraction of whole numbers)

= 612000 – 4080 = 607920

(iii) 2032 × 613

Since, 613 = (600 +13)

Therefore, 2032 × (600 + 13)

= (2032 × 600) + (2032 × 13)

= 1219200 + 26416 = 1245616

Question: 5

Find the values of each of the following using properties:

Solution:

(i) 493 × 8 + 493 × 2

= 493 × (8 + 2)

(Using distributivity of multiplication over addition of whole numbers)

= 493 × 10 = 4930

(ii) 24579 × 93 + 7 × 24579

= 24579 × (93 + 7)

(Using distributivity of multiplication over addition of whole numbers)

= 24579 × 100 = 2457900

(iii) 1568 × 184 – 1568 × 84

= 1568 × (184 – 84)

(Using distributivity of multiplication over subtraction of whole numbers)

= 1568 × 100 = 156800

(iv) 15625 × 15625 – 15625 × 5625

= 15625 × (15625 – 5625)

(Using distributivity of multiplication over subtraction of whole numbers)

= 15625 × 10000 = 156250000

Question: 6

Determine the product of:

(i) the greatest number of four digits and the smallest number of three digits.

(ii) the greatest number of five digits and the greatest number of three digits.

Solution:

(i) The largest four-digit number = 9999

The smallest three – digit number = 100

Therefore, Product of the smallest three-digit number and the largest four-digit number = 9999 × 100 = 999900

(ii) The largest five – digit number = 9999

The largest number of three digits = 999

Therefore, Product of the largest three-digit number and the largest five-digit number

= 9999 × 999

= 9999 × (1000 — 1)

= (9999 × 1000) — (9999 × 1)

= 9999000 – 9999

= 9989001

Question: 7

In each of the following, fill in the blanks, so that the statement is true:

Solution:

(i) (500 + 7) (300 – 1)

= 507 × 299

= 299 × 507 (Commutativity)

(ii) 888 + 777 + 555

= 111 (8 + 7 + 5)

= 111 × 20 (Distributivity)

(iii) 75 × 425

= (70 + 5) × 425

= (70 + 5) (340 + 85)

(iv) 89 × (100 – 2)

= 89 × 98

= 98 × 89

= 98 × (100 – 11) (Commutativity)

(v) (15 + 5) (15 – 5)

= 20 × 10

= 200

= 225 – 25

(vi) 9 × (10000 + 974)

= 98766

Question: 8

A dealer purchased 125 color television sets. If the cost of each set is Rs 19820, determine the cost of all sets together.

Solution:

Cost of 1 color television set = Rs 19820

Therefore, Cost of 125 color television sets = Rs (19820 × 125)

= Rs 19820 × (100 + 25)

= Rs (19820 × 100) + (19820 × 25)

= Rs 1982000 + 495500

= Rs 2477500

Question: 9

The annual fee charged from a student of class 6^th in a school is Rs 8880. If there are, in all, 235 students in class 6^th, find the total collection.

Solution:

Fees charged from 1 student = Rs 8880

Therefore, Fees charged from 235 students = Rs 8880 × 235

= 2086800

Thus, the total collection from class VI students is Rs 2086800.

Question: 10

A group housing society constructed 350 flats. If the cost of construction for each flat is Rs 993570, what is the total cost of construction of all the flats.

Solution:

Cost of construction of 1 flat = Rs 993,570

Total number of flats constructed = 350

Total cost of construction of 350 flats = Rs (993,570 × 350)

= Rs 347,749,500

Question: 11

The product of two whole numbers is zero. What do you conclude?

Solution:

If the product of two whole numbers is zero, then it means that either one of them is zero or both of them are zero.

Question: 12

What are the whole numbers which when multiplied with itself gives the same number?

Solution:

There are two numbers which when multiplied with themselves give the same numbers.

(i) 0 × 0 = 0

(ii) 1 × 1 = 1

Question: 13

In a large housing complex, there are 15 small buildings and 22 large building. Each of the large buildings has 10 floors with 2 apartments on each floor. Each of the small buildings has 12 floors with 3 apartments on each floor. How many apartments are there in all.

Solution:

Number of large buildings = 22

Number of small buildings = 15

Number of floors in 1 large building = 10

Number of apartments on 1 floor = 2

Therefore, Total apartments in 1 large building = 10 × 2 = 20

Similarly,

Total apartments in 1 small building = 12 × 3 = 36

Therefore, Total apartments in the entire housing complex = (22 × 20) + (15 × 36)

= 440 + 540

= 980

Exercise 4.4

Question: 1

Does there exists a whole number ‘a’ such that a/a = a?

