This chapter on ICSE Class 6 Maths Operations on Whole Numbers builds upon your understanding of numbers by exploring how we perform calculations like addition, subtraction, multiplication, and division. In this section from your RS Aggarwal Mathematics textbook, we will delve into the fundamental properties that govern these operations. You will learn about the commutative, associative, and distributive properties, which make calculations easier and more logical. We will also cover important concepts like the role of zero and one as identity elements for addition and multiplication, respectively. Understanding these rules is crucial as they form the foundation for more complex algebraic concepts you will study in higher classes.
If you are looking for clear, step-by-step solutions for the exercises in this chapter, you have come to the right place. This page provides detailed workings for all 43 questions from Exercise 2(A), Exercise 2(B), and the various sections of Exercise 2(C), including the Multiple Choice Questions and Mental Maths. Each solution is prepared by our subject experts and follows the exact method prescribed by the ICSE board, ensuring you learn the correct way to present your answers in exams. Here, you will find reliable guidance to verify your work and understand the logic behind each step.
Exercise 2(A)
Question 1
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 ……………)
(i) Addition feels the same whether you start from the left number or the right — that’s the Commutative Property at work. So 168 + 259 simply swaps places with 259 + 168.
(ii) Zero is special in addition. Drop it beside any number and nothing shifts. That’s why 0 + 317 stays exactly 317.
(iii) When three numbers team up for addition, the way you group them doesn’t change the final sum — the Associative Property. Here, (37 + 68) + 56 lines up perfectly with 37 + (68 + 56).
(iv) DMAS keeps things orderly: multiplication before addition. So 8 + 3 × 4 becomes 8 + 12, landing cleanly on 20.
(v) The Distributive Property lets us pull a common factor outside brackets. So 18 × (17 + 23) breaks into (18 × 17) + (18 × 23), giving the same result either way.
Question 2
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 = ……………
(i) Every number recognises itself when multiplied by 1 — that’s the Identity Property. So 237 × 1 naturally gives 237.
(ii) Multiply anything by zero and the result collapses to nothing. Hence 56 × 0 equals 0.
(iii) Zero shared equally among non-zero parts still leaves nothing. So 0 ÷ 53 is simply 0.
(iv) The Commutative Property of Multiplication means swapping the order of two factors doesn’t change the product. Thus 37 × 59 is the same as 59 × 37.
(v) Again, any number multiplied by zero becomes zero. So 0 × 138 is 0.
(vi) A non-zero number divided by itself always leaves exactly one whole part. Therefore 73 ÷ 73 equals 1.
Question 3
Divide 3605 by 29 and verify the division algorithm.
Dividend = 3605
Divisor = 29
Quotient = 124
Remainder = 9
To be absolutely sure, feed these values back into the division-algorithm formula: Dividend = (Divisor × Quotient) + Remainder.
Substituting: (29 × 124) + 9 = 3596 + 9 = 3605.
Since L.H.S. matches R.H.S.,
Hence, the result is verified by the division algorithm.
Question 4
Find the number which when divided by 45 gives 16 as quotient and 9 as remainder.
We know the divisor is 45, the quotient is 16, and the remainder is 9. The division algorithm gives us a direct path:
Dividend = (Divisor × Quotient) + Remainder
Plugging in: (45 × 16) + 9 = 720 + 9 = 729.
Hence, the number = 729.
Question 5
Find the largest number of 5-digits which is exactly divisible by 57.
The largest 5-digit number we can write is 99,999. To find the biggest multiple of 57 sitting just below it, divide 99,999 by 57 and strip away whatever remains.
\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 is 21. Subtracting it: 99,999 − 21 = 99,978.
Hence, the largest 5-digit number which is exactly divisible by 57 = 99,978.
Question 6
Find the smallest 6-digit number which is exactly divisible by 63.
The smallest 6-digit number is 1,00,000. To reach the first multiple of 63 at or above this, divide 1,00,000 by 63 and see what’s left over.
\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}We need to bridge the gap: 63 − 19 = 44. Add this to 1,00,000: 1,00,000 + 44 = 1,00,044.
