MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 4

Solve the following simultaneous equations :

$\displaystyle \frac{16}{x + y}\, +\, \frac{2}{x - y}\, =\, 1;\quad \frac{8}{x + y}\, -\, \frac{12}{x - y}\, =\, 7$

0%
$x = 2 , y = 3$
0%
$x = 6 , y = 1$
0%
$x = 7 , y = 4$
0%
$x = 3 , y = 5$

Explanation

$Let \dfrac { 1 }{ x+y } = a; \dfrac { 1 }{ x-y } =b$

So the equations are

$16a + 2b = 1$ ...(1)

$8a - 12b = 7$ ...(2)

Eqn (2)

$\times 2$

$\Rightarrow 16a - 24b = 14$ ....(3)

Subtracting eq(1) and eq(3) we get

$26b = -13 \Rightarrow b = -\dfrac12$

Eqn (1)

$\times 6 \Rightarrow 96a +12b = 6$ ....(4)

Adding eqn(4) and eq(2)

$104a = 13\\ \therefore a = \dfrac{1}{8}\\ \dfrac { 1 }{ x+y } = \dfrac { 1 }{ 8 }$

$x+y = 8$ ...(5)

$\dfrac { 1 }{ x-y } = \dfrac { 1 }{ -2 } \\ x-y = -2$ ...(6)

Adding eq(5) and eq(6)

$2x = 6 \Rightarrow x = 3$

Substitute for

$x$ in eqn(5)

$\therefore y = 8-3 = 5$

Solve the following simultaneous equations by substitution method.

$2x +3y= -4; \, \, x-5y =11$

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

Explanation

$2x +3y= -4$ ..........(1)

$x-5y =11$

$\therefore x=11+5y$ .............(2)
Substituting eq. (2) in eq. (1), we get,

$2(11+5y)+3y = -4$

$22+10y+3y=-4$

$22+4 = -13y$

$\therefore y = -\dfrac {26}{13}$

$y=-2$
Substituting

$y=-2$ in (1) we get,

$x=1$

Find the values of

$x$ and

$y$ , if

$\displaystyle \frac{2}{x}\, +\, \frac{6}{y}\, =\, 13;\quad \frac{3}{x}\, +\, \frac{4}{y}\, =\, 12$

0%
$x\, =\,\displaystyle \frac{1}{6}\, ,\, y\, =\, \frac{2}{3}$
0%
$x\, =\,\displaystyle \frac{1}{3}\, ,\, y\, =\, \frac{2}{3}$
0%
$x\, =\,\displaystyle \frac{1}{7}\, ,\, y\, =\, \frac{2}{3}$
0%
$x\, =\,\displaystyle \frac{1}{2}\, ,\, y\, =\, \frac{2}{3}$

Explanation

$\dfrac { 2 }{ x } +\dfrac { 6 }{ y } = 13$

$\dfrac { 3 }{ x } +\dfrac { 4 }{ y } = 12$

Lets assume, $\dfrac { 1 }{ x } = a, \dfrac { 1 }{ y } = b$ , then the above equations reduces to

$2a + 6b = 13$ ...(1)

$3a + 4b = 12$ ...(2)

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

$6a +18b = 39$ ...(3)

$6a + 8b = 24$ ...(4)

Subtracting (3) and (4), we get

$10b = 15\Rightarrow b = \dfrac { 3 }{ 2 }$

Now, $2a = 13 - 6\times \dfrac { 3 }{ 2 }= 13 - 9 = 4$

$\Rightarrow a = 2$

So, $x = \dfrac { 1 }{ 2 } ; y = \dfrac { 2 }{ 3 }$

Hence, option D is the correct answer.

