Class 6 CHAPTER 1 KNOWING OUR NUMBERS...

Sardhna: Kaland Chungi, Sardhana Meerut-250342 Meerapur: Padaw Chock, Meerapur, Muzaffarnagar- 251315

Class 6

CHAPTER – 1 KNOWING OUR NUMBERS

INTRODUCTORY QUESTIONS:

Ques.1 What are the Natural Numbers?

Ans. When we begin to court the numbers 1,2,3,4,5,…………. Come naturally. Hence, these are

called Natural Numbers.

Q1. Which is the first smallest Natural Numbers?

Ans The first or the smallest natural number is 1. 0

Q2. What are the whole numbers?

Ans The number ‘0’ together with natural numbers are called whole numbers.

Q3. What is numeration?

Ans Expressing a number is words is called numeration.

Q4. Write ‘Eight thousand fifty six’ is numerals

Ans Th Hun Tens Ones

8 0 5 6

Thus, the number is written as 8056.

NCERT SOLUTIONS:

1. Insert commas suitably and write the names according to International system of numeration: (a) 78921092 (b)7452283 (c)99985102 (d) 48049831

Solution: (a) 78,921,092 ----- Seventy eight million nine hundred twenty one thousand ninety two (b) 7,452,283 ----- Seven million four hundred fifty two thousand two hundred eighty three (c) 99,985,102 ------ Ninety nine million nine hundred eighty five thousand one hundred two.


(d) 48,049,831 ------ Forty eight million forty nine thousand eight hundred thirty one.

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

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

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

Solution: (a) 8,75,95,762 ---- Eight crore seventy five lakh ninety five thousand seven hundred sixty two.

(b) 85,46,283 ----- Eighty five lakh forty six thousand two hundred and eighty three. (c) 9,99,00,046 ----- Nine crore ninety nine lakh forty six.

(d) 9,84,32,701 ----- Nine crore eighty four lakh thirty two thousand seven hundred and one.

4. 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 days was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all the four days.

Solution: Number of tickets sold on the first day = 1094 Number of tickets sold on the second day = 1812 Number of tickets sold on the third day = 2050 Number of tickets sold on the final day = 2751

Total number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7707.


5. Shekhar is a famous cricket player. He has so far scored 6,980 runs in test matches. He wishes to complete 10,000 runs. How many more runs does he need?

Solution: Runs scored so far = 6,980 Runs wished to be scored = 10,000

Runs needed more = 10,000 - 6,980 = 3,020.

6. 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?

Solution: Votes registered by the successful candidate = 5,77,500 Votes secured by the nearest rival = 3,48,700 Margin by which the successful candidate won the election = 5,77,500 - 3,48,700 = 2,28,800.

7. Kirti Bookstore sold books worth Rs. 2,85,891 in the first week of June. The bookstore sold 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?

Solution: Sale of books in the first week = Rs. 2,85,891 Sale of books in the second week = Rs. 4,00,768

Sale for the two weeks together = Rs. 2,85,891 + Rs. 4,00,768 = Rs. 6,86,659. The sale was greater in the second week by Rs. 4,00,768 - Rs. 2,85,891 i.e., by Rs. 1,14,877.

8. Find the difference between the greatest and least numbers that can be written using the digits 6, 2, 7, 4, 3 each only once.

Solution: Greatest number that can be written using the digits 6, 2, 7, 4, 3 each only once = 76,432 Least number that can be written using the digits 6, 2, 7, 4, 3 each only once = 23,467

Difference between the greatest and least numbers that can be written using the digits 6, 2, 7, 4, 3 each only once = 76,432 - 23,467 = 52,965.

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

Solution: Number of screws manufactured by the machine a day on an average = 2,825.


Number of days in the month of January, 2006 = 31

Number of screws produced by the machine in the month of January, 2006 = 2,825 x 31 = 87,575.

10. 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 remainwith her after thepurchase ?

Solution: Money which the merchant had = Rs. 78,592 Cost of one radio sets = Rs. 1234

Cost of 40 radio sets = Rs. 1234 40 = Rs. 49,360 Money that will remain with the merchant after the purchase = Rs. 78,592 - Rs. 49,360 = Rs. 29,232.

11. A student multiplied 7236 by 65 instead of multiplying by 56. How much was his answer greater than the correct answer?

Solution:

Wrong answer = 7236 65 = 4,70,340

Correct answer = 7236 56 = 4,05,216 Wrong answer was greater than the correct answer by = 4,70,340 - 4,05,216 = 65,124.

12. To stitch a shirt 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain?

Solution: Cloth required to stitch a shirt = 2 m 15 cm

Convert m to cm as follows:

2 m 15 cm = 2 100 cm + 15 cm = 200 cm + 15 cm = 215 cm. Available cloth = 40 m

Convert m to cm as follows:

40 m = 40 100 cm = 4000 cm Hence, 18 shirts can be stitched and 130 cm cloth will remain.

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


Solution: Weight of medicines = 4 kg 500 g

= 4 1000 g + 500 g = 4000 g + 500 g = 4500 g.

Weight carried by van = 800 kg = 800 1000 g = 800000 g.

Hence, 177 such boxes can be loaded.

14. The distance between the school and the house of a student is 1 km 875 m. Everyday she walks both ways between her school and home. Find the total distance covered by her in six days.

Solution: Distance between the school and the house = 1 km 875 m

= 1 1000 m + 875 m = 1000 m + 875 m = 1875 m. Distance covered by her in a day in walking both ways between school and home

= 1875 2 m = 3750 m.

Total distance covered by her in six days in walking both ways between school and

home = 3750 6 cm = 22500 m

= 22000 m + 500 m = 22 1000 m + 500 m = 22 km + 500 m = 22 km 500 m.

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

Solution:

Capacity of the vessel = 41 500 ml = 4 1000 ml + 500 ml = 4000 ml + 500 ml = 4500 ml

Hence, it can be distributed in 180 glasses.


16. Estimate (a) 730 + 998 (b) 796 – 314 (c) 12,904 + 2,888 (d) 28,292 – 21,496 Make ten more of such examples of addition and subtraction and estimation of their outcome.

Solution: (a) 730 + 998 = 700 + 1000 = 1700 (b) 796 - 314 = 800 - 300 = 500 (c) 12,904 + 2,888 = 13,000 + 3,000 = 16,000 (d) 28,292 – 21,496 = 28,000 – 21,000 = 7,000

Few examples of addition and subtraction are as follows: (i) 712 + 270 = 700 + 300 = 1000 (ii) 1,165 - 325 = 1,200 - 300 = 900 (iii) 3,987 + 297 = 4,000 + 300 = 4300 (iv) 15,092 + 398 = 15,000 + 400 = 15,400 (v) 27,670 - 730 = 27,000 - 700 = 26,300 (vi) 10,450 + 298 = 10,000 + 300 = 10,300 (vii) 2,985 - 123 = 3,000 - 100 = 2900 (viii) 5,987 + 199 = 6,000 + 200 = 6200 (ix) 38,759 + 598 = 39,000 + 600 = 39,600 (x) 11,987 + 1,897 = 12,000 + 2,000 = 14,000.

17. 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 - 48,365 Make 4 more of such examples.

Solution: (a) (i) Rough estimate: 439 + 334 + 4317 = 400 + 300 + 4300 = 5000 (ii) Closer estimate: 439 + 334 + 4317 = 440 + 330 + 4320 = 5090.

(b) (i) Rough estimate: 1,08,734 - 47,599 = 1,00,000 - 50,000 = 50,000 (ii) Closer estimate: 1,08,734 - 47,599 = 1,08,730 - 47,600 = 61,130.

(c) (i) Rough estimate: 12,904 + 2,888 = 13,000 + 3,000


=16,000 (ii) Closer estimate: 12,904 + 2,888 = 12,900 + 2,890 = 15,790.

