MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 15

Two lines

$x+ 2y + 7 = 0$ and

$2x + ky + 18 = 0$ do not intersect each other. Find the value of

$k.$

0%
$3$
0%
$2$
0%
$1$
0%
$4$

Explanation

$x + 2y + 7 = 0$ and

$2x + ky + 18 = 0$ do not intersect each other, this means that the pair of equations is inconsistent.

On comparing

$x + 2y + 7 = 0$ with

$a_1x+b_1y+c_1=0$ and

$2x+ky+18=0$ with

$a_2x+b_2y+c_2=0$ we get,

$a_1=1, b_1=2, c_1=7, a_2=2,b_2=k, c_2=18$

Thus, we have

$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$

$\Rightarrow \dfrac{1}{2} = \dfrac{2}{k}$

$\Rightarrow k = 4$

Hence, if

$k=4$ then the pair of equations will be inconsistent.

For what value of

$'k'$ the system of equation

$kx + 2y = 5$ and

$3x + y = 1$ has no solution?

0%
$k = 3$
0%
$k = 6$
0%
$k \neq 6$
0%
$k = 4$

Explanation

$kx + 2y = 5 \dots (1)$

$3x + y = 1\ \dots (2)$

First multiply the second equation by

$2$

We get:

$6x + 2y = 2$

Now subtracting

$(2)$ from

$(1)$ , we get:

$kx + 2y - 6x - 2y = 5 -2$

$(k-6)x = 3$

$x =\dfrac{3}{k-6}$

So, here if we put any value of

$k$ , then we will get a corresponding value of

$x$ but if we put

$k= 6$ , then we will not get any value of

$x$

So, for

$k = 6$ this equation has no solution.

$8x-9y=20$ and

$7x-10y=9$ , then what is

$2x-y$ equal to?

0%
$10$
0%
$11$
0%
$12$
0%
$13$

Explanation

Given:

$8x-9y=20$ ...........

$(1)$

$7x-10y=9$ .........

$(2)$

Multiplying

$(1)$ by

$7$ and

$(2)$ by

$8$ and then subtract

$(2)$ from

$(1)$ , we get

$(56x-63y)-(56x-80y)=140-72$

$\Rightarrow 17y=68$

$\Rightarrow y=4$

Now, put

$y=4$ in (1)

$8x-9\times 4=20$

$\implies 8x=20+36=56$

$\implies x=7$

Now

$2x-y=2\times 7-4=10$

Sahil has Rs.12 less than three times what Sohan has. If both have a total of Rs.88, how many rupees each must have got?

0%
Sohan Rs.25 , Sahil-Rs.63
0%
Sohan Rs.25 , Sahil-Rs.33
0%
Sohan Rs.55 , Sahil-Rs.63
0%
Sohan Rs.25 , Sahil-Rs.55

Explanation

Let Sahil have rupees

$x$ and Sohan has rupees

$y$ .

Sahil has Rs.12 less than 3 times Sohan

$\implies x=3y-12$

$\implies x-3y=-12 ---eqn(1)$

Total both have Rs.88

$x+y=88 --- eqn(2)$

Substracting equation 1 from 2 we get

$y=25$

Then substituting in any one of the equation we get

$x=63$

Solve the following puzzles using the equations:

The present age of Beena is 2 years more than Reeta.The present age of Teena is 3 years more than Beena.If the sum of the ages of Reeta , Beena , and Teena is 79 years, find the present age of all the three of them.

0%
Reeta 24 years , Beena 26 years , Teena 29 years.
0%
Reeta 24 years , Beena 21 years , Teena 29 years.
0%
Reeta 24 years , Beena 29 years , Teena 29 years.
0%
Reeta 24 years , Beena 26 years , Teena 30 years.