Find the values of the

$x+y$ and

$x-y$ from the examples given below without solving for x and y

$4x+3y=24;\, \, 3x+4y=25$

0%
$7$ , $-1$
0%
$1$ , $-7$
0%
$1$ , $7$
0%
$-7$ , $-1$

Explanation

$\text{The given equations are}$

$4x+3y=24$ ...(i)

$3x+4y=25$ ...(ii)

On adding (i) and (ii), we get

$7x+7y=49\Rightarrow x+y=7$

On subtracting (ii) from (i), we get

$x-y=-1$

$\therefore x+y=7 \space and \space x-y= -1$

Sum of two numbers isIf the larger number is divided by the smaller, the quotient is 7 and the remainder isFind the numbers.

0%
$75, 14$
0%
$90, 14$
0%
$75, 18$
0%
$85, 12$

Explanation

$\textbf{Step-1: Assume the numbers to be equal to some variables& then apply all information's given.}$

$\text{Let the two numbers be}$

$x$

$\text{and}$

$y,$

$\text{where}$

$x$

$\text{is the larger number}$

$\text{Given, Sum of two numbers is 97}$

$x + y = 97$

$\text{...(1)}$

$\text{As per the given condition,}$

$\text{If the larger number is divided by the smaller, the quotient is 7 and the remainder is 1}$

$x = 7y +1$

$\text{...(2)}$

$\textbf{Step-2: Solve above equations.}$

$\text{Substitute eq(2) in eq(1) we get,}$

$7y+1 + y = 97\\ 8y = 96\\ y = 12\\ \therefore x = 7(12) +1 = 85$

$\textbf{Hence, option - D is the answer.}$

Solve the following simultaneous equations :

$\displaystyle \frac{1}{3x}\, +\, \frac{1}{5y}\, =\, \frac{1}{15};\quad \frac{1}{2x}\, +\, \frac{1}{3y}\, =\, \frac{1}{12}$

0%
$x = 5, y = 3$
0%
$x = 6, y = 6$
0%
$x = 2, y = -2$
0%
$x = 1, y = -5$

Explanation

On putting

$\dfrac1x=X$ and

$\dfrac1y=Y$ in given equations, we get

$\dfrac X3+\dfrac Y5=\dfrac{1}{15}\Rightarrow 5X+3Y=1$ ...(i)

$\dfrac X2+\dfrac Y3=\dfrac{1}{12}\Rightarrow 6X+4Y=1$ ...(ii)
On multiplying (i) by 6 and (ii) by 5, we get

$30X+18Y=6$

$\underline {\underset {-}30X\underset {-}{+}20Y=\underset{-}5}$

$-2Y=1$

$\therefore Y=-\dfrac12$
On putting

$Y=-\dfrac12$ in (i), we get

$5X+3\times-\dfrac12=1\Rightarrow 5X=\dfrac52\Rightarrow X=\dfrac12$

$\therefore x=\dfrac1X=2,y=\dfrac1Y=-2$

A man starts his job with a certain monthly salary and a fixed increment every year. If his salary will be Rs.

$11000$ after

$2$ years and Rs.

$14000$ after

$4$ years of his service. What is his starting salary and what is the annual increment?

0%
Starting salary is Rs. $7500$ and increment is Rs. $2500$
0%
Starting salary is Rs. $5000$ and increment is Rs. $1800$
0%
Starting salary is Rs. $8000$ and increment is Rs. $1500$
0%
Starting salary is Rs. $7000$ and increment is Rs. $1300$

Explanation

Let starting salary

$=\ x$

Increment every year

$=\ y$

$x\ + 2y\ =\ 11,000\quad$ ....... eq(1)

$x\ +\ 4y\ =\ 14,000$ ......eq(2)

Substract eq(1) from eq(2)

$2y\ =\ 3,000\\ y\ =\ 1500$

Substitute

$y =\ 1500$ in eq(1), we get

$x\ =\ 11,000\ -\ 3,000$

$x\ =\ 8,000$

starting salary

$\ =\ 8,000$

Increment

$\ = \ 1500$

Solve following pair of equations by equating the coefficient method:

$2x\, -\, y\, =\, 9$

$3x\, -\, 7y\, =\, 19$

0%
$x=2$ , $y=1$
0%
$x=4$ , $y=-1$
0%
$x=3$ , $y=2$
0%
$x=7$ , $y=-8$

Explanation

Multiply the equation

$2x-y=9$ by

$7$ to make the coefficients of

$y$ equal. Then we get the equations:

$14x-7y=63.........(1)$

$3x-7y=19.........(2)$

Subtract Equation (2) from Equation (1) to eliminate

$y$ , because the coefficients of

$y$ are the same. So, we get

$(14x-3x)+(-7y+7y)=63-19$

i.e.

$11x = 44$

i.e.

$x = 4$

Substituting this value of

$x$ in (2), we get

$12-7y=19$

i.e.

$-7y=19-12$

i.e.

$-7y=7$

i.e.

$y=-1$

Hence, the solution of the equations is

$x = 4, y = -1$ .

Solve the given pair of equations by substitution method:

$x\, +\, 5y\, =\, 18$

$3x\, +\, 2y\, =\, 41$

0%
$(13 , 1)$
0%
$(1 , 3)$
0%
$(12 , 17)$
0%
$(9 , 16)$

Explanation

We have,

$x+5y =18\Rightarrow x = 18-5y$ ..........(1)

and

$3x+2y=41$ .......(2)

Now substitute value x in terms of y from (1) to (2)

$\Rightarrow 3(18-5y)+2y=41$

$\Rightarrow 54-15y+2y=41$

$\Rightarrow 15y-2y=54-41$

$\Rightarrow 13y=13\Rightarrow y=\dfrac{13}{13}=1$

$x =18-5y=18-5(1)=13$

Hence solution set is

$(x,y)=(13,1)$

Solve the following pair of simultaneous equations:

$4x\, -\, 3y\, =\, 8$

$3x\, -\, 4y\, =\, - 1$

0%
$x=3,y=-1$
0%
$x=5,y=4$
0%
$x=-4,y=2$
0%
$x=7,y=-6$

Explanation

Multiply the equation

$4x-3y=8$ by

$3$ and equation

$3x-4y=-1$ by

$4$ to make the coefficients of

$x$ equal. Then we get the equations:

$12x-9y=24.........(1)$

$12x-16y=-4.........(2)$

Subtract Equation (2) from Equation (1) to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(12x-12x)+(-9y+16y)=24+4$

i.e.

$7y =28$

i.e.

$y =4$

Substituting this value of

$y$ in the equation

$4x-3y=8$ , we get

$4x-12=8$

i.e.

$4x=20$

i.e.

$x=5$

Hence, the solution of the equations is

$x = 5, y =4$ .

Draw the graph of

$2x - y -1=0$ and

$2x + y= 9$ on the same graph. Use

$2\ \mathrm{ cm} = 1$ unit on both axes.
Write down the co-ordinates of the point of intersection of the two lines.

0%
$x=3.5,\:y=4$
0%
$x=4.5,\:y=4$
0%
$x=1.5,\:y=4$
0%
$x=2.5,\:y=4$

Explanation

Plotting

$2x-y-1=0$ and

$2x+y=9$ in the graph, we get
the point of intersection as

$(2.5,4)$

Solve the following pair of simultaneous equations:

$\displaystyle \frac{x}{3}\, =\, \frac{y}{2}\,;\, \frac{2x}{3}\, -\, \frac{y}{2}\, =\, 2$

0%
$(-3 , 4)$
0%
$(1 , 9)$
0%
$(6 , 4)$
0%
$(3 , 8)$

Explanation

The equation

$\frac { x }{ 3 } =\frac { y }{ 2 }$ can be rewritten as:

$\frac { x }{ 3 } =\frac { y }{ 2 } \\ \Rightarrow 2x=3y\\ \Rightarrow 2x-3y=0\quad .........(1)$

The equation

$\frac { 2x }{ 3 } -\frac { y }{ 2 }=2$ can be solved as:

$\frac { 2x }{ 3 } -\frac { y }{ 2 } =2\\ \Rightarrow \frac { 4x-3y }{ 6 } =2\\ \Rightarrow 4x-3y=12\quad .........(2)$

Subtract Equation 2 from equation 1 to eliminate

$y$ , because the coefficients of

$y$ are the same. So, we get

$(4x-2x)+(-3y+3y)=12-0$

i.e.

