MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 2

Given that

$3x-y=-11$ and

$4x+y=4$ . Calculate the value of

$x$ .

0%
$-7$
0%
$-1$
0%
$8$
0%
$15$

Explanation

$3x-y=-11$ ...........

$(i)$

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

$(ii)$

Adding

$(i)$ and

$(ii),$ we get

$(3x-y)+(4x+y)=-11+4$

$\Rightarrow 3x-y+4x+y=-11+4$

$\Rightarrow 3x+4x=-11+4$

$\Rightarrow 7x=-7$

$\Rightarrow x=\dfrac{-7}{7}=-1$

Given that

$4x-3y=6$ and

$x-3y=-3$ , the value of

$y$ is

0%
$-3$
0%
$-2$
0%
$\dfrac { 4 }{ 3 }$
0%
$2$

Explanation

$4x-3y=6$ ...........

$(i)$

$x-3y=-3$ ...........

$(ii)$

Multiply

$(ii)$ by

$4,$ we get

$x\times 4-3y\times 4=-3\times 4$

$\Rightarrow 4x-12y=-12$ ...........

$(iii)$

Subtract

$(iii)$ with

$(i)$ , we get

$(4x-3y)-(4x-12y)=6-(-12)$

$\Rightarrow 4x-3y-4x+12y=6+12$

$\Rightarrow -3y+12y=18$

$\Rightarrow 9y=18$

$\Rightarrow y=\dfrac{18}{9}=2$

The sum of two numbers is equal to

$20$ and their difference is

$2.5$ . Find the ratio of the numbers

0%
$9 : 7$
0%
$7 : 9$
0%
$3 : 5$
0%
$2 : 7$

Explanation

Let the numbers be

$x$ and

$y$ , then

$x+y = 20...(1)$

$x-y = 2.5...(2)$

$(1)+(2)\Rightarrow 2x = 22.5$

$\Rightarrow x = 11.25$

$\Rightarrow y = 8.75$

$\therefore x:y = 11.25:8.75$

$\Rightarrow x:y = 9:7$

Given that

$2y=-6$ and

$x-3y=13$ , the value of

$x=$ ________

0%
$\dfrac { 13 }{ 9 }$
0%
$4$
0%
$19$
0%
$22$

Explanation

Given

$2y=-6$

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

$\Rightarrow y=-3$ ............

$(i)$

$x-3y=13$ .............

$(ii)$ (given )

Put the value of

$y$ in

$(ii)$ , we get

$x-3(-3)=13$

$\Rightarrow x+9=13$

$\Rightarrow x=13-9=4$

Then

$x=4$

Fill in the blank: _______ system of linear equation has no solution.

0%
Inconsistent
0%
Consistent
0%
Dependent
0%
Empty

The ratio of income of two persons is

$9 : 7$ and the ratio of their expenditure is

$4 : 3$ . If each of them manages to save Rs.

$2000$ per month, find their monthly income.

0%
$Rs. 18,000 , Rs. 14,000$
0%
$Rs. 1,000 , Rs. 14,000$
0%
$Rs. 18,000 , Rs. 1,000$
0%
None of these

Explanation

Let their salaries be

$9x$ and

$7x$ .

Let their expenditure be

$4y$ and

$3y$ .

According to the question,

$9x-4y=2000$

$—(1)$

$7x-3y=2000$

$—(2)$

From

$(1)$

$x=\dfrac{2000+4y}{9}$

$—(3)$

On putting

$x$ in

$(2)$ , we get

$7\times \dfrac{(2000+4y)}{9}-3y=2000$

$\dfrac{(14000+28y)}{9}-3y=2000$

$\dfrac{14000+28y-27y}{9}=2000$

$14000+y=18000$

$y=4000$

Now, put

$y$ in (3)

$x=\dfrac{2000+4\times 4000}{9}$

$x=\dfrac{2000+16000}{9}$

$x=\dfrac{18000}{9}=2000$

So,

Salary of first person

$=9\times 2000=Rs.\ 18000$

Salary of second person

$=7\times 2000=Rs.\ 14000$

Hence, this is the answer.

$(a+b,1)=(5,a-b)$ , then the value of

$2a+3b=$ ?

