RD Sharma 2022 Mcqs Solutions for Class 10 Maths Chapter 3 - Pair Of Linear Equations In Two Variables

Answer:

The given system of equations are

for unique solution

Here

By cross multiply we get

Hence, the correct choice is.

Answer:

The given system of equations are

For the equations to have infinite number of solutions,

Here,

Therefore

By cross multiplication of we get,

And

Therefore the value of k is 6

Hence, the correct choice is .

Answer:

The given system of equations are

For the equations to have no solutions,

If we take

Therefore the value of k is10.

Hence, correct choice is.

Answer:

The given system of equations are,

Here,

By cross multiply we get

Therefore the value of k is 6,

Hence, the correct choice is.

Answer:

The given system of equation is

If then the equation have no solution.

By cross multiply we get

Hence, the correct choice is.

Answer:

If a pair of linear equations in two variables is consistent, then its solution exists.

∴The lines represented by the equations are either intersecting or coincident.

Hence, correct choice is.

Answer:

The given system of equations

For the equations to have infinite number of solutions

If we take

And

Therefore, the value of k is 6.

Hence, the correct choice is .

Answer:

The given systems of equations are

If

Here

Hence, the correct choice is .

Answer:

The given equations are 
$$\begin{gathered} x-y=2 .....\left(1\right) \\ x+y=4 .....\left(2\right) \end{gathered}$$
Adding (1) and (2), we get
2x = 6
⇒ x = 3
Putting x = 3 in (1), we get
3 + y = 4
⇒ y = 1
So, x = a = 3 and y = b = 1.
Thus, the values of a and b are 3 and 1, respectively.

Hence, the correct answer is option (a).

Answer:

The given system of equations is
$$\begin{gathered} 3x-y+8=0 \\ 6x-ky+16=0 \end{gathered}$$
We know that the lines 
$$\begin{gathered} a_{1}x+b_{1}y+c_1=0 \\ {a}_{2}x+b_{2}y+c_2=0 \end{gathered}$$
are coincident iff
$\frac{a_1}{a_2}=\frac{{\displaystyle b_1}}{{\displaystyle b_2}}=\frac{{\displaystyle c_1}}{{\displaystyle c_2}}$
$$\begin{gathered} \therefore \frac{3}{6}=\frac{{\displaystyle -1}}{{\displaystyle -k}}=\frac{8}{16} \\ \Rightarrow \frac{1}{2}=\frac{1}{k}=\frac{1}{2} \\ \Rightarrow k=2 \end{gathered}$$
Thus, the value of k = 2.

Hence, the correct answer is option (c).

Answer:

The given pair of equations are y = 0 and y = $-$5.
The equation y = 0 represents x-axis and y = $-$5 is a line parallel to x-axis (no x-intercept).



We can see this graphically, both lines are parallel and thus will have no solution.

Hence, the correct answer is option (d).

Answer:

Let the cost of each chair be ₹x and for each table be ₹y.

Then,
$$\begin{gathered} 8x+5y=10500 .....\left(1\right) \\ 5x+3y=6450 .....\left(2\right) \end{gathered}$$

Multiplying (1) by 3 and (2) by 5, we get

$$\begin{gathered} 24x+15y=31500 .....\left(3\right) \\ 25x+15y=32250 .....\left(4\right) \end{gathered}$$

Subtracting (3) from (4), we get
$$\begin{gathered} 25x-24x=32250-31500 \\ \Rightarrow x=750 \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

By the property of rectangle, we know that opposite sides are equal i.e. CD = AB.
So, x + y = 12 .....(1)

Similarly, AD = BC.
So, x − y = 8 .....(2)

On adding (1) and (2), we get
2x = 20
$\Rightarrow$ x = 10

On substituting x = 10 in (1), we get
y = 2

Hence, the correct answer is option (a).

Answer:

The pair of linear equations $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$, the condition for no solution is given by $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$.

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get:

$$\begin{gathered} a_1=3, b_1=5, c_1=-3 \\ {a}_2=6, b_2=k, c_2=-8 \end{gathered}$$

$$\begin{gathered} \begin{array}{ccc}\Rightarrow \frac{3}{6}& =& \frac{5}{k}\ne \frac{-3}{-8}\end{array} \\ \begin{array}{ccc}\Rightarrow \frac{1}{2}& =& \frac{5}{k}\ne \frac{3}{8}\end{array} \end{gathered}$$
$\Rightarrow k=10 \mathrm{and} k\ne \frac{40}{3}$

Hence, the correct answer is option (b).