$2x=12$

i.e.

$x=6$

Substituting this value of

$x$ in the equation 1, we get

$12-3y=0$

i.e.

$3y=12$

i.e.

$y=4$

Hence, the solution of the equations is

$x =6, y =4$ .

Solve the given pair of equations by substitution method:

$x\, +\, y\, =\, 11$

$x\, -\, y\, =\, -3$

0%
$(4 , 6)$
0%
$(3 , 11)$
0%
$(4 , 7)$
0%
$(6 , 2)$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$x-y = -3$ and write it as

$x = -3+y...........(1)$

Substitute the value of

$x$ in Equation

$x+y = 11$ . We get

$-3+y+y = 11$

i.e.

$-3+y+y = 11$

i.e.

$-3+2y = 11$

i.e.

$2y = 14$

Therefore,

$y = 7$

Substituting this value of

$y$ in Equation (1), we get

$x = -3+7=4$

Hence, the solution is

$x = 4, y = 7$ .

Solve the given pair of equations by substitution method:

$4a\, -\, b\, =\, 10$

$2a\, +\, 3b\, =\, 12$

0%
$a = 0, b = 5$
0%
$a = 7, b = 6$
0%
$a = 1, b = 7$
0%
$a = 3, b = 2$

Explanation

We have,

$4a-b=10\Rightarrow b=4a-10$ .........(1)

and

$2a+3b=12$ ........(2)

Now substitute value of b in terms of a in (2)

$\Rightarrow 2a+3(4a-10)=12$

$\Rightarrow 2a+12a-30=12$

$\Rightarrow 14a=12+30=42\Rightarrow a=\dfrac{42}{14}=3$

$b=4a-10=4(3)-10=12-10=2$

Hence option 'D' is correct choice

Solve the given pair of equations by substitution method:

$x\, -\, 4y\, =\, -8$

$x\, -\, 2y\, =\, 0$

0%
$(2 , -1)$
0%
$(7, -6)$
0%
$(8 , 4)$
0%
$(-3 , 6)$

Explanation

We have,

$x-4y =-8\Rightarrow x = 4y-8$ ..........(1)

and

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

Now substitute value of x in terms of y from (1) to (2)

$\Rightarrow 4y-8-2y=0$

$\Rightarrow 2y=8$

$\Rightarrow y=\dfrac{8}{2}=4$

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

Hence solution set is

$(x,y)=(8,4)$

Solve the given pair of equations by substitution method:

$2a\, +\, 3b\, =\, 6$

$3a\, +\, 5b\, =\, 15$

0%
$a = 3, b = 2$
0%
$a = -15, b = 12$
0%
$a = 8, b = 15$
0%
$a = -7, b = 3$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$2a+3b=6$ and write it as

$b=\frac { 6-2a }{ 3 } .........(1)$

Substitute the value of

$b$ in Equation

$3a+5b=15$ . We get

$3a+5\left( \frac { 6-2a }{ 3 } \right) =15\\ \Rightarrow 9a+5\left( 6-2a \right) =45\\ \Rightarrow 9a+30-10a=45\\ \Rightarrow -a=45-30\\ \Rightarrow a=-15$

Therefore,

$a=-15$

Substituting this value of

$a$ in Equation (1), we get

$b=\frac { 6-(-30) }{ 3 }$

$b=\frac { 36 }{ 3 }=12$

Hence, the solution is

$a=-15, b=12$ .