Solution:

Yes, there exists a whole number ‘a’ such that a/a = a.

The whole number is 1 such that,

1/1 = 1

Question: 2

Find the value of:

Solution:

(i) 23457 / 1 = 23457

(ii) 0 / 97 = 0

(iii) 476 + (840 / 84) = 476 + 10 = 486

(iv) 964 – (425 / 425) = 964 – 1 = 963

(v) (2758 / 2758) – (2758 + 2758) = 1 – 1 = 0

(vi) 72450 / (583 – 58) = 72450 + 525 = 138

Question: 3

Which of the following statements are true:

Solution:

(i) False

LHS: 10 / (5 × 2)

= 10 / 10

=1

RHS: (10 / 5) × (10 / 2)

= 2 × 5 = 10

(ii) True

LHS: (35 – 14) / 7

= 21 / 7

=3

RHS: 35 / 7 – 14 / 7

= 5 – 2 = 3

(iii) False

LHS: 35 – 14 / 7

= 35 – 2 = 33

RHS: 35 / 7 – 14 / 7

= 5 – 2

= 3

(iv) False

LHS: (20 – 5) / 5

= 15 / 5

= 3

RHS: 20 / 5 – 5

= 4 – 5 = -1

(v) False

LHS: 12 × (14 / 7)

= 12 × 2

= 24

RHS: (12 × 14) / (12 × 7)

= 168 / 84

=2

(vi) True

LHS: (20 / 5) / 2

= 4 / 2

=2

RHS : ( 20 / 2 ) / 5

= 10 / 5

= 2

Question: 4

Divide and check the quotient and remainder:

Solution:

(i) 7777 / 58 = 134

Verification: [Dividend = Divisor × Quotient + Remainder ]

7772 = 58 × 134 + 0

7772 = 7772

LHS = RHS

(ii) 6906/35 gives quotient = 197 and remainder = 11

Verification: [Dividend = Divisor × Quotient + Remainder ]

6906 = 35 × 197 + 11

6906 = 6895 + 11

6906 = 6906

LHS = RHS

(iii) 16135 / 875 gives quotient = 18 and remainder = 385.

Verification: [Dividend = Divisor × Quotient + Remainder ]

16135 = 875 × 18 + 385

16135 = 15750 + 385

16135 = 16135

LHS = RHS

(iv) 16025/1000 gives quotient and remainder = 25

Verification: [Dividend = Divisor × Quotient + Remainder ]

16025 = 1000 × 16 + 25

16025 = 16000 + 25

16025 = 16025

LHS = RHS

Question: 5

Find a number which when divided by 35 gives the quotient 20 and remainder 18.

Solution:

Dividend = Divisor × Quotient + Remainder

Dividend = 35 × 20 + 18

= 700 + 18

= 718

Question: 6

Find the number which when divided by 58 gives a quotient 40 and remainder 31.

Solution:

Dividend = Divisor × Quotient + Remainder

Dividend = 58 × 40 + 31

= 2320 + 31

= 2351

Question: 7

The product of two numbers is 504347. If one of the numbers is 1591, find the other.

Solution:

Product of two numbers = 504347

One of the two numbers = 1591

Let the number be A.

Therefore, A × 1591 = 504347

A = 5043471591 = 317

Question: 8

On dividing 59761 by a certain number, the quotient is 189 and the remainder is 37. Find the divisor.

Solution:

Dividend = 59761

Quotient = 189

Remainder = 37

Divisor = A

Now, Dividend = Divisor × Quotient + Remainder

59761 =A × 189 + 37

59761 – 37 = A × 189

59724 = A × 18

Therefore, A = 59724189

= 316

Question: 9

On dividing 55390 by 299, the remainder is 75. Find the quotient.

Solution:

Dividend = 55390

Divisor = 299

Remainder = 75

Quotient = A

Dividend = Divisor × Quotient + Remainder

55390 = 299 × A + 75

55390 – 75 = A × 299

55315 = A × 299

Therefore, A = 55315299 = 185

Exercise 4.5

Question: 1

Without drawing a diagram, find:

Solution:

(i) 10th square number:

A square number can easily be remembered by the following rule

N^th square number = n × n

10th square number = 10 × 10 = 100

(ii) 6th triangular number:

A triangular number can easily be remembered by the following rule

N^th triangular number = n × ( n + 1 )2

Therefore, 6^th triangular number = 6 × (6 + 1)2 = 21

Question: 2

(i) Can a rectangle number also be a square number?

(ii) Can a triangular number also be a square number?

Solution:

(i) Yes, a rectangular number can also be a square number; for example, 16 is a square number also a rectangular number.

