Category: CBSE Class 10

RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two
Variables Ex 3.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.8
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.9
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.10
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.11
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables VSAQS
• RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables MCQS

Question 1.
Akhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs. 3, and a game of Hoopla costs Rs. 4. If she spend Rs. 20 in the fair, represent this situation algebraically and graphically.
Solution:
Let number of rides on the wheel = x
and number of play of Hoopla = y
According to the given conditions x = 2y ⇒ x – 2y = 0 ….(i)
and cost of ride on wheel at the rate of Rs. 3 = 3x
and cost on Hoopla = 4y
and total cost = Rs. 20
3x + 4y = 20 ….(ii)
Now we shall solve these linear equations graphically as under
We take three points of each line and join them to get a line in each case the point of intersection will be the solution
From equation (i)
x = 2y

y = 2, then x = 2 x 2 = 4
y = 0, then x = 2 x 0 = 0
y = 3, then x = 2 x 3 = 6
Now, we plot these points on the graphs and join them to get a line
Similarly in equation (ii)
3x + 4y = 20 ⇒ 3x = 20 – 4y

Now we plot these points and get another line by joining them
These two lines intersect eachother at the point (4, 2)
Its solution is (4, 2)
Which is a unique Hence x = 4, y = 2

Question 2.
Aftab tells his daughter, “Seven years ago, I w as seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Is not this interesting ? Represent this situation algebraically and graphically.
Solution:
Seven years ago
Let age of Aftab’s daughter = x years
and age of Aftab = y years
and 3 years later
Age of daughter = x + 10 years
and age of Aftab = y + 10 years
According to the conditions,
y = 7x ⇒ 7x – y = 0 ……….(i)
y + 10 = 3 (x + 10)
=> y + 10 = 3x + 30
3x – y = 10 – 30 = -20
3x – y = -20 ….(ii)
Equations are
7x – y = 0
3 x – y = -20
Now we shall solve these linear equations graphically as under
7x – y = 0 ⇒ y = 7x

If x = 0, y = 7 x 0 = 0
If x = 1, y = 7 x 1=7
If x = -1, y = 7 x (-1) = -7
Now plot these points on the graph and join
then
3x – y = -20
y = 3x + 20

If x = -1, y = 3 x (-1) + 20 = -3 + 20= 17
If x = -2, y = 3 (-2) + 20 = -6 + 20 = 14
If x = -3, y = 3 (-3) + 20 = -9 + 20= 11
Now plot the points on the graph and join them we see that lines well meet at a point on producing at (5, 35).

Question 3.
The path of a train A is given by the equation 3x + 4y – 12 = 0 and the path of another train B is given by the equation 6x + 8y – 48 = 0. Represent this situation graphically.
Solution:
Path of A train is 3x + 4y – 12 = 0
and path of B train is 6x + 8y – 48 = 0
Graphically, we shall represent these on the graph as given under 3x + 4y- 12 = 0

Question 4.
Gloria is walking along the path joining (-2, 3) and (2, -2), while Suresh is walking along the path joining (0, 5) and (4, 0). Represent this situation graphically.
Solution:
Plot the points (-2, 3) and (2, -2) and join them to get a line
and also plot the points (0, 5), (4, 0) and joint them to get another line as shown on the graph
We see that these two lines are parallel to each other

Question 5.
On comparing the ratios , and without drawing them, find out whether the lines representing following pairs of linear equations intersect at a point, are parallel or coincide :
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
(ii) 9x + 3y +12 = 0
18x + 6y + 24 = 0
(iii) 6x – 3y +10 = 0
2x – y + 9 = 0
Solution:

Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines.
Solution:

Question 7.
The cost of 2kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4kg of apples and 2kg of grapes is Rs. 300. Represent the situation algebraically and geo-metrically.
Solution:
Let cost of 1kg of apples = Rs. x
and cost of 1kg of grapes = Rs. y
Now according to the condition, the system of equation will be
2x + y = 160
4x + 2y = 300
Now 2x + y = 160
y = 160 – 2x

If x = 20, then y = 160 – 2 x 20 = 160 – 40 = 120
If x = 40, then y = 160 – 2 x 40 = 160 – 80 = 80
If x = 60, then y = 160 – 2 x 60 = 160 – 120 = 40
Now plot the points and join them and 4x + 2y = 300
=> 2x + y = 150
=> y = 150 – 2x

If x = 40, then y = 150 – 2 x 40 = 150 – 80 = 70
If x = 50, then y = 150 – 2 x 50 = 150 – 100 = 50
If x = 60, then y = 150 – 2 x 60 = 150 – 120 = 30
Now plot the points and join them We see that these two lines are parallel

Hope given RD Sharma Class 10 Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS. You must go through NCERT Solutions for Class 10 Maths to get better score in CBSE Board exams along with RS Aggarwal Class 10 Solutions.

Mark the correct alternative in each of the following :
Question 1.
If α, β are the zeros of the polynomial f(x) = x² + x + 1, then \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta }\) =
(a) 1
(b) -1
(c) 0
(d) None of these
Solution:
(b)

Question 2.
If α, β are the zeros of the polynomial p(x) = 4x² + 3x + 7, then \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta }\) is equal to
(a) \(\frac { 7 }{ 3 }\)
(b) – \(\frac { 7 }{ 3 }\)
(c) \(\frac { 3 }{ 7 }\)
(d) – \(\frac { 3 }{ 7 }\)
Solution:
(d)

Question 3.
If one zero of the polynomial f(x) = (k² + 4) x² + 13x + 4k is reciprocal of the other, then k =
(a) 2
(b) -2
(c) 1
(d) -1
Solution:
(a) f (x) = (k² + 4) x² + 13x + 4k
Here a = k² + 4, b = 13, c = 4k
One zero is reciprocal of the other
Let first zero = α

k = 2

Question 4.
If the sum of the zeros of the polynomial f(x) = 2x³ – 3kx² + 4x – 5 is 6, then value of k is
(a) 2
(b) 4
(c) -2
(d) -4
Solution:
(b)

