Rd Sharma 2018 Solutions for Class 6 Math Chapter 4 Operations On Whole Numbers are provided here with simple step-by-step explanations. These solutions for Operations On Whole Numbers are extremely popular among Class 6 students for Math Operations On Whole Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2018 Book of Class 6 Math Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2018 Solutions. All Rd Sharma 2018 Solutions for class Class 6 Math are prepared by experts and are 100% accurate.

#### Question 1:

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

(i) 785 × 0=....

(ii) 4567 × 1=...

(iii) 475 × 129=129 × ....

(iv) .... × 8975=8975 × 1243

(v) 10 × 100 × ...=10000

(vi) 27 × 18=27 × 9+27 × ...+27 × 5

(vii) 12 × 45=12 × 50 − 12 × ...

(viii) 78 × 89=78 × 100 − 78...+78 × 5

(ix) 66 × 85=66 × 90 − 66 × ... − 66

(x) 49 × 66+49 × 34=49 × (...+...)

(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 × + 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:

(i) 2 × 1497 × 50

(ii) 4 × 358 × 25

(iii) 495 × 625 × 16

(iv) 625 × 20 × 8 × 50

(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:

(i) 736 × 103
(ii) 258 × 1008
(iii) 258 × 1008

(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.

(i) 736 × 93
(ii) 816 × 745
(iii) 2032 × 613

(i) 736 × 93
∵ 93 = (100 − 7)
∴ 736 × (100 − 7)
= (736 × 100) − ( 736 × 7)
(Using distributivity of multiplication over subtraction of whole numbers)
= 73600 − 5152 = 68448

(ii) 816 × 745
∵ 745 = (750 − 5)
∴ 816 × (750 − 5)
= (816 × 750) − ( 816 × 5)
(Using distributivity of multiplication over subtraction of whole numbers)
= 612000 − 4080 = 607920

(iii) 2032 × 613
∵ 613 = (600 +13)
∴ 2032 × (600 + 13)
= (2032 × 600) + ( 2032 × 13)
= 1219200 + 26416 = 1245616

#### Question 5:

Find the values of each of the following using properties:

(i) 493 × 8 + 493 × 2

(ii) 24579 × 93 + 7 × 24579

(iii) 1568 × 184 − 1568 × 84

(iv) 15625 × 15625 − 15625 × 5625

(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.

(i) The largest four−digit number = 9999
The smallest three−digit number = 100
∴ 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
∴ 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:

(i) (500 + 7) (300 − 1) = 299 × ...

(ii) 888 + 777 + 555 = 111 × ....

(iii) 75 × 425 = (70 + 5) (... + 85)

(iv) 89 × (100 − 2) × 98 × (100 − ....)

(v) (15 + 5) (15 − 5) = 225 − ....

(vi) 9 × (10000 + ...) = 98766

(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 colour television sets. If the cost of each set is Rs 19820, determine the cost of all sets together.

Cost of 1 colour television set =  Rs 19820
∴ Cost of  125 colour 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 VI in a school is Rs 8880. If there are, in all, 235 students in class VI, find the total collection.

Fees charged from 1 student = Rs 8880
∴ 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.

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?

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?

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 floor 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.

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
∴ Total apartments in 1 large building = 10 × 2 = 20
Similarly,
​Total apartments in 1 small building = 12 × 3 = 36

∴​ Total apartments in the entire housing complex = (22 × 20) + (15 × 36)
= 440 + 540
= 980

#### Question 1:

Does there exist a whole number a such that a $÷$ a = a?

Yes, there exists a whole number a such that a $÷$ a = a.
The whole number is 1 such that .

