Yearly Archives: 2022

Chapter 22 Brief Review of Cartesian System of Rectangular Coordinates Exercise Ex. 22.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Chapter 22 Brief Review of Cartesian System of Rectangular Coordinates Exercise Ex. 22.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 22 Brief Review of Cartesian System of Rectangular Coordinates Exercise Ex. 22.3

Question 1

Solution 1

---

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 7(i)

Solution 7(i)

Question 7(ii)

Solution 7(ii)

Question 7(iii)

Solution 7(iii)

Question 8

Solution 8

Chapter 21 Some Special Series Exercise Ex. 21.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 8(iv)

Solution 8(iv)

Question 9

Solution 9

Chapter 21 Some Special Series Exercise Ex. 21.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 20 Geometric Progressions Exercise Ex. 20.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Chapter 20 Geometric Progressions Exercise Ex. 20.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 20 Geometric Progressions Exercise Ex. 20.3

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Find the sum of the geom etric series:

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

1, -a, a², – a³ , ….. to n terms (a ≠ 1)Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 4(v)

Solution 4(v)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

A person has 2 parents, 4 grandparents, 8 great grand parents,  and so on. Find the number his ancestors during the ten generations preceding his own.Solution 18

Question 19

(n – 1) S_n = 1ⁿ + 2ⁿ + 3ⁿ + ….+ nⁿSolution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Chapter 20 Geometric Progressions Exercise Ex. 20.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8(i)

Solution 8(i)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 8(iv)

Solution 8(iv)

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Chapter 20 Geometric Progressions Exercise Ex. 20.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8(I)

Solution 8(I)

Question 8(ii)

Solution 8(ii)

Question 8(iii)

Solution 8(iii)

Question 8(iv)

Solution 8(iv)

Question 8(v)

Solution 8(v)

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 9(iii)

Solution 9(iii)

Question 10(i)

Solution 10(i)

Question 10(ii)

Solution 10(ii)

Question 10(iii)

Solution 10(iii)

Question 11(i)

Solution 11(i)

Question 11(ii)

Solution 11(ii)

Question 11(iii)

Solution 11(iii)

Question 11(iv)

Solution 11(iv)

Question 12

Solution 12

Question 13

Solution 13

Question 14

If the 4^th, 10^th, and 16^th terms of a G.P. are x, y, and z respectively. Prove that x, y, z are in G.P.Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

If p^th, q^th, and r^th terms of a_n A.P. and G.P. are both a, b, and c respectively, show that a^b-c b^c-a c^a-b = 1.Solution 23

Chapter 20 Geometric Progressions Exercise Ex. 20.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Chapter 19 Arithmetic Progressions Exercise Ex. 19.1

Question 1

Solution 1

Question 2

Solution 2

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 7

Solution 7

Question 8

Solution 8

Question 3

Find the first four terns of the sequence defined by a₁ = 3 and, a_n = 3a_{n– 1} + 2, for all n > 1Solution 3

Chapter 19 Arithmetic Progressions Exercise Ex. 19.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Chapter 19 Arithmetic Progressions Exercise Ex. 19.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Chapter 19 Arithmetic Progressions Exercise Ex. 19.4

Question 1

Solution 1

( vii ) 

            Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 15

Find the rth term of an A.P., the sum of whose first n terms is 3n² + 2n.Solution 15

Chapter 19 Arithmetic Progressions Exercise Ex. 19.5

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Show that x² + xy + y², z² + zx + x² and y² + yz + z² are consecutive terms of an A.P., if x, y and z are in A.P.Solution 7

Chapter 19 Arithmetic Progressions Exercise Ex. 19.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.Solution 9

Chapter 19 Arithmetic Progressions Exercise Ex. 19.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

A carpenter was hired to build 192 window frames. The first day he made five frames and each day thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?Solution 12

Question 13

We know that the sum of the interior angles of a triangle is 180^o. Show that the sums of the interior angles of polygons with 3, 4, 5, 6,…. sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.Solution 13

Question 14

In a potato race 20 potatoes are placed in a line at intervals of 4 meters with the first potato 24 meters from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?Solution 14

Question 15(i)

A man accepts a position with an initial salary of Rs. 5200 per month. It is understood that he will receive an automatic increase of Rs. 320 in the very next month and each month thereafter.

i. Find his salary for the tenth month.Solution 15(i)

Question 15(ii)

A man accepts a position with an initial salary of Rs. 5200 per month. It is understood that he will receive an automatic increase of Rs. 320 in the very next month and each month thereafter.

What is his total earnings during the first year?Solution 15(ii)

Question 16

A man saved Rs. 66000 in 20 years. In each succeeding year after the first year he saved Rs. 200 more then what he saved in the previous year. How much did he save in the first year?Solution 16

Question 17

In a cricket team tournament 16 teams participated. A sum of Rs. 8000 is to be awarded among themselves as prize money. If the last place team is awarded Rs. 275 in prize money and the award increases by the same amount for successive finishing places, how much amount will the first place team receive?Solution 17

Chapter 18 Binomial Theorem Exercise Ex. 18.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 1(x)

Solution 1(x)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 2(x)

Solution 2(x)

Question 3

Solution 3

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

Solution 5(iv)

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Show that 2^{4n + 4} – 15n – 16, where n Î N is divisible by 225.Solution 12

Chapter 18 Binomial Theorem Exercise Ex. 18.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9(i)

Solution 9(i)

Question 9(ii)

Solution 9(ii)

Question 9(iii)

Solution 9(iii)

Question 9(iv)

Solution 9(iv)

Question 9(v)

Solution 9(v)

Question 9(vi)

Solution 9(vi)

Question 9(vii)

Solution 9(vii)

Question 9(viii)

Find the coefficient of x in the expansion of

 (1 – 3x + 7x²) (1 – x)¹⁶.Solution 9(viii)

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13(i)

Solution 13(i)

Question 13(ii)

Solution 13(ii)

Question 13(iii)

Solution 13(iii)

Question 13(iv)

Solution 13(iv)

Question 14(i)

Solution 14(i)

Question 14(ii)

Solution 14(ii)

Question 14(iii)

Solution 14(iii)

Question 14(iv)

Solution 14(iv)

Question 15(i)

Solution 15(i)

Question 15(ii)

Solution 15(ii)

Question 15(iii)

Solution 15(iii)

Question 15(iv)

Solution 15(iv)

Question 15(v)

Solution 15(v)

Question 15(vi)

Solution 15(vi)

Question 15(vii)

Solution 15(vii)

Question 15(viii)

Find the middle term (s) in expansion of:

Solution 15(viii)

Question 15(ix)

Find the middle term (s) in expansion of:

Solution 15(ix)

Question 15(x)

Find the middle term (s) in expansion of:

Solution 15(x)

Question 16(i)

Solution 16(i)

Question 16(ii)

Solution 16(ii)

Question 16(iii)

Solution 16(iii)

Question 16(iv)

Solution 16(iv)

Question 16(v)

Solution 16(v)

Question 16(vi)

Solution 16(vi)

Question 16(vii)

Solution 16(vii)

Question 16(viii)

Solution 16(viii)

Question 16(ix)

Solution 16(ix)

Question 16(x)

Solution 16(x)

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

If the seventh term from the beginning and end in the binomial expansion of  are equal, find n.Solution 39

Chapter 17 Combinations Exercise Ex. 17.1

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

If ^{n + 2}C₈ : ^{n – 2}P₄ = 57 : 16, find n.Solution 10

Question 11

Solution 11

Question 12

If ⁿC₂, ⁿC₅ and ⁿC₆ are in A.P., then find nSolution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20 (i)

Solution 20 (i)

Question 20(ii)

Solution 20(ii)

Question 20(iii)

Solution 20(iii)

Question 20(iv)

Solution 20(iv)

Chapter 17 Combinations Exercise Ex. 17.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Chapter 17 Combinations Exercise Ex. 17.3

Question 1

Solution 1

Question 3

Solution 3

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 2

Solution 2

Question 4

Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.Solution 4

Chapter 16 Permutations Exercise Ex. 16.1

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4(i)

Solution 4(i)

Question 4(ii)

Solution 4(ii)

Question 4(iii)

Solution 4(iii)

Question 4(iv)

Solution 4(iv)

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11(i)

Solution 11(i)

Question 11(ii)

Solution 11(ii)

Question 12

Solution 12

Chapter 16 Permutations Exercise Ex. 16.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

How many three-digit numbers are there whit no digit repeated?Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19(i)

Solution 19(i)

Question 19(ii)

Solution 19(ii)

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed?Solution 24Any number less than 1000 may be any of a number from one-digit number, two-digit number and three-digit number.