Hence, the smallest 6-digit number exactly divisible by 63 = 1,00,044.
Question 7
On dividing 1653 by a certain number, we get 45 as quotient and 33 as remainder. Find the divisor.
Given:
Dividend = 1653
Quotient = 45
Remainder = 33
The division-algorithm formula ties everything together:
Dividend = (Divisor × Quotient) + Remainder
Substituting: 1653 = (Divisor × 45) + 33
Shift the remainder across: Divisor × 45 = 1653 − 33 = 1620
So Divisor = \dfrac{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.
Question 8
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
(i) 576 × 285 + 576 × 115
Both terms carry 576 as a common factor. Pull it out using the distributive law: a × b + a × c = a × (b + c).
⇒ 576 × (285 + 115) = 576 × 400 = 2,30,400.
Hence, 576 × 285 + 576 × 115 = 2,30,400.
(ii) 385 × 178 − 385 × 78
Again, 385 is common. The distributive law for subtraction reads a × b − a × c = a × (b − c).
⇒ 385 × (178 − 78) = 385 × 100 = 38,500.
Hence, 385 × 178 − 385 × 78 = 38,500.
(iii) 365 × 645 + 135 × 645
Rearrange to spotlight the common factor: 645 × 365 + 645 × 135.
⇒ 645 × (365 + 135) = 645 × 500 = 3,22,500.
Hence, 365 × 645 + 135 × 645 = 3,22,500.
(iv) 407 × 168 − 307 × 168
Factor out 168: 168 × 407 − 168 × 307.
⇒ 168 × (407 − 307) = 168 × 100 = 16,800.
Hence, 407 × 168 − 307 × 168 = 16,800.
Question 9
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
(i) 5 × 648 × 20
Grouping 5 and 20 first gives a neat century: 648 × (5 × 20) = 648 × 100 = 64,800.
Hence, 5 × 648 × 20 = 64,800.
(ii) 8 × 329 × 25
8 teamed with 25 produces 200: 329 × (8 × 25) = 329 × 200 = 65,800.
Hence, 8 × 329 × 25 = 65,800.
(iii) 8 × 12 × 25 × 7
Pair 8 with 25 to get 200, then cruise through the rest: (8 × 25) × 12 × 7 = 200 × 12 × 7 = 2400 × 7 = 16,800.
Hence, 8 × 12 × 25 × 7 = 16,800.
(iv) 125 × 40 × 8 × 25
Lock 8 with 25 for 200, then let 125 meet 8000: 125 × 40 × (8 × 25) = 125 × 40 × 200 = 125 × 8000 = 10,00,000.
Hence, 125 × 40 × 8 × 25 = 10,00,000.
Question 10
Divide and verify the answer by division algorithm :
(i) 3680 ÷ 87
(ii) 17368 ÷ 327
(iii) 32679 ÷ 265
(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.
Check: (87 × 42) + 26 = 3,654 + 26 = 3,680. L.H.S. = R.H.S.
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.
Check: (327 × 53) + 37 = 17,331 + 37 = 17,368. L.H.S. = R.H.S.
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.
Check: (265 × 123) + 84 = 32,595 + 84 = 32,679. L.H.S. = R.H.S.
Hence, the result is verified by the division algorithm.
Question 11
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)
(i) 2867 + 986 = 986 + 2867
The Commutative Property of Addition tells us a + b = b + a.
L.H.S. = 2867 + 986 = 3,853
R.H.S. = 986 + 2867 = 3,853
Both sides match.
Hence, proved that 2867 + 986 = 986 + 2867.
(ii) 368 × 215 = 215 × 368
Multiplication also commutes: a × b = b × a.
L.H.S. = 368 × 215 = 79,120
R.H.S. = 215 × 368 = 79,120
Both sides match.
Hence, proved that 368 × 215 = 215 × 368.
(iii) (156 + 273) + 74 = 156 + (273 + 74)
Addition is associative: (a + b) + c = a + (b + c).
