IntroductionIntroduction

Counting things is easy for us now. We can count object in large numbers, for example, the number of student in the school, and represent them through numerals. We can also communicate large number using suitable number names.

Find the greatest and smallest number.

Greatest Smallest 4892 437015800 1507325286 2521066895 24569

Multiple choice Question Multiple choice Question [MCQ][MCQ] While writing number from 1to 100

the digit 0 appear how many times?(a)10 times (b)11 times (c) 9 times

(d) 20 timesThe greatest number of 3 digit is(a) 9000 (b) 901 (c) 899 (d) 999

EXERCISE 1.1EXERCISE 1.1

Fill in the blanks.1 lake = 10 ten thousand.1million= 10 hundred thousand.1 core = 10 ten lake.1 core = 10 million.1 million= 10 lakh.

Question 2:Place commas correctly and write the numerals:(a). Seventy three lakh seventy five thousand three hundred seven.(b). Nine crore five lakh forty one.(c). Seven crore fifty two lakh twenty one thousand three hundred two.(d). Fifty eight million four hundred twenty three thousand two hundred two.(e). Twenty three lakh thirty thousand ten.

Answer 2:

(a) 73,75,307(b) 9,05,00,041(c) 7,52,21,302(d) 58,423,202(e) 23,30,010

Question 3:Insert commas suitably and write the names according to Indian System of Numeration:(a). 87595762 (b). 8546283(c). 99900046 (d). 98432701Answer 3:

(a) 8,75,95,762Eight crore seventy five lakh ninety five thousand seven hundred sixty two(b) 85,46,283Eighty five lakh forty six thousand two hundred eighty three(c) 9,99,00,046Nine crore ninety nine lakh forty six(d) 9,84,32,701Nine crore eighty four lakh thirty two thousand seven hundred one

Question 4:Insert commas suitably and write the names according to International System of Numeration:(a). 78921092 (b). 7452283(c). 99985102 (c). 48049831Answer 4:

(a) 78,921,092Seventy eight million nine hundred twenty one thousand ninety two(b) 7,452,283Seven million four hundred fifty two thousand two hundred eighty three(c) 99,985,102Ninety nine million nine hundred eighty five thousand one hundred two(d) 48, 049,831Forty eight million forty nine thousand eight hundred thirty one

EXERCISE EXERCISE 1.21.2Question 1:

A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third, and final day was respectively 1094, 1812, 2050, and 2751. Find the total number of tickets sold on all the four days.Answer1:

Tickets sold on 1st day = 1094Tickets sold on 2nd day = 1812Tickets sold on 3rd day = 2050Tickets sold on 4th day = 2751Total tickets sold = 1094 + 1812 + 2050 + 2751

275120501812

+1094---------------------------- 7707∴ Total tickets sold = 7,707

Question 2:Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10, 000 runs. How many more runs does he need? Answer 2:

Runs scored so far = 6980Runs Shekhar wants to score = 10,000More runs required = 10,000 − 6980

10000 - 6980 ----------- 3020 ∴ Shekhar requires 3,020 more runs.

In an election, the successful candidate registered 5, 77, 500 votes and his nearest rival secured 3, 48, 700 votes. By what margin did the successful candidate win the election?Answer3:

Votes secured by successful candidate = 5,77,500Votes secured by rival = 3,48,700Margin = 5,77,500 − 3,48,700

577500 - 348700 -------------- 228800∴ Margin = 2,28,800

Question 4:Kirti bookstore sold books worth Rs 2,85,891 in the first week of June and books worth Rs 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?Answer 4:

Value of Books sold in 1st week = Rs 2,85,891Value of books sold in 2nd week = Rs 4,00,768Total sale = Sale in 1st week + Sale in 2nd week= 2,85,891 + 4,00,768

285891 + 400768 -------------- 686659The sale for the two weeks together was 6,86,659.Since 4,00,768 > 2,85,891, sale in 2nd week was greater than 1st week.

400768 - 285891 ------------- 114877 ∴ The sale in 2nd week was larger than the sale in 1st week by Rs 1,14,877.

Question 5:Find the difference between the greatest and the least number that can be written using the digits 6, 2, 7, 4, 3 each only once.Answer 5:

Greatest number = 76432Smallest number = 23467Difference = 76432 − 23467

76432 - 23467 ----------------- 52965Therefore, the difference between the greatest and the least number that can be written using the digits 6, 2, 7, 4, 3 each only once is 52,965.

Question 6:A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?Answer 6:

Screws produced in one day = 2,825Days in January = 31Screws produced in 31 days = 2825 × 31

Therefore, screws produced during Jan, 06 = 87,575

Question 7:A merchant had Rs 78,592 with her. She placed an order for purchasing 40 radio sets at Rs 1200 each. How much money will remain with her after the purchase?Answer 7:

Cost of one radio set = Rs 1200Cost of 40 radio sets = 1200 × 40 = Rs 48000Money with Merchant = Rs 78,592Money spent = Rs 48,000Money left = 78592 − 48000 78592 - 48000 ---------------- 30592Therefore, Rs 30,592 will remain with her after the purchase.

Question 8:A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the correct answer? (Hint: Do you need to do both the multiplications?)Answer8:

Difference between 65 and 56 = 9Difference in the answer = 7236 × 9 7236 x 9 -------------- 65124Therefore, his answer was greater than the correct answer by 65,124.

Question 9:To stitch a shirt, 2m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain? (Hint: convert data in cm.)Answer9:

2 m 15 cm = 215 cm (1 m = 100 cm)40 m = 40 × 100= 4000 cmCloth required for one shirt = 215 cmNumber of shirts that can be stitched out of 4000 cm = 4000 ÷ 215

Therefore, 18 shirts can be made. 130 cm, i.e. 1 m 30 cm, cloth will remain.

