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

Draw the graphs of the equations

$x = 3, x = 5$ and

$2x - y - 4 = 0$ . Also find the area of the quadrilateral formed by the lines and the x-axis

0%
$2$ sq. unit
0%
$8$ sq. unit
0%
$15$ sq. unit
0%
$20$ sq. unit

Explanation

Clearly the quadrilateral is trapezium
Thus it's area

$=\cfrac{1}{2}($ AD

$+$ BC

$)\times$ AB

$=\cfrac{1}{2}\times (2+6)\times 2=8$ sq. units

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:

$\dfrac {3x-2}{3y+7}=\dfrac {5x-1}{5y+16}; \dfrac {3x-15}{x-9}=\dfrac {6y-5}{2y+3}$

0%
$x = 1, y = 4$
0%
$x = 3, y = 2$
0%
$x = 7, y = 4$
0%
$x = 6, y = 5$

Explanation

$\Rightarrow \dfrac{3x-2}{3y+7}=\dfrac{5x-1}{5y+16}$
$\Rightarrow 15xy-10y+48x-32=15xy+35x-3y-7$
$\Rightarrow 13x-7y=25.........eq1$
$\Rightarrow \dfrac{3x-15}{x-9}=\dfrac{6y-5}{2y+3}$
$\Rightarrow 6xy-30y+9x-45=6xy-54y-5x+45$
$\Rightarrow 14x+24y=90.......eq2$
Multiply eq1 by 24 and eq2 by 7
$\Rightarrow 312x-168y=600.......eq3$
$\Rightarrow 98x+168y=630.......eq4$
Add eq3 and eq4
$\Rightarrow 410x=1230$
$\Rightarrow x=3$
Put x=3 in eq 1
$\Rightarrow 13\times3-7y=25$
$\Rightarrow -7y=-14$
$\Rightarrow y=2$
Hence, x=3 and y=2

Aftab tells his daughter, "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." (Isn't this interesting?) What are their present ages? Solve graphically introducing necessary varaibles.

0%
$12, 42$
0%
$13, 47$
0%
$18, 51$
0%
None of these

Explanation

Let the present age of Aftab's daughter

$= x$ years. and the present age of Aftab

$= y$ years,

$(y > x)$

According to the given conditions, we have
Seven years ago,

$(y-7)=7\times (x-7)$

$\Rightarrow y - 7 = 7x - 49$

$\Rightarrow 7x - y - 42 = 0$ ...(i)
Three years later,

$(y+ 3) = 3 \times (x + 3)$

$\Rightarrow y+ 3 = 3x + 9$

$\Rightarrow 3x- y + 6 = 0$ ...(ii)
Thus, the algebraic relations are

$7x - y - 42 = 0, 3x - y + 6 = 0$ .
Now, we represent the problem graphically as below:

$7x - y - 42 = 0$ ...(i)

$3x - y + 6 = 0$ ...(ii)
From the graph, we find that

$x = 12$
and

$y = 42$
Thus, the present age of Aftab's daugther

$= 12$ years and the present age of Aftab

$= 42$ years.

A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.

$27$ for a book kept for seven days, while Susy paid Rs.

$21$ for the book she kept for five days. Find the fixed charge and the charge for each extra day.

0%
Fixed charge: Rs. $12$ , Charge for each extra day: Rs. $9$
0%
Fixed charge: Rs. $15$ , Charge for each extra day: Rs. $3$
0%
Fixed charge: Rs. $18$ , Charge for each extra day: Rs. $3$
0%
Data insufficient

Explanation

Let the fixed charge for

$3$ days

$=x$ and additional charge

$=y$
According to the question

$\Rightarrow x+4y=27....eq1$

$\Rightarrow x+2y=21....eq2$
Subtract

$eq1$ and

$eq2$

$\Rightarrow \left ( 3x+4y=27 \right )-\left ( 3x+2y=21 \right )\Rightarrow 2y=6\Rightarrow y=3$
put

$y=3$ in

$eq1$

$\Rightarrow x+4\times3=27\Rightarrow x=15$
Fixed charges

$=$ Rs.

$15$
The charge for each extra day

$=$ Rs.

$3$

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.

