MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 7

Solve the following pairs of linear (simultaneous) equation by the method of elimination:

$x + y = 7$ ,

$5x + 12y = 7$

0%
$x =13$ and $y=5$
0%
$x=11$ and $y=-4$
0%
$x=7$ and $y =-2$
0%
$x=5$ and $y =7$

Explanation

Multiply the equation

$x+y=7$ by

$5$ to make the coefficients of

$x$ equal. Then we get the equations:

$5x+5y=35.........(1)$

$5x+12y=7.........(2)$

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

$x$ , because the coefficients of

$x$ are the same. So, we get

$(5x-5x)+(5y-12y)=35-7$

i.e.

$7y=-28$

i.e.

$y = -4$

Substituting this value of

$y$ in the equation

$x+y=7$ , we get

$x-4=7$

i.e.

$x=11$

Hence, the solution of the equations is

$x = 11, y = -4$ .

Solve the following pair of equations:

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

$\displaystyle \frac{2x}{3}\, +\, 4y\, +\, \displaystyle \frac{1}{2}\, =\, 0$

0%
$x\, =\, \displaystyle \frac{1}{2}\, ;\, y\, =\, \displaystyle -\frac{7}{6}$
0%
$x\, =\, \displaystyle \frac{21}{20}\, ;\, y\, =\, \displaystyle -\frac{3}{10}$
0%
$x\, =\, \displaystyle \frac{14}{5}\, ;\, y\, =\, \displaystyle -\frac{6}{11}$
0%
$x\, =\, \displaystyle \frac{18}{19}\, ;\, y\, =\, \displaystyle -\frac{16}{10}$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$2x-3y-3=0$ and write it as

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

Substitute the value of

$y$ in the equation

$\cfrac { 2x }{ 3 } +4y+\cfrac { 1 }{ 2 } =0$ . We get

$\cfrac { 2x }{ 3 } +4y+\cfrac { 1 }{ 2 } =0\\ \Rightarrow \cfrac { 2x }{ 3 } +4\left( \cfrac { 2x-3 }{ 3 } \right) +\dfrac { 1 }{ 2 } =0\\ \Rightarrow \cfrac { 2x }{ 3 } +\cfrac { 8x-12 }{ 3 } +\cfrac { 1 }{ 2 } =0\\ \Rightarrow \cfrac { 2x+8x-12 }{ 3 } +\dfrac { 1 }{ 2 } =0\\ \Rightarrow \cfrac { 10x-12 }{ 3 } +\dfrac { 1 }{ 2 } =0\\ \Rightarrow \cfrac { 2(10x-12)+3 }{ 6 } =0\\ \Rightarrow \cfrac { 20x-24+3 }{ 6 } =0\\ \Rightarrow 20x=21\\ \Rightarrow x=\cfrac { 21 }{ 20 }$

Substituting this value of

$x$ in equation 1, we get

$y=\dfrac { 2\left( \dfrac { 21 }{ 20 } \right) -3 }{ 3 } =\cfrac { 42-60 }{ 60 } =-\cfrac { 18 }{ 60 } =-\cfrac { 3 }{ 10 }$

Hence, the solution is

$x=\cfrac { 21 }{ 20 } ,y=-\cfrac { 3 }{ 10 }$ .

Solve the following pair of equations :

$7x + 6y = 71$

$5x - 8y = -23$

0%
$x = 2; y = 7$
0%
$x = 5; y = 6$
0%
$x = 4; y = 9$
0%
$x = 1; y = 8$

Explanation

Given equations are

$7x + 6y = 71$ --- (1)

and

$5x - 8y = -23$ --- (2)

Multiplying equation $(1)$ with $4$ we get, $28x + 24y = 284$ ----- equation $(3)$

Multiplying equation $(2)$ with $3$ we get, $15x - 24y = - 69$ ----- equation $(4)$

