Solve the following pair of equations by reducing them to a pair of linear equations: $$\displaystyle \frac {1}{2x}+\frac {1}{3y}=2, \frac {1}{3x}+\frac {1}{2y}=\frac {13}{6}$$

$$x=\dfrac {1}{2}$$ and $$y=\dfrac {1}{3}$$

$$x=\dfrac {1}{4}$$ and $$y=\dfrac {1}{7}$$

$$x=\dfrac {2}{5}$$ and $$y=\dfrac {1}{3}$$

$$x=\dfrac {3}{7}$$ and $$y=\dfrac {1}{7}$$

For what value of $$k$$ does the system of equations $$\displaystyle 2x+ky=11\:and\:5x-7y=5$$ has no solution?

$$\displaystyle \frac{-14}{5}$$

$$\displaystyle \frac{-11}{5}$$

$$\displaystyle \frac{-14}{9}$$

In the system of equations $$\dfrac {12}{x+y}+\dfrac {8}{x-y}=8$$ and $$\dfrac {27}{x+y}-\dfrac {12}{x-y}=3$$, the values of $$x$$ and $$y$$ will be

$$\dfrac {5}{2}$$ and $$\dfrac {1}{2}$$

$$\dfrac {5}{3}$$ and $$\dfrac {1}{3}$$

$$\dfrac {5}{4}$$ and $$\dfrac {1}{4}$$

MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 11

Solve the following pair of equations by reducing them to a pair of linear equations:

\frac{10}{(x + y)} + \frac{2}{(x - y)} = 4, \frac{15}{(x + y)} - \frac{5}{(x - y)} = - 2

$\dfrac {10}{(x+y)}+\dfrac {2}{(x-y)}=4, \dfrac {15}{(x+y)}-\dfrac {5}{(x-y)}=-2$

0%
$x=1,\,\,y=3$
0%
$x=3,\,\,y=2$
0%
$x=8,\,\,y=1$
0%
$x=0,\,\,y=3$

Explanation

Given:

$\dfrac{10}{x+y}+\dfrac{2}{x-y}=4$

$\dfrac{15}{x+y}+\dfrac{5}{x-y}=-2$

let $\dfrac{1}{x+y}=a \ and \ \dfrac{1}{x-y}=b$

$10a+2b=4.......eq1$
$15a-5b=-2........eq2$

Multiply eq1 with 3 and eq2 with 2 we get
$30a+6b=12......eq3$
$30a-10b=-4......eq4$

subtracting eq3 and eq4 we get
$16b=16$
$b=1$

Substituting $b=1$ in eq1 we get
$10a+2\times 1=4$
$10a=2$
$a=\dfrac{1}{5}$

As $\dfrac{1}{x+y}=a=\dfrac{1}{5}$

$x+y=5............eq5$

and $\dfrac{1}{x-y}=b=1$

$x-y=1...........eq6$

subtracting eq5 and eq6 we get
$2x=6$
$x=3$

putting $x=3$ in eq 5 we get
$3+y=5$
$y=2$

hence $x=3$ and $y=2$

Solve the following pair of equations by reducing them to a pair of linear equations:

6 x + 3 y = 6 x y, 2 x + 4 y = 5 x y

$6x + 3y = 6xy, 2x + 4y = 5xy$

0%
$x=1,\,\,y=1$
0%
$x=1,\,\,y=2$
0%
$x=0,\,\,y=1$
0%
None of these

Explanation

$6x+3y=6xy$ .........Given

Multiply by $\dfrac{1}{xy}$ we get

$\dfrac{6x}{xy}+\dfrac{3y}{xy}=\dfrac{6xy}{xy}$

$\dfrac{6}{y}+\dfrac{3}{x}=6............eq1$

$2x+4y=5xy$ ...........Given

Multiply with $\dfrac{1}{xy}$ we get

$\dfrac{2x}{xy}+\dfrac{4y}{xy}=\dfrac{5xy}{xy}$

$\dfrac{2}{y}+\dfrac{4}{x}=5........eq2$

Let $\dfrac{1}{x}=a \ and \ \dfrac{1}{y}=b$

$a+2b=2.......eq3$
$4a+2b=5.....eq4$

Subtracting eq3 and eq4 we get
$-3a=-3$
$a=1$

Substituting a=1 in eq3 we get
$1+2b=2$
$2b=1$
$b=\dfrac{1}{2}$

Thus if $a=\dfrac{1}{x}=1\Rightarrow x=1$

If $b=\dfrac{1}{y}=\dfrac{1}{2}\Rightarrow y=2$

Hence $x=1$ and $y=2$

Solve the following pair of equations by reducing them to a pair of linear equations:

\frac{1}{2 x} + \frac{1}{3 y} = 2, \frac{1}{3 x} + \frac{1}{2 y} = \frac{13}{6}

$\displaystyle \frac {1}{2x}+\frac {1}{3y}=2, \frac {1}{3x}+\frac {1}{2y}=\frac {13}{6}$

0%
$x=\dfrac {1}{4}$ and $y=\dfrac {1}{7}$
0%
$x=\dfrac {2}{5}$ and $y=\dfrac {1}{3}$
0%
$x=\dfrac {3}{7}$ and $y=\dfrac {1}{7}$
0%
$x=\dfrac {1}{2}$ and $y=\dfrac {1}{3}$

Explanation

Given pair of equations,

\frac{1}{2 x} + \frac{1}{3 y} = 2, \frac{1}{3 x} + \frac{1}{2 y} = \frac{13}{6}

$\displaystyle \frac {1}{2x}+\frac {1}{3y}=2, \ \ \frac {1}{3x}+\frac {1}{2y}=\frac {13}{6}$

Putting

\frac{1}{x} = u

$\dfrac {1}{x}=u$ and

\frac{1}{y} = v

$\dfrac {1}{y}=v$

We get ,

\frac{1}{2} u + \frac{1}{3} v = 2 3 u + 2 v = 12 . . . . . . . . (1)