#### Question 2:

Find the value of:

(i) 23457 $÷$ 1
(ii) 0 $÷$ 97
(iii) 476 + (840 $÷$ 84)
(iv) 964 − (425 $÷$ 425)
(v) (2758 $÷$ 2758) − (2758 $÷$ 2758)
(vi) 72450 $÷$ (583 − 58)

(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:

(i) 10 $÷$ (5 × 2) = (10 $÷$ 5) × (10 $÷$ 2)
(ii) (35 − 14) $÷$ 7 = 35 $÷$ 7 − 14 $÷$ 7
(iii) 35 − 14 $÷$ 7 = 35 $÷$ 7 − 14 $÷$ 7
(iv) (20 − 5) $÷$ 5 = 20 $÷$ 5 − 5
(v) 12 × (14 $÷$ 7) = (12 × 14) $÷$ (12 × 7)
(vi) (20 $÷$ 5) $÷$ 2 = (20 $÷$ 2) $÷$ 5

(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:

(i) 7772 $÷$ 58
(ii) 6906 $÷$ 35
(iii) 16135 $÷$ 875
(iv) 16025 $÷$ 1000

(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 = 16 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.

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.

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.

Product of two numbers = 504347
One of the two numbers = 1591
Let the number be A.
∴ A × 1591 = 504347

A = $\frac{504347}{1591}$ = 317 #### Question 8:

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

Dividend = 59761
Quotient = 189
Remainder = 37
Divisor = A
Now, Dividend = Divisor × Quotient + Remainder
59761 = A × 189 + 37
59761 − 37 = A × 189
59724 = A × 18

A = $\frac{59724}{189}$ = 316

#### Question 9:

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

Dividend = 55390
Divisor = 299
Remainder = 75
Quotient = A
Dividend = Divisor × Quotient + Remainder
55390 = 299 × A + 75
55390 - 75 = A × 299
55315 = A × 299

A = $\frac{55315}{299}$ = 185

#### Question 1:

Without drawing a diagram, find

(i) 10th square number
(ii) 6th triangular number.

(i) 10th square number: A square number can easily be remembered by the following rule:
nth square number = n2
∴ 10th square number = 102 = 100

(ii) 6th triangular number: A triangular number can easily be remembered by the following rule:
nth triangular number = $\frac{n×\left(n+1\right)}{2}$
∴ 6th triangular number = $\frac{6×\left(6+1\right)}{2}$ = 21

#### Question 2:

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

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

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:

9 × 9 + 7              = 88
98 × 9 + 6            = 888
987 × 9 + 5          = 8888
9876 × 9 + 4        = ...............
98765 × 9 + 3      = ...............
987654 × 9 + 2    = ...............
9876543 × 9 + 1  = ...............

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:

6 × 2 −  5        = 7
7 × 3 − 12       = 9
8 × 4 − 21       = 11
9 × 5 − 32       = 13
.... × .... − ..... = ......
.... × .... − ..... = ......
.... × .... − ..... = ......

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

(i) 1 + 3 + 5 + 7 + 9 + 11
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
(iii) 21 + 23 + 25 + ........ + 51

(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:

1 × 1 + 2 × 2                                = $\frac{2×3×5}{6}$
1 × 1 + 2 × 2 + 3 × 3                   = $\frac{3×4×7}{6}$
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4      = $\frac{4×5×9}{6}$
By observing the above pattern, write next two steps.

The next two steps are as follows:
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 = $\frac{5×6×11}{6}$ = 55
1 × 1 + 2 × 2 + 3 × 3 + 4 × 4 + 5 × 5 + 6 × 6 = $\frac{6×7×13}{6}$ = 91

#### Question 8:

Study the following pattern:

1                        = $\frac{1×2}{2}$
1 + 2                 = $\frac{2×3}{2}$
1 + 2 + 3           = $\frac{3×4}{2}$
1 + 2 + 3 + 4    = $\frac{4×5}{2}$
By observing the above pattern, find

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

(ii) 50 + 51 + 52 + ................ + 100

(iii) 2 + 4 + 6 + 8 + 10 + ............... + 100

(i) 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = $\frac{10×11}{2}$ = 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 ) = $\frac{100×101}{2}$
and, (1 + 2 + 3 + 4 + ...+ 47 + 49 )  = $\frac{49×50}{2}$
So, (50 + 51 + 52 + ...+ 100 ) = $\frac{100×101}{2}$ - $\frac{49×50}{2}$ = 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 ) = $\frac{50×51}{2}$ = 1275
∴ (2 + 4 + 6 + 8 + 10 + ...+ 100)  = 2 × 1275 = 2550

#### Question 1:

Which one of the following is the smallest whole number?
(a) 1                     (b) 2                         (c) 0                                     (d) None of these

