MCQ Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 3

Solve the following simultaneous equations:

$37x+29y=13;\, \, 29x+37y=53$

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

Explanation

Consider the equations:

$37x+29y=13........(1)$ and

$29x+37y=53........(2)$

Multiplying equation 1 by

$29$ and equation 2 by

$37$ , we get:

$1073x+841y=377........(3)$ and

$1073x+1369y=1961........(4)$

Now subtracting equations 1 from equation 2 we get:

$1073x+1369y-1073x-841y=1961-377$

$528y=1584$

$y=3$

Substitute the value of

$y$ in equation 1:

$37x+87=13$

$37x=-74$

$x=-2$

Hence,

$x=-2,y=3$ .

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

$x-y=-1$

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

Find the values of the

$x+y$ and

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

$7x-5y=-1; \, \, 5x-7y=-11$

0%
$- 1$ , $-5$
0%
$1$ , $-5$
0%
$5$ , $1$
0%
$5$ , $-1$

Explanation

$\text{The given equations are}$

$7x-5y=-1$ ...(i)

$5x-7y=-11$ ...(ii)

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

$12x-12y=-12\Rightarrow x-y=-1$

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

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

$\therefore x+y=5 \space and \space x-y=-1$

Solve the following simultaneous equations.

$3x+4y=18; \, \, 4x+3y=17$

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

Explanation

The given equations are

$3x+4y=18$ ...(i)

$4x+3y=17$ ...(ii)

On multiplying (i) by 4 and (ii) by 3 and subtracting, we get

$12x+16y=72$

$\underline {\underset {-}12x\underset {-}{+}9y=\underset{-}51}$

$7y=21$

$\Rightarrow y=3$

On putting

$y=3$ in (i), we get

$3x+12=18\Rightarrow 3x=6\Rightarrow x=2$

$\therefore x=2,y=3$

Solve the following simultaneous equations:

$15x-17y=28;\, \, 15y-17x+36=0$

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

Explanation

The given equations are

$15x-17y=28$ ...(i)

$17x-15y=36$ ...(ii)
On multiplying (i) by 17 and (ii) by 15 and subtracting, we get

$255x-289y=476$

$\underline {\underset {-}255x\underset {+}{-}225y=\underset{-}540}$

$-64y=-64$

$\Rightarrow y=1$
On putting

$y=1$ in (i), we get

$15x-17=28\Rightarrow 15x=45\Rightarrow x=3$

$\therefore x=3,y=1$

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 following simultaneous equation:

$4x+5y = 10; \, \, 5x+4y=17$

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

Explanation

$4x+5y=10$ .....(1)

$5x+4y=17$ .......(2)

Adding (1) and (2). we get

$9x+9y=27$

$\Rightarrow x+y=3$ ......(3)

Now, subtracting (1) from (2). we get,

$x-y=7$ .......(4)

Adding (3) and (4). we get

$2x=10$

$\Rightarrow x=5$

Now, subtracting (3) from (4). we get,

$-2y=4$

$\Rightarrow y=-2$

$\therefore x=5 , y=-2$

Option D is correct.

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 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 following pair of equations by equating the coefficient method:

$8x\, =\, 5y$

$13x\, =\, 8y\, +\, 1$

0%
$x=3$ and $y=1$
0%
$x=5$ and $y=8$
0%
$x=6$ and $y=2$
0%
$x=9$ and $y=-5$

Explanation

Rewriting the given equations, we get

$8x-5y=0$ ...(i)

$13x-8y=1$ ...(ii)
On multiplying (i) by 13 and (ii) by 8, we get

$104x-65y=0$

$\underline {\underset {-}104x\underset {+}{-}64y=\underset{-}8}$

$-y=-8$

$\therefore y=8$
On putting

$y=8$ in (i), we get

$x=\frac{5y}8\Rightarrow x=5$

$\therefore x=5,y=8$

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 following pairs of equations:

