Natural Numbers & Whole Numbers

ICSE Class 6 Mathematics · Selina Solutions Concise Mathematics Class 6 ICSE · 39 solved questions

Exercise
Marks
1Exercise 3(A)

Fill in the blanks:

(i) Smallest natural number is ...............

(ii) Smallest whole number is ...............

(iii) Largest natural number is ...............

(iv) Largest whole number is ...............

(v) All natural numbers are ...............

(vi) All whole numbers are not ...............

(vii) Successor of 4099 is ...............

(viii) Predecessor of 4330 is ...............

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(i) The smallest natural number is 1.

(ii) The smallest whole number is 0.

(iii) The largest natural number is not possible (natural numbers are infinite).

(iv) The largest whole number is not possible (whole numbers are infinite).

(v) All natural numbers are whole numbers.

(vi) All whole numbers are not natural numbers.

(vii) Successor of 4099 = 4099 + 1 = 4100.

(viii) Predecessor of 4330 = 4330 - 1 = 4329.

2Exercise 3(A)

State true or false:

(i) Whole numbers are closed for addition.

(ii) If a and b are any two whole numbers, then a + b is not a whole number.

(iii) If a and b are any two whole numbers, then a + b = b + a.

(iv) 0 + 18 = 18 + 0.

(v) Addition of whole numbers is associative.

(vi) 10 + 12 + 16 = (10 + 12) + 16 = 10 + (12 + 16).

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(i) True. The sum of two whole numbers is always a whole number, so whole numbers are closed for addition.

(ii) False. By the closure property of addition, a + b is always a whole number.

(iii) True. Addition of whole numbers is commutative, so a + b = b + a.

(iv) True. By the commutative property, 0 + 18 = 18 + 0 = 18.

(v) True. Addition of whole numbers is associative.

(vi) True. By the associative property, (10 + 12) + 16 = 10 + (12 + 16) = 38.

3Exercise 3(A)

Fill in the blanks:

(i) 54 + 234 = 234 + ...............

(ii) 332 + 497 = ............... + 332

(iii) 286 + 0 = ...............

(iv) 286 × 1 = ...............

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

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(i) By the Commutative Property of Addition: a + b = b + a.

54 + 234 = 234 + 54.

(ii) By the Commutative Property of Addition: a + b = b + a.

332 + 497 = 497 + 332.

(iii) By the Additive Identity property, adding 0 to a whole number gives the number itself.

286 + 0 = 286.

(iv) By the Multiplicative Identity property, multiplying a whole number by 1 gives the number itself.

286 × 1 = 286.

(v) By the Associative Property of Addition: a + (b + c) = (a + b) + c.

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

4Exercise 3(A)

Verify that:

(i) 3 + (5 + 4) = (3 + 5) + 4

(ii) 8 × (8 + 0) = 8 × 8 + 8 × 0

(iii) (7 + 6) × 10 = 7 × 10 + 6 × 10

(iv) (15 - 12) × 18 = 15 × 18 - 12 × 18

(v) 16 + 0 = 16

(vi) 23 + (-23) = 0

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(i) 3 + (5 + 4) = (3 + 5) + 4

L.H.S. = 3 + (5 + 4) = 3 + 9 = 12.

R.H.S. = (3 + 5) + 4 = 8 + 4 = 12.

Since L.H.S. = R.H.S., the result is verified.

(ii) 8 × (8 + 0) = 8 × 8 + 8 × 0

L.H.S. = 8 × (8 + 0) = 8 × 8 = 64.

R.H.S. = 8 × 8 + 8 × 0 = 64 + 0 = 64.

Since L.H.S. = R.H.S., the result is verified.

(iii) (7 + 6) × 10 = 7 × 10 + 6 × 10

L.H.S. = (7 + 6) × 10 = 13 × 10 = 130.

R.H.S. = 7 × 10 + 6 × 10 = 70 + 60 = 130.

Since L.H.S. = R.H.S., the result is verified.

(iv) (15 - 12) × 18 = 15 × 18 - 12 × 18

L.H.S. = (15 - 12) × 18 = 3 × 18 = 54.

R.H.S. = 15 × 18 - 12 × 18 = 270 - 216 = 54.

Since L.H.S. = R.H.S., the result is verified.

(v) 16 + 0 = 16

L.H.S. = 16 + 0 = 16 = R.H.S.

By the Additive Identity property, the result is verified.