Question 5.
If α and β are the zeros of the polynomial f(x) = x² + px + q, then a polynomial having α and β is its zeros is
(a) x² + qx + p
(b) x² – px + q
(c) qx² + px + 1
(d) px² + qx + 1
Solution:
(c)

Question 6.
If α, β are the zeros of polynomial f(x) = x² – p (x + 1) – c, then (α + 1) (β + 1) =
(a) c – 1
(b) 1 – c
(c) c
(d) 1 + c
Solution:
(b)

Question 7.
If α, β are the zeros of the polynomial f(x) = x² – p(x + 1) – c such that (α + 1) (β + 1) = 0, then c =
(a) 1
(b) 0
(c) -1
(d) 2
Solution:
(a)

Question 8.
If f(x) = ax² + bx + c has no real zeros and a + b + c < 0, then
(a) c = 0
(b) c > 0
(c) c < 0
(d) None of these
Solution:
(d) f(x) = ax² + bx + c
Zeros are not real
b² – 4ac < 0 ….(i)
but a + b + c < 0
b < – (a + c)
Squaring both sides b² < (a + c)²
=> (a + c)² – 4ac < 0 {From (i)}
=> (a – c)² < 0
=> a – c < 0
=> a < c

Question 9.
If the diagram in figure shows the graph of the polynomial f(x) = ax² + bx + c, then
(a) a > 0, b < 0 and c > 0
(b) a < 0, b < 0 and c < 0
(c) a < 0, b > 0 and c > 0
(d) a < 0, b > 0 and c < 0

Solution:
(a) Curve ax² + bx + c intersects x-axis at two points and curve is upward.
a > 0, b < 0 and c> 0

Question 10.
Figure shows the graph of the polynomial f(x) = ax² + bx + c for which
(a) a < 0, b > 0 and c > 0
(b) a < 0, b < 0 and c > 0
(c) a < 0, b < 0 and c < 0
(d) a > 0, b > 0 and c < 0

Solution:
(b) Curve ax² + bx + c intersects x-axis at two points and curve is downward.
a < 0, b < 0 and c > 0

Question 11.
If the product of zeros of the polynomial f(x) = ax³ – 6x² + 11x – 6 is 4, then a =
(a) \(\frac { 3 }{ 2 }\)
(b) – \(\frac { 3 }{ 2 }\)
(c) \(\frac { 2 }{ 3 }\)
(d) – \(\frac { 2 }{ 3 }\)
Solution:
(a) f(x) = ax³ – 6x² + 11x – 6

Question 12.
If zeros of the polynomial f(x) = x³ – 3px² + qx – r are in AP, then
(a) 2p³ = pq – r
(b) 2p³ = pq + r
(c) p³ = pq – r
(d) None of these
Solution:
(a) f(x) = x³ – 3px² + qx – r
Here a = 1, b = -3p, c = q, d= -r
Zeros are in AP
Let the zeros be α – d, α, α + d

Question 13.
If the product of two zeros of the polynomial f(x) = 2x³ + 6x² – 4x + 9 is 3, then its third zero is
(a) \(\frac { 3 }{ 2 }\)
(b) – \(\frac { 3 }{ 2 }\)
(c) \(\frac { 9 }{ 2 }\)
(d) – \(\frac { 9 }{ 2 }\)
Solution:
(b)

Question 14.
If the polynomial f(x) = ax² + bx – c is divisible by the polynomial g(x) = ax² + bx + c, then ab =
(a) 1
(b) \(\frac { 1 }{ c }\)
(c) – 1
(d) – \(\frac { 1 }{ c }\)
Solution:
(a)

Question 15.
In Q. No. 14, ac =
(a) b
(b) 2b
(c) 2b2
(d) -2b
Solution:
(b) In the previous questions
Remainder = 0
(b – ac + ab²) = 0
b + ab² = ac
=> ac = b (1 + ab) = b (1 + 1) = 2b

Question 16.
If one root of the polynomial f(x) = 5x² + 13x + k is reciprocal of the other, then the value of k is
(a) 0
(b) 5
(c) \(\frac { 1 }{ 6 }\)
(d) 6
Solution:
(b)

Question 17.
If α, β, γ are the zeros of the polynomial f(x) = ax³ + bx² + cx + d, then \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } +\frac { 1 }{ \gamma }\) =
(a) – \(\frac { b }{ d }\)
(b) \(\frac { c }{ d }\)
(c) – \(\frac { c }{ d }\)
(d) \(\frac { c }{ a }\)
Solution:
(c)

Question 18.
If α, β, γ are the zeros of the polynomial f(x) = ax³ + bx² + cx + d, then α² + β² + γ² =

Solution:
(d)

Question 19.
If α, β, γ are the zeros of the polynomial f(x) = x³ – px² + qx – r, then \(\frac { 1 }{ \alpha \beta } +\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha }\) =
(a) \(\frac { r }{ p }\)
(b) \(\frac { p }{ r }\)
(c) – \(\frac { p }{ r }\)
(d) – \(\frac { r }{ p }\)
Solution:
(b)

Question 20.
If α, β are the zeros of the polynomial f(x) = ax² + bx + c, then \(\frac { 1 }{ { \alpha }^{ 2 } } +\frac { 1 }{ { \beta }^{ 2 } }\) =


Solution:
(b)

Question 21.
If two of the zeros of the cubic polynomial ax³ + bx² + cx + d are each equal to zero, then the third zero is
(a) \(\frac { -d }{ a }\)
(b) \(\frac { c }{ a }\)
(c) \(\frac { -b }{ a }\)
(d) \(\frac { b }{ a }\)
Solution:
(c) Two of the zeros of the cubic polynomial ax³ + bx² + cx + d are each equal to zero
Let α, β and γ are its zeros, then

Third zero will be \(\frac { -b }{ a }\)

Question 22.
If two zeros of x³ + x² – 5x – 5 are √5 and – √5 then its third zero is
(a) 1
(b) -1
(c) 2
(d) -2
Solution:
(b)

Question 23.
The product of the zeros of x³ + 4x² + x – 6 is
(a) – 4
(b) 4
(c) 6
(d) – 6
Solution:
(c)