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

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

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

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 of 999 and the predecessor of 1001 is
(a) one lakh                            (b) one billion                               (c) one million                                (d) one crore

Successor of 999 = 999 + 1 = 1000
Predecessor of 1001 = 1001 − 1 = 1000
Now
Product = (Successor of 999) $\text{×}$ (Predecessor of 1001)
= 1000 $\text{×}$ 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

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

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.
∴ 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

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

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

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

Here, Divisor = 53, Quotient = 8 and Remainder = 5.
Now, using the relation Dividend = Divisor $\text{×}$ Quotient + Remainder, we get
Dividend = 53 $\text{×}$ 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

Here, n + 35 = 101.
Adding − 35 on both sides, we get
n + 35 + (− 35)= 101 + (− 35)
$\text{⇒}$ n + 0 = 66
$\text{⇒}$ n = 66
Hence, the correct option is (d).

#### Question 13:

The 4 $\text{×}$ 378 $\text{×}$ 25 is

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

By regrouping, we get
4 $\text{×}$ 378 $\text{×}$ 25 = 4 $\text{×}$ 25 $\text{×}$ 378
= 100 $\text{×}$ 378
= 37800
Hence, the correct option is (a).

#### Question 14:

The value of 1735 $\text{×}$ 1232 − 1735 $\text{×}$ 232 is

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

Using distributive law of multiplication over subtraction, we get
1735 $\text{×}$ 1232 − 1735 $\text{×}$ 232 = 1735(1232 − 232)
= 1735 $\text{×}$ 1000
= 1735000
Hence, the correct option is (c).

#### Question 15:

The value of 47 $\text{×}$ 99 is

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

∵  99 = 100 − 1
∴ 47 $\text{×}$ 99 = 47 $\text{×}$ (100 − 1)
= 47$\text{×}$ 100 − 47
= 4700 − 47
= 4653
Thus, the value of 47 $\text{×}$ 99 is 4653.
Hence, the correct option is (b).

#### Question 1:

Which one of the following is not a natural numbers ?

(a) 1                 (b) 10                       (c) 0                              (d) 20

The numbers 1, 2, 3, 4, 5, ... form the collection of natural numbers.
Thus, 0 is not a natural number.
Hence, the correct option is (c).

#### Question 2:

If n is an odd natural number greater than 1, then the product of its successor and predecessor

(a) is an odd natural number                 (b) is an even natural number

(c) can be even or odd                           (d) None of these

The successor and predecessor of an odd natural number are both even.
Here, n = odd. So
Successor of n = n + 1                     (Even natural number)
Predecessor of n = n  − 1                 (Even natural number)
Product = (n + 1) $\text{×}$ (n  − 1) = even $\text{×}$ even = even
Thus, the product of the successor and predecessor of n is an even number.
Hence, the correct option is (b).

#### Question 3:

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

(a) 1000                             (b) 999                                (c) 998                                    (d) None of these

The smallest whole number is 0.
Greatest three digit number = 999
The number of whole numbers from 0 to 999 is 1000.
So, the number of whole numbers between 0 and 999 (excluding 0 and 999) is 1000 − 2 = 998.
Hence, the correct option is (c).

#### Question 4:

If x + 12 = 12 + 7, then by commutativity of addition x =

(a) 12                             (b) 7                                (c) 19                                   (d) 5

Here, x + 12 = 12 + 7.
But, by commutative of addition, we have
7 + 12 = 12 + 7
So, comparing x + 12 = 12 + 7 and 7 + 12 = 12 + 7, we have
x = 7
Hence, the correct option is (b).

#### Question 5:

If (31 + 15) + x = 31 + (15 + 23), then by associativity of addition x =

(a) 46                             (b) 38                                (c) 23                                   (d) 69

Here, (31 + 15) + x = 31 + (15 + 23).
But, by associativity of addition, we have
(31 + 15) + 23 = 31 + (15 + 23)
So, comparing (31 + 15) + x = 31 + (15 + 23) and (31 + 15) + 23 = 31 + (15 + 23), we have
x = 23
Hence, the correct option is (c).

#### Question 6:

What is the multiplicative identity element in the set of whole numbers?