$\displaystyle \frac {1}{2}u+\frac {1}{3}v=2\\ 3u+2v =12 \ ........(1)$

\frac{1}{3} u + \frac{1}{2} v = \frac{13}{6} 2 u + 3 v = 13 . . . . . . . . . (2)

$\dfrac {1}{3}u+\dfrac {1}{2}v=\dfrac {13}{6}\\ 2u+3v = 13\ .........(2)$

Multiplying

(1)

$(1)$ by

3

$3$ and

(2)

$(2)$ by

2

$2$ ,

3 (3 u + 2 v) = 3 \times 12 . . . . . . . . . (3)

$3 (3u + 2v)=3 \times 12\ .........(3)$

2 (2 u + 3 v) = 2 \times 13 . . . . . . . . . (4)

$2(2u+ 3v)=2 \times 13\ .........(4)$

then subtract

(4)

$(4)$ from

(3)

$(3)$ , we get

3 (3 u + 2 v) - 2 (2 u + 3 v) = 3 \times 12 - 2 \times 13

$3 (3u + 2v) - 2(2u+ 3v) = 3 \times 12 - 2 \times 13$

9 u - 4 u = 36 - 26 5 u = 10 \Rightarrow u = 2

$9u - 4u = 36 - 26\\ 5u=10\Rightarrow u = 2$

Then substituting

u = 2

$u = 2$ in

(1)

$(1)$ , we get

\Rightarrow 6 + 2 v = 12

$\Rightarrow 6 + 2v = 12$

\Rightarrow v = 3

$\Rightarrow v=3$

Now,

u = 2

$u = 2$ and

v = 3

$v = 3$

\Rightarrow \frac{1}{x} = 2

$\Rightarrow \dfrac {1}{x}=2$ and

\frac{1}{y} = 3 \Rightarrow x = \frac{1}{2}

$\dfrac {1}{y}=3\Rightarrow x=\dfrac {1}{2}$ and

y = \frac{1}{3}

$y=\dfrac {1}{3}$

For which value of k will the following pair of linear equations have no solution?

3 x + y = 1

$3x + y = 1$

(2 k - 1) x + (k - 1) y = 2 k + 1

$(2k - 1) x + (k -1) y = 2k + 1$

0%
$k=4$
0%
$k=1$
0%
$k=2$
0%
$k=5$

Explanation

The given pair of equations are:
$3x+y = 1...(1)$
$\left ( 2k-1 \right )x +\left ( k-1 \right )y = 2k+1..(2)$
Now rearranging eq1 and eq2 will get
$3x+y-1=0...(3)$
$\left ( 2k-1 \right )x +\left ( k-1 \right )y -\left( 2k+1 \right )= 0..(4)$
Now compare with
$a_{1} = 3, b_{1} = 1, c_{1} = -1$
$a_{2} = 2k-1, b_{2} = k-1, c_{2} = -\left( 2k+1 \right )$
Now we get
$\dfrac{a_{1}}{a_{2}} =\dfrac{3}{2k-1} , \dfrac{b_{1}}{b_{2}} = \dfrac{1}{k-1},\dfrac{c_{1}}{c_{2}} = \dfrac{-1}{-\left( 2k+1 \right )}$
Now will take
$\dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}}$
$\Rightarrow \dfrac{3}{2k-1} = \dfrac{1}{k-1}$
$\Rightarrow 3k-3= 2k-1$
$\Rightarrow 3k-2k = -1+3$
$\Rightarrow k = 2$
Hence $k =2$ is the value.

Solve the following pair of equations by reducing them to a pair of linear equations:

\frac{4}{x} + 3 y = 14, \frac{3}{x} - 4 y = 23

$\dfrac {4}{x}+3y=14, \dfrac {3}{x}-4y=23$

0%
$x=7,\,\,y=1$
0%
$x=-3,\,\,y=-1$
0%
$x=\dfrac{1}{5},\,\,y=-2$
0%
$x=\dfrac{7}{2},\,\,y=7$

Explanation

Given

\frac{4}{x} + 3 y = 14

$\dfrac{4}{x}+3y=14$

\frac{3}{x} - 4 y = 23

$\dfrac{3}{x}-4y=23$

Let

\frac{1}{x} = a

$\dfrac{1}{x}=a$

4 a + 3 y = 14....... e q 1

$4a+3y=14.......eq1$

3 a - 4 y = 23........ e q 2

$3a-4y=23........eq2$

multiply eq1 by 4 and eq2 by 3 we get

16 a + 12 y = 56...... e q 3

$16a+12y=56......eq3$

9 a - 12 y = 69......... e q 4

$9a-12y=69.........eq4$

Adding eq3 and eq4 we get

25 a = 125

$25a=125$

a = 5

$a=5$

Substituting

a = 5

$a=5$ in eq1 we get

4 \times 5 + 3 y = 14

$4\times 5+3y=14$

3 y = 14 - 20

$3y=14-20$

3 y = - 6

$3y=-6$

y = - 2

$y=-2$

Hence

\frac{1}{x} = 5

$\dfrac{1}{x}=5$ then

x = \frac{1}{5}, y = - 2

$x=\dfrac{1}{5},y=-2$

Given

3 x - 4 y = 7

$3x - 4y = 7$ and

x + c y = 13

$x + cy = 13$ , for what value of

c

$c$ will the two equations not have a solution ?

0%
$\displaystyle \frac{3}{4}$
0%
$\displaystyle \frac{4}{3}$
0%
$-4$
0%
$\displaystyle \frac{-4}{3}$

Explanation

3 x - 4 y = 7

$3x - 4y = 7$

\Rightarrow a_{1} = 3, b_{1} = - 4, c_{1} = 7

$\Rightarrow a_1=3,\,b_1=-4,\,c_1=7$

x + c y = 13 \Rightarrow a_{2} = 1, b_{2} = c, c_{2} = 13

$x + cy = 13\Rightarrow a_2=1,\,b_2=c,\,c_2=13$
As we know the system of linear equation has no solution if

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

$\dfrac{a_{1}}{a_{2}}= \dfrac{b_{1}}{b_{2}}\neq \dfrac{c_{1}}{c_{2}}$

∴

$\therefore$

\frac{3}{1} = \frac{- 4}{c} \neq \frac{7}{13}

$\dfrac{3}{1}= \dfrac{-4}{c} \ne\dfrac{7}{13}$

c = \frac{- 4}{3}

$c = \dfrac{-4}{3}$
Hence, option D is correct.