One-digit odd number:

3 possible ways are there. These numbers are 3 or 5 or 7.

Two-digit odd number:

Tens place can be filled up by 3 ways (using any of the digit among 3, 5 and 7) and then the ones place can be filled in any of the remaining 2 digits.

So, there are 3  2 = 6 such 2-digit numbers.

Three-digit odd number:

Ignore the presence of zero at ones place for some instance.

Hundreds place can be filled up in 3 ways (using any of any of the digit among 3, 5 and 7), then tens place in 3 ways by using remaining 3 digits (after using a digit, there will be three digits) and then the ones place in 2 ways.

So, there are a total of 3  3  2 = 18 numbers of 3-digit numbers which includes both odd and even numbers (ones place digit are zero). In order to get the odd numbers, it is required to ignore the even numbers i.e. numbers ending with zero.

To obtain the even 3-digit numbers, ones place can be filled up in 1 way (only 0 to be filled), hundreds place in 3 ways (using any of the digit among 3, 5, 7) and then tens place in 2 ways (using remaining 2 digits after filling up hundreds place).

So, there are a total of 1  3  2 = 6 even 3-digit numbers using the digits 0, 3, 5 and 7 (repetition not allowed)

So, number of three-digit odd numbers using the digits 0, 3, 5 and 7 (repetition not allowed) = 18 –  6 = 12.

Therefore, odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed are 3 + 6 + 12 = 21.Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Serial numbers for an item produced in a factory are to be made using two letters followed by four digits (0 to 9). If the letters are to be are to be taken from six letters of English alphabet without repetition and the digits are also not repeated in a serial number, how many serial numbers are possible?Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

How many for digit natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3, and 4, if the digits can repeat?Solution 33

The given digits are 1, 2, 3 and 4. These digits can be repeated while forming the numbers. So, number of required four digit natural numbers can be found as follows.

Consider four digit natural numbers whose digit at thousandths place is 1.

Here, hundredths place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Similarly, tens place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Ones place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)

Number of four digit natural numbers whose digit at thousandths place is 1 = 4  4  4 = 64

Similarly, number of four digit natural numbers whose digit at thousandths place is 2 = 4  4  4  = 64

Now, consider four digit natural numbers whose digit at thousandths place is 4:

Here, if the digit at hundredths place is 1, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.

If the digit at hundredths place is 2, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.

If the digit at hundredths place is 3 and the digit at tens place is 1, then ones place can be filled in 4 ways.

If the digit at hundredths place is 3 and the digit at tens place is 2, then ones place can be filled only in 1 way so that the number formed is not exceeding 4321.

Number of four digit natural numbers not exceeding 4321 and digit at thousandths place is 3 = 4  4 + 4  4 + 4 + 1 = 37

Thus, required number of four digit natural numbers not exceeding 4321 is 64 + 64 + 64 + 37 = 229.Question 34

How many numbers of six digits can be formed from the digits 0, 1, 3, 5, 7, and 9 when on digit is repeated? How many of them are divisible by 10?Solution 34

Question 35

Solution 35

Question 36

Solution 36

Question 37

Solution 37

Question 38

Solution 38

Question 39

Solution 39

Question 40

Solution 40

Question 41

Solution 41

Question 42

Solution 42

Question 43

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

In how many ways can 5 different balls be distributed among three boxes?Solution 46

Question 47(i)

Solution 47(i)

Question 47(ii)

Solution 47(ii)

Question 47(iii)

Solution 47(iii)

Question 48

There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.Solution 48

Each lamps has two possibilities either it can be switched on or off.

There are 10 lamps in the hall.

So the total numbers of possibilities are 2¹⁰.

To illuminate the hall we require at least one lamp is to be switched on.

There is one possibility when all the lamps are switched off. If all the bulbs are switched off then hall will not be illuminated.

So the number of ways in which the hall can be illuminated is 2¹⁰-1.

Chapter 16 Permutations Exercise Ex. 16.3

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(i)

Solution 1(i)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

All the letters of the word ‘EAMCOT’ are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other.Solution 32

Chapter 16 Permutations Exercise Ex. 16.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 16 Permutations Exercise Ex. 16.5

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 1(vi)

Solution 1(vi)

Question 1(vii)

Solution 1(vii)

Question 1(viii)

Solution 1(viii)

Question 1(ix)

Solution 1(ix)

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

How many permulations of the letters of the world ‘MADHUBANI’ do not begain with M but end with I?Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Chapter 15 Linear Inequations Exercise Ex. 15.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Chapter 15 Linear Inequations Exercise Ex. 15.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solve the system of equation Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solve the system of equation 

Solution 21

Chapter 15 Linear Inequations Exercise Ex. 15.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Chapter 15 Linear Inequations Exercise Ex. 15.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Chapter 15 Linear Inequations Exercise Ex. 15.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 15 Linear Inequations Exercise Ex. 15.6

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 1(v)

Solution 1(v)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 6(iv)

Solution 6(iv)

Question 7

Show that the following system of linear equations has no solution:

X + 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1.Solution 7

Question 8

Show that the solution set of the following system of linear inequalities is an unbounded region 2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0.Solution 8

Chapter 14 Quadratic Equations Exercise Ex. 14.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Chapter 14 Quadratic Equations Exercise Ex. 14.2

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 2(v)

Solution 2(v)

Question 2(vi)

Solution 2(vi)

Question 2(vii)

Solution 2(vii)

Question 2(viii)

Solution 2(viii)

Question 2(ix)

Solution 2(ix)

Question 2(x)

Solution 2(x)

Question 2(xi)

Solution 2(xi)

Question 2(xii)

Solution 2(xii)

Exercise 3A

Page No 45:

Question 1:

Write the next three whole numbers after 30999.

ANSWER:

The next three whole numbers after 30999 are 31000, 31001, and 31002.

Page No 45:

Question 2:

Write the three whole numbers occurring just before 10001.

ANSWER:

Three whole numbers occurring just before 10001 are as follows:

10001 − 1 = 10000
10000 − 1 = 9999
9999 − 1 = 9998

∴ The three whole numbers just before 10001 are 10000, 9999 and 9998.

Page No 45:

Question 3:

How many whole numbers are there between 1032 and 1209?

ANSWER:

Number of whole numbers between 1032 and 1209 = (1209 − 1032) − 1
                                                                                      = 177 − 1
                                                                                        = 176

Page No 45:

Question 4:

Which is the smallest whole number?

ANSWER:

0 (zero) is the smallest whole number.

All the natural numbers along with 0 are called whole numbers.