(d) (i) Rough estimate: 28,292 - 21,496 = 28,000 - 21,000 = 7,000 (ii) Closer estimate: 28,292 - 21,496 = 28,290 - 21,500 = 6,790. Few examples for rough and closer estimate: (i) 476 + 337 Rough estimate: 476 + 337 = 500 + 300 = 800 Closer estimate: 476 + 337 = 480 + 340 = 820

(ii) 894 + 1579 Rough estimate: 894 + 1579 = 900 + 1600 = 2500 Closer estimate: 894 + 1579 = 900 + 1580 = 2480

(iii) 413 - 187 Rough estimate: 413 - 187 = 400 - 200 = 200 Closer estimate: 413 - 187 = 410 - 190 = 223

(iv) 595 - 379 Rough estimate: 595 - 379 = 600 - 400 = 200 Closer estimate: 595 - 379 = 600 - 380 = 220.

HOTS (High Order thinking skills):-

Q1. Fill in the blanks:-

a. 909090909 comes just before 909090910

b. 57325499 comes just after 54325498

c. 33465800 comes just before 33465801

d. 52547900 comes just after 52547899


Q2. Write the smallest 7-digit number having four different digits.

Four digits are different.

The smallest 7-digit number is 1000023

UNSOLVED

Q3. I as a Roman numeral, am CMXCIX break me up and then can you recognize me?

Q4. At a rock concert there 550 spectors out of which 750 were invites who were given free

tickets. The total collection from the sale of ticket was Rs 5,93,750/-. What was the cost

of 1 ticket.

VALUE BASED QUESTIONS

Q1. Move one matchstick to correct each statement.

a. III – III = IV

b. VI – X = IV

c. IX + V = III

We have learnt that we can use the matchsticks to develop the Roman numbers. We

have learnt about the Indian and the international systems of numeration.

The oldest system of numeration is developed by the Romans and is still is common use.

We have learnt the value that matchsticks are help find solution of Roman numbers.

ACTIVITY – Complete the Number Puzzle

Across

a) Seventy two thousand four thousand five hundred sixty one.

b) Predecessor of 9741

c) Successor of 123456

d) CCLXVI + CXIX

e) Smallest 7-digits number

a b c

d

e f

g

h

i


CHAPTER – 2

(WHOLE NUMBERS)

INTRODUCTORY QUESTIONS:

Q1. What is a number line?

Ans A straight line representing numbers is called a number line.

Q2. What are the whole numbers?

Ans If we include 0 in the set of natural numbers, we have the set of whole numbers.

Q3. Write the consecutive successors of 999?

Ans 1000, 1001, 1002

Q4. Find the sum?

5 6 2 8 + 3 9 7 8 4

Ans 5 6 2 8 + 3 9 7 8 4 = 4 5 4 1 2

NCERT SOLUTIONS

1. Write the next three natural numbers after 10999. Solution:

The next three natural numbers after 10999 are 11000, 11001 and 11002.

2. Write the three whole numbers occurring just before 10001.

Solution: The three whole numbers occurring just before 10001 are 10000, 9999 and 9998.

3. Which is the smallest whole number.

Solution: 0 is the smallest whole number.

4. How many whole numbers are there between 32 and 53?

Solution: There are 20 whole numbers between 32 and 53. There are 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51 and 52.


5. Write successor of (a)2440701 (b) 100199 (c) 1099999 (d) 2345670.

Solution: (a) The successor of 2440701 is 2440702. (b) The successor of 100199 is 100200. (c) The successor of 1099999 is 1100000. (d) The successor of 2345670 is 2345671.

6. Write the predecessor of (a) 94 (b) 10000 (c) 208090 (d) 7654321.

Solution: (a) The predecessor of 94 is 93. (b) The predecessor of 10000 is 9999. (c) The predecessor of 208090 is 208089. (d) The predecessor of 7654321 is 7654320.

7. In each of the following pairs of numbers, state which whole number is on the left of the other number on the number line. Also write them with the appropriate sign (>, <) between them. (a) 530,503 (b) 370,307(c) 98765, 56789 (d) 9830415, 10023001.

Solution: (a) The whole number 503 is on the left of the whole number 530 on the number line. So, 503 < 530. (b) The whole number 307 is on the left of the whole number 370 on the number line. So, 307 < 370. (c) The whole number 56789 is on the left of the whole number 98765 on the number line. So, 56789 < 98765. (d) The whole number 9830415 is on the left of, the whole number 10023001 on the number line. So, 9830415 < 10023001.

8. Find the sum by suitable rearrangement : (a) 837 + 208 + 363 (b) 1962 + 453 + 1538 + 647.

Solution: (a) 837 + 208 + 363 = 837 + 363 + 208 = (837 + 363) + 208 = 1200 + 208 =1408

(b) 1962 + 453 + 1538 + 647 = 1962 + 1538 + 453 + 647 = (1962 + 1538) + (453 + 647) = 3500 + 1100 = 4600.


9. Find the product by a suitable rearrangement:

(a) 2 1768 50 (b) 4 166 25

(c) 8 291 125 (d) 625 279 16

(e) 285 5 60 (f) 125 40 8 25. Solution:

(a) 2 1768 50 2 x 50 x 1768 = (2 x 50) x 1768 = 100 x 1768 = 1,76,800. (b) 4 x 166 x 25 4 x 25 x 166 = (4 x 25) x 166 = 100 x 166 = 16,600.

(c) 8 x 291 x 125 8 x 125 x 291 = (8 x 125) x 291 = 1000 x 291 = 2,91,000

(d) 625 x 279 x 16 = 625 x 16 x 279 = (625 x 16) x 279 = 10000 x 279 = 27,90,000

(e) 285 x 5 x 60 = 285 x (5 x 60) = 285 x 300 = 85,500

(f) 125 x 40 x 8 x 25 = (125 x 40) x (8 x 25) = 5000 x 200 = 10,00,000

10. Find the value of the following: (a) 297 x 17 + 297 x 3 (b) 54279 x 92 + 8 x 54279 (c) 81265 x 169 – 81265 x 69 (d) 3845 x 5 x 782 + 769 x 25 x 218.

Solution: (a) 297 x 17 + 297 x 3 = 297 x (17 + 3) = 297 x 20 = 5940

(b) 54279 x 92 + 8 x 54279 = 54279 x 92 + 54279 x 8 = 54279 x (92 + 8) = 54279 x 100 = 54,27,900


(c) 81265 x 169 - 81265 x 69 = 81265 x (169 - 69) = 81265 x 100 = 81,26,500

(d) 3845 x 5 x 782 + 769 x 25 x 218 = 3845 x 5 x 782 + 769 x 5 x 5 x 218 = 3845 x 5 x 782 + (769 x 5) x 5 x 218 = 3845 x 5 x 782 + 3845 x 5 x 218 = 3845 x 5 x (782 + 218) = 3845 x 5 x 1000 = 19225 x 1000 = 1,92,25,000.

11. Find the product, using suitable properties: (a) 738 x 103 (b) 854 x 102 (c) 258 x 1008 (d) 1005 x 168.

Solution: (a) 738 x 103 = 738 x (100 + 3) = 738 x 100 + 738 x 3 = 73,800 + 2,214 = 76,014 (b) 854 x 102 = 854 x (100 + 2) = 854 x 100 + 854 x 2 = 85,400 + 1,708 = 87,108 (c) 258 x 1008 = 258 x (1000 + 8) = 258 x 1000 + 258 x 8 = 2,58,000 + 2,064 = 2,60,064 (d) 1005 x 168 = 168 x 1005 = 168 x (1000 + 5) = 168 x 1000 + 168 x 5 = 1,68,000 + 840 = 1,68,840.

12. A taxi-driver filled his car petrol tank with 40 litres of petrol on Monday. The next day, he filled the tank with 50 litres of petrol. If the petrol costs Rs. 44 per litre, how much did he spend in all on petrol?

Solution: Petrol filled on Monday = 40 litres Petrol filled the next day = 50 litres

Total petrol filled on the two days = 40 litres + 50 litres = 90 litres Cost of petrol per litre = Rs. 44 Cost of 90 litres petrol = Rs. 44 x 90 = Rs. 3960.


13. A vendor supplies 32 litres of milk to a hotel in the morning and 68 litres of milk in the evening. If the milk costs Rs. 15 Per litre, how much money is due to the vendor per day.

Solution: Milk supplied in the morning = 32 litres Milk supplied in the evening = 68 litres

Milk supplied per day = 32 litres + 68 litres = 100 litres. Cost of milk per litre = Rs 15

Money due to the vendor per day = Cost of 100 litres of milk = Rs. 15 x 100 = Rs. 1500.