Explanation

Let Beena's present age be

$b$ , Reeta's present age be

$r$ and Teena's present age be

$t$

Present age of Beena is 2 years more than of Reeta

$\implies b=r+2$

$\implies b-r=2 --eqn(1)$

Present age of Teena is 3years more than Beena

$\implies t=b+3$

$\implies t-b=3 --eqn(2)$

Sum of all the present ages is 79

$\implies b+t+r=79--eqn(3)$

Adding Equation 1 and 2 we get

$t-r=5 --eqn(4)$

Adding equation 3 and 4 we get

$b+2t=84--eqn(5)$

Adding equation 2 and 5 we get

$3t=87$

$t=29$

Thus substituting

$t$ in equation 2 and equation 4 we will get

$b=26$

and

$r=24$

Solve the following puzzles using the equations:

In a village , the number of women is 89 more than the number of men.The number of children is 400 more than the number of men.If the total population of the village isFind the number of men , women and children.

0%
Men-1500 , Women 1589 , Children 1900.
0%
Men-1500 , Women 1600 , Children 1900.
0%
Men-1500 , Women 1589 , Children 1200.
0%
Men-1400 , Women 1589 , Children 1900.

Solve for

$x$ and

$y$ in the following question:

$\displaystyle \frac{2}{x + 2y} + \frac{1}{2x - y} + \frac{5}{9} = 0$ ,

$\displaystyle \frac{9}{x + 2y} + \frac{6}{2x - y} + 4 = 0$

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

Explanation

$\dfrac{2}{x+2y} + \dfrac{1}{2x-y} + \dfrac{5}{9} = 0......(1)$

$\dfrac{9}{x+2y} + \dfrac{6}{2x-y} + 4 = 0......(2)$

Let

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

$\dfrac{1}{2x-y} = b$

Substituting the assumed values in equations

$1$ and

$2$ , we get:

$2a+b=\dfrac{-5}{9}$ or

$18a+9b=-5......(3)$

and

$9a+6b=-4......(4)$

Solving equations

$(3)$ and

$(4)$ using elimination method:

Multiplying equation

$(4)$ by

$2$ and subtracting

$(3)$ from it, we get

$3b=-3$ or

$b=-1$

$\implies 2x-y = -1 ......(5)$

Substituting

$b=-1$ in

$(4)$

$9a-6 = -4$

$a = \dfrac{2}{9}$

$\Rightarrow x+2y = \dfrac{9}{2}......(6)$

Solving equations

$(5)$ and

$(6)$ using elimination method:

Multiplying equation

$(6)$ by

$2$ and subtracting

$(5)$ from it, we get

$2x+4y-2x+y=9+1$

$\therefore y = 2$

Substituting in

$(5)$ , we get

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

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

When a bucket is half full, the weight of the bucket and the water is

$10\text{ kg}$ . When the bucket is two-thirds full, the total weight is

$11\text{ kg}$ . What is the total weight, in kg, when the bucket is completely full?

0%
$12 \text{ kg}$
0%
$12\displaystyle\frac{1}{2} \text{ kg}$
0%
$12\displaystyle\frac{2}{3}\text{ kg}$
0%
$13\text{ kg}$

Explanation

Let the weight of the empty bucket be

$B$ kg and total weight of water it can accommodate be

$W$ kg.

$\therefore B + \cfrac{W}{2} = 10 \ \ ...(1)$

Also,

$B + \cfrac{2W}{3} = 11 \ \ ...(2)$

Subtracting (1) from (2), we have

$\cfrac{2W}{3} - \cfrac{W}{2} = 1$

$\therefore \cfrac{4W - 3W}{6} = 1$ or

$W = 6$

Substituting

$W=6$ in (1)

$\implies B+\dfrac{6}{2}=10$

$\therefore B = 7$

Total weight when completely filled

$= B + W = 13$ kg

Let

$f(x)$ be a quadratic polynomial with

$f(2)=10$ and

$f(-2)=-2$ . Then the coefficient of x in

$f(x)$ is

$?$

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Let the quadratic equation be

$f(x)=x^{2}+ax+b$

Given

$f(2)=10$ and

$f(-2)=-2$

$\Rightarrow 4+2a+b=10$ and

$4-2a+b=-2$

$\Rightarrow 2a+b=6$ and

$2a-b=6$

on solving both equations , we get

$a=3$ and

$b=0$

Therefore, the coefficient of

$x$ in

$f(x)$ is

$3$ .