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

A multiplicative identity is a number which when multiplied with any number, the number
is unchanged.
For example :1 $\text{×}$ 7 = 7, 1 $\text{×}$ 9 = 9, 1 $\text{×}$ 100 = 100 etc.
Thus, 1 is multiplicative identity in the set of whole numbers is 1.
Hence, the correct option is (c).

#### Question 7:

What is the additive identity element in the set of whole numbers?

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

An additive identity is a number which when added to any number, the number is unchanged.
For example : 0 + 7 = 7, 0 + 9 = 9, 0 + (− 11) = − 11 etc.
Thus, 0 is the additive identity in the set of whole numbers.
Hence, the correct option is (a).

#### Question 8:

If n is an even number, then the product of its successor and predecessor
(a) is an even natural number                             (b) is an odd natural number
(c) can be even or odd                                        (d) None of these

Here, n is an even number.
Successor of n = n + 1
Predecessor of n = n − 1
Product = (n + 1) $\text{×}$ (n − 1) = odd $\text{×}$ odd = odd
Thus, the product of the successor and the predecessor of n is an odd natural number.
Hence, the correct option is (b).

#### Question 9:

(a) $\frac{1}{28}$                            (b) 0                                (c) − 28                            (d) 82

Here, the given number is 28.
Additive inverse of a given number is a number which when added to the given number gives zero.
Now
28 + (− 28) = 0
Thus, −28 is the additive inverse of 28.
Hence, the correct option is (c).

#### Question 10:

Which of the following is not zero?

(a) 0 $\text{×}$ 0                            (b) $\frac{\text{0}}{\text{2}}$                                (c) $\frac{\left(\text{6-6}\right)}{\text{2}}$                            (d) 4 + 0

In the product of two numbers, if one of the numbers (or both) is (are) zero, then the product is zero.

In a division, if the numerator is zero, then the answer is zero.

Similarly, $\frac{\left(\text{6-6}\right)}{\text{2}}=\frac{\text{0}}{\text{2}}\text{=0}$
Here, 0 is an additive identity in the set of whole numbers.
∴     4 + 0 =4
Thus, 4 + 0 is not zero.
Hence, the correct option is (d).

#### Question 11:

Determine the product of the greatest number of three digits and the smallest number of two digits.

Greatest 3-digit number = 999
Smallest 2-digit number = 10
Product = 999 $\text{×}$ 10 = 9990
Hence, the required product is 9990.

#### Question 12:

Simplify 754 $\text{×}$ 845 + 754 $\text{×}$ 155 by using distributivity of multiplication over addition.

Distributive law of multiplication over addition: a (b + c) = a $\text{×}$ b + a $\text{×}$ c
∴    754 $\text{×}$ 845 + 754 $\text{×}$ 155 = 754(845 + 155)
= 754 $\text{×}$ 1000
= 754000
Hence, 754 $\text{×}$ 845 + 754 $\text{×}$ 155 = 754000.

#### Question 13:

Match the following:

(i) 625 $\text{×}$ 436 = 625 $\text{×}$ 400 + 625 $\text{×}$ 30 + 625 $\text{×}$ 6            (a) Commutative under addition
(ii) 25 $\text{×}$ 69 $\text{×}$ 8 = 25 $\text{×}$ 8 $\text{×}$ 69                                          (b) Commutative under multiplication
(iii) 60 + 19758 + 840 = 60 + 940 + 19758                        (c) Distributive of multiplication over addition

(i)
625 $\text{×}$ 436 = 625 $\text{×}$ (400 + 30 + 6)
= 625 $\text{×}$ 400 + 625 $\text{×}$ 30 + 625 $\text{×}$ 6
This is the distributive law of multiplication over addition.
(ii)
25 $\text{×}$ 69 $\text{×}$ 8 = 25 $\text{×}$ 8 $\text{×}$ 69
This is the commutative law of multiplication.
(iii)
60 + 19758 + 840 = 60 + 940 + 19758
This is the commutative law of addition.
Hence, the match is (i) $\text{→}$ (c), (ii) $\text{→}$ (b), (iii) $\text{→}$ (a).

