Ncert Solutions For Class 8 Maths Ex 6.1

Ncert Solutions For Class 8 Maths Chapter 6 Ex 6.1

Question-1) Determine which digit will be at the unit’s place in the squares of the numbers given below:

  1. 36
  2. 273
  3. 798
  4. 3864
  5. 58637
  6. 63545
  7. 16542
  8. 45640
  9. 98231
  10. 89999

Solution:

“Say x is at the unit’s place in a number, then its square will have unit digit = x*x.”

1)36

Digit at unit’s place = 6

Unit digit in 362 = 6*6 = 36

So, the unit digit in 362 is 6.

2) 273

Digit at unit’s place = 3

Unit digit in 2732 = 3*3 = 9

So, the unit digit in 2732 is 9.

3) 798

Digit at unit’s place = 8

Unit digit in 7982 = 8*8 = 64

So, the unit digit in 7982 is 4.

4) 3864

Digit at unit’s place = 4

Unit digit in 38642 = 4*4 = 16

So, the unit digit in 38642 is 6.

5) 58637

Digit at unit’s place = 7

Unit digit in 586372 = 7*7 = 49

So, the unit digit in 586372 is 9.

6) 63545

Digit at unit’s place = 5

Unit digit in 635452 = 5*5 = 25

So, the unit digit in 635452 is 5.

7) 16542

Digit at unit’s place = 2

Unit digit in 165422 = 2*2 = 4

So, the unit digit in 165422 is 4.

8) 45640

Digit at unit’s place = 0

Unit digit in 456402 = 0*0 = 0

So, the unit digit in 456402 is 0.

9) 98231

Digit at unit’s place = 1

Unit digit in 982312 = 1*1 = 1

So, the unit digit in 982312 is 1.

10) 89999

Digit at unit’s place = 9

Unit digit in 899992 = 9*9 = 81

So, the unit digit in 899992 is 1.

 

Question-2) Explain why following numbers are not perfect squares.

  1. 1263
  2. 654657
  3. 25000
  4. 23438
  5. 888080
  6. 895352
  7. 35500000
  8. 798657

Solution:

“The square of numbers generally ends with 0,1,5,6, or 9. Perfect square always ends with even numbers of zeros.”

1) 1263

Digit at unit’s place = 3

∴ this number is not a perfect square.

2) 654657

Digit at unit’s place = 7

∴ this number is not a perfect square.

3) 25000

Digit at unit’s place = 0

But the given number contains three 0’s and that is odd number and as a perfect square cannot end with odd numbers of zeros.

∴ this number is not a perfect square.

4) 23438

Digit at unit’s place = 8

∴ this number is not a perfect square.

5) 888080

Digit at unit’s place = 0

But the given number contains one 0 and that is odd number and as a perfect square cannot end with odd numbers of zeros.

∴ this number is not a perfect square.

6) 895352

Digit at unit’s place = 2

∴ this number is not a perfect square.

7) 35500000

Digit at unit’s place = 0

But the given number contains five 0’s and that is odd number and as a perfect square cannot end with odd numbers of zeros.

∴ this number is not a perfect square.

8) 798657

Digit at unit’s place = 7

∴ this number is not a perfect square.

 

Question-3) From the numbers given below which number’s square would be the odd number?

  1. 541
  2. 667
  3. 2558
  4. 3250

Solution:

We know that, “the square of any odd number will be odd and the square of any even number will be even.”

From the numbers given in question 541 and 667 are odd numbers and 2558 and 3250 are even numbers.

So, the square of 541 and 667 will be an odd number.

 

Question-4) Find out the missing number by observing the pattern given below.

212 = 441

2012 = 40401

20012 = 4004001

200012 = 400040001

20000000012 = ________

Solution:

It can be seen from the pattern that in a square of a given number there is equal number of 0’s both the sides of the middle digit 4.

So, it can be said that

 20000000012 = 4000000004000000001

This is the missing number.

 

Question-5) Find out the value of x by observing the pattern given below.

92 = 81

992 = 9801

9992 = 998001

 99992 = 99980001

x2 = 99999980000001

Solution:

It can be seen from the pattern that if a number contain n number of nines than the square of that number is of the form,

(n – 1) numbers of nines then 8 then (n – 1) numbers of zeros then 1

i.e. (n- 1)9’s 8 (n – 1)0’s 1

In the question the x2 = 99999980000001 is given.