Choose the correct answer from the alternatives given.
If

$\displaystyle \frac{x \, + \, y \, - \, 8}{2} \, = \, \frac{x \, + \, 2y \, - \, 14}{3} \, = \, \frac{3x \, + \, y \, - \, 12}{11}$ then find the values of x and y, respectively.

0%
$2,6$
0%
$4,8$
0%
$3,5$
0%
$4,5$

Explanation

Given that

$\displaystyle \frac{x \, + \, y \, - \, 8}{2} \, = \, \frac{x \, + \, 2y \, - \, 14}{3}$

$3x + 3y - 24 = 2x + 4y - 28$

$x - y = -4$ ....... (1)
and

$\displaystyle \frac{x \, + \, y \, - \, 8}{2} \, = \, \frac{3x \, + \, y \, - \, 12}{11}$

$11x + 11y - 88 = 6x + 2y - 24$

$5x + 9y = 64$ ....... (2)

Multiply (1)by 9

$9x-9y=-36$ ..................(3)

adding (2) and (3)

$14x=28$

$x=\dfrac{28}{14}=2$

put value of x in (1)

$2-y=-4$

$-y=-4-2$

$-y=-6$

$y=6$

we get

$x = 2, y = 6$

Solve the following equations:

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

$\dfrac {2}{x} + \dfrac {5}{y} = \dfrac {5}{6}$ .

0%
$x=6; y=10$
0%
$x=5; y=6$
0%
$x=4; y=15$
0%
$x=3; y=9$

Explanation

Given equations are $\dfrac { x }{ 2 } +\dfrac { y }{ 5 } =5$ and $\dfrac { 2 }{ x } +\dfrac { 5 }{ y } =\dfrac { 5 }{ 6 }$

Let $\dfrac { x }{ 2 } =u$ and $\dfrac { y }{ 5 } =v$

$u+v=5$ .....(i)

$\dfrac { 1 }{ u } +\dfrac { 1 }{ v } =\dfrac { 5 }{ 6 }$ .....(ii)

$\Rightarrow \dfrac { u+v }{ uv } =\dfrac { 5 }{ 6 } \\ \Rightarrow \dfrac { 5 }{ uv } =\dfrac { 5 }{ 6 } \\ \Rightarrow uv=6$ .........(iii)

From (i) $v=5-u$

$(5-u)u=6\\ \Rightarrow 5u-{ u }^{ 2 }=6\\ \Rightarrow { u }^{ 2 }-5u+6=0\\ \Rightarrow { u }^{ 2 }-2u-3u+6=0\\ \Rightarrow u(u-2)-3(u-2)=0\\ \Rightarrow (u-3)(u-2)=0\\ \Rightarrow u=2,3$

But $\dfrac { x }{ 2 } =u$

$\Rightarrow x=2u\\ \Rightarrow x=4,6$

Also, $\dfrac { y }{ 5 } =v$

$\Rightarrow y=5v\\ \Rightarrow y=15,10$

The sum of a two digit number and the number obtained by reversing the order of its digits is

$121$ , and the two digits differ by

$3$ . Find the number.

0%
$47$
0%
$74$
0%
$44$
0%
$34$

Explanation

Let the digit in the unit's place be

$x$ and the digit at the ten's place be

$y$ . Then,

Number

$=10y+x$

The number obtained by reversing the order of the digits is

$10x+y$ .

According to the given conditions, we have

$(10y+x)+(10x+y)=121$

$\Rightarrow 11(x+y)=121$

$\Rightarrow x+y=11$

and,

$x-y=\pm 3$ [

$\because$ Difference of digits is

$3$ ]

Thus, we have the following sets of simultaneous equations and,

$\left.\begin{matrix} x+y=11 & (i)\\ x-y=3 & (ii)\end{matrix}\right.$ or

$\left\{\begin{matrix} x+y=11 & .(iii) \\ x-y=-3 & (iv) \end{matrix}\right.$

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

$2x=14,x=7$ , Putting this in (ii)

$y=7-3=4$

$x=7, y=4$

On adding equations (iii) and (iv), we get

$2x=8,x=4$ ,

Putting this in (iv)

$y=4+3=7$

$x=4, y=7$

When

$x=7, y=4$ , we have

Number

$=10y+x=10\times 4+7=47$

When

$x=4, y=7$ , we have

Number

$=10y+x=10\times 7+4=74$

Hence, the required number is either

$47$ or

$74$ .