L.H.S. = (156 + 273) + 74 = 429 + 74 = 503
R.H.S. = 156 + (273 + 74) = 156 + 347 = 503
Both sides match.
Hence, proved that (156 + 273) + 74 = 156 + (273 + 74).
(iv) (86 × 55) × 110 = 86 × (55 × 110)
Multiplication is associative too: (a × b) × c = a × (b × c).
L.H.S. = (86 × 55) × 110 = 4730 × 110 = 5,20,300
R.H.S. = 86 × (55 × 110) = 86 × 6050 = 5,20,300
Both sides match.
Hence, proved that (86 × 55) × 110 = 86 × (55 × 110).
Question 12
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
(i) 39 − 18 ÷ 3 + 2 × 3
DMAS guides the way: division and multiplication first.
= 39 − 6 + 2 × 3
= 39 − 6 + 6
= 33 + 6
= 39.
Hence, 39 − 18 ÷ 3 + 2 × 3 = 39.
(ii) 8 + 2 × 5
Multiply before adding:
= 8 + 10
= 18.
Hence, 8 + 2 × 5 = 18.
(iii) 5 × 8 − 6 ÷ 2
Division and multiplication share priority, left to right:
= 5 × 8 − 3
= 40 − 3
= 37.
Hence, 5 × 8 − 6 ÷ 2 = 37.
(iv) 19 − 9 × 2
Multiplication first:
= 19 − 18
= 1.
Hence, 19 − 9 × 2 = 1.
(v) 15 ÷ 5 × 4 ÷ 2
Division and multiplication march left to right:
= 3 × 4 ÷ 2
= 3 × 2
= 6.
Hence, 15 ÷ 5 × 4 ÷ 2 = 6.
Exercise 2(B)
Question 1(i)
Study the following pattern. In each case write the next three steps :
111 ÷ 3 = 37
222 ÷ 6 = 37
333 ÷ 9 = 37
444 ÷ 12 = 37
111 ÷ 3 = 37
222 ÷ 6 = 37
333 ÷ 9 = 37
444 ÷ 12 = 37
The top number marches up by 111 each time, while the divisor climbs by 3 — keeping the quotient locked at 37.
Carrying the pattern forward:
555 ÷ 15 = 37
666 ÷ 18 = 37
777 ÷ 21 = 37
Question 1(ii)
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
999999 × 1 = 0999999
999999 × 2 = 1999998
999999 × 3 = 2999997
999999 × 4 = 3999996
The multiplier rises by 1 each step, and the answer follows its own tidy rhythm.
Continuing:
999999 × 5 = 4999995
999999 × 6 = 5999994
999999 × 7 = 6999993
Question 1(iii)
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
12345679 × 9 = 111111111
12345679 × 18 = 222222222
12345679 × 27 = 333333333
12345679 × 36 = 444444444
The multiplier grows by 9 each turn, producing strings of identical digits.
Next three steps:
12345679 × 45 = 555555555
12345679 × 54 = 666666666
12345679 × 63 = 777777777
Question 2
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?
(i) Look closely at the sums:
1 + 2 + 3 + 4 + 5 = 15. The middle number is 3, and 3 × 5 = 15.
2 + 3 + 4 + 5 + 6 = 20. The middle number is 4, and 4 × 5 = 20.
3 + 4 + 5 + 6 + 7 = 25. The middle number is 5, and 5 × 5 = 25.
The rule is clear: multiply the middle term by 5 to get the sum.
(ii) For a total of 75:
Middle number = 75 ÷ 5 = 15.
The five consecutive numbers centred on 15 are 13, 14, 15, 16, 17.
Hence, the row is 13 + 14 + 15 + 16 + 17 = 75.
(iii) Trying 92:
Middle number = 92 ÷ 5 = 18.4.
Since the pattern demands whole numbers, 18.4 won’t do. 92 leaves a remainder when divided by 5, so no such row exists.
Hence, there will not be any row of the above pattern which gives the sum as 92.