Question 10:Medicine is packed in boxes, each weighing 4 kg 500 g. How many such boxes can be loaded in a van which cannot carry beyond 800 kg?Answer10:

1 kg = 1000 g4 kg 500 g = 4500 g800 kg = 800 × 1000 = 800000 gNumber of boxes that can be loaded in the van = 800000 ÷ 4500

Hence, 177 boxes at maximum can be loaded in the van.

Question 11:The distance between the school and the house of a student’s house is 1 km 875 m. Everyday she walks both ways. Find the total distance covered by her in six days.Answer11:

Distance between school and house = 1 km 875 mNow, 1 km = 1000 m1 km 875 m = 1875 mDistance covered each day = 1875 × 2 = 3750 mDistance covered in 6 days = 3750 × 6

Therefore, distance covered in 6 days = 22,500 m= 22.5 km or 22 km 500 m

Question 12:A vessel has 4 litres and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?Answer12:

Capacity of vessel = 4 l 500 ml= 4500 ml (1 l = 1000 ml)Capacity of a glass = 25 mlNumber of glasses that can be filled = 4500 ÷ 25

∴ 180 glasses can be filled.

EXERCISE EXERCISE 1.31.3

Question 1:Estimate each of the following using general rule:(a) 730 + 998 (b) 796 − 314 (c) 12, 904 + 2, 888(d) 28, 292 − 21, 496Make ten more such examples of addition, subtraction and estimation of their outcome.Answer1:

(a) 730 + 998By rounding off to hundreds, 730 rounds off to 700 and 998 rounds off to 1000.

(b) 796 − 314By rounding off to hundreds, 796 rounds off to 800 and 314 rounds off to 300.

(c) 12904 + 2822By rounding off to thousands, 12904 rounds off to 13000 and 2822 rounds off to 3000.

(d) 28,296 − 21,496By rounding off to nearest thousands, 28296 rounds off to 28000 and 21496 rounds off to 21000.

Question 2:Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):(a) 439 + 334 + 4, 317 (b) 1,08, 734 − 47, 599 (c) 8325 − 491(d) 4, 89, 348 − 48, 365Make four more such examples.Answer2:

(a) 439 + 334 + 4317Rounding off to nearest hundreds, 439, 334, and 4317 may be rounded off to 400, 300, and 4300 respectively.

Rounding off to nearest tens, 439, 334, and 4317 may be rounded off to 440, 330, and 4320 respectively.

(b) 1,08,734 − 47,599Rounding off to hundreds, 1,08,734 and 47,599 may be rounded off to 1,08,700 and 47,600 respectively.

Rounding off to tens, 1,08,734 and 47,599 may be rounded off to 1,08,730 and 47,600 respectively.

(c) 8325 − 491Rounding off to hundreds, 8325 and 491 may be rounded off to 8300 and 500 respectively.

Rounding off to tens, 8325 and 491 may be rounded off to 8330 and 490 respectively.

(d) 4,89,348 − 48,365Rounding off to hundreds, 489348 and 48365 may be rounded off to 489300 and 48400 respectively.

Rounding off to tens, 489348 and 48365 may be rounded off to 489350 and 48370 respectively.

Question 3:Estimate the following products using general rule:(a) 578 × 161 (b) 5281 × 3491(c) 1291 × 592 (d) 9250 × 29Answer3:

(a) 578 × 161Rounding off by general rule, 598 and 161 may be rounded off to 600 and 200 respectively.

(b) 5281 × 3491Rounding off by general rule, 5281 and 3491 may be rounded off to 5000 and 3000 respectively.

(c) 1291 × 592Rounding off by general rule, 1291 and 592 may be rounded off to 1000 and 600 respectively.

(d) 9250 × 29Rounding off by general rule, 9250 and 29 may be rounded off to 9000 and 30 respectively.

IntroductionIntroductionAs we know, we use 1,2,3,4…. When

we begin to count. They come naturally when we start counting. Hence, mathematicians call the counting number as Natural number.

Natural numbers = The number starting from 1 are natural number or counting number.

Whole number = The natural number along with zero from of collection of whole number these 0,1,2,3 are whole number.

Properties of whole number = CLOSURE = If a and b are two whole number ,

then A+B , A-P is a also a whole number.

COMMUTATIVE= If a and b are two whole number , then a+b = b+a, a X b = b X a is also whole number . These addition and multiplication are commutative for whole number.

EXERCISE 2.1EXERCISE 2.1

Find the sum by suitable rearrangements

1932 + 453 + 1538 + 647 [ 1962 + 1538 ] + [453 + 647 ] [ 3500 + 1100 ] 4600 837 + 208 + 363 837 +[ 208 + 363 ] 837 + [ 571 ] 1408

Find the product by suitable rearrangement ? 2 X 1768 X 50 1768 X [ 2 X 50 ] 1768 X [ 100 ] 176800 4 X 166 X 25 166 X [ 25 X 4 ] 166 X [ 100 ] 16600

EXERCISE 2.2EXERCISE 2.2

Rajesh has 6 marbles with him. He wants to arrange them in rows in such a way that each row has the same number of marbles. He arranges them in the following ways and matches the total number of marbles.