Page No 45:

Question 5:

Write the successor of:
(i) 2540801
(ii) 9999
(iii) 50904
(iv) 61639
(v) 687890
(vi) 5386700
(vii) 6475999
(viii) 9999999

ANSWER:

(i) Successor of 2540801 = 2540801 + 1 = 2540802
(ii) Successor of 9999 = 9999 + 1 = 10000
(iii) Successor of 50904 = 50904 + 1 = 50905 
(iv) Successor of 61639 = 61639 + 1 = 61640
(v) Successor of 687890 = 687890 + 1 = 687891
(vi) Successor of 5386700 = 5386700 + 1 = 5386701
(vii) Successor of 6475999 = 6475999 + 1 = 6476000
(viii) Successor of 9999999 = 9999999 + 1 = 10000000

Page No 46:

Question 6:

Write the predecessor of:
(i) 97
(ii) 10000
(iii) 36900
(iv) 7684320
(v) 1566391
(vi) 2456800
(vii) 100000
(viii) 1000000

ANSWER:

(i) Predecessor of 97 = 97 − 1 = 96
(ii) Predecessor of 10000 = 10000 − 1 = 9999
(iii) Predecessor of 36900 = 36900 − 1 = 36899
(iv) Predecessor of 7684320 = 7684320 − 1 = 7684319
(v) Predecessor of 1566391 = 1566391 − 1 = 1566390
(vi) Predecessor of 2456800 = 2456800 − 1 = 2456799
(vii) Predecessor of 100000 = 100000 − 1 = 99999
(viii) Predecessor of 1000000 = 1000000 − 1 = 999999

Page No 46:

Question 7:

Write down three consecutive whole numbers just preceding 7510001.

ANSWER:

The three consecutive whole numbers just preceding 7510001 are as follows:

7510001 − 1 = 7510000
7510000 − 1 = 7509999
7509999 − 1 = 7509998

∴ The three consecutive numbers just preceding 7510001 are 7510000, 7509999 and 7509998.

Page No 46:

Question 8:

Write (T) for true and (F) for false against each of the following statements:
(i) Zero is the smallest natural number.
(ii) Zero is the smallest whole number.
(iii) Every whole number is a natural number.
(iv) Every natural number is a whole number.
(v) 1 is the smallest whole number.
(vi) The natural number 1 has no predecessor.
(vii) The whole number  1 has no predecessor.
(viii) The whole number 0 has no predecessor.
(ix) The predecessor of a two-digit number is never a single-digit number.
(x) The successor of a two-digit number is always a two-digit number.
(xi) 500 is the predecessor of 499.
(xii) 7000 is the successor of 6999.

ANSWER:

(i) False. 0 is not a natural number.1 is the smallest natural number.
(ii) True.
(iii) False. 0 is a whole number but not a natural number.
(iv) True. Natural numbers include 1,2,3 …, which are whole numbers.
(v) False. 0 is the smallest whole number.
(vi) True. The predecessor of 1 is 1 − 1 = 0, which is not a natural number.
(vii) False. The predecessor of 1 is 1 − 1 = 0, which is a whole number.
(viii) True. The predecessor of 0 is 0 − 1 = −1, which is not a whole number.
 (ix) False. The predecessor of a two-digit number can be a single digit number. For example, the predecessor of 10 is 10 − 1, i.e., 9.
(x) False. The successor of a two-digit number is not always a two-digit number. For example, the successor of 99 is 99 + 1, i.e., 100.
(xi) False. The predecessor of 499 is 499 − 1, i.e., 498.
(xii) True. The successor of 6999 is 6999 + 1, i.e., 7000.

Page No 48:

Exercise 3B

Question 1:

Fill in the blanks to make each of the following a true statement:
(i) 458 + 639 = 639 + ……
(ii) 864 + 2006 = 2006 + ……
(iii) 1946 + …… = 984 + 1946
(iv) 8063 + 0 = ……
(v) 53501 + (574 + 799) = 574 + (53501 + ……)

ANSWER:

(i) 458 + 639 = 639 + 458
(ii) 864 + 2006 = 2006 + 864
​(iii) 1946 + 984 = 984 + 1946
(iv) 8063 + 0 = 8063
(v) 53501 + (574 + 799) = 574 + (53501 + 799)

Page No 48:

Question 2:

Add the following numbers and check by revershing the order of the addends:
(i) 16509 + 114
(ii) 2359 + 548
(iii) 19753 + 2867

ANSWER:

(i) 16509 + 114 = 16623
 By reversing the order of the addends, we get:
  114 + 16509 = 16623  
∴ 16509 + 114 = 114 + 16509

(ii) 2359 + 548 = 2907 
 By reversing the order of the addends, we get:
     548 + 2359 = 2907
∴ 2359 + 548 = 548 + 2359

(iii) 19753 + 2867 = 22620
   By reversing the order of the addends, we get:
     2867 + 19753 = 22620
∴ 19753 + 2867 = 2867 + 19753
   

Page No 48:

Question 3:

Find the sum: (1546 + 498) + 3589.
Also, find the sum: 1546 + (498 + 3589).
Are the two sums equal?
State the property satisfied.

ANSWER:

We have:

(1546 + 498) + 3589 = 2044 + 3589 = 5633

Also, 1546 + (498 + 3589) = 1546 + 4087 = 5633

Yes, the two sums are equal.

The associative property of addition is satisfied.

Page No 48:

Question 4:

Determine each of the sums given below using suitable rearrangement.
(i) 953 + 707 + 647
(ii) 1983 + 647 + 217 + 353
(iii) 15409 + 278 + 691 + 422
(iv) 3259 + 10001 + 2641 + 9999
(v) 1 + 2 + 3 + 4 + 96 + 97 + 98 + 99
(vi) 2 + 3 + 4 + 5 + 45 + 46 + 47 + 48

ANSWER:

(i) 953 + 707 + 647
953 + (707 + 647)                                   (Using associative property of addition)
= 953 + 1354  
= 2307   
 
(ii) 1983 + 647 + 217 + 353
(1983 + 647)  + (217 +353)                    (Using associative property of addition)
= 2630 + 570
=  3200


(iii) 15409 + 278 + 691 + 422
(15409 + 278) + (691 + 422)                     (Using associative property of addition)
= 15687 + 1113
= 16800

(iv) 3259 + 10001 + 2641 + 9999
(3259 + 10001) + (2641 +  9999)             (Using associative property of addition)
= 13260 + 12640
= 25900

(v)1 + 2 + 3 + 4 + 96 + 97 + 98 + 99
(1 + 2 + 3 + 4) + (96 + 97 + 98 + 99)       (Using associative property of addition)
= (10) + (390)
=  400

(vi) 2 + 3 + 4 + 5 + 45 + 46 + 47 + 48
(2 + 3 + 4 + 5) + (45 + 46 + 47 + 48)                 (Using associative property of addition)
= 14 + 186  
= 200  

Page No 48:

Question 5:

Find the sum by short method:
(i) 6784 + 9999
(ii) 10578 + 99999

ANSWER:

(i)  6784 + 9999
=  6784 + (10000 − 1)
=  (6784 + 10000) − 1                              (Using associative property of addition)
= 16784 − 1
= 16783

(ii) 10578 + 99999
= 10578 + (100000 − 1)
= (10578 + 100000) − 1                         (Using associative property of addition)
= 110578 − 1
= 110577

Page No 48:

Question 6:

For any whole numbers a, b, c, is it true that (a + b) + c = a + (c + b)? Give reasons.

ANSWER:

For any whole numbers a, b and c, we have:
(a + b) + c = a + ( b + c​) 

Let a = 2, b = 3 and c = 4 [we can take any values for a, b and c]

LHS = (a + b​) + c
= (2 + 3) + 4
= 5 + 4
= 9


RHS = a + (c + b)
= a + (b + c)       [∵ Whole numbers follow the commutative law]                    
= 2 + (3 + 4)
= 2 + 7
= 9

∴ This shows that associativity (in addition) is one of the properties of whole numbers.

Page No 48:

Question 7:

Complete each one of the following magic squares by supplying the missing numbers:
(i)

	9	2
	5
8

(ii)

16	2
	10
		4

(iii) 

2	15	16
9	12
		7	10
14			17

(iv) 

	18	17	4
		14	11
	9	10
19			16

ANSWER:

  In a magic square, the  sum of each row is equal to the sum of each column and the sum of each main diagonal.  By using this concept, we have:
(i)

4	9	2
3	5	7
8	1	6

(ii)

16	2	12
6	10	14
8	18	4

(iii)

2	15	16	5
9	12	11	6
13	8	7	10
14	3	4	17

(iv)

7	18	17	4
8	13	14	11
12	9	10	15
19	6	5	16

Page No 48:

Question 8:

Write (T) for true and (F) for false for each of the following statements:
(i) The sum of two odd numbers is an odd number.
(ii) The sum of two even numbers is an even number.
(iii) The sum of an even number and an odd number is an odd number.