$ax + 2y = 5$

$3x - 6y = 20$
In the system of equations above,

$a$ is a constant. if the system has one solution, which of the following CANNOT be the value of

$a$ ?

0%
$-1$
0%
$4$
0%
$1$
0%
$3$

Explanation

The given equations are :

$ax + 2y +(-5) = 0\quad\quad\quad\dots(i)$

$3x+(-6)y+(-20)=0\quad\quad\quad\dots(ii)$

As we can see they are of the form

$ax + by +c =0$

Also the condition for some given system of equations to have solution is,

$\Rightarrow \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$

Here

$a_1, a_2, b_1, b_2$ all are constants, and their values are respectively:

$a_1 = a$

$a_2 = 3$

$b_1 = 2$

$b_2 = -6$

So if the given system of equations has exactly one solution then,

$\Rightarrow \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$

$\Rightarrow \dfrac{a}{3} \neq \dfrac{2}{-6}$

$\Rightarrow a \neq -1$

Hence out of all the values given in the options

$a$ can not be equal to

$-1.$

Hence option

$A$ is correct.

Ravish tells his daughter Aarushi, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be. If present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically and find their present ages.

0%
$x=12,y=49$
0%
$x=12,y=42$
0%
$x=10,y=49$
0%
$x=10,y=42$

Explanation

Present Age of Ravish

$= y$ years

Present Age of Aarushi

$= x$ years

According to Question

$7$ Years ago,

$y-7=7(x-7)\Rightarrow 7x-y-42=0.............(1)$

and

$3$ Years from now

$y+3=3(x+3)\Rightarrow 3x-y+6=0............(2)$

From (1) and (2)

$(1)-(2)\Rightarrow 7x-3x-y+y-42-6=0\Rightarrow 4x=48\Rightarrow x=12$

putting

$x=12$ in Equation (2)

we get,

$3 \times 12-y+6=0\Rightarrow y=42$

$\dfrac{22}{x+y}+\dfrac{15}{x-y}=5$

$\dfrac{55}{x+y}+\dfrac{45}{x-y}=14$

0%
$x=-8$ , $y=3$
0%
$x=-3$ , $y=8$
0%
$x=8$ , $y=3$
0%
$x=8$ , $y=-3$

Explanation

$let\>(\frac{1}{x+y})=u\>and\>(\frac{1}{x-y})=v\\then\>22u+165v=5\rightarrow\>(1.)\\and\>55+45v=14\rightarrow\>(2.)\\multiply\>equation\>(1.)by\>3,we\>get\\66u+45v=15\rightarrow\>(1.)\\+55u+45v=14\rightarrow\>(2.)\\upon\>subtraction\>of\>(1.)with(2.),we\>get\>\\\>11u+0=1\\\therefore\>u=(\frac{1}{11})or\>x+y\>=11\rightarrow\>(3.)\\and\>then\>\\v=\>(\frac{5-22u}{15})=(\frac{5-2}{15})=(\frac{1}{5})\\\therefore\>x-y=5\rightarrow\>(4.)\\add\>(3.)and\>(4.)equation\>we\>get\>\\2x=16\\\therefore\>x=8\\and\>y\>=11-x=11-8=3$

Find the solution of pair of equations

$\dfrac{x}{10}+\dfrac{y}{5}-1=0$ and

$\dfrac{x}{8}+\dfrac{y}{6}=15.$ Hence, find

$\lambda$ , if

$y=\lambda x+5$ .