14. Which of the following will not represent zero:

(a) 1 + 0 (b) 0 x 0 (c) (d) Solution:

(a) 1 + 0 will not represent zero. It will represent 1.

15. If the product of two whole numbers is zero can we say that one or both of them will be zero? Justify through examples.

Solution: One of them is essentially zero. For example: 2 x 0 = 0 0 x 3 = 0. Both of them can be zero as 0 x 0 = 0.

16. If the product of two whole numbers is 1, can we say that one or both of them will be 1? Justify through examples.

Solution: Both of them will be essentially 1 as 1 x 1 = 1.

17. Find by distributivity method : (a) 728 x 101 (b) 5437 x 1001 (c) 824 x 25 (d) 4275 x 125 (e) 504 x 35 (f) 5040 + 35.

Solution: (a) 728 x 101 = 728 x (100 + 1) = 728 x 100 + 728 x 1 = 72,800 + 728 = 73,528 (b) 5437 x 1001 = 5437 x (1000 + 1)


= 5437 x 1000 + 5437 x 1 = 54,37,000 + 5437 = 54,42,437

(c) 824 x 25 = = = 20,600

(e) 504 x 35 = 35 x 504 = 35 x (500 + 4) = 35 x 500 + 35 x 4 = 17500 + 140 = 17,640

(f ) 5040 35 = 1008 7 = 144.

18. Study the pattern:

1 x 8 + 1 = 9 12 x 8 + 2 = 98

123 x 8 + 3 = 987 1234 x 8 + 4 = 9876

12345 x 8 + 5 = 98765

Write the next two steps. Can you say how the pattern works? (Hint: 12345 = 11111 + 1111 + 111 + 11 + 1).

Solution: Next two steps are as follows:

123456 8 6 = 987654

1234567 8 7 = 9876543.

HOTS (High Order Thinking Skills):-

Q1. Use simple method to complete the following show the steps involved.

a. 9 + 99 – 999 + 9999

Ans= (10 – 1) + (100 – 1) – (1000 – 1) + (10000 – 1)

= 11110 – 4

= 11106

b. 333 222 666


111666

666111

666666111

6661112333

UNSOLVED

Q2. Find a whole number ‘P’ such that P + P = P

Q3. Find the product of the largest 2-digits number and the largest 3-digits number using

suitable property.

VALUE BASED QUESTIONS:

Q1. Use numbers 1 to 8 to fill the circles

What value do you from this?

If we will ‘2’ & ‘1’

We have seen that all division, multiplication, addition and subtraction are used from

this. The value of number is very powerful. This sum is develop the BODMAS-rule.

ACTIVITY

To verify that addition is commutative for whole numbers by paper cutting and pasting.


CHAPTER – 3

(PLAYING WITH NUMBERS)

INTRODUCTORY QUESTIONS

Q1. Define BODMAS?

Ans Acording to BODMAS a calculation involves a combination of the fundamental

operations (+, -, , )

a) Division

b) Multiplication

c) Addition

d) Substraction

“Bracket of DMAS”

Q2. Which operation should you perform first when you evaluate 65 – (8 + 355) 4

Ans Division

Q3. What are the factors?

Ans A factor of a number is a number which divides that number exactly that is, without a

Remainder.

Q4. What are the numbers?

Ans A multiple of any natural number is a number formed by multiplying that no. by any

whole number.

NCERT SOLUTIONS

1. Write all the factors of the following numbers: (i) 24 (ii) 75 (iii) 21 (iv) 27 (v) 12 (vi) 20 (vii) 18 (viii) 23 (ix) 36.

Solution: (i) 24 24 = 1 x 24 24 = 2 x 12 24 = 3 x 8 24 = 4 x 6 Thus, all the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.


(ii) 15 15 = 1 x 15 15 = 3 x 5 Thus, all the factors of 15 are 1, 3, 5 and 15. (iii) 21 21 = 1 x 21 21 = 3 x 7 Thus, all the factors of 21 are 1, 3, 7 and 21. (iv) 27 27 = 1 x 27 27 = 3 x 9 Thus, all the factors of 27 are 1, 3, 9 and 27. (v) 12 12 = 1 x 12 12 = 2 x 6 12 = 3 x 4 Thus, all the factors of 12 are 1, 2, 3, 4, 6 and 12.

(vi) 20 20 = 1 x 20 20 = 2 x 10 20 = 4 x 5 Thus, all the factors of 20 are 1, 2, 4, 5, 10 and 20.

(vii) 18 18 = 1 x 18 18 = 2 x 9 18 = 3 x 6 Thus, all the factors of 18 are 1, 2, 3, 6, 9 and 18.

(viii) 23 23 = 1 x 23 Thus, all the factors of 23 are 1 and 23.

(ix) 36 36 = 1 x 36 36 = 2 x 18 36 = 3 x 12 36 = 4 x 9


36 = 6 x 6 Thus, all the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

2. Write first five multiples of (i) 5 (ii) 8 (iii) 9.

Solution: (i) 5 First five multiples of 5 are 5 x 1, 5 x 2, 5 x 3, 5 x 4 and 5 x 5 i.e., 5, 10, 15, 20 and 25. (ii) 8 First five multiples of 8 are 8 x 1, 8 x 2, 8 x 3, 8 x 4 and 8 x 5 i.e., 8, 16, 24, 32 and 40. (iii) 9 First five multiples of 9 are 9 x 1, 9 x 2, 9 x 3, 9 x 4 and 9 x 5 i.e., 9, 18, 27, 36 and 45.

3. What is the sum of two: (a) Odd numbers (b) Even numbers.

Solution: (a) The sum of any two odd numbers is an even number. (b) The sum of any two even numbers is an even number.

4. The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3. Find such pairs of prime numbers upto 100.

Solution: The other such pairs of prime numbers upto 100 are as follows: 17, 71; 37, 73; 79, 97.

5. Write down separately the prime and composite numbers less than 20.

Solution:

Prime numbers less than 20 Composite numbers less than 20

2 4

3 6

5 8

7 9

11 10


13 12

17 14

19 15

16

18

6. What is the greatest prime number between 1 and 10?

Solution: The greatest prime number between 1 and 10 is 7.

7. Express the following as the sum of two odd primes: (a) 44 (b) 36 (c) 24 (d) 18

Solution: (a) 44 = 7 + 37 (b) 36 = 5 + 29 (c) 24 = 5 + 19 (d)18 = 5 + 13.

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

Solution: The three pairs of prime numbers, whose difference is 2, are as follows: 3 and 5 ; 5 and 7 ; 11 and 13.

9. Which of the following numbers are prime? (a) 23 (b) 51 (c) 37 (d) 26.

Solution: (a) 23 and (c) 37 are prime numbers.

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

Solution: 84, 85, 86, 87, 88, 89, 90.

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

Solution: (a) 21 = 3 + 7 + 11 (b) 31 = 7 + 11 + 13

.


(c) 53 = 11 + 13 + 29 (d) 61 = 11 + 13 + 37.

12. Write five pairs of prime numbers below 20 whose sum is divisible by 5. (Hint 3 + 7 = 10).

Solution: 2 and 3 ; 2 and 13 ; 3 and 7 ; 3 and 17 ; 11 and 19.

13. Fill in the blanks in the following : (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 ………….

Solution: (a) A number which has only two factors is called a prime number.

(b) A number which has more than two factors is called a composite number.

(c) 1 is neither prime nor composite.

(d) The smallest prime number is 2.

(e) The smallest composite number is 4.

(f) The smallest even number is 2.

