MCQ Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 8

Six years hence, a man's age will be three times the age of his son and three years ago he was nine times as old as his son. The present age of the man is

0%
$28$ years
0%
$30$ years
0%
$32$ years
0%
$34$ years

Explanation

Let the present age of man be

$x$ years and the present age of son be

$y$ years

After six years:
Man's age =

$(x+6)\ years$

Son's age =

$(y+6)\ years$

According to the condition

$\Rightarrow x+6=3\left (y+6 \right )$

$\Rightarrow x-3y=12$

$\Rightarrow x=12+3y$

$...(1)$

Three years ago:
Man's age =

$(x-3)\ years$
Son's age =

$(y-3)\ years$

According to the condition

$\Rightarrow x-3=9\left (y-3 \right )$

$\Rightarrow x-9y=-24$

$...(2)$

Put the value of

$x$ in equation

$\left ( 2 \right )$

$\Rightarrow 12+3y-9y=-24$

$\Rightarrow y=6$

Put the value of

$y$ in equation

$\left ( 1\right )$

$x-3\times6=12$

$\Rightarrow x=30$

So, Man's age = 30 years

The pair of linear equations

$x + 2y = 5, 7x + 3y = 13$ has a unique solution

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

Explanation

Given equations are:

$x + 2y = 5$ ....(1)

$7x+3y = 13$ .....(2)
Multiplying equation (1) by 7

$7x+14y = 35$ .....(3)
Subtracting equation (2) from equation (3), we have

$11y = 22$

$y = 2$
Now putting

$y$ value in equation (1)

$\Rightarrow x + 2\left ( 2 \right ) = 5$

$\Rightarrow x = 5 - 4$

$\Rightarrow x = 1$

$x = 1 , y = 2$

The pair of linear equations

$x+y=3, 2x+5y=12$ has a unique solution

$x=x_1, y=y_1$ then value of

$x_1$ is?

0%
$1$
0%
$2$
0%
$-1$
0%
$-2$

Explanation

Given equations are

$x + y = 3$

$2x + 5y = 12$
Since, the system has unique solution at

$x = x_1$ and

$y=y_1$
That means the lines intersect at only one point

$(x_{1},y_{1})$
By putting

$x=x_1$ and

$y=y_1$ in equations, we get

$x_{1} + y_{1} = 3$ ....

$(1)$

$2 x_{1} +5y_{1} = 12$ ....

$(2)$
From equation

$(1)$ , we have

$y_1 = 3 - x_{1}$
Putting the value of

$y_{1}$ in equation

$(2)$ ,

$\Rightarrow 2x_{1} + 5\left ( 3 - x_{1} \right ) = 12$

$\Rightarrow 2x_{1} + 15 - 5x_{1} = 12$

$\Rightarrow 2x_{1} - 5x_{1} = 12 - 15$

$\Rightarrow - 3x_{1} = -3$

$\Rightarrow x_{1} = \displaystyle \frac{-3}{-3}$

$\Rightarrow x_{1} = 1$

From Delhi station, if we buy 2 tickets for station A and 3 tickets for station B, the total cost is Rs.But if we buy 3 tickets for station A and 5 tickets for station B, the total cost is Rs.What are the fares from Delhi to station A and to station B?

0%
$A = Rs. 12 ; B = Rs. 17$
0%
$A = Rs. 13 ; B = Rs. 17$
0%
$A = Rs. 18 ; B = Rs. 17$
0%
$A = Rs. 19 ; B = Rs. 17$

Explanation

${\textbf{Step -1: Writing mathematical equations according to conditions given in question}}$

${\text{Let the price of ticket for station A be Rs}}{\text{. x and for station B be Rs}}{\text{. y,}}$