(vi) 23 + (-23) = 0

L.H.S. = 23 + (-23) = 23 - 23 = 0 = R.H.S.

The result is verified.

5Exercise 3(A)

State true or 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 odd number.

(vi) Every whole number is a natural number.

(vii) Every natural number is a whole number.

(viii) Every whole number + 0 = The whole number itself.

(ix) Every whole number × 1 = The whole number itself.

(x) Commutativity and associativity are properties of natural numbers and whole numbers both.

(xi) Commutativity and associativity are properties of addition for natural numbers and whole numbers both.

(xii) If x is a whole number then -x is also a whole number.

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(i) False. The sum of two odd numbers is an even number. For example, 3 + 5 = 8.

(ii) True. The sum of two odd numbers is an even number. For example, 7 + 9 = 16.

(iii) True. The sum of two even numbers is an even number. For example, 4 + 6 = 10.

(iv) False. The sum of two even numbers is an even number, not an odd number.

(v) True. The sum of an even number and an odd number is an odd number. For example, 4 + 3 = 7.

(vi) False. 0 is a whole number but it is not a natural number, so every whole number is not a natural number.

(vii) True. Every natural number is a whole number.

(viii) True. Adding 0 to any whole number gives the number itself (additive identity).

(ix) True. Multiplying any whole number by 1 gives the number itself (multiplicative identity).

(x) False. Commutativity and associativity are not properties of every operation (for example, they do not hold for subtraction and division). The statement is incomplete as it does not mention the operation.

(xi) True. For the operation of addition, both commutativity and associativity hold for natural numbers as well as whole numbers.

(xii) False. If x is a whole number, then -x is a negative number, and negative numbers are not whole numbers.

1Exercise 3(B)

Consider two whole numbers a and b such that a is greater than b.

(i) Is a - b a whole number? Is this result always true?

(ii) Is b - a a whole number? Is this result always true?

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(i) Since a is greater than b, the difference a - b is a positive number.

So, a - b is a whole number, and yes, this result is always true (whenever a is greater than b).

(ii) Since a is greater than b, the difference b - a is a negative number, and negative numbers are not whole numbers.

So, b - a is not a whole number, and yes, this result is always true (whenever a is greater than b, b - a is always negative).

2Exercise 3(B)

Write the identity number, if possible for subtraction of whole numbers.

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For any whole number x, we have x - 0 = x, but 0 - x ≠ x.

An identity element must give the same number when applied from both sides, but for subtraction 0 works only from the right side and not from the left.

Hence, the identity number for subtraction of whole numbers is not possible (it does not exist).

3Exercise 3(B)

Fill in the blanks:

(i) 12 × (9 - 6) = ............... = ...............

(ii) 12 × 9 - 12 × 6 = ............... = ...............

(iii) Is 12 × (9 - 6) = 12 × 9 - 12 × 6? ...............

(iv) Is this type of result always true? ...............

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(i) 12 × (9 - 6) = 12 × 3 = 36.

(ii) 12 × 9 - 12 × 6 = 108 - 72 = 36.

(iii) Both results are equal to 36, so yes, 12 × (9 - 6) = 12 × 9 - 12 × 6.

(iv) Yes. By the Distributive Law of Multiplication over Subtraction, this type of result is always true.

4Exercise 3(B)

Fill in the blanks:

(i) (16 - 8) × 24 = ............... = ...............

(ii) 16 × 24 - 8 × 24 = ............... - ............... = ...............

(iii) Is (16 - 8) × 24 = 16 × 24 - 8 × 24? ...............

(iv) Is this type of result always true? ...............

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(i) (16 - 8) × 24 = 8 × 24 = 192.

(ii) 16 × 24 - 8 × 24 = 384 - 192 = 192.

(iii) Both results are equal to 192, so yes, (16 - 8) × 24 = 16 × 24 - 8 × 24.

(iv) Yes. By the Distributive Law of Multiplication over Subtraction, this type of result is always true.

5Exercise 3(B)

Find the difference between the largest number of four digits and the smallest number of six digits.

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The largest number of four digits = 9999.

The smallest number of six digits = 100000.

Difference = 100000 - 9999 = 90001.

Hence, the required difference is 90001.

6Exercise 3(B)

Find the difference between the smallest number of eight digits and the largest number of five digits.

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The smallest number of eight digits = 10000000.

The largest number of five digits = 99999.