(i) 1 marble in each row Number of rows = 6 Total number of marbles = 1 × 6 = 6

Question 1:Write all the factors of the following numbers:(a) 24 (b) 15 (c) 21(d) 27 (e) 12 (f) 20(g) 18 (h) 23 (i) 36

Answer 1:(a) 2424 = 1 × 24 24 = 2 × 12 24 = 3 × 824 = 4 × 6 24 = 6 × 4∴Factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24

(b) 1515 = 1 × 15 15 = 3 × 5 15 = 5 × 3∴Factors of 15 are 1, 3, 5, and 15

(c) 2121 = 1 × 21 21 = 3 × 7 21 = 7 × 3∴Factors of 21 are 1, 3, 7, and 21

(d) 2727 = 1 × 27 27 = 3 × 9 27 = 9 × 3∴Factors of 27 are 1, 3, 9, and 27

(e) 1212 = 1 × 12 12 = 2 × 6 12 = 3 × 4 12 = 4 × 3∴Factors of 12 are 1, 2, 3, 4, 6, and 12

(f) 2020 = 1 × 20 20 = 2 × 10 20 = 4 × 5 20 = 5 × 4∴Factors of 20 are 1, 2, 4, 5, 10, and 20

(g) 1818 = 1 × 18 18 = 2 × 9 18 = 3 × 6 18 = 6 × 3∴Factors of 18 are 1, 2, 3, 6, 9, and 18

(h) 2323 = 1 × 23 23 = 23 × 1

∴ Factors of 23 are 1 and 23

(i) 3636 = 1 × 36 36 = 2 × 18 36 = 3 × 12 36 = 4 × 936 = 6 × 6∴Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36

Question 2:Write first five multiplies of:(a)5 (b) 8 (c) 9

Answer 2:(a) 5 × 1 = 5 5 × 2 = 10 5 × 3 = 15 5 × 4 = 20 5 × 5 = 25∴ The required multiples are 5, 10, 15, 20, and 25.

(b) 8 × 1 = 8 8 × 2 = 16 8 × 3 = 24 8 × 4 = 32 8 × 5 = 40∴ The required multiples are 8, 16, 24, 32, and 40.

(c) 9 × 1 = 9 9 × 2 = 18 9 × 3 = 27 9 × 4 = 36 9 × 5 = 45∴ The required multiples are 9, 18, 27, 36, and 45.

Column 1 Column 2(i) 35 (b) Multiple of 7

(ii) 15 (d) Factor of 30

(iii) 16 (a) Multiple of 8

(iv) 20 (f) Factor of 20

(v) 25 (e) Factor of 50

Question 3: Match the items in column 1 with the items in column 2.

Answer 3:Column 1 Column 2

(i) 35 (a) Multiple of 8(ii) 15 (b) Multiple of 7(iii) 16 (c) Multiple of 70(iv) 20 (d) Factor of 30(v) 25 (e) Factor of 50

- (f) Factor of 20

2. State whether the following statements are True or False:

(a) The sum of three odd numbers is even.( T )(b) The sum of two odd numbers and one even number is even. ( T )(c) The product of three odd numbers is odd. ( F )(d) If an even number is divided by 2, the quotient is always odd. ( F )(e) All prime numbers are odd. ( F )(f) Prime numbers do not have any factors. ( F )(g) Sum of two prime numbers is always even. ( T )(h) 2 is the only even prime number. ( T )(i) All even numbers are composite numbers. ( F ) (j) The product of two even numbers is always even. ( T )

1. What is the sum of any two (a) Odd numbers? (b) Even numbers? Answer : (b) Even numbers

Question 3:The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers up to 100.Answer:17, 7137, 7379, 97

Question 4:Write down separately the prime and composite numbers less than 20.Answer 4:Prime numbers less than 20 are2, 3, 5, 7, 11, 13, 17, 19Composite numbers less than 20 are4, 6, 8, 9, 10, 12, 14, 15, 16, 18

Question 5:What is the greatest prime number between 1 and 10?Answer 5:Prime numbers between 1 and 10 are 2, 3, 5, and 7. Among these numbers, 7 is the greatest.Question 6:Express the following as the sum of two odd primes.(a)44 (b) 36 (c) 24 (d) 18

Answer 6:(a) 44 = 37 + 7(b) 36 = 31 + 5(c) 24 = 19 + 5(d) 18 = 11 + 7

Question 7:

Give three pairs of prime numbers whose difference is 2.[Remark: Two prime numbers whose difference is 2 are called twin primes].

Answer 7:3, 541, 4371, 73

Question 8:Which of the following numbers are prime?(a)23 (b) 51 (c) 37 (d) 26

Answer 8:(a) 23 23 = 1 × 23 23 = 23 × 123 has only two factors, 1 and 23. Therefore, it is a prime number.(b) 51 51 = 1 × 51 51 = 3 × 1751 has four factors, 1, 3, 17, 51. Therefore, it is not a prime number. It is a composite number.(c) 37It has only two factors, 1 and 37. Therefore, it is a prime number.(d) 2626 has four factors (1, 2, 13, 26). Therefore, it is not a prime number. It is a composite number.

Question 9:Write seven consecutive composite numbers less than 100 so that there is no prime number between them.

Answer 9:Between 89 and 97, both of which are prime numbers, there are 7 composite numbers. They are90, 91, 92, 93, 94, 95, 96Numbers Factors90 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 9091 1, 7, 13, 9192 1, 2, 4, 23, 46, 9293 1, 3, 31, 9394 1, 2, 47, 9495 1, 5, 19, 9596 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

Question 10:

Express each of the following numbers as the sum of three odd primes:(a)21 (b) 31 (c) 53 (d) 61

Answer 10:(a) 21 = 3 + 7 + 11(b) 31 = 5 + 7 + 19(c) 53 = 3 + 19 + 31(d) 61 = 11 + 19 + 31

Question 11:Write five pairs of prime numbers less than 20 whose sum is divisible by 5.(Hint: 3 + 7 = 10)

Answer 11:2 + 3 = 52 + 13 = 153 + 17 = 207 + 13 = 2019 + 11 = 30

Question 12:Fill in the blanks:(a) A number which has only two factors is called a _______.(b) A number which has more than two factors is called a _______.(c) 1 is neither _______ nor _______.(d) The smallest prime number is _______.(e) The smallest composite number is _______.(f) The smallest even number is _______.

