Operations on Whole Numbers

ICSE Class 6 Mathematics · Solutions for Class 6 RS Aggarwal ICSE Foundation Mathematics · 43 solved questions

Exercise
Marks
1Exercise 2(A)

Fill in the blanks :

(i) 168 + 259 = ............... + 168

(ii) ....... + 317 = 317

(iii) (37 + 68) + ............... = 37 + (............... + 56)

(iv) 8 + 3 x 4 = ...............

(v) 18 x (............... + 23) = (18 x 17) + (18 x ...............)

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(i) According to the Commutative Property of Addition: (a + b = b + a)

168 + 259 = 259 + 168.

(ii) If 0 is added to any number, the number remains unchanged.

0 + 317 = 317.

(iii) According to the Associative Property of Addition: (a + b) + c = a + (b + c).

(37 + 68) + 56 = 37 + (68 + 56).

(iv) According to the DMAS rule, multiplication is performed before addition.

⇒ 8 + (3 x 4)

⇒ 8 + 12

⇒ 20

8 + 3 x 4 = 20.

(v) According to the Distributive property of multiplication over addition:

[a x (b + c) = (a x b) + (a x c)].

18 x (17 + 23) = (18 x 17) + (18 x 23).

2Exercise 2(A)

Fill in the blanks :

(i) 237 x 1 = ...............

(ii) 56 x ............... = 0

(iii) 0 ÷ 53 = ...............

(iv) 37 x 59 = 59 x ...............

(v) 0 x 138 = ...............

(vi) 73 ÷ 73 = ...............

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(i) According to the Identity Property of Multiplication, any number multiplied by 1 equals itself.

237 x 1 = 237.

(ii) According to the Zero Property of Multiplication, any number multiplied by 0 equals 0.

56 x 0 = 0.

(iii) Zero divided by any non-zero number is 0.

0 ÷ 53 = 0.

(iv) According to the Commutative Property of Multiplication, the order in which you multiply two numbers does not change the result.

37 x 59 = 59 x 37.

(v) According to the Zero Property of Multiplication, any number multiplied by 0 equals 0.

0 x 138 = 0.

(vi) Any non-zero number divided by itself is 1.

73 ÷ 73 = 1.

3Exercise 2(A)

Divide 3605 by 29 and verify the division algorithm.

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$$\begin{array}{l} \phantom{x^2 }{\quad 124} \\ 29\overline{\smash{\big)}3605} \\ \phantom{x}\phantom{}\underline{-29} \\ \phantom{{x^2 }+} 70 \\ \phantom{{x}+}\underline{-58} \\ \phantom{{x^2 } +2} 125 \\ \phantom{{x} +}\underline{-116} \\ \phantom{{x^2 + 3x}} 9 \\ \end{array}$$

Dividend = 3605

Divisor = 29

Quotient = 124

Remainder = 9

Verification: Dividend = (Divisor × Quotient) + Remainder

Substituting values we get :

(Divisor × Quotient) + Remainder

= (29 x 124) + 9

= 3596 + 9

= 3605.

Since L.H.S. = R.H.S.

Hence, the result is verified by the division algorithm.

4Exercise 2(A)

Find the number which when divided by 45 gives 16 as quotient and 9 as remainder.

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Given, Divisor = 45

Quotient = 16

Remainder = 9

Using formula, Dividend = (Divisor × Quotient) + Remainder

= (45 × 16) + 9

= 720 + 9

= 729

Hence, the number = 729.

5Exercise 2(A)

Find the largest number of 5-digits which is exactly divisible by 57.

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The largest 5-digit number is 99,999.

