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KC Sinha Mathematics Solution Class 10 Chapter 1 Real numbers Exercise 1.1

Exercise 1.1

Exercise 1.2

Exercise 1.3

Exercise 1.4

Exercise 1.1


Question 1 A 

Using Euclid’s division algorithm, find the HCF of 156 and 504

Sol :

Given numbers are 156 and 504

Here, 504 > 156

So, we divide 504 by 156

By using Euclid’s division lemma, we get

504 = 156 × 3 + 36

Here, r = 36 ≠ 0.

On taking 156 as dividend and 36 as the divisor and we apply Euclid’s division lemma, we get

156 = 36 × 4 + 12

Here, r = 12 ≠ 0

So, on taking 36 as dividend and 12 as the divisor and again we apply Euclid’s division lemma, we get

36 = 12 × 3 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 12, the HCF of 156 and 504 is 12.


Question 1 B 

Using Euclid’s division algorithm, find the HCF of 135 and 225

Sol :

Given numbers are 135 and 225

Here, 225 > 135

So, we divide 225 by 135

By using Euclid’s division lemma, we get

225 = 135 × 1 + 90

Here, r = 90 ≠ 0.

On taking 135 as dividend and 90 as the divisor and we apply Euclid’s division lemma, we get

135 = 90 × 1 + 45

Here, r = 45 ≠ 0

So, on taking 90 as dividend and 45 as the divisor and again we apply Euclid’s division lemma, we get

90 = 45 × 2 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 45, the HCF of 135 and 225 is 45.


Question 1 C 

Using Euclid’s division algorithm, find the HCF of 455 and 42

Sol :

Given numbers are 455 and 42

Here, 455 > 42

So, we divide 455 by 42

By using Euclid’s division lemma, we get

455 = 42 × 10 + 35

Here, r = 35 ≠ 0.On taking 42 as dividend and 35 as the divisor and we apply Euclid’s division lemma, we get

42 = 35 × 1 + 7

Here, r = 7 ≠ 0

So, on taking 35 as dividend and 7 as the divisor and again we apply Euclid’s division lemma, we get

35 = 7 × 5 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 7, the HCF of 455 and 42 is 7.


Question 1 D 

Using Euclid’s division algorithm, find the HCF of 8840 and 23120

Sol :

Given numbers are 8840 and 23120

Here, 23120 > 8840

So, we divide 23120 by 8840

By using Euclid’s division lemma, we get

23120 = 8840 × 2 + 5440

Here, r = 5440 ≠ 0.

On taking 8840 as dividend and 5440 as the divisor and we apply Euclid’s division lemma, we get

8840 = 5440 × 1 + 3400

Here, r = 3400 ≠ 0

On taking 5440 as dividend and 3400 as the divisor and again we apply Euclid’s division lemma, we get

5440 = 3400 × 1 + 2040

Here, r = 2040 ≠ 0.

On taking 3400 as dividend and 2040 as the divisor and we apply Euclid’s division lemma, we get

3400 = 2040 × 1 + 1360

Here, r = 1360 ≠ 0

So, on taking 2040 as dividend and 1360 as the divisor and again we apply Euclid’s division lemma, we get

2040 = 1360 × 1 + 680

Here, r = 680 ≠ 0

So, on taking 1360 as dividend and 680 as the divisor and again we apply Euclid’s division lemma, we get

1360 = 680 × 2 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 680, the HCF of 8840 and 23120 is 680.


Question 1 E 

Using Euclid’s division algorithm, find the HCF of 4052 and 12576

Sol :

Given numbers are 4052 and 12576

Here, 12576 > 4052

So, we divide 12576 by 4052

By using Euclid’s division lemma, we get

12576 = 4052 × 3 + 420

Here, r = 420 ≠ 0.

On taking 4052 as dividend and 420 as the divisor and we apply Euclid’s division lemma, we get

4052 = 420 × 9 + 272

Here, r = 272 ≠ 0

On taking 420 as dividend and 272 as the divisor and again we apply Euclid’s division lemma, we get

420 = 272 × 1 + 148

Here, r = 148 ≠ 0

On taking 272 as dividend and 148 as the divisor and again we apply Euclid’s division lemma, we get

272 = 148 × 1 + 124

Here, r = 124 ≠ 0.

