Home NCERT Solu. class 10

Use Euclid's division algorithm to find the HCF of:

(i) 135 and 225 (ii)196 and 38220 (iii)867 and 255

Answer:

135 and 225

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain

225 = 135 × 1 + 90

Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain

135 = 90 × 1 + 45

We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain

90 = 2 × 45 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain

38220 = 196 × 195 + 0

Since the remainder is zero,the process stops.

Since the divisor at this stage is 196,

Therefore, the HCF of 196 and 38220 is 196.

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain

867 = 255 × 3 + 102

Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain

255 = 102 × 2 + 51

We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain

102 = 51 × 2 + 0

Since the remainder is zero, the process stops.

Since the divisor at this stage is 51,

Therefore, the HCF of 867 and 255 is 51

Show that any positive odd integer is of the form 6q+1, or 6q+3, or 6q+5, where q is some integer.

Answer:

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,

a = 6q + r for some integer q ≥ 0 , and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 × 3q + 1 = 2k₁ + 1, where k₁ is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k₂ + 1, where k₂ is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k₃ + 1, where k₃ is an integer

Clearly, 6q + 1, 6q +3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.

Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1,

or 6q + 3,

or 6q + 5

An army contingent of 616 members is to march behind an army band

of 32 members in a parade. The two groups are to march in the same

number of columns. What is the maximum number of columns in

which they can march?

Answer:

HCF (616, 32) will give the maximum number of columns in which they can march.

We can use Euclid’s algorithm to find the HCF.

616 = 32 × 19 + 8

32 = 8 × 4 + 0

The HCF (616, 32) is 8.

Therefore, they can march in 8 columns each.

Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m.

Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

Answer:

Let a be any positive integer and b = 3.

Then a = 3q + r for some integer q ≥ 0

And r = 0, 1, 2 because 0 ≤ r < 3

Therefore, a = 3q or 3q + 1 or 3q + 2

Or,

a²=(3q)² or (3q+1)² or (3q+2)²

a²=(9q²)or 9q²+6q+1 or 9q²+12q+4

=3*(3q²) or 3(3q²+2q)+1 or 3(3q²+4q+1)+1

=3k₁ or 3k₂+1 or 3k₃+1

Where k₁, k₂, and k₃ are some positive integers

Hence, it can be said that the square of any positive integer is either of

the form 3m or 3m + 1.

Use Euclid’s division lemma to show that the cube of any positive

integer is of the form 9m, 9m + 1 or 9m + 8.

Answer:

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

∴a=3q or 3q+1 or 3q+2

Therefore,every number can be represented as these three forms.

There are three cases.

a³=(3q)³=27q³=9(3q³)=9m,

Where m is an integer such that m = 3q³

a³ = (3q +1)³

a³ = 27q³ + 27q² + 9q + 1

a³ = 9(3q³ + 3q² + q) + 1

a³ = 9m + 1

Where m is an integer such that m = (3q³ + 3q² + q)

a³ = (3q +2)³

a³ = 27q³ + 54q² + 36q + 8

a³ = 9(3q³ + 6q² + 4q) + 8

a³ = 9m + 8

Where m is an integer such that m = (3q³ + 6q² + 4q)

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Express each number as a product of its prime factors:

(i) 140 (ii) 156 (iii)3825 (iv) 5005 (v) 7429

Answer:

(i) 140=2x2x5x7=2²x5x7

(ii) 156=2x2x3x13=2²x3x13

(iii) 3825=3x3x5x5x17=3²x5²x17

(iv) 5005= 5x7x11x13

(v) 7429 =17x19x23

Find the LCM and HCF of the following pairs of integers and verify that

LCM × HCF = product of the two numbers.

(i) 26 and 91 (ii) 510 and 92 (iii)336 and 54

Answer:

26=2x13

91=7x13

HCF=13

LCM=2x7x13=182

Product of two numbers= 26x91=2366

HCFxLCM=13x182=2366

Hence, product of two numbers=HCFxLCM

510=2x3x5x17

92=2x2x23

HCF=2

LCM=2x2x3x5x17x23=23460

Product of two numbers= 510x92= 46920

HCFxLCM= 2x23460=46920

Hence, product of two numbers=HCFxLCM

336=2x2x2x2x3x7

336=2⁴4x3x7

54=2x3x3x3

54=2x3³

HCF=2x3=6

LCM=2x3³x7=3024

Product of two numbers= 336x54=18144

HCFxLCM=6x3024=18144

Hence, product of two numbers=HCFxLCM

Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12,15 and 21 (ii)17,23 and 29 (iii) 8,9 and 25

12=2²x3

15=3x5

21=3x7

HCF=3

LCM=2²x3x5x7=420

17=1x17

23=1x23

29=1x29

HCF=1

LCM=17x23x29=11339

8=2x2x2

9=3x3

25=5x5

HCF=1

LCM=2x2x2x3x3x5x5=1800

Given that HCF (306, 657) = 9, find LCM (306, 657).

Answer:

HCF(306,657)=9

We know that, LCMxHCF =Product of two numbers

∴LCMxHCF=306x657

LCM=306x657/HCF = 306x657/9

LCM=22338

Check whether 6ⁿ can end with the digit 0 for any natural number n.

Answer:

If any number ends with the digit 0, it should be divisible by 10 or in

other words, it will also be divisible by 2 and 5 as 10 = 2 × 5

Prime factorisation of 6ⁿ = (2 ×3)ⁿ

It can be observed that 5 is not in the prime factorisation of 6ⁿ.

Hence, for any value of n, 6ⁿ will not be divisible by 5.

Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.

Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Answer:

Numbers are of two types - prime and composite. Prime numbers can be

divided by 1 and only itself, whereas composite number

have factors other than 1 and itself.