This number contains six 9’s and six 0’s.

The number of nines in a square should be (n – 1)

So, here (n – 1) = 6

n = 6 + 1

n = 7

So, the required number is

x = 9999999

 

Question-6) Find out the missing number X, Y and Z by observing the pattern given below.

62+422+72=432

92+902+102=912

 132+X2+142=1832

232+5522+242=5532

362+13322+372=13332

162+Y2+172=Z2

 Solution:

It can be seen from the pattern that,

  • The middle number in L.H.S is product of the first and third number.
  • The number in the R.H.S is equal to one plus the value of middle number in the L.H.S.

Hence the missing numbers are:

132+X2+142=1832

Here, X = 13*14 or (183 – 1) = 182

162+Y2+172=Z2

Here, Y = 16*17 = 272

And Z = 272 + 1 = 273

Thus, the required numbers are

X = 182

Y = 272

Z = 273

 

Question-7) Without adding find the sum of the following series.

  1. 1 + 3 + 5 + 7 + 9 + 11 + 12
  2. 25 + 27 + 29 + 31
  3. 43 + 45 + 47 + 49

Solution:

Now the “sum of first n odd numbers is n2”.

1) 1 + 3 + 5 + 7 + 9 + 11 + 12

Here first six number are six consecutive odd numbers so there sum is

62 = 36

Thus, the sum of the given series = 36 +12 = 48

2) 25 + 27 + 29 + 31

Here the numbers given are the 13th,14th,15th and 16th odd numbers

So, there sum is

= 162122

= 256 – 144 = 112

3) 43 + 45 + 47 + 49

Here the numbers given are the 22th, 23rd, 24th and 25th odd numbers

So, there sum is

= 252212

= 625 – 441 = 184

 

Question-8) 

a) Show that 81 as a sum of 9 odd numbers.

b) Show 196 as sum of 14 odd numbers

Solution:

a) 81

Now, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81

i.e.

i.e. 81 = 92

Thus 81 is the sum of first 9 odd numbers.

b) 196

Now, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 + 25 + 27 = 196

i.e. 196 = 142

Thus 196 is the sum of first 14 odd numbers.

 

Question-9) How many numbers would be present between the squares of the numbers given below?

  1. 14 and 15
  2. 27 and 28
  3. 43 and 44
  4. 101 and 102

Solution:

As we know that “there will be 2x numbers in between the squares of the numbers x and (x + 1).

1) Count of numbers between 142and152, there will be

= 2*14 = 28 numbers.

2) Count of numbers between 272and282, there will be

= 2*27 = 54 numbers.

3) Count of numbers between 432and442, there will be

= 2*43 = 86 numbers.

4) Count of numbers between 1012and1022 there will be

= 2*101 = 202 numbers.

 

Question-10) Find out the square root of the numbers given below by division method:

  1. 2209
  2. 4624
  3. 3721
  4. 576
  5. 2809

Solution:

1) 2209

47
4 22¯¯¯¯¯09¯¯¯¯¯

-16

87 609

609

0

∴  220947

 

2) 4624

68
6 46¯¯¯¯¯24¯¯¯¯¯

-36

128 1024

1024

0

462468

 

3) 3721

61
6 37¯¯¯¯¯21¯¯¯¯¯

-36

121 121

121

0

372161

 

4) 576

64
2 5¯¯¯76¯¯¯¯¯

-4

44 176

176

0

 

57624

 

5) 2809

53
5 28¯¯¯¯¯09¯¯¯¯¯

-25

103 309

309

0

∴  280953

 

Question-11) Find the number of digits in the square root of the numbers given below (without doing any calculation).

  1. 81
  2. 121
  3. 6084
  4. 15129
  5. 328329

Solution:

1) 81

On keeping bars on the given number, we get

81 = 81¯¯¯¯¯

Here, as there is only single bar,

Therefore, the square root of 81 contain only one digit.

2) 121

On keeping bars on the given number, we get

121 =  1¯¯¯21¯¯¯¯¯

Here, as there are two bars available,

Therefore, the square root of 121 contains only two digits.

3) 6084

On keeping bars on the given number, we get

6084 =  60¯¯¯¯¯84¯¯¯¯¯

Here, as there are two bars available,

Therefore, the square root of 6084 contains only two digits.