Solve the following pair of linear equations by the substitution method

$x + y = 14, x - y = 4$

0%
$x = 8, y = 5$
0%
$x = 6, y = 9$
0%
$x = 7, y = 10$
0%
None of these

Explanation

$x + y = 14$ ....(i)

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

$y=x - 4$ ...(iii)
Substituting y from (iii) in (i), we get

$x + x - 4 = 14$

$\Rightarrow 2x = 18$

$\Rightarrow x =9$
Substituting

$x = 9$ in (iii), we get

$y = 9 - 4 = 5$ ,
i.e,

$y = 5$

$x = 9, y = 5$

On comparing the ratios

$\dfrac {a_1}{a_2}, \dfrac {b_1}{b_2}$ and

$\dfrac {c_1}{c_2}$ , find out whether the following pairs of linear equations are consistent, or inconsistent.

$5x - 3y = 11; - 10x + 6y = 22$

0%
Consistent
0%
Inconsistent
0%
Ambiguous
0%
Data insufficient

Explanation

The given linear equation are

$\Rightarrow 5x-3y-11=0....eq1$

$\Rightarrow a_{1}=5,b_{1}=-3,c_{1}=-11$

$\Rightarrow -10x+6y-22=0...eq2$

$\Rightarrow a_{2}=-10,b_{2}=6,c_{2}=-22$

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{5}{-10} \Rightarrow- \frac{1}{2}$

$\Rightarrow \frac{b_{1}}{b_{2}}=\frac{-3}{6} \Rightarrow -\frac{1}{2}$

$\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-11}{-22} \Rightarrow \frac {1}{2}$
comparing

$\Rightarrow \frac{a_{1}}{a_{2}},\frac{b_{1}}{b_{2}},\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Hence, the following pairs of linear equations are inconsistent.

On comparing the ratios

$\dfrac {a_1}{a_2}, \dfrac {b_1}{b_2}$ and

$\dfrac {c_1}{c_2}$ , find out whether the following pairs of linear equations are consistent, or inconsistent.

$2x - 3y = 8; 4x - 6y = 9$

0%
Consistent
0%
Inconsistent
0%
Ambiguous
0%
Data insufficient

Explanation

The given linear equation are

$\Rightarrow 2x-3y-8=0....eq1$

$\Rightarrow a_{1}=2,b_{1}=-3,c_{1}=-8$

$\Rightarrow 4x-6y-9=0...eq2$

$\Rightarrow a_{2}=4,b_{2}=-6,c_{2}=-9$

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{4} \Rightarrow \frac{1}{2}$

$\Rightarrow \frac{b_{1}}{b_{2}}=\frac{-3}{-6} \Rightarrow \frac{1}{2}$

$\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-8}{-9}$
comparing

$\Rightarrow \frac{a_{1}}{a_{2}},\frac{b_{1}}{b_{2}},\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
Hence, the following pairs of linear equations are inconsistent.

Determine whether the following pair of linear equations are consistent/inconsistent.

$2x - 2y - 2 = 0, 4x - 4y - 5 = 0$

0%
The graph is coincident straight line.
0%
The graph is intersecting lines.
0%
Inconsistent
0%
Data insufficient

Draw the graphs of the equations

$x - y + 1 = 0$ and

$3x + 2y - 12 = 0$ . Determine the co-ordinates of the vertices of the triangle formed by these lines and the

$x$ -axis, and shade the triangular region.

0%
$A (1,3), B (-1, 6)$ and $C (4, 0)$
0%
$A (2,5), B (-1, 6)$ and $C (2, 0)$
0%
$A (-2,5), B (-1, 6)$ and $C (2, 0)$
0%
$A (2,3), B (-1, 0)$ and $C (4, 0)$

Explanation

For equation

$x – y + 1 = 0$ or

$x=y-1$ , we have following points which lie on the line

Plug

$y =0,1$ we get

$x=0-1=-1$

$x=1-1=0$

Therefore, the required points are

$(-1,0), (0,1)$ .

For equation

$3x + 2y – 12 = 0$ or

$y=\frac { 12-3x }{ 2 }$ , we have following points which lie on the line

Plug

$x= 0 ,4$ we get

$y=\frac { 12-0 }{ 2 }=6$

$y=\frac { 12-12 }{ 2 }=0$

Therefore, the required points are

$(0,6), (4,0)$ .