A fraction becomes

\frac{1}{3}

$\dfrac {1}{3}$ when

1

$1$ is subtracted from the numerator and it becomes

\frac{1}{4}

$\dfrac {1}{4}$ when

8

$8$ is added to its denominator. Find the fraction.

0%
$\dfrac{1}{4}$
0%
$\dfrac{4}{17}$
0%
$\dfrac{5}{12}$
0%
None of these

Explanation

Let the fraction be

\frac{x}{y}

$\dfrac {x}{y}$

Now as per the question we will have

\frac{x - 1}{y} = \frac{1}{3} . . (1)

$\dfrac {x-1}{y} = \dfrac {1}{3}..(1)$

\frac{x}{y + 8} = \frac{1}{4} . . (2)

$\dfrac {x}{y+8} = \dfrac {1}{4}..(2)$

From eq1 will get

3 x - y = 3... (3)

$3x-y = 3...(3)$

From eq2 will get

4 x - y = 8... (4)

$4x-y = 8...(4)$

Now subtract eq4 from eq3

\Rightarrow 3 x - y - 4 x + y = 3 - 8

$\Rightarrow 3x-y-4x+y = 3-8$

\Rightarrow - x = - 5

$\Rightarrow -x = -5$

\Rightarrow x = 5

$\Rightarrow x = 5$

Now put x in eq 4 will get

\Rightarrow 4 \times 5 - y = 8

$\Rightarrow 4 \times 5-y = 8$

\Rightarrow 20 - y = 8

$\Rightarrow 20-y =8$

\Rightarrow - y = 8 - 20

$\Rightarrow -y = 8-20$

\Rightarrow - y = - 12

$\Rightarrow -y = -12$

\Rightarrow y = 12

$\Rightarrow y = 12$

Hence fraction will be

\frac{5}{12}

$\dfrac {5}{12}$

For what value of

k

$k$ does the system of equations

2 x + k y = 11 a n d 5 x - 7 y = 5

$\displaystyle 2x+ky=11\:and\:5x-7y=5$ has no solution?

0%
$\displaystyle \frac{-14}{5}$
0%
$\displaystyle \frac{-11}{5}$
0%
$\displaystyle \frac{-14}{9}$
0%
$\displaystyle \frac{14}{5}$

Explanation

The equations are

2 x + k y = 11

$2x+ky=11$ and

5 x - 7 y = 5

$5x-7y=5$

2 x + k y - 11 = 0

$2x+ky-11=0$ and

5 x - 7 y - 5 = 0

$5x-7y-5=0$

They will have no solution if

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} . . . . . . . . (1)

$\dfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\dfrac { { b }_{ 1 } }{ { b }_{ 2 } } \neq \dfrac { { c }_{ 1 } }{ { c }_{ 2 } }. .......(1)$

Here

a_{1} = 2, a_{2} = 5, b_{1} = k, b_{2} = - 7, c_{1} = - 11, c_{2} = - 5

${ a }_{ 1 }=2,\quad { a }_{ 2 }=5,\quad { b }_{ 1 }=k,\quad { b }_{ 2 }=-7,\quad { c }_{ 1 }=-11,\quad { c }_{ 2 }=-5$

∴ \frac{a_{1}}{a_{2}} = \frac{2}{5}, \frac{b_{1}}{b_{2}} = \frac{k}{- 7}, \frac{c_{1}}{c_{2}} = \frac{- 11}{- 5} = \frac{11}{5}

$\therefore \dfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\dfrac { 2 }{ 5 } , \dfrac { { b }_{ 1 } }{ { b }_{ 2 } } =\dfrac { k }{ -7 } ,\quad \dfrac { { c }_{ 1 } }{ { c }_{ 2 } } =\dfrac { -11 }{ -5 } =\dfrac { 11 }{ 5 }$

Now from

(1)

$(1)$

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$

∴ \frac{2}{5} = \frac{k}{- 7} \Rightarrow k = \frac{- 14}{5}

$\therefore \quad \dfrac { 2 }{ 5 } =\dfrac { k }{ -7 } \\ \Rightarrow k=\dfrac { -14 }{ 5 }$

Also the first two ratios

\neq \frac{c_{1}}{c_{2}}

$\neq \dfrac { { c }_{ 1 } }{ { c }_{ 2 } }$

∴ k = \frac{- 14}{5}

$\therefore k=\dfrac { -14 }{ 5 }$

Hence, the value of

k

$k$ is

\frac{- 14}{5}

$\dfrac { -14 }{ 5 }$

Sum of two numbers is

35

$35$ and their difference is

13

$13$ . Then the numbers are

0%
$10, 25$
0%
$22, 13$
0%
$24, 11$
0%
$20, 15$

Explanation

${\textbf{Step -1: Framing equations}}$

$\text{Without loss of generality, suppose,}$ $x$ ${\text{and}}$ $y$ ${\text{be two numbers and }} x > y.$

${\text{Given that, sum of two numbers is 35}}{\text{.}}$

$\Rightarrow x + y = 35$ $\ldots \left( 1 \right)$

${\text{Difference between these two numbers is 13}}{\text{.}}$

$\Rightarrow x - y = 13$ $\ldots \left( 2 \right)$

${\textbf{Step -2: Adding equation }}\left( \mathbf 1 \right)$ ${\textbf{and equation }}\left(\mathbf 2 \right)\textbf .$

$\Rightarrow x + y + \left( {x - y} \right) = 35 + 13$

$\Rightarrow 2x = 48$

$\Rightarrow x = 24$

${\text{Substituting the value of}}$ $x$ ${\text{in equation }}\left( 1 \right)$ ${\text{we get,}}$

$\Rightarrow 24 + y = 35$

$\Rightarrow y = 35 - 24$

$\Rightarrow y = 11$

${\text{Hence, these two numbers are 24 and 11}}{\text{.}}$

${\textbf{ Hence, option (C) is correct answer.}}$

Solve the following system of linear equations graphically:

2 x + y = 6, x - 2 y + 2 = 0

$2x + y = 6, x - 2y + 2 = 0$ . Find the vertices of the triangle formed by the above two lines and the

x

$x$ -axis. Also find the area of the triangle.