4) 15129

On keeping bars on the given number, we get

15129 = 1¯¯¯51¯¯¯¯¯29¯¯¯¯¯

Here, as there are three bars available,

Therefore, the square root of 15129 contains only three digits.

5) 328329

On keeping bars on the given number, we get

328329 = 32¯¯¯¯¯83¯¯¯¯¯29¯¯¯¯¯

Here, as there are three bars available,

Therefore, the square root of 328329 contains only three digits.

 

Question-12) Find the square root of the following numbers (decimal numbers):

  1. 89
  2. 25
  3. 89
  4. 96
  5. 16

Solution.

1) 89

1.7
1 2¯¯¯.89¯¯¯¯¯—-1
27 189

189

0

2.891.7

 

2) 25

2.5
2 6¯¯¯.25¯¯¯¯¯

-4

45 225

225

0

 

6.252.5

 

3) 89

8.3
8 68¯¯¯¯¯.89¯¯¯¯¯—64
163 489

489

0

68.898.3

 

4) 96

6.4
6 40¯¯¯¯¯.96¯¯¯¯¯—36
124 496

496

0

40.966.4

 

5) 16

5.4
1 29¯¯¯¯¯.16¯¯¯¯¯—25
104 416

416

0

29.165.4

 

Question-13) Find the least number that can be subtracted from the numbers given below in order to get the perfect square. And also find the square root of that perfect square:

  1. 124
  2. 2049
  3. 2213
  4. 630
  5. 2824

Solution:

1) 124

11
1 1¯¯¯24¯¯¯¯¯

-1

21 24

21

03

Here, the remainder is 3, which represents that the square of 11 is 3 less than 124.

Hence, we will get perfect square by subtracting 3 from the 124.

Thus, the required number is = 124 – 3 = 121

Now, 12111

 

2) 2049

45
4 20¯¯¯¯¯49¯¯¯¯¯

-16

85 449

425

24

Here, the remainder is 24, which represents that the square of 45 is 24 less than 2049.

Hence, we will get perfect square by subtracting 24 from the 2049.

Thus, the required number is = 2049 -24 = 2045

Now, 204545

 

3) 2213

47
4 22¯¯¯¯¯13¯¯¯¯¯

-16

87 613

609

04

Here, the remainder is 4, which represents that the square of 47 is 4 less than 2213.

Hence, we will get perfect square by subtracting 4 from the 2213.

Thus, the required number is = 2213 – 4 =2209

Now, 220947

 

4) 630

25
2 6¯¯¯30¯¯¯¯¯

-4

45 230

225

05

Here, the remainder is 5, which represents that the square of 25 is 05 less than 630.

Hence, we will get perfect square by subtracting 5 from the 630.

Thus, the required number is = 630 – 5 = 625

Now, 62525

 

5) 2824

53
5 28¯¯¯¯¯24¯¯¯¯¯

-25

103 324

309

15

Here, the remainder is 15, which represents that the square of 53 is 15 less than 2824.

Hence, we will get perfect square by subtracting 15 from the 2824.

Thus, the required number is = 2824 – 15 = 2809

Now, 280953

 

Question-14) Find the least number that can be added to the numbers given below in order to get the perfect square. And also find the square root of that perfect square.

  1. 670
  2. 1840
  3. 355
  4. 1518
  5. 6230

Solution:

1) 670

25
2 6¯¯¯70¯¯¯¯¯

-4

45 270

225

45

Here, the remainder is 45.

This represents that the square of 25 is less than 670.

The next number is 26 and its square is i.e. 262 = 676.

Thus, the required number to be added to 670 = 262 – 670 = 6.

Thus, the required perfect square is

67626

 

2) 1840

42
2 18¯¯¯¯¯40¯¯¯¯¯

-16

82 240

164

76

Here, the remainder is 76.

This represents that the square of 42 is less than 1840.

The next number is 43 and its square is i.e. 432 = 1849.

Thus, the required number to be added to 1840 = 432 – 1840 = 9.

Thus, the required perfect square is

184943

 

3) 355

18
1 3¯¯¯55¯¯¯¯¯

-1

28 255

224

31

Here, the remainder is 31.

This represents that the square of 18 is less than 670.