We can see from the graphs that points of intersection of the lines with the x–axis are

$(–1, 0), (2, 3)$ and

$(4, 0)$ .

Determine whether the following pairs of linear equations are consistent/inconsistent.

$x-y=8, 3x-3y= 16$

If consistent, obtain the solution graphically.

0%
The graph is coincident straight line.
0%
The graph is intersecting lines.
0%
Inconsistent
0%
None of these

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

Meena went to a bank to withdraw Rs.

$2000.$ She asked the cashier to give her Rs.

$50$ and Rs.

$100$ notes only. Meena got

$25$ notes in all. Find how many notes of Rs.

$50$ and Rs.

$100$ she received

0%
$15$ notes of Rs. $50$ and $18$ notes of Rs. $100.$
0%
$12$ notes of Rs. $50$ and $18$ notes of Rs. $100$ .
0%
$8$ notes of Rs. $50$ and $25$ notes of Rs. $100.$
0%
$10$ notes of Rs. $50$ and $15$ notes of Rs. $100$ .

Explanation

Let

$x$ and

$y$ be the number of '

$50$ and '

$100$ notes.respectively.

Then,

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

$50x + 100y = 2000$ ...(2)
Now multiplying (1) from

$100$ and (2) from (1) subtracting (2) from (1) will get

$\Rightarrow 100\left ( x +y \right )-1\left ( 50x -100y \right ) = 100\times 25 -1\times 2000$

$\Rightarrow 100x+100y-50x-100y = 2500-2000$

$\Rightarrow 50x = 500$

$\Rightarrow x = 10$
Now putting

$x$ in (1)

$\Rightarrow 10+y = 25$

$\Rightarrow y = 25-10$

$\Rightarrow y = 15$
Hence Meena Received

$10$ notes of Rs.

$50$ and

$15$ notes of Rs.

$100$ .

$10$ students of class

$X$ took part in a Mathematics quiz. If the number of girls is

$4$ more than the number of boys, find the number of boys and girls who took part in the quiz by graphical method.

0%
$2, 7$
0%
$6, 9$
0%
$3, 12$
0%
None of these

Explanation

Let number of boys be

$x$ .

Number of girls be

$y$ .

Given that total number of student is

$10$ so that

$x + y= 10$

Subtract

$y$ both side we get

$x= 10 – y$

Plug

$y = 0 , 5, 10$ we get

$x= 10 – 0= 10$

$x= 10 – 5= 5$

$x= 10 – 10= 0$

Therefore, the required points are

$(0,10),(5,5),(10,0)$ .

Given that If the number of girls is

$4$ more than the number of boys

So that

$y= x + 4$

Plug

$x = -4, 0, 4$ , and we get

$y= - 4 + 4= 0$

$y= 0 + 4= 4$

$y= 4 + 4= 8$

Therefore, the required points are

$(-4,0),(0,4),(4,8)$ .

The graph is as shown above:

Since the point of intersection in the above graph is

$(3,7)$ ,

Hence, the number of boys are

$3$ and the number of girls are

$7$ .

$5$ pencils and

$7$ pens together cost Rs.

$50$ , whereas

$7$ pencils and

$5$ pens together cost Rs.

$46$ . Find the cost of one pencil and that of one pen by graphical method.

0%
Cost of one pencil $=$ Rs. $2$ and that of one pen $=$ Rs. $4$
0%
Cost of one pencil $=$ Rs. $3$ and that of one pen $=$ Rs. $5$
0%
Cost of one pencil $=$ Rs. $5$ and that of one pen $=$ Rs. $6$
0%
Cost of one pencil $=$ Rs. $2$ and that of one pen $=$ Rs. $6$

Explanation

Let cost of pencil be Rs.

$x$

Cost of pens be Rs.

$y$

$5$ pencils and

$7$ pens together cost Rs.