${\text{According to question, }}$

${\text{2x + 3y = 77 }} \to {\text{ Equation 1}}$

${\text{3x + 5y = 124 }} \to {\text{ Equation 2}}$

${\textbf{Step -2: Calculating fares for station A and station B .}}$

${\text{Solving Equation 1 and Equation 2,}}$

${\text{Multiplying Equation 1 by 3,}}$

${\text{6x + 9y = 231 }} \to {\text{ Equation 3}}$

${\text{Multiplying Equation 2 by 2,}}$

${\text{6x + 10y = 248 }} \to {\text{ Equation 4}}$

${\text{Subtracting Equation 3 from Equation 4,}}$

$\Rightarrow {\text{ y = 248 - 231}}$

$\Rightarrow {\text{ y = 17}}$

${\text{Substituting value of y in Equation 1,}}$

$\Rightarrow {\text{ 2x + 3(17) = 77}}$

$\Rightarrow {\text{ 2x + 51 = 77}}$

$\Rightarrow {\text{ 2x = 26}}$

$\Rightarrow {\text{ x = 13}}$

${\textbf{Thus, the fare to station A is Rs}}{\textbf{. 13 and to station B is Rs}}{\textbf{.17 .}}$

Three chairs and two tables cost Rs1,Five chairs and three tables cost Rs 2,Then the total cost of one chair and one table is

0%
Rs 800
0%
Rs 850
0%
Rs 900
0%
Rs 950

Explanation

Let cost of one chair be Rs

$x$ and cost of one table be Rs

$y$
Now as per the question,

$3x+2y = 1850$ .....(1)

$5x + 3y = 2850$ ......(2)
Now mutiplying equation (1) by 5 and equation (2) by 3

$15x+10y = 9250$ .....(3)

$15x+9y = 8550$ ......(4)
Now subtracting equation (4) from equation (3)

$\Rightarrow y = 700$
Put this value in eq (1),

$\Rightarrow 3x+2\left ( 700 \right ) = 1850$

$3x+1400 = 1850$

$3x = 450$

$x = 150$
Now total cost of 1 chair and 1 table =

$x + y$

$=150+700= 850$
So, total cost of 1 chair and 1 table is Rs 850

Solve the following equations by the substitution method

$\dfrac {1}{2}(9x+10y)=23, \dfrac {5x}{4}-2y=3$

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

Explanation

The equations are

$\Rightarrow \frac{1}{2}\left ( 9x+10y \right )=23\Rightarrow 9x+10y=46....eq\left ( 1 \right )$

$\Rightarrow x=\frac{46-10y}{9}$

$\Rightarrow \frac{5x}{4}-2y=3\Rightarrow 5x-8y=12....eq\left ( 2 \right )$
Substitute the value of x in eq (2 )

$\Rightarrow \frac{5\left (46-10y \right )}{9}-8y=12$

$\Rightarrow 230-50y-72y=108\Rightarrow -122y=-122\Rightarrow y=1$
Substitute the value of y in eq (2 )

$\Rightarrow 5x-8\left (1 \right )=12\Rightarrow 5x=20\Rightarrow x=4$

$\Rightarrow x=4,y=1$

$2x + y = 23$ and

$4x - y = 19$ , find the values of

$5y - 2x$ and

$\dfrac{y}{x} - 2$ .

0%
$36, -\dfrac13$
0%
$31, -\dfrac57$
0%
$38, \dfrac67$
0%
None of these

Explanation

Given System of Equations are:

$2x+y = 23$ ...

$(1)$

$4x-y = 19$ ...

$(2)$
Now adding both the equation will have:

$6x = 42$

$x = 7$
Now substituting

$x = 7$ in in equation

$(1)$

$\Rightarrow 2\left ( 7 \right ) + y = 23$

$\Rightarrow 14 + y = 23$

$\Rightarrow y = 9$
Now substituting

$x = 7$ and

$y = 9$ in

$5y - 2x$

$\Rightarrow 5\left ( 9 \right ) - 2\left ( 7 \right )$

$\Rightarrow 45-14$

$\Rightarrow 31$
Now Substituting Now substituting

$x = 7$ and

$y = 9$ in

$\frac{y}{x}-2$

$\Rightarrow \dfrac{9}{7}-2$

$\Rightarrow \dfrac{9-14}{7}$

$\Rightarrow \dfrac{-5}{7}$

Solve the following equations by the substitution method

$0.04 x + 0.02y = 5, 0.5x - 0.4y = 30$

0%
$x=27, y=61$
0%
$x=100, y=50$
0%
$x=200, y=39$
0%
$x=54, y=122$

Explanation

The equations are

$\Rightarrow .04x+.02y=5....eq\left ( 1 \right )$

$\Rightarrow x=\frac{5-.02y}{.04}$

$\Rightarrow .5x-.4y=30....eq\left (2 \right )$
Substitute the value of x in eq (2)

$\Rightarrow \frac{.5\left (5-.02y \right )}{.04}-.4y=30$

$\Rightarrow 2.5-.01y-.016y=1.2\Rightarrow -.026y=-1.3\Rightarrow y=50$
Substitute the value of y in eq (2)

$\Rightarrow .5x-.4\left (50 \right )=30\Rightarrow .5x=50\Rightarrow x=100$

$\Rightarrow x=100,y=50$

Solve the following equations by the substitution method

$x = 3y - 19, y = 3x - 23$

0%
$x = 5, y = 7$
0%
$x = 11, y = 10$
0%
$x = 13, y = 7$
0%
$x = 3, y = 11$

Explanation

The equations are

$\Rightarrow x=3y-19....eq\left ( 1 \right )$

$\Rightarrow y=3x-23....eq\left ( 2 \right )$
Substitute the value of x in eq (2 )

$\Rightarrow y=3\left (3y-19 \right )-23$

$\Rightarrow y=9y-57-23\Rightarrow -8y=-80\Rightarrow y=10$
Substitute the value of y in eq (1 )

$\Rightarrow x=3\left (10 \right )-19\Rightarrow x=30-19\Rightarrow x=11$

$\Rightarrow x=11,y=10$

Solve the following pairs of linear equations by elimination method:

$78x + 91y = 39$ and

$65x + 117y = 42$

0%
$x = \dfrac{3}{13}, y = \dfrac{3}{13}$
0%
$x = \dfrac{1}{11}, y = \dfrac{2}{11}$
0%
$x = \dfrac{5}{17}, y = \dfrac{4}{17}$
0%
Cannot be determined

Explanation

Given system of equations are:

$78x+91y = 39 ...(1)$

$65x+117y =42 ...(2)$
Now using elimination method will first eliminate

$x$ .
now multiplying eq

$(1)$ by

$65$ and eq

$(2)$ by

$-78$ .

$\Rightarrow \left ( 65 \right )78x + \left ( 65 \right )91y = \left ( 65 \right )39$

$\Rightarrow \left ( -78 \right )65x + \left ( 78 \right )117y = \left ( 65 \right )42$

$\Rightarrow 5070x + 5915y = 2535 ...(3)$

$\Rightarrow -5070x - 9126y = -3276 ...(4)$
From equation

$(3)$ and

$(4)$ , we get

$-3211y = -741$

$\Rightarrow y = \dfrac{-741}{-3211}$

$y = \dfrac{3}{13}$
Now substituting

$y = \dfrac{3}{13}$ in eq

$(1)$

$\Rightarrow 78x + 91\left ( \dfrac{3}{13} \right )= 39$

$\Rightarrow 1014x +273 = 507$

$\Rightarrow 1014x = 507 - 273$

$\Rightarrow 1014x = 234$

$\Rightarrow x = \dfrac{234}{1014}$

$\Rightarrow x = \dfrac{3}{13}$

$\Rightarrow x = \dfrac{3}{13} , y = \dfrac{3}{13}$

The equations

$x-y=0.9$ and

$\displaystyle \frac {11}{x+y}=2$ have the solution

0%
$x=5, y=1$
0%
$x=3.2$ and $y=2.3$
0%
$x=3$ and $y=2$
0%
none of these

Explanation

The equation are

$x-y=0.9$

$\Rightarrow x=0.9+y$ .....(1)