Question 24.
What should be added to the polynomial x² – 5x + 4, so that 3 is the zero of the resulting polynomial ?
(a) 1
(b) 2
(c) 4
(d) 5
Solution:
(b) 3 is the zero of the polynomial f(x) = x² – 5x + 4
x – 3 is a factor of f(x)
Now f(3) = (3)² – 5 x 3 + 4 = 9 – 15 + 4 = 13 – 15 = -2
-2 is to be subtracting or 2 is added

Question 25.
What should be subtracted to the polynomial x² – 16x + 30, so that 15 is the zero of the resulting polynomial ?
(a) 30
(b) 14
(b) 15
(d) 16
Solution:
(c) 15 is the zero of polynomial f(x) = x² – 16x + 30
Then f(15) = 0
f(15) = (15)² – 16 x 15 + 30 = 225 – 240 + 30 = 255 – 240 = 15
15 is to be subtracted

Question 26.
A quadratic polynomial, the sum of whose zeroes is 0 and one zero is 3, is
(a) x² – 9
(b) x² + 9
(c) x² + 3
(d) x² – 3
Solution:
(a) In a quadratic polynomial
Let α and β be its zeros
and α + β = 0
and one zero = 3
3 + β = 0 ⇒ β = -3 .
Second zero = -3
Quadratic polynomial will be
(x – 3) (x + 3) ⇒ x² – 9

Question 27.
If two zeroes of the polynomial x³ + x² – 9x – 9 are 3 and -3, then its third zero is
(a) -1
(b) 1
(c) -9
(d) 9
Solution:
(a)

=> γ = -1
Third zero = -1

Question 28.
If √5 and – √5 are two zeroes of the polynomial x³ + 3x² – 5x – 15, then its third zero is
(a) 3
(b) – 3
(c) 5
(d) – 5
Solution:
(b)

Question 29.
If x + 2 is a factor x² + ax + 2b and a + b = 4, then
(a) a = 1, b = 3
(b) a = 3, b = 1
(c) a = -1, b = 5
(d) a = 5, b = -1
Solution:
(b) x + 2 is a factor of x² + ax + 2b and a + b = 4
x + 2 is one of the factor
x = – 2 is its one zero
f(-2) = 0
=> (-2)² + a (-2) + 2b = 0
=> 4 – 2a + 2b = 0
=> 2a – 2b = 4
=> a – b = 2
But a + b = 4
Adding we get, 2a = 6 => a = 3
and a + b = 4 => 3 + b = 4 => b = 4 – 3 = 1
a = 3, b = 1

Question 30.
The polynomial which when divided by – x² + x – 1 gives a quotient x – 2 and remainder 3, is
(a) x³ – 3x² + 3x – 5
(b) – x³ – 3x² – 3x – 5
(c) – x³ + 3x² – 3x + 5
(d) x³ – 3x² – 3x + 5
Solution:
(c) Divisor = – x² + x – 1, Quotient = x – 2 and
Remainder = 3, Therefore
Polynomial = Divisor x Quotient+Remainder
= (-x² + x – 1) (x – 2) + 3
= – x³ + x² – x + 2x² – 2x + 2 + 3
= – x³ + 3x² – 3x + 5

Question 31.
The number of polynomials having zeroes -2 and 5 is
(a) 1
(b) 2
(c) 3
(d) more than 3
Solution:
(d)

Hence, the required number of polynomials are infinite i.e., more than 3.

Question 32.
If one of the zeroes of the quadratic polynomial (k – 1)x² + kx + 1 is -3, then the value of k is
(a) \(\frac { 4 }{ 3 }\)
(b) – \(\frac { 4 }{ 3 }\)
(c) \(\frac { 2 }{ 3 }\)
(d) – \(\frac { 2 }{ 3 }\)
Solution:
(a)

Question 33.
The zeroes of the quadratic polynomial x² + 99x + 127 are
(a) both positive
(b) both negative
(c) both equal
(d) one positive and one negative
Solution:
(b)

Question 34.
If the zeroes of the quadratic polynomial x² + (a + 1) x + b are 2 and -3, then
(a) a = -7, b = -1
(b) a = 5, b = -1
(c) a = 2, b = -6
(d) a = 0, b = -6
Solution:
(d)

Question 35.
Given that one of the zeroes of the cubic polynomial ax³ + bx² + cx + d is zero, the product of the other two zeroes is
(a) – \(\frac { c }{ a }\)
(b) \(\frac { c }{ a }\)
(c) 0
(d) – \(\frac { b }{ a }\)
Solution:
(b)

Question 36.
The zeroes of the quadratic polynomial x² + ax + a, a ≠ 0,
(a) cannot both be positive
(b) cannot both be negative
(c) area always unequal
(d) are always equal
Solution:
(a) Let p(x) = x² + ax + a, a ≠ 0
On comparing p(x) with ax² + bx + c, we get
a = 1, b = a and c = a

So, both zeroes are negative.
Hence, in any case zeroes of the given quadratic polynomial cannot both the positive.

Question 37.
If one of the zeroes of the cubic polynomial x³ + ax² + bx + c is -1, then the product of other two zeroes is
(a) b – a + 1
(b) b – a – 1
(c) a – b + 1
(d) a – b – 1
Solution:
(a)



=> α β = -a + b + 1
Hence, the required product of other two roots is (-a + b + 1)

Question 38.
Given that two of the zeroes of the cubic polynomial ax³ + bx² + cx + d are 0, the third zero is
(a) – \(\frac { b }{ a }\)
(b) \(\frac { b }{ a }\)
(c) \(\frac { c }{ a }\)
(d) – \(\frac { d }{ a }\)
Solution:
(a) Two of the zeroes of the cubic polynomial
ax³ + bx² + cx + d = 0, 0
Let the third zero be d
Then, use the relation between zeroes and coefficient of polynomial, we have
d + 0 + 0 = – \(\frac { b }{ a }\)
⇒ d = – \(\frac { b }{ a }\)

Question 39.
If one zero of the quadratic polynomial x² + 3x + k is 2, then the value of k is
(a) 10
(b) -10
(c) 5
(d) -5
Solution:
(b) Let the given quadratic polynomial be P(x) = x² + 3x + k
It is given that one of its zeros is 2
P(2) = 0
=> (2)² + 3(2) + k = 0 => 4 + 6 + k = 0
=> k + 10 = 0 => k = -10

Question 40.
If the zeroes of the quadratic polynomial ax² + bx + c, c ≠ 0 are equal, then
(a) c and a have opposite signs
(b) c and b have opposite signs
(c) c and a have the same sign
(d) c and b have the same sign
Solution:
(c) The zeroes of the given quadratic polynomial ax² + bx + c, c ≠ 0 are equal. If coefficient of x² and constant term have the same sign
i.e., c and a have the same sign. While b i.e., coefficient of x can be positive/negative but not zero.