Answer:

Let the ages of a father and his son in years be x and y respectively.

According to the question,
$$\begin{gathered} x+y=65 .....\left(1\right) \\ 2\left(x-y\right)=50 \\ \Rightarrow x-y=25 .....\left(2\right) \end{gathered}$$

Adding (1) and (2), we get
x = 45 and y = 20

Thus, the age of father is 45 years.

Hence, the correct answer is option (b).

Answer:

The given systems of equations are

For the equations to have infinite number of solutions,

Here ,

Let us take

By cross multiplication we get,

Now take

By cross multiplication we get,

Substitute in the above equation

Substitute in equation we get,

Therefore and.

Hence, the correct choice is.

Answer:

The given system of equations is inconsistent,

If the system of equations is in consistent, we have

Therefore, the value of k is2.

Hence, the correct choice is .

Answer:

Given the system of equations has

We know that intersecting lines have unique solution

Here

Therefore intersecting lines, have unique solution

Hence, the correct choice is

Answer:

Given the system of equations are

For the equations to have infinite number of solutions,

By cross multiplication we have

Divide both sides by 2. we get

Hence, the correct choice is .

Answer:

The given system of equations are

For coincident lines , infinite number of solution

Option A.:

Option B:

$$\begin{gathered} 5a+b=0 \\ 5\left(-5\right)+\left(-1\right) = -25-1 = -26\ne 0 \end{gathered}$$

Option.C:

a - b = 0

-5 - (-1) = -4 $\ne$0

None of the option satisfies the values.

Answer:

Given the area of the triangle formed by the line

If in the equation either A and B approaches infinity, The line become parallel to either x axis or y axis respectively,

Therefore

Area of triangle

Hence, the correct choice is .

Answer:

Given and

We have plotting points as when

Therefore, area of

Area of triangle is square units

Hence, the correct choice is _.

Answer:

Given and

We have plotting points as when

Therefore, area of

Area of triangle is square units

Hence, the correct choice is

Answer:

Let the digits at the tens and the ones place be x and y, respectively. So, the two digit number is 10x + y.
Now,
x + y = 9         .....(i)
Also,
10x + y + 27 = 10y + x
⇒ 9x − 9y = −27
⇒ x − y = −3            .....(ii)
Adding (i) and (ii), we get
2x = 6
⇒ x = 3
Putting x = 3 in (i), we get
3 + y = 9
⇒ y = 6 
Thus, the required number is 10 × 3 + 6 = 36.

Hence, the correct answer is option (d).

Answer:

Let the number of ₹1 coins be x and that of ₹2 coins be y.
Now,
Total number of coins = 50
So, x + y = 50            .....(i)
Also,
₹1 × x + ₹2 × y = ₹75
∴ x + 2y = 75            .....(ii)
Subtracting (i) from (ii), we get
y = 25
Putting y = 25 in (i), we get
x + 25 = 50
⇒ x = 25
So, the number of ₹1 coins and ₹2 coins are 25 and 25, respectively. 

Hence, the correct answer is option (d).

Disclaimer: The answer given in the book does not match with the one obtained.

Answer:

37x + 43y = 123 ....(1)
43x + 37y = 117 ....(2)

Subtracting (1) from (2), we get
6x − 6y = −6
⇒ x = y − 1

Substitute in (1), we get
37(y − 1) + 43y = 123
⇒ 80y = 123 + 37 = 160
⇒ y = 2 = b
⇒ x = 1 = a

∴ a³ + b³ = 1³ + 2³ = 9

Hence, the correct answer is option (c).

Answer:

Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get:

$$\begin{gathered} a_1=5, b_1=7, c_1=-3 \\ {a}_2=15, b_2=21, c_2=-k \end{gathered}$$
As $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, which means that lines are coincident 

$$\begin{gathered} \begin{array}{ccc}\Rightarrow \frac{5}{15}& =& \frac{7}{21}=\frac{-3}{-k}\end{array} \\ \begin{array}{ccc}\Rightarrow \frac{1}{3}& =& \frac{1}{3}=\frac{3}{k}\end{array} \end{gathered}$$
Solving we get k = 9.
 
