MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 14

$2a = b$ , the pair of equations

$x + 2y = 2a - 6b, ax + by = 2a^{2} - 3b^{2}$ possess _____ solution(s)

0%
No
0%
Only one
0%
Only two
0%
An infinite number of slutions

Explanation

Given

$2a=b\Rightarrow a=\dfrac{b}{2}$

$x+2y=2a-6b$ [first given equation]

$\Rightarrow x+2y=b-6b=-5b$ ........eq(i)

$ax+by=2a^{2}-3b^{2}$ [second given equation]

$\Rightarrow \dfrac{b}{2}x+by=\dfrac{2b^{2}}{4}-3b^{2}$

$\Rightarrow \dfrac{bx}{2}+by=\dfrac{-5b^{2}}{2}$

$\Rightarrow bx+2by=-5b^{2}$ ..........eq(ii)

From (i) and (ii), Both represents same equations.

So there exits infinite number of solutions.

If the sum of

$2$ real numbers is

$20$ and their difference is

$6$ , find the value of their product.

0%
$51$
0%
$64$
0%
$75$
0%
$84$
0%
$91$

Explanation

Let the two numbers are

$x$ and

$y$

Then according to the question,

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

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

Add

$(1)$ and

$(2)$ , we get

$\Rightarrow 2x=26$

$\Rightarrow x=\dfrac{26}{2}=13$

Put the value of

$x$ in

$(1)$

$\Rightarrow 13+y=20$

$\Rightarrow y=20-13=7$

Product of

$x$ and

$y=$

$x\times y=13\times 7=91$

If the system of linear equations

$\displaystyle\begin{cases}\dfrac12x - \dfrac23y = 7\\ax-8y = -1\end{cases}$ has no solution then the value of constant

$a$ is.

0%
$-2$
0%
$-\displaystyle\frac{1}{2}$
0%
$2$
0%
$6$

Explanation

If a system of linear equation has no solution then,

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

Therefore

$\dfrac{\dfrac{1}{2}}{a}=\dfrac{-\dfrac{2}{3}}{-8}$

$\Rightarrow \dfrac{1}{2a}=\dfrac{1}{12}$

$\Rightarrow 2a=12$

Hence

$a=6$

$3x+y=19$ , and

$x+3y=1$ . Find the value of

$2x+2y$ .

0%
$20$
0%
$18$
0%
$11$
0%
$10$
0%
None of the above

$\displaystyle \begin{cases} \dfrac{3}{4}x-\dfrac{1}{2}y=12\\ \\ kx-2y=22 \end{cases}$
If the above system of equations has no solution, find the value of

$k$ .

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

Explanation

Given:

$\dfrac 34 x - \dfrac 12 y -12=0$ and

$kx-2y-22 = 0$

$\therefore a_1 = \dfrac 34, b_1 = -\dfrac 12, c_1 = -12$

And,

$a_2 = k, b_2 = -2, c_2 = -22$

Since the given system of equations has no solution,

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

$\dfrac {3}{4k} = \dfrac {1}{4} \neq \dfrac {12}{22}$

Hence,

$k = 3$

$\left| x \right| +x+y=10$ and

$x+\left| y \right| -y=12$ then find the value of '

$x+y$ '.

0%
$\dfrac { 18 }{ 5 }$
0%
$\dfrac { 17 }{ 5 }$
0%
$\dfrac { 8 }{ 5 }$
0%
None of these

Explanation

Assuming

$y>0$ leads to

$x = 12$ from 2nd equation

as:

$x+y-y = 12$

$x= 12$

and then from 1st Equation

$|x|+x+y = 10$

$2x+y = 10$ or

$y = 14$

which is contradiction as

$y>0$

Assuming

$y \leq 0$ leads to two cases

(i) if

$x\leq 0$ then from 1st Equation

$y = 10$ and

that contradictory again

(2) if

$x>0$ then

$2x+y = 10...(1), x-2y = 12 ...(2)$

$(1)\times 2+(2)$

$4x+2y=20$

$x-2y = 12$

__________

$5x = 32$

$x = \dfrac{32}{5}$

So on subtituting the value of

$x$ in equ (2)

$\dfrac{32}{5}-2y = 12$

$2y = \dfrac{32}{5}-12$

$y = \dfrac{-14}{5}$

$\therefore x+y=\dfrac{32}{5}+\dfrac{-14}{5}$

$\therefore x+y=\dfrac{18}{5}$

Some students are divided into two groups A & B. If

$10$ students are sent from A to B, the number in each is the same. But if

$20$ students are sent from B to A, the number in A is double the number in B. Find the number of students in each group A & B.