Solve the given pair of equations by substitution method:

$x\, +\, y\, =\, 0$

$y\, -\, x\, =\, 6$

0%
$(2 , 7)$
0%
$(-3 , 3)$
0%
$(4 , 9)$
0%
$(-6 , 2)$

Explanation

We have,

$x+y =0\Rightarrow x = -y$ ..........(1)

and

$y-x=6$ .......(2)

Now substitute value of x in terms of y from (1) to (2)

$\Rightarrow y-(-y)=6$

$\Rightarrow y+y=6$

$\Rightarrow 2y=6$

$\Rightarrow y=\dfrac{6}{2}=3$

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

Hence solution set is

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

Solve the following pair of simultaneous equations:

$\displaystyle\, y\, -\, \frac{3}{x}\, =\, 8\, ;\, 2y\, +\, \frac{7}{x}\, =\, 3$

0%
$x = - 7$ and $y = 3$
0%
$x = 0$ and $y = 3$
0%
$x = - 5$ and $y = 9$
0%
$x = - 1$ and $y = 5$

Explanation

The equation

$y-\frac { 3 }{ x } =8$ can be solved as:

$y-\frac { 3 }{ x } =8\\ \Rightarrow \frac { xy-3 }{ x } =8\\ \Rightarrow xy-3=8x\quad .........(1)$

The equation

$2y+\frac { 7 }{ x } =3$ can be solved as:

$2y+\frac { 7 }{ x }=3\\ \Rightarrow \frac { 2xy+7 }{ x } =3\\ \Rightarrow 2xy+7=3x\quad .........(2)$

Multiply the equation 1 by

$2$ . Then we get:

$2xy-6=16x.........(3)$

Subtract equation 3 from equation 1 to eliminate

$xy$ , because the coefficients of

$xy$ are same. So, we get

$(2xy-2xy)+(7+6)=3x-16x$

i.e.

$-13x=13$

i.e.

$x=-1$

Substituting this value of

$x$ in the equation 1, we get

$-y-3=-8$

i.e.

$-y=-5$

i.e.

$y=5$

Hence, the solution of the equations is

$x=-1,y=5$ .

Solve the following pair of simultaneous equations:

$\displaystyle \frac{8}{x}\, -\, \frac{9}{y}\, =\, 1;\,\frac{10}{x}\, +\, \frac{6}{y}\, =\, 7$

0%
$x= 2;y= 3$
0%
$x= 7;y= 5$
0%
$x= 4;y= 6$
0%
$x= 1;y= 7$

Explanation

Let

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

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

We have,

$8a -9b = 1$ ...

$(1)$

$10a +6b = 7$ ...

$(2)$

Multiplying

$(1)$ by

$10$ and

$(2)$ by

$8$

We get,

$80a - 90b = 10$

$80a + 48b = 56$

Subtracting, we get

$-138b = -46$

$\therefore b = \dfrac{1}{3}$ .

Substituting in

$(1)$ or

$(2)$ , we get

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

So,

$x =2$ and

$y = 3$ .

Option

$A$

Solve the following pair of simultaneous equations:

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

0%
$x= \displaystyle \frac{3}{2};y= \displaystyle \frac{3}{4}$
0%
$x= \displaystyle \frac{1}{2};y= \displaystyle \frac{2}{3}$
0%
$x= \displaystyle \frac{1}{3};y= \displaystyle \frac{1}{2}$
0%
$x= \displaystyle \frac{2}{3};y= \displaystyle \frac{1}{2}$

Explanation

Let

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

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

So, it becomes

$u + v = 5$ and

$u - v = 1$

Adding the both equations,

$2u = 6$ and

$u = 3$ .

Substituting

$u$ in either of the equations, we get

$v = 2$ .