0%
$5$
0%
$12$
0%
$8$
0%
$6$

Explanation

According to question

$a + b = 5$

$1 = a - b$

Adding above equations, we get

$2a = 6 \Rightarrow a = 3$

Put

$a=3$ in

$a+b = 5 \Rightarrow 3+b = 5 \Rightarrow b = 2$

$a=3$ and

$b=2$

So value of

$2a + 3b = 6 + 6 = 12$

To eliminate x from equations

$x + y + 3x = 12 \rightarrow$ ....... (1) &

$8x + 3y = 17 \rightarrow$ ..... (2), equation (2) is to be multiplied by ..............( from the given options)

0%
8
0%
$- \dfrac{1}{2}$
0%
3
0%
None of the above

Explanation

The given equations are

$x + y + 3x = 12$ ........ (1)
and

$8x + 3y = 17$ ....... (2)
Equation (i) can be further simplified as

$4x + y = 12$
So, to eliminate x, equation (2) should be multiplied by

$- \dfrac{1}{2}$ .

Determine graphically the vertices of the triangle, the equations of whose side are give below:

$y=x,y =0$ and

$3x+3y=10$

0%
$(0,0) \, \left(\dfrac{10}{3},0\right)\, \left(\dfrac{5}{3}, \dfrac{5}{3}\right)$
0%
$(0,0) \, \left(\dfrac{17}{3},0\right)\, \left(\dfrac{5}{3}, \dfrac{5}{3}\right)$
0%
$(0,0) \, \left(\dfrac{14}{3},0\right)\, \left(\dfrac{5}{3}, \dfrac{5}{3}\right)$
0%
$(0,0) \, \left(\dfrac{10}{3},0\right)\, \left(\dfrac{5}{3}, \dfrac{8}{3}\right)$

Explanation

As shown in the graph , to find vertices of required triangle,

We can find intersection of lines

$3x+3y=10$ and

$y=x$ which is

$(\dfrac{5}{3},\frac{5}{3})$

$y=0$ and

$3x+3y =10$ which is

$(\dfrac{10}{3},0)$

$y=x$ and

$y=0$ which is origin

$(0,0)$

1394214_986202_ans_d0d152ea8f0a4aec8c7a3906bcf6d2f0.png

The solution of

$8x+5y=9$ ,

$3x+2y=4$ lie on

$x+y=k$ , then

$k=$

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

Explanation

$8x+5y=9 ---(1) \times 2$

$3x+2y=4 ---(2) \times 5$

subtracting both

$(16x+10y) - (15x+10y)=18-20$

$x=-2$

Putting

$x=-2$ in

$(2)$

$3x+2y=4$

$\Rightarrow 3(-2)+2y=4$

$\Rightarrow 2y=10$

$\Rightarrow y=5$

Since point

$(-2,5)$ lies on

$x+y=k$

So,

$-2+5=k$

$\Rightarrow k=3$

The solution of

$\sqrt{2}x+\sqrt{3}y=0$ and

$\sqrt{3}x+\sqrt{8}y=0$ is

0%
$(1, 2)$
0%
$(\sqrt{2}, \sqrt{8})$
0%
$(0, 0)$
0%
$(\sqrt{3}, \sqrt{2})$

Explanation

$\sqrt 2 x + \sqrt 3 y = 0 - - - \,\,\,\,\,\left( 1 \right) \times \sqrt 3$

$\sqrt 3 x + \sqrt 8 y = 0 - - - \,\,\,\,\,\left( 2 \right) \times \sqrt 2$

subtracting both

$\left( {\sqrt 6 x + 3y} \right) - \left( {\sqrt 6 x + 4y} \right) = 0 - 0$

$- y = 0 \Rightarrow y = 0$

From (1)

$\sqrt 2 x + \sqrt 3 y = 0$

$\Rightarrow \sqrt 2 x + 0 = 0$

$\Rightarrow x = 0$

So, point of intersection is

$(0,0)$

$7$ pens and

$3$ books cost

$Rs 1395$ , while

$5$ pens and

$4$ books cost

$Rs 1665$ . Find the cost of a book

0%
$360$
0%
$45360$
0%
$35350$
0%
$none\ of\ these$

Explanation

Let the cost of

$1$ pen be

$x$

and the cost of

$1$ book be

$y$

Then

$7x+3y=1395$ -----

$(1) \times 4$

$5x+4y=1665--(2) \times 3$

$4\times (1)$ --------

$3 \times (2)$

$(28x+12y)-(15x+12y)=5580-4995$

$13x=585$

$x=45$

Putting

$x=45$ in (1)

$7 \times 45+3y=1395$

$\Rightarrow 315+3y=1395$

$\Rightarrow 3y=1395-315$

$\Rightarrow 3y=1080$

$\Rightarrow y=360$

Therefore Cost of one pen is Rs

$45$ and cost of one book is Rs

$360$

The cost of

$4$ pens and

$4$ pencil boxes is Rs

$100$ . Three times the cost of a pen is Rs

$15$ more than the cost of a pencil box. The cost of a pen and a pencil box respectively are

0%
Rs $15$ and Rs $8$
0%
Rs $10$ and Rs $15$
0%
Rs $12$ and Rs $10$
0%
Rs $16$ and Rs $12$

Explanation

Let the cost of a pen and a pencil be

$x$ and

$y$ respectively.