$\displaystyle \frac{11}{x+y}=2$

$\Rightarrow 2x+2y=11$ .....(2)
Put the value of

$x$ in eq(2)

$\Rightarrow 2\left (.9+y \right )+2y=11$

$\Rightarrow 1.8+2y+2y=11$

$\Rightarrow 4y=9.2$

$\Rightarrow y=2.3$

$\Rightarrow x=.9+y\Rightarrow x=.9+2.3=3.2$

$\therefore x=3.2 ,y=2.3$

A students walks from his house at less 4 km per hour and reaches his school late by 5 minutes. If his speed has been increase by 5 km per hour then he would have reached 10 minutes early. The distance of the school from his house is

0%
$\displaystyle \frac{3}{5}km$
0%
$5$ km
0%
$6$ km
0%
$4$ km

Explanation

Let the actual time be 't'
and distance be x
Now according to first condition, we have

$\dfrac{x}{4}=t+\dfrac{5}{60}$
And according to Second condition, we have

$\dfrac{x}{5}=t-\dfrac{10}{60}$

$60x=4t+20$

$60x=5t-50$
Subtracting equation 1 from 2, we have

$0=t-70$

$\therefore t=70$
Putting the value of t we get

$x=5km$

Solve:

$\frac {4}{9}x+\frac {1}{3}y=1, 5x+2y=13$

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

Explanation

Given system of equations are:

$\frac{4}{9}x + \frac{1}{3}y = 1$ ... (1)

$5x+2y = 13$ ... (2)
From equation 1 will get:

$4x + 3y = 9$
Now subtracting eq1 and eq2

$4x +3y = 9$

$5x +2y = 13$
First multiply eq1 from 2 and eq2 from 2 will get:

$8x + 6y = 18$

$15x+6y = 39$
now will subtract the equation

$\Rightarrow 8x +6y = 18$

$-15x -6y = -39$

$\Rightarrow -7x = -21$

$\Rightarrow x = 3$
now substituting x = 3 in eq 2

$\Rightarrow 15\left ( 3 \right )+ 6y = 39$

$\Rightarrow 45 + 6y = 39$

$\Rightarrow 6y = 39-45$

$\Rightarrow 6y = -6$

$\Rightarrow y = -1$

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

Solve the following equations by the substitution method

$\dfrac {x+11}{7}+2y=10, 3x=8+\dfrac {y+7}{11}$

0%
$x = 1, y = -2$
0%
$x = 8, y = -7$
0%
$x = 9, y = 2$
0%
$x = 3, y = 4$

Explanation

The equations are

$\Rightarrow \dfrac{x+11}{7}+2y=10\Rightarrow x+11+14y=70$

$\Rightarrow x+14y=59 ....eq\left ( 1 \right )$

$\Rightarrow x=59-14y$

$\Rightarrow 3x=8+\dfrac{y+7}{11}\Rightarrow 33x=88+y+7$

$\Rightarrow 33x-y=95....eq\left ( 2 \right )$
Substitute the value of x in eq (2 )

$\Rightarrow 33\left (59-14y \right )-y=95$

$\Rightarrow 1947-462y-y=95\Rightarrow 463y=1853\Rightarrow y=4$

$\Rightarrow x=59-14y\Rightarrow x=59-14\left (4 \right ) \Rightarrow x=3$

$\Rightarrow x=3,y=4$

Solve the following equations by the substitution method

$11x-8y=27, 3x+5y=-7$

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

Explanation

The equations are

$\Rightarrow 11x-8y=27....eq\left ( 1 \right )$

$\Rightarrow x=\frac{27+8}{11}$

$\Rightarrow 3x+5y=-7....eq\left ( 2 \right )$
Substitute the value of x in eq(2 )

$\Rightarrow \frac{3\left (27+8y \right )}{11}+5y=-7$

$\Rightarrow 81+24y+55y=-77\Rightarrow 79y=-158\Rightarrow y=-2$
Substitute the value of y in eq (2 )