Question 3(i)
Complete the following magic square by supplying the missing numbers :
Question 3(ii)
Complete the following magic square by supplying the missing numbers :
Question 3(iii)
Complete the following magic square by supplying the missing numbers :
Question 3(iv)
Complete the following magic square by supplying the missing numbers :
Exercise 2(C) — Multiple Choice Questions
Question 1
The smallest whole number is :
- (a) 1
- (b) 2
- (c) 10
- (d) none of these
The smallest whole number is 0.
Hence, option 4 is the correct option.
Question 2
The least number of 4 digit which is exactly divisible by 7 is :
- (a) 1,015
- (b) 1,008
- (c) 1,001
- (d) 1,022
The smallest 4-digit number is 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 hit the next multiple of 7, add what’s missing: 7 − 6 = 1.
Next multiple = 1000 + 1 = 1001.
Hence, option 3 is the correct option.
Question 3
The largest number of 4 digits exactly divisible by 13 is
- (a) 9,996
- (b) 9,997
- (c) 9,995
- (d) 9,984
The largest 4-digit number is 9,999.
\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}Remainder = 2. Knock it off: 9,999 − 2 = 9,997.
Hence, option 2 is the correct option.
Question 4
What least number should be subtracted from 10,003 to get a number exactly divisible by 11?
- (a) 7
- (b) 6
- (c) 5
- (d) 4
Given number = 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}Remainder = 4. That’s exactly what must be removed.
Hence, option 4 is the correct option.
Question 5
What least number should be added to 6,000 to get a number exactly divisible by 19?
- (a) 9
- (b) 8
- (c) 4
- (d) 6
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}Remainder = 15. The shortfall is 19 − 15 = 4.
Hence, option 3 is the correct option.
Question 6
What whole number is nearest to 457 which is divisible by 11?
- (a) 462
- (b) 460
- (c) 451
- (d) 450
Given number = 457.
\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 just below 457 is 11 × 41 = 451.
The next multiple above is 11 × 42 = 462.
Distance to 451: |457 − 451| = 6.
Distance to 462: |457 − 462| = 5.
462 is closer.
Hence, option 1 is the correct option.
Question 7
How many whole numbers are there between 1,036 and 1,263?
- (a) 227
- (b) 228
- (c) 226
- (d) 225
Counting whole numbers between two values means excluding both endpoints.
(1263 − 1036) − 1 = 227 − 1 = 226.
Hence, option 3 is the correct option.
Question 8
A number when divided by 43 gives 12 as quotient and 24 as remainder. The number is
- (a) 547
- (b) 545
- (c) 543
- (d) 540
Given: Divisor = 43, Quotient = 12, Remainder = 24.
Dividend = Divisor × Quotient + Remainder
= 43 × 12 + 24
= 516 + 24
= 540.
Hence, option 4 is the correct option.
Question 9
In a division sum, the dividend is 398, quotient is 15 and the remainder is 8. What is the divisor?
- (a) 31
- (b) 26
- (c) 17
- (d) 23
Given: Dividend = 398, Quotient = 15, Remainder = 8.
398 = Divisor × 15 + 8
Divisor × 15 = 398 − 8 = 390
Divisor = \dfrac{390}{15} = 26.
Hence, option 2 is the correct option.
Question 10
8 x 273 x 125 = ?
- (a) 27,300
- (b) 2,70,300
- (c) 2,73,000
- (d) 27,30,000
8 × 273 × 125
Group 8 and 125 first: 273 × (8 × 125) = 273 × 1,000 = 2,73,000.
Hence, option 3 is the correct option.
Question 11
4 x 346 x 25 = ?
- (a) 28,400
- (b) 32,500
- (c) 33,800
- (d) 34,600
4 × 346 × 25
4 and 25 make a handy 100: 346 × (4 × 25) = 346 × 100 = 34,600.
Hence, option 4 is the correct option.
Question 12
13,729 x 93 + 13,729 x 7 = ?
- (a) 23,62,900
- (b) 13,72,900
- (c) 6,86,450
- (d) 4,57,620
13,729 × 93 + 13,729 × 7
Factor out the common term: 13,729 × (93 + 7) = 13,729 × 100 = 13,72,900.
Hence, option 2 is the correct option.