Answer 12:(a) Prime number(b) Composite number(c) Prime number, composite number(d) 2(e) 4(f) 2

Question 1: What is the sum of any two (a) Odd numbers? (b) Even numbers?Answer1:

(a) The sum of two odd numbers is even.e.g., 1 + 3 = 413 + 19 = 32(b) The sum of two even numbers is even.e.g., 2 + 4 = 610 + 18 = 28

Question 2 :

Using divisibility tests, determine which of the following numbers are divisible by4; by 8:(a) 572 (b) 726352 (c) 5500 (d) 6000 (e) 12159(f) 14560 (g) 21084 (h) 31795072 (i) 1700 (j) 2150

Answer2: a) 572The last two digits are 72. Since 72 is divisible by 4, the given numberis also divisible by 4.The last three digits are 572. Since 572 is not divisible by 8, the given number is also not divisible by 8.(b) 726352The last two digits are 52. As 52 is divisible by 4, the given number is also divisible by 4.The last three digits are 352. Since 352 is divisible by 8, the given number is also divisible by 8.

(c) 5500Since last two digits are 00, it is divisible by 4.The last 3 digits are 500. Since 500 is not divisible by 8, the given number is also not divisible by 8.(d) 6000Since the last 2 digits are 00, the given number is divisible by 4.Since the last 3 digits are 000, the given number is divisible by 8.(e) 12159The last 2 digits are 59. Since 59 is not divisible by 4, the given number is also not divisible by 4.The last 3 digits are 159. Since 159 is not divisible by 8, the given number is not divisible by 8.(f) 14560The last two digits are 60. Since 60 is divisible by 4, the given number is divisible by 4.The last 3 digits are 560. Since 560 is divisible by 8, the given number is divisible by 8.(g) 21084The last two digits are 84. Since 84 is divisible by 4, the given number is divisible by 4.The last three digits are 084. Since 084 is not divisible by 8, the given number is not divisible by 8.

(h) 31795072The last two digits are 72. Since 72 is divisible by 4, the given number is divisible by 4.The last three digits are 072. Since 072 is divisible by 8, the given number is divisible by 8.(i) 1700The last two digits are 00. Since 00 is divisible by 4, the given number is divisible by 4.The last three digits are 700. Since 700 is not divisible by 8, the given number is not divisible by 8.(j) 2150The last two digits are 50. Since 50 is not divisible by 4, the given number is not divisible by 4.The last three digits are 150. Since 150 is not divisible by 8, the given number is not divisible by 8.

Question 3:

Using divisibility tests, determine which of following numbers are divisible by 6:(a) 297144 (b) 1258 (c) 4335 (d) 61233(e) 901352 (f) 438750 (g) 1790184 (h) 12583(i)639210 (j) 17852

Answer3:

(a) 297144Since the last digit of the number is 4, it is divisible by 2.On adding all the digits of the number, the sum obtained is 27. Since 27 is divisible by 3, the given number is also divisible by 3.As the number is divisible by both 2 and 3, it is divisible by 6.(b) 1258Since the last digit of the number is 8, it is divisible by 2.On adding all the digits of the number, the sum obtained is 16. Since 16 is not divisible by 3, the given number is also not divisible by 3.As the number is not divisible by both 2 and 3, it is not divisible by 6.(c) 4335

(c) 4335The last digit of the number is 5, which is not divisible by 2. Therefore, the given number is also not divisible by 2.On adding all the digits of the number, the sum obtained is 15. Since 15 is divisible by 3, the given number is also divisible by 3.As the number is not divisible by both 2 and 3, it is not divisible by 6.

(d) 61233The last digit of the number is 3, which is not divisible by 2. Therefore, the given number is also not divisible by 2.On adding all the digits of the number, the sum obtained is 15. Since 15 is divisible by 3, the given number is also divisible by 3.As the number is not divisible by both 2 and 3, it is not divisible by 6.

(e) 901352Since the last digit of the number is 2, it is divisible by 2.On adding all the digits of the number, the sum obtained is 20. Since 20 is not divisible by 3, the given number is also not divisible by 3.As the number is not divisible by both 2 and 3, it is not divisible by 6.

(f) 438750Since the last digit of the number is 0, it is divisible by 2.On adding all the digits of the number, the sum obtained is 27. Since 27 is divisible by 3, the given number is also divisible by 3.As the number is divisible by both 2 and 3, it is divisible by 6.

(g) 1790184Since the last digit of the number is 4, it is divisible by 2.On adding all the digits of the number, the sum obtained is 30. Since 30 is divisible by 3, the given number is also divisible by 3.As the number is divisible by both 2 and 3, it is divisible by 6.

(h) 12583Since the last digit of the number is 3, it is not divisible by 2.On adding all the digits of the number, the sum obtained is 19. Since 19 is not divisible by 3, the given number is also not divisible by 3.As the number is not divisible by both 2 and 3, it is not divisible by 6.

(i) 639210Since the last digit of the number is 0, it is divisible by 2.On adding all the digits of the number, the sum obtained is 21. Since 21 is divisible by 3, the given number is also divisible by 3.As the number is divisible by both 2 and 3, it is divisible by 6.