It can be observed that

7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)

= 13 × 78

= 13 ×13 × 6

The given expression has 6 and 13 as its factors. Therefore, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 ×(7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × (1008 + 1)

= 5 ×1009

1009 cannot be factorised further. Therefore, the given expression has

5 and 1009 as its factors. Hence, it is a composite number.

There is a circular path around a sports field. Sonia takes 18 minutes

to drive one round of the field, while Ravi takes 12 minutes for the

same. Suppose they both start at the same point and at the same

time, and go in the same direction. After how many minutes will they

meet again at the starting point?

Answer:

It can be observed that Ravi takes lesser time than Sonia for

completing 1 round of the circular path. As they are going in the same

direction, they will meet again at the same time when Ravi will have

completed 1 round of that circular path with respect to Sonia.

And the total time taken for completing this 1 round of circular path will be the

LCM of time taken by Sonia and Ravi for completing 1 round of circular

path respectively i.e., LCM of 18 minutes and 12 minutes

18 = 2 × 3 × 3

And, 12 = 2 × 2 × 3

LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36

Therefore, Ravi and Sonia will meet together at the starting pointafter

36 minutes.

Prove that √5 is irrational.

Answer:

Let √5 is a rational number.

Therefore, we can find two integers a, b (b ≠ 0) such that√5= a/b

Let a and b have a common factor other than 1. Then we can divide

them by the common factor, and assume that a and b are co-prime.

a = √5b

a² = 5b²

Therefore, a² is divisible by 5 and it can be said that a is divisible by 5.

Let a = 5k, where k is an integer

(5k)²= 5b²

b² = 5k² This means that b² is divisible by 5 and hence, b is divisible

by 5.

This implies that a and b have 5 as a common factor.

And this is a contradiction to the fact that a and b are co-prime.

Hence, √5 cannot be expressed as p/q or it can be said that √5 is

irrational.

Prove that 3+2√5 is irrational.

Answer:

Let 3+2√5 is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

3+2√5=a/b

2√5=a/b - 3

√5=1/2(a/b-3)

Since a and b are integers,1/2(a/b-3) will also be rational and

therefore,√5 is rational.

This contradicts the fact that is √5 irrational. Hence, our assumption

that 3+2√5 is rational is false. Therefore,3+2√5 is irrational.

Prove that the following are irrationals:

(i) 1/√2 (ii)7√5 (iii)6+√2

Answer:

(i) 1/√2

Let 1/√2 is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

1/√2=a/b

√2=b/a

b/a is rational as a and b are integers.

Therefore,√2 is rational which contradicts to the fact that √2 is

irrational.

Hence, our assumption is false and 1/√2 is irrational.

(ii)7√5

Answer:

Let 7√5 is rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

7√5=a/b for some integers a and b

∴√5=a/7b

a/7b is rational as a and b are integers.

Therefore,√5 should be rational.

This contradicts the fact that √5 is irrational. Therefore, our assumption

that 7√5 is rational is false. Hence,7√5 is irrational.

(iii) 6+√2

Answer:

Let 6+√2 be rational.

Therefore, we can find two integers a, b (b ≠ 0) such that

6+√2=a/b

√2=a/b-6

Since a and b are integers,a/b-6 is also rational and hence,√2 should

be rational. This contradicts the fact that √2 is irrational. Therefore,

our assumption is false and hence,6+√2 is irrational.

Without actually performing the long division, state whether the

following rational numbers will have a terminating decimal expansion

or a non-terminating repeating decimal expansion:

(i)13/3125 (ii)17/8 (iii)64/455 (iv)15/1600

(v)29/343 (vi)23/2³5² (vii)129/2²5⁷ 7⁵ (viii)6/15

(ix)35/50 (x)77/210

Answer:

(i)13/3125

3125=5

The denominator is of the form 5

Hence, the decimal expansion of

(ii)17/8

8=2³

The denominator is of the form 2

Hence, the decimal expansion of

(iii)64/455

455 = 5 × 7 × 13

Since the denominator is not in the form 2

7 and 13 as its factors, its decimal expansion will be non-terminating

repeating.

(iv)15/1600

1600 = 2⁶× 5²

The denominator is of the form 2

Hence, the decimal expansion of

(v) 23/2³5²

The denominator is of the form 2

Hence, the decimal expansion of 23/2³5² is terminating.

(vi)23/2²5⁵

Since the denominator is not of the form 2

its factor, the decimal expansion of 23/2²5⁵ is non-terminating repeating.

(vii) 129/2

since the denominator is not of the form 2

its factor ,the decimal expansion of 129/2²5

repeating.

(viii) 6/15=2x3/3x5=2/5

The denominator is of the form 5

Hence, the decimal expansion of 6/15 is terminating.

(ix) 35/50=7x5/10x5=7/10

10=2x5

The denominator is of the form 2

Hence, the decimal expansion of 35/50 is terminating.

(x)77÷210=11x7/30x7=11/30

30=2x3x5

Since the denominator is not of the form 2

its factors, the decimal expansion of 77/210 is non-terminating repeating.

Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

Answer:

(i)13/3125=0.00416 3125⟌

The following real numbers have decimal expansions as given below.

In each case, decide whether they are rational or not. If they are

rational, and of the form , what can you say about the prime factor

of q?

(i) 43.123456789 (ii) 0.120120012000120000… (iii) 43.123456789

Answer:

(i) 43.123456789

Since this number has a terminating decimal expansion, it is a rational

number of the form p/q and q is of the form 2

i.e., the prime factors of q will be either 2 or 5 or both.

(ii)0.120120012000120000…

The decimal expansion is neither terminating nor recurring. Therefore,

the given number is an irrational number.

(iii)43.123456789 Since the decimal expansion is non-terminating recurring, the given

number is a rational number of the form p/q and q is not of the form

2