$\displaystyle \dfrac{x+1}{4}= \displaystyle \dfrac{2}{3}\left ( 1-2y \right )$

$\displaystyle \dfrac{2+5y}{3}= \displaystyle \dfrac{x}{7} -2$

0%
$x= 8,\:y= -1$
0%
$x= 3,\:y= -1$
0%
$x= 7,\:y= -1$
0%
$x= 9,\:y= -1$

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:

$\displaystyle \frac{x}{3}\, +\, \frac{x\, +\, y}{6}\, =\,3;\, \frac{y}{3}\, -\, \frac{x\, -\, y}{2}\, =\, 6$

0%
$x=2,y=-5$
0%
$x=7,y=4$
0%
$x=6,y=3$
0%
$x=3,y=9$

Explanation

$\frac{x}{3}+\frac{x+y}{6}=3\Rightarrow \frac{2x+x+y}6=3\Rightarrow 3x+y=18$ ...(i)

$\frac{y}{3}-\frac{x-y}{2}=6\Rightarrow \frac{2y-3x+3y}6=6\Rightarrow -3x+5y=36$ ...(ii)
On adding (i) and (ii), we get

$3x+y=18$

$\underline {-3x+5y=36}$

$6y=54$

$\therefore y=9$
On putting

$y=9$ in (i), we get

$3x+9=18\Rightarrow 3x=9\Rightarrow x=3$

$\therefore x=3,y=9$

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

$x = 5, y =4$ .

Solve the following pair of simultaneous equations:

$8a\, -\, 7b\, =\, 1$

$4a\, =\, 3b\, +\, 5$

0%
$a=8 , b=9$
0%
$a=3 , b=-8$
0%
$a=7 , b=-6$
0%
$a=1 , b=-1$

Explanation

Multiply the equation

$4a=3b+5$ or

$4a-3b=5$ by

$2$ to make the coefficients of

$a$ equal. Then we get the equations:

$8a-7b=1.........(1)$

$8a-6b=10.........(2)$

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

$a$ , because the coefficients of

$a$ are the same. So, we get

$(8a-8a)+(-7b+6b)=1-10$

i.e.

$-b=-9$

i.e.

$b=9$

Substituting this value of

$b$ in the equation

$8a-7b=1$ , we get

$8a-63=1$

i.e.

$8a=64$

i.e.

$a=8$

Hence, the solution of the equations is

$a=8, b=9$ .

Solve the following pair of linear (simultaneous) equations by the method of elimination:

$y= 4x-7$

$16x-5y= 25$

0%
$\displaystyle x= \displaystyle \frac{3}{2},\displaystyle y= 6$
0%
$\displaystyle x= \displaystyle \frac{5}{2},\displaystyle y= 3$
0%
$\displaystyle x= \displaystyle \frac{7}{4},\displaystyle y= 2$
0%
$\displaystyle x= \displaystyle \frac{9}{2},\displaystyle y= 0$

Explanation

$4x - y = 7$ ...

$(1)$

$16x - 5y = 25$ ...

$(2)$

Multiplying

$(1)$ by

$4$ , we get

$16x - 4y = 28$ ...

$(3)$

$16x - 5y = 25$ ...

$(2)$

Subtracting

$(3)$ and

$(2)$ ,

$y = 3$

Substituting

$y = 3$ in

$(1)$ , we get

$x$ as

$4x - 3 = 7$

$4x = 10$

$x = \dfrac{10}{4}$ or

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

$x = \dfrac{5}{2}, y = 3$

Option

$B$

Solve the following pair of simultaneous equations:

$5x\, -\, 6y\, =\, 8$

$7y\, -\, 15x\,=\, 9$

0%
(1 , 3)
0%
(6 , -1)
0%
(-2 , -3)
0%
(3 , -5)

Explanation

Multiply the equation

$5x-6y=8$ by

$3$ to make the coefficients of

$x$ equal. Another equation is

$7y-15x=9$ or

$-15x+7y=9$ . Then we get the equations:

$15x-18y=24.........(1)$

$-15x+7y=9.........(2)$

Add Equations 1 and 2 to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(15x-15x)+(-18y+7y)=24+9$

i.e.