Hence, the correct answer is option (a).

Answer:

The condition for pair of equation to be dependent linear equations is $\frac{a_1}{a_2} = \frac{b_1}{b_2} =\frac{c_1}{c_2}$.
Comparing the coefficients of the first equation, with general equation of the form ax + by + c = 0, we get 
$a_1=-5, b_1=7, c_1=-2$
Let the second equation be $a_{2}x+b_{2}y+c_2=0$.
$\Rightarrow \frac{-5}{a_2}=\frac{7}{b_2}=\frac{-2}{c_2}=k$, where k is any arbitrary constant.

Putting $k=\frac{-1}{2}$, we get 
$a_2=10, b_2=-14, c_2=4$

$\therefore$The required equation becomes $10x\mathit{-}14y+4=0$.

Hence, the correct answer is option (d).

Answer:

217x + 131y = 913 .....(1)
131x + 217y = 827 .....(2)

Adding (1) and (2), we get
348x + 348y = 1740
⇒ x + y = 5 ....(3)

Subtracting (2) from (1), we get
86x − 86y = 86
⇒ x − y = 1 .....(4)

Adding (3) and (4), we get
⇒ x = 3 and y = 2
⇒ x + y = 3 + 2 = 5

Hence, the correct answer is option (a).

Answer:

3^{x + y} = 243
⇒ 3^{x + y} = 3⁵
⇒ x + y = 5 .....(1)

And, 243^{x – y} = 3
⇒ (3⁵)^{x – y} = 3¹
⇒ 5(x – y) = 1
⇒ x – y = $\frac{1}{5}$..... (2)

Adding (1) and (2), we get
⇒ $x=\frac{13}{5}, y=\frac{12}{5}$

Thus, the number of solutions is 1.

Hence, the correct answer is option (b).

Answer:

Given and

If We have plotting points as

Therefore, area of

Area of triangle is square units

Hence, the correct choice is

Answer:

Let the intersection point of line CD and x-axis be E.



Thus, EB = 3 units and the perpendicular distance from point C on EB is 2 units.

Area of triangle ECB = $\frac{1}{2}\times \mathrm{EB}\times \mathrm{height} \left(\because \mathrm{Area} \mathrm{of} \mathrm{a} \mathrm{triangle}=\frac{1}{2}\times \mathrm{base}\times \mathrm{height}\right)$
                                  $$\begin{gathered} =\frac{1}{2}\times 3\times 2 \\ =3 \mathrm{sq}. \mathrm{units} \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

(i) Consider x and y be the number of birds and deer respectively.
Since each bird and deer has 2 eyes.
And, total animals have 1000 eyes.

Thus, the equation will be
2x + 2y = 1000
⇒ x + y = 500 .....(1)

Hence, the correct answer is option (b).

(ii) Since, each bird has 2 legs and each deer has 4 legs.
And, total animals have 1400 legs.

Thus, equation will be
2x + 4y = 1400
⇒ x + 2y = 700 .....(2)

Hence, the correct answer is option (c).

(iii) From (1) and (2),
x + y = 500 .....(1)
x + 2y = 700 .....(2)

Multiplying (1), by 2 we get
2x + 2y = 1000 ....(3)

Subtracting (2) from (3), we get
x = 300

Hence, the correct answer is option (c).

(iv) Substituting x = 300 in (1), we get
300 + y = 500
y = 200

Hence, the correct answer is option (b).

(v)
Total number of animals in the zoo are total number of birds and deer, i.e., 300 + 200 = 500.

Hence, the correct answer is option (c).

Answer:

(i) Speed of boat is 5 km/hr and speed of stream is 2 km/hr.
Speed of the boat in downstream = Speed of boat + speed of stream
                                                     = 5 + 2 km/hr
                                                     = 7 km/hr

Hence, the correct answer is option (c).

(ii) Speed of boat is 5 km/hr and speed of stream is 2 km/hr.
Speed of the boat in upstream = Speed of boat − speed of stream
                                                     = 5 − 2 km/hr
                                                     = 3 km/hr

Hence, the correct answer is option (d).