So,

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

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

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

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

Option

$C$

Solve the following pair of simultaneous equations:

$\displaystyle\,3x\, +\, \frac{1}{y}\, =\, 13\, ;\, \frac{2}{y}\, -\, x\, =\, 5$

0%
$\displaystyle \frac{2}{5}$ and $4$
0%
$3$ and $\displaystyle \frac{1}{4}$
0%
$7$ and $\displaystyle \frac{1}{7}$
0%
$3$ and $\displaystyle \frac{3}{4}$

Explanation

The equation

$3x +\frac { 1 }{ y } =13$ can be solved as:

$3x+\frac { 1 }{ y } =13\\ \Rightarrow \frac { 3xy+1 }{ y } =13\\ \Rightarrow 3xy+1=13y\quad .........(1)$

The equation

$\frac { 2 }{ y }-x =5$ can be solved as:

$\frac { 2 }{ y }-x =5\\ \Rightarrow \frac { 2-xy }{ y } =5\\ \Rightarrow 2-xy=5y\quad .........(2)$

Multiply the equation 2 by

$3$ . Then we get:

$-3xy+6=15y.........(3)$

Add equations 1 and 3 to eliminate

$xy$ , because the coefficients of

$xy$ are opposite. So, we get

$(3xy-3xy)+(1+6)=13y+15y$

i.e.

$28y=7$

i.e.

$y=\frac { 1 }{ 4 }$

Substituting this value of

$y$ in the equation 1, we get

$3x\left( \frac { 1 }{ 4 } \right) +1=13\left( \frac { 1 }{ 4 } \right) \\ \Rightarrow \frac { 3 }{ 4 } x+1=\frac { 13 }{ 4 } \\ \Rightarrow \frac { 3 }{ 4 } x=\frac { 13 }{ 4 } -1\\ \Rightarrow \frac { 3 }{ 4 } x=\frac { 9 }{ 4 } \\ \Rightarrow 3x=9\\ \Rightarrow x=3$

Hence, the solution of the equations is

$x=3,y=\frac { 1 }{ 4 }$ .

A man's age is three times that of his son and in twelve years he will be twice as old as his son would be. What are their present ages?

0%
Present age of man is $60$ years and

Present age of son is $20$ years
0%
Present age of man is $45$ years and

Present age of son is $15$ years
0%
Present age of man is $54$ years and

Present age of son is $18$ years
0%
Present age of man is $36$ years and

Present age of son is $12$ years

Explanation

Let the present age of son be

$x$ years

As per question the present age of father is three times

then present age of father is

$3x$

As per question after

$12$ years the age of father will be twice the age of son

$3x+12=2(x+12)$

$\Rightarrow 3x+12=2x+24$

$\Rightarrow 3x-2x=24-12$

$\Rightarrow x=12$

Then present age of son is

$12$ years

So present age of father

$=3\times12=36$ years

Solve the following pair of simultaneous equations:

$\displaystyle \frac{6}{x}\, -\, \frac{2}{y}\, =\, 1\,;\, \frac{9}{x}\, -\, \frac{6}{y}\,=\, 0$

0%
$x = 7, y = -4$
0%
$x = 6, y = -4$
0%
$x = =1, y = 2$
0%
$x = 3, y = 2$

Explanation

The equation

$\cfrac { 6 }{ x } -\cfrac { 2 }{ y } =1$ can be solved as:

$\cfrac { 6 }{ x } -\cfrac { 2 }{ y } =1\\ \Rightarrow \cfrac { 6y-2x }{ xy } =1\\ \Rightarrow 6y-2x=xy\quad .........(1)$

The equation

$\cfrac { 9 }{ x } -\cfrac { 6 }{ y } =0$ can be solved as:

$\cfrac { 9 }{ x } -\cfrac { 6 }{ y } =0\\ \Rightarrow \cfrac { 9y-6x }{ xy } =0\\ \Rightarrow 9y-6x=0\quad .........(2)$

Multiply the equation 1 by

$3$ .Then we get:

$18y-6x=3xy.........(3)$

Subtract equation 2 from equation 3 to eliminate

$x$ , because the coefficients of

$x$ are same. So, we get

$(18y-9y)+(-6x+6x)=3xy-0$

i.e.