#### Question 14:

Use the following sum by suitable rearrangement:
2 + 3 + 4 + 5 + 45 + 46 + 47 + 48

Thus sum 2 + 3 + 4 + 5 + 45 + 46 + 47 + 48 has to be evaluated using rearrangement.
We observe that
2 + 48 = 50
3 + 47 = 50
4 + 46 = 50
5 + 45 = 50
Therefore
2 + 3 + 4 + 5 + 45 + 46 + 47 + 48
= (5 + 45) (4 + 46) + (3 + 47) + (2 + 48)
= 50 + 50 + 50 + 50
= 200
Hence, 2 + 3 + 4 + 5 + 45 + 46 + 47 + 48 = 200.

#### Question 15:

Complete the following magic square by supplying the missing numbers: In magic square the numbers in each vertical, horizontal, and diagonal row add up to the same value.
Sum of diagonal elements = 2 + 12 + 7 + 17 = 38
Now
Missing element in the first column = 38 − (2 + 9 + 14) = 38 − 25 = 13
Missing element in the first row = 38 − (2 + 15 + 16) = 38 − 33 = 5

Missing element in the second row and fourth column = 38 − (5 + 10 + 17) = 38 − 32 = 6
Missing element in the second row and third column = 38 − (9 + 12 + 6) = 38 − 27 = 11
Missing element in the fourth row and third column = 38 − (16 + 11 + 7) = 38 − 34 = 4
Missing element in the third row and third column = 38 − (13 + 7 + 10) = 38 − 30 = 8
Missing element in the fourth row and second column = 38 − (14 + 4 + 17) = 38 − 35 = 3

Hence, the complete magic square is #### Question 16:

Replace * by correct digit in the following: The complete subtraction is shown below: Thus, the the digits replaced are circled as below: #### Question 17:

Shikha withdrew ₹1,00,000 from her bank account. She purchased a TV set for ₹38,750, a
refrigerator for ₹23,890 and jewellery worth ₹35,560. How much money was left with her.

Cost of TV set = ₹38,750
Cost of referigerator = ₹23,890
Cost of jewellery = ₹35,560
Cost of all purchased items = ₹38,750 + ₹23,890 + ₹35,560 = ₹98,200
$\begin{array}{cccccc}& 3& 8& 7& 5& 0\\ & 2& 3& 8& 9& 0\\ +& 3& 5& 5& 6& 0\\ & 9& 8& 2& 0& 0\end{array}$

Money left with Shikha = ₹1,00,000 − ₹98,200 = ₹1,800
Hence, money left with Shikha is ₹1,800.

#### Question 18:

Find:

The product is shown beow:
$\begin{array}{cccccccc}& & & 1& 5& 9& 0& 8\\ & & & & ×& 5& 4& 2\\ & & & 3& 1& 8& 1& 6\\ & & 6& 3& 6& 3& 2& ×\\ +& 7& 9& 5& 4& ×& ×& ×\\ & 8& 6& 2& 2& 1& 3& 6\end{array}$

#### Question 19:

Divide 57086 by 247 and check the result by division algorithm.

Let us divide 57086 by 247 using long division method. Here, divisor = 247, dividend = 57086, quotient = 231 and remainder = 29.
Now

 dividend = divisor $\text{×}$ quotient + remainder                = 247 $\text{×}$ 231 + 29                = 57057 + 29                = 57086 #### Question 20:

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

Here, divisor = 299, dividend = 55390 and remainder = 75.

 dividend = divisor $\text{×}$ quotient + remainder $\text{⇒}$  55390 = 299 $\text{×}$ quotient + 75 $\text{⇒}$  299$\text{×}$ quotient = 55390 − 75 = 55315 $\text{⇒}$ quotient = 55315 $\text{÷}$ 299 =  185 Hence, the quotient is 185.

#### Question 21:

The sum of two odd numbers is an ............ number.

The numbers 3, 7 , 11, 35 etc. are odd numbers. Now observe the following:
3 + 7 = 10, 3 + 11 = 14, 3 + 35 = 38, 7 + 11 = 18, 7 + 35 = 42 etc.
Here, the last digits of 10, 14, 38, 18, 42 ... ends with 0, 4, 8 and 2.
We know that, the number whose last digits is 0, 2, 4, 6 or 8 is an even number.
Thus, 10, 14, 38, 18, 42 etc. are all even numbers. Hence,
The sum of two odd numbers is an even number.