(ii) Yes, there e×ists only one triangular number that is both a triangular number and a square number, and that number is 1.

Question: 3

Write the first four products of two numbers with difference 4 starting from in the following order:

1 , 2 , 3 , 4 , 5 , 6 , ………..

Identify the pattern in the products and write the next three products.

Solution:

1 × 5 = 5 (5 – 1 = 4)

2 × 6 = 12 (6 – 2 =4)

3 × 7 = 21 (7 – 3 = 4)

4 × 8 = 32 (8 – 4 = 4)

Question: 4

Observe the pattern in the following and fill in the blanks:

Solution:

9 × 9 + 7 =88

98 × 9 + 6 = 888

987 × 9 + 5 = 8888

9876 × 9 + 4 = 88888

98765 × 9 + 3 = 888888

987654 × 9 + 2 = 8888888

9876543 × 9 + 1 = 88888888

Question: 5

Observe the following pattern and extend it to three more steps:

Solution:

6 × 2 – 5 = 7

7 × 3 – 12 = 9

8 × 4 – 21 = 11

9 × 5 – 32 = 13

10 × 6 – 45 = 15

11 × 7 – 60 = 17

12 × 8 – 77 = 19

Question: 6

Study the following pattern:

1 + 3 = 2 × 2

1 + 3 + 5 = 3 × 3

1 + 3 + 5 + 7 = 4 × 4

1 + 3 + 5 + 7 + 9 = 5 × 5

By observing the above pattern, find:

Solution:

(i) 1 + 3 + 5 + 7 + 9 + 11

= 6 × 6

= 36

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

= 8 × 8

= 64

(iii) 21 + 23 + 25 + … + 51

= (21 + 23 + 25 +…+ 51) can also be written as

(1 + 3 + 5 + 7 +…+ 49 + 51) – (1 + 3 + 5 +…+ 17 + 19)

(1 + 3 + 5 + 7+…+ 49 + 51) = 26 × 26 = 676

and, (1 + 3 + 5 +…+ 17 + 19 ) = 10 × 10 = 100

Now,

(21 + 23 + 25 +…+ 51 ) =  676  – 100 = 576

Question: 7

Study the following pattern:

By observing the above pattern, write next two steps.

Solution:

The next two steps are as follows:

1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5

= 5 × 6 × 116

= 55

1 × 1+ 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6

= 6 × 7 × 136

= 91

Question: 8

Study the following pattern:

By observing the above pattern, find:

Solution:

(i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10

= 10 × 112

= 55

(ii) 50 + 51 + 52 + …+ 100

This can also be written as

(1 + 2 + 3 + …+ 99 + 100) – (1 + 2 + 3 + 4 + …+ 47 + 49)

Now,

(1 + 2 + 3 + …+ 99 + 100 ) = 100 × 1012

and, (1 + 2 + 3 + 4 +…+ 47 + 49 ) = 49 × 502

So, (50 + 51 + 52 + …+ 100 ) = 100 × 1012 – 49 × 502

= 5050 – 1225

= 3825

(iii) 2 + 4 + 6 + 8 + 10 +…+ 100

This can also be written as 2 × (1 + 2 + 3 + 4 + …+ 49 + 50)

Now,

(1 + 2 + 3 + 4 + …+ 49 + 50 ) = 50 × 512

= 1275

Therefore, (2 + 4 + 6 + 8 + 10 + …+ 100) = 2 × 1275 = 2550

Exercise 4.6

Question: 1

Which one of the following is the smallest whole number?

(a) 1  (b)  2   (c) 0  (d) None of these

Solution:

The set of whole numbers is {0 , 1, 2, 3, 4, …}.

So, the smallest whole number is 0.

Hence, the correct option is (c).

Question: 2

Which one of the following is the smallest even whole number?

(a) 0 (b) 1 (c) 2 (d) None of these

Solution:

The natural numbers along with 0 form the collection of whole numbers.

So, the numbers 0, 1, 2, 3, 4, … form the collection of whole numbers.

The number which is divisible by 2 is an even number.

So, in the collection “0, 1, 2, 3, 4, …”, 2 is the smallest even number.

Hence, the correct option is (c).

Question: 3

Which one of the following is the smallest odd whole number?

(a) 0 (b) 1 (c) 3 (d) 5

Solution:

The natural numbers along with 0 form the collection of whole numbers.

So, the numbers 0, 1, 2, 3, 4, … form the collection of whole numbers.

A natural number which is not divisible by 2 is called an odd whole number.

So, in the collection “0, 1, 2, 3, 4, …”, 1 is the smallest odd whole number.