Difference = 10000000 - 99999 = 9900001.

Hence, the required difference is 9900001.

1Exercise 3(C)

Fill in the blanks:

(i) 42 × 0 = ...............

(ii) 592 × 1 = ...............

(iii) 328 × 573 = ............... × 328

(iv) 229 × ............... = 578 × 229

(v) 32 × 15 = 32 × 6 + 32 × 7 + 32 × ...............

(vi) 23 × 56 = 20 × 56 + ............... × 56

(vii) 83 × 54 + 83 × 16 = 83 × (...............) = 83 × ............... = ...............

(viii) 98 × 273 - 75 × 273 = (...............) × 273 = ............... × 273

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(i) Any number multiplied by 0 gives 0.

42 × 0 = 0.

(ii) Any number multiplied by 1 gives the number itself.

592 × 1 = 592.

(iii) By the Commutative Property of Multiplication: a × b = b × a.

328 × 573 = 573 × 328.

(iv) By the Commutative Property of Multiplication: a × b = b × a.

229 × 578 = 578 × 229.

(v) Since 6 + 7 + 2 = 15, we have:

32 × 15 = 32 × 6 + 32 × 7 + 32 × 2.

(vi) Since 23 = 20 + 3, we have:

23 × 56 = 20 × 56 + 3 × 56.

(vii) Using the Distributive Law of Multiplication over Addition:

83 × 54 + 83 × 16 = 83 × (54 + 16) = 83 × 70 = 5810.

(viii) Using the Distributive Law of Multiplication over Subtraction:

98 × 273 - 75 × 273 = (98 - 75) × 273 = 23 × 273.

2Exercise 3(C)

Evaluate (using distributive property):

(i) 984 × 102

(ii) 385 × 1004

(iii) 446 × 10002

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(i) 984 × 102

⇒ 984 × (100 + 2)

Using the Distributive Law of Multiplication over Addition:

⇒ 984 × 100 + 984 × 2

⇒ 98400 + 1968

⇒ 100368.

Hence, 984 × 102 = 100368.

(ii) 385 × 1004

⇒ 385 × (1000 + 4)

Using the Distributive Law of Multiplication over Addition:

⇒ 385 × 1000 + 385 × 4

⇒ 385000 + 1540

⇒ 386540.

Hence, 385 × 1004 = 386540.

(iii) 446 × 10002

⇒ 446 × (10000 + 2)

Using the Distributive Law of Multiplication over Addition:

⇒ 446 × 10000 + 446 × 2

⇒ 4460000 + 892

⇒ 4460892.

Hence, 446 × 10002 = 4460892.

3Exercise 3(C)

Evaluate using properties:

(i) 548 × 98

(ii) 924 × 988

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(i) 548 × 98

⇒ 548 × (100 - 2)

Using the Distributive Law of Multiplication over Subtraction:

⇒ 548 × 100 - 548 × 2

⇒ 54800 - 1096

⇒ 53704.

Hence, 548 × 98 = 53704.

(ii) 924 × 988

⇒ 924 × (1000 - 12)

Using the Distributive Law of Multiplication over Subtraction:

⇒ 924 × 1000 - 924 × 12

⇒ 924000 - 11088

⇒ 912912.

Hence, 924 × 988 = 912912.

4Exercise 3(C)

Evaluate using properties:

(i) 679 × 8 + 679 × 2

(ii) 284 × 12 - 284 × 2

(iii) 55873 × 94 + 55873 × 6

(iv) 7984 × 15 - 7984 × 5

(v) 8324 × 1945 - 8324 × 945

(vi) 3333 × 987 + 13 × 3333

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(i) 679 × 8 + 679 × 2

Using the Distributive Law of Multiplication over Addition: a × b + a × c = a × (b + c).

⇒ 679 × (8 + 2)

⇒ 679 × 10

⇒ 6790.

Hence, 679 × 8 + 679 × 2 = 6790.

(ii) 284 × 12 - 284 × 2

Using the Distributive Law of Multiplication over Subtraction: a × b - a × c = a × (b - c).

⇒ 284 × (12 - 2)

⇒ 284 × 10

⇒ 2840.

Hence, 284 × 12 - 284 × 2 = 2840.

(iii) 55873 × 94 + 55873 × 6

Using the Distributive Law of Multiplication over Addition:

⇒ 55873 × (94 + 6)

⇒ 55873 × 100

⇒ 5587300.