0%
$\dfrac{-165}{17}$
0%
$\dfrac{-1}{2}$
0%
$\dfrac{361}{696}$
0%
$\dfrac{340}{17}$

Explanation

$\dfrac{x}{10}+\dfrac{y}{5}-1=0$

$\dfrac{x}{10}+\dfrac{y}{5}=1$

$\dfrac{x+2y}{10}=1$

$x+2y=10$ .....................(1)

now consider, $\dfrac{x}{8}+\dfrac{y}{6}=15$

$\dfrac{6x+8y}{48}=15$

$3x+4y=24\times{15}=360$ .................................(2)

now multiply (1)by 3

$3x+6y=30$ ........................(3)

subtracting (2) and (3)

$2y=-330$

$y=-165$

put in(1)

$x+2(-165)=10$

$x-330=10$

$x=340$

now, $y=\lambda{x}+5$

$\implies{-165=\lambda{(340)}+5}$

$-170=\lambda{(340)}$

$\lambda{=\dfrac{-170}{340}=\dfrac{-1}{2}}$

Solve by using substitution method.

$3x +4y =10$ &

$2x - 2y =2$

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

Explanation

$3x+4y= 10 \ .....(1)$

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

equation $(2)$ can be written as

$\Rightarrow x-y=1$

$\Rightarrow x=y+1\ .....(3)$

substitute $x$ value in equation $(2)$

$\Rightarrow 3x+4y= 10$

$\Rightarrow 3(y+1)+4y=10$

$\Rightarrow 3y+3+4y=10$

$\Rightarrow 7y=7$

$\Rightarrow y=1$

then, substitute $y$ value in equation $(3)$

$\Rightarrow x=y+1=1+1=2$

$\therefore x=2, y=1$

Solve the following pair of equations graphically;

$2x -3y =1,$ &

$4x -3y +1=0$

0%
$x=1$ $y=-1$
0%
$x=-1$ $y=-1$
0%
$x=-1$ $y=1$
0%
$x=1$ $y=1$

Explanation

Plot each line by taking two points on them.
$2x-3y=1$
$x=0\implies y =\frac{-1}{3}$ and $y=0\implies x=\frac{1}{2}$
Join the two points $(0,\frac{-1}{3})$ and $(\frac{1}{2},0)$ by drawing line.

$4x-3y+1=0$

$x=0\implies y =\frac{-1}{3}$ and $y=0\implies x=\frac{1}{4}$

Join the two points $(0,\frac{-1}{3})$ and $(\frac{1}{4},0)$ by drawing line.

From the graph, the point of intersection of both lines is $(-1,-1)$ and hence solution of given system of linear equation is $x=-1, y=-1$

1192404_1089679_ans_14695cbe041f4077a08a22dda9dc8f62.PNG

Solve each pair of equation by using the substitution method.

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

$3x-\dfrac{8}{y}=5$

0%
$x=3$ and $y=-2$
0%
$x=30$ and $y=2$
0%
$x=3$ and $y=2$
0%
None of these

Explanation

$x+\dfrac {6}{y}=6......(i)$

$3x-\dfrac {8}{y}=5.....(ii)$

From equation

$(i)$ , we get

$\Rightarrow \ x=6-\dfrac {6}{y}...(iii)$

Substituting value of

$x$ in equation

$(ii)$ , we get

$3\left (6-\dfrac {6}{y}\right)-\dfrac {8}{y}=5$

$\Rightarrow \ 18-\dfrac {18}{y}-\dfrac {8}{y}=5$

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

$\Rightarrow \ y=2$

Substituting the value of

$y$ in (iii)

$\ x=6-\dfrac {6}{2}=3$

$\therefore x = 3, y = 2$

Solve each pair of equation by using the substitution method.

$0.2x+0.3{y}=1.3$

$0.4x+0.5{y}=2.3$

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

Explanation

$0.2x+0.3y=1.3.......(i)$

$0.4x+0.5y=2.3......(ii)$

From equation

$(i)$ , we get

$0.2x=1.3-0.3y$

$\Rightarrow \ x=\dfrac {13-3y}{2}.....(iii)$

From equation

$(iii)$ and equation

$(ii)$

$\therefore \ 0.4\left(\dfrac {13-3y}{2}\right)+0.5y=2.3$

$\Rightarrow \ 2.6-0.6y+0.5y=2.3$

$\Rightarrow \ -0.1y=-0.3\\\ \Rightarrow y=3$

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

Solve each pair of equation by using the substitution method.