0%
Vertices of the triangle are $A(1,2), 8(5, 0)$ and $C(-2,0)$ and Area: $10$ square units
0%
Vertices of the triangle are $A(4,2), 8(2, 0)$ and $C(-2,0)$ and Area: $15$ square units
0%
Vertices of the triangle are $A(0,2), 8(1, 0)$ and $C(-2,0)$ and Area: $7$ square units
0%
Vertices of the triangle are $A(2,2), 8(3, 0)$ and $C(-2,0)$ and Area: $5$ square units

Explanation

2 x + y = 6

$2x+y=6$

⟹ y = 6 - 2 x

$\implies y=6-2x$

Let,

x = 1

$x=1$ , then

y = 6 - 2 (1) = 4

$y=6-2(1)=4$

Similarly, when

x = 2, y = 2

$x=2, y=2$

x - 2 y + 2 = 0

$x-2y+2=0$

⟹ x = 2 y - 2

$\implies x=2y-2$

Let,

y = 0,

$y=0,$ then

x = 2 (0) - 2 = - 2

$x=2(0)-2=-2$

Similarly, when

y = 2, x = 2

$y=2, x=2$

Plotting the above points on the graph, we get, the vertices of the traingle, i.e.

A (2, 3), B (3, 0), C (- 2, 0)

$A(2,3), B(3,0), C(-2,0)$

∴

$\therefore$ Area of the triangle

= \frac{1}{2} \times 5 units \times 2 units

$=\dfrac{1}{2}\times 5 \text{ units} \times 2 \text{ units}$

= 5

$=5$ sq. units

Rs. 58 is divided among 150 children such that each girl gets 25p and each boy get 50p. How many boys are there ?

0%
$52$
0%
$54$
0%
$68$
0%
$62$

Explanation

Let number of boys be

x

$x$ and number of girls be

y

$y$
Since, Total number of children's are

150

$150$

∴

$\therefore$

x + y = 150

$x+y=150$ ....(1)
Rs. 58 is divided among 150 children such that each girl gets 25p and each boy get 50p

∴

$\therefore$ Out of

R s . 58

$Rs. 58$ Boys will get Rs.

0.25 x

$0.25 x$ and girls will get Rs.

0.5 y

$0.5y$ .

∴

$\therefore$

\frac{25}{100} x + \frac{50}{100} y = 58

$\dfrac{25}{100}x+\dfrac{50}{100}y=58$

∴

$\therefore$

x + 2 y = 232

$x+2y=232$ ...(2)
Subtract equation (1) from equation (2), we get

y = 82

$y = 82$

And put

y = 82

$y = 82$ in equation (1)

x = 68

$x=68$ and

y = 82

$y=82$
Number of boys are

68

$68$

Solve graphically the following system of linear equations:

3 x + y + 1 = 0

$3x + y+ 1 = 0$

2 x - 3 y + 8 = 0

$2x - 3y + 8 = 0$

0%
$x = 0,\, y = 6$
0%
$x = - 1,\, y = 2$
0%
$x = 1,\, y = 1$
0%
$x = - 2,\, y = 3$

Explanation

3 x + y + 1 = 0

$3x+y+1=0$

⟹ y = - 1 - 3 x

$\implies y=-1-3x$

Let,

x = - 1

$x=-1$ , then

y = - 1 - 3 (- 1) = 2

$y=-1-3(-1)=2$

Similarly, when

x = 1, y = - 4

$x=1, y=-4$

And,

2 x - 3 y + 8 = 0

$2x-3y+8=0$

⟹ x = \frac{3 y - 8}{2}

$\implies x=\dfrac{3y-8}{2}$

Let,

y = 2,

$y=2,$ then

x = \frac{3 (2) - 8}{2} = - 1

$x=\dfrac{3(2)-8}{2}=-1$

Similarly, when

y = 0, x = - 4

$y=0, x=-4$

Plotting the above points on the graph, we get, the intersecting point at

(- 1, 2)

$(-1,2)$

∴ x = - 1, y = 2

$\therefore x=-1, y=2$

Albert buys

4

$4$ horses and

9

$9$ cows for Rs.

13400

$13400$ . If he sells the horses at

10 %

$10\%$ profit and the cows at

20 %

$20\%$ profit, then he earns a total profit of Rs.

1880

$1880$ . The cost of a horse is

0%
Rs. $1000$
0%
Rs. $2000$
0%
Rs. $2500$
0%
Rs. $3000$

Solve the following pair of equations by reducing them to a pair of linear equations:

\frac{1}{(3 x + y)} + \frac{1}{(3 x - y)} = \frac{3}{4}, \frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = \frac{- 1}{8}

$\dfrac {1}{(3x+y)}+\dfrac {1}{(3x-y)}=\dfrac {3}{4},\ \dfrac {1}{2(3x+y)}-\dfrac {1}{2(3x-y)}=\dfrac {-1}{8}$

0%
$x=2,\,\,y=7$
0%
$x=5,\,\,y=0$
0%
$x=1,\,\,y=1$
0%
$x=2,\,\,y=3$

Explanation

Given:

$\dfrac {1}{(3x+y)}+\dfrac {1}{(3x-y)}=\dfrac {3}{4}, \dfrac {1}{2(3x+y)}+\dfrac {1}{2(3x-y)}=\dfrac {-1}{8}$

$let \dfrac{1}{3x+y}=a and \dfrac{1}{3x-y}=b$

Then $a+b=\dfrac{3}{4}\Rightarrow 4a+4b=3........eq1$

$\dfrac{a}{2}-\dfrac{b}{2}=\dfrac{-1}{8}$

$\dfrac{a-b}{2}=\dfrac{-1}{8}$

$a-b=\dfrac{-2}{8}\Rightarrow a-b=\dfrac{-1}{4}$

$4a-4b=-1..............eq2$

Adding eq1 and eq2 we get

$8a=2$

$a=\dfrac{2}{8}\Rightarrow \dfrac{1}{4}$

Substituting the value of a in eq1 we get

$4\times \dfrac{1}{4}+4b=3$

$4b=2$

$b=\dfrac{2}{4}\Rightarrow \dfrac{1}{2}$

Then $\dfrac{1}{3x+y}=\dfrac{1}{4}$

$3x+y=4................eq3$

$\dfrac{1}{3x-y}=\dfrac{1}{2}$

$3x-y=2...............eq4$

Adding eq3 and eq4 we get
$6x=6$
$x=1$

Substituting $x=1$ in eq3 we get
$3\times 1 +y=4$
$y=1$

Thus $x=1$ and $y=1$

r + s = t

$\displaystyle r+s=t$ and

x + t = y - 2 s

$\displaystyle x+t=y-2s$ , then which of the following must be true ?

0%
$\displaystyle r+2s-x=y-t$
0%
$\displaystyle r+2s-t=x+y$
0%
$\displaystyle x+r=y+s$
0%
$\displaystyle x+r=y-3s$

Explanation

r + s = t

$\displaystyle r+s=t$ ---eq1 and

x + t = y - 2 s

$\displaystyle x+t=y-2s$ ----eq2
putting the value of t in eq 2

x + r + s = y - 2 s

$\displaystyle x+r+s=y-2s$

x + r = y - 3 s

$\displaystyle x+r=y-3s$
Answer (D)

x + r = y - 3 s

$\displaystyle x+r=y-3s$

Solve for

x

$x$ and

y

$y$

\frac{2}{3 x + 2 y} + \frac{3}{3 x - 2 y} = \frac{17}{5}; \frac{5}{3 x + 2 y} + \frac{1}{3 x - 2 y} = 2

$\displaystyle \frac{2}{3x+2y}+\frac{3}{3x-2y}=\frac{17}{5};\, \, \frac{5}{3x+2y}+\frac{1}{3x-2y}=2$

0%
$x = 1, y = 3$
0%
$x = -2, y = 1$
0%
$x = 1, y = 1$
0%
$\displaystyle x=\frac{1}{5},y=\frac{1}{5}$

Two years ago, a father was five times as old as his son. Two years later, his age will be

8

$8$ more than

3

$3$ times the age of the son. Find the present age of father.

0%
$22$ years
0%
$60$ years
0%
$39$ years
0%
$42$ years

Explanation

Let the present age of the father be

x

$x$ and that of the son be

y

$y$

Two years ago their ages were

(x - 2)

$(x-2)$ and

(y - 2)

$(y-2)$ respectively

Then, according to question, we have

(x - 2) = 5 (y - 2)

$(x-2)=5(y-2)$

\Rightarrow x - 2 = 5 y - 10

$\Rightarrow x-2=5y-10$

\Rightarrow x - 5 y = - 8...... (1)

$\Rightarrow x-5y=-8......(1)$

Two years later their ages will be

(x + 2)

$(x+2)$ and

(y + 2)

$(y+2)$ respectively

Then again,

(x + 2) = 3 (y + 2) + 8

$(x+2)=3(y+2)+8$

\Rightarrow x + 2 = 3 y + 6 + 8

$\Rightarrow x+2=3y+6+8$

\Rightarrow x - 3 y = 12..... (2)

$\Rightarrow x-3y=12.....(2)$

Subtracting equation (1) from (2), we get

\Rightarrow (x - 3 y) - (x - 5 y) = 12 - (- 8)

$\Rightarrow (x-3y)-( x-5y)=12-(-8)$

\Rightarrow x - 3 y - x + 5 y = 12 + 8

$\Rightarrow x-3y- x+5y=12+8$

\Rightarrow 2 y = 20

$\Rightarrow 2y=20$

\Rightarrow y = 10

$\Rightarrow y=10$

Putting

y = 10

$y = 10$ in equation (2), we get

x - 3 \times 10 = 12

$x-3\times 10=12$

\Rightarrow x = 12 + 30

$\Rightarrow x=12+30$

\Rightarrow x = 42

$\Rightarrow x=42$
The present ages of father and son are

42

$42$ years and

10

$10$ years respectively.

So,

O p - D

$Op-D$

Mr. Joshi has

430

$430$ cabbage-plants which he wants to plant out. Some

25

$25$ to a row and the rest

20

$20$ to a row. If there are to be

18

$18$ rows in all how many rows of

25

$25$ will there be?

0%
$10$
0%
$14$
0%
$8$
0%
$12$

Explanation

Let '

x

$x$ ' rows of

25

$25$ and '

y

$y$ ' rows of

20

$20$ .

Given,

25 x + 20 y = 430

$25x + 20y = 430$

5 x + 4 y = 86

$5x+4y=86$ ....(1)

Also

x + y = 18

$x + y = 18$ ....(2)

Multiply equation (1) by

4

$4$ , we get

4 x + 4 y = 72

$4x+4y=72$ ....(3)

Subtract equations (1) and (3), we get

x = 14

$x=14$

Hence, answer is

14

$14$ .

The difference between two numbers is

7

$7$ and their sum is

35

$35$ . What will be their product?