Adding equations $(4)$ and $(3)$ , we get $43x = 215 => x = 5$

Substituting $x = 5$ in the equation $(2)$ , we get

$5(5) - 8y = -23 => y = 6$

Solve the following pair of equations :

$y = 2x - 6$

$y = 0$

0%
$x =3$
0%
$x =4$
0%
$x =-6$
0%
$x =-4$

Explanation

Given equations are

$y=2x-6$ ....(1)

and

$y=0$ ....(2)

Put the value of equation (2) in equation (1), we get

$0=2x-6$

$\Rightarrow 2x=6$

$\Rightarrow x=3$

Hence, option A is the correct answer.

Solve :

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

$x\, -\, y\, =\, 6xy$

0%
$x\, =\, -\displaystyle \frac{1}{2}$ and $y\, =\, \displaystyle \frac{1}{7}$
0%
$x\, =\, -\displaystyle \frac{1}{2}$ and $y\, =\, \displaystyle \frac{1}{4}$
0%
$x\, =\, -\displaystyle \frac{1}{2}$ and $y\, =\, \displaystyle \frac{1}{2}$
0%
$x\, =\, -\displaystyle \frac{1}{2}$ and $y\, =\, \displaystyle \frac{1}{5}$

Explanation

Let

$x + y = 2xy$ ---

$(1)$

$x - y = 6xy$ -----

$(2)$
Adding equations

$(1)$ and

$(2)$ , we get

$2x = 8xy$

$\Rightarrow y = \dfrac {1}{4}$

Substituting $y = \dfrac {1}{4}$ in the equation $(1)$ , we get

$x + \dfrac {1}{4} = 2(x)( \dfrac {1}{4})$

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

Solve graphically, the following pairs of equations:

$3x + 7y = 27$

$8 - y = \displaystyle \frac{5}{2}x$

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

Explanation

$3x+7y=27$

$\implies 3x=27-7y$

$\implies x=\dfrac{27-7y}{3}$

Let,

$y=3$ , then

$x=\dfrac{27-7(3)}{3}=2$

Similarly, when

$y=6, x=-5$

$\therefore$ the coordinates are

$(2,3),(-5,6)$

$8-y=\dfrac{5}{2}x$

$\implies 8-\dfrac{5}{2}x=y$

Let,

$x=2$ , then

$y=8-\dfrac{5}{2}\times 2=3$

Similarly, when

$x=4,y=-2$

$\therefore$ the coordinates are

$(2,3),(4,-2)$

$\therefore x=2,y=3$

Solve the following pair of equations:

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

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

0%
$x\, =\, \displaystyle \frac{16}{7}\, ;\, y\, =\, \displaystyle -\frac{4}{5}$
0%
$x\, =-\, \displaystyle \frac{4}{5}\, ;\, y\, =\, \displaystyle -\frac{17}{8}$
0%
$x\, =\, \displaystyle \frac{18}{5}\, ;\, y\, =\, \displaystyle -\frac{4}{5}$
0%
$x\, =\, \displaystyle \frac{26}{3}\, ;\, y\, =\, \displaystyle -\frac{8}{3}$

Explanation

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$3-(x-5)=y+2$ and write it as

$3-x+5=y+2$

$-x-y=-6$

$x+y=6$

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

Now the equation

$2(x+y)=4-3y$ can be rewritten as:

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

$2x+5y=4$

Substitute the value of

$y$ in the equation

$2x+5y=4$ . We get

$2x+5(6-x)=4$

i.e.

$2x+30-5x=4$

i.e.

$-3x=-26$

i.e.

$x=\dfrac { 26 }{ 3 }$

Therefore,

$x=\dfrac { 26 }{ 3 }$

Substituting this value of

$x$ in equation 1, we get

$y=6-\dfrac { 26 }{ 3 } =\dfrac { 18-26 }{ 3 } =-\dfrac { 8 }{ 3 }$

Hence, the solution is

$x=\dfrac { 26 }{ 3 } ,y=-\dfrac { 8 }{ 3 }$ .