$9y=3xy$

i.e.

$9=3x$

i.e.

$x=3$

Substituting this value of

$x$ in the equation 1, we get

$6y-6=3y$

i.e.

$3y=6$

i.e.

$y=2$

Hence, the solution of the equations is

$x=3 ,y=2$ .

Find two numbers whose sum is

$15$ and difference is

$3$ .

0%
$13$ and $2$
0%
$7$ and $8$
0%
$9$ and $6$
0%
$3$ and $12$

Explanation

Let the two numbers be

$x$ and

$y$ .

It is given that the sum of the numbers is

$15$ that is

$x+y=15..........(1)$

Also the difference of the numbers is

$3$ that is

$x-y=3..........(2)$ .

Add equations 1 and 2 to eliminate

$y$ , because the coefficients of

$y$ are opposite. So, we get

$(x+x)+(y-y)=15+3$

i.e.

$2x=18$

$\implies x=9$

Substituting this value of

$x$ in (1), we get

$9+y=15$

i.e.

$y=15-9=6$

Hence, the two numbers are

$9$ and

$6$ .

Solve the following pair of simultaneous equations:

$\displaystyle \frac{a}{4}\, -\, \frac{b}{3}\, =\, 0\,;\, \frac{3a\, +\, 8}{5}\, =\, \frac{2b\, -\, 1}{2}$

0%
$a = -4.5, b = 12$
0%
$a = 4, b = -5$
0%
$a = 14, b = 10.5$
0%
$a = 12, b = 11.5$

Explanation

$\dfrac { a }{ 4 } -\dfrac { b }{ 3 } =0\\ \Rightarrow a =\dfrac { 4b }{ 3 }$

Substituting for '

$a$ ' in the other equation.

$\dfrac { 3a+8 }{ 5 } =\dfrac { 2b-1 }{ 2 } \\ 6a+16 =10b - 5\\ 6*\dfrac { 4b }{ 3 } + 16 = 10b - 5\\ 8b +16 = 10b - 5\\ 21 = 2b\\ b = 10.5$

So,

$a = \dfrac { 4\times10.5 }{ 3 } = 14$

$\therefore a = 14, b = 10.5$

Option

$C$

Solve the following pair of simultaneous equations:

$\displaystyle \frac{3}{a}\, +\, \frac{4}{b}\, =\, 2\,;\, \frac{9}{a}\, -\, \frac{4}{b}\, =\, 2$

0%
$a= 3, b = 4$
0%
$a = 1, b = -2$
0%
$a = 5, b = -3$
0%
$a = 0, b = 6$

Explanation

$\displaystyle \frac{3}{a}\, +\, \frac{4}{b}\, =\, 2\,;\, \frac{9}{a}\, -\, \frac{4}{b}\, =\, 2$

Let

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

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

So,

$3x + 4y = 2$ and

$9x - 4y = 2$

Adding the above equations, we get

$12x = 4$

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

Substituting

$x = \dfrac{1}{3}$ in either equation, we get

$y =\dfrac{1}{4}$

So,

$a = \dfrac{1}{x} = 3$ and

$b = \dfrac{1}{y} = 4$

$a = 3, b = 4$

Option

$A$ .

Find two numbers, which differ by

$7$ , such that twice the greater added to five times the smaller gives

$42$ .

0%
$19$ and $12$
0%
$12$ and $5$
0%
$11$ and $4$
0%
$16$ and $9$

Explanation

Let the two numbers be

$x$ and

$y$ .