Hence, 55873 × 94 + 55873 × 6 = 5587300.

(iv) 7984 × 15 - 7984 × 5

Using the Distributive Law of Multiplication over Subtraction:

⇒ 7984 × (15 - 5)

⇒ 7984 × 10

⇒ 79840.

Hence, 7984 × 15 - 7984 × 5 = 79840.

(v) 8324 × 1945 - 8324 × 945

Using the Distributive Law of Multiplication over Subtraction:

⇒ 8324 × (1945 - 945)

⇒ 8324 × 1000

⇒ 8324000.

Hence, 8324 × 1945 - 8324 × 945 = 8324000.

(vi) 3333 × 987 + 13 × 3333

⇒ 3333 × 987 + 3333 × 13

Using the Distributive Law of Multiplication over Addition:

⇒ 3333 × (987 + 13)

⇒ 3333 × 1000

⇒ 3333000.

Hence, 3333 × 987 + 13 × 3333 = 3333000.

5Exercise 3(C)

Find the product of the:

(i) greatest number of three digits and smallest number of five digits.

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

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(i) The greatest number of three digits = 999.

The smallest number of five digits = 10000.

Product = 999 × 10000 = 9990000.

Hence, the required product is 9990000.

(ii) The greatest number of four digits = 9999.

The greatest number of three digits = 999.

Product = 9999 × 999

⇒ 9999 × (1000 - 1)

⇒ 9999 × 1000 - 9999 × 1

⇒ 9999000 - 9999

⇒ 9989001.

Hence, the required product is 9989001.

6Exercise 3(C)

Fill in the blanks:

(i) (437 + 3) × (400 - 3) = 397 × ............... = ...............

(ii) 66 + 44 + 22 = 11 × (...............) = 11 × ............... = ...............

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(i) Here, 437 + 3 = 440 and 400 - 3 = 397.

So, (437 + 3) × (400 - 3) = 440 × 397 = 397 × 440 [by commutative property]

⇒ 397 × 440 = 174680.

(ii) Since 66 = 11 × 6, 44 = 11 × 4 and 22 = 11 × 2, using the distributive property:

⇒ 66 + 44 + 22 = 11 × 6 + 11 × 4 + 11 × 2

⇒ 11 × (6 + 4 + 2) = 11 × 12 = 132.

1Exercise 3(D)

Show that:

(i) division of whole numbers is not closed.

(ii) any whole number divided by 1, always gives the number itself.

(iii) every non-zero whole number divided by itself gives 1 (one).

(iv) zero divided by any non-zero number is zero only.

(v) a whole number divided by 0 is not defined.

For each part, given above, give two suitable examples.

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(i) Division of whole numbers is not closed because the result of dividing one whole number by another is not always a whole number.

Examples: 5 ÷ 8 and 12 ÷ 24 are not whole numbers, etc.

(ii) Any whole number divided by 1, always gives the number itself.

Examples: 5 ÷ 1 = 5, and 16 ÷ 1 = 16, etc.

(iii) Every non-zero whole number divided by itself gives 1.

Examples: 8 ÷ 8 = 1, and 12 ÷ 12 = 1, etc.

(iv) Zero divided by any non-zero whole number gives zero only.

Examples: 0 ÷ 6 = 0, and 0 ÷ a = 0, if a ≠ 0, etc.

(v) A whole number divided by 0 is not defined.

Examples: 7 ÷ 0 is not defined, and 16 ÷ 0 is not defined, etc.

2Exercise 3(D)

If x is a whole number such that x ÷ x = x; state the value of x.

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For every non-zero whole number, dividing it by itself gives 1, i.e. x ÷ x = 1.

So, x ÷ x = x means x = 1.

Hence, the value of x is 1.

3Exercise 3(D)

Fill in the blanks:

(i) 987 ÷ 1 = ...............

(ii) 0 ÷ 987 = ...............

(iii) 336 - (888 ÷ 888) = ...............

(iv) (23 ÷ 23) - (437 ÷ 437) = ...............

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(i) Any whole number divided by 1 gives the number itself.

987 ÷ 1 = 987.

(ii) Zero divided by any non-zero whole number gives 0.

0 ÷ 987 = 0.

(iii) Any non-zero whole number divided by itself gives 1.

⇒ 888 ÷ 888 = 1

⇒ 336 - (888 ÷ 888) = 336 - 1 = 335.