$\Rightarrow 3x+5\left (-2 \right )=-7\Rightarrow 3x=3\Rightarrow x=1$

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

Find the values of x and y in the following rectangle

0%
$x=7, y=-8$
0%
$x=1, y=-5$
0%
$x=2, y=0$
0%
$x=1, y=4$

Explanation

${\textbf{Step-1: Use rectangle property.}}$

${\text{As we know that opposite sides of a rectangle are equal.}}$

${\text{According to the question equations are}}$

$\Rightarrow x+3y=13$

$...(1)$

$\Rightarrow x=13-3y$

$\Rightarrow 3x+y=7$

$...(2)$

${\textbf{Step-2: Solve the equations to find the values of x and y}.}$

${\text{Put the value of}}$

$x$

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

$\Rightarrow 3(13-3y)+y=7$

$\Rightarrow 39 -9y + y = 7$

$\Rightarrow -8y =-32$

$\Rightarrow y = 4$

${\text{Put the value of}}$

$y$

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

$\Rightarrow x=13-3y$

$\Rightarrow x=13-3 \times 4$

$\Rightarrow x=13-12$

$\Rightarrow x=1$

${\textbf{Hence, the value of x = 1 and y = 4. So, option D is correct.}}$

Solve the following pair of linear equations by elimination method

$0.5x + 0.7y = 0.74$ and

$0.3x + 0.5y = 0.5$

0%
$x = 1, y = 1.2$
0%
$x = 0.7, y = 0.1$
0%
$x = 0.5, y = 0.7$
0%
$x = 1.5, y = 0.2$

Explanation

The equations are

$\Rightarrow 0.5x+0.7y=0.74 eq\left ( 1 \right )$

$\Rightarrow 0.3x+0.5y=0.5 eq\left ( 2\right )$

multiply 0.3 in eq ( 1 ) and 0.5 in eq (2 )

$\Rightarrow 0.15x+0.21y=0.222 eq\left (3 \right )$

$\Rightarrow 0.15x+0.25y=0.25 eq\left ( 4 \right )$

$subtracting \, eq\left ( 3 \right ) and \, eq\left ( 4 \right )$

$\Rightarrow 0.04y=0.028\Rightarrow y=0.7$

put the value of y in eq( 1 )

$\Rightarrow 0.5x+0.7\times 0.7=0.74\Rightarrow 0.5x=0.74-0.49\Rightarrow x=0.5$

$\Rightarrow x=0.5,y=0.7$

Solve the following pairs of linear equations by elimination method:

$47x + 31y = 63$ and

$31x + 47y = 15$

0%
$x = 1, y = 0$
0%
$x = 2, y = - 1$
0%
$x = 3, y = 4$
0%
$x = 9, y = - 7$

Explanation

Multiply the equation

$47x+31y=63$ by

$31$ and equation

$31x+47y=15$ by

$47$ to make the coefficients of

$x$

equal. Then we get the equations:

$1457x+961y=1953.........(1)$

$1457x+2209y=705.........(2)$

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

$x$ , because the coefficients of

$x$ are the same. So, we get

$(1457x-1457x)+(2209y-961y)=32-27$

i.e.

$1248y =-1248$

i.e.

$y =-1$

Substituting this value of

$y$ in the equation

$47x+31y=63$ , we get

$47x-31=63$

i.e.

$47x=94$

i.e.

$x=2$

Hence, the solution of the equations is

$x = 2, y =-1$ .

A and B each have a certain number of mangoes. A says to B," if you give 30 of your mangoes, I will have twice as many as left with you."B replies,"if you give me 10, I will have thrice as many as left with you." How many mangoes does each have?

