CBSE Solutions For Class 10 Mathematics Chapter 1 Real Numbers

Real Numbers Exercise – 1.1

Question .1 Use Euclid division algorithm, to find the H.C.F of the following:

1. 70 and 40
2. 180 and 45
3. 165 and 225
4. 155 and 1385
5. 105 and 135
6. 272 and 1032

Solution:

1. 70 and 40

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Here 40 > 70

7o = 4O * 1 + 30

4o = 30 * 1 + 10

3o = 10 * 1 + 0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F (40,70)

2. 180 and 45

Hare 18 > 45

45 = 18 * 2 + 9

18 = 9 * 2 + 0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F ( 18,45) = 9

3. 165 and 225

Here 165 > 225

225 = 165 * 1 + 90

165 = 90 * 1 + 75

90 = 75 * 1 + 15

75 = 15 * 5 + 0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F ( 165,225) = 15

4. 155 and 1385

Hare 155 and 1385

1385 = 155 * 8 + 145

155 = 145 * 1 + 0

145 = 10 * 14 + 5

10 = 5 * 2 + 0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F ( 155,1385) = 15

5. 105 and 135

Here 105 and 135

135 = 105 * 1 + 30

105 = 30 * 30 + 15

30 = 15 * 2 + 0

Since remainder = 0

The recent divisor is the H.C.F

H, C.F ( 105, 135)

6. 272 and 1032

Here 272 and 1032

1032 = 272 * 3 + 216

272 = 216 * 1 + 56

216 = 56 * 3 + 48

48 = 8 * 6 + 0

Since remainder = 0

The recent divisor is the H.C.

H, C.F ( 272, 1032)

Question: 2. The H.C.F of 408 and 1032 is expressible in the form of 1032×2-408xy, then find the Value of y.
Solution:

First, we will find the H-CF of 408 and 1032

Here 408 >1032

1032 = 408 x 2 + 216

408 = 216 x 1 + 192

216= 192 x 1 + 24

192=24 x 8 + 0

Since remainder = 0

The recent divisor is the H-C.F

H.C.F. of (408, 1032) = 24

Now, 1032×2-408xy = 24

= -408xy = 24-1032X 2

-y = \(\frac{24-2064}{408}\)

-y = \(\frac{-2040}{408}\)

y = 5

Question 3. If the H.C.F of 56 and 72 is expressible in the form of 56x+72×53, then find the Value of x.
Solution:

First, we will find the H.C.F of 56 and 72

Here 56 > 372

72 = 56 x 146

56= 16 x 3 + 8

16 = 8 x 2 + 0

Since remainder = 0

The recent divisor is the H·C.F

H.C.F. of (56,72)=8

Now, 56x+72×53 = 8

56x = 8 – 72 x 53

56x= 8-3,816

x = \(\frac{-3,808}{56}\)

x = – 68

Question 4. Express the H.C.F of 18 and 24 in the form
solution:

Here 18>24

24 = 18× 1+6

18=6×2+6

6=6×1+0

Since remainder =0

The recent divisor is the H.C.F

H.C.F. (18,24)=6

Now,

6=18-6×2

6=18-(24-18X1)

= 18-24 +18 x |

18×2-24 = 18x +244

where x = 2, y=-1

Question 5. Express the H.CF of 30 and 36 in the form of 30x + 36y.

Solution:

Here 30> 36

30= 200 6×4+6

6= 6X1+0

Since remainder = 0

The recent divisor is the H.C.F

H.C.F. (30,36) = 6

Now, 6=30-6×4

6=30-(36-30×1)

= 30-36+30X1

= 30X2-36

= 30x+364

where, x=2 and y=-1

Question 6. Find the largest number that divides 189 and 249 9 in each case.

Solution: We have to find a number, which divides the other numbers

Means H.C.F.

It is then that the required Number, when divided between 189 and 249 leaves the remainder 9; 9 is extra in each number. It means that if these numbers are 6 less, then there is no remainder in each case.