Question 41.
If one of the zeroes of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it
(a) has no linear term and constant term is negative.
(b) has no linear term and the constant term is positive.
(c) can have a linear term but the constant term is negative.
(d) can have a linear term but the constant term is positive.
Solution:
(a)

Given that, one of the zeroes of a quadratic polynomial p(x) is negative of the other.
αβ < 0
So, b < 0 [from Eq. (i)]
Hence, b should be negative Put a = 0, then,
p(x) = x² + b = 0 => x² = – b
=> x = ± √-b [ b < 0]
Hence, if one of the zeroes of quadratic polynomial p(x) is the negative of the other, then it has no linear term i.e., a = 0 and the constant term is negative i.e., b < 0. Alternate Method Let f(x) = x² + ax + b and by given condition the zeroes are a and -a. Sum of the zeroes = α – α = a => a = 0
f(x) = x² + b, which cannot be linear and product of zeroes = α (-α) = b
=> – α² = b
which is possible when, b < 0.
Hence, it has no linear term and the constant tenn is negative.

Question 42.
Which of the following is not the graph of a quadratic polynomial?




Solution:
(d) For any quadratic polynomial ax²+ bx + c, a 0, the graph of the corresponding equation y = ax² + bx + c has one of the two shapes either open upwards like ∪ or open downwards like ∩ depending on whether a > 0 or a < 0. These curves are called parabolas. So, option (d) cannot be possible.
Also, the curve of a quadratic polynomial crosses the X-axis on at most two points but in option (d) the curve crosses the X-axis on the three points, so it does not represent the quadratic polynomial.

Hope given RD Sharma Class 10 Solutions Chapter 2 Polynomials MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5

Other Exercises

• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQS

Question 1.
Show that the following numbers are irrational
(i) \(\frac { 1 }{ \surd 2 }\)
(ii) 7 √5
(iii) 6 + √2
(iv) 3 – √5
Solution:



But it contradics that because √5 is irrational
3 – √5 is irrational

Question 2.
Prove that following numbers are irrationals :
(i) \(\frac { 2 }{ \surd 7 }\)
(ii) \(\frac { 3 }{ 2\surd 5 }\)
(iii) 4 + √2
(iv) 5 √2
Solution:




5 √2 is an irrational number

Question 3.
Show that 2 – √3 is an irrational number. [C.B.S.E. 2008]
Solution:
Let 2 – √3 is not an irrational number

√3 is a rational number
But it contradicts because √3 is an irrational number
2 – √3 is an irrational number
Hence proved.

Question 4.
Show that 3 + √2 is an irrational number.
Solution:
Let 3 + √2 is a rational number

and √2 is irrational
But our suppositon is wrong
3 + √2 is an irrational number

Question 5.
Prove that 4 – 5√2 is an irrational number. [CBSE 2010]
Solution:
Let 4 – 5 √2 is not are irrational number
and let 4 – 5 √2 is a rational number
and 4 – 5 √2 = \(\frac { a }{ b }\) where a and b are positive prime integers

√2 is a rational number
But √2 is an irrational number
Our supposition is wrong
4 – 5 √2 is an irrational number

Question 6.
Show that 5 – 2 √3 is an irrational number.
Solution:
Let 5 – 2 √3 is a rational number
Let 5 – 2 √3 = \(\frac { a }{ b }\) where a and b are positive integers

and √3 is a rational number
Our supposition is wrong
5 – 2 √3 is a rational number

Question 7.
Prove that 2 √3 – 1 is an irrational number. [CBSE 2010]
Solution:
Let 2 √3 – 1 is not an irrational number
and let 2 √3 – 1 a ration number
and then 2 √3 – 1 = \(\frac { a }{ b }\) where a, b positive prime integers

√3 is a rational number
But √3 is an irrational number
Our supposition is wrong
2 √3 – 1 is an irrational number

Question 8.
Prove that 2 – 3 √5 is an irrational number. [CBSE 2010]
Solution:
Let 2 – 3 √5 is not an irrational number and let 2 – 3 √5 is a rational number
Let 2 – 3 √5 = \(\frac { a }{ b }\) where a and b are positive prime integers

\(\Longrightarrow \frac { 2b-a }{ 3b } =\surd 5\)
√5 is a rational
But √5 is an irrational number
Our supposition is wrong
2 – 3 √5 is an irrational

Question 9.
Prove that √5 + √3 is irrational.
Solution:
Let √5 + √3 is a rational number
and let √5 + √3 = \(\frac { a }{ b }\) where a and b are co-primes

√3 is a rational number
But it contradics as √3 is irrational number
√5 + √3 is irrational

Question 10.
Prove that √2 + √3 is an irrational number.
Solution:
Let us suppose that √2 + √3 is rational.
Let √2 + √3 = a, where a is rational.
Therefore, √2 = a – √3
Squaring on both sides, we get

which is a contradiction as the right hand side is a rational number while √3 is irrational.
Hence, √2 + √3 is irrational.

Question 11.
Prove that for any prime positive integer p, √p is an irrational number.
Solution:
Suppose √p is not a rational number
Let √p be a rational number
and let √p = \(\frac { a }{ b }\)
Where a and b are co-prime number


But it contradicts that a and b are co-primes
Hence our supposition is wrong
√p is an irrational

Question 12.
If p, q are prime positive integers, prove that √p + √q is an irrational number
Solution:


Hence proved.