The next number is 19 and its square is i.e. 192 = 361.

Thus, the required number to be added to 355 = 192 – 355 = 6.

Thus, the required perfect square is 6.

36119

 

4) 1518

38
3 15¯¯¯¯¯18¯¯¯¯¯

-9

68 618

544

74

 

Here, the remainder is 74.

This represents that the square of 38 is less than 1518.

The next number is 39 and its square is i.e. 392 = 1521.

Thus, the required number to be added to 1518 = 392 – 1518 = 3.

Thus, the required perfect square is

152139

 

5) 6230

78
7 62¯¯¯¯¯30¯¯¯¯¯

-49

148 1330

1184

146

 

Here, the remainder is 146.

This represents that the square of 78 is less than 6230.

The next number is 79 and its square is i.e. 792 = 6241.

Thus, the required number to be added to 6230 = 792 – 6230 = 11.

Thus, the required perfect square is

624179

 

Question-15) If the area of a square is 841m2 is given find out the length of a side.

Solution:

Say, p m is the length of a side of a square.

Given that, area of a square = p2 = 841m2

∴ p =  841

29
2 8¯¯¯41¯¯¯¯¯

-4

49 441

441

0

841 = 29

∴ p = 29 m

Thus, the length of a side of a square is 29 m.

 

Question -16) A right angled triangle XYZ, Y=90.

  1. Given that, XY = 3 mm, YZ = 4 mmthen XZ = ______.
  2. Given that, XZ= 13 mm, YZ = 5 mmthen XY = ______.

 Solution:

1) In XYZ, Y=90 is given.

∴ By using “Pythagoras theorem”, we get

XZ2=XY2+YZ2 XZ2=(3mm)2+(4mm)2 XZ2=(9+16)mm=25mm XZ=(25mm2)=5mm

2) In XYZ, Y=90 is given.

∴ By using “Pythagoras theorem”, we get

XZ2=XY2+YZ2 (13mm)2=XY2+(5mm)2 (13mm)2(5mm)2=XY2 XY2=(16925)mm=(144mm)2 XY=(144mm2)=12mm

 

Question-17) A school has 1400 books in the library. The librarian wants to arrange it in such a way that number of Horizontal lines of the books and number of vertical lines of the books are same. Find out the minimum number of books that a librarian will require to add, to make this Horizontal and vertical lines same.

Solution:

Here, it is given that there are 1400 books in the library and the numbers of horizontal and vertical lines of books are same.

For finding minimum number of books that a librarian will require to add, to make this Horizontal and vertical lines same,

We need to find the number of books that should be added to 1400 to get it done.

So, calculating the square root of 1400 and finding the perfect square out of it.

  37
3 14¯¯¯¯¯00¯¯¯¯¯

-9

67 500

469

  31

Here, the remainder is 31.

This represents that the square of 37 is less than 1400.

The next number is 38 and its square is i.e. 382 = 1444.

Thus, the required number to be added to 1400 = 382 – 1400 = 44.

Thus, the required perfect square is

1444 = 38

Thus, the required number of books to be added is 44 and there will be 38 horizontal and 38 vertical lines made.

 

Question-18) There are 820 students in a ground. Teacher instructed students to stand in such an order that the number of rows and number of columns remain same. Calculate the number of students that would left out of this order or arrangement.

Solution:

Here, there are 820 students in a ground. Teacher instructed students to stand in such an order that the number of rows and number of columns remain same.

For finding number of students that would left out of this order or arrangement,

We need to find the square root of 820 by long division method.

  28
2 8¯¯¯20¯¯¯¯¯

-4

48 420

384

  36

Here, the remainder is 36, which represents that the square of 28 is 36 less than 1820.

Hence, we will get perfect square by subtracting 36 from the 820.

Thus, the required number is = 820 – 36 = 1784

Now, 1784 = 28

So, the students will form 28 rows and 28 columns.

The number of student that left out of the arrangement is 36.

 

Question-19) Find out the possible number at the unit’s place in the square root of the number given below:

  1. 99980001
  2. 106276
  3. 6241
  4. 625

Solution.

1) Here, the 1 is at the units place in the given number.

From this possible number at the unit’s place in the square root of may be 1 or 9.

∴ unit digit of the square root of 99980001 is either 1 or 9.