14. Using divisibility tests, determine which the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10; by 11; (say yes or no);

Number Divisible by

2 3 4 5 6 8 9 10 11

128 Yes No Yes No No Yes No No No 990 --- --- --- --- --- --- --- --- --- 1586 --- --- --- --- --- --- --- --- --- 275 --- --- --- --- --- --- --- --- ---


6686 --- --- --- --- --- --- --- --- --- 639210 --- --- --- --- --- --- --- --- --- 42971 --- --- --- --- --- --- --- --- --- 2856 --- --- --- --- --- --- --- --- --- 3060 --- --- --- --- --- --- --- --- --- 406839 --- --- --- --- --- --- --- --- ---

Solution:

Number Divisible by

2 3 4 5 6 8 9 10 11

128 Yes No Yes No No Yes No No No 990 Yes Yes No Yes Yes No Yes Yes Yes 1586 Yes No No No No No No No No 275 No No No Yes No No No No Yes 6686 Yes No No No No No No No No 639210 Yes Yes No Yes Yes No No Yes Yes 429714 Yes Yes No No Yes No Yes No No 2856 Yes Yes Yes No Yes Yes No No No 3060 Yes Yes Yes Yes Yes No Yes Yes No 406839 No Yes No No No No No No No

15. Using divisibility tests, determine which of the following numbers are divisible by 4 ; by 8: (a) 572 (b) 726352 (c) 5500 (d) 6000 (e) 12159 (f) 14560 (g) 21084 (h) 31795072 (i) 1700

Solution: (a) 572 (i) Divisibility by 4 The number formed by last two digitis = 72

Remainder is 0

72 is divisible by 4


(ii) Divisibility by 8 The number formed by last three digits = 572

Remainder is not 0


572 is not divisible by 8. (b) 726352 (i) Divisibility by 4 The number formed by last two digits = 52

Remainder is 0




Remainder is 0



(c) 5500 (i) Divisibility by 4 The number formed by last two digits = 00, which is divisible by 4



Reminder is not '0'

500 is not divisible by 8

5500 is not divisble by 8.

(d) 6000 (i) Divisibility by 4 The number formed by last two digits = 00,which is divisible by 4

6000 is divisible by 4.


(ii) Divisibility by 8 The number formed by last three digits = 000, which is divisible by 8


(e) 12159 (i) Divisibility by 4 The number formed by last two digits = 59

Remainder is 0


12159 is not divisible by 4 (ii) Divisibility by 8 The number formed by last three digits = 159

Remainder is not 0



(f) 14560 (i) Divisibility by 4 The number formed by last two digits = 60

Remainder is 0

14560 is divisible by 4 (ii) Divisibility by 8 The number formed by last three digits = 560


Remainder is 0



(g) 21084 (i) Divisibility by 4 The number formed by last two digits = 84

Remainder is 0



(ii) Divisibility by 8 The number formed by last three digits = 084 = 84

Remainder is not 0


21084 is not divisible by 8 (h) 31795072 (i) Divisibility by 4 The number formed by two digits = 72

Remainder is 0




(ii) Divisibility by 8 The number formed by last three digits = 072 = 72

Remainder is 0


31795072 is divisible by 8 (i) 1700 (i) Divisibility by 4 The number formed by last two digits = 00, which is divisible by 4



Remainder is not 0



(j) 2150 (i) Divisibility by 4 The number formed by last two digits = 50

Remainder is not 0





Remainder is not 0



16. 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) 12853 (i) 17852.

Solution: (a) 297144 (i) Divisibility by 2

Unit’s digit = 4


(ii) Divisibility by 3 Sum of the digits = 2 + 9 + 7 + 1 + 4 + 4 = 27, which is divisible by 3

297144 is divisible by 3 Since 297144 is divisible by 2 and 3 both, so it is divisible by 6. (b) 1258 (i) Divisibility by 2 Unit's digit = 8


(ii) Divisibility by 3 Sum of the digits =1 + 2 + 5 + 8 = 16, which is not divisible by 3

1258 is not divisible by 3. Since 1258 is divisible by 2 but not by 3, so 1258 is not divisible by 6. (c) 4335 (i) Divisibility by 2

Unit's digit = 5, which is not any of the digits 0, 2, 4, 6 or 8


4335 is not divisible by 6.


(d) 61233 (i) Divisibility by 2 Unit's digit = 3, which is not any of the digits 0, 2,4, 6 or 8



(e) 901352 (i) Divisibility by 2

Unit's digit = 2


(ii) Divisibility by 3 Sum of the digits = 9 + 0 + 1 + 3 + 5 + 2 = 20, which is not divisible by 3

901352 is not divisible by 3 Since 901352 is divisible by 2 but not by 3, so it is not divisible by 6. (f) 438750 (i) Divisibility by 2

Unit's digit = 0



438750 is divisible by 3 Since 438750 is divisible by both 2 and 3 , so it is divisible by 6.

(g) 1790184 (i) Divisibility by 2 Unit's digit = 4


(ii) Divisibility by 3 Sum of the digits = 1+ 7 + 9 + 0 + 1 + 8 + 4 = 30, which is divisible by 3

1790184 is divisible by 3 Since 1790184 is divisible by 2 and 3 both, so it is divisible by 6. (h) 12583 (i) Divisibility by 2

Unit's digit = 3, which is not any of the digits 0, 2, 4, 6 or 8

12583 is not divisible by 2 12583 is not divisible by 6.


(i) 639210 (i) Divisibility by 2 Unit's digit = 0



639210 is divisible by 3 Since 639210 is divisible by both 2 and 3, so it is divisible by 6. (i) 17852 (i) Divisibility by 2 Unit's digit = 2


(ii) Divisibility by 3 Sum of the digits = 1+ 7 + 8 + 5 + 2 = 23, which is not divisible by 3

17852 is not divisible by 3 Since 17852 is divisible by 2 but not by 3, so it is not divisible by 6.

17. Using divisibility tests, determine which of the following numbers are divisible by 11: (a) 5445 (b) 10824 (c)7138965 (d) 70169308 (e) 10000001 (f) 901153.

Solution: (a) 5445 Sum of the digits (at odd places) from the right = 5 + 4 = 9 Sum of the digits (at even places) from the right = 4 + 5 = 9 Difference of these sums = 9 - 9 = 0


5445 is divisible by 11. (b) 10824 Sum of the digits (at odd places) from the right = 4 + 8 + 1 = 13 Sum of the digits (at even places) from the right = 2 + 0 = 2 Difference of these sums =13 – 2 = 11 11 is divisible by 11


(c) 7138965 Sum of the digits (at odd places) from the right = 5 + 9 + 3 + 7 = 24 Sum of the digits (at even places) from the right = 6 + 8 + 1 = 15 Difference of these sums = 24 - 15 = 9




(d) 70169308 Sum of the digits (at odd places) from the right = 8 + 3 + 6 + 0 = 17 Sum of the digits (at even places) from the right = 0 + 9 + 1+7 = 17 Difference of these sums = 17 - 17 = 0 0 is divisible by 11 70169308 is divisible by 11.

(e) 10000001 Sum of the digits (at odd places) from the right = 1 + 0 + 0 + 0 = 1 Sum of the digits (at even places) from the right = 0 + 0 + 0 + 1 = 1 Difference of these sums = 1 - 1 = 0



(f) 901153 Sum of the digits (at odd places) from the right = 3 + 1 + 0 = 4 Sum of the digits (at even places) from the right = 5 + 1+ 9 =15 Difference of these sums =15 - 4 =11 11 is divisible by 11


18. Write the smallest digit and the largest digit in the blank space of each of the following numbers so that the number is divisible by 3: (a) _6724 (b) 4765_2.

Solution: (a) __ 6724 (i) Smallest digit Sum of the given digits = 6 + 7 + 2 + 4 = 19


Smallest digit (non-zero) is 2.

(ii) Largest digit The largest digit is 8.

(b) 4765 _ 2 (i) Smallest digit Sum of the given digits = 4 + 7 + 6 + 5 + 2 = 24


Smallest digit is 0.


(ii) Largest digit The largest digit is 9.

19. Write digit in the blank space of each of the following numbers so that the number is divisible by 11: (a) 92 _ 389 (b) 8 _ 9484.

Solution: (a) 92 _ 389 Sum of the given digits (at odd places) from the right = 9 + 3 + 2 = 14 Sum of the given digits (at even places) from the right = 8 + 9 = 17 The above sums show that for the difference of the sums of the digits at odd places and at even places to be divisible by 11, the digit in the blank space must be 8. Hence, the required number is 928389. (b) 8 _ 9484 Sum of the given digits (at odd places) from the right = 4 + 4 = 8 Sum of the given digits (at even places) from the right = 8 + 9 + 8 = 25

The above sums show that for the difference of the sums of the digits at odd places and at even places to be divisible by 11, the digit in the blank space must be 6. Hence, the required number is 869484.