89-9=180 and 249-9=240 are completely divisible by the required number.

H.C.F. (180,240) = 60

Hence, the required number =  60.

Question 7. Find the largest number that divides 280 and 1248 u and 6 respectively. leaving the remainder solution.

Solution: We have to find a number, which divides the other numbers means → H.C.F.

It is given that the required number, when divided between 280 and 1248, leaves the remaining 4 and 6 respectively. It means that if 280 is 4 less than, 1248 is 6 less, then on division, gives no remainder.

280-4=276 and 1248-6=1242 are Completely divisible to the required number.

First, we will find the H.C.F of 276 and 1242

H.C.F. (2.76, 1242) = 138

Hence, the required number =138.

Question 8. Find the greatest number that divides 699, 572, and 442 leaving remainders 6, 5, and 1 respectively.

Solution: We have to find a number, which divides the other numbers means → H.C·F.

It is given that the required numbers when divided into 699, 572, and 442, leave the remaining 6,5 and I respectively. It means that if 699 is 6 less, 572 is s less, and 442 is I less, then on division, gives no remainder.

699-6=693, 572-5=567 and 442-1=441 are and 442-1=441 are completely by the required number .

First, we will find the H.C.F of 693 and 567.

693 = 567×1 +126

567 = 126×4 +63

126 = 63X2  + o

Now, we filled the H.C.F of 63 and 441.

441=63×7 +0

H.C.F. (63, 441) = 63

Required Number = 63

Question 9. A sweet Seller has 420 kaju burfis and 130 badam burfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray what is the number of Sweets that Can be Placed In each stack for this purpose? Also, find the number of stacks.

Solution: Maximum number of burfis in each stack = H.C.F of 420 and 130

420 = 2X2 X3 X5 X 7

130 = 2X5X 13

H·CF = 2×5 = 10

Maximum number of burfis in each stack = 10

Also, number of stacks = \(=\frac{420}{10}+\frac{130}{10}=42+13=55 .\)

Question 10. Three sets of English, Hindi, and Mathematics books have to be stacked In such a way that all the books are stored topicwise and the height of each stack is the Same. The number of English books is 96, the number of Hindi books is 240 and the number of mathematics books is 336. Assuming that the books are of the Same thickness, determine the number of Stacks of English, Hindi, and Mathematics books and hence the total number of Stacks.

Solution: Maximum number of books in each stack H.C.F. of

96,240 and 336

96= 2x2x2x 2 x 2 x 3

240 = 2x2x2x2x3x5

336 = 2x2x2 × 2 × 3 × 7

H·C.F = 2x2x2x2x3

Maximum number of books in each stack = 48
Also, number of stacks = 96

Also number Stacks = \(\frac{96}{48}+\frac{240}{48}+\frac{336}{48}\)

Real Numbers Exercise 1. 2

Question 1. Express each of the following as a product of prime factors:

1. 96
2. 48
3. 150
4. 3072

Solution:

96 = 2 * 2 * 2 * 2 * 2 * 3

96 = 25 * 3

84 = 2 * 2 * 3 * 7

84 = 22 * 3 * 7

150=2 x 3 x 5 x 5

150 = 2 x 3 x 52

3072 = 2× 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

3072= 210 X 3

Question 2. Find the H.C.F and L.C.M of the following pairs using the prime factorization method:

1. 12 and 25
2. 20 and 25
3. 96 and 404
4. 336 and 56

Solution:

1. 12 and 25

Now

H.C.F = 3

and L.C.M= 2 x 2 x 3 x 5

= 60

2. 20 and 25

Now

H.C.F = 5

and L.C.M = 2 x 2 x 5 x 5

= 100

3. 96 and 404

Now

H.C.F = 2 x 2= 4

and L.C.M = 2 x 2 x 2 x 2 x 2 x 3 x 101

= 9696

4. 336 and 56

Now

H.C.F = 2 x 2 x 2 x 7 = 56

and L.c.M = 2 x 2 x 2 x 2 x 3 x 7 = 336

Question 3. Using the prime factorization Method, find the HC.F and L.C.M of the following Pairs. Hence Verify H-C.F. XL.C.M= Product of two numbers.