Solve:

$\displaystyle \frac{20}{x\, +\, y}\, +\, \displaystyle \frac{3}{x\, -\,y}\, =\, 7$ and

$\displaystyle \frac{8}{x\, -\, y}\, -\, \displaystyle \frac{15}{x\, +\, y}\, =\, 5$

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

Explanation

$\dfrac {20}{x+y} + \dfrac {3}{x-y} = 7$ ...(1)

$\dfrac {8}{x-y} - \dfrac {15}{x+y} = 5$ ...(2)

Multiplying eq(1) with $3$ we get,

$\dfrac {60}{x+y} + \dfrac {9}{x-y} = 21$ ...(3)

Multiplying eq(2) with $4$ we get,

$\dfrac {32}{x-y} - \dfrac {60}{x+y} = 20$ ...(4)

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

$\dfrac {41}{x-y} = 41 => x - y = 41$ ...(5)

Substituting $x-y = 41$ in the eq(1), we get

$\dfrac {20}{x+y} + \dfrac {3}{1} = 7 => \dfrac {20}{x+y} = 4 => x +y = 5$ ... (6)

Adding eq(5) and eq(6), we get

$2x = 6 => x = 3$

Substituting $x = 3$ in the eq(5), we get

$3-y = 1 => y = 2$

Solve for

$x$ and

$y$ :

$mx \, -\, ny\, =\, m^{2}\, +\, n^{2}$ ,

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

0%
$x\, =\, m\, +\, n\,;\, y\, =\, m\, -\, n$
0%
$x\, =\, m\, -\, n\, ;\, y\, =\, mn\, -\, n$
0%
$x\, =\, m\, +\, mn\, ;\, y\, =\, m\, +\, n$
0%
$x\, =\, mn\, -\, n\, ;\, y\, =\, m\, -\, n$

Explanation

The equations given are:

$mx-ny=m^2+n^2...........(1)$

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

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation 2 and write it as

$x=2n+y$

Substitute the value of

$x$ in equation 1. We get

$m\left( 2n+y \right) -ny=m^{ 2 }+n^{ 2 }\\ \Rightarrow 2mn+my-ny=m^{ 2 }+n^{ 2 }\\ \Rightarrow y\left( m-n \right) =m^{ 2 }+n^{ 2 }-2mn\\ \Rightarrow y\left( m-n \right) =\left( m-n \right) ^{ 2 }\\ \Rightarrow y=m-n$

Therefore,

$y = m-n$

Substituting this value of

$y$ in the equation

$x=2n+y$ , we get

$x=2n+y=2n+m-n=m+n$

Hence, the solution is

$x = m+n, y = m-n$ .

Divide 80 into two numbers, such that 5 times one number is equal to 3 times the other number.

0%
$30 \ and \ 50$
0%
$35 \ and \ 52$
0%
$40 \ and \ 60$
0%
$22 \ and \ 49$

Explanation

Let the two numbers be

$x$ and

$y$
Given,

$x+y=80$ (i) and

$5x=3y$

$\implies \displaystyle x=\frac{3y}{5}$ (ii)
Putting the value of

$x$ in equation (i) we get,

$x+y=80$

$\implies \displaystyle \frac{3y}{5}+y=80$

$\implies \displaystyle \frac{3y+5y}{5}=80$

$\implies \displaystyle 8y=400$

$\implies \displaystyle y=50$
Putting the value of

$y$ in equation (i) we get,

$x+y=80$

$\implies \displaystyle x+50=80$

$\implies \displaystyle x=80-50$

$\implies \displaystyle x=30$

$\therefore x=30, y=50$

Solve, graphically, the following pairs of equations:

$2x + y = 23$

$4x - y = 19$

0%
$x = 7, y = 7$
0%
$x = 7, y = 3$
0%
$x = 7, y = 6$
0%
$x = 7, y = 9$

A and B both have some pencils. If A gives

$10$ pencils to B, then B will have twice as many as A. And if B gives

$10$ pencils to A, then they will have the same number of pencils. How many pencils does each has?