To find the largest 5-digit number exactly divisible by 57, divide 99,999 by 57 and subtract the remainder from 99,999.

$$\begin{array}{l} \phantom{x^2 }{\quad1754} \\ 57\overline{\smash{\big)}99999} \\ \phantom{x)}\phantom{}\underline{-57} \\ \phantom{{x^2 }+} 429 \\ \phantom{{x} )))}\underline{-399} \\ \phantom{{x^2 } + 2x } 309 \\ \phantom{{x} +1)}\underline{-285} \\ \phantom{{x^2 + 3x +)}} 249 \\ \phantom{{x} + 1x +}\underline{-228} \\ \phantom{{x^2 + 3x + 1)}} 21\ \end{array}$$

The remainder when 99,999 is divided by 57 is 21.

Therefore, 99,999 − 21 = 99,978.

Hence, the largest 5-digit number which is exactly divisible by 57 = 99,978.

6Exercise 2(A)

Find the smallest 6-digit number which is exactly divisible by 63.

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The smallest 6-digit number = 1,00,000.

To find the smallest 6-digit number exactly divisible by 63, we divide 1,00,000 by 63 and add the difference between the divisor and remainder to 1,00,000.

$$\begin{array}{l} \phantom{x^2 }{\quad 1587} \\ 63\overline{\smash{\big)}100000} \\ \phantom{x}\phantom{)}\underline{-63} \\ \phantom{x^2+} 370 \\ \phantom{{x}+}\underline{-315} \\ \phantom{{x^2 }2x + } 550 \\ \phantom{{x}+ 11}\underline{-504} \\ \phantom{{x^2 + 3x))}} 460 \\ \phantom{{x} + 1x)}\underline{-441} \\ \phantom{{x^2 + 3x)44}} 19\ \end{array}$$

Required number to be added = 63 − 19 = 44

Number = 1,00,000 + 44 = 1,00,044.

Hence, the smallest 6-digit number exactly divisible by 63 = 1,00,044.

7Exercise 2(A)

On dividing 1653 by a certain number, we get 45 as quotient and 33 as remainder. Find the divisor.

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Given:

Dividend = 1653

Quotient = 45

Remainder = 33

Using the formula,

Dividend = (Divisor × Quotient) + Remainder

Substituting the values, we get :

⇒ 1653 = (Divisor × 45) + 33

⇒ Divisor × 45 = 1653 - 33

⇒ Divisor × 45 = 1620

⇒ Divisor = $\frac{1620}{45}$

$$\begin{array}{l} \phantom{x^2 }{\quad 36} \\ 45\overline{\smash{\big)}1620} \\ \phantom{x}\phantom{)}\underline{-135} \\ \phantom{{x^2 }+} 270 \\ \phantom{{x} +}\underline{-270} \\ \phantom{{x^2 a}+} 0\ \end{array}$$

⇒ Divisor = 36

Hence, the divisor = 36.

8Exercise 2(A)

Use distributive law and evaluate :

(i) 576 x 285 + 576 x 115

(ii) 385 x 178 - 385 x 78

(iii) 365 x 645 + 135 x 645

(iv) 407 x 168 - 307 x 168

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(i) 576 x 285 + 576 x 115

Using distributive law: a x b + a x c = a x (b + c)

⇒ 576 x (285 + 115)

⇒ 576 x 400

⇒ 2,30,400

Hence, 576 x 285 + 576 x 115 = 2,30,400.

(ii) 385 x 178 - 385 x 78

Using distributive law: a x b - a x c = a x (b - c)

⇒ 385 x (178 - 78)

⇒ 385 x 100

⇒ 38,500

Hence, 385 x 178 - 385 x 78 = 38,500.

(iii) 365 x 645 + 135 x 645

⇒ 645 x 365 + 645 x 135

Using distributive law: a x b + a x c = a x (b + c)

⇒ 645 x (365 + 135)

⇒ 645 x 500

⇒ 3,22,500

Hence, 365 x 645 + 135 x 645 = 3,22,500.

(iv) 407 x 168 - 307 x 168

⇒ 168 x 407 - 168 x 307

Using distributive law: a x b - a x c = a x (b - c)

⇒ 168 x (407 - 307)

⇒ 168 x 100

⇒ 16,800

Hence, 407 x 168 - 307 x 168 = 16,800.