(iv) Any non-zero whole number divided by itself gives 1.

⇒ 23 ÷ 23 = 1 and 437 ÷ 437 = 1

⇒ (23 ÷ 23) - (437 ÷ 437) = 1 - 1 = 0.

4Exercise 3(D)

Which of the following statements are true?

(i) 12 ÷ (6 × 2) = (12 ÷ 6) × (12 ÷ 2)

(ii) a ÷ (b - c) = $\frac{a}{b} - \frac{a}{c}$

(iii) (a - b) ÷ c = $\frac{a}{c} - \frac{b}{c}$

(iv) (15 - 13) ÷ 8 = (15 ÷ 8) - (13 ÷ 8)

(v) 8 ÷ (15 - 13) = $\frac{8}{15} - \frac{8}{13}$

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(i) L.H.S. = 12 ÷ (6 × 2) = 12 ÷ 12 = 1.

R.H.S. = (12 ÷ 6) × (12 ÷ 2) = 2 × 6 = 12.

Since L.H.S. ≠ R.H.S., this statement is false.

(ii) Division is not distributive over subtraction in this form, so a ÷ (b - c) ≠ $\frac{a}{b} - \frac{a}{c}$.

This statement is false.

(iii) (a - b) ÷ c = $\frac{a - b}{c} = \frac{a}{c} - \frac{b}{c}$.

This statement is true.

(iv) L.H.S. = (15 - 13) ÷ 8 = 2 ÷ 8 = $\frac{2}{8} = \frac{1}{4}$.

R.H.S. = (15 ÷ 8) - (13 ÷ 8) = $\frac{15}{8} - \frac{13}{8} = \frac{2}{8} = \frac{1}{4}$.

Since L.H.S. = R.H.S., this statement is true.

(v) L.H.S. = 8 ÷ (15 - 13) = 8 ÷ 2 = 4.

R.H.S. = $\frac{8}{15} - \frac{8}{13}$, which is a negative number.

Since L.H.S. ≠ R.H.S., this statement is false.

Hence, statements (iii) and (iv) are true.

1Exercise 3(E)

For each pattern, given below, write the next three steps:

(i) 1 × 9 + 1 = 10

12 × 9 + 2 = 110

123 × 9 + 3 = 1110

(ii) 9 × 9 + 7 = 88

98 × 9 + 6 = 888

987 × 9 + 5 = 8888

(iii) 1 × 8 + 1 = 9

12 × 8 + 2 = 98

123 × 8 + 3 = 987

(iv) 111 ÷ 3 = 37

222 ÷ 6 = 37

333 ÷ 9 = 37

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(i) The next three steps are:

1234 × 9 + 4 = 11110

12345 × 9 + 5 = 111110

123456 × 9 + 6 = 1111110

(ii) The next three steps are:

9876 × 9 + 4 = 88888

98765 × 9 + 3 = 888888

987654 × 9 + 2 = 8888888

(iii) The next three steps are:

1234 × 8 + 4 = 9876

12345 × 8 + 5 = 98765

123456 × 8 + 6 = 987654

(iv) The next three steps are:

444 ÷ 12 = 37

555 ÷ 15 = 37

666 ÷ 18 = 37

2Exercise 3(E)

Complete each of the following magic squares:

(i)

67
59
84

(ii)

48
7
10

(iii)

162
10
4
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In a magic square, the sum of the numbers in each row, each column and each diagonal is the same. This common sum is called the magic sum, and for a 3 × 3 magic square it equals 3 times the centre number.

(i) The centre number is 5, so the magic sum = 3 × 5 = 15.

Row 1: 6 + 7 + ............... = 15 ⇒ missing number = 2.

Row 2: ............... + 5 + 9 = 15 ⇒ missing number = 1.

Row 3: 8 + ............... + 4 = 15 ⇒ missing number = 3.

The missing numbers (row-wise) are 2, 1 and 3. The completed magic square is:

672
159
834

(ii) The centre number is 7, so the magic sum = 3 × 7 = 21.

Diagonal: 8 + 7 + ............... = 21

⇒ Bottom-left number = 6.

Row 1: 4 + ............... + 8 = 21

⇒ Missing number = 9.

Diagonal 2: 8 + 7 + ............... = 21

⇒ Missing number = 6.

Column 1: 4 + ............... + 6 = 21

⇒ Middle-left number = 11.