ANSWER:

(i)  F (false). The sum of two odd numbers may not be an odd number. Example: 3 + 5 = 8, which is an even number.
  
(ii) T (true). The sum of two even numbers is an even number. Example: 2 + 4 = 6, which is an even number.
  
(iii) T (true). The sum of an even and an odd number is an odd number. Example: 5 + 4 = 9, which is an odd number.

Page No 49

Exercise 3C

Question 1:

Perform the following subtractions. Check your results by the corresponding additions.
(i) 6237 − 694
(ii) 21205 − 10899
(iii) 100000 − 78987
(iv) 1010101 − 656565

ANSWER:

(i) Subtraction: 6237 − 694 = 5543
   Addition: 5543 + 694 = 6237


(ii) Subtraction: 21205 − 10899 = 10306
  Addition: 10306 + 10899 = 21205


(iii) Subtraction: 100000 − 78987 = 21013
  Addition: 21013 + 78987 = 100000 


(iv) Subtraction: 1010101 − 656565 = 353536
 Addition: 353536 + 656565 = 1010101 

Page No 49:

Question 2:

Replace each * by the correct digit in each of the following:
(i) 
(ii) 
(iii) 
(iv) 

ANSWER:

(i)   917 − *5* =  5*8

  
  ⇒ 917 − 359 =  558

(ii) 6172 − **69 =  29**
    

⇒ 6172 − 3269 = 2903

(iii) 5001003 − **6987 =  484****
  

 ⇒ 5001003 − 155987 = 4845016

(iv)  1000000 − ****1 = *7042*
    

  ⇒ 1000000 − 29571 = 970429

Page No 49:

Question 3:

Find the difference:
(i) 463 − 9
(ii) 5632 − 99
(iii) 8640 − 999
(iv) 13006 − 9999

ANSWER:

(i) 463 − 9
= 463 − 10 + 1
= 464 − 10
= 454

(ii) 5632 − 99
= 5632 − 100 + 1
= 5633 − 100
=  5533

(iii) 8640 − 999
= 8640 − 1000 + 1
= 8641 − 1000
= 7641

(iv) 13006 − 9999
= 13006 − 10000 + 1
= 13007 − 10000
= 3007

Page No 50:

Question 4:

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

ANSWER:

Smallest seven-digit number = 1000000
Largest four-digit number = 9999              
∴ Their difference = 1000000 − 9999
                               =1000000 − 10000 + 1
                                =1000001 − 10000
                                 =990001

Page No 50:

Question 5:

Ravi opened his account in a bank by depositing Rs 136000. Next day he withdrew Rs 73129 from it. How much money was left in his account?

ANSWER:

Money deposited by Ravi = Rs 1,36,000
Money withdrawn by Ravi= Rs 73,129
Money left in his account  =  money deposited − money withdrawn
                                           = Rs (136000 − 73129)
                                             = Rs 62871

∴ Rs 62,871 is left in Ravi’s account.

Page No 50:

Question 6:

Mrs Saxena withdrew Rs 100000 from her bank account. She purchased a TV set for Rs 38750, a refrigerator for Rs 23890 and jewellery worth Rs 35560. How much money was left with her?

ANSWER:

Money withdrawn by Mrs Saxena  = Rs 1,00,000
Cost of the TV set = Rs 38,750
Cost of the refrigerator = Rs 23,890
Cost of the jewellery = Rs 35,560
Total money spent = Rs (38750 + 23890 + 35560) = Rs 98200

Now, money left = money withdrawn − money spent
                            = Rs (100000 − 98200)
                            = Rs 1800

∴ Rs 1,800 is left with Mrs Saxena.

Page No 50:

Question 7:

The population of a town was 110500. In one year it increased by 3608 due to new births. However, 8973 persons died or left the town during the year. What was the population at the end of the year?

ANSWER:

Population of the town = 110500
Increased population = 110500 + 3608 = 114108
Number of persons who died or left the town = 8973
Population at the end of the year = 114108 − 8973 = 105135

∴ The population at the end of the year will be 105135.

Page No 50:

Question 8:

Find the whole number n when:
(i) n + 4 = 9
(ii) n + 35 = 101
(iii) n − 18 = 39
(iv) n − 20568 = 21403

ANSWER:

(i) n + 4 = 9
⇒ n = 9 − 4 = 5

(ii) n + 35 = 101
⇒ n = 101 − 35 = 66

(iii) n – 18 = 39
 ⇒ n =  18 + 39 = 57

(iv) n − 20568 = 21403
  ⇒ n  = 21403 + 20568 = 41971

Page No 53:

Exercise 3D

Question 1:

Fill in the blanks to make each of the following a true statement:
(i) 246 × 1 = ……
(ii) 1369 × 0 = …….
(iii) 593 × 188 = 188 × …….
(iv) 286 × 753 = …… × 286
(v) 38 × (91 × 37) = …… × (38 × 37)
(vi) 13 × 100 × …… = 1300000
(vii) 59 × 66 + 59 × 34 = 59 × (…… + ……)
(viii) 68 × 95 = 68 × 100 − 68 × …….

ANSWER:

(i) 246 × 1 = 246
(ii) 1369 × 0 = 0
(iii) 593 × 188  = 188 × 593
(iv) 286 × 753 = 753 × 286
(v) 38 × (91 × 37) = 91 × (38 × 37)
(vi) 13 × 100 × 1000 = 1300000
(vii) 59 × 66 + 59  ×  34 = 59 × ( 66 + 34)
(viii) 68 × 95 = 68 × 100 − 68 × 5 

Page No 53:

Question 2:

State the property used in each of the following statements:
(i) 19 × 17 = 17 × 19
(ii) (16 × 32) is a whole number
(iii) (29 × 36) × 18 = 29 × (36 × 18)
(iv) 1480 × 1 = 1480
(v) 1732 × 0 = 0
(vi) 72 × 98 + 72 × 2 = 72 × (98 + 2)
(vii) 63 × 126 − 63 × 26 = 63 × (126 − 26)

ANSWER:

(i) Commutative law in multiplication
(ii) Closure property
(iii) Associativity of multiplication
(iv) Multiplicative identity 
(v) Property of zero
(vi) Distributive law of multiplication over addition
(vii) Distributive law of multiplication over subtraction

Page No 53:

Question 3:

Find the value of each of the following using various properties:
(i) 647 × 13 + 647 × 7
(ii) 8759 × 94 + 8759 × 6
(iii) 7459 × 999 + 7459
(iv) 9870 × 561 − 9870 × 461
(v) 569 × 17 + 569 × 13 + 569 × 70
(vi) 16825 × 16825 − 16825 × 6825

ANSWER:

(i) 647 × 13 + 647 × 7
=  647 × (13 + 7)
= 647 ×  20
= 12940                                 (By using distributive property)

(ii)  8759 × 94 + 8759 × 6
= 8759 × (94 + 6)
= 8759 ×  100
= 875900                              (By using distributive property)

(iii) 7459 × 999 + 7459
= 7459× (999 + 1)
= 7459 × 1000
= 7459000                         (By using distributive property)

(iv) 9870 × 561 − 9870 × 461
= 9870 × (561 − 461)
= 9870 × 100                           
= 987000                           (By using distributive property)

(v)  569 × 17 + 569 × 13 + 569 × 70
    = 569 × (17+ 13+ 70)
   = 569  × 100
   = 56900                            (By using distributive property)

(vi) 16825 × 16825 − 16825 × 6825
= 16825 × (16825 − 6825)
=  16825 × 10000
= 168250000                        (By using distributive property)

Page No 53:

Question 4:

Determine each of the following products by suitable rearrangements:
(i) 2 × 1658 × 50
(ii) 4 × 927 × 25
(iii) 625 × 20 × 8 × 50
(iv) 574 × 625 × 16
(v) 250 × 60 × 50 × 8
(vi) 8 × 125 × 40 × 25