9Exercise 2(A)

Using the most convenient grouping, find each of the following products :

(i) 5 x 648 x 20

(ii) 8 x 329 x 25

(iii) 8 x 12 x 25 x 7

(iv) 125 x 40 x 8 x 25

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(i) 5 x 648 x 20

⇒ 648 x (5 x 20)

⇒ 648 x 100

⇒ 64,800.

Hence, 5 x 648 x 20 = 64,800.

(ii) 8 x 329 x 25

⇒ 329 x (8 x 25)

⇒ 329 x 200

⇒ 65,800.

Hence, 8 x 329 x 25 = 65,800.

(iii) 8 x 12 x 25 x 7

⇒ (8 x 25) x 12 x 7

⇒ 200 x 12 x 7

⇒ 2400 x 7

⇒ 16,800.

Hence, 8 x 12 x 25 x 7 = 16,800.

(iv) 125 x 40 x 8 x 25

⇒ 125 x 40 x (8 x 25)

⇒ 125 x 40 x 200

⇒ 125 x (40 x 200)

⇒ 125 x 8000

⇒ 10,00,000.

Hence, 125 x 40 x 8 x 25 = 10,00,000.

10Exercise 2(A)

Divide and verify the answer by division algorithm :

(i) 3680 ÷ 87

(ii) 17368 ÷ 327

(iii) 32679 ÷ 265

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(i) 3680 ÷ 87

$$\begin{array}{l} \phantom{x^2 }{\quad 42} \\ 87\overline{\smash{\big)}3680} \\ \phantom{x}\phantom{)}\underline{-348} \\ \phantom{{x^2 }+} 200 \\ \phantom{{x} +}\underline{-174} \\ \phantom{{x^2 a}+} 26\ \end{array}$$

Dividend = 3680

Divisor = 87

Quotient = 42

Remainder = 26

Verification: Dividend = (Divisor × Quotient) + Remainder

Substituting the values in R.H.S. of the equation :

(Divisor × Quotient) + Remainder

= (87 x 42) + 26

= 3,654 + 26

= 3,680.

Since, L.H.S. = R.H.S. = 3,680

Hence, the result is verified by the division algorithm.

(ii) 17368 ÷ 327

$$\begin{array}{l} \phantom{x^2 ++}{\quad 53} \\ 327\overline{\smash{\big)}\quad 17368} \\ \phantom{x^ + )}\phantom{))}\underline{-1635} \\ \phantom{{x^2 } + 2a)} 1018 \\ \phantom{{x} +1a )}\underline{-981} \\ \phantom{{x^2 a} + 2x)} 37\ \end{array}$$

Dividend = 17368

Divisor = 327

Quotient = 53

Remainder = 37

Verification: Dividend = (Divisor × Quotient) + Remainder

Substituting the values in R.H.S. of the equation :

(Divisor × Quotient) + Remainder

= (327 x 53) + 37

= 17,331 + 37

= 17,368

Since, L.H.S. = R.H.S. = 17,368

Hence, the result is verified by the division algorithm.

(iii) 32679 ÷ 265

$$\begin{array}{l} \phantom{x^2 ))}{\quad 123} \\ 265\overline{\smash{\big)}\quad 32679} \\ \phantom{x^ + )}\phantom{))}\underline{-265} \\ \phantom{{x^2 } + 2a)} 617 \\ \phantom{{x} +1)}\underline{-530} \\ \phantom{{x^2 a} + 2x)} 879 \\ \phantom{{x} +1a ))}\underline{-795} \\ \phantom{{x^2 a} + 2x))} 84\ \end{array}$$

Dividend = 32679

Divisor = 265

Quotient = 123

Remainder = 84

Verification: Dividend = (Divisor × Quotient) + Remainder

Substituting the values in R.H.S. of the equation :

(Divisor × Quotient) + Remainder

= (265 x 123) + 84

= 32,595 + 84

= 32,679.