0%
$100, 80$
0%
$80, 100$
0%
$110, 70$
0%
$70, 110$
0%
None of these

Explanation

Let the number of students in

$A$ and

$B$ be

$a \& b$ respectively.
As per the question:

$a-10 = b + 10$

$a- b = 20 ... (i)$
and

$a + 20 = 2 (b -20)$

$a - 2b = -60..... (ii)$
Since coefficient of

$a$ is same in both equations, we can subract

$(ii)$ from

$(i)$ .

$(i)-(ii) :$

$a- b -(a-2b) = 20-(-60)$

$\implies b =80$

$a =80+20 =100$

Two bus tickets from city A to B and three tickets from city A to C cost Rs. 77 but three tickets from city A to B and two tickets from city A to C cost Rs.What are the fares for cities B and C from A ?

0%
Rs. $4$ , Rs. $23$
0%
Rs. $13$ , Rs. $17$
0%
Rs. $15$ , Rs. $14$
0%
Rs. $17$ , Rs. $13$

Explanation

Let Rs.

$x$ be the fare of city B from city A and Rs.

$y$ be the fare of city C from city A.
Then,

$2x + 3y = 77$ ...(i)

and

$3x + 2y = 73$ ...(ii)
Multiplying (i) by

$3$ and (ii) by

$2$ and subtracting,

we get:

$5y = 85$ or

$y = 17.$
Putting

$y = 17$ in (i), we get:

$x = 13.$

Henry is three times as old as Truman. Two years ago, Henry was five times as old as Truman. How old is Henry now?

0%
4
0%
8
0%
12
0%
16
0%
20

A sum of

$Rs. 6.25$ is made up of

$80$ coins which are either

$10$ paise or

$5$ paise coins. The number of

$10$ paise and

$5$ paise coins are _____ and ____ respectively.

0%
$45 , 35$
0%
$42 , 38$
0%
$39, 41$
0%
$37 , 43$

Draw the graph of each of the following linear equations and answer the following question.
(i) y = x (ii) y = 2x (iii) y = -2x (iv) y = 3x (v) y = -3x
these equations are of the form y = mx, where m is a real number.
State True or False : All these graphs pass through origin.

0%
True
0%
False

Explanation

Yes, these equations

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

$y=-3x$ all pass through origin.

Hence, the statement is true.

1936154_570479_ans_77f636b592c04566a693da7f31b251f2.png

Srikant bought two cows for Rs.

$30000$ . By selling one at a loss of

$15\%$ and the other at a gain of

$19\%$ , he found that selling price of both cows is the same. Find the cost price of each ( in Rs).

0%
$10000, 20000$
0%
$15000, 15000$
0%
$17500, 12500$
0%
$17000, 12500$

Explanation

Let

$x,y$ be the

$C.P$ of the

$2$ cows

$x+y=3000-1$ (given)

One is sold at

$15$ % loss and other at gain of

$19$ % S.P of these

$2$ are equal

$x-\cfrac { 15 }{ 100 } x=y+\cfrac { 19 }{ 100 } y$

$85x=119y\Rightarrow x=\cfrac { 119 }{ 85 } y$

Using

$1$ and above result

$\cfrac { 119 }{ 85 } y+y=30000$

$204y=85\times 30000$

$y=12,500$

$x=30000-12500=17500$

Cost price of each cow is

$17500,12500$

The graph of the equations

$x + y = 5$ and

$x -y = 1$ will be.....

0%
Parallel lines
0%
Intersecting lines
0%
Lines coincide
0%
Concurrent lines

Explanation

Take few points on bothe lines and draw the graphs of both lines.

For

$x + y = 5$ :

x	0	3	5
y	5	2	0

For

$x - y = 1$ :

x	0	1	3
y	-1	0	2

From the graphs, we can say that the lines are intesecting at

$(3,2)$ .

Solve the following pairs of equations.

$4x-y-5=0$ ;

$x+y-5=0$

0%
$\left( 2,3 \right)$
0%
$\left( 3,3 \right)$
0%
$\left( 4,3 \right)$
0%
None of these

Explanation

$x+y=5$ .......eq1;

$4x-y=5$ ......eq2

Add both the equations, we will get

$x=2$

Put

$x= 2$ in eq1, we get

$y=3$

Solve the following pairs of equations.

$2x-4=0$ ;

$4x+y+4=0$

0%
$\left( 2,-12 \right)$
0%
$\left( 1,-12 \right)$
0%
$\left( 2,12 \right)$
0%
None of these

Explanation

$2x-4= 0 \implies x=2$

and

$4x+y =-4$

$4(2)+y=-4$

$8+y =-4$

$y =-12$

$\Rightarrow x=2, y = -12$

Solve the following pairs of equations.