0%
$324$
0%
$294$
0%
$79$
0%
$245$

Explanation

x + y = 35

$x + y = 35$

x - y = 7

$x - y = 7$

Add both the equations

\Rightarrow

$\displaystyle \Rightarrow$

2 x = 42

$2x = 42$

\Rightarrow

$\displaystyle \Rightarrow$

x = 21

$x = 21$

∴

$\displaystyle \therefore$

21 + y = 35

$21 + y = 35$

\Rightarrow

$\displaystyle \Rightarrow$

y = 14

$y = 14$

∴

$\displaystyle \therefore$

x

$x$

\times

$\displaystyle \times$

y = 21

$y = 21$

\times

$\displaystyle \times$

14 = 294

$14 = 294$

\frac{x + y}{x - y} = \frac{5}{3} a n d \frac{x}{(y + 2)} = 2

$\displaystyle \frac{x+y}{x-y}=\frac{5}{3}\: and\: \frac{x}{\left ( y+2 \right )}=2$ the value of (x , y) is

0%
(4, 1)
0%
(2, 8)
0%
(1, 4)
0%
(8, 2)

Explanation

\frac{x}{(y + 2)} = 2

$\dfrac{x}{\left ( y+2 \right )}=2$

x = 2 y + 4

$x=2y+4$ .....

(1)

$(1)$

\frac{x + y}{x - y} = \frac{5}{3}

$\displaystyle \dfrac{x+y}{x-y}=\frac{5}{3}$ .....

(2)

$(2)$
From eqs

(1)

$(1)$ and

(2)

$(2)$

\frac{2 y + 4 + y}{2 y + 4 - y} = \frac{5}{3}

$\dfrac { 2y+4+y }{ 2y+4-y } =\dfrac { 5 }{ 3 }$

\frac{3 y + 4}{y + 4} = \frac{5}{3}

$\dfrac { 3y+4 }{ y+4 } =\dfrac { 5 }{ 3 }$

9 y + 12 = 5 y + 20

$9y+12=5y+20$

4 y = 8

$4y=8$

y = 2

$y=2$
and

x = 2 y + 4 = 2 \times 2 + 4 = 8

$x=2y+4=2\times 2+4=8$
Hence,

(x, y) = (8, 2)

$(x , y)=(8,2)$

In covering a distance of

30

$30$ km, Abhay takes

2

$2$ hours more than Sameer. If Abhay doubles his speed, then he would take

1

$1$ hour less than Sameer. What is Abhay's speed? (in km/hr)

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

Let Abhay's speed be

x

$x$ km/hr and Sameer's speed by

y

$y$ km/hr
Then

\frac{30}{x} - \frac{30}{y} = 2

$\displaystyle \frac{30}{x}-\frac{30}{y}=2$ ..... (i)

and

\frac{30}{y} - \frac{30}{2 x} = 1

$\displaystyle \frac{30}{y}-\frac{30}{2x}=1$ ......(ii)
Adding equation (i) and (ii), we get

\frac{30}{x} - \frac{30}{2 x} = 3

$\displaystyle \frac{30}{x}-\frac{30}{2x}=3$

\Rightarrow \frac{30}{2 x} = 3

$\Rightarrow \displaystyle \frac{30}{2x}=3$

\Rightarrow 2 x = 10

$\Rightarrow 2x=10$

\Rightarrow x = 5

$\Rightarrow x=5$ km/hr

Solve the following pair of equations:

\frac{25}{x + y} - \frac{3}{x - y} = 1

$\displaystyle \frac{25}{x+y}-\frac{3}{x-y}=1$ and

\frac{40}{x + y} + \frac{2}{x - y} = 5

$\dfrac{40}{x+y}+\dfrac{2}{x-y}=5$ ,

0%
$x = 8 , y = 6$
0%
$x = 4 , y = 6$
0%
$x = 6, y = 4$
0%
None of these

Explanation

\frac{25}{x + y} - \frac{3}{x - y} = 1, \frac{40}{x + y} + \frac{2}{x - y} = 5

$\displaystyle \frac{25}{x+y}-\frac{3}{x-y}=1,\frac{40}{x+y}+\frac{2}{x-y}=5$

Let,

\frac{1}{x + y} = m

$\dfrac{1}{x +y} = m$ and

\frac{1}{x - y} = n

$\dfrac{1}{x -y} = n$

25 m - 3 n = 1

$25 m - 3n = 1$ ...(1)

40 m + 2 n = 5

$40 m + 2n = 5$ ...(2)
Multiply eq(1) by 2 and eq(2) by 3 and add

50 m + 120 m = 17

$50 m + 120 m= 17$

m = \frac{1}{10}

$m = \dfrac{1}{10}$
Hence,

n = \frac{1}{2}

$n = \dfrac{1}{2}$
Thus,

x + y = 10

$x + y = 10$
an

x - y = 2

$x - y = 2$
or,

x = 6, y = 4

$x = 6, y = 4$

If Rs.

50

$50$ is distributed amount

150

$150$ children giving

50

$50$ p to each boy and

25

$25$ p to each girl. Then the number of boys is

0%
$25$
0%
$40$
0%
$36$
0%
$50$

Explanation

Let number of boys

= x

$= x$
Number of girls

= y

$= y$
Then,

x + y = 150

$x + y = 150$ .....(1)
and

0.5 x + 0.25 y = 50

$0.5x + 0.25 y = 50$

\Rightarrow 2 x + y = 200

$\Rightarrow 2x + y = 200$ ......(2)
Subtracting (1) from (2), we get

x = 50

$x = 50$
Thus, number of boys are

50

$50$ .