Since, L.H.S. = R.H.S. = 32,679

Hence, the result is verified by the division algorithm.

11Exercise 2(A)

Verify each of the following :

(i) 2867 + 986 = 986 + 2867

(ii) 368 x 215 = 215 x 368

(iii) (156 + 273) + 74 = 156 + (273 + 74)

(iv) (86 x 55) x 110 = 86 x (55 x 110)

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(i) 2867 + 986 = 986 + 2867

According to the Commutative Property of Addition: a + b = b + a.

Taking L.H.S. = 2867 + 986

= 3,853

Taking R.H.S. = 986 + 2867

= 3,853

Since, L.H.S. = R.H.S.

Hence, proved that 2867 + 986 = 986 + 2867.

(ii) 368 x 215 = 215 x 368

According to the Commutative Property of Multiplication: a x b = b x a.

Taking L.H.S. = 368 x 215

= 79,120

Taking R.H.S. = 215 x 368

= 79,120

Since, L.H.S. = R.H.S.

Hence, proved that 368 x 215 = 215 x 368.

(iii) (156 + 273) + 74 = 156 + (273 + 74)

According to the Associative Property of Addition: (a + b) + c = a + (b + c).

Taking L.H.S. = (156 + 273) + 74

= 429 + 74

= 503

Taking R.H.S. = 156 + (273 + 74)

= 156 + 347

= 503

Since, L.H.S. = R.H.S.

Hence, proved that (156 + 273) + 74 = 156 + (273 + 74).

(iv) (86 x 55) x 110 = 86 x (55 x 110)

According to the Associative Property of Multiplication: (a x b) x c = a x (b x c).

Taking L.H.S. = (86 x 55) x 110

= 4730 x 110

= 5,20,300

Taking R.H.S. = 86 x (55 x 110)

= 86 x 6050

= 5,20,300

Since, L.H.S. = R.H.S.

Hence, proved that (86 x 55) x 110 = 86 x (55 x 110).

12Exercise 2(A)

Simplify :

(i) 39 - 18 ÷ 3 + 2 x 3

(ii) 8 + 2 x 5

(iii) 5 x 8 - 6 ÷ 2

(iv) 19 - 9 x 2

(v) 15 ÷ 5 x 4 ÷ 2

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(i) 39 - 18 ÷ 3 + 2 x 3

= 39 - 6 + 2 x 3

= 39 - 6 + 6

= 33 + 6 [Addition and subtraction from left to right]

= 39.

Hence, 39 - 18 ÷ 3 + 2 x 3 = 39.

(ii) 8 + 2 x 5

= 8 + 10

= 18.

Hence, 8 + 2 x 5 = 18.

(iii) 5 x 8 - 6 ÷ 2

= 5 x 8 - 3

= 40 - 3

= 37.

Hence, 5 x 8 - 6 ÷ 2 = 37.

(iv) 19 - 9 x 2

= 19 - 18

= 1.

Hence, 19 - 9 x 2 = 1.

(v) 15 ÷ 5 x 4 ÷ 2

= 3 x 4 ÷ 2

= 3 x 2

= 6.

Hence, 15 ÷ 5 x 4 ÷ 2 = 6.

1(i)Exercise 2(B)

Study the following pattern. In each case write the next three steps :

111 ÷ 3 = 37

222 ÷ 6 = 37

333 ÷ 9 = 37

444 ÷ 12 = 37

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111 ÷ 3 = 37

222 ÷ 6 = 37

333 ÷ 9 = 37

444 ÷ 12 = 37

The numerator increases by 111 and the divisor increases by 3.

Thus, next three steps are :

555 ÷ 15 = 37

666 ÷ 18 = 37

777 ÷ 21 = 37

1(ii)Exercise 2(B)

Study the following pattern. In each case write the next three steps :

999999 x 1 = 0999999

999999 x 2 = 1999998

999999 x 3 = 2999997

999999 x 4 = 3999996

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999999 x 1 = 0999999

999999 x 2 = 1999998

999999 x 3 = 2999997

999999 x 4 = 3999996

The multiplier increases by 1 and the pattern continues accordingly.