$-11y=33$

i.e.

$y=-3$

Substituting this value of

$y$ in the equation

$5x-6y=8$ , we get

$5x+18=8$

i.e.

$5x=-10$

i.e.

$x=-2$

Hence, the solution of the equations is

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

Solve following pair of equations by equating the coefficient method:

$3x\, -\, 7y\, =\, 35$

$2x\, +\, 5y\, =\, 4$

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

Explanation

Multiply the equation

$3x-7y=35$ by

$2$ and equation

$2x+5y=4$ by

$3$ to make the coefficients of

$x$ equal. Then we get the equations:

$6x-14y=70.........(1)$

$6x+15y=12.........(2)$

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

$x$ , because the coefficients of

$x$ are the same. So, we get

$(6x-6x)+(-14y-15y)=70-12$

i.e.

$-29y =58$

i.e.

$y = -2$

Substituting this value of

$y$ in (2), we get

$6x-30=12$

i.e.

$6x=42$

i.e.

$x=7$

Hence, the solution of the equations is

$x = 7, y = -2$ .

Solve following pair of equations by equating the coefficient method:

$x\, +\, 2y\, =\, 11$

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

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

Explanation

Multiply the equation

$x+2y=11$ by

$2$ to make the coefficients of

$x$ equal. Then we get the equations:

$2x+4y=22.........(1)$

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

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

$x$ , because the coefficients of

$x$ are the same. So, we get

$(2x-2x)+(4y+y)=22-2$

i.e.

$5y =20$

i.e.

$y = 4$

Substituting this value of

$y$ in (2), we get

$2x-4=2$

i.e.

$2x=6$

i.e.

$x=3$

Hence, the solution of the equations is

$x = 3, y = 4$ .

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 following pair of simultaneous equations:

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

0%
$x = \displaystyle 24\frac{2}{7}, y = \displaystyle 30\frac{3}{7}$
0%
$x = \displaystyle 12\frac{1}{3}, y = \displaystyle 10\frac{2}{3}$
0%
$x = \displaystyle 16\frac{7}{6}, y = \displaystyle 13\frac{4}{3}$
0%
$x = \displaystyle 2\frac{22}{7}, y = \displaystyle 12\frac{13}{17}$

Explanation

The equation

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

$\cfrac { x }{ 2 } -\cfrac { y }{ 3 } =2\\ \Rightarrow \cfrac { 3x-2y }{ 6 } =2\\ \Rightarrow 3x-2y=12\quad .........(1)$

The equation

$\cfrac { x }{ 5 } +\cfrac { y }{ 3 } =15$ can be solved as:

$\cfrac { x }{ 5 } +\cfrac { y }{ 3 } =15\\ \Rightarrow \cfrac { 3x+5y }{ 15 } =15\\ \Rightarrow 3x+5y=225\quad .........(2)$

Subtract Equation 2 from equation 1 to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(3x-3x)+(-2y-5y)=12-225$

i.e.

$-7y=-213$

i.e.

$y=\cfrac { 213 }{ 7 } =30\cfrac { 3 }{ 7 }$

Substituting this value of

$y$ in the equation 1, we get

$3x-2\left( \cfrac { 213 }{ 7 } \right) =12\\ \Rightarrow 3x-\cfrac { 426 }{ 7 } =12\\ \Rightarrow 21x-426=84\\ \Rightarrow 21x=510\\ \Rightarrow x=\cfrac { 510 }{ 21 } =\cfrac { 170 }{ 7 } =24\cfrac { 2 }{ 7 }$

Hence, the solution of the equations is

$x=24\cfrac { 2 }{ 7 } ,y=30\cfrac { 3 }{ 7 }$ .