$2x+3y=9$

$3x+4y=5$

0%
$x=-21$ and $y=7$
0%
$x=1$ and $y=17$
0%
$x=-21$ and $y=17$
0%
None of these

Explanation

$2x+3y=9---(i)$

$3x+4y=5--(ii)$

From equation

$(i)$ , we get

$\Rightarrow \ 2x=9-3y\ \Rightarrow x=\dfrac {9-3y}{2}---(iii)$

Substituting equation

$(iii)$ in equartion

$(ii)$ , we get

$\Rightarrow \ 3\left(\dfrac {9-3y}{2}\right)+4y=5$

$\Rightarrow \ 27-9y+8y=10$

$\Rightarrow \ y=27-10=17$

$\therefore \ x=\dfrac {9-3\times 17}{2}=\dfrac {-42}{2}=-21$

Classes A and B have

$35$ students each. If seven girls shift from class

$A$ to class

$B$ , then the number of girls in the classes would interchange. If four girls shift from class

$B$ to class

$A$ , then the number of girls in class

$A$ would be twice the original number of girls in class

$B$ . What is the number of boys in Class

$A$ and in Class

$B$ ?

0%
$18$ and $11$
0%
$24$ and $19$
0%
$18$ and $27$
0%
$17$ and $24$

Explanation

$\textbf{Step -1: Form the required equations.}$

$\text{Let the number of girls in class }A\text{ be }x.$

$\text{Let the number of girls in class }B\text{ be }y.$

$\text{Let the number of boys in class }A\text{ be }35-x.$

$\text{Let the number of boys in class }B\text{ be }35-y.$

$\text{If seven girls are shifted from class }A\text{ to }B,$

$\text{Number of girls in class }A\text{ would become }x-7.$

$\text{Number of girls in class }B\text{ would become }y+7.$

$\text{But according to the question,}$

$y=x-7$

$\Rightarrow x-y=7\ldots(i)$

$\text{If four girls are shifted from class }B\text{ to }A,$

$\text{Number of girls in class }A\text{ would become }x+4.$

$\text{Number of girls in class }B\text{ would become }y-4.$

$\text{But according to the question,}$

$x+4=2y$

$\Rightarrow x-2y=-4\ldots(ii)$

$\textbf{Step -2: Solve the above formed two equations simultaneously.}$

$\text{On subtracting equation }(ii)\text{ from }(i),\text{ we get}$

$-y+2y=7+4$

$\Rightarrow y=11$

$\text{On putting the value of }y\text{ in equation }(i),\text{ we get}$

$x-11=7$

$\Rightarrow x=18$

$\textbf{Step -3: Find the number of boys in class A and B.}$

$\text{Number of boys in class }A=35-18$

$=17$

$\text{Number of boys in class }B=35-11$

$=24$

$\textbf{Therefore, option D is correct.}$

A number is

$\dfrac{2}5{}$ times another number. If their sum is

$70$ , Find the numbers.

0%
$70$ and $50$
0%
$20$ and $30$
0%
$40$ and $50$
0%
$20$ and $50$

Explanation

Let the numbers be

$x$ and

$y$

According to the question

$x=\dfrac{2}{5}y............(i)$

$x+y=70...............(ii)$

$\dfrac{2}{5}y+y=70$

$\dfrac{2y+5y}{5}=70$

$7y=350$

$y=50$

$x+50=70$

$x=20$

So the numbers are

$20$ and

$50$

Sum of two numbers is

$407$ . The sum and difference of their LCM and HCF are

$925$ and

$851$ respectively. The difference of two numbers is

0%
$518$
0%
$185$
0%
$158$
0%
$175$

Explanation

Let the two numbers: be

$x$ &

$y$

$\therefore x+y=407$ .........

$(1)$

Acc to the question,

$LCM+HCF=925$ ..........

$(2)$

$LCM-HCF=851$ .........