Thus, next three steps are :

999999 x 5 = 4999995

999999 x 6 = 5999994

999999 x 7 = 6999993

1(iii)Exercise 2(B)

Study the following pattern. In each case write the next three steps :

12345679 x 9 = 111111111

12345679 x 18 = 222222222

12345679 x 27 = 333333333

12345679 x 36 = 444444444

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12345679 x 9 = 111111111

12345679 x 18 = 222222222

12345679 x 27 = 333333333

12345679 x 36 = 444444444

The multiplier increases by 9 each time and the product forms repeated digits.

Thus, next three steps are :

12345679 x 45 = 555555555

12345679 x 54 = 666666666

12345679 x 63 = 777777777

2Exercise 2(B)

Observe the following pattern and answer the questions that follow :

1 + 2 + 3 + 4 + 5 = 15

2 + 3 + 4 + 5 + 6 = 20

3 + 4 + 5 + 6 + 7 = 25

(i) By which number should we multiply the middle number to get the sum?

(ii) Write the row of the pattern which gives the sum as 75.

(iii) Can there be any row of the pattern which gives the sum as 92?

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(i) Observing the pattern:

1 + 2 + 3 + 4 + 5 = 15. The middle number is 3 and 3 x 5 = 15.

2 + 3 + 4 + 5 + 6 = 20. The middle number is 4 and 4 x 5 = 20.

3 + 4 + 5 + 6 + 7 = 25. The middle number is 5 and 5 x 5 = 25.

The pattern shows that we should multiply the middle number by 5 to get the sum.

(ii) Using the rule from part (i), we can find the middle number for a sum of 75:

Middle number = 75 ÷ 5 = 15.

The row consists of five consecutive numbers with 15 as the middle number.

The two numbers before 15 are 13 and 14.

The two numbers after 15 are 16 and 17.

Hence, the row is 13 + 14 + 15 + 16 + 17 = 75.

(iii) For a sum of 92, the middle number would have to be:

Middle number = 92 ÷ 5 = 18.4

Since the numbers in the pattern are always integers, the middle number must also be an integer. As 92 is not exactly divisible by 5 (it has a remainder of 2), there cannot be a row of this pattern that gives a sum of 92.

Hence, there will not be any row of the above pattern which gives the sum as 92.

3(i)Exercise 2(B)

Complete the following magic square by supplying the missing numbers :

Complete the following magic square by supplying the missing numbers. Operations on Whole Numbers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.
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Complete the following magic square by supplying the missing numbers. Operations on Whole Numbers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.
3(ii)Exercise 2(B)

Complete the following magic square by supplying the missing numbers :

Complete the following magic square by supplying the missing numbers. Operations on Whole Numbers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.
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Complete the following magic square by supplying the missing numbers. Operations on Whole Numbers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.
3(iii)Exercise 2(B)

Complete the following magic square by supplying the missing numbers :

Complete the following magic square by supplying the missing numbers. Operations on Whole Numbers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.
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Complete the following magic square by supplying the missing numbers. Operations on Whole Numbers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.
3(iv)Exercise 2(B)

Complete the following magic square by supplying the missing numbers :

Complete the following magic square by supplying the missing numbers. Operations on Whole Numbers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.
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Complete the following magic square by supplying the missing numbers. Operations on Whole Numbers, R.S. Aggarwal Mathematics Solutions ICSE Class 6.
11 Mark · Exercise 2(C) — Multiple Choice Questions

The smallest whole number is :

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The smallest whole number is 0.

Hence, option 4 is the correct option.