(j) 17852Since the last digit of the number is 2, it is divisible by 2.On adding all the digits of the number, the sum obtained is 23. Since 23 is not divisible by 3, the given number is also not divisible by 3.As the number is not divisible by both 2 and 3, it is not divisible by 6.

Question 4:Using divisibility tests, determine which of the following numbers are divisible by 11:(a) 5445 (b) 10824 (c) 7138965 (d) 70169308(e) 10000001 (f) 901153Answer4:(a) 5445Sum of the digits at odd places = 5 + 4 = 9Sum of the digits at even places = 4 + 5 = 9Difference = 9 − 9 = 0As the difference between the sum of the digits at odd places and the sum of the digits at even places is 0, therefore, 5445 is divisible by 11.

(b) 10824Sum of the digits at odd places = 4 + 8 + 1 = 13Sum of the digits at even places = 2 + 0 = 2Difference = 13 − 2 = 11The difference between the sum of the digits at odd places and the sum of the digits at even places is 11, which is divisible by 11. Therefore, 10824 is divisible by 11.(c) 7138965Sum of the digits at odd places = 5 + 9 + 3 + 7 = 24Sum of the digits at even places = 6 + 8 + 1 = 15Difference = 24 − 15 = 9The difference between the sum of the digits at odd places and the sum of digits at even places is 9, which is not divisible by 11. Therefore, 7138965 is not divisible by 11.

(d) 70169308Sum of the digits at odd places = 8 + 3 + 6 + 0 = 17Sum of the digits at even places = 0 + 9 + 1 + 7 = 17Difference = 17 − 17 = 0As the difference between the sum of the digits at odd places and the sum of the digits at even places is 0, therefore, 70169308 is divisible by 11.

(e) 10000001Sum of the digits at odd places = 1Sum of the digits at even places = 1Difference = 1 − 1 = 0As the difference between the sum of the digits at odd places and the sum of the digits at even places is 0, therefore, 10000001 is divisible by 11.

(f) 901153Sum of the digits at odd places = 3 + 1 + 0 = 4Sum of the digits at even places = 5 + 1 + 9 = 15Difference = 15 − 4 = 11The difference between the sum of the digits at odd places and the sum of the digits at even places is 11, which is divisible by 11. Therefore, 901153 is divisible by 11.

Question 5:

Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3:(a)___6724 (b) 4765 ___2

Answer5:

(a) _6724Sum of the remaining digits = 19To make the number divisible by 3, the sum of its digits should be divisible by 3.The smallest multiple of 3 which comes after 19 is 21.Therefore, smallest number = 21 − 19 = 2Now, 2 + 3 + 3 = 8However, 2 + 3 + 3 + 3 = 11If we put 8, then the sum of the digits will be 27 and as 27 is divisible by 3, the number will also be divisible by 3.Therefore, the largest number is 8.

(b) 4765_2

Sum of the remaining digits = 24To make the number divisible by 3, the sum of its digits should be divisible by 3. As 24 is already divisible by 3, the smallest number that can be placed here is 0.Now, 0 + 3 = 33 + 3 = 63 + 3 + 3 = 9However, 3 + 3 + 3 + 3 = 12If we put 9, then the sum of the digits will be 33 and as 33 is divisible by 3, the number will also be divisible by 3.Therefore, the largest number is 9.

Question 6:

Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11:(a)92 ___ 389 (b) 8 ___9484

Answer6:

(a) 92_389Let a be placed in the blank.Sum of the digits at odd places = 9 + 3 + 2 = 14Sum of the digits at even places = 8 + a + 9 = 17 + aDifference = 17 + a − 14 = 3 + aFor a number to be divisible by 11, this difference should be zero or a multiple of 11.If 3 + a = 0, thena = − 3However, it cannot be negative.A closest multiple of 11, which is near to 3, has to be taken. It is 11itself.3 + a = 11a = 8Therefore, the required digit is 8.

(b) 8_9484

Let a be placed in the blank.Sum of the digits at odd places = 4 + 4 + a = 8 + aSum of the digits at even places = 8 + 9 + 8 = 25Difference = 25 − (8 + a)= 17 − aFor a number to be divisible by 11, this difference should be zero or a multiple of 11.If 17 − a = 0, thena = 17This is not possible.A multiple of 11 has to be taken. Taking 11, we obtain17 − a = 11a = 6Therefore, the required digit is 6.

Question 1:

Find the common factors of:(a) 20 and 28 (b) 15 and 25(c) 35 and 50 (d) 56 and 120Answer1:

(a) Factors of 20 = 1, 2, 4, 5, 10, 20Factors of 28 = 1, 2, 4, 7, 14, 28Common factors = 1, 2, 4(b) Factors of 15 = 1, 3, 5, 15Factors of 25 = 1, 5, 25Common factors = 1, 5(c) Factors of 35 = 1, 5, 7, 35Factors of 50 = 1, 2, 5, 10, 25, 50Common factors = 1, 5(d) Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120Common factors = 1, 2, 4, 8

Question 2:Find the common factors of:(a)4, 8 and 12 (b) 5, 15 and 25

Answer2:

(a) 4, 8, 12Factors of 4 = 1, 2, 4Factors of 8 = 1, 2, 4, 8Factors of 12 = 1, 2, 3, 4, 6, 12Common factors = 1, 2, 4

(b) 5, 15, and 25Factors of 5 = 1, 5Factors of 15 = 1, 3, 5, 15Factors of 25 = 1, 5, 25Common factors = 1, 5