Hope given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4

Other Exercises

• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQS

Question 1.
Find the L.C.M. and H.C.F. of the following pairs of integers and verify that L.C.M. x H.C.F. = Product of the integers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
(i) 26 and 91
26 = 2 x 13
91 = 7 x 13
H.C.F. = 13
and L.C.M. = 2 x 7 x 13 = 182
Now, L.C.M. x H.C.F. = 182 x 13 = 2366
and 26 x 91 = 2366
L.C.M. x H.C.F. = Product of integers

Question 2.
Find the L.C.M. and H.C.F. 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
(iv) 40, 36 and 126
(v) 84, 90 and 120
(vi) 24,15 and 36
Solution:

Question 3.
Given that HCF (306, 657) = 9, Find LCM (306, 657). [NCERT]
Solution:
HCF of 306, 657 = 9

Question 4.
Can two numbers have 16 as their H.C.F. and 380 as their L.C.M. ? Give reason.
Solution:
H.C.F. of two numbers = 16
and their L.C.M. = 380

We know the H.C.F. of two numbers is a factor of their L.C.M. but 16 is not a factor of 380 or 380 is not divisible by 16
It can not be possible.

Question 5.
The H.C.F. of two numbers is 145 and their L.C.M. is 2175. If one number is 725, find the other.
Solution:
First number = 725
Let second number = x
Their H.C.F. = 145
and L.C.M. = 2175

Second number = 435

Question 6.
The H.C.F. of two numbers is 16 and their product is 3072. Find their L.C.M.
Solution:
H.C.F. of two numbers = 16
and product of two numbers = 3072

Question 7.
The L.C.M. and H.C.F. of two numbers are 180 and 6 respectively. If one of the number is 30, find the other number.
Solution:
First number = 30
Let x be the second number
Their L.C.M. = 180 and H.C.F. = 6
We know that
first number x second number = L.C.M. x H.C.F.
30 x x = 180 x 6
⇒ x = \(\frac { 180\times 6 }{ 30 }\) = 36
Second number = 36

Question 8.
Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.
Solution:
L.C.M. of 520 and 468

= 2 x 2 x 9 x 10 x 13 = 4680
The number which is increased = 17
Required number 4680 – 17 = 4663

Question 9.
Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.
Solution:
Dividing by 28 and 32, the remainders are 8 and 12 respectively
28 – 8 = 20
32 – 12 = 20
Common difference = 20
Now, L.C.M. of 28 and 32

= 2 x 2 x 7 x 8 = 224
Required smallest number = 224 – 20 = 204

Question 10.
What is the smallest number that, when divided by 35, 56 and 91 leaves remainders of 7 in each case ?
Solution:
L.C.M. of 35, 56, 91

= 5 x 7 x 8 x 13 = 3640
Remainder in each case = 7
The required smallest number = 3640 + 7 = 3647

Question 11.
A rectangular courtyard is 18 m 72 cm long and 13 m 20 cm broad. It is to be paved with square tiles of same size. Find the least possible number of such tiles.
Solution:
Length of rectangle = 18 m 72 cm = 1872 cm
and breadth = 13 m 20 cm = 1320 cm
Side of the greatest size of square tile = H.C.F. of 1872 and 1320

Question 12.
Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.
Solution:
Greatest number of 6 digits = 999999
Now L.C.M. of 24, 15 and 36

We get quotient = 2777
and remainder = 279
Required number = 999999 – 279 = 999720

Question 13.
Determine the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.
Solution:

Required number will be = 110000 – 800 =109200

Question 14.
Find the least number that is divisible by ail the numbers between 1 to 10 (both inclusive).
Solution:
The required least number which is divisible by 1 to 10 will be the L.C.M. of 1 to 10
L.C.M. of 1 to 10

Question 15.
A circular field has a circumference of 360 km. Three cyclists start together and can cycle 48, 60 and 72 km a day, round the field. When will they meet again ?
Solution:
Circumference of a circular field = 360 km
Three cyclist start together who can cycle 48, 60 and 72 km per day round the field
L.C.M. of 48, 60, 72

They will meet again after 720 km distance

Question 16.
In a morning walk, three persons step off together, their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that they can cover the distance in complete steps ?
Solution:
Measures of steps of three persons = 80 cm, 85 cm and 90 cm
Minimum required distance covered by them = L.C.M. of 80 cm, 85 cm, 90 cm

= 2 x 5 x 8 x 9 x 17
= 12240 cm
= 122.40 m
= 122 m 40 cm

Hope given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3

Other Exercises

• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQS

Question 1.
Express each of the following integers as a product of its prime factors :
(i) 420
(ii) 468
(iii) 945
(iv) 7325
Solution:
(i) 420
=2 x 2 x 3 x 5 x 7
= 2² x 3 x 5 x 7

Question 2.
Determine the prime factorization of each of the following positive integer :
(i) 20570
(ii) 58500
(iii) 45470971
Solution:
(i) 20570

20570 = 2 x 5 x 11 x 11 x 17 = 2 x 5 x 11² x 17

Question 3.
Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers ?
Solution:
We know that a composite number is that number which can be factorize. It has more factors other than itself and one
Now, 7 x 11 x 13 + 13 = 13 (7 x 11 + 1) = 13 x 78
Which is composite number
Similarly,
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5(7 x 6 x 4 x 3 x 2 x 1 + 1)
= 5 x 1009
Which is a composite number
Hence proved

Question 4.
Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solution:
No, 6ⁿ can’t end with the digit 0 as the number ending 0 can be factorise of the type
2ⁿ x 5^m only but 6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ
Which does not has 5^m as factors.

Question 5.
Explain why 3 x 5 x 7 + 7 is a composite number. [NCERT Exemplar]
Solution:
We have, 3 x 5 x 7 + 7 = 105 + 7 = 112
Now, 112 = 2 x 2 x 2 x 2 x 7 = 2⁴ x 7
So, it is the product of prime factors 2 and 7. i.e., it has more than two factors.
Hence, it is a composite number.