Solve for

$x$ and

$y$ :

$4x= 17-\displaystyle \frac{x-y}{8}$

$2y+x= 2+\displaystyle \frac{5y+2}{3}$

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

Explanation

Given equations are,

$4x=17-\dfrac {x-y}{8}$

$\Rightarrow 32x=136-x+y$

$\Rightarrow 33x-y=136$ ....(1)

and

$2y+x=2+\dfrac {5y+2}{3}$

$\Rightarrow 3(2y+x)=6+5y+2$

$\Rightarrow 6y+3x=6+5y+2$

$\Rightarrow 3x+y=8$ ....(2)

Add equations (1) and (2), we get

$36x=144$

$\Rightarrow x=4$

Put this value in equation (1), we get

$33(4)-y=136$

$\Rightarrow 132-y=136$

$\Rightarrow y=-4$

Therefore, the solution is

$x=4, y=-4$ .

Solve the following pair of simultaneous equations:

$3x\, +\, 5(y\, +\, 2)\, =\, 1$

$3x\, +\, 8y\, =\, 0$

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

Explanation

The equation

$3x+5(y+2)=1$ can be written as

$3x+5y+10=1$ or

$3x+5y=-9$ . Another equation is

$3x+8y=0$ . Then we get the equations:

$3x+5y=-9.........(1)$

$3x+8y=0.........(2)$

Subtract Equation 2 from equation 1 to eliminate

$x$ , because the coefficients of

$x$ are the same. So, we get

$(3x-3x)+(5y-8y)=-9-0$

i.e.

$-3y=-9$

i.e.

$y=3$

Substituting this value of

$y$ in the equation

$3x+8y=0$ , we get

$3x+24=0$

i.e.

$3x=-24$

i.e.

$x=-8$

Hence, the solution of the equations is

$x = -8, y =3$ .

Solve the following pair of simultaneous equations

$3x\, +\,2y\, =\, -1$

$6y\, =\, 5(1\, -\, x)$

0%
$x=- 2$ and $y=2.5$
0%
$x=- 1$ and $y=3$
0%
$x=6$ and $y=1.5$
0%
$x=- 7$ and $y=-4$

Explanation

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

$6y=5(1 - x)\Rightarrow 5x+6y=5$ ...(ii)
On multiplying (i) by 5 and (ii) by 3, we get

$15x+10y=-5$

$\underline {\underset {-}15x\underset {-}{+}18y=\underset{-}15}$

$-8y=-20$

$\therefore y=\frac{20}8=2.5$
On putting

$y=2.5$ in (i), we get

$3x+2(2.5)=-1\Rightarrow 3x=-6\Rightarrow x=-2$

$\therefore x=-2,y=2.5$

Solve the following pair of simultaneous equations:

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

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

Explanation

$\displaystyle \frac{x-1}{2}+ \frac{y+1}{5}=4\frac{1}{5}\Rightarrow \frac{5x-5+2y+2}{10}=\frac{21}5\Rightarrow 5x+2y=45$ ...(i)

$\displaystyle \frac{x+y}{3}=y-1\Rightarrow x+y=3y-3\Rightarrow x-2y=-3$ ...(ii)

On multiplying (i) by 1 and (ii) by 5, we get

$5x+2y=45$

$\underline {\underset {-}5x\underset {+}{-}10y=\underset{+}{-}15}$

$12y=60$

$\therefore y=5$

On putting

$y=5$ in (ii), we get

$x-2(5)=-3\Rightarrow x=7$

$\therefore x=7,y=5$

Find two numbers such that twice of the first added to the second gives

$21$ , and twice the second added to the first gives

$27$ .

0%
$5$ and $11$
0%
$9$ and $16$
0%
$3$ and $18$
0%
$3$ and $17$

Explanation

Let the first number is

$x$ and second is

$y$

Given twice of first added with second gives

$21$

$\therefore 2x+y=21$ ...

$(1)$

And given twice of second added with first gives

$27$

$\therefore x+2y=27$ ...

$(2)$

Multiply

$(1)$ by

$2$

Then

$4x+2y=42$ ...

$(3)$

Subtract

$(3)$ with

$(2)$

Then

$3x=15$

$x=5$

Put the value of

$x=5$ in

$(1)$

Then

$10+y=21$

$y=21-10$

$y=11$

Then first number is

$5$ and second is

$11$ .

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

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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