ANSWER:

(i) 2 × 1658 × 50
= (2 × 50) × 1658
= 100 × 1658
= 165800

(ii) 4 × 927 × 25
= (4 × 25) × 927
= 100 × 927
= 92700

(iii) 625 × 20 × 8 × 50  
= (20  × 50) ×  8 × 625
= 1000 ×  8 × 625
= 8000 × 625
= 5000000

(iv) 574 × 625 × 16
= 574 × (625 × 16)
=  574 × 10000
= 5740000

(v)  250 × 60 × 50 × 8
= (250 × 8) × (60 × 50)
=  2000  × 3000
=  6000000

(vi)  8 × 125 × 40 × 25
=  (8 × 125) × (40 × 25)
= 1000 × 1000
= 1000000

Page No 53:

Question 5:

Find each of the following products, using distributive laws:
(i) 740 × 105
(ii) 245 × 1008
(iii) 947 × 96
(iv) 996 × 367
(v) 472 × 1097
(vi) 580 × 64
(vii) 439 × 997
(viii) 1553 × 198

ANSWER:

(i)  740 × 105
= 740 × (100 + 5)
= 740 × 100 + 740 × 5                    (Using distributive law of multiplication over addition)
= 74000 + 3700
= 77700

(ii) 245 × 1008
= 245 × (1000 + 8)
= 245 × 1000 + 245 × 8                  (Using distributive law of multiplication over addition)
= 245000 + 1960
= 246960

(iii) 947 × 96
= 947 × ( 100 − 4)
=  947 × 100 − 947 × 4                    (Using distributive law of multiplication over subtraction)
= 94700 − 3788
= 90912

(iv)  996 × 367
=  367 × (1000 − 4)
=   367 × 1000 − 367 × 4             (Using distributive law of multiplication over subtraction)
= 367000 × 1468
= 365532

(v) 472 × 1097
= 472 × ( 1000 + 97)  
= 472 × 1000 + 472 × 97                 (Using distributive law of multiplication over addition)
= 472000 + 45784
 = 517784

(vi)  580 × 64
=  580 × (60 + 4)
=  580 × 60 + 580 × 4                        (Using distributive law of multiplication over addition)
= 34800 + 2320
= 37120

(vii) 439 × 997
= 439 × (1000 − 3)
= 439 × 1000 − 439 × 3                   (Using distributive law of multiplication over subtraction)
= 439000 − 1317
= 437683

(viii) 1553 × 198
= 1553 × (200 − 2)
= 1553 × 200 − 1553 × 2                 (Using distributive law of multiplication over subtraction)
= 310600 − 3106
= 307494

Page No 53:

Question 6:

Find each of the following products, using distributive laws:
(i) 3576 × 9
(ii) 847 × 99
(iii) 2437 × 999

ANSWER:

Distributive property of multiplication over addition states that a (b + c) = ab + ac
Distributive property of multiplication over subtraction  states that a (b − c) = ab – ac
(i) 3576  ​×  9
= 3576 × (10 − 1)
= 3576 ​× 10 − 3576 × 1
= 35760 − 3576
= 32184

(ii) 847 ×  99
= 847 × (100 − 1)
= 847 × 100 − 847 × 1
=  84700 − 847
= 83853

(iii) 2437 × 999
= 2437 × (1000 − 1)
= 2437 × 1000 − 2437 × 1
=  2437000 − 2437
= 2434563

Page No 54:

Question 7:

Find the products:
(i) 

(ii) 

(iii) 

(iv) 

ANSWER:

(i)

458 × 67 = 30686

(ii)

3709 × 89 = 330101

(iii)


4617 × 234 = 1080378

(iv)

15208 × 542 = 8242736

Page No 54:

Question 8:

Find the product of the largest 3-digit number and the largest 5-digit number.

ANSWER:

Largest three-digit number = 999
Largest five-digit number = 99999
∴ Product of the two numbers = 999 × 99999
                                                    = 999 × (100000 − 1)                  (Using distributive law)
                                                     = 99900000 − 999
                                                      = 99899001

Page No 54:

Question 9:

A car moves at a uniform speed of 75 km per hour. How much distance will it cover in 98 hours?

ANSWER:

Uniform speed of a car = 75 km/h

Distance = speed × time
               = 75 × 98
               =75 × (100 − 2)                     (Using distributive law)
               =75 × 100 − 75 × 2
               =7500 − 150
              = 7350 km

∴ The distance covered in 98 h is 7350 km.

Page No 54:

Question 10:

A dealer purchased 139 VCRs. If the cost of each set is Rs 24350, find the cost of all the sets together.

ANSWER:

Cost of 1 VCR set = Rs 24350 
Cost of 139 VCR sets = 139 × 24350
                                  =24350 × (140 − 1)                 (Using distributive property)
                                  =24350 × 140 − 24350
                                  =3409000 − 24350
                                   = Rs. 3384650 

∴ The cost of all the VCR sets is Rs 33,84,650.

Page No 54:

Question 11:

A housing society constructed 197 houses. If the cost of construction for each house is Rs 450000, what is the total cost for all the houses?

ANSWER:

Cost of construction of 1 house = Rs 450000
Cost of construction of 197 such houses = 197 × 450000
                                                                = 450000 × (200 − 3)
                                                                = 450000 × 200 − 450000 × 3               [Using distributive property of multiplication over subtraction]
                                                                = 90000000 − 1350000
                                                                = 88650000 

∴ The total cost of construction of 197 houses is Rs 8,86,50,000.

Page No 54:

Question 12:

50 chairs and 30 blackboards were purchased for a school. If each chair costs Rs 1065 and each blackboard costs Rs 1645, find the total amount of the bill.

ANSWER:

Cost of a chair = Rs 1065
Cost of a blackboard = Rs 1645
Cost of 50 chairs = 50 × 1065 = Rs 53250
Cost of 30 blackboards = 30 × 1645 = Rs 49350
∴ Total amount of the bill = cost of 50 chairs + cost of 30 blackboards 
                                           = Rs (53250 + 49350)
                                            = Rs 1,02,600

Page No 54:

Question 13:

There are six sections of Class VI in a school and there are 45 students in each section. If the monthly charges from each student be Rs 1650, find the total monthly collection from Class VI.

ANSWER:

Number of student in 1 section = 45
Number of students in 6 sections = 45 × 6 = 270
Monthly charges from 1 student = Rs 1650
∴ Total monthly collection from class VI = Rs 1650 × 270 = Rs 4,45,500

Page No 54:

Question 14:

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

ANSWER:

If the product of two whole numbers is zero, then one of them is definitely zero.
Example: 0 × 2 = 0 and 0 × 15 = 0

If the product of whole numbers is zero, then both of them may be zero.
i.e., 0 × 0 = 0

Now, 2 × 5 = 10. Here, the product will be non-zero because the numbers to be multiplied are not equal to zero.

Page No 54:

Question 15:

Fill in the blanks:
(i) Sum of two odd numbers is an …… number.
(ii) Product of two odd numbers is an …… number.
(iii) a ≠ 0a ≠ 0 and a × a = a ⇒⇒ a = ?