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

Solution: (a) 20 and 28 Factors of 20 are 1, 2, 4, 5, 10 and 20. Factors of 28 are 1, 2, 4, 7, 14 and 28. Hence, the common factors of 20 and 28 are 1, 2 and 4.

(b) 15 and 25 Factors of 15 are 1, 3, 5 and 15. Factors of 25 are 1, 5 and 25. Hence, the common factors of 15 and 25 are 1 and 5. (c) 35 and 50 Factors of 35 are 1, 5, 7 and 35. Factors of 50 are 1, 5, 10, 25 and 50. Hence, the common factors of 35 and 50 are 1 and 5. (d) 56 and 120


Factors of 56 are 1, 2, 4, 7, 8, 14, 28 and 56. Factors of 120 are 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120. Hence, the common factors of 56 and 120 are 1, 2, 4 and 8.

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

Solution: (a) 4, 8 and 12 Factors of 4 are 1, 2 and 4. Factors of 8 are 1, 2, 4 and 8. Factors of 12 are 1, 2, 3, 4, 6 and 12. Hence, the common factors of 4, 8 and 12 are 1, 2 and 4.

(b) 5, 15 and 25 Factors of 5 are 1 and 5. Factors of 15 are 1, 3 and 5. Factors of 25 are 1, 5 and 25. Hence, the common factors of 5, 15 and 25 are 1 and 5.

22. Find first three common multiples of: (a) 6 and 8 (b) 12 and 18.

Solution: (a) 6 and 8 Multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96. Multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96.

Common multiples of 6 and 8 are 24, 48, 72, 96,

First three common multiples of 6 and 8 are 24, 48 and 72. (b) 12 and 18 Multiples of 12 are 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, 132, 144 Multiples of 18 are 18, 36, 54, 72, 90, 108, 126, 144

Common multiples of 12 and 18 are 36, 72, 108, 144

First three common multiples of 12 and 18 are 36, 72 and 108.

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

Solution: Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, …


Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, …

Common multiples of 3 and 4 are 12, 24, 36, 48, 60, 72, 84, 96, 108, …

All the numbers less than 100 which are common multiples of 3 and 4 are 12, 24, 36, 48, 60, 72, 84 and 96.

24. 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 16.

Solution: (a) 18 and 35 Factors of 18 are 1, 2, 3, 6, 9 and 18. Factors of 35 are 1, 5, 7 and 35.

Common factor of 18 and 35 is 1.

18 and 35 have only 1 as the common factor

18 and 35 are co-prime numbers. (b) 15 and 37 Factors of 15 are 1, 3, 5 and 15. Factors of 37 are 1 and 37.



15 and 37 are co-prime numbers. (c) 30 and 415 Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. Factors of 415 are 1, 5, 83 and 415.

Common factors of 30 and 415 are 1 and 5. 30 and 415 have two common factors

30 and 45 are not co-prime numbers.

(d) 17 and 68 Factors of 17 are 1 and 17. Factors of 68 are 1,2, 4, 17, 34 and 68.

Common factors of 17 and 68 are 1 and 17.

17 and 68 have two common factors.

17 and 68 are not co-prime numbers. (e) 216 and 215 Factors of 216 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 72, 108 and 216. Factors of 215 are 1, 5 and 43




216 and 215 are co-prime numbers.

(f) 81 and 16 Factors of 81 are 1, 3, 9, 27 and 81. Factors of 16 are 1, 2, 4, 8 and 16.


81 and 16 have only 1 as the common factor 81 and 16 are co-prime numbers.

25. A number is divisible by both 5 and 12. By which other number will that number be always divisible?

Solution: The number will be divisible by 5 and 12 is 60. The number 60 is also divisible by 1, 2, 3, 4, 6, 10, 15, 20, 30.

26. A number is divisible by 12. By what other numbers will that number be divisible?

Solution: Factors of 12 are 1, 2, 3, 4, 6 and 12. So, that number will be divisible by 1, 2, 3, 4 and 6 also.

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

Solution: (a) This statement is false. (b) This statement is true. (c) This statement is false. (d) This statement is true. (e) This statement is false. (f) This statement is false.


(g) This statement is true. (h) This statement is true. (i) This statement is true. (j) This statement is false.

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

(b)

Solution:


(a)

(b)

29. Which factors are not included in the prime factorisation of a composite number?

Solution: 1 and composite factors are not included in the prime factorization of a composite number.

30. Write the greatest four digit number and express it in terms of its prime factors.

Solution: The greatest four digit number is 9999

9999 = 3 3 X 11 101.


31. Write the smallest five digit number and express it into the form of prime factors. Solution:

The smallest five digit number is 10000.

10000 = 2 2 2 2 5 5 5 5

32. Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any between two consecutive prime factors.

Solution:

1729 = 7 13 19. All the prime factors of 1729 are 7, 13 and 19. When arranged in ascending order, these are : 7, 13, 19. We observe that 13 - 7 = 6 19 - 13 = 6 Relation between two consecutive prime factors is that their difference is the same.


33. The product of three consecutive numbers is always divisible by 6. Explain this statement with the help of some examples.

Solution: Example 1 Take three consecutive numbers 21 , 22 and 23. 21 is divisible by 3. 22 is divisible by 2.

21 22 is divisible by 3 2 (= 6)

21 22 23 is divisible by 6. Example 2 Take three consecutive numbers 47, 48 and 49. 48 is divisible by 2 and 3 both.

48 is divisible by 2 x 3 (= 6)

47 x 48 x 49 is divisible by 6.

34. In which of the following expressions, prime factorisation has been done: (a) 24 = 2 x 3 x 4

(b) 56 = 1 x 7 x 2 2 x 2 x 2 (c) 70 = 2 x 5 x 7 (d) 54 = 2 x 3 x 9

Solution: (a) Prime factorisation has not been done. (b) Prime factorisation has been done. (c) Prime factorisation has been done. (d) Prime factorisation has not been done.

35. Write the prime factorisation of 15470.

Solution:

15470 = 2 x 5 x 7 x 221.


36. 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].

Solution: Divisibility of 25110 by 5

Number in the unit's place of 25110 = 0

25110 is divisible by 5. Divisibility of 25110 by 9 Sum of the digits of the number 25110 =2 + 5 + 1 + 1 + 0 = 9


25110 is divisible by 9 As 25110 is divisible by both 5 and 9 and as 5 and 9 are co-prime numbers, 25110 is divisible by 5 x 9 = 45.

37. 18 is divisible by both 2 and 3. It is also divisible by 2 x 3 = 6. Similarly, a number is divisible by both 4 and 6. Can we say that the number must also be divisible by 4 x 6 = 24? If not, give an example to justify your answer.

Solution: No ! we cannot say that the number will be divisible by 4 x 6 = 24, if it is divisible by both 4 and 6 because 4 and 6 are not co-prime numbers (they have two common factors are 1 and 2). Example: 36 is divisible by both 4 and 6. But, 36 is not divisible by 24.

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

Solution: 210 is the smallest number, having four different prime factors 2, 3, 5 and 7.

39. Find the H.C.F. of the following numbers. (a) 18, 48 (b) 30, 42 (c) 18, 60 (d) 27, 63 (e) 36, 84 (f) 34, 102 (g) 70, 105, 175 (h) 91, 112, 49 (i) 18, 54, 81 (j) 12, 45, 75.

Solution: (a) 18, 48 Factors of 18 are 1, 2, 3, 6, 9 and 18. Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.

Common factors of 18 and 48 are 1, 2, 3 and 6. Highest of these common factors is 6.

H.C.F. of 18 and 48 is 6.


(b) 30, 42 Factors of 30 are 1, 2, 3, 5, 6, 10, 15 and 30. Factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.



(c) 18, 60 Factors of 18 are 1, 2, 3, 6, 9 and 18. Factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.



(d) 27, 63 Factors of 27 are 1, 3, 9 and 27. Factors of 63 are 1, 3, 7, 9, 21 and 63.

Common factors of 27 and 63 are 1, 3 and 9. Highest of these common factors is 9.


(e) 36, 84 Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36. Factors of 84 are 1, 2, 3, 4, 7, 12, 14, 21, 42 and 84.