Row 2: 11 + 7 + ............... = 21

⇒ Middle-right number = 3.

Row 3: 6 + ............... + 10 = 21

⇒ Missing number = 5.

The missing numbers (row-wise) are 9, 11, 3, 6 and 5. The completed magic square is:

498
1173
6510

(iii) The centre number is 10, so the magic sum = 3 × 10 = 30.

Row 1: 16 + 2 + ............... = 30

⇒ Missing number = 12.

Diagonal: 12 + 10 + ............... = 30

⇒ Bottom-left number = 8.

Column 1: 16 + ............... + 8 = 30

⇒ Middle-left number = 6.

Row 2: 6 + 10 + ............... = 30

⇒ Middle-right number = 14.

Row 3: 8 + ............... + 4 = 30

⇒ Missing number = 18.

The missing numbers (row-wise) are 12, 6, 14, 8 and 18. The completed magic square is:

16212
61014
8184
3Exercise 3(E)

See the following pattern carefully:

See the following pattern carefully: Mathematics Solutions ICSE Class 7.

(i) If n denotes the figure number and S denotes the number of matchsticks, find S in terms of n.

(ii) Find how many matchsticks are required to make the:

(1) 15th figure

(2) 40th figure

(iii) Write a description of the pattern in words.

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From the pattern, the number of matchsticks for the first few figures is:

Figure number (n)Number of matchsticks (S)
17
210
313
416

(i) Each new figure is formed by adding one square, and each added square shares one edge with the previous square.

So, each new figure requires 3 more matchsticks.

Therefore, for the nth figure,

S = 7 + 3(n - 1)

S = 7 + 3n - 3

S = 3n + 4

Hence, S = 3n + 4.

(ii) Calculating,

(1) For the 15th figure, n = 15:

S = 3 × 15 + 4 = 45 + 4 = 49 matchsticks.

(2) For the 40th figure, n = 40:

S = 3 × 40 + 4 = 120 + 4 = 124 matchsticks.

(iii) The number of matchsticks (S) is equal to 4 more than three times the figure number (n).

4Exercise 3(E)

(i) In the following pattern, draw the next two figures.

In the following pattern, draw the next two figures. Mathematics Solutions ICSE Class 7.

(ii) Construct a table to describe the figures in the above pattern.

(iii) If n denotes the number of figures and L denotes the number of matchsticks, find L in terms of n.

(iv) Find how many matchsticks are required to make the:

(1) 12th figure (2) 20th figure

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(i) The next two figures in the pattern are:

In the following pattern, draw the next two figures. Mathematics Solutions ICSE Class 7.

(ii) The table describing the figures is:

Number of figures (n)12345
Number of matchsticks (L)246810

(iii) Each time the number of figures (n) increases by 1, the number of matchsticks (L) increases by 2.

For n = 1, 2n = 2 and L = 2.

Hence, L = 2n.

(iv) (1) For the 12th figure, n = 12:

L = 2 × 12 = 24 matchsticks.

(2) For the 20th figure, n = 20:

L = 2 × 20 = 40 matchsticks.

11 Mark · Multiple Choice Question

The whole number that does not have its predecessor is:

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The predecessor of a whole number is obtained by subtracting 1 from it.

The predecessor of 0 would be 0 - 1 = -1, which is not a whole number.

So, 0 is the only whole number that does not have a predecessor.

Hence, option 2 is the correct option.

21 Mark · Multiple Choice Question

The predecessor of ten thousand is:

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Ten thousand = 10000.

Predecessor of 10000 = 10000 - 1 = 9999.

Hence, option 2 is the correct option.

31 Mark · Multiple Choice Question

The value of 43 × 27 + 43 × 73 is equal to:

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Using the Distributive Law of Multiplication over Addition:

43 × 27 + 43 × 73 = 43 × (27 + 73)

= 43 × 100 = 4300.

Hence, option 1 is the correct option.

41 Mark · Multiple Choice Question

484 × 101 is equal to:

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484 × 101 = 484 × (100 + 1)

Using the Distributive Law of Multiplication over Addition:

= 484 × 100 + 484 × 1

= 48400 + 484 = 48884.

Hence, option 3 is the correct option.

51 Mark · Multiple Choice Question

57 × 13 - 57 × 3 is equal to:

Show SolutionHide Solution↓

Using the Distributive Law of Multiplication over Subtraction:

57 × 13 - 57 × 3 = 57 × (13 - 3) = 57 × 10.