ANSWER:

(i) Sum of two odd numbers is an even number. Example: 3 + 5 = 8, which is an even number.
(ii) Product of two odd numbers is an odd number. Example: 5 × 7 = 35, which is an odd number.
(iii)  a ≠ 0 anda × a = a 
     Given: a × a = a
                ⇒ a = aa=1aa=1, a ≠ 0             

Page No 56:

Exercise 3E

Question 1:

Divide and check your answer by the corresponding multiplication in each of the following:
(i) ​1936 ​÷ 36
(ii) 19881 ​÷ 47
(iii) 257796 ​÷ 341
(iv) 612846 ​÷ 582
(v) 34419 ​÷ 149
(vi) 39039 ​÷ 1001

ANSWER:

(i)


Dividend = 1936, Divisor = 36 , Quotient = 53 , Remainder = 28
Check: Divisor × Quotient + Remainder  =  36 × 53 + 28
                                                              = 1936
                                                              =Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.
(ii) 19881 ​÷ 47
    
 Dividend = 19881, Divisor = 47 , Quotient = 423, Remainder = 0
Check: Divisor ×Quotient + Remainder= 47 × 423 + 0
                                                           = 19881
                                                           =Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

(iii)

Dividend = 257796 , Divisor = 341 , Quotient = 756 , Remainder = 0
Check : Divisor × Quotient + Remainder = 341 × 756 + 0
                                                              = 257796
                                                             = Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

(iv) 612846 ​÷ 582 
   

Dividend = 612846 , Divisor = 582, Quotient = 1053 , Remainder = 0
Check :  Divisor × Quotient + Remainder= 582 × 1053 + 0
                                                              = 612846
                                                              =Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

(v) 34419 ​÷ 149
    
Dividend = 34419, Divisor = 149 , Quotient = 231, Remainder = 0
Check : Divisor × Quotient + Remainder  = 149 × 231 + 0
                                                               = 34419
                                                               =Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.
 
(vi) 39039 ​÷ 1001

  
Dividend = 39039 , Divisor = 1001 , Quotient = 39 , Remainder = 0
Check : Divisor × Quotient + Remainder = 1001 × 39 + 0
                                                              = 39039
                                                              =Dividend
Hence, Dividend = Divisor × Quotient + Remainder
Verified.

Page No 56:

Question 2:

Divide, and find out the quotient and remainder. Check your answer.
(i) 6971 ​÷ 47
(ii) 4178 ​÷ 35
(iii) 36195 ​÷ 153
(iv) 93575 ​÷ 400
(v) 23025 ​÷ 1000
(vi) 16135 ​÷ 875

ANSWER:

(i)  6971 ​÷ 47
     
Quotient = 148 and Remainder = 15
Check: Divisor × Quotient + Remainder = 47 × 148 + 15
                                                                  = 6971
                                                                 = Dividend
∴ Dividend = Divisor × Quotient + Remainder 
Verified.

(ii)  4178 ​÷ 35

       
Dividend = 119 and Remainder = 13
Check: Divisor × Quotient + remainder = 35 ×  119 + 13 
                                                                    = 4178
                                                                   = Dividend

∴ Dividend= Divisor × Quotient + Remainder
Verified.
(iii) 36195 ​÷ 153
     
Quotient = 236 and Remainder = 87
Check: Divisor × Quotient + Remainder =  153 × 236 + 87
                                                            = 36195
                                                            = Dividend
∴ Dividend= Divisor × Quotient +Remainder
Verified.
(iv) 93575 ​÷ 400
      

Quotient = 233 and Remainder = 375
Check: Divisor × Quotient + Remainder =  400 ×  233 + 375
                                                              = 93575
                                                              = Dividend
∴ Dividend= Divisor × Quotient + Remainder
Verified.

(v)  23025 ​÷ 1000 
        
Quotient = 23 and remainder = 25
Check: Divisor × Quotient + Remainder =1000  × 23 + 25
                                                                = 23025
                                                                = Dividend
∴ Dividend= Divisor × Quotient +Remainder
Verified.
(vi) 16135 ​÷ 875
        
Quotient = 18 and Remainder = 385
Check: Divisor × Quotient + Remainder =875  ×  18 + 385
                                                                 = 16135
                                                                = Dividend
∴ Dividend= Divisor × Quotient +Remainder
Verified.

Page No 56:

Question 3:

Find the value of
(i) 65007 ​÷ 1
(ii) 0 ​÷ 879
(iii) 981 + 5720 ​÷ 10
(iv) 1507 − (625 ÷ 25)
(v) 32277 ÷ (648 − 39)
(vi) (1573 ÷ 1573) − (1573 ÷ 1573)

ANSWER:

(i) 65007 ​÷ 1 = 65007

(ii) 0 ​÷ 879  = 0

(iii) 981 + 5720 ​÷ 10
= 981 + (5720 ​÷ 10)                               (Following DMAS property)
= 981 + 572
= 1553

(iv) 1507 − (625 ÷ 25)                             (Following BODMAS property)
= ​​1507 − 25
= 1482

(v) 32277 ÷ (648 − 39)                                 (Following BODMAS property)
= ​32277 ÷ (609)
=  53

(vi)  (1573 ÷ 1573) − (1573 ÷ 1573)            (Following BODMAS property)
= 1 − 1
= 0 

Page No 56:

Question 4:

Find a whole number n such that n ÷ n = n.

ANSWER:

Given:  n ÷ n = n
⇒  nnnn= n 
​⇒  n = n^​2

i.e., the whole number n is equal to n^​2.

∴ The given whole number must be 1.                  

Page No 56:

Question 5:

The product of two numbers is 504347. If one of the numbers is 317, find the other.

ANSWER:

Let x and y be the two numbers.

Product of the two numbers = x × y = 504347

If x = 317, we have:

317 × y = 504347
⇒ y = 504347 ÷ 317
      


 y=  1591

∴ The other number is 1591.

Page No 56:

Question 6:

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

ANSWER:

Dividend = 59761, quotient = 189, remainder = 37 and divisor = ?
  
Dividend = divisor × quotient + remainder
⇒ 59761 = divisor × 189 + 37
⇒ 59761 − 37 = divisor × 189
⇒ 59724 = divisor × 189
⇒ Divisor = 59724 ​÷ 189                  
              
                                                  
Hence, divisor =316                                    
                                                            

Page No 56:

Question 7:

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

ANSWER:

Here, Dividend = 55390, Divisor = 299 and Remainder = 75
We have to find the quotient.
Now, Dividend = Divisor × Quotient + Remainder 
⇒ 55390 = 299 × Quotient + 75
⇒ 55390 − 75 = 299 × Quotient
⇒ 55315 = 299 × Quotient 
⇒ Quotient = 55315 ​÷ 299                                 
                                                         
                                                                           
Hence, quotient =185       

Page No 56:

Question 8:

What least number must be subtracted from 13601 to get a number exactly divisible by 87?

ANSWER:

First, we will divide 13601 by 87.



Remainder = 29
So, 29 must be subtracted from 13601 to get a number exactly divisible by 87.
i.e., 13601 − 29 = 13572

Now, we have:

  
∴ 29 must be subtracted from 13601 to make it divisible by 87.

Page No 56:

Question 9:

What least number must be added to 1056 to get a number exactly divisible by 23?

ANSWER:

First, we will divide 1056 by 23.
        

Required number = 23 − 21 = 2
So, 2 must be added to 1056 to make it exactly divisible by 23.
i.e., 1056 + 2 = 1058

Now, we have:



∴ 1058 is exactly divisible by 23.
                         

Page No 56:

Question 10:

Find the largest 4-digit number divisible by 16.

ANSWER:

We have to find the largest four digit number divisible by 16 .
The largest four-digit number = 9999
Therefore, dividend =9999
Divisor =16


Here, we get remainder =15
Therefore, 15 must be subtracted from 9999 to get the largest four digit number that is divisible by 16.
i.e., 9999 − 15 = 9984

Thus, 9984 is the largest four-digit number that is divisible by 16.

Page No 56:

Question 11:

Divide the largest 5 digit number by 653. Check your answer by the division algorithm.

ANSWER:

Largest five-digit number =99999
  
Dividend = 99999, Divisor = 653, Quotient = 153 and Remainder = 90
Check: Divisor ×Quotient + Remainder
         = 653 × 153 + 90
         = 99909 + 90
         = 99999
         = Dividend

∴ Dividend = Divisor × Quotient + Remainder                  
   Verified.

Page No 56:

Question 12:

Find the least 6-digit number exactly divisible by 83.

ANSWER:

Least six-digit number = 100000
Here, dividend = 100000 and divisor = 83


 
In order to find the least 6-digit number exactly divisible by 83, we have to add 83 − 68 = 15 to the dividend.
I.e., 100000 + 15 = 100015

So, 100015 is the least six-digit number exactly divisible by 83.