$4x+4y=100$

$x+y=25$ ...(1)

$3x=y+15$

$3x-y=15$ ...(2)
Adding (1) and (2), we get

$4x=40$

$x=10$
Substitute

$x=10$ in equation (1) to get

$y=15$ .

The sum of two numbers is

$2$ and their difference is

$1$ . Find the numbers.

0%
$2,0$
0%
$1.5,0.5$
0%
$1,1$
0%
$-0.5,2.5$

Explanation

Let two numbers be

$x$ and

$y$

According to question sum of

$x$ and

$y$ is

$2$ and difference is

$1$

$\therefore x+y=2......(1)$ and

$x-y=1......(2)$

Now adding equation

$(1)$ and

$(2)$

$x+y=2$

$\underline{x-y=1}$

$2x\quad=3$

$\Rightarrow x=\dfrac{3}{2}$

$\Rightarrow x=1.5$

Now substitute the value of

$x$ in equation

$(1),$

$1.5+y=2$

$\Rightarrow y=2-1.5$

$\Rightarrow y=0.5$

Solve:

$4x + \frac{6}{y} = 15$

$6x - \frac{8}{y} = 14$

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

Explanation

Given:

$4x+\frac{6}{y}=15......(1)$

$6x-\frac{8}{y}=14......(2)$

Multiplying equation (1) by 4 and equation (2) by 3 and then adding them,

$16x+\frac{24}{y}=60$

$\underline{18x-\frac{24}{y}=42}$

$34x\quad=102\Rightarrow x=3$

Now putting the value of x in equation (1)

$4\times3+\frac{6}{y}=15\Rightarrow\frac{6}{y}=15-12\Rightarrow\frac{6}{y}=3\Rightarrow y=\dfrac{6}{3}=2$

To eliminate x, from

$3x+y=7$ and

$-x+2y=2$ second equation is multiplied by ______

0%
$1$
0%
$3$
0%
$2$
0%
$-1$

Explanation

$by\>multiplying\>second\>equation\>with\>3\>,\>we\>get\\-3x+6y=6\\now\>by\>adding\>it\>with\>equation\\3x+y=7,\>values\>of\>unknowns\>can\>be\>calculated$

Solve for

$x$ and

$y$ by using method of substitution :

$0.2x+0.3y=1.3$ ;

$0.4x+0.5y=2.3$

0%
$-2,-3$
0%
$2,-3$
0%
$2,3$
0%
$-2,3$

Explanation

$0.2x+0.3y=1.3 \Rightarrow 2x+3y=13$

$\Rightarrow x=\dfrac{13-3y}{2}$

and

$0.4x+0.5y=2.3\Rightarrow 4x+5y=23$

$\Rightarrow 4\left ( \dfrac{13-3y}{2} \right )+5y=23$

$\Rightarrow 26-6y+5y=23$

$\Rightarrow 3=y$

$\therefore x=\dfrac{13-3\times 3}{2}=2$

Solve

$5x-4y+8=0$

$7x+6y-9=0$ .

0%

$y = -\cfrac{101}{58}$
0%

$y = \cfrac{101}{58}$
0%

$y = \cfrac{101}{58}$
0%
None of these

Explanation

$5x-4y+8 = 0$ ...

$\times 7$

$7x+6y-9 = 0$ ....