Common factors of 36 and 84 are 1, 2, 3, 4 and 12. Highest of these common factors is 12.

H.C.F. of 36 and 84 is 12. (f) 34, 102 Factors of 34 are 1, 2, 17 and 34. Factors of 102 are 1, 2, 3, 6, 17, 34, 51 and 102.


H.C.F. of 34 and 102 is 34.

(g) 70, 105, 175 Factors of 70 are 1, 2, 5, 7, 10, 14, 35 and 70. Factors of 105 are 1, 3, 5, 7, 15, 21, 35 and 105. Factors of 175 are 1, 5, 7, 25, 35 and 175.

Common factors of 70, 105 and 175 are 1, 5 and 35. Highest of these common factors is 35.

H.C.F. of 70, 105 and 175 is 35.


(h) 91, 112, 49 Factors of 91 are 1, 7, 13 and 91. Factors of 112 are 1, 2, 4, 7, 8, 14, 16, 28, 56 and 112. Factors of 49 are 1, 7 and 49.

Common factors of 91, 112 and 49 are 1 and 7. Highest of these common factors is 7.

H.C.F. of 91, 112 and 49 is 7.

(i) 18, 54, 81 Factors of 18 are 1, 2, 3, 6, 9 and 18. Factors of 54 are 1, 2, 3, 6, 9, 18, 27 and 54. Factors of 81 are 1, 3, 9, 27 and 81.

Common factors of 18, 54 and 81 are 1, 3 and 9. Highest of these common factors is 9.

H.C.F. of 18, 54 and 81 is 9.

(j) 12, 45, 75 Factors of 12 are 1, 2, 3, 4, 6 and 12. Factors of 45 are 1, 3, 5, 9, 15 and 45. Factors of 75 are 1, 3, 5, 15, 25 and 75.

Common factors of 12, 45 and 75 are 1 and 3. Highest of these common factors is 3.

H.C.F. of 12,45 and 75 is 3.

40. What is the H.C.F of two consecutive : (a) numbers ? (b) even numbers ? (c) odd numbers?

Solution: (a) The H.C.F. of two consecutive numbers is 1. (b) The H.C.F. of two consecutive even numbers is 2. (c) The H.C.F. of two consecutive odd numbers is 1.

41. H.C.F. of co-prime numbers 4 and 15 was found as follows by factorisation: 4 = 2 x 2 and 15 = 3 x 5 since there is no common prime factor, so H.C.F of 4 and 15 is 0. Is the answer correct? If not, what is the correct H.C.F

Solution: No ! the answer is not correct. The correct answer is as follows: H.C.F. of 4 and 15 is 1.


42. Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Solution: Factors of 75 are 1, 3, 5, 15, 25 and 75. Factors of 69 are 1, 3, 23 and 69.

Common factors of 75 and 69 are 1 and 3. Highest of these common factors is 3.

H.C.F. of 75 and 69 is 3. Hence, the maximum capacity of weight which can measure the weight of the fertiliser exact number of times is 3 kg.

43. 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?

Solution:

L.C.M. of 63, 70 and 77 = 2 x 3 x 3 x 5 x 7 x 11

= 6930.

Hence, the minimum distance each should cover so that all cover the distance in complete steps is 6930 cm or 69 m 30 cm.

44. 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.

Solution: Factors of 825 are 1, 3, 5, 11, 15, 25, 33, 55, 75, 165, 275 and 825. Factors of 675 are 1, 3, 5, 9, 15, 25, 27, 45, 75, 135, 225 and 675. Factors of 450 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 25, 30, 45, 50, 75, 90, 150, 225 and 450.

Common factors of 825, 675 and 450 are 1, 3, 5, 15, 25 and 75. Highest of these common factors is 75. Hence, the length of the longest tape which can measure the three dimensions of the room exactly is 75 cm.


45. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

Solution:

L.C.M. of 6, 8 and 12 = 2 x 2 x 2 x 3 = 24. Multiples of 24 are 24, 48, 72, 96, 120, 144, … Hence, the smallest 3-digit number which is exactly divisible by 6, 8 and 12 is 120.

46. Determine the largest 3-digit number exactly divisible by 8, 10 and 12.

Solution:

L.C.M. of 8, 10 and 12 = 2 x 2 x 2 x 3 x 5 = 120.

Multiples of 120 are : 120 x 1 = 120, 120 x 2 = 240, 120 x 3 = 360, 120 x 4 = 480, 120 x 5 = 600, 120 x 6 = 720, 120 x 7 = 840, 120 x 8 = 960, 120 x 9 = 1080

Hence, the largest 3-digit number exactly divisible by 8, 10 and 12 is 960.

47. 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 am at what time they will change simultaneously again?

Solution: L.C.M. of 48, 72 and 108 = 2 x 2 X 2 x 2 x 3 x 3 x 3= 432. 432 seconds = 7 min 12 seconds. Hence, they will change simultaneously again 7 min 12 seconds after 7 am.


48. 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.

Solution: Factors of 403 are 1, 13, 31 and 403. Factors of 434 are 1, 2, 7, 14, 31,62, 217 and 434. Factors of 465 are 1, 3, 5, 15, 31,93,155 and 465. Common factors of 403, 434 and 465 are 1 and 31. Highest of these common factors is 31.

H.C.F. of 403, 434 and 465 is 31.

Hence, the maximum capacity of the container that can measure the diesel of the three containers exact number of times is 31 litres.

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

Solution:

L.C.M. of 6, 15 and 18 = 2 x 3 x 3 x 5 = 90. Hence, the required number is 90 + 7 i.e., 97.

50. Find the smallest four digit number which is divisible by 18, 24 and 32.

Solution:


L.C.M. = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288. Multiples of 288 are: 288 x 1 = 288, 288 x 2 = 576, 288 x 3 = 864, 288 x 4 = 1152 Hence, the smallest four digit number which is divisible by 18, 24 and 32 is 1152.

51. Find the L.C.M. of the following numbers in which one number is always a multiple of 3: (a) 9 and 4 (b) 12 and 5 (c) 6 and 5 (d) 15 and 4 . Observe a common property in the obtained LC.Ms. Is L.C.M. the product of two numbers in each case? Is L.C.M. always a multiple of 3?

Solution: (a) 9 and 4

L.C.M. of 9 and 4 = 2 x 2 x 3 x 3 = 36 (= 9 x 4). (b) 12 and 5

L.C.M. of 12 and 5 = 2 x 2 x 3 x 5 = 60 (= 12 x 5).


(c) 6 and 5

L.C.M. of 6 and 5 = 2 x 3 x 5 = 30 (= 6 x 5).

(d) 15 and 4

L.C.M. of 15 and 4 = 2 x 2 x 3 x 5 = 60 (= 15 x 4). We observe a common property in the obtained L.C.M.s that L.C.M. is the product of two numbers in each case. Also, L.C.M. is always a multiple of 3.

52. Find the L.C.M. of the following numbers in which one number is the factor of the other. (a) 5, 20 (b) 6, 18 (c) 12, 48 (d) 9, 45. What do you observe in the results obtained?

Solution: (a) 5, 20 Prime factorisations of 5 and 20 are as follows: 5 = 5 20 = 2 x 2 x 5

L.C.M. of 5 and 20 = 2 x 2 x 5 = 20. (b) 6, 18 Prime factorisations of 6 and 18 are as follows: 6 = 2 x 3 18 = 2 x 3 x 3

L.C.M. of 6 and 18 = 2 x 3 x 3 = 18.


(c) 12, 48 Prime factorisations of 12 and 48 are as follows: 12 = 2 x 2 x 3 48 = 2 x 2 x 2 x 2 x 3

L.C.M. of 12 and 48 = 2 x 2 x 2 x 2 x 3 = 48. (d) 9, 45 Prime factorisations of 9 and 45 are as follows: 9 = 3 x 3 45 = 3 x 3 x 5

L.C.M. of 9 and 45 = 3 x 3 x 5 = 45.

In the results obtained, we observe that L.C.M. of the two numbers in which one number is the factor of the other is the greater number.

53. Find all the multiples of 9 upto 100. Solution: The multiples of 9 upto 100 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.

HOTS (High Order Thinking Skills)

Q1. What is the value of 24864

1

22

248

248

64

Q2. If the 8-digits number 136x5785 is divisible by 15, then find the least possible value of x.