Page No 56:

Question 13:

1 dozen bananas cost Rs 29. How many dozens can be purchased for Rs 1392?

ANSWER:

Cost of 1 dozen bananas = Rs 29
Number of dozens purchased for Rs 1392 = 1392 ÷ 29




Hence, 48 dozen of bananas can be purchased with Rs. 1392.

Page No 56:

Question 14:

19625 trees have been equally planted in 157 rows. Find the number of trees in each row.

ANSWER:

Number of trees planted in 157 rows = 19625
Trees planted in 1 row = 19625 ÷ 157



∴ 125 trees are planted in each row.                                  

Page No 56:

Question 15:

The population of a town is 517530. If one out of every 15 is reported to be literate, find how many literate persons are there in the town.

ANSWER:

Population of the town = 517530
(115)115 of the population is reported to be literate, i.e., (115)115 × 517530 = 517530 ÷÷ 15


∴ There are 34502 illiterate persons in the given town.

Page No 56:

Question 16:

The cost price of 23 colour television sets is Rs 570055. Determine the cost price of each TV set if each costs the same.

ANSWER:

Cost price of 23 colour TV sets = Rs 5,70,055
Cost price of 1 TV set  = Rs 570055 ÷ 23

                        
∴ The cost price of one TV set is Rs 24,785.

Page No 56:

Exercise 3F

Question 1:

The smallest whole number is
(a) 1
(b) 0
(c) 2
(d) none of these

ANSWER:

(b) 0

The smallest whole number is 0.

Page No 56:

Question 2:

The least number of 4 digits which is exactly divisible by 9 is
(a) 1018
(b) 1026
(c) 1009
(d) 1008

ANSWER:

(d) 1008


(a)

Hence, 1018 is not exactly divisible by 9.

(b)

Hence, 1026 is exactly divisible by 9.
(c)

Hence, 1009 is not exactly divisible by 9.

(d)

Hence, 1008 is exactly divisible by 9.

(b) and (d) are exactly divisible by 9, but (d) is the least number which is exactly divisible by 9.

Page No 57:

Question 3:

The largest number of 6 digits which is exactly divisible by 16 is
(a) 999980
(b) 999982
(c) 999984
(d) 999964

ANSWER:

(c) 999984

(a)

Hence, 999980 is not exactly divisible by 16.
(b)

Hence, 999982 is not exactly divisible by 16.
(c)

Hence, 999984 is exactly divisible by 16.
(d)

Hence, 999964 is not exactly divisible by 16.

The largest six-digit number which is exactly divisible by 16 is 999984.

          

Page No 57:

Question 4:

What least number should be subtracted from 10004 to get a number exactly divisible by 12?
(a) 4
(b) 6
(c) 8
(d) 20

ANSWER:

(c) 8

Here we have to tell what least number should be subtracted from 10004 to get a number exactly divisible by 12
So, we will first divide 10004 by 12.


Remainder = 8
So, 8 should be subtracted from 10004 to get the number exactly divisible by 12.
i.e., 10004 − 8 = 9996


  Hence, 9996 is exactly divisible by 12.

Page No 57:

Question 5:

What least number should be added to 10056 to get a number exactly divisible by 23?
(a) 5
(b) 18
(c) 13
(d) 10

ANSWER:

(a) 18

Here , we have to tell that what least number must be added to 10056 to get a number exactly divisible by 23
So, first we will divide 10056 by 23


Remainder = 5
Required number = 23 − 5 = 18

So, 18 must be added to 10056 to get a number exactly divisible by 23.
i.e., 10056 + 18 = 10074
  
Hence, 10074 is exactly divisible by 23 .

Page No 57:

Question 6:

What whole number is nearest to 457 which is divisible by 11?
(a) 450
(b) 451
(c) 460
(d) 462

ANSWER:

(d) 462

(a)

Hence, 450 is not divisible by 11.
(b)

Hence, 451 is divisible by 11.
(c)

Hence, 460 is not divisible by 11.
(d)

Hence, 462 is divisible by 11.

Here, the numbers given in options (b) and (d) are divisible by 11. However, we want a whole number nearest to 457 which is divisible by 11.
So, 462 is whole number nearest to 457 and divisible by 11.    

Page No 57:

Question 7:

How many whole numbers are there between 1018 and 1203?
(a) 185
(b) 186
(c) 184
(d) none of these

ANSWER:

(c) 184

Number of whole numbers = (1203 − 1018) − 1
                                             = 185 − 1
                                              =  184

Page No 57:

Question 8:

A number when divided by 46 gives 11 as quotient and 15 as remainder. The number is
(a) 491
(b) 521
(c) 701
(d) 679

ANSWER:

(b) 521

Divisor = 46
Quotient = 11
Remainder = 15
Dividend = divisor × quotient + remainder
             = 46 × 11 + 15
             = 506 + 15
             = 521

Page No 57:

Question 9:

In a division sum, we have dividend = 199, quotient = 16 and remainder = 7. The divisor is
(a) 11
(b) 23
(c) 12
(d) none of these

ANSWER:

(c) 12

Dividend = 199
Quotient = 16
Remainder = 7
According to the division algorithm, we have: 
Dividend = divisor × quotient + remainder
⇒ 199 = divisor × 16 + 7
⇒ 199 − 7 = divisor × 16
⇒ Divisor = 192 ÷ 16
            

Page No 57:

Question 10:

7589 − ? = 3434
(a) 11023
(b) 4245
(c) 4155
(d) none of these

ANSWER:

(a) 11023

7589 − ? = 3434
⇒ 7589 − x = 3434
⇒ x = 7589 + 3434
⇒ x = 11023

Page No 57:

Question 11:

587 × 99 = ?
(a) 57213
(b) 58513
(c) 58113
(d) 56413

ANSWER:

(c) 58113

587 × 99
= 587 × (100 − 1)
= 587 × 100 − 587 × 1         [Using distributive property of multiplication over subtraction]
= 58700 − 587
= 58113

Page No 57:

Question 12:

4 × 538 × 25 = ?
(a) 32280
(b) 26900
(c) 53800
(d) 10760

ANSWER:

(c) 53800

4 × 538 × 25
= (4 × 25) × 538
=  100 × 538  
= 53800

Page No 57:

Question 13:

24679 × 92 + 24679 × 8 = ?
(a) 493580
(b) 1233950
(c) 2467900
(d) none of these

ANSWER:

(c) 2467900

By using the distributive property, we have:
24679 × 92 + 24679 × 8  
= 24679 ×  (92 + 8)
= 24679 × 100
= 2467900

Page No 57:

Question 14:

1625 × 1625 − 1625 × 625 = ?
(a) 1625000
(b) 162500
(c) 325000
(d) 812500

ANSWER:

(a) 1625000

By using the distributive property, we have:

1625 × 1625 − 1625 × 625
= 1625 × (1625 − 625)
=1625 × 1000
= 1625000

Page No 57:

Question 15:

1568 × 185 − 1568 × 85 = ?
(a) 7840
(b) 15680
(c) 156800
(d) none of these

ANSWER:

(c) 156800

By using the distributive property, we have:
1568 × 185 − 1568 × 85
= 1568 × (185 − 85)
= 1568 × 100
= 156800

Page No 57:

Question 16:

(888 + 777 + 555) = (111 × ?)
(a) 120
(b) 280
(c) 20
(d) 140

ANSWER:

(c) 20

(888 + 777 + 555) = (111 × ?)
⇒ (888 + 777 + 555) = 111 × (8 + 7 + 5)          [By taking 111 common]
                                   = ​111 × (20) = 2220

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Question 17:

The sum of two odd numbers is
(a) an odd number
(b) an even number
(c) a prime number
(d) a multiple of 3

ANSWER:

(b) an even number

The sum of two odd numbers is an even number.

Example: 5 + 3 = 8

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Question 18:

The product of two odd numbers is
(a) an odd number
(b) an even number
(c) a prime number
(d) none of these

ANSWER:

(a) an odd number

The product of two odd numbers is an odd number.