2) 106276

From this possible number at the unit’s place in the square root of may be 4 or 6.

∴ unit digit of the square root of 106276 is either 4 or 6.

3) 6241

From this possible number at the unit’s place in the square root of may be 1 or 9.

∴ unit digit of the square root of 6241 is either 1 or 9.

4) 625

From this possible number at the unit’s place in the square root will be 5.

∴unit digit of the square root is 5.

 

Question-20) Find out which number is not perfect square from the numbers given below. (Without doing any calculations)

  1. 163
  2. 267
  3. 418
  4. 625

Solution:

“The square of numbers generally ends with 0,1,5,6, or 9. Perfect square always ends with even numbers of zeros.”

1) 163

Digit at unit’s place = 3

∴this number is not a perfect square.

2) 267

Digit at unit’s place = 7

∴ this number is not a perfect square.

3) 418

Digit at unit’s place = 8

∴this number is not a perfect square.

4) 625

Digit at unit’s place = 5

∴  this number is a perfect square.

 

Question-21) Calculate the square roots of 225 and 289 by the method of repeated subtraction.

Solution:

Now the “sum of first n odd numbers is n2”.

For 225

  • 225
  • 225 – 1 = 224
  • 224 – 3 = 221
  • 221 – 5 = 216
  • 216 – 7 = 209
  • 209 – 9 = 200
  • 200 – 11 = 189
  • 189 – 13 = 176
  • 176 – 15 = 161
  • 161 – 17 = 144
  • 144 – 19 = 125
  • 125 – 21 = 104
  • 104 – 23 = 81
  • 81 – 25 = 56
  • 56 – 27 = 29
  • 29 – 29 = 0

Here, as we get zero at 15th step

Thus, 225 = 15

 

For 289

  • 289
  • 289 – 1 = 288
  • 288 – 3 = 285
  • 285 – 5 = 280
  • 280 – 7 = 273
  • 273 – 9 = 264
  • 264 – 11 = 253
  • 253 – 13 = 240
  • 240 – 15 = 225
  • 225 – 17 = 208
  • 208 – 19 = 189
  • 189 – 21 = 168
  • 168 – 23 = 145
  • 145 – 25 = 120
  • 120 – 27 = 93
  • 93 – 29 = 64
  • 64 – 31 = 33
  • 33 – 33 = 0

Here, as we get zero at 17th step

Thus, 289 = 17

 

Question-22) Using Prime Factorisation method find the square roots of the numbers given below.

  1. 625
  2. 900
  3. 1521
  4. 3364
  5. 2304
  6. 9801
  7. 1089
  8. 7396
  9. 484
  10. 6400

Solution:

1) 625

5 625
5 125
5 25
5 5
1

625 = 5*5 5*5

625=55=25

 

2) 900

2 900
2 450
3 225
3 75
5 25
5 5
1

900 = 2*2 * 3*3 * 5*5

900=235=30

 

3) 1521

3 1521
3 507
13 169
13 13
1

1521 = 3*3 * 13*13

1521=313=39

 

4) 3364

2 3364
2 1682
29 841
29 29
1

3364 = 2*2 * 29*29

3364=229=58

 

5) 2304

2 2304
2 1152
2 576
2 288
2 144
2 72
2 36
2 18
3 9
3 3
1

2304 = 2*2 * 2*2 * 2*2 * 2*2 * 3*3

2304=22223=48

 

6) 9801

3 9801
3 3267
3 1089
3 363
11 121
11 11
1

9801 = 3*3 * 3*3 * 11*11

9801=3311=99

 

7) 1089

3 1089
3 363
11 121
11 11
1

1089 = 3*3 * 11*11

1089=311=33

 

8) 7396

2 7396
2 3698
43 1849
43 43
1

7396 = 2*2 * 43*43

7396=243=86

 

9) 484

2 484
2 242
11 121
11 11
1

484 = 2*2 * 11*11

484=211=22

 

10) 6400

2 6400
2 3200
2 1600
2 800
2 400
2 200
2 100
2 50
5 25
5 5
1

6400 = 2*2 * 2*2 * 2*2 * 2*2 * 5*5

6400=22225=80

 

Question-23) Find the smallest integer by which the numbers given below should be multiplied in order to get a perfect square. Also find the square root of that perfect square.