0%
A = $20$ and B = $55$
0%
A = $20$ and B = $60$
0%
A = $30$ and B = $50$
0%
A = $50$ and B = $70$

Explanation

Let A has

$x$ pencils and B has

$y$ .

As per the statement, "If A gives 10 pencils to B, then B will have twice as many as A":

$\implies 2(x - 10) = y + 10$

$2x - y = 30$ --- (1)
Also, as per the statement, "if B gives 10 pencils to A, then they will have the same number of pencils"

$\implies x + 10 = y - 10$

$x - y = -20$ --- (2)

Subtracting equation $(2)$ from $(1)$ , we get:

$2x-y-x+y=30+20$

$x = 50$

Substituting $x = 50$ in equation $(2)$ , we get:

$50 -y = -20$

$\implies y = 70$

So, A has $50$ pencils and B has $70$ pencils.

Solve the following pair of equations:

$\displaystyle \frac{a}{x}\, -\, \displaystyle \frac{b}{y}\, =\, 0$

$\displaystyle \frac{ab^{2}}{x}\, +\, \displaystyle \frac{a^{2}b}{y}\, =\, a^{2} \, +\, b^{2}$

0%
$x\, =\, ab$ and $y\, =\, b$
0%
$x\, =\, a$ and $y\, =\, b$
0%
$x\, =\, b$ and $y\, =\, a$
0%
$x\, =\, b-a$ and $y\, =\, ab$

Explanation

Given equations are

$\dfrac {a}{x}-\dfrac {b}{y}=0$ ....(1)

and

$\dfrac {ab^2}{x}+\dfrac {a^2b}{y}=a^2+b^2$ ....(2)

Multiply equation (1) by

$a^2$ , we get

$\dfrac {a^3}{x}-\dfrac {a^2b}{y}=0$ ....(3)

Add equations (2) and (3),

$\dfrac {1}{x}\left (ab^2+a^3\right)=a^2+b^2$

$\Rightarrow \dfrac {1}{x}[a(a^2+b^2)]=a^2+b^2$

$\Rightarrow x=a$

Substitute this value in equation (1), we get

$\dfrac {a}{a}-\dfrac {b}{y}=0$

$\Rightarrow 1-\dfrac {b}{y}=0$

$\Rightarrow y=b$

Therefore, solution is

$x=a, y=b$ .

The sum of the numerator and the denominator of a fraction is equal to

$7$ . Four times the numerator is

$8$ less than

$5$ times the denominator. Find the fraction.

0%
$\displaystyle \frac{1}{6}$
0%
$\displaystyle \frac{3}{4}$
0%
$\displaystyle \frac{4}{7}$
0%
$\displaystyle \frac{7}{11}$

Explanation

Let the fraction be

$\dfrac {x}{y}$

Given, the sum of the numerator and the denominator of a fraction is equal to

$7$

$\Rightarrow x + y = 7$

$...(1)$

Also, four times the numerator is

$8$ less than

$5$ times the denominator

$\Rightarrow 4x = 5y - 8$

$4x -5y = -8$

$...(2)$

Multiplying equation $(1)$ with $5$

We get $5x + 5y = 35$ $...(3)$

Adding equations $(2)$ and $(3),$

$4x-5y=-8$

$5x+5y=35$
______________

$9x\ \ \ \ \ \ \ \ =27$

$\Rightarrow 9x = 27$

$\Rightarrow x = 3$

Substituting $x = 3$ in the equation $(1)$

We get $3 + y = 7$

$\Rightarrow y = 4$

Hence, the fraction is $\dfrac34$

Solve the following pair of equations:

$\displaystyle \frac{x}{a}\, -\, \displaystyle \frac{y}{b}\, =\,0$ ,

$ax\, +\, by\, =\, a^{2}\, +\, b^{2}$

0%
$x\, =\, b-a$ and $y\, =\, b$
0%
$x\, =\, 0$ and $y\, =\, ba$
0%
$x\, =\, ba$ and $y\, =\, ba$
0%
$x\, =\, a$ and $y\, =\, b$

Explanation

The equation

$\frac { x }{ a }-\frac { y }{ b }=0$ can be rewritten as:

$bx-ay=0...........(1)$

The other equation given is:

$ax+by=a^2+b^2..........(2)$

We pick either of the equations and write one variable in terms of the other.

Let us consider the equation

$bx-ay=0$ and write it as

$x=\frac { ay }{ b }$

Substitute the value of

$x$ in equation

$ax+by=a^2+b^2$ . We get

$a\left( \frac { ay }{ b } \right) +by=a^{ 2 }+b^{ 2 }\\ \Rightarrow \frac { a^{ 2 }y }{ b } +by=a^{ 2 }+b^{ 2 }\\ \Rightarrow \frac { a^{ 2 }y+b^{ 2 }y }{ b } =a^{ 2 }+b^{ 2 }\\ \Rightarrow \left( a^{ 2 }+b^{ 2 } \right) y=b\left( a^{ 2 }+b^{ 2 } \right) \\ \Rightarrow y=b$

Therefore,

$y = b$

Substituting this value of

$y$ in the equation

$x=\frac { ay }{ b }$ , we get

$x=\frac { ab }{ b } =a$

Hence, the solution is

$x = a, y = b$ .

Rohit says to Ajay "Give me a hundred, I shall then become twice as rich as you." Ajay replies "If you give me ten, I shall be six times as rich as you." How much does each have originally?

0%
Rohit has Rs. $20$ and Ajay has Rs. $90$
0%
Rohit has Rs. $50$ and Ajay has Rs. $200$
0%
Rohit has Rs. $40$ and Ajay has Rs. $170$
0%
Rohit has Rs. $56$ and Ajay has Rs. $190$

Explanation

$\\ Let\quad the\quad numbers\quad be\quad x,\quad w,\quad z\quad and\quad y.\\ 2(x-100)=y+100;\\ x+10=6(y-10);\\ x=6y-70;\\ Substituting\quad the\quad value\quad in\quad the\quad 1st\quad equation:\\ 2(6y-70-100)=y+100;\\ Therefore:\\ y=40\\ x=170\\$

The sum of two-digit numbers and the number obtained by reversing the order of the digit is

$121$ . Find the number, if the digits differ by

$3$ .

0%
$47$ or $74$
0%
$36$ or $63$
0%
$67$ or $76$
0%
$94$ or $49$

Explanation

Let the two digit number be

$10x + y$

On reversing its digits, we get the number

$10y + x$

Given,

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

$11x + 11y = 121$
=>

$x + y = 11$ ---

$(1)$

Given,

$x - y = 3$ --

$(2)$
Adding

$(1)$ and

$(2),$ we get

$2x = 14$ =>

$x = 7$

Substituting

$x = 7$ in the equation

$(2)$ , we get

$7 -y = 3 \Rightarrow y = 4$

Hence, the number is

$74$ or

$47$

$A$ 's age is twice as

$B$ 's age.

$4$ years ago,

$A$ was three times as old as

$B$ . Find their present ages.

0%
$30$ years and $15$ years
0%
$20$ years and $10$ years
0%
$10$ years and $5$ years
0%
$16$ years and $8$ years

Explanation

$\\ Let\quad the\quad numbers\quad be\quad x\quad and\quad y.\\ x=2y;\\ x-4=3(y-4);\\ Substituting\quad the\quad value\quad in\quad the\quad 2nd\quad equation:\\ 2y-4=3(y-4);\\ Therefore:\\ y=8\\ x=16\\$

Divide

$32$ into two parts such that if the larger is divided by the smaller, the quotient is

$2$ and the remainder is

$5$ .