$\times 5$

$35x-28y+56 = 0$

$35x+30y-45 = 0$

$58y = 101$

$y = \cfrac{101}{58}$

$x = -\cfrac{6}{29}$

Solve :

$ax + by = a - b$

$bx - ay = a + b$

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

Explanation

$ax+by=a-b$ ............ (1)

$bx-ay=a+b$ .............. (2)

Multiply equation (1) with

$a$ and equation (2) with

$b$ ,

$a^{2}x+aby=a^{2}-ab$ ................. (3)

$b^{2}x-aby=ab+b^{2}$ ................. (4)

Adding equation (3) and (4), we get

$a^{2}x+b^{2}x=a^{2}+b^{2}$

$\left ( a^{2}+b^{2} \right )x=a^{2}+b^{2}$

$\Rightarrow x=1$

Substitute

$x=1$ in equation (1),

$a(1)+by=a-b$

$by=-b$

$y=-\cfrac{b}{b}=-1$

Solve

$0.2x+0.3y=1.3$

$0.4x+0.5y=2.3$

0%
$x=-2$ and $y=3$
0%
$x=3$ and $y=3$
0%
$x=2$ and $y=3$
0%
None of these

Explanation

Let

$0.2x+0.3y=1.3$ ....

$\left(1\right)$

$0.4x+0.5y=2.3$ ....

$\left(2\right)$

Multiply eqn

$\left(1\right)$ by

$10$ we get

$2x+3y=13$ ....

$\left(3\right)$

Multiply eqn

$\left(2\right)$ by

$10$ we get

$4x+5y=23$ ....

$\left(4\right)$

Solving equations

$\left(3\right)$ and

$\left(4\right)$

$5\times \left(3\right)\Rightarrow 10x+15y=65$

$3\times \left(4\right)\Rightarrow 12x+15y=69$

Subtracting we get

$12x+15y-10x-15y=69-65=4$

$2x=4$ or

$x=2$

Substituting

$x=2$ in eqn

$\left(3\right)$ we get

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

$3y=13-4=9$ or

$y=3$

Hence

$x=2,y=3$

Given two straight lines

$x + y - 7 = 0$ and

$x - y + 3 = 0$ . Find point of intersection of them.

0%
$(2,5)$
0%
$(-2,-5)$
0%
$(2,-5)$
0%
$(-2,5)$

Explanation

$x+y-7=0.......eq1\\x-y+3=0......eq2$

Add both the equations

Thus we get,

$2x-4=0\\2x=4\\x=\dfrac{4}{2}\\x=2\\2+y-7=0\\y=7-2=5\\(x,y)=(2,5)$

The graph of the linear equation

$2x + 3y = 6$ is a line which meets the

$x-$ axis at the point

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

Explanation

$2x + 3y =6$ will meet the

$x-$ axis at

$y=0$ .
For

$y=0$ ,

$x =3$ .

$\therefore$ The point is

$(3,0)$

$2x + y = 23$ and

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

$5y - 2x$

0%
$31$
0%
$22$
0%
$65$
0%
$10$

Explanation

The given equations are

$2x+y=23$ ...(i)

$4x-y=19$ ...(ii)

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

$6x=42\Rightarrow x=7$

On putting

$x=7$ in (i), we get

$14+y=23\Rightarrow y=9$

Now,

$5y-2x=5\times9-2\times7=31$

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 respectively are

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

Explanation

$\textbf{Step1: Convert given data into equations .}$

$\text{Let, the present age of father = x years}$

$\text{Let, the present age of son = y years}$

$\text{Given that fathers Age is $$6$$ time of sons age}$

$\Rightarrow x=6y \ldots (1)$

$\text{After four years,}$

$\text{Age of the father will be four times of his son's age}$

$\Rightarrow x+4=4 (y+4)$

$\Rightarrow x+4=4y+16$

$\Rightarrow x=4y+16-4$

$\Rightarrow x=4y+12 \ldots(2)$

$\textbf{Step2: Solving the equations.}$

$\text{Substitute equation $$(1)$$ in equation $$(2)$$}$

$\Rightarrow 6y= 4y+12$

$\Rightarrow 2y=12$

$\Rightarrow y=6$

$\text{Substitute $$y=6$$ in equation (1)}$

$\Rightarrow x=36$

$\textbf{Hence, the solution of the father and his son ages are $$x=36, y=6$$.}$

The graph of the linear equation

$2x + 3y = 6$ intersects the

$y-$ axis at the point

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

Explanation

$2x + 3y =6$
Since , the line intersects at

$y-$ axis.
Put

$x =0$ in the given equation

$\Rightarrow y =2$
Hence, the line intersects

$y-$ axis at

$(0,2)$

Determine the vertices of the triangle formed by the lines

$y = x, 3y = x$ and

$x + y = 8$ .

0%
$(0, 0), (0, 4), (6, 2)$
0%
$(0, 0), (4, 4), (0, 2)$
0%
$(0, 0), (7, 7), (5, 2)$
0%
$(0, 0), (4, 4), (6, 2)$

Explanation

The vertices of the triangle will be the point of intersection of given lines taken two at a time. So, we need to find the points of intersection of the lines:

A. Taking

$y = x$

$\dots (1)$

$3y = x$

$\dots (2)$

Let, the point of intersection of above two lines is

$A$ .