Hence, the correct option is (b).

Question: 4

How many whole numbers are between 437 and 487?

(a) 50 (b) 49 (c) 51 (d) None of these

Solution:

The whole numbers between 437 and 487 are 438, 439, 440, 441, … , 484, 485 and 486. To find the required number of whole numbers,

We need to subtract 437 from 487 and then subtract again 1 from the result.

Thus, there are (487 – 437) – 1 whole numbers between 437 and 487.

Now, (487 – 437) – 1 = 50 – 1 = 49

Hence, the correct option is (b).

Question: 5

The product of the successor 999 and predecessor of 1001 is:

(a) one lakh (b) one billion (c) one million (d) one crore

Solution:

Successor of 999 = 999 + 1 = 1000

Predecessor of 1001 = 1001 – 1 = 1000

Now,

Product = (Successor of 999) × (Predecessor of 1001)

= 1000 × 1000

= 1000000

= one million

Hence, the correct option is (c).

Question: 6

Which one of the following whole numbers does not have a predecessor?

(a) 1 (b) 0 (c) 2 (d) None of these

Solution:

The numbers 0, 1, 2, 3, 4, …. form the collection of whole numbers.

The smallest whole number is 0.

So, 0 does not have a predecessor.

Hence, the correct option is (b).

Question: 7

The number of whole numbers between the smallest whole number and the greatest 2 digit number is:

(a) 101 (b) 100  (c) 99 (d) 98

Solution:

Smallest whole number = 0

Greatest 2-digit whole number = 99

The whole numbers between 0 and 99 are 1, 2, 3, 4 …… 97, 98.

To find the number of whole numbers between 0 and 99,

Subtract 1 from the difference of 0 and 99.

Therefore, Number of whole numbers between 0 and 99 = (99 – 0) – 1

= 99 – 1

= 98

Hence, the correct option is (d).

Question: 8

If n is a whole number such that n + n = n, then n =?

(a) 1 (b) 2 (c) 3 (d) None of these

Solution:

Here, 0 + 0 = 0, 1 + 1 = 2 , 2 + 2 = 4 …..

So, the statement n + n = n is true only when n = 0.

Hence, the correct option is (d).

Question: 9

The predecessor of the smallest 3 digit number is:

(a) 999 (b) 99 (c) 100 (d) 101

Solution:

Smallest 3-digit number = 100

Predecessor of 3-digit number = 100 — 1 = 99

Hence, the correct option is (b).

Question: 10

The least number of 4 digits which is exactly divisible by 9 is:

(a)1008  (b)1009  (c)1026  (d)1018

Solution:

Least 4-digit number = 1000

The least 4-digit number exactly divisible by 9 is 1000 + (9 – 1) = 1008.

Hence, the correct option is (a).

Question: 11

The number which when divided by 53 gives 8 as quotient and 5 as remainder is:

(a) 424 (b) 419 (c) 429 (d) None of these

Solution:

Here, Divisor = 53, Quotient = 8 and Remainder = 5.

Now, using the relation Dividend = Divisor x Quotient + Remainder

We get

Dividend = 53 x 8 + 5

= 424 + 5

= 429

Thus, the required number is 429.

Hence, the correct option is (c).

Question: 12

The whole number n satisfying n + 35 = 101 is:

(a) 65 (b) 67 (c) 64 (d) 66

Solution:

Here, n+ 35 = 101.

Adding – 35 on both sides, we get

n + 35 + (- 35) = 101 + (- 35)

n + 0 = 66

n = 66

Hence, the correct option is (d).

Question: 13

The value of 4 x 378 x 25 is:

(a) 37800 (b) 3780 (c) 9450 (d) 30078

Solution:

By regrouping, we get

4 × 378 × 25 = 4 × 25 × 378

= 100 × 378

= 37800

Hence, the correct option is (a).

Question: 14

The value of 1735 x 1232 – 1735 x 232 is:

(a) 17350 (b) 173500 (c) 1735000 (d) 173505

Solution:

Using distributive law of multiplication over subtraction, we get

1735 × 1232 – 1735 × 232 = 1735 (1232 – 232)

= 1735 × 1000

= 1735000

Hence, the correct option is (c).

Question: 15

The value of 47 × 99 is:

(a) 4635 (b) 4653 (c) 4563 (d) 6453

Solution:

Since, 99 = 100 — 1

Therefore, 47 × 99 = 47 × (100 — 1)

= 47 × 100 — 47

= 4700 — 47

= 4653

Thus, the value of 47 × 99 is 4653.

Hence, the correct option is (b).