0%
$13$ and $5$
0%
$23$ and $9$
0%
$17$ and $39$
0%
$28$ and $45$

Explanation

Let the larger part be

$a$ and smaller part be

$b$

Given,

$a+b=32$ ....(i)

and

$\dfrac{a}{b}=2+\dfrac{5}{b}$

Multiplying both sides by

$b$ , we get

$\Rightarrow a=2b+5$ .....(ii)

Substituting the value of

$a$ in equation (i), we get

$a+b=32$

$\Rightarrow 2b+5+b=32$

$\Rightarrow 3b=32-5$

$\Rightarrow b=\dfrac{27}{3}$

$\Rightarrow b=9$

Putting the value of

$b$ in equation (i), we get

$a+b=32$

$\Rightarrow a+9=32$

$\Rightarrow a=32-9$

$\Rightarrow a=23$

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 meI will have thrice as many as left with you." How many mangoes does each have ?

0%
A has 34 Mangoes and B has 62 Mangoes
0%
A has 14 Mangoes and B has 35 Mangoes
0%
A has 70 Mangoes and B has 148 Mangoes
0%
A has 96 Mangoes and B has 182 Mangoes

Explanation

$\\ Let\quad the\quad numbers\quad be\quad x\quad and\quad y.\\ x+30=2(y-30);\\ 3(x-10)=(y+10);\\ x=2y-90;\\ Substituting\quad the\quad value\quad in\quad the\quad 1st\quad equation:\\ 3(2y-90-10)=(y+10);\\ Therefore:\\ y=62\\ x=34\\$

Divide 184 into two parts such that one- third of one part may exceed one-seventh of the other part by 4.

0%
61.6 and 120.4
0%
62.6 and 120.4
0%
63.6 and 120.4
0%
64.6 and 120.4

Explanation

Let the two parts be

$x$ and

$y$

=>

$x + y = 184$ --- (1)

Given, one- third of one part may exceed one-seventh of the other part by

$4$

$=> \frac {x}{3} - \frac {y}{7} = 4$

$=> 7x -3y = 84$ --- (2)

Multiplying equation $(1)$ with $3$ we get, $3x + 3y = 552$ ----- equation $(3)$

Adding equations $2$ and $3$ , we get $10x = 636 => x = 63.6$

Substituting

$x = 63.6$ in the equation $(2)$ , we get $63.6 + y = 184 =>y = 120.4$

Thus , the parts are $63.6 ; 120.4$

Graphically, the pair of equations

$6x - 3y + 10 = 0, 2x - y + 9 = 0$ represents two lines which are

0%
Intersecting at exactly one point
0%
Intersecting at exactly two points
0%
Coincident
0%
Parallel

Explanation

The linear equations are

$6x-3y+10=0$

$2x-y+9=0$
Here,

$a_{1}=6,b_{1}=-3,c_{1}=10$

$a_{2}=2,b_{2}=-1,c_{2}=9$

$\therefore \displaystyle \frac{a_{1}}{a_{2}}=\frac{6}{2}=3$

$\displaystyle \frac{b_{1}}{b_{2}}=\frac{-3}{-1}=3$

$\displaystyle \frac{c_{1}}{c_{2}}=\frac{10}{9}$

$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{2}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$

$\therefore$ The lines represented by the given equations are parallel.

Two articles

$A$ and

$B$ are sold for Rs.

$1,167$ making

$5\%$ profit on

$A$ and

$7\%$ profit on B. If the two articles are sold for Rs.

$1,165$ , a profit of

$7\%$ is made on

$A$ and a profit of

$5\%$ is made on

$B$ . Find the cost price of each article.