Question 3:Find first three common multiples of:(a)6 and 8 (b) 12 and 18Answer3:

(a) 6 and 8Multiple of 6 = 6, 12, 18, 24, 30…..Multiple of 8 = 8, 16, 24, 32……3 common multiples = 24, 48, 72(b) 12 and 18Multiples of 12 = 12, 24, 36, 48Multiples of 18 = 18, 36, 54, 723 common multiples = 36, 72, 108

Question 4:Write all the numbers less than 100 which are common multiples of 3 and 4.Answer4:

Multiples of 3 = 3, 6, 9, 12, 15…Multiples of 4 = 4, 8, 12, 16, 20…Common multiples = 12, 24, 36, 48, 60, 72, 84, 96

Question 5:Which of the following numbers are co-prime?(a) 18 and 35 (b) 15 and 37 (c) 30 and 415(d) 17 and 68 (e) 216 and 215 (f) 81 and 16Answer5:

(a) Factors of 18 = 1, 2, 3, 6, 9, 18Factors of 35 = 1, 5, 7, 35Common factor = 1Therefore, the given two numbers are co-prime.(b) Factors of 15 = 1, 3, 5, 15Factors of 37 = 1, 37Common factors = 1Therefore, the given two numbers are co-prime.(c) Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30Factors of 415 = 1, 5, 83, 415Common factors = 1, 5As these numbers have a common factor other than 1, the given two numbers are not co-prime.

(d) Factors of 17 = 1, 17Factors of 68 = 1, 2, 4, 17, 34, 68Common factors = 1, 17As these numbers have a common factor other than 1, the given two numbers are not co-prime.

(e) 216 and 215Factors of 216 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108, 216Factors of 215 = 1, 5, 43, 215Common factors = 1Therefore, the given two numbers are co-prime.

(f) 81 and 16Factors of 81 = 1, 3, 9, 27, 81Factors of 16 = 1, 2, 4, 8, 16Common factors = 1Therefore, the given two numbers are co- prime.

Question 6:A number is divisible by both 5 and 12. By which other number will that number be always divisible?Answer6:

Factors of 5 = 1, 5Factors of 12 = 1, 2, 3, 4, 6, 12As the common factor of these numbers is 1, the given two numbers are co- prime and the number will also be divisible by their product, i.e. 60, and the factors of 60, i.e., 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

Question 7:A number is divisible by 12. By what other number will that number be divisible?Answer7:

Since the number is divisible by 12, it will also be divisible by its factors i.e., 1, 2, 3, 4, 6, 12. Clearly, 1, 2, 3, 4, and 6 are numbers other than 12 by which this number is also divisible.

Question 1:

Which of the following statements are true?(a) If a number is divisible by 3, it must be divisible by 9. (T)(b) If a number is divisible by 9, it must be divisible by 3. (T)(c) A number is divisible by 18, if it is divisible by both 3 and 6. (F)(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90. (T)(e) If two numbers are co-primes, at least one of them must be prime. (F)(f) All numbers which are divisible by 4 must also be divisible by 8. (F)(g) All numbers which are divisible by 8 must also be divisible by 4. (T)(h) If a number exactly divides two numbers separately, it must exactly divide their sum. (T)(i)If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately. (F)

Question 2:Here are two different factor trees for 60. Write the missing numbers.(a)

(b)

Answer2:(a) As 6 = 2 × 3 and 10 = 5 × 2

(b) As 60 = 30 × 2, 30 = 10 × 3, and 10 = 5 × 2

Question 3:

Which factors are not included in the prime factorization of a composite number? Answer3:

1 and the number itself

Question 4:Write the greatest 4-digit number and express it in terms of its prime factors. Answer4:

Greatest four-digit number = 99999999 = 3 × 3 × 11 × 101

Question 5:Write the smallest 5-digit number and express it in the form of its prime factors.Answer5:

Smallest five-digit number = 10,00010000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

Question 6:Find all prime factors of 1729 and arrange them in ascending order. Now state the relation, if any; between two consecutive prime factors.Answer6:

7 1729

13 247

19 19

1

1729 = 7 × 13 × 1913 − 7 = 6, 19 − 13 = 6The difference of two consecutive prime factors is 6.

Question 8:

The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.

Answer8:

3 + 5 = 8, which is divisible by 415 + 17 = 32, which is divisible by 419 + 21 = 40, which is divisible by 4

Question 9:

In which of the following expressions, prime factorization has been done?(a) 24 = 2 × 3 × 4 (b) 56 = 7 × 2 × 2 × 2(c) 70 = 2 × 5 × 7 (d) 54 = 2 × 3 × 9

Answer9:

(a) 24 = 2 × 3 × 4Since 4 is composite, prime factorisation has not been done.(b) 56 = 7 × 2 × 2 × 2Since all the factors are prime, prime factorisation has been done.(c) 70 = 2 × 5 × 7Since all the factors are prime, prime factorisation has been done.(d) 54 = 2 × 3 × 9Since 9 is composite, prime factorisation has not been done.

Question 10:

Determine if 25110 is divisible by 45.[Hint: 5 and 9 are co-prime numbers. Test the divisibility of the number by 5 and 9].

Answer10:

45 = 5 × 9Factors of 5 = 1, 5Factors of 9 = 1, 3, 9Therefore, 5 and 9 are co-prime numbers.Since the last digit of 25110 is 0, it is divisible by 5.Sum of the digits of 25110 = 2 + 5 + 1 + 1 + 0 = 9As the sum of the digits of 25110 is divisible by 9, therefore, 25110 is divisible by 9.Since the number is divisible by 5 and 9 both, it is divisible by 45.