0%
A : $34$ mangoes; B : $62$ mangoes
0%
A : $22$ mangoes; B : $48$ mangoes
0%
A : $38$ mangoes; B : $90$ mangoes
0%
A : $56$ mangoes; B : $92$ mangoes

Explanation

let A has

$x$ mangoes and B has

$y$ mangoes
According to the question

$\Rightarrow \left (x+30 \right )=2\left ( y-30 \right )$

$\Rightarrow x+30=2y-60\Rightarrow x-2y=-90.....eq1$

$\Rightarrow 3\left (x-10 \right )=y+10$

$\Rightarrow 3x-30=y+10\Rightarrow 3x-y=40....eq2$

Multiply

$eq2$ by

$2$

$\Rightarrow 6x-2y=80......eq3$

Subtract

$eq1$ and

$eq2$

$\Rightarrow 5x=170\Rightarrow x=34$
put

$x=34$ in

$eq1$

$\Rightarrow 34-2y=-90\Rightarrow-2y=-124\Rightarrow y=62$
A has

$34$ mangoes
B has

$62$ mangoes

Solve the following pairs of linear equations by elimination method:

$217x + 131y = 913$ and

$131x + 217y = 827$

0%
$x = 0, y = 3$
0%
$x = -9, y = -4$
0%
$x = -8, y = -1$
0%
$x = 3, y = 2$

Explanation

The equations are

$\Rightarrow 217x+131y=913$ .....eq (1)

$\Rightarrow 131x+217y=827$ .....eq (2)
Multiply

$131$ in eq (1) and

$217$ in eq (2)

$\Rightarrow 28427x+17161y=119603$ ... (3)

$\Rightarrow 28427x+47089y=179459$ ....(4)
Subtracting eq (3) and eq (,4)

$\Rightarrow 29928y=59856$

$\Rightarrow y=2$

Put the value of

$y$ in eq (1)

$\Rightarrow 217x+131\times 2=913$

$\Rightarrow 217x=913-262$

$\Rightarrow x=3$

Hence, option D is correct.

Solve the following pairs of linear equations by elimination method:

$2(ax - by)+ (a + 4b) = 0$ and

$2(bx + ay)+ (b - 4a) =0$

0%
$x = 6, y = 3$
0%
$x =-1, y = 3$
0%
$x =-\dfrac12, y = 2$
0%
None of these

Solve the following pair of linear equations by elimination method

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

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

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

Explanation

Given system of equations are:

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

$x - \frac{y}{3} = 3$ ................(2)
From eq 1 will have

$\Rightarrow 3x+4y = - 6 .$ ........(3)
From eq 2 will have

$\Rightarrow 3x-y = 9$ ...........(4)

Now using elimination method frist we will eliminate y :
now multiplying eq 4 from 4 will have

$\Rightarrow 12x-4y = 36$
Now from eq 3 and 4 will get x:

$3x+4y = -6$

$12x-4y = 36$

$\Rightarrow 15x = 30$

$\Rightarrow x = 2$
Now substituting

$x = 2$ in eq 2 will get y

$\Rightarrow \left ( 2 \right )3 - y = 9$

$\Rightarrow 6 - y = 9$

$\Rightarrow -y = 9-6$

$\Rightarrow y = -3$

$\Rightarrow x = 2 , y = -3$

Solve the following pair of linear equations by elimination method

$\sqrt 7x+\sqrt {11}y=0$ and

$\sqrt 3x-\sqrt 5y=0$

0%
$x = 0, y = 0$
0%
$x = 1, y = 1$
0%
$x = 7\sqrt{11}, y =5\sqrt{3}$
0%
Data insufficient

Explanation

$\sqrt 7x + \sqrt {11}y = 0$ ......(i)

$\sqrt 3x - \sqrt 5y = 0$ ......(ii)
On multiplying equation (i) by

$\sqrt 3$ and equation (ii) by

$\sqrt 7$ . we get,

$\sqrt {21}x + \sqrt {33}y = 0$ ......(iii)

$\sqrt {21}x - \sqrt {35}y = 0$ ......(iv)
On subtracting equation (iv) from equation (iii). we get,

$\sqrt {21}x + \sqrt {33}y = 0$

$\underline {\underset {-} {\sqrt {21}}x - \underset {+} {\sqrt {35}}y = \underset {-} {0}}$

$(\sqrt {33} + \sqrt {35})y=0$

$\Rightarrow y =0$
On substituting

$y=0$ in equation (i). we get,

$\sqrt 7x + \sqrt {11}\times 0 = 0$

$\Rightarrow \sqrt 7x=0$

$\Rightarrow x=0$

$\therefore x=0 , y=0$
Option A is correct.