(iii) Distance travelled by boat along downstream is 21 km.
Speed of the boat in downstream = $\frac{\mathrm{Distance} \mathrm{travelled} \mathrm{by} \mathrm{boat} \mathrm{in} \mathrm{downstream}}{\mathrm{Time} \mathrm{required}}$
$\begin{array}{rcl}& \Rightarrow & \mathrm{Time} \mathrm{required}=\frac{\mathrm{Distance}}{\mathrm{Speed}}\\ & =& \frac{21}{7}\\ & =& 3 \mathrm{hr}\end{array}$

Hence, the correct answer is option (d).

(iv) Distance travelled by boat along upstream is 12 km.
Speed of the boat in upstream = $\frac{\mathrm{Distance} \mathrm{travelled} \mathrm{by} \mathrm{boat} \mathrm{in} \mathrm{downstream}}{\mathrm{Time} \mathrm{required}}$
$\begin{array}{rcl}& \Rightarrow & \mathrm{Time} \mathrm{required}=\frac{\mathrm{Distance}}{\mathrm{Speed}}\\ & =& \frac{12}{3}\\ & =& 4 \mathrm{hr}\end{array}$

Hence, the correct answer is option (a).

(v) If speed of boat and stream be x km/hr and y km/hr respectively.
Speed of the boat in downstream = Speed of boat + speed of stream
                                                     = (x + y) km/hr

Speed of the boat in downstream = $\frac{\mathrm{Distance} \mathrm{travelled} \mathrm{by} \mathrm{boat} \mathrm{in} \mathrm{downstream}}{\mathrm{Time} \mathrm{required}}$
$\begin{array}{rcl}& \Rightarrow & \mathrm{Distance} \mathrm{travelled}=\mathrm{speed}\times \mathrm{time}\\ & =& \left(x+y\right)\times t\\ & =& t\left(x+y\right) \mathrm{km}\end{array}$

Hence, the correct answer is option (b).

Answer:

The pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 represents parallel lines, if $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}.$
Thus, statement-2 is true.


3x + 5y – 4 = 0 and 15x + 25y – 25 = 0
Comparing the coefficients of two equations, with the general equation of the form ax + by + c = 0, we get
$$\begin{gathered} a_1=3, b_1=5, c_1=-4 \\ {a}_2=15, b_2=25, c_2=-25 \end{gathered}$$

$$\begin{gathered} \frac{a_1}{a_2}=\frac{3}{15}=\frac{1}{5} \\ \frac{b_1}{b_2}=\frac{5}{25}=\frac{1}{5} \\ \frac{c_1}{c_2}=\frac{-4}{-25}=\frac{4}{25} \\ \Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2} \end{gathered}$$

As $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$ which means that lines are parallel, so the system of linear equations is inconsistent.
Statement-1 is true, and Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

The graph of x = 8 is a line parallel to y-axis at a distance of 8 units to the right of it.
So, the line l is the graph of x = 8.
The graph of y = 6 is a line parallel to the x-axis at a distance of 6 units above it.
So, the line m is the graph of y = 6.



The figure enclosed by the lines x = 8, y = 6, the x-axis and the y-axis is OABC, which is a rectangle.

The vertices of the rectangle OABC are O (0, 0), A (8, 0), B (8, 6), C (0, 6).
The length and breadth of this rectangle are 8 units and 6 units, respectively.
As the area of a rectangle = length × breadth,
the area of rectangle OABC = 8 × 6 = 48 sq. units

Thus, statement-1 is false.

From the graph, x = 8, y = 6 has a unique solution (8, 6) and thus a consistent system of equations.
Thus, statement-2 is true.

Hence, the correct answer is option (d).

Answer:

A pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 represents coincident lines if $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.$

Thus, statement-2 is true.

If a pair of linear equations represent coincident lines, then the equations are consistent and have infinitely many solutions.

Thus, statement-1 is false.

Hence, the correct answer is option (d).

Answer:

The system of equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 is inconsistent if $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}.$

Thus, statement-2 is false.

The system of equations 3x + 6y = 10 and 2x – ky + 5 = 0 is inconsistent, then
$$\begin{gathered} \frac{3}{2}=\frac{6}{-k}\ne \frac{-10}{5} \\ \Rightarrow k=-4 \end{gathered}$$

Thus, statement-1 is true.
Hence, the correct answer is option (c).