  1. 1584
  2. 3825
  3. 720
  4. 3380
  5. 1872
  6. 6

Solution:

1) 1584

2 1584
2 792
2 396
2 198
3 99
3 33
11 11
1

1594 = 2*2 * 2*2 * 3*3 * 11

This prime factor 11 is not having a pair.

As 11 is not having pair the given number cannot be a perfect square. So, we need to multiply with 11 in order to make a pair.

1594*11 = 2*2 * 2*2 * 3*3 * 11*11

∴ 17534 = 2*2 * 2*2 * 3*3 * 11*11

17534=22311=132

 

2) 3825

3 3825
3 1275
5 425
5 85
17 17
1

 

3825 = 3*3 * 5*5 * 17

This prime factor 17 is not having a pair.

As 17 is not having pair the given number cannot be a perfect square. So, we need to multiply with 17 in order to make a pair.

3825*17 = 3*3 * 5*5 * 17*17

∴ 65025 = 3*3 * 5*5 * 17*17

65025=3517=255

 

3) 720

2 720
2 360
2 180
2 90
3 45
3 15
5 5
1

720 = 2*2 * 2*2 * 3*3 * 5

This prime factor 5 is not having a pair.

As 5 is not having pair the given number cannot be a perfect square. So, we need to multiply with 5 in order to make a pair.

720*5 = 2*2 * 2*2 * 3*3 * 5*5

∴ 3600 = 2*2 * 2*2 * 3*3 * 5*5

3600=2235=60

 

4) 3380

2 3380
2 1690
5 845
13 169
13 13
1

 

3380 = 2*2 * 5 * 13*13

This prime factor 5 is not having a pair.

As 5 is not having pair the given number cannot be a perfect square. So, we need to multiply with 5 in order to make a pair.

3380*5 = 2*2 * 13*13 * 5*5

∴ 16900 = 2*2 * 13*13 * 5*5

16900=2513=130

 

5) 1872

2 1872
2 936
2 468
2 234
3 117
3 39
13 13
1

1872 = 2*2 * 2*2 * 3*3 * 13

This prime factor 13 is not having a pair.

As 13 is not having pair the given number cannot be a perfect square. So, we need to multiply with 13 in order to make a pair.

1872*13 = 2*2 * 2*2 * 3*3 * 13*13

∴ 24336 = 2*2 * 2*2 * 3*3 * 13*13

24336=22313=156

 

6) 2268

2 2268
2 1134
3 567
3 189
3 63
3 21
7 7
1

2268 = 2*2 * 3*3 * 3*3 * 7

This prime factor 7 is not having a pair.

As 7 is not having pair the given number cannot be a perfect square. So, we need to multiply with 7 in order to make a pair.

2268*7 = 2*2 * 3*3 * 3*3 * 7*7

∴15876 = 2*2 * 3*3 * 3*3 * 7*7

15876=2337=126

 

Question-24) The employees of “XYZ” company has done a charity of Rs. 4096 in all, for an orphanage. The amount donated by the single person is equal to the number of employees in the company. Find the number of employees in the company.

Solution:

Here, it is given that the amount donated by the single person is equal to the number of employees in the company.

So, the number of employees in the company can be calculated by calculating the square root of the amount of charity.

∴Number of employees in the company = 4096

2 4096
2 2048
2 1024
2 512
2 256
2 128
2 64
2 32
2 16
2 8
2 4
2 2
1

4096 = 2*2 * 2*2 2*2 * 2*2 * 2*2 * 2*2

4096 = 2*2*2*2*2*2

4096 = 64

Therefore, there are 64 employees in the Company.

 

Question-25) 7225 books are to be kept in a bookshelf of a library in such a way that each column contains as many books as the number of columns. Find out the number of column and number of books in each column.

Solution:

Here, it is given that each column contains as many books as the number of columns.

So, it can be said from the above statement that,

Number of books in each column = Number of Columns

∴ Total number of books = Number of columns * Number of books in each column

∴ Number of columns * Number of books in each column = 7225

(Numberofbooksineachcolumn)2 = 7225

∴ Number of books in each column = 7225

5 7225
5 1445
17 289
17 17
1

7225 = 5*5 * 17*17

7225 = 5*17

7225 = 85

∴ Number of books in each column = Number of Columns = 85

So, the number of columns and the number of books in each column is 85.