The electricity bill of a certain establishment is partly fixed and partly varies as the number of units of electricity consumed. When in a certain month 540 units are consumed, the bill is Rs.In another month 620 units are consumed and the bill is Rs.In yet another month 500 units are consumed then, the bill for that month would be

0%
Rs. 1560
0%
Rs. 1680
0%
Rs. 1840
0%
Rs. 1950

Explanation

Let the fixed amount be Rs.

x

$x$ and the cost of each unit be Rs.

y

$y$ . Then,

540 y + x = 1800

$540y + x = 1800$ ...(i) and

620 y + x = 2040

$620y + x = 2040$ ...(ii)
On subtracting (i) from (ii), we get

80 y = 240 \Rightarrow y = 3

$80y = 240 \Rightarrow y = 3$
Putting y = 3 in (i), we get :

540 \times 3 + x = 1800 \Rightarrow x = (1800 - 1620) = 180

$540 \times 3 + x = 1800 \Rightarrow x = \left(1800 - 1620\right) = 180$

∴

$\therefore$ Fixed charges

= R s . 180,

$= Rs. 180,$ Charge per unit

= R s . 3

$= Rs. 3$
Total charges for consuming

500

$500$ units

= R s . (180 + 500 \times 3) = R s . 1680

$= Rs.\left (180 + 500 \times 3\right) = Rs. 1680$

The equations

2 x - 3 y + 5 = 0

$2x - 3y + 5 = 0$ and

6 y - 4 x = 10

$6y - 4x = 10$ , when solved simultaneously has

0%
Only one solution
0%
No solution
0%
Only two solutions
0%
Infinite number of solutions

Explanation

Given equations:

2 x - 3 y + 5 = 0

$2x-3y+5=0$ and

6 y - 4 x = 10

$6y-4x=10$

Rewriting the equations, we get

2 x - 3 y + 5 = 0

$2x - 3y + 5=0$ ...

(i)

$(i)$

- 4 x + 6 y - 10 = 0

$-4x+6y-10=0$ ...

(i i)

$(ii)$

Comparing the coefficients from

(i)

$(i)$ and

(i i)

$(ii)$ gives

\frac{a_{1}}{a_{2}} = \frac{2}{- 4}, \frac{b_{1}}{b_{2}} = \frac{- 3}{6}, \frac{c_{1}}{c_{2}} = \frac{5}{- 10}

$\dfrac{a_1}{a_2} =\dfrac2{-4},\, \dfrac{b_1}{b_2} =\dfrac{-3}6,\, \dfrac{c_1}{c_2}=\dfrac5{-10}$

i.e,

\frac{a_{1}}{a_{2}} = - \frac{1}{2}, \frac{b_{1}}{b_{2}} = - \frac{1}{2}, \frac{c_{1}}{c_{2}} = - \frac{1}{2}

$\dfrac{a_1}{a_2}=-\dfrac12,\, \dfrac{b_1}{b_2}=-\dfrac12,\, \dfrac{c_1}{c_2}=-\dfrac12$

\Rightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

$\Rightarrow \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

Therefore, equations have infinite numbers of solutions.

Hence, Option

D

$D$ is correct.

A certain two digits number is equal to five times the sum of its digits. If

9

$9$ were added to the number, its digits would be reversed. The sum of the digits of the number is:

0%
$6$
0%
$7$
0%
$8$
0%
$9$

Explanation

Let the ones and tens digit of the number be

y

$y$ and

x .

$x.$

According to question, we have

10 x + y = 5 (x + y)

$10x+y=5\left( x+y \right)$

\Rightarrow 5 x - 4 y = 0 \dots (i)

$\Rightarrow 5x-4y=0\quad\quad\quad\dots(i)$

And,

10 x + y + 9 = 10 y + x

$10x+y+9=10y+x$

\Rightarrow 9 x - 9 y = - 9

$\Rightarrow 9x-9y=-9$

\Rightarrow x - y = - 1 \dots (i i)

$\Rightarrow x-y=-1\quad\quad\quad\dots(ii)$

Multiply

(i i)

$(ii)$ by

- 4

$-4$ we get,

\Rightarrow - 4 x + 4 y = 4 \dots (i i i)

$\Rightarrow -4x+4y=4\quad\quad\quad\dots(iii)$

Add equations

(i)

$(i)$ and

(i i i),

$(iii),$

(5 x - 4 y) + (- 4 x + 4 y) = 0 + 4

$\left( {5x - 4y} \right) + \left( { - 4x + 4y} \right) = 0 + 4$

\Rightarrow 5 x - 4 x - 4 y + 4 y = 4

$\Rightarrow 5x - 4x - 4y + 4y = 4$

\Rightarrow x = 4

$\Rightarrow x = 4$

Substitute

x = 4

$x=4$ in equation

(i i),

$(ii),$

4 - y = - 1

$4 - y = - 1$

\Rightarrow - y = - 5

$\Rightarrow - y = - 5$

\Rightarrow y = 5

$\Rightarrow y=5$

Thus, the sum of the digits of the number is equal to

x + y = 9.

$x+y=9.$ .

The father's age is six times his son's age. Four years hence, the age of the father will be four times his son's age. The present ages, in years, of the son and the father are respectively,

0%
$4$ and $24$
0%
$5$ and $30$
0%
$6$ and $36$
0%
$3$ and $24$

Explanation

Let the present age of father's be

x

$x$ years and present age of son's be

y

$y$ years.
According to the problem

x = 6 y

$x=6y$
After

4

$4$ years

x + 4 = 4 (y + 4)

$x+4 = 4(y+4)$
Hence we get two equations

x = 6 y

$x=6y$ .........

(1)

$(1)$

x + 4 = 4 (y + 4)

$x+4 = 4(y+4)$ .....