Question 11:

18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 × 6 = 24? If not, give an example to justify our answer:

Answer11:

No. It is not necessary because 12 and 36 are divisible by 4 and 6 both, but are not divisible by 24.

Question 12:

I am the smallest number, having four different prime factors. Can you find me?

Answer12:

Since it is the smallest number of such type, it will be the product of 4 smallest prime numbers.2 × 3 × 5 × 7 = 210

Question 1:What is the HCF of two consecutive(a)Numbers? (b) Even numbers? (c) Odd numbers?

Answer1:

(i) 1 e.g., HCF of 2 and 3 is 1.(ii) 2 e.g., HCF of 2 and 4 is 2.(iii) 1 e.g., HCF of 3 and 5 is 1.

Question 2:

HCF of co-prime numbers 4 and 15 was found as follows by factorization:4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factors, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Answer2:

No. The answer is not correct. 1 is the correct HCF.

Question 1: Rena purchases two bags of fertilizer of weight 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer exact number of times.Answer1:

Weight of the two bags = 75 kg and 69 kgMaximum weight = HCF (75, 69)

3 75

5 25

5 5

1

3 69

23 23

1

75 = 3 × 5 × 569 = 3 × 23HCF = 3Hence, the maximum value of weight, which can measure the weight of the fertilizer exact number of times, is 3 kg.

Question 2: Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?Answer2:Step measure of 1st Boy = 63 cmStep measure of 2nd Boy = 70 cmStep measure of 3rd Boy = 77 cmLCM of 63, 70, 77

2 63, 70, 77

3 63, 35 , 77

3 21,35,77

5 7,35,77

7 7,7,77

11

1,1,11

1, 1, 1Hence, the minimum distance each should cover so that all can cover the distance in complete steps is 6930 cm.

LCM = 2 × 3 × 3 × 5 × 7 × 11 = 6930

Question 3: The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.Answer3:Length = 825 cm = 3 × 5 × 5 × 11Breadth = 675 cm = 3 × 3 × 3 × 5 × 5Height = 450 cm = 2 × 3 × 3 × 5 × 5Longest tape = HCF of 825, 675, and 450 = 3 × 5 × 5 = 75 cmTherefore, the longest tape is 75 cm.

Question 4:Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.Answer4:

Smallest number = LCM of 6, 8, 12

2 6, 8, 12

2 3,4,6

2 3,2,3

3 3,1,3

1,1,1

LCM = 2 × 2 × 2 × 3 = 24We have to find the smallest 3-digit multiple of 24.It can be seen that 24 × 4 = 96 and 24 × 5 = 120.Hence, the smallest 3-digit number which is exactly divisible by 6, 8, and 12 is 120.

Question 5:Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.Answer5:

LCM of 8, 10, and 12

2 8,10,12

2 4,5,6

2 2,5,3

3 1,5,3

5 1,5,1

1,1,1LCM = 2 × 2 × 2 × 3 × 5 = 120We have to find the greatest 3-digit multiple of 120.It can be seen that 120 ×8 = 960 and 120 × 9 = 1080.Hence, the greatest 3-digit number exactly divisible by 8, 10, and 12 is 960.

Question 6:The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again? Answer 6:

Time period after which these lights will change = LCM of 48, 72, 108 2

48,72,108

2

24,36,54

2

12,18,27

2

6,9,27

3

3,9,27

3

1,3,9

3

1,1,3

1,1,1

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432They will change together after every 432 seconds i.e., 7 min 12 seconds.Hence, they will change simultaneously at 7:07:12 am.

Question 7:Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.Answer 7:

Maximum capacity of the required tanker = HCF of 403, 434, 465403 = 13 × 31434 = 2 × 7 × 31465 = 3 × 5 × 31HCF = 31∴ A container of capacity 31 l can measure the diesel of 3 containers exact number of times

Question 8:Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.Answer8:

LCM of 6, 15, 18

2 6,15,18

3 3,5,9

3 1,5,3

5 1,5,1

1,1,1LCM = 2 × 3 × 3 × 5 = 90Required number = 90 + 5 = 95

Question 9:Find the smallest 4-digit number which is divisible by 18, 24 and 32.Answer9:

LCM of 18, 24, and 322 18,24,

32

2 9,12,16

2 9,6,8

2 9,3,4

2 9,3,2

3 9,3,1

3 3,1,1

1,1,1LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288We have to find the smallest 4-digit multiple of 288.It can be observed that 288 ×3 = 864 and 288 ×4 = 1152.Therefore, the smallest 4-digit number which is divisible by 18, 24, and 32 is1152.

Question 10:Find the LCM of the following numbers:(a) 9 and 4 (b) 12 and 5(c) 6 and 5 (d) 15 and 4Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?Answer10:

(a)

LCM = 2 × 2 × 3 × 3 = 36

LCM = 2 × 2 × 3 × 5 = 60

LCM = 2 × 3 × 5 = 30

LCM = 2 × 2 × 3 × 5 = 60Yes, it can be observed that in each case, the LCM of the given numbers is the product of these numbers. When two numbers are co-prime, their LCM is the product of those numbers. Also, in each case, LCM is a multiple of 3.