It is given that two numbers differ by

$7$

$\Rightarrow x-y=7$

$...(1)$

Also, twice the greater number added to five times the smaller gives

$42$

$\Rightarrow 2x+5y=42$

$...(2)$

Multiply equation

$(1)$ by

$2$

$2x-2y=14$

$...(3)$

Let us subtract equation

$(3)$ from equation

$(2)$ to eliminate

$x$ , because the coefficients of

$x$ are the same.

$(2x+5y)-(2x-2y)=42-14$

$\Rightarrow (2x-2x)+(5y+2y)=28$

$\Rightarrow 7y=28$

$\Rightarrow y=4$

Substituting this value of

$y$ in

$(1),$

we get

$x-4=7$

i.e.

$x=11$

Hence, the two numbers are

$11$ and

$4$

Solve the following pair of simultaneous equations:

$\displaystyle\, 4x\, +\, \frac{3}{y}\, =\, 1\,; 3x\, -\, \frac{2}{y}\, =\, 5$

0%
$x=-6,y=7$
0%
$x=-3,y=1$
0%
$x=1,y=-1$
0%
$x=2,y=0$

Explanation

On putting

$\dfrac1y=Y$ , we get

$4x+3Y=1$ ...(i)

$3x-2Y=5$ ...(ii)
On multiplying (i) by 3 and (ii) by 4 and subtracting, we get

$12x+9Y=3$

$\underline {\underset {-}12x\underset {+}{-}8Y=\underset{-}20}$

$17Y=-17$

$\therefore Y=-1$
On putting

$Y=-1$ in (i), we get

$4x+3(-1)=1\Rightarrow 4x=4\Rightarrow x=1$

$\therefore x=1,y=\dfrac1Y=\dfrac1{-1}=-1$

Solve the following pairs of equations, graphically:

$x = 0$ and

$x + 3y = 6$

0%
$(0,2)$
0%
$(2,0)$
0%
$(0,-2)$
0%
$(-2,0)$

Explanation

The solution set of equation

$x+3y=6$ is as follows:

$x=0$ then

$0+3y=6\\ \Rightarrow 3y=6\\ \Rightarrow y=2$

$x=3$ then

$3+3y=6\\ \Rightarrow 3y=6-3=3\\ \Rightarrow y=1$

The required points are

$(0,2),(3,1)$ .

Another equation of line given is

$x=0$ .

Plot all the above points in the graph. Since the point of intersection in the graph is

$(0,2)$ .

Hence the solution is

$x=0,y=2$ .

Solve the following pairs of equations, graphically:

$3x - 4y = 1$ and

$x - 2y + 1 = 0$

0%
$(-3, -2)$
0%
$(-3, 2)$
0%
$(3, -2)$
0%
$(3, 2)$

Explanation

The solution set of equation

$3x-4y=1$ is as follows:

$x=-1$ then

$3\times (-1)-4y=1\\ \Rightarrow -3-4y=1\\ \Rightarrow -4y=1+3\\ \Rightarrow -4y=4\\ \Rightarrow y=-1$

$x=3$ then

$3\times (3)-4y=1\\ \Rightarrow 9-4y=1\\ \Rightarrow -4y=1-9\\ \Rightarrow -4y=-8\\ \Rightarrow y=2$

The required points are

$(-1,-1),(3,2)$ .

Also, the solution set of equation

$x-2y+1=0$ or

$x-2y=-1$ is as follows:

$x=1$ then

$1-2y=-1\\ \Rightarrow -2y=-1-1\\ \Rightarrow -2y=-2\\ \Rightarrow y=1$

$x=3$ then

$3-2y=-1\\ \Rightarrow -2y=-1-3\\ \Rightarrow -2y=--4\\ \Rightarrow y=2$

The required points are

$(1,1),(3,2)$ .

Plot all the above points in the graph. Since the point of intersection in the graph is

$(3,2)$ .

Hence the solution is

$x=3,y=2$ .

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 4 - 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 4 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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