Ans The 8 digit number = 136x5485 if the divisible is 15.

The least possible value of x is 1.

UNSOLVED

Q3. I am 2 digit prime No. my tens digit is 5 and my ones digit is a Prime Number?

Q4. What least Number should be subtracted from 26492518 so that the resulting Number

is divisible by 3, but not by 9?



Q1. A merchant has 600 litres of oil of one kind, 864 l of another kind and 312 l of third kind.

He wants to sell the oil by mixing the three kinds of oil in tins of equal capacity. What

should be the greatest capacity of such tin?

a. Is the merchant correct in mixing the oils?

b. Which value is opposite of what the merchant is practicing?

Ans.

a. Yes, the merchant correct is mixing the oils

ACTIVITY

To find the L.C.M. of three given numbers.


CHAPTER – 4

(INTEGERS)

INTRODUCTORY QUESTIONS

Q1. What are whole numbers?

Ans If we include 0 in the set of natural Numbers, we have the set of whole Numbers.

Q2. What are positive integers?

Ans the numbers +1, +2, +3, +4, +5, +6,…… can be written as 1,2,3,4,5,6,….. the positive sign

is omitted and understood these are called (+)ve integers.

Q3. What are the negative integers?

Ans That sign is used is the placement of a minus sign to the number. This indicates that no.

with a(-)ve signs are less than zero. These are called (-)ve integers.

Q4. Can you represent the integers on number line?

NCERT SOLUTIONS

1. Write opposite of the following: (a) Increase in weight (b) 30 km north (c) 326 BC (d) Loss of Rs. 700 (e) 100 m above sea level.

Solution: (a) Decrease in weight (b) 30 km south (c) 326 AC (d) Profit of Rs. 700 (e) 100 m below sea level.

2. Represent the following numbers as integers with appropriate signs. (a) An aeroplane is flying at a height two thousand metre above the ground. (b) A submarine is moving at a depth, eight hundred metre below the sea level. (c) A deposit of rupees two hundred. (d) Withdrawal of rupees seven hundred.


Solution: (a) +2000 metre (b) -800 metre (c) +Rs. 200 (d) -Rs. 700.

3. Represent the following numbers on a number line: (a) + 5 (b) -10 (c) + 8 (d) - 1 (e) - 6.

Solution: (a) + 5 (b) - 10

(c) + 8

(d) - 1

(e) - 6

4. Adjacent is a vertical number line, representing integers. Observe it and locate the following points:


(a) If point D is + 8, then which point is - 8? (b) Is point G a negative integer or a positive integers? (c) Write integers for points B and E. (d) Which point marked on this number line has the least value? (e) Arrange all the points in decreasing order of value.

Solution: (a) Point F (b) negative integer (c) Integers for points B and E are + 4 and - 10 respectively. (d) Point E (e) D, C, B, A, O, H, G, F, E.

5. Following is the list of temperature of five places in India, on a particular day of the year. Place Temperature Siachin 10°C below 0°C ........... Shimla 2°C below 0°C ........... Ahmedabad 30°C above 0°C ........... Delhi 20°C above 0°C ........... Srinagar 5°C below 0°C ........... (a) Write the temperature of these places in the form of integers in the blank column.


(b) Following is the number line representing the temperature in degree Celsius. Plot the name of the city against its temperature.

(c) Which is the coolest place ? (d) Write the names of the places whose temperature are above 10°C.

Solution: (a)Place Temperature Siachin 10°C below 0°C - 10°C Shimla 2°C below 0°C - 2°C Ahmedabad 30°C above 0°C + 30°C Delhi 20°C above 0°C + 20°C Srinagar 5°C below 0°C - 5°C

(b)

(c) Siachin is the coolest place. (d) Delhi and Ahmedabad.

6. In each of the following pairs, which number is to the right of the other on the number line? (a) 2, 9 (b) - 3, - 8 (c) 0, - 1 (d) - 11, 10 (e) - 6, 6 (f) 1, - 100.

Solution: (a) 2, 9 The number 9 is to the right of the number 2.

(b) - 3, - 8 The number - 3 is to the right of the number - 8.

(c) 0, -1 The number 0 is to the right of the number - 1.


(d) - 11, 10 The number 10 is to the right of the number -11.

(e) - 6, 6 The number 6 is to the right of the number - 6.

(f) 1, -100 The number 1 is to the right of the number - 100.

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7. Write all the integers between the given pairs (write them in the increasing order.) (a) O and -7 (b) – 4 and 4 (c) – 8 and -15 (d) – 30 and - 23.

Solution: (a) 0 and - 7 The integers between 0 and - 7 in increasing order are - 6, - 5, - 4, - 3, - 2 and - 1. (b) - 4 and 4 The integers between - 4 and 4 in increasing order are - 3, - 2, - 1, 0, 1, 2 and 3. (c) - 8 and -15 The integers between - 8 and - 15 in increasing order are - 14, - 13, - 12, - 11,- 10 and –9 (d) - 30 and - 23 The integers between - 30 and - 23 in increasing order are - 29, - 28, - 27, - 26, - 25 and - 24.

8. (a) Write four negative integers greater than - 20. (b) Write four negative integers less than –10.

Solution: (a) Four negative integers greater than - 20 are - 19, - 18, - 17 and - 16. (b) Four negative integers less than - 10 are - 11, - 12, - 13 and - 14.

9. For the following statements write True (T) or False (F). If the statement is false, correct the statement. (a) - 8 is to the right of-10 on a number line. (b) -100 is to the right of- 50 on a number line. (c) Smallest negative integer is — 1. (d) - 26 is larger than - 25.


Solution:

(a) True (T) (b) False (F); - 50 is to the right of - 100 on a number line. (c) False (F); Largest negative integer is - 1. (d) False (F); - 26 is smaller than - 25.

10. Draw a number line and answer the following: (a) Which number will we reach if we move 4 numbers to the right of - 2. (b) Which number will we reach if we move 5 numbers to the left of 1. (c) If we are at - 8 on the number line, in which direction should we move to reach -13? (d) If we are at - 6 on the number line, in which direction should we move to reach -1?

Solution: (a) We will reach the number 2.

(b) We will reach the number - 4.

(c) We should move in the left direction.

(d) We should move in the right direction.

.

11. Using number line write the integer which is: (a) 3 more than 5 (b) 5 more than - 5 (c) 6 less than 2 (d) 3 less than - 2.

Solution: (a) 3 more than 5 We start from 5 and proceed 3 steps to the right of 5 to reach 8 as shown below:

Therefore, 3 more than 5 is 8. (b) 5 more than - 5 We start from - 5 and move to the right by 5 steps and obtain 0 as shown below:


Therefore, 5 more than - 5 is 0.

(c) 6 less than 2

We start from 2 and proceed 6 steps to the left of 2 to reach - 4 as shown below:

(d) 3 less than - 2 We start from - 2 and move to the left by 3 steps and obtain - 5 as shown below:

Therefore, 3 less than - 2 is - 5.

12. Use number line and add the following integers: (a) 9 + (- 6) (b) 5 + (-11) (c) (-1) + (-7) (d) (- 5)+ 10 (e) (- 1) + (- 2)+ (- 3) (f) (- 2)+ 8 + (- 4).

Solution: (a) 9 + (- 6) On the number line we first move 9 steps to the right from 0 reaching 9 and then we move 6 steps to the left of 9 and reach 3. Thus, 9 + (- 5) = 3.

(b) 5 + (-11) On the number line we first move 5 steps to the right from 0 reaching 5 and then we move 11 steps to the left of 5 reach –6. Thus, 5 +(-11) = -6


(c) (-1) + (-7) On the number line we first move 1 step to the left of 0 reaching –1, then we move 7 steps to the left of –1 and reach –8. Thus, (-1) + (+7) = -8

(d) (-5) + 10 First we move 5 steps to the left of 0 reaching 05, then from this point we move 10 steps to the rights. We reach the point +5. Thus, (-5) + 10 = 5

(e) (-1) + (-2) + (-3) First we move 1 step to the left of 10 reaching –1, then from this point we move 2 steps to the left to reach –3 and finally from –3, we move 3 steps to the left. We reach the point –6. Thus, (-1) + (-2) + (-3) = -6

(f) (-2) + 8 +(-4) First we move 2 steps to the left of 0 reaching - 2, then from this point we move 8 steps to the right to reach + 6 and finally from + 6 we move 4 steps to the left. We reach the point Thus, (-2) + 8 + (-4) = 2.