(2)

$(2)$
Simplifying eq

(2)

$(2)$

x + 4 = 4 y + 16

$x+4 = 4y+16$

x - 4 y = 12

$x-4y = 12$
Put

x = 6 y

$x=6y$ in eq

(2)

$(2)$ , we get

6 y - 4 y = 12

$6y-4y = 12$

2 y = 12

$2y=12$

y = 6

$y=6$ years
and

x = 6 y = 36

$x=6y=36$ years
Present age of son

= 6

$=6$ years
and present age of father

= 36

$=36$ years

The expression

a x + b

$ax + b$ is equal to

13

$13$ when

x

$x$ is

5

$5$ and

a x + b

$ax + b$ is equal to

29

$29$ when

x

$x$ is

13

$13$ . The value of expression when

x

$x$ is

0.5

$0.5$

0%
$2.9$
0%
$\dfrac {29}{5}\times 13$
0%
$4$
0%
$\dfrac {29\times 5}{13}$

Explanation

5 a + b = 13

$5a+ b= 13$ .......eq1 &

13 a + b = 29

$13a+ b= 29$ ......eq2

Subtract eq1 from eq2, we get

a = 2

$a=2$

b = 3

$b = 3$
when

x = 0.5 \times 2 + 3 = 1 + 3 = 4

$x=0.5\times 2+3=1+3=4$ .

In the system of equations

\frac{12}{x + y} + \frac{8}{x - y} = 8

$\dfrac {12}{x+y}+\dfrac {8}{x-y}=8$ and

\frac{27}{x + y} - \frac{12}{x - y} = 3

$\dfrac {27}{x+y}-\dfrac {12}{x-y}=3$ , the values of

x

$x$ and

y

$y$ will be

0%
$\dfrac {5}{3}$ and $\dfrac {1}{3}$
0%
$\dfrac {5}{2}$ and $\dfrac {1}{2}$
0%
2 and $\dfrac {1}{3}$
0%
$\dfrac {5}{4}$ and $\dfrac {1}{4}$

Explanation

\frac{1}{x + y} = a

$\dfrac {1}{x+y}=a$ and

\frac{1}{x - y} = b

$\dfrac {1}{x-y}=b$ , then the equations will be:

12 a + 8 b = 8

$12a + 8b = 8$ ........ (i)

and

27 a - 12 b = 3

$27a -12b = 3$ ........... (ii)

On multiplying equation (i) by 3 and equation (ii) by 2 and adding the two equations,

3 (12 a + 8 b) + 2 (27 a - 12 b) = 8 \times 3 + 3 \times 2

$3(12a + 8b) + 2(27a -12b) = 8\times 3+3 \times 2$

∴ 36 a + 24 b + 54 a - 24 b = 24 + 6

$\therefore 36a + 24b + 54a -24b = 24 + 6$

∴ 90 a = 30 \Rightarrow a = \frac{30}{90} = \frac{1}{33}

$\therefore 90a=30\Rightarrow a=\dfrac {30}{90}=\dfrac {1}{33}$

On substituting this value of a in equation (i),

12 \times \frac{1}{3} + 8 b = 8 ∴ 4 + 8 b = 8

$12\times \dfrac {1}{3}+8b=8\therefore 4+8b=8$

\Rightarrow 8 b = 4 \Rightarrow b = \frac{1}{2}

$\Rightarrow 8b=4\Rightarrow b=\dfrac {1}{2}$

∵ \frac{1}{x + y} = a = \frac{1}{3} \Rightarrow x + y = 3

$\because \dfrac {1}{x+y}=a=\dfrac {1}{3}\Rightarrow x+y=3$ ......(iii)

and

\frac{1}{x + y} = b = \frac{1}{2} \Rightarrow x - y = 2

$\dfrac {1}{x+y}=b=\dfrac {1}{2}\Rightarrow x-y=2$ ......(iv)

On adding equation (iii) and (iv),

2 x = 5 \Rightarrow x = \frac{5}{2}

$2x=5\Rightarrow x=\dfrac {5}{2}$

On substituting this value of x in equation (iii),

\frac{5}{2} + y = 3 \Rightarrow y = 3 - \frac{5}{2}

$\dfrac {5}{2}+y=3\Rightarrow y=3-\dfrac {5}{2}$

\Rightarrow y = \frac{1}{2}

$\Rightarrow y=\dfrac {1}{2}$

∴ x = \frac{5}{2}

$\therefore x=\dfrac {5}{2}$ and

y = \frac{1}{2}

$y=\dfrac {1}{2}$

When one is added to each of two given numbers, their ratio becomes

3 : 4

$3 : 4$ and when

5

$5$ is subtracted from each, the ratio becomes

7 : 10

$7:10$ . The numbers are

0%
$8, 11$
0%
$11, 15$
0%
$26, 35$
0%
$27, 36$

Explanation

Let the two numbers be

x

$x$ and

y

$y$ .

According to the given condition,

\frac{x + 1}{y + 1} = \frac{3}{4}

$\dfrac {x+1}{y+1}=\dfrac {3}{4}$

\Rightarrow 4 x + 4 = 3 y + 3

$\Rightarrow 4x+4=3y+3$

4 x - 3 y = - 1

$4x-3y=-1$ ........

(i)

$(i)$

Also,

\frac{x - 5}{y - 5} = \frac{7}{10}

$\dfrac {x-5}{y-5}=\dfrac {7}{10}$

\Rightarrow 10 x - 50 = 7 y - 35

$\Rightarrow 10x-50=7y-35$

\Rightarrow 10 x - 7 y = 15

$\Rightarrow 10x-7y=15$ .......

(i i)

$(ii)$

Multiply equation

(1)

$(1)$ and

(2)

$(2)$ by

7

$7$ and

3

$3$ respectively.

[4 x - 3 y = - 1] \times 7

$[4x -3y = -1] \times 7$

[10 x - 7 y = 15] \times 3

$[10x -7y = 15] \times 3$

28 x - 21 y = - 7

$28x -21y =-7$ .............

(4)

$(4)$

30 x - 21 y = 45

$30x-21y=45$ ...............

(5)

$(5)$

Subtract equation

(4)

$(4)$ from

(5)

$(5)$

2 x = 52

$2x = 52$

\Rightarrow x = 26

$\Rightarrow x=26$

Putting the value of

x

$x$ in

(i)

$(i)$

4 (26) - 3 y = - 1

$4(26) -3y =-1$

- 3 y = - 105

$-3y = -105$

y = 35

$y = 35$

∴

$\therefore$ Numbers are

26, 35

$26, 35$ .

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 11 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 11 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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