$(3)$

On subtracting

$(2)$ &

$(3)$

$LCM+HCF=925\\ -LCM+HCF=-851\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \ \ \ \ \ 2HCF=74\\ \quad \quad \quad \ \ \ \ \ \ \ HCF=37$

$\therefore LCM -(37)=857$

$LCM=857+37$

$=888$

As we know,

$LCM\times HCF=$ product of two numbers

$\therefore 888\times 37=x xy$ .............

$(4)$

From

$(1)$ ,

$x=(407-y)$

$\therefore 888\times 37 = (407-y)y$

$\Rightarrow y^{2}-407y+832856$

$(y-111)(y-296)$

$\therefore x=296$ and

$y=111$

$\therefore$ Difference of two numbers

$=296-111$

$=185$

$\therefore$ Option

$B$ is correct.

The liquids

$X$ and

$Y$ are mixed in the ratio of

$3:2$ and the mixture is sold at

$Rs\;11$ per liter at a profit of

$10\%$ . If the liquid

$X$ costs

$Rs\;2$ more per liter than

$Y$ , the cost of

$X$ per liter is (in Rs.):

0%
$9.50$
0%
$10.80$
0%
$11.75$
0%
$11$

Explanation

Let Cost of

$X$ be

$x$

Cost of

$Y$ be

$y$

and their volumes be

$3K$ and

$2K$

Now,

$x=y+2$

Let original price be

$P$

So,

$P+10$ % of

$P=11$

$P+0.1P=11$

$1.1P=11$

$P=10$

So,

$\dfrac { 3Kx+2Ky }{ 3K+2K } =\dfrac { 3x+2y }{ 5 } =10$

$3x+2y=50$ ....(1)

and

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

Using (2) in (1), we get

$\Rightarrow 3y+6+2y=50\Rightarrow 5y=44\Rightarrow y=\dfrac { 44 }{ 5 } =8.8$ Rs.

$x=10.8$ Rs.

Solve the following pair of linear equations in two variables (by graph) :

$2x+3y=5,\ x+6y=25$

0%
$(1,2)$
0%
$(1,3)$
0%
$(1,6)$
0%
None of these

At a zoo, there were parrots and rabbits in the same enclouser. Sravanthi counted 30 heads and 100 legs altogether. how many animals of each type were in the enclouser

0%
$9$ parrots $21$ rabbits
0%
$10$ parrots and $20$ rabbits
0%
$20$ parrots and $19$ rabbits
0%
$11$ rabbits and $19$ parrots

$x - 2y = -1$ ,

$y=-\dfrac { 1 }{ 2 }$ , then find the value of

$x$

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

Explanation

Given:

$x-2y=-1$ .....................[1]

$y=-\dfrac12$ ..................[2]

Putting

$y=-\dfrac12$ in equation

$1$ :

$x-2\times -\dfrac12=-1$

$\implies x+1=-1$

$\implies x=-2$

$ax+by=c$ and

$mx+ny=d$ where

$an\neq bm$ then these simultaneous equation have -

0%
Only one common solution
0%
No solution
0%
Infinite number of solution
0%
Only two solutions.

The system of linear equations

$5x+my=10$ and

$4x+ny=8$ have infinitely many solutions, where m and n are positive integers. Then the minimum possible value of

$(m+n)$ is equal to

0%
$9$
0%
$5$
0%
$6$
0%
$10$

Explanation

We know, if two linear equations are

$a_1x+b_1y+c_1=0$ and

$a_2x+b_2y+c_2=0$

For infinity many solutions :

$\Rightarrow$

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

Here, the two linear equations are,

$5x+my=10$

$\Rightarrow$

$5x+my-10=0$

$4x+ny=8$

$\Rightarrow$

$4x+ny-8=0$

For, infinity solutions,

$\Rightarrow$

$\dfrac{5}{4}=\dfrac{m}{n}=\dfrac{-10}{-8}$

$\Rightarrow$

$\dfrac{5}{4}=\dfrac{m}{n}=\dfrac{5}{4}$

So, the least value of

$m$ is

$5$ and

$n$ is

$4$ .

$\Rightarrow$ Minimum possible value of

$(m+n)=5+4=9$

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 15 - MCQExams.com

0:0:2

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 15 - MCQExams.com

0:0:2

Practice Class 10 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.