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13. Add without using number line: (a) 11 + (- 7) (b) (- 13) + (+18) (c) (- 10) + (+ 19) (d) (- 250) + (+150) (e) (- 380) + (- 270) (f) (- 217) + (- 100).

Solution: (a) 11+ (-7) = 11+ (- 7) = 4 + 7 + (- 7) = 4 + 0 = 4.


(b) (-13) + (+18) = (-13) + (+18) = (-13) + (+13) + (+5) = 0 + (+ 5) = 5. (c) (- 10) + (+ 19) = (-10) + (+19) = (- 10) + (+ 10) + (+ 9) = 0 + (+ 9) = 9. (d) (- 250) + (+ 150) = (-250)+ (+ 150) = (- 100) + (- 150) + (+ 150) = (- 100) + 0 = -100. (e) (- 380) + (- 270) = (- 380) + (- 270) = - 650. (f) (- 217)+ (- 100) = - 317.

14. Find the sum of: (a) 137 and - 354 (b) - 52 and 52 (c) - 312, 39 and 192 (d) - 50, - 200 and 300.

Solution: (a)137+ (-354) = 137 + (- 137) + (- 217) = 0 + (-217) = -217

(b) - 52 and 52 = - 52 + (52) = 0

(c) - 312 + 39 + 192 = (- 312) + (39) + (192) = (- 312)+ (231) = (- 81)+ (- 231) + (231) = (- 81) + 0 = - 81

(d) - 50 + (- 200) + 300 = (- 50)+ (- 200)+ (300) = (- 250) + (300) = (- 250) + (250) + (50) = 0 + (50) = 50.

15. Find the value of: (a) (-7) + (- 9) + 4 + 16 (b) (37) + (- 2) + (- 65) + (- 8)

Solution: (a) (- 7) + (- 9) + 4 + 16 = (- 7) + (- 9) + 4 + 16 = (- 16) + 20


= (- 16) + 16 + 4 = 0 + 4 = 4 (b) (37) + (- 2) + (- 65) + (- 8) = (37)+ (- 75) = -38.

16. Subtract : (a) 35 - (20) (b) 72 - (90) (c) (-15) - (-18) (d) (- 20) - (13) (e) 23 - (-12) (f) (- 32) - (- 40).

Solution: (a) 35 - (20) = 35 - (20) = 35 + (additive inverse of 20) = 35 + (- 20) = 15 + 20 + (- 20) = 15 + 0 = 15

(b) 72 - 90 72 - 90 = 72 + (additive inverse of 90) = 72 + (- 90) = 72 + (- 72) + (- 18) = 0 + (- 18) = - 18

(c) (- 15) - (- 18) = (- 15) + (additive inverse of - 18) = (- 15) + (18) = (- 15) + (15) + (3) = 0 + (3) = 3

(d)(- 20) - (13) = (-20) - (13) = (- 20) + (additive inverse of 13) = (-20)+ (-13) = -33

(e) 23 - (- 12) = 23 - (- 12) = 23 + (additive inverse of - 12) = 23 + 12 = 35

(f) (- 32) - (- 40) = (-32) - (- 40) = (- 32) + (additive inverse of - 40) = (- 32)+ (+40) = (- 32)+ (+32)+ (+8) = 0 + (+8) = 8.


17. Fill in the blanks with >, < or = sign: (a) (- 3) + (- 6).........(- 3) - (- 6) (b) (- 21) - (-10).........(- 31) + (-11) (c) 45 - (-11).........57 + (- 4) (d) (- 25) - (- 42).........(- 42) - (- 25).

Solution: (a) L.H.S. = (- 3) + (- 6) = - 9

R.H.S. = (- 3) - (- 6) = (- 3) + (additive inverse of- 6) = (-3) + 6 = (- 3) + 3 + 3 = 0+ 3 = 3

(-3) + (-6) < (-3) - (-6) (b) L.H.S. = (- 21) - (-10) = (- 21) + (additive inverse of - 10) = (- 21) + 10 = (- 11)+ (- 10)+10 = (- 11) + 0 = - 11

R.H.S. = (- 31) + (- 11) = - 42

(- 21) - (- 10) > (- 31) + (- 11)

(c) L.H.S. = 45 - (- 11) = 45 + (additive inverse of - 11) = 45 + 11= 56 R.H.S. = 57 + (- 4) = 53 + 4 + (- 4) = 53 + 0 = 53

45 - (- 11) > 57 + (- 4)

(d) L.H.S. = (- 25) - (- 42) = (- 25) + (additive inverse of - 42) = (- 25)+ (+ 42) = (- 25)+ (+ 25)+ (+17) = 0 + (+17) = 17 R.H.S.= (- 42) - (- 25) = (- 42) + (additive inverse of - 25) = (- 42)+ (+25) = (- 17) + (- 25) + (+25) = (- 17) + 0 = - 17

-25 - (- 42) > (- 42) - (- 25).


18. Fill in the blanks: (a) (- 8) +......= 0 (b) 13 +.....=0 (c) 12 + (- 12) =...... (d) (- 4) +......+ = -12 (e) ...... - 15 = - 10.

Solution: (a) (- 8) + 8 = 0 (b) 13 + - 13 = 0 (c) 12 + (- 12) = 0. (d) (- 4) + (- 8) = -12 (e) (+5) - 15 = -10.

19. Find the value of: (a) (- 7) - 8 - (-25) (b) (- 13) + 32 - 8 - 1 (c) (- 7)+ (- 8)+ (- 90) (d) 50 - (- 40) - (- 2)

Solution: (a) (- 7) - 8 - (- 25) = (- 7) + (additive inverse of 8) - (- 25) = (- 7)+ (- 8) - (- 25) = - 15 - (- 25) = - 15 + (additive inverse of - 25) = - 15+ (+25) = - 15 + (+15) + (+10) = 0 + (+10) = 10

(b)(- 13) + 32 - 8 - 1 = (- 13) + 32 - 8 - 1 = (- 13) + 32 - 9 = (- 13) + 32 + (additive inverse of 9) = (- 13) + 32 + (- 9) = (- 13) + 23 + 9 + (- 9) = (- 13) + 23 + 0 = (- 13) + 23 = (- 13) + 13 + 10 = 0 + 10 = 10. (c) (- 7) + (- 8) + (- 90) = (- 7) + (- 8) + (- 90) = (- 15) + (- 90) = - 105. (d) 50 - (- 40) – (- 2) = 50 - (- 40) – (- 2) = 50 + (additive inverse of - 40) – (- 2)


= 50 + (40) – (- 2) = 90 – (- 2) = 90 + (additive inverse of – 2) = 90 + 2 = 92.

HOTS (High Order Thinking Skills) Q1. Find the least integer that could replace the in order to make the sentence.

true67816

Ans

Q2. Evaluate : 1+(-2)+3+(-4)+5+(-6)+7+(-8)+9+(-10)

Ans 1-2+3-4+5-6+7-8+9-10

-1-1-1-1-1

= - 5

Q3. Find two integers whose sum is 2 and differences is 8.

Q4. If + means – and – means + , then what is the value of +9 – 8 – 7 + 6 – 5 – 4 + 3 + 2 + 1?


Q1. A shopkeeper had a profit of Rs 8000/- on Monday, a loss of Rs 3200/- on Tuesday, a

loss of Rs 1560/- on Wednesday and a profit of Rs 1275/- on Thursday. But he was

happy.

What value do you learn from this transaction?

Ans Shopkeeper had a profit on Monday & Thursday = Rs 8000 + Rs 1275

= Rs 9275/-

Shopkeeper had a loss on Tuesday & Wednesday = Rs 3200 + Rs 1560

6

71

732

73121

73816


= Rs 4760/-

He was happy b’coz he had more profit than loss.

We have learnt from this transaction the value of profit is more than loss.

ACTIVITY

To find the product of two integers.

Class 6 CHAPTER 1 KNOWING OUR NUMBERS...

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