$3x-y=0$ ;

$x-2=0$

0%
$x=2,y=6$
0%
$x=3,y=6$
0%
$x=4,y=6$
0%
None of these

Explanation

As we are give

$3x-y=0$ &

$x-2= 0$

$\therefore$

$x=2$ &, on substituting the value of

$x$ in

$3x-y=0$

$\therefore y=6$

Solve the following pairs of equations.

$x-y=0$ ;

$y+3=0$

0%
$\left( -3,-3 \right)$
0%
$\left( 3,-3 \right)$
0%
$\left( 2,-3 \right)$
0%
None of these

Explanation

Consider

$x-y=0$

$\Rightarrow x=y$

$y+3=0 \implies y =-3$

$\Rightarrow x= y =-3$

Hence solution for given equations is

$(-3,-3)$

Solve the following pairs of equations.

$2x=y+1$ ;

$x+2y-8=0$

0%
$x= 2,y=3$
0%
$x= 3,y=3$
0%
$x= 2,y=2$
0%
None of these

Explanation

$2x=y+1$ .....(ii) ;

and

$x+2y=8$

$x=8-2y$ .....(i)

Put (i) in (ii)

$16-4y=y+1$

$y=3$

$y=3$ &

$x=2$

Solve the following pairs of equations.

$x+y=5$ ;

$x-y=1$

0%
$\left( 3,2 \right)$
0%
$\left( 2,2 \right)$
0%
$\left( 4,2 \right)$
0%
None of these

Explanation

$x-y = 1$ ....eq1

$x+y=5$ .....eq2

Add & subtract both the equations, we will get

$2X=6$

$\implies x=3$ &

Put the value of x in equ 1

$y=2$

Solve the following pairs of equations.

$y=2x+1$ ;

$y+3x-6=0$

0%
$\left( 1,3 \right)$
0%
$\left( 2,3 \right)$
0%
$\left( 4,3 \right)$
0%
None of these

Explanation

Here given equations are

$y-2x= 1$ ;

$y+3x=6$

Subtract both the equations, we will get

$x=1$ .

On substituting the value of

$x$ in any given equations we get

$y=3$

Solve the following equations by substitution method.

$x + 3y = 10; 2x + y = 5$

0%
$x = 1, y = 3$
0%
$x = 2, y = 3$
0%
$x = 1, y = 1$
0%
None of these

Explanation

$x+3y=10$ .....

$(1)$ ;

$2x+y=5$ ........

$(2)$

$y=5-2x$ ......from

$(2)$

putting in

$(1)$ , we get

$\Rightarrow x+3(5-2x)=10$

$\Rightarrow x+15-6x=10$

$\Rightarrow -5x=-5$

$\Rightarrow x=1$

$1+3y=10$

$y=3$

Solve the system of equations

$65x-33y=97,$ and

$33x-65y=1$ by substitution method.

0%
$(2, 1)$
0%
$(1, 1)$
0%
$(2, 2)$
0%
None of these

Solve the following equations by substitution method.

$\dfrac{1}{x} + \dfrac{2}{y} = 9; \dfrac{2}{x} + \dfrac{1}{y} = 12 (x \neq 0, y \neq 0)$

0%
$x = \dfrac{1}{5} , y = \dfrac{1}{2}$
0%
$x = \dfrac{2}{5} , y = \dfrac{1}{2}$
0%
$x = \dfrac{1}{5} , y = \dfrac{-1}{2}$
0%
None of these

Explanation

Given equations,

$\dfrac{1}{x} + \dfrac{2}{y} = 9$ ..............(1)

$\dfrac{2}{x} + \dfrac{1}{y} = 12$ ..............(2)

$\dfrac{1}{x}=9-\dfrac{2}{y}$ .........from (1)

putting value of

$\dfrac {1}{x}$ in (2), we get

$2\left[9-\dfrac{2}{y}\right]+\dfrac{1}{y}=12$

$18-\dfrac{3}{y}=12$

$6=\dfrac{3}{y}$ ......

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

putting value of

$y$ in (1), we get

$\dfrac {1}{x}=9-4$

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

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

Solve the following equations by substitution method.