 

Question-26) For the numbers given below find out the smallest square number is divisible by each of them.

16, 27 and 40

Solution:

To, find the smallest square number  is divisible by 16, 27 and 40 we need to find the L.C.M of these three numbers.

2 16 36 40
2 8 18 20
2 4 9 10
2 2 9 5
3 1 9 5
3 1 3 5
5 1 1 5
1 1 1

Thus, L.C.M of 16, 27 and 40 = 2*2 * 2*2 * 3*3 * 5 = 720

This prime factor 5 is not having a pair.

As 5 is not having pair the given number cannot be a perfect square. So, we need to multiply with 5 in order to make a perfect square.

720*5 = 3600

Thus, the required number is 3600.

 

Question-27) For the numbers given below find out the smallest square number is divisible by each of them.

9, 14 and 24

Solution:

To, find the smallest square number is divisible by 16, 27 and 40 we need to find the L.C.M of these three numbers.

2 9 14 24
2 9 7 12
2 9 7 6
3 9 7 3
3 3 7 1
5 1 7 5
7 1 7 1
1 1 1

Thus, L.C.M of 9, 14 and 24 = 2*2 * 2 * 3*3 * 5 * 7 = 2520

As 2, 5 and 7 are not having pair the given number cannot be a perfect square. So, we need to multiply with 2, 5 and 7 in order to make a perfect square.

2520*2*5*7 = 176400

Thus, the required number is 176400.

 

Question-28) Find out the square of the numbers given below;

  1. 42
  2. 45
  3. 96
  4. 83
  5. 61
  6. 56

Solution:

1) 42

422=(40+2)2

= 40(40 + 2) + 2(40 + 2)

= 402 + 40*2 + 2*40 + 22

= 1600 + 80 + 80 + 4

= 1764

2) 45

452=(40+5)2

= 40(40 + 5) + 5(40 + 5)

= 402 + 40*5 + 5*40 + 52

= 1600 + 200 + 200 + 25

= 2025

3) 96

962=(90+6)2

= 90(90 + 6) + 6(90 + 6)

= 902 + 90*6 + 6*90 + 62

= 8100 + 540 + 540 + 36

= 9216

4) 83

832=(80+3)2

= 80(80 + 3) + 3(80 + 3)

= 802 + 80*3 + 3*80 + 32

= 6400 + 240 + 240 + 9

= 6889

5) 61

612=(60+1)2

= 60(60 + 1) + 1(60 + 1)

= 602 + 60*1 + 1*60 + 12

= 3600 + 60 + 60 1

= 3721

6) 56

562=(50+6)2

= 50(50 + 6) + 6(50 + 6)

= 502 + 50*6 + 6*50 + 62

= 2500 + 300 + 300 + 36

= 3136

 

Question-29) Write a Pythagorean triplet whose one member is

  1. 12
  2. 22
  3. 36
  4. 28

Solution:

x>1,2x,x21,x2+1 forms a Pythagorean Triplet

Where, xN

1) 12

Let us assume x2+1=12,thenx2=11

Thus, the value of x will be non-integer.

So, let us assume x21=12,thenx2=13

Thus, the value of x will be non-integer.

So, let us assume 2x = 12

∴ x = 6

the Pythagorean triplets are 26,621,62+1 i.e. 12,35,37.

 2) 22

Let us assume x2+1=22,thenx2=21

Thus, the value of x will be non-integer.

So, let us assume x21=22,thenx2=23

Thus, the value of x will be non-integer.

So, let us assume 2x = 22

∴ x = 11

the Pythagorean triplets are 211,1121,112+1 i.e. 22,120,122.

3) 28

Let us assume x2+1=28,thenx2=27

Thus, the value of x will be non-integer.

So, let us assume x21=28,thenx2=29

Thus, the value of x will be non-integer.

So, let us assume 2x = 28

∴ x = 14

the Pythagorean triplets are 214,1421,142+1 i.e. 28,195,197.

4) 36

Let us assume x2+1=36,thenx2=35

Thus, the value of x will be non-integer.

So, let us assume x21=36,thenx2=37

Thus, the value of x will be non-integer.

So, let us assume 2x = 36

∴ x = 18

the Pythagorean triplets are 218,1821,182+1 i.e. 36,323,325.

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