#### Question 22:

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

The numbers 3, 7 , 11 etc. are odd numbers and the numbers 8, 10, 12 etc. are even numbers.
Now observe the following:
$\text{×}$ 8 = 24, 3 $\text{×}$ 12 = 36, 7 $\text{×}$ 2 = 14, 7 $\text{×}$ 8 = 56, 11 $\text{×}$ 12 = 132, 11 $\text{×}$ 10 = 110 etc.
Here, the last digits of 24, 36, 14, 56, 132, 110 ... ends with 0, 2, 4 or 6.
We know that, the number whose last digits is 0, 2, 4, 6 or 8 is an even number.
Thus, 24, 14, 36, 14, 56, 132, 110 etc. are all even numbers. Hence,
The product of an odd number and an even number is an even number.

#### Question 23:

When a whole number a is divided by a non-zero whole number b, then there exists whole numbers q and r
such that a = bq + r, where r = ......... or r < .............

Let a be dividend and b is the divisor. If q is the quotient and r is the remainder, then by division algorithm
a = bq + r, where r = 0 or r < b
Hence, r = 0 or r < b.

#### Question 24:

If a whole number a is divided by a non-zero whole number b such that the quotient is q, then
the remainder is .............

Let a be dividend and b is the divisor. If q is the quotient and r is the remainder, then by division algorithm
a = bq + r, where r = 0 or r < b.
Now
a = bq + r $\text{⇒}$ r = abq
Hence, r = abq.

#### Question 25:

A number when divided by 12 gives 7 as quotient and 9 as remainder. The number is ..............

By division algorithm, we have
dividend = divisor $\text{×}$ quotient + remainder
Here, divisor = 12, quotient = 7 and remainder = 9.
Therefore
dividend = divisor $\text{×}$ quotient + remainder
= 12 $\text{×}$ 7 + 9
= 84 + 9
= 93
Hence, the required number is 93.

#### Question 1:

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

(i) 359+476=476+....
(ii) ... + 1952=1952+2008
(iii) 90758 + 0=...
(iv) 54321 +(489+699)=489+(54321+...)

(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:

(i) 5628 + 39784
(ii) 923584 + 178
(iii) 15409 + 112
(iv) 2359 + 641

(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:

(i) 953+407+647
(ii) 15409+178+591+322
(iii) 2359+10001+2641+9999
(iv) 1+2+3+4+1996+1997+1998+1999
(v) 10+11+12+13+14+15+16+17+18+19+20

(i) 953 + 407 + 647
∵ 53 + 47 = 100
∴ (953 + 647) + 407 = 1600 + 407 = 2007

(ii) 15409 + 178 + 591 + 322
∵ 409 + 91 = 500
and
78+22 = 100
∴ (15409 + 591) + (178 + 322)
= (16000) + (500)
= 16500

(iii) 2359 + 10001 + 2641 + 9999
∵ 59 + 41 = 100
and
99 + 01 = 100
∴ (2359 + 2641) + (10001 + 9999)
= (5000) + (20000)
= 25000

(iv) 1 + 2 + 3 + 4 + 1996 + 1997 + 1998 + 1999
∵ 99 + 1 = 100
98 + 2 = 100
97 + 3 = 100
and
96 + 4 = 100
∴ (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
∴ (10 + 20) + (11 + 19) + (12 + 18) + (13 + 17) + (14 + 16) + 15
= 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 two even number 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)  Commutativity and associativity are properties of whole numbers.
(xii) Commutativity and associativity are properties of addition of whole numbers.