$\dfrac{3}{x} + \dfrac{1}{y} = 7; \dfrac{5}{x} - \dfrac{4}{y} =(x \neq 0, y \neq 0)$

0%
$x = \dfrac{1}{2}, y = 1$
0%
$x = \dfrac{-1}{2}, y = 1$
0%
$x = \dfrac{1}{2}, y = -1$
0%
None of these

Explanation

Given equations are,

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

$\dfrac{5}{x}-\dfrac{4}{y}=6$

$\dfrac{1}{x}=\dfrac{1}{3}\left(7-\dfrac{1}{y}\right)$ .................from (1)

Putting value of

$\dfrac {1}{x}$ in (1), we get

$5\left[\dfrac{1}{3}(7-\dfrac{1}{y})\right]-\dfrac{4}{y}=6$

$\dfrac{35}{3}-\dfrac{5}{3y}-\dfrac{4}{y}=6$

$\dfrac{17}{3}=\dfrac{5+12}{3y}$

$\boxed {y=1}$

Putting value of

$y$ in (1)

$\dfrac{1}{x}=\dfrac{1}{3}\times 6\Rightarrow \dfrac {1}{x}=2=\boxed{x=\dfrac{1}{2}}$

Solve following system of equations by elimination method.

$x +2y=7, x-2y=1$

0%
$\left ( 4, \dfrac{3}{2} \right )$
0%
$\left ( 4, \dfrac{2}{2} \right )$
0%
$\left ( 3, \dfrac{3}{2} \right )$
0%
None of these

Explanation

Given equations are

$x+2y=7$ .......(i)

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

Subtract equation (ii) from (i), we get

$2x=8$

$\Rightarrow x=4$

put value of

$x$ in equation (ii)

$4-1=2y$

$3=2y$

$\therefore y=\dfrac{3}{2}$

$\therefore x=4 , \ y=\dfrac{3}{2}$

Solve each of the following system of equations by substitution method.

$13x +11y=70, 11x+13y=74$

0%
$(2, 4)$
0%
$(3, 4)$
0%
$(2, 2)$
0%
None of these

Explanation

Since,

$13x+11y=70$ ...eq1

$11x+13y=74$ ......eq2

$x=\dfrac{70-11y}{13}$ , put this value in eq2, we get

$11\left[\dfrac{70-11y}{13}\right]+13y=74$

$770-121y+169y=74\times 13$

$\Rightarrow y=4$

and

$x=\dfrac{70-11(4)}{13}$

$x=2$

Solve the following equations by substitution method.

$5x + 3y = 21; 2x - y = 4$

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

Explanation

$5x+3y=21$ .........(1)

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

$y=2x-4$

$5x+3(2x-4)=21$ .........putting value of

$y$ in (1)

$\Rightarrow 5x+6x-12=21$

$\Rightarrow 11x=33$ .......................

$\boxed{x=3}$

Putting value of

$x$ in (2)

$6-y=4$ ..........................

$\boxed{y=2}$

Solve the following equations by substitution method.

$2x + y = 1; 3x - 4y = 18$

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

Explanation

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

$3x-4y=18$ ........(2)

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

putting value of

$y$ in (1)

$\Rightarrow 3x-4(1-2x)=18$

$\Rightarrow 3x-4+8x=18$

$\Rightarrow 11x=22$

$\Rightarrow x=2$

Using value of

$x$

$\Rightarrow 4+y=1$

$\Rightarrow y=-3$

Solve the following system of equations by substitution method.

$\dfrac{15}{x}+\dfrac{2}{y}=17, \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{36}{5}, x\neq 0, y\neq 0$

0%
$\left ( 5, \dfrac{1}{7} \right )$
0%
$\left ( 2, \dfrac{1}{7} \right )$
0%
$\left ( 5, \dfrac{2}{7} \right )$
0%
None of these

Explanation

Given equations are

$\dfrac{15}{x}+\dfrac{2}{y}=17,$

$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{36}{5}$

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

$\therefore \ 15a+2b=17$ ......(i)

and

$a+b=\dfrac{36}{5}$

$\Rightarrow a=\dfrac{36}{5}-b$ ....(ii)

Putting equation (ii) in (i) we get,

$15\left[\dfrac{36}{5}-b\right]+2b=17$

$\Rightarrow 108-13b=17$

$\Rightarrow b=7$

$\therefore \ y=\dfrac{1}{7}$

Putting value of

$b$ in equation (i), we get,

$15a+(2\times7)=17$

$\Rightarrow a=\dfrac{1}{5}$

$\therefore \ x=5$

Hence,

$\left(5,\dfrac17\right)$ is the solution of the given system of equations.

The sum of two numbers is

$25$ and their difference is

$7$ , then the numbers are.

0%
$20$ and $5$
0%
$18$ and $7$
0%
$15$ and $10$
0%
$9$ and $16$

Explanation

let the two numbers be

$A$ and

$B$

so it is given that

$A + B = 25$ ...(1)

$A - B = 7$ ...(2)

Adding the equations we get

$A + B + A - B = 25 +7$

$2A = 32$

$A = 16$

Using the above value of A in (1)

$16 + B = 25$

$=>B = 9$

Thus

$A =16$ and

$B =9$

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 14 - 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 14 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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