21 Mark · Exercise 2(C) — Multiple Choice Questions

The least number of 4 digit which is exactly divisible by 7 is :

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The smallest 4-digit number = 1000.

$$\begin{array}{l} \phantom{x^2 }{142} \\ 7\overline{\smash{\big)}1000} \\ \phantom{x}\phantom{}\underline{-7} \\ \phantom{{x^2))}} 30 \\ \phantom{{x} )}\underline{-28} \\ \phantom{{x^2 } + } 20 \\ \phantom{{x} ))}\underline{-14} \\ \phantom{{x^2 +)}} 6 \\ \end{array}$$

Quotient = 142

Remainder = 6

“To get the next multiple of 7, we add (7 − remainder) to 1000.”

Next multiple = 1000 + (7 − 6)

= 1000 + 1

= 1001.

Therefore, the least 4-digit number exactly divisible by 7 is 1,001.

Hence, option 3 is the correct option.

31 Mark · Exercise 2(C) — Multiple Choice Questions

The largest number of 4 digits exactly divisible by 13 is

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The largest 4-digit number = 9,999.

To find the largest 4-digit number exactly divisible by 13, we divide 9,999 by 13 and subtract the remainder to get the largest multiple.

$$\begin{array}{l} \phantom{x^2}{769} \\ 13\overline{\smash{\big)}9999} \\ \phantom{x}\phantom{)}\underline{-91} \\ \phantom{{x^2 } +} 89 \\ \phantom{{x))} }\underline{-78} \\ \phantom{{x^2 } + 2} 119 \\ \phantom{{x} +)}\underline{-117} \\ \phantom{{x^2 + 3)}} 2 \\ \end{array}$$

The remainder when 9,999 is divided by 13 is 2.

Therefore, 9,999 − 2 = 9,997.

Hence, option 2 is the correct option.

41 Mark · Exercise 2(C) — Multiple Choice Questions

What least number should be subtracted from 10,003 to get a number exactly divisible by 11?

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Given number = 10,003.

To make 10003 exactly divisible by 11, we divide 10,003 by 11 and subtract the remainder from 10,003.

$$\begin{array}{l} \phantom{x^2 +}{\quad 909} \\ 11\overline{\smash{\big)}\quad 10003} \\ \phantom{x^ + )}\phantom{)}\underline{-99} \\ \phantom{{x^2 } + 2a} 10 \\ \phantom{{x} +1a}\underline{-00} \\ \phantom{{x^2 } + 2xa } 103 \\ \phantom{{x} +1xa}\underline{-99} \\ \phantom{{x^2 + 3x +)}} 4 \\ \end{array}$$

The remainder when 10003 is divided by 11 is 4.

Least number to be subtracted = remainder = 4.

Hence, option 4 is the correct option.

51 Mark · Exercise 2(C) — Multiple Choice Questions

What least number should be added to 6,000 to get a number exactly divisible by 19?

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Given number = 6,000.

$$\begin{array}{l} \phantom{x^21}{\quad 315} \\ 19\overline{\smash{\big)}\quad 6000} \\ \phantom{x^ + )}\phantom{)}\underline{-57} \\ \phantom{{x^2 } + 2a} 30 \\ \phantom{{x} +1a}\underline{-19} \\ \phantom{{x^2 } + 2xa } 110 \\ \phantom{{x} +1xa}\underline{-95} \\ \phantom{{x^2 + 3x +)}} 15 \\ \end{array}$$

The remainder when 6000 is divided by 19 is 15.

Least number to be added to make the number divisible = 19 − 15 = 4.

Hence, option 3 is the correct option.

61 Mark · Exercise 2(C) — Multiple Choice Questions

What whole number is nearest to 457 which is divisible by 11?

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Given number = 457.

To find the nearest number divisible by 11 :

$$\begin{array}{l} \phantom{x^21}{\quad 41} \\ 11\overline{\smash{\big)}\quad 457} \\ \phantom{x^ + )}\phantom{)}\underline{-44} \\ \phantom{{x^2 } + 2a} 17 \\ \phantom{{x} +1a}\underline{-11} \\ \phantom{{x^2 } + 2xa } 6\ \end{array}$$

The multiple of 11 just below 457 is 11 x 41 = 451.