0%
$A =$ Rs. $200$ and $B =$ Rs. $300$
0%
$A =$ Rs. $600$ and $B =$ Rs. $700$
0%
$A =$ Rs. $100$ and $B =$ Rs. $200$
0%
$A =$ Rs. $500$ and $B =$ Rs. $600$

Explanation

Let the cost price of

$A =$ Rs.

$x$ and that of

$B =$ Rs.

$y$
Given,
They are sold for Rs.

$1,167$ making

$5\%$ profit on

$A$ and

$7\%$ profit on

$B$

$\Rightarrow 1.05x + 1.07y = 1167$ ------

$(1)$

Also, when the two articles are sold for Rs.

$1,165$ a profit of

$7\%$ is made on

$A$ and a profit of

$5\%$ is made on

$B$ .

$\Rightarrow 1.07x + 1.05y = 1165$ ------

$(2)$

Multiplying equation

$(1)$ with

$1.07$ we get,

$1.1235x + 1.1449y = 1248.69$ ----- equation

$(3)$

Multiplying equation $(2)$ with $1.05$ we get, $1.1235x + 1.1025y = 1223.25$ ----- equation $(4)$

Subtracting equation $(4)$ from $(3)$ , we get $0.0424y = 25.44 \Rightarrow y = 600$

Substituting $y = 600$ in the equation $(2)$ , we get $1.07x + 1.05(600) = 1165 \Rightarrow x = 500$

Hence, cost price of $A$ is Rs. $500$ and of $B$ is Rs. $600$

The class XI students of a school wanted to give a farewell party to the out going students of class XII. They decided to purchase two kinds of sweets, one costing Rs. 250 per kg and the other costing Rs. 350 per kg. They estimated that 40 kg of sweets were needed. if the total budget for the sweets was Rs. 11,800; find how much sweets of each kind were bought ?

0%
20 kg and 18 kg
0%
22 kg and 18 kg
0%
28 kg and 18 kg
0%
26 kg and 18 kg

Explanation

Let

$x$ kg of one sweet be bought and

$y$ kg of other.

Given,

$x + y = 40$ --- (1)

And

$250x + 350y = 11800$
=>

$5x + 7y = 236$ --- (2)

Multiplying equation

$(1)$ with

$5$ we get,

$5x + 5y = 200$ ----- equation

$(3)$

Subtracting equation

$(3)$ from

$(2)$ , we get

$2y = 36 => y = 18$

Substituting $y = 18$ in the equation $(1)$ , we get $x + 18 = 40 => x = 22$

Hence $22 kg$ and $18 kg$ of two types of sweets were bought.

The difference between two whole numbers is

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

0%
$45$ and $15$
0%
$48$ and $12$
0%
$60$ and $20$
0%
$39$ and $13$

Explanation

Let the numbers be

$x$ and

$y$

Given,

$x - y = 26$ ---

$(1)$
Also,

$x = 3y$ ---

$(2)$

From equations

$(1)$ and

$(2)$ , we get

$3y - y = 26 \Rightarrow 2y = 26 \Rightarrow y = 13$
So,

$x = 3y = 3 \times 13 = 39$

Numbers are

$39, 13$

$1250$ persons went to see a circus-show. Each adult paid Rs.

$75$ and each child paid Rs.

$25$ for the admission ticket. Find the number of adults and number of children, if the total collection from them amounts to Rs.

$61,250$ .

0%
Adults $=600$ and children $=650$
0%
Adults $=300$ and children $=450$
0%
Adults $=800$ and children $=700$
0%
Adults $=640$ and children $=800$

Explanation

Let

$x$ be the number of adults and

$y$ children.

Given,

$x + y = 1250$ ---

$(1)$

And

$75x + 25y = 61250$

$\Rightarrow 3x + y = 2450$ ---

$(2)$

Subtracting equation

$(1)$ from

$(2)$ , we get

$2x = 1200 \Rightarrow x = 600$

Substituting $x = 600$ in the equation $(1)$ , we get $600 + y = 1250 \Rightarrow y = 650$

Hence,

$600$ adults and

$650$ children visited.