Solve the following pair of linear equations by elimination method

$\dfrac {bx}{a}-\dfrac {ay}{b}+a+b=0$ and

$bx-ay+2ab=0$

0%
$x = - a, y = b$
0%
$x = a, y = ab$
0%
$x = - ab, y = b-a$
0%
$x = 1, y = \dfrac{b}{a}$

Explanation

The equations are

$\dfrac { bx }{ a } -\dfrac { ay }{ b } +a+b=0$ ..........(i)

and

$bx-ay+2ab=0$ .........(ii)

Multiplying (i) by

$a$ , we get

$bx-\dfrac { { a }^{ 2 } }{ b } y+{ a }^{ 2 }+ab=0$ ......(iii)

Subtracting (iii) from (ii), we get

$y\left( \dfrac { ab-{ a }^{ 2 } }{ b } \right) =ab-{ a }^{ 2 }$

$\Rightarrow y=b$

Substituting

$y=b$ in (ii), we have

$bx-ab+2ab=0$

$\Rightarrow bx=-ab$

$\Rightarrow x=-a$

$\therefore (x,y)=(-a,b)$

$4$ tables and

$3$ chairs together cost Rs.

$2250$ and

$3$ tables and

$4$ chairs cost Rs.

$1950$ . Find the cost of

$2$ chairs and

$1$ table.

0%
Rs. $150$
0%
Rs. $750$
0%
Rs. $350$
0%
Rs. $850$

Explanation

Suppose the cost of $1$ table $=x$ and cost of $1$ chair $=y$
Then according to the question
$\Rightarrow 4x+3y=2250$ ...........(1)
$\Rightarrow 3x+4y=1950$ .............(2)

Multiply (1) by $4$ and (2) by $3$ and Subtract both
$\Rightarrow (16x+12y=9000)- (9x+12y=5850)$

$\Rightarrow 7x=3150\Rightarrow x=450$

Put $x=450$ in $eq2$

$\Rightarrow 450\times4+3y=2250\Rightarrow 3y=2250-1800\Rightarrow 3y=450$

$\Rightarrow y=150$

Cost of $1$ table $=450$
Cost of $1$ chair $=150$
$\therefore$ Cost of $1$ table and $2$ chair $=450+150\times2=750$

One says," give me a hundred, friend! I shall then become twice as rich as you," The other replies," If you give me ten, I shall be six times as rich as you." Tell me what is the amount of their respective capital?

0%
$70 \ and \ 160$
0%
$66\ and \ 190$
0%
$40 \ and \ 170$
0%
$30 \ and \ 120$

Explanation

let one has x Rs. and others has y Rs.
Then according to the question

$\Rightarrow \left (x+100 \right )=2\left ( y-100\right )$

$\Rightarrow x+100=2y-200\Rightarrow x-2y=-300.....eq1$

$\Rightarrow 6\left (x-10 \right )=y+10$

$\Rightarrow 6x-60=y+10\Rightarrow 6x-y=70....eq2$

Multiply eq 2 by 2

$\Rightarrow 12x-2y=140......eq3$
subtract eq1 and eq3

$\Rightarrow -11x=-440\Rightarrow x=40$
put

$x=40$ in eq1

$\Rightarrow 40-2y=-300\Rightarrow-2y=-340\Rightarrow y=170$
One has 40 Rs and others has 170 Rs.

Form the pair of linear equations for the following problem and find their solution by substitution method.

The difference between the two numbers is

$26$ and one number is three times the other.