The multiple of 11 just above 457 is 11 x 42 = 462.

Distance from 457 to 451: ∣457 − 451∣ = 6.

Distance from 457 to 462: ∣457 − 462∣ = 5.

The smaller distance is 5, which corresponds to the number 462.

Therefore, the whole number nearest to 457 that is divisible by 11 is 462.

Hence, option 1 is the correct option.

71 Mark · Exercise 2(C) — Multiple Choice Questions

How many whole numbers are there between 1,036 and 1,263?

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Number of whole numbers = (1263 − 1036) − 1

= 227−1

= 226

Hence, option 3 is the correct option.

81 Mark · Exercise 2(C) — Multiple Choice Questions

A number when divided by 43 gives 12 as quotient and 24 as remainder. The number is

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Given:

Divisor = 43

Quotient = 12

Remainder = 24

By formula,

Dividend = Divisor x Quotient + Remainder

Substitute the values into the formula:

Dividend = 43 x 12 + 24

= 516 + 24

= 540.

Hence, option 4 is the correct option.

91 Mark · Exercise 2(C) — Multiple Choice Questions

In a division sum, the dividend is 398, quotient is 15 and the remainder is 8. What is the divisor?

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Given:

Dividend = 398

Quotient = 15

Remainder = 8

By formula; Dividend = Divisor x Quotient + Remainder

Substitute the values into the formula:

⇒ 398 = Divisor x 15 + 8

⇒ Divisor x 15 = 398 - 8

⇒ Divisor x 15 = 390

⇒ Divisor = $\frac{390}{15}$

⇒ Divisor = 26

Hence, option 2 is the correct option.

101 Mark · Exercise 2(C) — Multiple Choice Questions

8 x 273 x 125 = ?

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Given :

⇒ 8 x 273 x 125

⇒ 273 x (8 x 125)

⇒ 273 x 1,000

⇒ 2,73,000

Hence, option 3 is the correct option.

111 Mark · Exercise 2(C) — Multiple Choice Questions

4 x 346 x 25 = ?

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Given :

⇒ 4 x 346 x 25

⇒ 346 x (4 x 25)

⇒ 346 x 100

⇒ 34,600

Hence, option 4 is the correct option.

121 Mark · Exercise 2(C) — Multiple Choice Questions

13,729 x 93 + 13,729 x 7 = ?

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Given, 13,729 x 93 + 13,729 x 7

Using the Distributive law of multiplication over addition : a x b + a x c = a x (b + c)

⇒ 13,729 x (93 + 7)

⇒ 13,729 x 100

⇒ 13,72,900

Hence, option 2 is the correct option.

131 Mark · Exercise 2(C) — Multiple Choice Questions

2,563 x 187 - 2,563 x 87 = ?

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Given,

2,563 x 187 - 2,563 x 87

Using the Distributive law of multiplication over subtraction: a x b - a x c = a x (b - c)

⇒ 2,563 x (187 - 87)

⇒ 2,563 x 100

⇒ 2,56,300

Hence, option 1 is the correct option.

141 Mark · Exercise 2(C) — Multiple Choice Questions

546 x 98 = ?

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Given,

⇒ 546 x 98

⇒ 546 x (100 - 2)

⇒ 546 x 100 - 546 x 2

⇒ 54,600 - 1,092

⇒ 53,508

Hence, option 2 is the correct option.

151 Mark · Exercise 2(C) — Multiple Choice Questions

8,456 - ? = 3,580

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Given,

8,456 - ? = 3,580

Let missing number be x.

⇒ 8,456 - x = 3,580

⇒ x = 8,456 - 3,580 = 4,876.

Hence, option 3 is the correct option.

1Exercise 2(C) — Mental Maths

Fill in the blanks :

(i) 156 x 48 - 156 x 38 = ...............

(ii) 76 x 53 + 76 x 47 = ...............

(iii) 138 x 67 + 62 x 67 = ...............

