Practice Questions Class 10 NCERT Chapter 1 Real Numbers

Question 1: Use Euclid’s division algorithm to find the HCF of:

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

Solution:

i) 135 and 225

We need to apply Euclid’s algo, to find HCF, The first step is to take the Divisor, we always take

Divisor is smaller than the Dividend. 

Divisor = 135 

Dividend = 225 

Quotient = 225/135 = 1(quotient is an integer value) 

Euclid’s Div. Algo: 225 = 135×1 + 90 

Since 90!=0 that means We have to again apply the division lemma for remainder:90,

NowRemainder Will become Divisor and Divisor will become Dividend. 

Euclid’s Div. Algo: 135 = 90×1 + 45 

Again we need to apply the Division lemma because Yet remainder is not equal to 0

Divisor = 45 and Dividend = 90 

Euclid’s Div. Algo: 90 = 45×2 + 0 

Here Remainder = 0 that means We need to stop here, 

When Remainder = 0

Then HCF = Divisor

=> HCF(225, 135) = HCF(135, 90) 

= HCF(90,45) 

= 45

ii)196 and 38220

Divisor = 196 

Dividend = 38220 

Quotient = 38220/196 = 195 

Euclid’s Div. Algo: 38220 = 196×195 + 0 

Remainder = 0,

We don’t need to apply Division lemma further, 

HCF(38220, 196) = 196

iii) 867 and 255

Divisor = 255 

Dividend = 867 

Quotient = 867/225 = 3 

Euclid’s Div. algo: 867 = 255×3 + 102 

Remainder = 102

That means Again we need to apply Division lemma method, Divisor = 102, Dividend = 255, Quotient = 255/102 = 2 

Euclid’s Div. algo: 255 = 102×2 + 51 

Again Remainder = 51

We need to apply Division lemma method again, Divisor = 51, Dividend = 102, Quotient = 102/51 = 2 

Euclid’S Div. algo: 102 = 51×2 + 0 

Remainder = 0,

We need to stop here

HCF(867, 255) = HCF(255, 102) 

= HCF(102,51) 

= 51

Question 2: Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Solution:

We know that any odd no integer is not divisible by 2, 

When we divide 6q by 2 then it is perfectly divisible by 2 and give result 3q without any remainder,=>6q is even number 

let take any positive integer x and y=6 

as Per Euclid’s Div algo x=6q+r, q>=0 & 0<=r<6 

if r=0 => x = 6q, we earlier concluded that 6q is divisible by 2 => x = 6q is positive even integer 

if r=2, 4 => x = 6q+2, x = 6q+4 also An positive even integer because 6q is even as well as 2,4 also even and even divided by even gives even 

if r = 1 

Dividing 6q+1 by 2:

Dividend = 6q+1, Divisor = 2 

Quotient = (6q+1)/2 =3q 

as per Euclid’s div algo 6q + 1 = 3q×2 + 1 

As We Got Remainder = 1 after dividing by 2 that means 6q+1 is odd number, As q >= 0 that means 6q+1 is positive odd integer 

if r = 3 

Dividing 6q+3 by 2:

Dividend=6q+3, Divisor=2 

Quotient = (6q+3)/2 = 3q+1 

as per Euclid’s div algo 6q+3 = (3q+1)×2 + 1

As We Got Remainder = 1 after dividing by 2 that means 6q+3 is odd number, As q>=0 that means 6q+3 is positive odd integer 

if r = 5 

Dividing 6q+5 by 2:

Dividend = 6q+5, Divisor = 2 

Quotient = (6q+5)/2 = 3q+2 

as per Euclid’s div algo 6q+5 = (3q+2)×2 + 1 

As We Got Remainder = 1 after dividing by 2 that means 6q+5 is odd number, As q>=0 that means 6q+5 is positive odd integer 

Overall Conclusion: Any positive integer x can be in form 6q+1, 6q+3 or 6q+5 if it is odd otherwise it is in form 6q, 6q+2, 6q+4 

Question 3: 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?

Solution:

Given,

Total number of army contingent members = 616

Total number of army band members = 32

Given two groups are to march in same number of column then maximum number of columns will be Highest Common Factor between two groups i.e. HCF (616, 32)

Finding HCF (616, 32)
———————

Divisor = 32, Dividend = 616, quotient = 616/32 = 19

Euclid’s Div. Algo: 616 = 32×19 + 8

Remainder = 8, again Apply Division lemma

Divisor = 8,Dividend = 32,Quotient = 32/8 = 4

Euclid’s Div. Algo: 32 = 8×4 + 0

Remainder = 0, We have to stop here

HCF(616, 32) = 8

Therefore, maximum number of columns will be 8 in which both groups can march.

Question 4: Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solution:

Let x be any positive integer, and y = 3,

As per Euclid’s Div algo. x = 3q+r; q >= 0 && 0 <= r < 3 => r = (0, 1, 2)

If r = 0: x = 3q

Squaring both sides: x² = 3×3×q²

let assume m₁ = 3q², Here m₁ will be any positive integer because q >= 0

=>x² = 3m₁

if r = 1: x = 3q+1

Squaring both sides:x² = (3q+1)²

=>x² = 3(3q² + 2q)+1

let assume m₂ = 3q²+2q

=> x² = 3m₂ + 1, Here m₂ will be any positive integer because q >= 0

if r = 2: x = 3q+2

Squaring both sides:x² = (3q+2)²

=>x² = 3(3q² + 4q+1)+1

let assume m₃ = 3q²+4q+1

=> x² = 3m₃ + 1 ,Here m₃ will be any positive integer because q >= 0

Since m₁, m₂, m₃ are positive integer ,therefore we can Conclude that x² = 3m or 3m+1, Where m is some integer

This proved that the square of any positive integer(x) is either of the form 3m or 3m + 1 for some integer m

Question 5.Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution:

Let x be any positive integer, and y = 3 

As per Euclid’s Div algo. x = 3q+r; q >= 0 && 0 <= r < 3 => r = (0, 1, 2) 

if r = 0: x = 3q 

Cubing both sides: x³ = 9×3×q³

let assume m₁ = 3q³, Here m₁ will be any positive integer because q >= 0

=>x₂ = 9m₁ 

if r = 1: x = 3q+1 

Cubing both sides: x³ =(3q+1)³

=>x³ = 9(3q³+3q²+q)+1 

Let assume m₂ = 3q³+3q²+q 

=> x³ = 9m₂ + 1, Here m₂ will be any positive integer because q >= 0. 

if r = 2: x = 3q+2 

Cubing both sides: x³ = (3q+2)³

=>x³ = 9(3q³ + 6q²+4q)+8 

let assume m₃ = 3q³+6q²+4q 

=> x³ = 9m₃ + 8, Here m₃ will be any positive integer because q >= 0. 

Since m₁, m₂, m₃ are positive integer ,therefore we can Conclude that x³ = 9m, 9m+1 or 3m+8, where m is some integer. 

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