(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

#### Question 1:

A magic square is an array of numbers having the same number of rows and columns and the sum of numbers in each row, column or diagonal being the same. Fill in the blank cells of the following magic square:

(i)

 8 13 12 11

(ii)
 22 6 13 20 10 12 19 9 11 18 25 15 17 24 26 16 7 14

(i) It can be seen that diagonally, 13 + 12 + 11 = 36.
Thus,
Number in the first cell of the first row = 36 − (8 + 13) = 15
Number in the first cell of the second row =  36 − (15 + 11) = 10
Number in the third cell of the second row =  36 − (10 + 12) = 14
Number in the second cell of the third row = 36 − (8 + 12) = 16
Number in the third cell of the third row = 36 − (11 + 16) = 9

 15 8 13 10 12 14 11 16 9

(ii) It can be seen that diagonally, 20 + 19 + 18 + 17 + 16 = 90.
Thus,
Number in the second cell of the first row = 90 − (22 + 6 + 13 + 20) = 29
Number in the first cell of the second row = 90 − (22 + 9 + 15 + 16) = 28
Number in the fifth cell of the second row =  90 − (28 + 10 + 12 + 19) = 21
Number in the fifth cell of the third row =  90 − (9 + 11 + 18 + 25) = 27
Number in the fifth cell of the fourth row = 90 − (15 + 17 + 24 + 26) = 8
Number in the second cell of the fifth row = 90 − (29 + 10 + 11 + 17) = 23
Number in the third cell of the fifth row = 90 − (6 + 12 + 18 + 24) = 30

 22 29 6 13 20 28 10 12 19 21 9 11 18 25 27 15 17 24 26 8 16 23 30 7 14

#### Question 2:

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

(i) 57839 − 2983
(ii) 92507 − 10879
(iii) 400000 − 98798
(iv) 5050501 − 969696
(v) 200000 − 97531
(vi) 3030301 − 868686

(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 3:

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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(i) 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.

8 7 6
$-$   2 3 9
6 3 7

(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.

8 9 8 9
$-$   3 4 5 5
5 5 3 4

Thus, the correct answer is :
8 9 8 9
$-$  5 5 3 4
3 4 5 5

(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.
So, in order to get the addend, we will subtract 5061129 from 6000107.
6  0  0  0  1  0  7
$-$  5  0  6  1  1  2  9
0  9  3  8  9  7  8

6  0  0  0  1  0  7
$-$  0  9  3  8  9  7  8
5  0  6  1  1  2  9

(iv) In the units place, 10 $-$ 1 = 9. Also, in the lakhs place, 9 $-$ 0 = 9. So, clearly the addend (difference) is 970429.
To find out the other addend, we will subtract 970429 from 1000000.

1  0  0  0  0  0  0
$-$   0  9  7  0  4  2  9
0  0  2  9  5  7  1

1  0  0  0  0  0  0
$-$   0  0  2  9  5  7  1
0  9  7  0  4  2  9

(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.
So, to get the other addend, we will subtract 4844016 from 5001003.
5  0  0  1  0  0  3
$-$   4  8  4  4  0  1  6
0  1  5  6  9  8  7

Thus, the other addend is 156987.

5  0  0  1  0  0  3
$-$   0  1  5  6  9  8  7
4  8  4  4  0  1  6

(vi) It is clear from the units place that 11 $-$ 9 = 2. So, the addend (difference) is 54322.
So, to get the other addend, we will subtract 54322 from 111111.
1  1  1  1  1  1
$-$       5  4  3  2  2
5  6  7  8  9

Thus, the other addend is 56789.
1  1  1  1  1  1
$-$       5  6  7  8  9
5  4  3  2  2

#### Question 4:

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

The largest five-digit number is 99999.
The smallest six-digit number is 100000.
∴ Difference between them = 100000 − 99999 = 1

#### Question 5:

Find the difference between the largest number of 4 digits and the smallest number of 7 digits.

The largest four-digit number = 9999
The smallest seven-digit number = 1000000
∴ Difference between them = 1000000 − 9999 = 990001

#### Question 6:

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

Money deposited by Rohit  =  Rs 125000
Money withdrawn by Rohit =  Rs 35425
∴ Money left in the account = Rs (125000 − 35425) =  Rs 89575

#### Question 7:

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.

Total population of the town  = 96209
Number of men = 29642
Number of women  = 29167
Sum of men and women = (29642 + 29167 ) = 58809
∴ Number of children in the town = (Total population) − (Sum of men and women)
= 96209 − 58809 = 37400

#### Question 8:

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

Original number = 39460
New number =   − 36490
Difference =            2970

#### Question 9:

The population of a town was 59000. In one year it was increased by 4536 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?