(iv) 4 x 269 x 25 = ...............

(v) There are ............... whole numbers up to 40.

(vi) ............... is the whole number which has no predecessor.

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(i) 156 x 48 - 156 x 38 = 156 x (48 - 38) = 156 x 10 = 1,560.

(ii) 76 x 53 + 76 x 47 = 76 x (53 + 47) = 76 x 100 = 7,600.

(iii) 138 x 67 + 62 x 67 = 67 x (138 + 62) = 67 x 200 = 13,400.

(iv) 4 x 269 x 25 = 269 x (4 x 25) = 269 x 100 = 26,900.

(v) There are 41 whole numbers up to 40.

(vi) 0 is the whole number which has no predecessor.

2(i)Exercise 2(C) — Mental Maths

Write T for true and F for false statement :

The least natural number is 0.

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False

Reason

The set of natural numbers is defined as the set of positive integers: {1, 2, 3, 4, ...}. The least natural number is 1.

2(ii)Exercise 2(C) — Mental Maths

Write T for true and F for false statement :

Subtraction is associative on natural numbers.

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False

Reason

For an operation to be associative, the following property must hold for any three numbers a, b, and c:

(a − b) − c = a − (b − c)

Let us take an example with natural numbers (e.g., 5, 3, and 1):

L.H.S. = (5 − 3) − 1 = 2 − 1 = 1

R.H.S. = 5 − (3 − 1) = 5 − 2 = 3

Since, 1 ≠ 3, the associative property does not hold for subtraction.

2(iii)Exercise 2(C) — Mental Maths

Write T for true and F for false statement :

In whole numbers, the multiplicative identity is 1.

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True

Reason

The multiplicative identity is a number that, when multiplied by any other number, leaves the other number unchanged.

In the set of whole numbers, this number is 1, because for any whole number 'n': n × 1 = n and 1 × n = n.

2(iv)Exercise 2(C) — Mental Maths

Write T for true and F for false statement :

For whole numbers a, b, c, we always have (a + b).c = a.c + b.c.

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True

Reason

For whole numbers a, b, c, we always have (a + b).c = a.c + b.c.

This is known as the Distributive Property of Multiplication over Addition.

2(v)Exercise 2(C) — Mental Maths

Write T for true and F for false statement :

78 x 395 + 78 x 605 = 78000

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True

Reason

Given, 78 x 395 + 78 x 605 = 78000

Taking L.H.S. = 78 x 395 + 78 x 605

= 78 x (395 + 605)

= 78 x 1,000

= 78,000

Taking R.H.S. = 78,000

So, L.H.S. = R.H.S.

1Exercise 2(C) — Assertion-Reason Questions

Assertion (A): For any three whole numbers a, b and c, we have a x (b + c) = a x b + a x c.

Reason (R): The multiplication of whole numbers is associative.

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For any three whole numbers a, b and c, we have a x (b + c) = a x b + a x c.

This statement is the definition of the Distributive Property of Multiplication over Addition.

∴ Assertion (A) is true.

The multiplication of whole numbers is associative.

This statement is the definition of the Associative Property of Multiplication, which states that for any three whole numbers a, b, and c, we have (a × b) × c = a × (b × c). This is also a fundamental and true property of whole numbers.

∴ Reason (R) is true.

Both A and R are true but R is not the correct explanation of A.

Hence, option 2 is the correct option.

2Exercise 2(C) — Assertion-Reason Questions

Assertion (A): On simplifying 5 x 4 ÷ 2 - 1, we get 20.

Reason (R): For simplifying an expression we use DMAS rule.

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Given,

5 x 4 ÷ 2 - 1

= 5 x 2 - 1

= 10 - 1

= 9

∴ Assertion (A) is false.

For simplifying an expression we use DMAS rule.

This statement correctly identifies the standard order of operations (Division, Multiplication, Addition, Subtraction) used in arithmetic.

∴ Reason (R) is true.

Hence, option 4 is the correct option.

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