The pair of linear equations

$13x + ky = k, 39x + 6y = k +4$ has infinitely many solutions if

0%
$k=1$
0%
$k=2$
0%
$k=4$
0%
$k=6$

Explanation

The pair of linear equations

$13x+ky-k=0$

$39x+6y-(k+4)=0$

Here,

$a_{1}=13,b_{1}=k,c_{1}=-k$

$a_{2}=39,b_{2}=6,c_{2}=-(k+4)$

The system has infinitely many solutions if

$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \displaystyle \frac{13}{39}=\frac{k}{6}=\frac{-k}{-(k+4)}$

Taking,

$\displaystyle \frac{13}{39}=\frac{k}{6}$

$\Rightarrow \displaystyle \frac{1}{3}=\frac{k}{6}$

$\Rightarrow 3k=6$

$\Rightarrow k=2$

Taking,

$\displaystyle \frac{k}{6}=\frac{-k}{-(k+4)}$

$\displaystyle \Rightarrow \frac{1}{6}=\frac{1}{(k+4)}$

$\Rightarrow k+4=6 \Rightarrow k=2$

Thus,

$k=2$ is the correct answer

The pair of equations

$3x+4y=k, 9x+12y=6$ has infinitely many solutions if

0%
$k=2$
0%
$k=6$
0%
$k\neq 6$
0%
$k=3$

Explanation

The equations are

$3x+4y-k=0$

$9x+12y-6=0$
Here,

$a_{1}=3,b_{1}=4,c_{1}=-k$
and

$a_{2}=9,b_{2}=12,c_{2}=-6$
For the system to have infinite solutions condition is

$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \displaystyle \frac{3}{9}=\frac{4}{12}=\frac{-k}{-6}$

$\dfrac{4}{12}=\dfrac{-k}{-6}$

$\Rightarrow \dfrac{1}{3}=\dfrac{-k}{-6}$

$\Rightarrow 3k=6$

$\Rightarrow k=2$

The pair of linear equations

$2x+5y=k, kx+15y=18$ has infinitely many solutions if

0%
$k=3$
0%
$k=6$
0%
$k=9$
0%
$k=18$

Explanation

The pair of linear equations

$2x+5y=k$
and

$kx+15y=18$
Here,

$a_{1}=2,b_{1}=5,c_{1}=-k$
and

$a_{2}=k,b_{2}=15,c_{2}=-18$

The system of equation has infinitely many solutions if

$\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

$\Rightarrow \displaystyle \frac{2}{k}=\frac{5}{15}=\frac{-k}{-18}$

$\Rightarrow \displaystyle \frac{2}{k}=\frac{1}{3} =\frac{k}{18}$

Taking,

$\displaystyle \frac{2}{k}=\frac{1}{3} \Rightarrow k=6$

Taking,

$\displaystyle \frac{1}{3} =\frac{k}{18} \Rightarrow k=6$

So, answer is

$k=6$

The pair of linear equations

$3x+7y=k, 12x+2ky=4k+1$ do not have any solution if

0%
$k=7$
0%
$k=14$
0%
$k=21$
0%
$k=28$

Explanation

The pair of linear equations

$3x+7y-k=0$

$12x+2ky-(4k+1)=0$
Here,

$a_{1}=3,b_{1}=7,c_{1}=-k$
and

$a_{2}=12,b_{2}=2k,c_{2}=-(4k+1)$
The system has no solution if

$\displaystyle\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$

$\Rightarrow \displaystyle \frac{3}{12}=\frac{7}{2k}\neq \frac{-k}{-(4k+1)}$

$\Rightarrow \frac{1}{4}=\frac{7}{2k} \neq \frac{k}{4k+1}$

$\Rightarrow 2k=28$

$\Rightarrow k=14$

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 7 - 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 7 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.