Hence, option 2 is the correct option.

61 Mark · Multiple Choice Question

(17 ÷ 17) - (97 ÷ 97) is equal to:

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Every non-zero whole number divided by itself gives 1.

(17 ÷ 17) - (97 ÷ 97) = 1 - 1 = 0.

Hence, option 3 is the correct option.

71 Mark · Multiple Choice Question

The number line for the whole numbers between 3 and 7 is:

The number line for the whole numbers between 3 and 7 is: Mathematics Solutions ICSE Class 7.
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The whole numbers between 3 and 7 are 4, 5 and 6.

The correct number line is the one in which only the points 4, 5 and 6 are marked between 3 and 7.

Hence, option 2 is the correct option.

81 Mark · Multiple Choice Question

The product of the greatest number of two digits and the smallest two digit number is:

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The greatest number of two digits = 99.

The smallest number of two digits = 10.

Product = 99 × 10 = 990.

Hence, option 2 is the correct option.

91 Mark · Multiple Choice Question

If a ÷ a = a, then a is:

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For every non-zero whole number, a ÷ a = 1.

So, a ÷ a = a means a = 1.

Hence, option 3 is the correct option.

101 Mark · Multiple Choice Question

If 8 - (a - 7) = 8 + (4 - b), then:

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L.H.S. = 8 - (a - 7) = 8 - a + 7 = 15 - a.

R.H.S. = 8 + (4 - b) = 8 + 4 - b = 12 - b.

So, 15 - a = 12 - b, which gives b - a = -3.

Substituting values from option 2:

⇒ a = -4 and b = -7 gives b - a = -7 - (-4) = -3.

Hence, option 2 is the correct option.

11Statement I-II Type Questions

Statement 1: For a whole number x, x - 0 = x implies 0 is identity element for subtraction.

Statement 2: For a whole number x, 0 - x = x implies 0 is identity element for subtraction.

Which of the following options is correct?

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Statement 1: x - 0 = x shows that 0 is a right identity for subtraction. But 0 is an identity element only if it works from both the right and the left, i.e. 0 - x = x must also hold. Since 0 - x ≠ x, the number 0 is not the identity element for subtraction.

∴ Statement 1 is false.

Statement 2: 0 - x = x is not true for whole numbers (for example, 0 - 5 = -5 ≠ 5), so 0 is not the identity element for subtraction.

∴ Statement 2 is false.

Hence, option 2 is the correct option.

12Statement I-II Type Questions

Statement 1: On subtracting a whole number from another whole number, the result is a natural number.

Statement 2: 15 and 25 are whole numbers, but 15 - 25 is not a natural number.

Which of the following options is correct?

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Statement 1: Subtracting one whole number from another does not always give a natural number. For example, 5 - 5 = 0 (which is not a natural number) and 5 - 8 = -3 (which is not even a whole number).

∴ Statement 1 is false.

Statement 2: 15 - 25 = -10, which is a negative number and hence not a natural number. So the statement is correct.

∴ Statement 2 is true.

Hence, option 4 is the correct option.

13Assertion Reason Type Questions

Assertion (A): Associativity properties of addition of natural numbers holds good.

Reason (R): For natural numbers a, b and c, a + (b + c) = (a + b) + c shows addition of natural numbers is associative.

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Assertion (A): Addition of natural numbers is associative, i.e. for natural numbers a, b and c, a + (b + c) = (a + b) + c. So, the assertion is true.

Reason (R): The relation a + (b + c) = (a + b) + c is exactly the statement of the associative property of addition of natural numbers. So, the reason is true and it correctly explains the assertion.

Both A and R are true.

Hence, option 3 is the correct option.

14Assertion Reason Type Questions

Assertion (A): 656 × 42 - 656 × 17 = 656 × (42 - 17) = 9850

Reason (R): For any three numbers x, y and z : x × (y - z) = x × y - x × z the multiplication is distributive over subtraction.

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Assertion (A): Using the distributive law, 656 × 42 - 656 × 17 = 656 × (42 - 17) = 656 × 25 = 16400, not 9850. So, the assertion is false.

Reason (R): For any three numbers x, y and z, x × (y - z) = x × y - x × z, which is the distributive property of multiplication over subtraction. So, the reason is true.

A is false and R is true.

Hence, option 2 is the correct option.

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