Example: 5 × 3 = 15

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Question 19:

If a is a whole number such that a + a = a, then a = ?
(a) 1
(b) 2
(c) 3
(d) none of these

ANSWER:

(d) none of these

Given: a is a whole number such that a + a = a.

If a = 1, then 1+ 1 = 2 ≠ 1 
If a =2, then 2 + 2 = 4 ≠ 2
If a =3, then 3 + 3 = 6 ≠ 3

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Question 20:

The predecessor of 10000 is
(a) 10001
(b) 9999
(c) none of these

ANSWER:

(b) 9999

Predecessor of 10000 = 10000 − 1 = 9999

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Question 21:

The successor of 1001 is
(a) 1000
(b) 1002
(c) none of these

ANSWER:

(b) 1002

Successor of 1001 = 1001 + 1 = 1002

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Question 22:

The smallest even whole number is
(a) 0
(b) 2
(c) none of these

ANSWER:

(b) 2

The smallest even whole number is 2. Zero (0) is neither an even number nor an odd number.

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Exercise 3G

Question 1:

How many whole numbers are there between 1064 and 1201?

ANSWER:

Number of whole numbers between 1201 and 1064 = ( 1201 − 1064 ) − 1
                                                                                    = 137 − 1
                                                                                     = 136

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Question 2:

Fill in the blanks.
     1000000  −****1*7042*      1000000  -****1*7042* 

ANSWER:

   1000000 
−      ****1
                      
      *7042*  

Then, we have:

   1000000
−    29571
                      
     970429

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Question 3:

Use distributive law to find the value of            1063 × 128 − 1063 × 28.

ANSWER:

Using distributive law, we have:
 1063 × 128 − 1063 × 28
= 1063 × (128 − 28)
= 1063 × 100
= 106300

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Question 4:

Find the product of the largest 5-digit number and the largest 3-digit number using distributive law.

ANSWER:

Largest five-digit number = 99999
Largest three-digit number = 999

By using distributive law, we have:

Product = 99999 × 999               
= 99999 × (1000 − 1)                                  [By using distributive law]
= 99999 × 1000 −  99999 × 1
= 99999000 − 99999
= 99899001

OR

999 × 99999
= 999 × ( 100000 − 1)                                   [By using distributive law]
= 999 × 100000 − 999 × 1
= 99900000 − 999
= 99899001

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Question 5:

Divide 53968 by 267 and check the result by the division algorithm.

ANSWER:

 

Dividend = 53968, Divisor = 267, Quotient = 202 and Remainder = 34

Check: Quotient × Divisor + Remainder
         = 267  × 202 + 34
         =  53934 + 34
          = 53968
           = Dividend

∴ Dividend = Quotient × Divisor + Remainder

Verified.

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Question 6:

Find the largest 6-digit number divisible by 16.

ANSWER:

Largest six-digit number = 999999
  
 Remainder = 15

Largest six-digit number divisible by 16 = 999999 − 15 = 999984

∴ 999984 is divisible by 16.

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Question 7:

The cost price of 23 TV sets is Rs 570055. Find the cost of each such set.

ANSWER:

Cost price of 23 TV sets = Rs 5,70,055
Cost price of 1 TV set = 570055 ÷ 23



∴ The cost of one TV set is Rs 24,785.

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Question 8:

What least number must be subtracted from 13801 to get a number exactly divisible by 87?

ANSWER:

We have to find the least number that must be subtracted from 13801 to get a number exactly divisible by 87
So, first we will divide 13801 by 87
   

Remainder = 55
The number 55 should be subtracted from 13801 to get a number divisible by 87.
i.e., 13801 − 55 = 13746

∴ 13746 is divisible by 87.

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Question 9:

The value of (89 × 76 + 89 × 24) is
(a) 890
(b) 8900
(c) 89000
(d) 10420

ANSWER:

(b) 8900

(89 × 76 + 89 × 24)
= 89 × (76 + 24)       [Using distributive property of multiplication over addition]
= 89 × 100
= 8900   

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Question 10:

On dividing a number by 53 we get 8 as quotient and 5 as remainder. The number is
(a) 419
(b) 423
(c) 429
(d) none of these

ANSWER:

(c) 429

Divisor = 53, Quotient = 8, Remainder = 5 and Dividend = ?

Now, Dividend = Quotient × Divisor +Remainder
                         = 8 × 53 + 5 
                          = 429

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Question 11:

The whole number which has no predecessor is
(a) 1
(b) 0
(c) 2
(d) none of these

ANSWER:

(b) 0

The whole number which has no predecessor is 0.

i.e., 0 − 1 = −1, which is not a whole number.

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Question 12:

67 + 33 = 33 + 67 is an example of
(a) closure property
(b) associative property
(c) commutative property
(d) distributive property

ANSWER:

(c) Commutative property

67 + 33 = 33 + 67 is an example of​ commutative property of addition.

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Question 13:

Additive inverse of 36 is
(a) 136136
(b) 0
(c) −36
(d) none of these

ANSWER:

(c) -36
The additive inverse of 36 is −36.

i.e., 36 + (−36) = 0

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Question 14:

Which of the following is not zero?
(a) 0 × 0
(b) 0202
(c) (8 − 8)28 – 82
(d) 2 + 0

ANSWER:

(d) 2+0

(a) 0 × 0 = 0
(b) 0/2 = 0 
(c) (8−8)2=028-82=02 =0
(d) 2 + 0 = 2

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Question 15:

The predecessor of the smallest 3-digit number is
(a) 999
(b) 100
(c) 101
(d) 99

ANSWER:

(d) 99

Smallest three-digit number = 100
∴ Predecessor of 100 = 100 − 1 = 99

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Question 16:

The number of whole numbers between the smallest whole number and the greatest 2-digit number is
(a) 88
(b) 98
(c) 99
(d) 101

ANSWER:

(b) 98
Smallest whole number = 0
Greatest two-digit number = 99
Number of whole numbers between 0 and 99 = (99 − 0 ) − 1 = 98

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Question 17:

Fill in the blanks.
(i) The smallest natural number is …… .
(ii) The smallest whole number is …… .
(iii) Division by …… is not defined.
(iv) …… is a whole number which is not a natural number.
(v)…… is a whole number which is not a natural number.

ANSWER:

(i) The smallest natural number is 1.
(ii) The smallest whole number is 0.
(iii) Division by 0 is not defined.
(iv) 0 is a whole number which is not a natural number.
(v) 1 is the multiplicative identity for whole numbers.

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Question 18:

Write ‘T’ for true and ‘F’ for false in each of the following:
(i) 0 is the smallest natural number.
(ii) Every natural number is a whole number.
(iii) Every whole number is a natural number.
(iv) 1 has no predecessor in whole numbers.

ANSWER:

(i)  F (false). 0 is not a natural number.
​(ii) T (true).  
(iii) F (false). 0 is a whole number but not a natural number.
(iv) F (false). 1 − 1 = 0 is a predecessor of 1, which is a whole number.

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Question 19:

Match the following columns on whole numbers:

column A	column B
(a) 137 + 63 = 63 + 137	(i) Associativity of multiplication
(b) (16 × 25) is a number	(ii) Commutativity of multiplication
(c) 365 × 18 = 18 × 365	(iii) Distributive law of multiplication over addition
(d) (86 × 14) × 25 = 86 × (14 × 25)	(iv) Commutativity of addition
(e) 23 × (80 + 5) = (23 × 80) + (23 × 5)	(v) Closure property for multiplication

ANSWER:

Column A	Column B
(a) 137 + 63 = 63 + 137	(iv) Commutativity of addition
(b) (16 × 25) is a number	(v) Closure property for multiplication
(c) 365 × 18 = 18 × 365	(ii) Commutativity of multiplication
(d) (86 × 14) × 25 = 86 × (14 × 25)	(i) Associativity of multiplication
(e) 23 × (80 + 5) = (23 × 80) + (23 × 5)	(iii) Distributive law of multiplication over addition