Hope given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2

Other Exercises

• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQS

Question 1.
Define H.C.F. of two positive integers and find the H.C.F. of the following pairs of numbers.
(i) 32 and 54
(ii) 18 and 24
(iii) 70 and 30
(iv) 56 and 88
(v) 475 and 495
(vi) 75 and 243
(vii) 240 and 6552
(viii) 155 and 1385
(ix) 100 and 190
(x) 105 and 120
Solution:
Definition : The greatest among the common divisor of two or more integers is the Greatest Common Divisor (G.C.D.) or Highest Common Factor (H.C.F.) of the given integers.
(i) HC.F. of 32 and 54
Factors 32 = 1, 2, 4, 8, 16, 32
and factors of 54 = 1, 2, 3, 6, 9, 18, 27, 54
H.C.F. = 2
(ii) H.C.F. of 18 and 24
Factors of 18 = 1, 2, 3, 6, 9, 18
and factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Highest common factor = 6
H.C.F. = 6
(iii) H.C.F. of 70 and 30
Factors of 70 = 1, 2, 5, 7, 10, 14, 35, 70
and factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30
H.C.F. = 10
(iv) H.C.F. of 56 and 88
Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56
and factors of 88 = 1, 2, 4, 8, 11, 22, 44, 88
H.C.F. = 8
(v) H.C.F. of 475 and 495
Factors of 475 = 1, 5, 25, 19, 95, 475
and factors of 495 = 1, 3, 5, 9, 11, 15, 33, 45, 55, 99, 165, 495
H.C.F. = 5
(vi) H.C.F. of 75 and 243
Factors of 75 = 1, 3, 5, 15, 25, 75
Factors of 243 = 1, 3, 9, 27, 81, 243
H.C.F. = 3
(vii) H.C.F. of 240 and 6552
Factors of 240 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 48, 60, 80, 120, 240
Factors of 6552 = 1, 2, 3, 4, 6, 7, 8, 9, 12, 13, 14, 18, 21, 24, 26, 28, 36, 39, 42, 52, 56, 63, 72, 91, 104, 117, 126, 156, 168, 182, 234,252, 273, 312, 364, 488, 504, 546, 728, 819, 936, 1092, 1638, 2184, 3276, 6552
H.C.F. = 24
(viii) H.C.F. of 155 and 1385
Factors of 155 = 1, 5, 31, 155
Factors of 1385 = 1, 5, 277, 1385
H.C.F. = 5
(ix) 100 and 190

Question 2.
Use Euclid’s division algorithm to find the H.C.F. of
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
(iv) 184, 230 and 276
(v) 136,170 and 255
Solution:
(i) H.C.F. of 135 and 225
135 < 225

225 = 135 x 1 + 90
135 = 90 x 1 +45
45 = 45 x 2 + 0
Last remainder = 0
and last divisor = 45
H.C.F. = 45
(ii) H.C.F. of 196 and 38220
196 < 38220

38220 = 196 x 195 + 0
Last remainder = 0
and last divisor = 196
H.C.F. = 196
(iii) H.C.F. 867 and 255
255 < 867

867 = 255 x 3 + 102
255 = 102 x 2 + 51
102 = 51 x 2 + 0
Last remainder = 0
and last divisor = 51
H.C.F. = 51
(iv) H.C.F. of 184, 230 and 276
Let us find the highest common factor (H.C.F.) of 184 and 230

Hence, H.C.F. of 184 and 230 = 46
Now, find the H.C.F. of 276 and 46

Hence, H.C.F. of 276 and 46 = 46
Required H.C.F. of 184, 230 and 276 = 46
(v) H.C.F. of 136, 170 and 255
Let us find the highest common factor (H.C.F.) of 136 and 70

Hence, H.C.F. of 136 and 170 = 34
Now, find the H.C.F. of 34 and 255

Hence, highest common factor of 34 and 255 = 17
Required H.C.F. of 136, 170 and 255 = 17

Question 3.
Find the H.C.F. of the following pairs of integers and express it as a linear combinations of them.
(i) 963 and 657
(ii) 592 and 252
(iii) 506 and 1155
(iv) 1288 and 575
Solution:
(i) 963 and 657

(ii) HCF of 592 and 252

76x + 88y
Where x = 7, y = -6
(iii) 506 and 1155
H.C.F. = 11

(iv) 1288 and 575
H.C.F. = 23

= 575 x 9 + 1288 x (-4)
= ax + by
x = 9, y = -4

Question 4.
Find the largest number which divides 615 and 963 leaving remainder 6 in each case.
Solution:
The given numbers are 615 and 963
Remainder in each case = 6

615 – 6 = 609 and 963 – 6 = 957 are divisible by the required number which is the H.C.F. of 609 and 957 = 87
Hence the required largest number = 87

Question 5.
If the H.C.F. of 408 and 1032 is expressible in the form 1032m – 408 x 5, find m.
Solution:
408, 1032
H.C.F. = 24

Which is in the form of 1032m – 408 x 5 comparing, we get m = 2

Question 6.
If the H.C.F. of 657 and 963 is expressible in the form 657x + 963 x (-15), find x.
Solution:
657 and 963
H.C.F. = 9

Question 7.
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:
The required number of columns will be the H.C.F. of 616 and 32

Using Euclid’s division
We get H.C.F. = 4
Number of columns = 4

Question 8.
A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin ?
Solution:
Quantity of oil of one kind =120l
and quantity of second kind = 180l
and third kind of oil = 240l
Maximum capacity of oil in each tin = H.C.F. of 120l, 180l and 240l
H.C.F. of 120 and 180 = 60

Question 9.
During a sale, colour pencils were being sold in packs of 24 each and crayons in packs of 32 each. If you want full packs of both and the same number of pencils and crayons, how many of each would you need to buy ?
Solution:
Number of pencils in each pack = 24
and number of crayons pack = 32

Highest number of pencils and crayons in packs will be = H.C.F. of 24 and 32 = 8
Number of pencil’s pack = \(\frac { 24 }{ 8 }\) = 3
and number of crayon’s pack = \(\frac { 32 }{ 8 }\) = 4