0%
$x - y = 26$ and $x = 3y$ , $x=33$ and $,y=11$
0%
$x - y = 26$ and $x = 3y$ , $x=39$ and $y=13$
0%
$x - y = 13$ and $x = 3y$ , $x=75$ and $y=25$
0%
None of these

Explanation

Let the two numbers be

$x$ and

$y\ (x> y)$ .

In question given that the difference between two number is

$26$ .

$\Rightarrow x- y = 26$ .......(i)

And the one number is

$3$ times of the other number

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

Now putting

$x = 3y$ in equation (1)

$\Rightarrow 3y - y = 26$

$\Rightarrow 2y = 26$

$\Rightarrow y = 13$

Now substituting

$y$ value in (2),

$\Rightarrow x = 3 \times 13$

$\Rightarrow x = 39$

Hence,

$x = 39$ ,

$y = 13$

A man has only

$20$ paise coins and

$25$ paise coins in his purse. If he has

$50$ coins in all totaling Rs.

$11.25$ , how many coins of each kind does he have?

0%
$25$ coins of each kind
0%
$16$ coins of each kind
0%
$44$ coins of each kind
0%
$32$ coins of each kind

Explanation

Let no of $20$ p coins $=x$
No of $25$ p coins $=y$
According to the question
$x+y=50$
$\Rightarrow x=50-y.....eq1$
And $20x+25y=1125\Rightarrow 4x+5y=225....eq2$
Put the value of $x$ from $eq1$
$\Rightarrow 4\left ( 50-y \right )+5y=225......eq3$
$\Rightarrow 200-4y+5y=225\Rightarrow y=25$
Put $y=25$ in $eq1$
$\Rightarrow x+25=50\Rightarrow x=25$

No of $20$ p coins $=25$

No of $25$ p coins $=25$

The sum of two numbers is

$8$ . If their sum is

$4$ times their difference. Find the numbers.

0%
$4, 4$
0%
$6, 2$
0%
$5, 3$
0%
None of these

Explanation

Let the first number be $x$ and second number $y.$

According to question
$x+y=8\quad\quad\quad\dots(i)$

$4(x-y)=8$

$\Rightarrow x-y=2\quad\quad\quad\dots(ii)$

Add equations $(i)$ and $(ii),$
$\begin{aligned}{}\left( {x + y} \right) + \left( {x - y} \right)& = 8 + 2\\2x &= 10\\x &= 5\end{aligned}$

Substitute $x=5$ in $(i),$
$\begin{aligned}{}\left( 5 \right) + y &= 8\\y&=8-5\\y &= 3\end{aligned}$

So, the numbers are $5$ and $3.$

Solve the following pair of linear equations by the substitution method.

$\dfrac {3x}{2}-\dfrac {5y}{3}=-2,\,\, \dfrac {x}{3}+\dfrac {y}{2}=\dfrac {13}{6}$

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

Explanation

Given Pair of equations are:

$\frac{3x}{2}-\frac{5y}{3} = -2\Rightarrow 9x-10y=-12$ ...eq(i)

$\frac{x}{3}+\frac{y}{2} = \frac{13}{6}\Rightarrow 2x+3y=13$ ...eq(ii)
Now from eq2 will get

$2x+3y = 13$
Now using substitution method

$y = \frac{13-2x}{3}$
Now substituting y in eq (i) will get

$\Rightarrow 9x-\frac{10\left ( 13-2x \right )}{3} = -12$

$\Rightarrow 27x-130+20x= -36$

$\Rightarrow 47x = -36+130$

$\Rightarrow 47x = 94$

$\Rightarrow x = 2$
Now substituting x in eq (ii) will get

$\Rightarrow 2\times 2 +3y = 13$

$\Rightarrow 4 +3y = 13$

$\Rightarrow 3y = 13-4$

$\Rightarrow 3y = 9$

$\Rightarrow y = 3$

$\Rightarrow x = 2 , y = 3$

CBSE Questions for Class 12 Commerce Applied Mathematics Linear Equations Quiz 8 - 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 8 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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