On taking 148 as dividend and 124 as the divisor and we apply Euclid’s division lemma, we get

148 = 124 × 1 + 24

Here, r = 24 ≠ 0

So, on taking 124 as dividend and 24 as the divisor and again we apply Euclid’s division lemma, we get

124 = 24 × 5 + 4

Here, r = 4 ≠ 0

So, on taking 24 as dividend and 4 as the divisor and again we apply

Euclid’s division lemma, we get

24 = 4 × 6 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 4, the HCF of 4052 and 12576 is 4.


Question 1 F 

Using Euclid’s division algorithm, find the HCF of 3318 and 4661

Sol :

Given numbers are 3318 and 4661

Here, 4661 > 3318

So, we divide 4661 by 3318

By using Euclid’s division lemma, we get

4661 = 3318 × 1 + 1343

Here, r = 1343 ≠ 0.

On taking 3318 as dividend and 1343 as the divisor and we apply Euclid’s division lemma, we get

3318 = 1343 × 2 + 632

Here, r = 632 ≠ 0

So, on taking 1343 as dividend and 632 as the divisor and again we apply Euclid’s division lemma, we get

1343 = 632 × 2 + 79

Here, r = 79 ≠ 0

So, on taking 632 as dividend and 79 as the divisor and again we apply Euclid’s division lemma, we get

632 = 79 × 8 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 79, the HCF of 3318 and 4661 is 79.


Question 1 G 

Using Euclid’s division algorithm, find the HCF of 250, 175 and 425

Sol :

Given numbers are 250, 175 and 425

 425 > 250 > 175

On applying Euclid’s division lemma for 425 and 250, we get

425 = 250 × 1 + 175

Here, r = 175 ≠ 0.

So, again applying Euclid’s division lemma with new dividend 250 and new divisor 175, we get

250 = 175 × 1 + 75

Here, r = 75 ≠ 0

So, on taking 175 as dividend and 75 as the divisor and again we apply Euclid’s division lemma, we get

175 = 75 × 2 + 25

Here, r = 25 ≠ 0.

So, again applying Euclid’s division lemma with new dividend 75 and new divisor 25, we get

75 = 25 × 3 + 0

Here, r = 0 and divisor is 25.

So, HCF of 425 and 225 is 25.

Now, applying Euclid’s division lemma for 175 and 25, we get

175 = 25 × 7 + 0

Here, remainder = 0

So, HCF of 250, 175 and 425 is 25.


Question 1 H 

Using Euclid’s division algorithm, find the HCF of 4407, 2938 and 1469

Sol :

Given numbers are 4407, 2938 and 1469

 4407 > 2938 > 1469

On applying Euclid’s division lemma for 4407 and 2938, we get

4407 = 2938 × 1 + 1469

Here, r = 1469 ≠ 0.

So, again applying Euclid’s division lemma with new dividend 2938 and new divisor 1469, we get

2938 = 1469 × 2 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this stage is 1469, the HCF of 4407 and 2938 is 1469.

Now, applying Euclid’s division lemma for 1469 and 1469, we get

1469 = 1469 × 1 + 0

Here, remainder = 0

So, HCF of 4407, 2938 and 1469 is 1469.


Question 2 

Show that every positive even integer is of the form 2q and that every positive odd integer is of the form 2q + 1, where q is some integer.

Sol :

Let a and b be any two positive integers, such that a > b.

Then, a = bq + r, 0 ≤ r < b …(i) [by Euclid’s division lemma]

On putting b = 2 in Eq. (i), we get

a = 2q + r, 0 ≤ r < 2 …(ii)

 r = 0 or 1

When r = 0, then from Eq. (ii), a = 2q, which is divisible by 2

When r = 1, then from Eq. (ii), a = 2q + 1, which is not divisible by 2.

Thus, every positive integer is either of the form 2q or 2q + 1.

That means every positive integer is either even or odd. So, if a is a positive even integer, then a is of the form 2q and if a, is a positive odd integer, then a is of the form 2q + 1.


Question 3 

Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.

Sol :

Let a be any positive odd integer. We apply the division algorithm with a and b = 4.

Since 0 ≤ r < 4, the possible remainders are 0,1,2 and 3.

i.e. a can be 4q, or 4q + 1, or 4q + 2, or 4q + 3, where q is the quotient.

As we know a is odd, a can’t be 4q or 4q + 2 because they both are divisible by 2.

Therefore, any positive odd integer is of the form 4q + 1 or 4q + 3.


Question 4 

There are 250 and 425 liters of milk in two containers. What is the maximum capacity of the container which can measure completely the quantity of milk in the two containers?

Sol :

Given the capacities of the two containers are 250 L and 425 L.

Here, 425 > 250

Now, we divide 425 by 250.