Question 10.
144 cartons of Coke Cans and 90 Cartons of Pepsi Cans, are to be stacked in a Canteen. If each stack is of the same height and is to contain cartons of the same drink what would be the greatest number of cartons each stack would have?
Solution:
Number of Coke Cans Cartons = 144
and number Pepsi Cartons = 90

Required greatest number of cartons of each = H.C.F. of 144 and 90 = 18

Question 11.
Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively.
Solution:
The given numbers are 285 and 1249 Remainder are 9 and 7 respectively
285 – 9 = 276
and 1249 – 7 = 1242 are divisible by required number

Required number = H.C.F. of 276 and 1242
Now, H.C.F. of 276 and 1242 = 138
Required number = 138

Question 12.
Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.
Solution:
The given numbers are 280 and 1245 Remainder are 4 and 3 respectively
280 – 4 = 276 and 1245 – 3 = 1242 are divisible by a number
The required number = H.C.F. of 276 and 1242

H.C.F. of 276 and 1242 = 138
Hence required number =138

Question 13.
What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively.
Solution:
Given numbers are 626, 3127 and 15628 and remainders are 1, 2 and 3 respectively
626 – 1 = 625

3127 – 2 = 3125 and 15628 – 3 = 15625 are divisible by a required greatest number

The greatest number will be the H.C.F. of 625, 3125 and 15625
H.C.F. of 625 and 3125 = 625
and H.C.F. of 625 and 15625 = 625
The required number = 625

Question 14.
Find the greatest numbers that will divide 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.
Solution:
Given numbers are 445, 572 and 699
and remainders are 4, 5, 6 respectively
445 – 4 = 441
572 – 5 = 567

699 – 6 = 693 are exactly divisible by a certain number which is the H.C.F. of these numbers
H.C.F. of 441 and 567 = 63
and H.C.F. of 63 and 693 = 63

The required number = 63

Question 15.
Find the greatest number which divides 2011 and 2623 leaving remainders 9 and 5 respectively.
Solution:
The given numbers are 2011 are 2623 and remainders are 9 and 5 respectively
2011 – 9 = 2002 and 2623 – 5 = 2618 are divisible by a greatest number which is the H.C.F. of 2002 and 2618

H.C.F. = 2002 and 2618 = 154
The required number= 154

Question 16.
Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively. [NCERT Exemplar]
Solution:
Since, 1,2 and 3 are the remainders of 1251, 9377 and 15628, respectively.
Thus, after subtracting these remainders from the numbers.
We have the numbers, 1251 – 1 = 1250, 9377 – 2 = 9375 and 15628 – 3 = 15625
which is divisible by the required number.
Now, required number = HCF of 1250, 9375 and 15625 [for the largest number]
By Euclid’s division algorithm,

Question 17.
Two brands of chocolates are available in packs of 24 and 15 respectively. If I need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind I would need to buy ?
Solution:
Number of chocolates of first kind = 24 and of second kind = 15
Number of chocolates to be bought equally of both kinds = H.C.F. of 24 and 15

= 3 chocolates
Least number of boxes of first kind = \(\frac { 24 }{ 3 }\) = 8
and of second kind = \(\frac { 15 }{ 3 }\) = 5

Question 18.
A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10 ft. by 8 ft. What would be the size in inches of the tile required that has to be cut and how many such tiles are required ?
Solution:
Size of bathroom = 10 ft. x 8 ft.

Largest size of tile = H.C.F. of 10 ft. and 8 ft. = 2 ft.
= 2 x 12 = 24 inches (1 ft. = 12 inches)

Question 19.
15 pastries and 12 biscuit packets have been donated for a school fete. These are to be packed in several smaller identical boxes with the same number of pastries and biscuit packets in each. How many biscuit packets and how many pastries will each box contain ?
Solution:
Number of pastries = 15
and number of biscuit packets =12
The number of pastries and pack of biscuits to be packed in smaller identical boxes
H.C.F. of 15 and 12

H.C.F. = 3
Each box will contain = \(\frac { 15 }{ 3 }\) pastries and \(\frac { 12 }{ 3 }\) pack of biscuits
= 5 pastries and 4 pack of biscuits

Question 20.
105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip ?
Solution:
The required number of animals will be the H.C.F. of 105 goats, 140 donkeys, 175 cows

H.C.F. of 175 and 140 = 35
and H.C.F. of 35 and 105 = 35

The required number of animals = 35

Question 21.
The length, breadth and height of a room are 8m 25 cm, 6 m 75 cm and 4 m 50 cm respectively. Determine the longest rod which can measure the three dimensions of the room exactly.
Solution:
Length = 8m 25 cm = 825 cm
Breadth = 6 m 75 cm = 675 cm
Height = 4 m 50 cm = 450 cm

The required measure will be the H.C.F. of these three dimensions
H.C.F. of 825 and 675 = 75
and H.C.F. of 75 and 450 = 75

The required length = 75 cm

Question 22.
Express the H.C.F. of 468 and 222 as 468x + 222y where x, y are integers in two different ways.
Solution:
468 and 222
H.C.F. = 6

6 = (222 – 24 x 9)
= 222 – (468 – 222 x 2) x 9
= 222 – 468 x 9 + 222 x 18
= 222 x 19 + 468 x (-9)
= 468 (9) + 222 x 19
Which is in the form of 468x + 222y
Similarly we can write it in the following form also
6 = 468 x 213 +222 x (-449)

Hope given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1

Other Exercises

• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.2
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.3
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.4
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.5
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.6
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers VSAQS
• RD Sharma Class 10 Solutions Chapter 1 Real Numbers MCQS

Question 1.
If a and b are two odd positive integers such that a > b, then prove that one of the two numbers \(\frac { a+b }{ 2 }\) and \(\frac { a-b }{ 2 }\) is odd and the other is even.
Solution:
a and b are two odd numbers such that a > b
Let a = 2n + 1, then b = 2n + 3