Question 13
2,563 x 187 – 2,563 x 87 = ?
- (a) 2,56,300
- (b) 1,28,150
- (c) 3,84,450
- (d) none of these
2,563 × 187 − 2,563 × 87
Pull out 2,563: 2,563 × (187 − 87) = 2,563 × 100 = 2,56,300.
Hence, option 1 is the correct option.
Question 14
546 x 98 = ?
- (a) 52,518
- (b) 53,508
- (c) 54,108
- (d) 52,708
546 × 98
Write 98 as (100 − 2): 546 × 100 − 546 × 2 = 54,600 − 1,092 = 53,508.
Hence, option 2 is the correct option.
Question 15
8,456 – ? = 3,580
- (a) 2,698
- (b) 4,586
- (c) 4,876
- (d) 5,016
8,456 − ? = 3,580
Let the missing number be x.
8,456 − x = 3,580
x = 8,456 − 3,580 = 4,876.
Hence, option 3 is the correct option.
Exercise 2(C) — Mental Maths
Question 1
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.
(i) 156 × 48 − 156 × 38 = 156 × (48 − 38) = 156 × 10 = 1,560.
(ii) 76 × 53 + 76 × 47 = 76 × (53 + 47) = 76 × 100 = 7,600.
(iii) 138 × 67 + 62 × 67 = 67 × (138 + 62) = 67 × 200 = 13,400.
(iv) 4 × 269 × 25 = 269 × (4 × 25) = 269 × 100 = 26,900.
(v) There are 41 whole numbers up to 40.
(vi) 0 is the whole number which has no predecessor.
Question 2(i)
Write T for true and F for false statement :
The least natural number is 0.
False
Natural numbers start from 1: {1, 2, 3, 4, …}. The smallest among them is 1, not 0.
Question 2(ii)
Write T for true and F for false statement :
Subtraction is associative on natural numbers.
False
For subtraction to be associative, (a − b) − c must equal a − (b − c). Take 5, 3, and 1:
L.H.S. = (5 − 3) − 1 = 2 − 1 = 1
R.H.S. = 5 − (3 − 1) = 5 − 2 = 3
1 ≠ 3, so the property fails.
Question 2(iii)
Write T for true and F for false statement :
In whole numbers, the multiplicative identity is 1.
True
The multiplicative identity leaves every number untouched when multiplied. For whole numbers, that identity is 1, since n × 1 = n and 1 × n = n for any whole number n.
Question 2(iv)
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.
True
For whole numbers a, b, c, the Distributive Property of Multiplication over Addition guarantees (a + b) · c = a · c + b · c always holds.
Question 2(v)
Write T for true and F for false statement :
78 x 395 + 78 x 605 = 78000
True
L.H.S. = 78 × 395 + 78 × 605
= 78 × (395 + 605)
= 78 × 1,000
= 78,000
R.H.S. = 78000
So L.H.S. = R.H.S.
Exercise 2(C) — Assertion-Reason Questions
Question 1
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.
- (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
- (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
- (c) Assertion (A) is true but Reason (R) is false.
- (d) Assertion (A) is false but Reason (R) is true.
For any three whole numbers a, b and c, a × (b + c) = a × b + a × c defines the Distributive Property of Multiplication over Addition.
∴ Assertion (A) is true.
The multiplication of whole numbers is associative: (a × b) × c = a × (b × c).
∴ Reason (R) is also true.
Both statements are correct, yet R does not explain A.
Hence, option 2 is the correct option.
Question 2
Assertion (A): On simplifying 5 x 4 ÷ 2 – 1, we get 20.
Reason (R): For simplifying an expression we use DMAS rule.
- (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
- (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
- (c) Assertion (A) is true but Reason (R) is false.
- (d) Assertion (A) is false but Reason (R) is true.
5 × 4 ÷ 2 − 1
= 5 × 2 − 1
= 10 − 1
= 9
So Assertion (A) claiming the answer is something else is false.
Reason (R) correctly names the DMAS rule for simplifying expressions.
∴ R is true.
Hence, option 4 is the correct option.