Subtract

$(1)$ from

$(2)$ :

$2y = 0$

$y = 0$

Put this value of

$y$ in equation

$(1)$ :

$y=x=0$

Hence, the coordinates of A are

$(0,0)$ .

B. Taking

$y = x$

$\dots (2)$

$x+y = 8$

$\dots (3)$

Let, the point of intersection of above two lines is

$B$ .

Subtract

$(2)$ from

$(3)$ :

$x+y-y = 8-x$

$2x = 8$

$x= 4$

Put this value of

$x$ in equation

$(3)$ :

$y=x=4$

Hence, the coordinates of B are

$(4,4)$ .

C. Taking

$x+y = 8$

$\dots (3)$

$x = 3y$

$\dots (1)$

Let, the point of intersection of above two lines is

$C$ .

Subtract equation

$(1)$ from

$(3)$ :

$x+y-x = 8-3y$

$y=8-3y$

$4y = 8$

$y=2$

Put this value of

$y$ in equation

$(1)$ :

$x=3(2) = 6$

Hence, the coordinates of C are

$(6,2)$ .

The coordinates of the triangle are

$(0,0), (4,4)$ and

$(6,2).$

Are the following pair of linear equations consistent?

$-3x-4y=12$ ,

$4y+3x=12$

0%
No
0%
Yes
0%
Data Insufficient
0%
None of these

Explanation

The given pair of linear equations can be written as

$-3x-4y-12=0$ and

$3x+4y-12=0$
Here

$a_1=-3,b_1=-4,c_1=-12$
and

$a_2=3,b_2=4,c_2=-12$

$\therefore \frac{a_1}{a_2}=\frac{-3}{3}=-1$

$\frac{b_1}{b_2}=\frac{-4}{4}=-1$

$\frac{c_1}{c_2}=\frac{-12}{-12}=1$

$\Rightarrow \frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$
Hence the given pair of linear equations is inconsistent.

The values of x and y satisfying the two equations

$32x + 33y = 31$ ,

$33x + 32y = 34$ respectively will be

0%
$- 1, 2$
0%
$2, - 1$
0%
$0, 0$
0%
$2, 3$

Explanation

Multiplying

$32x+33y=31$ with

$33$ , we get

$1056x + 1089y = 1023$ ---- (1)

Multiplying

$33x+32y=34$ with

$32$ , we get

$1056x + 1024y = 1088$ ---- (2)

Subtracting (2) from (1),

$1056x + 1089y - 1056x - 1024y = 1023 - 1088$

$65y = - 65$

$y = -1$

Substituting

$y = - 1$ in

$32x+33y=31$ , we get $$ 32x - 33 = 31

$32x = 64$

$x = 2$

$x=a, y=b$ is the solution of the equations

$x-y=2$ and

$x+y=4$ , then the values of

$a$ and

$b$ are, respectively:

0%
$3$ and $5$
0%
$5$ and $3$
0%
$3$ and $1$
0%
$-1$ and $-3$

Explanation

Given,

$x-y=2$ and

$x+y=4$ .

Adding both the equations, we get

$2x=6$

$\Rightarrow x=3$

Putting the value of

$x$ in the first equation, we get

$3-y=2$

$\Rightarrow y=1$

As stated in the question,

$x=a$ and

$y=b$ .

Thus,

$a=3$ and

$b=1$

The solution of

$217x+131y=913$

$131x+217y=827$ is

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

Explanation

Let

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

and

$131x+217y=827$ ......(2)

Here, according to the given passage

$a=217, b=131, c=913$ and

$d=827$

$\therefore 2x= \left[ \dfrac { c+d }{ a+b } +\dfrac { c-d }{ a-b } \right]$

$=\dfrac { 1740 }{ 348 } +\dfrac { 86 }{ 86 } =\dfrac { 2088 }{ 348 }$

$\therefore x=\dfrac { 2088 }{ 348\times 2 } =3$

and

$2y= \left[ \dfrac { c+d }{ a+b } -\dfrac { c-d }{ a-b } \right]$

$=\dfrac { 1740 }{ 348 } -1=\dfrac { 1392 }{ 348 }$

$\therefore y=\dfrac { 1392 }{ 348\times 2 } =2$

$\therefore x=3, y=2$

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 2 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 2 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.