Question 2.
Prove that the product of two consecutive positive integers is divisible by 2.
Solution:
Let n and n + 1 are two consecutive positive integer
We know that n is of the form n = 2q and n + 1 = 2q + 1
n (n + 1) = 2q (2q + 1) = 2 (2q² + q)
Which is divisible by 2
If n = 2q + 1, then
n (n + 1) = (2q + 1) (2q + 2)
= (2q + 1) x 2(q + 1)
= 2(2q + 1)(q + 1)
Which is also divisible by 2
Hence the product of two consecutive positive integers is divisible by 2

Question 3.
Prove that the product of three consecutive positive integer is divisible by 6.
Solution:
Let n be the positive any integer Then
n(n + 1) (n + 2) = (n² + n) (n + 2)

Which is also divisible by 6
Hence the product of three consecutive positive integers is divisible by 6

Question 4.
For any positive integer n, prove that n³ – n is divisible by 6.
Solution:


Which is divisible by 6
Hence we can similarly, prove that n² – n is divisible by 6 for any positive integer n.
Hence proved.

Question 5.
Prove that if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for some integer q, but not conversely.
Solution:
Let n = 6q + 5, where q is a positive integer
We know that any positive integer is of the form 3k or 3k + 1 or 3k + 2, 1
q = 3k or 3k + 1 or 3k + 2
If q = 3k, then n = 6q + 5

Question 6.
Prove that the square of any positive integer of the form 5q + 1 is of the same form.
Solution:
Let a be any positive integer
Then a = 5m + 1
a² = (5m + 1 )² = 25m² + 10m + 1
= 5 (5m² + 2m) + 1
= 5q + 1 where q = 5m² + 2m
Which is of the same form as given
Hence proved.

Question 7.
Prove that the square of any positive, integer is of the form 3m or, 3m + 1 but not of the form 3m + 2.
Solution:
Let a be any positive integer
Let it be in the form of 3m or 3m + 1
Let a = 3q, then

Hence proved.

Question 8.
Prove that the square of any positive integer is of the form 4q or 4q + 1 for some integer q.
Solution:
Let a be the positive integer and
Let a = 4m

Hence proved.

Question 9.
Prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 for some integer q.
Solution:
Let a be the positive integer, and
Let a = 5m, then

Question 10.
Show that the square of an odd positive integer is of the form 8q + 1, for some integer q.
Solution:
Let n is any positive odd integer
Let n = 4p + 1, then
(4p + 1)² = 16p² + 8p + 1
n² = 8p (2p + 1) + 1
= 8q + 1 where q = p(2p + 1)
Hence proved.

Question 11.
Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
Solution:
Let n be any positive odd integer and
let n = 6q + r
=> 6q + r, b = 6, and 0 ≤ r < 6
or r = 0, 1, 2, 3, 4, 5
If n = 6q = 2 x 3q
But it is not odd
When n = 6q + 1 which is odd
When n = 6q + 2 which is not odd = 2 (3q+ 1)
When n = 6q + 3 which is odd
When n = 6q + 4 = 2 (3q + 2) which is not odd
When n = 6q + 5, which is odd
Hence 6q + 1 or 6q + 3 or 6q + 5 are odd numbers.

Question 12.
Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m. [NCERT Exemplar]
Solution:
Let a be an arbitrary positive integer, then by Euclid’s division algorithm, corresponding to the positive integers a and 6, there exist non-negative integers q and r such that


Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Question 13.
Show that the cube of a positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5. [NCERT Exemplar]
Solution:
Let a be an arbitrary positive integer. Then, by Euclid’s division algorithm, corresponding to the positive integers ‘a’ and 6, there exist non-negative integers q and r such that



Hence, the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the forms 6m, 6m + 1, 6m + 3, 6m + 3, 6m + 4 and 6m + 5 i.e., 6m + r.

Question 14.
Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.
[NCERT Exemplar]
Solution:
Given numbers are n, (n + 4), (n + 8), (n + 12) and (n + 16), where n is any positive integer.
Then, let n = 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4 for q ∈N [By Euclid’s algorithm]
Then, in each case if we put the different values of n in the given numbers. We definitely get one and only one of given numbers is divisible by 5.
Hence, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.
Alternate Method
On dividing on n by 5, let q be the quotient and r be the remainder.
Then n = 5q + r, where 0 ≤ r < 5. n = 5q + r, where r = 0, 1, 2, 3, 4
=> n = 5q or 5q + 1 or 5q + 2 or 5q + 3 or 5q + 4
Case I: If n = 5q, then n is only divisible by 5. .
Case II: If n = 5q + 1, then n + 4 = 5q + 1 + 4 = 5q + 5 = 5(q + 1), which is only divisible by 5.
So, in this case, (n + 4) is divisible by 5.
Case III : If n = 5q + 3, then n + 2 = 5q + 3 + 12 = 5q + 15 = 5(q + 3), which is divisible by 5.
So, in this case (n + 12) is only divisible by 5.
Case IV : If n = 5q + 4, then n + 16 = 5q + 4 + 16 = 5q + 20 = 5(q + 4), which is divisible by 5.
So, in this case, (n + 16) is only divisible by 5.
Hence, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

Question 15.
Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer ? [NCERT Exemplar]
Solution:
We know that any positive integer can be of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4 or 6m + 5, for some integer m.
Thus, an odd positive integer can be of the form 6m + 1, 6m + 3, or 6m + 5 Thus we have:

Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.

Question 16.
A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, 3m or 3m + 2 for some integer m? Justify your answer.
Solution:
No, by Euclid’s Lemma, b = aq + r, 0 ≤ r < a
Here, b is any positive integer

Question 17.
Show that the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.
Solution:
By Euclid’s lemma, b = aq + r, 0 ≤ r ≤ a
Here, b is any positive integer,
a = 3, b = 3q + r for 0 ≤ r ≤ 2
So, any positive integer is of the form 3k, 3k + 1 or 3k + 2

Which is in the form of 3m + 1. Hence, square of any positive number cannot be of the form 3m + 2.

Hope given RD Sharma Class 10 Solutions Chapter 1 Real Numbers Ex 1.1 are helpful to complete your math homework.

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