A B A

B

A BC D

Geometry has a long and rich history. The term ‘Geometry’ is the Englishequivalent of the Greek word ‘Geometron’. ‘Geo’ means Earth and ‘metron’means Measurement. According tohistorians, the geometrical ideas shaped upin ancient times, probably due to the needin art, architecture and measurement. Theseinclude occasions when the boundaries ofcultivated lands had to be marked withoutgiving room for complaints. Construction ofmagnificent palaces, temples, lakes, damsand cities, art and architecture propped upthese ideas. Even today geometrical ideasare reflected in all forms of art,measurements, architecture, engineering, cloth designing etc. You observeand use different objects like boxes, tables, books, the tiffin box you carryto your school for lunch, the ball with which you play andso on. All such objects have different shapes. The ruler which you use, thepencil with which you write are straight. The pictures of a bangle, the onerupee coin or a ball appear round.Here, you will learn some interesting facts that will help you know moreabout the shapes around you.

1. A point determines a location. It is usually denoted by a capital letter.

2. A line segment corresponds to the shortest distance between two points.

3. A line is obtained when a line segment like AB is extended on both sides indefinitely; it is denoted by AB or sometimes by a single small letter like l.

4. Two distinct lines meeting at a point are called intersecting lines.5. Two lines in a plane are said to be parallel if they do not meet.6. A ray is a portion of line starting at a point and going in one

direction endlessly.7. Any drawing (straight or non-straight) done without lifting the

pencil may be called a curve. 8. A simple curve is one that does not cross itself.9. A curve is said to be closed if its ends are joined; otherwise it is

said to be open.10. A polygon is a simple closed curve made up of line segments.

Here,(i) The line segments are the sides of the polygon.(ii) Any two sides with a common end point are adjacent sides.(iii) The meeting point of a pair of sides is called a vertex.(iv) The end points of the same side are adjacent vertices.(v) The join of any two non-adjacent vertices is a diagonal.

11. An angle is made up of two rays starting from a common end point.

12. A triangle is a three-sided polygon.

13. A quadrilateral is a four-sided polygon. (It should be named cyclically).

14. A circle is the path of a point moving at the same distance from a fixed point.

15.The fixed point is the centre, the fixed distance is the radius and the distance

around the circle is the circumference.16. A chord of a circle is a line segment joining any two points on the circle.

17. A diameter is a chord passing through the centre of the circle.

18. A sector is the region in the interior of a circle enclosed by an arc on one side

and a pair of radii on the other two sides.

19. A segment of a circle is a region in the interior of the circle enclosed by an arc and a chord. 20.The diameter of a circle divides it into two semi-circles.

Question 1:Use the figure to name:

(a) Five points(b) A line(c) Four rays(d) Five line segmentsAnswer1:

(a) The five points are D, E, O, B, and C.(b) (c) (d)

Question 2:Name the line given in all possible (twelve) ways, choosing only two letters at a time from the four given.

Answer 2:

Question 3:Use the figure to name:

(a) Line containing point E.(b) Line passing through A.(c) Line on which O lies(d) Two pairs of intersecting lines.Answer 3:

Question 4:How many lines can pass through (a) one given point? (b) Two given points?Answer4:

(a) Infinite number of lines can pass through a single point.(b) Only one line can pass through two given points.

Question 5:Draw a rough figure and label suitably in each of the following cases:

Answer 5:

Question 1:Classify the following curves as (i) Open or (ii) Closed.

Answer1:

(a) Open(b) Closed(c) Open(d) Closed(e) Closed

Question 2:Draw rough diagrams to illustrate the following:(a)Open curve (b) Closed curve.

Answer 2:

Question 3:Draw any polygon and shade its interior.Answer 3:

Question 4:Consider the given figure and answer the questions:(a) Is it a curve? (b) Is it closed?

Answer 4:(a) Yes(b) Yes

Question 5:Illustrate, if possible, each one of the following with a rough diagram:(a) A closed curve that is not a polygon.(b) An open curve made up entirely of line segments.(c) A polygon with two sides.Answer 5:

Question 1:Name the angles in the given figure.

Answer1:

∠BAD, ∠ADC, ∠DCB, ∠CBA

Question 2:In the given diagram, name the point (s)

(a) In the interior of DOE∠(b) In the exterior of EOF∠(c) On EOF∠

Answer2:

(a) A(b) C, A, D(c) B, E, O, F

Question 1:Draw a rough sketch of a triangle ABC. Mark a point P in its interior and a point Q in its exterior. Is the point A in its exterior or in its interior?

Answer 1:

Point A lies on the given Δ ABC.

Question 2:

(a) Identify three triangles in the figure.(b) Write the names of seven angles.(c) Write the names of six line segments.(d) Which two triangles have ∠B as common?

Question 1:Draw a rough sketch of a quadrilateral PQRS. Draw its diagonals. Name them. Is the meeting point of the diagonals in the interior or exterior of the quadrilateral?

Answer1:

Diagonals are PR and QS. They meet at point O which is in the interior of &mnSq1PQRS.

Question 2:Draw a rough sketch of a quadrilateral KLMN. State,(a) Two pairs of opposite sides,(b) Two pairs of opposite angles,(c) Two pairs of adjacent sides,(d) Two pairs of adjacent angles.Answer 2:

Question 1:From the figure, identify:

(a) The centre of circle (e) Two points in the interior(b) Three radii (f) a point in the exterior(c) a diameter (g) a sector(d) a chord (h) a segmentAnswer 1:

Question 2:(a) Is every diameter of a circle also a chord?(b) Is every chord of circle also a diameter?Answer2:

(a) Yes. The diameter is the longest possible chord of the circle.(b) No

Question 3:Draw any circle and mark(a) Its centre (e) a segment(b) a radius (f) a point in its interior(c) a diameter (g) a point in its exterior(d) a sector (h) an arcAnswer 3:

Question 4:Say true or false:(a) Two diameters of a circle will necessarily intersect.(b) The centre of a circle is always in its interior.Answer4:

(a) True. They will always intersect each other at the centre of the circle.(b) True