1. 96 and 120
2. 16 and 20
3. 396 and 1080
4. 144 and 192

Solution:

1. 96 and 120

96=2x2x2 x 2 x 2 x 3

120 = 2x 2 x 2 × 3 × 5

Now, H.C.F. = 2×2×2×3=24

and LCM2 = 2 X 2 X 2 X 2 X 2 X 3 X 5 = 480

Now H.C.F X L.C. M = 12×144=24×480 = 11,520

and product of two numbers = 96X120 = 11,520

Hence, H-C.F X L.C.M = Product of two numbers

2. 16 and 20

16 = 2 X 2 X 2 x 2

L.C.M = 2 * 2 * 2 * 5 = 80

Now H.C.F x L .C.F = 4 * 80 = 320

and product of two numbers = 16 * 20

Hence H.C.F= product of two numbers

3. 396 and 1080

396 = 2×2 × 3 × 3 × 11

1080 = 2x2x2 × 3 × 3 × 3 × 5

Now, H.C.F. = 2 x2 x3x3 =36

and L.C.M. = 2X2 X2 X3 X3X3 X5x11

=11880

Now H.C.FXL.CM = 36X11880 = 4,27,680 and Product of two numbers = 4,27,680

Hence, H.C.F. XL.C.M = Product of two numbers

4. 144 and 192

144 = 2x2x 2×2×3×3

192 = 2x2x2 × 2 × 2 × 2 × 3

Now, H.C.F=2x2x2x2 x3 = 48

and L.C.M= 2×2 X2 X2 X2 X2 X3 X3

= 576

Now H.C.FXL.CM = 48X576 = 27,648 and Product of two numbers = 144 X 192

= 27,648

Hence, H.C.F. X  L.C.M.= product of two numbers

Question 4. The H.C.F and LCM of the two numbers are 145 and 2175 respectively. If the first number is 435, find the second number.

Solution:

Here, H.GF = 145

L.C.M = 2175

Now, First no. x Second no. = H.C.F. X L.C.M.

Second no = \(\frac{\text { H.C.F. XLC.M }}{\text { Firstno. }}\)

Second no = \(\frac{145 \times 2175}{435}\)

Second no = \(\frac{315375}{435}\)

Question 5. check whether 18 ⁿ can end with the digit o for the natural number n.

Solution:

18= 2 x 3 x 3 = 22 x 32

18n = (2×32) ⁿ = 2 ⁿ x32 ⁿ

It has no term containing

No Value of MEN for which 18 ⁿ ends with digit 0.

Question 6. On a morning walk, three persons step off together and their steps are 40cm, 42cm, and us cm respectively, what is the difference minimum distance each should walk so that each Can Cover the Same distance In Complete Steps?

Solution: We have to find a number (distance) that is divided by each number Completely, which means → L.CM

we have to find the L.C.M. of 400m, 42cm, and our cm to get the required distance.

Now, L. C.M = 2 x 2 x 2 x 5×3×7 × 3 = 2520

Minimum distance each should walk = 2520 Cm

Question 7. Write the missing numbers in the following factor tree:

Solution:

1. The upper box, on 7 and 13 is filled by the Product of 7 and 13, 9.
2. The upper next box, on Sand 91 is filled by the product of 5 and 91, 1.e., 455
3. The upper next box, on 3 and USS is filled by the product of 3 and USS, i.e., 1365
4. The topmost box, on 3 and 1365 will be filled by the product of 3 and 1365 i.e, 4095

Question 8. State whether the given statements are true or false:
Solution:

1. The Sum of two rationals is always rational. (True)
2. The Sum of two irrationals is always irrational. (False)
3. The product of two rationals is always rational. (True)
4. The product of two irrationals is always irrational, (False)
5. The Sum of a rational and an irrational is always rational. (False)

The product of a rational and an irrational is always rational. (True)