We used Euclid’s division lemma.

425 = 250 × 1 + 175

Here, remainder r = 175 ≠ 0

So, the new dividend is 250 and the new divisor is 175, again we apply Euclid division algorithm.

250 = 175 × 1 + 75

Here, remainder r = 75 ≠ 0

On taking the new dividend is 175 and the new divisor is 75, we apply Euclid division algorithm.

175 = 75 × 2 + 25

Here, remainder r = 25 ≠ 0

On taking new dividend is 75 and the new divisor is 25, again we apply Euclid division algorithm.

75 = 25 × 3 + 0

Here, remainder is zero and divisor is 25.

So, the HCF of 425 and 250 is 25.

Hence, the maximum capacity of the required container is 25 L.


Question 5 

A rectangular surface has length 4661 meters and breadth 3318 meters. On this area, square tiles are to be put. Find the maximum length of such tiles.

Sol :

Given length and breadth are 4661 m and 3318 m respectively.

Here, 4661 > 3318

So, we divide 4661 by 3318

By using Euclid’s division lemma, we get

4661 = 3318 × 1 + 1343

Here, r = 1343 ≠ 0.

On taking 3318 as dividend and 1343 as the divisor and we apply Euclid’s division lemma, we get

3318 = 1343 × 2 + 632

Here, r = 632 ≠ 0

So, on taking 1343 as dividend and 632 as the divisor and again we apply Euclid’s division lemma, we get

1343 = 632 × 2 + 79

Here, r = 79 ≠ 0

So, on taking 632 as dividend and 79 as the divisor and again we apply Euclid’s division lemma, we get

632 = 79 × 8 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 79, the HCF of 3318 and 4661 is 79.

Hence, the maximum length of such tiles is 79 meters.


Question 6 

Find the least number of square tiles which can the floor of a rectangular shape having length and breadth 16 meters 58 centimeters and 8 meters 32.

Sol :

Firstly, we find the length of the largest tile so for that we have to find the HCF of 1658 and 832.

Here, 1658 > 832

So, we divide 1658 by 832

By using Euclid’s division lemma, we get

1658 = 832 × 1 + 826

Here, r = 826 ≠ 0.

On taking 832 as dividend and 826 as the divisor and we apply Euclid’s division lemma, we get

832 = 826 × 1 + 6

Here, r = 6 ≠ 0

So, on taking 826 as dividend and 6 as the divisor and again we apply Euclid’s division lemma, we get

826 = 6 × 137 + 4

Here, r = 4 ≠ 0

So, on taking 6 as dividend and 4 as the divisor and again we apply Euclid’s division lemma, we get

6 = 4 × 1 + 2

Here, r = 2 ≠ 0

So, on taking 4 as dividend and 2 as the divisor and again we apply Euclid’s division lemma, we get

4 = 2 × 2 + 0

The remainder has now become 0, so our procedure stops. Since the divisor at this last stage is 79, the HCF of 1658 and 832 is 2.

So, the length of the largest tile is 2 cm

Area of each tile = 2 × 2 = 4cm2

The required number of tiles = 

= 

= 344864

Least number of square tiles are required are 344864


S.noChaptersLinks
1Real numbersExercise 1.1

Exercise 1.2

Exercise 1.3

Exercise 1.4

2PolynomialsExercise 2.1

Exercise 2.2

Exercise 2.3

3Pairs of Linear Equations in Two VariablesExercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

Exercise 3.5

4Trigonometric Ratios and IdentitiesExercise 4.1

Exercise 4.2

Exercise 4.3

Exercise 4.4

5TrianglesExercise 5.1

Exercise 5.2

Exercise 5.3

Exercise 5.4

Exercise 5.5

6StatisticsExercise 6.1

Exercise 6.2

Exercise 6.3

Exercise 6.4

7Quadratic EquationsExercise 7.1

Exercise 7.2

Exercise 7.3

Exercise 7.4

Exercise 7.5

8Arithmetic Progressions (AP)Exercise 8.1

Exercise 8.2

Exercise 8.3

Exercise 8.4

9Some Applications of Trigonometry: Height and DistancesExercise 9.1
10Coordinates GeometryExercise 10.1

Exercise 10.2

Exercise 10.3

Exercise 10.4

11CirclesExercise 11.1

Exercise 11.2

12ConstructionsExercise 12.1
13Area related to CirclesExercise 13.1
14Surface Area and VolumesExercise 14.1

Exercise 14.2

Exercise 14.3

Exercise 14.4

15ProbabilityExercise 15.1

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