$50$ ,

So we get

$5x + 7y = 50$

Subtract

$7y$ both side we get

$5x= 50 – 7 y$

Divide by

$5$ we get

$x=10-\frac { 7 }{ 5 }y$

Plug value of

$y$ which is factor of

$5$ to get whole number so plug

$y = 5 , 10, 15$ we get

for

$y=5$

$x=10-\frac { 7 }{ 5 }y=10-7=3$

for

$y=10$

$x=10-\frac { 7 }{ 5 }y=10-14=-4$

for

$y=15$

$x=10-\frac { 7 }{ 5 }y=10-21=-11$

Therefore, the required points are

$(3,5), (-4,10),(-11,15)$ .

Given that

$7$ pencils and

$5$ pens together cost Rs.

$46$

$7x + 5y= 46$

Subtract

$7x$ both side we get

$5y= 46 – 7x$

Divide by

$5$ we get

$y= 9.2 – 1.4 x$

Plug

$x= 0 , 2, 4$ we get

for

$x=0$

$y= 9.2 – 0 = 9.2$

for

$x=2$

$y= 9.2 – 2.8 = 6.4$

for

$x=4$

$y=9.2 – 5.6 = 3.6$

Therefore, the required points are

$(0,9.2), (2,6.4),(4,3.6)$ .

The graph is as shown below:

Since the point of intersection is

$(3,5)$ ,

Hence, the cost of one pencil is Rs.

$3$ and the cost of one pen is Rs.

$5$

On comparing the ratios

$\dfrac {a_1}{a_2}, \dfrac {b_1}{b_2}$ and

$\dfrac {c_1}{c_2}$ , find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.

$6x - 3y + 10 = 0; 2x - y + 9 = 0$

0%
Intersect at a point
0%
Parallel
0%
Coincident
0%
Data insufficient

Explanation

The given linear equation are

$\Rightarrow 6x-3y+10=0....eq1$

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

$\Rightarrow 2x-y+9=0...eq2$

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

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{6}{3} \Rightarrow=\frac{3}{1}$

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

$\Rightarrow \frac{c_{1}}{c_{2}}=\frac{10}{9}$
comparing

$\Rightarrow \frac{a_{1}}{a_{2}},\frac{b_{1}}{b_{2}},\frac{c_{1}}{c_{2}}$

$\Rightarrow \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$
Hence,the line represented by eq1 and eq2 are parallel.

On comparing the ratios

$\frac {a_1}{a_2}, \frac {b_1}{b_2}$ and

$\frac {c_1}{c_2}$ , find out whether the lines representing the following pairs of linear equations intersect at a point, or are parallel or coincident.

$9x + 3y + 12 = 0; 18x + 6y + 24 = 0$

0%
Intersect at a point
0%
Parallel
0%
Coincident
0%
Cannot be determined

Explanation

The given linear equation are

$\Rightarrow 9x+3y+12=0....(1)$

$\Rightarrow a_{1}=9,b_{1}=3,c_{1}=12$

$\Rightarrow 18x+6y+24=0...(2)$

$\Rightarrow a_{2}=18,b_{2}=6,c_{2}=24$

$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}}=\frac{9}{18} = \frac{1}{2}$

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

$\Rightarrow \displaystyle \frac{c_{1}}{c_{2}}=\frac{12}{24}=\frac{1}{2}$

comparing

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

We have

$\ \displaystyle \frac{a_{1}}{a_{2}}= \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
Hence, the lines represented by eq1 and eq2 are coincident

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs.What are the fixed charges and the charge per kilometer? How much does a person have to pay for traveling a distance of 25 km?

0%
Fixed charge is Rs. $5$ and the charge per kilometer is Rs. $10$ For $25$ km person have to pay Rs. $255$ .
0%
Fixed charge is Rs. $10$ . and the charge per kilometer is Rs. $5$ . For $25$ km person have to pay Rs $135$ .
0%
Fixed charge is Rs. $15$ and the charge per kilometer is Rs. $5$ . For $25$ km person have to pay Rs. $140$ .
0%
Fixed charge is
Rs. $50$ . and the charge per kilometer is Rs $20$ . For $25$ km person have to pay Rs $50$

Explanation

Let fixed charge be Rs x and charge per km be Rs y.

$\Rightarrow x + 10y = 105 ...(1)$

$\Rightarrow x + 15y = 155 ..(2)$ .

Now From eq 2

$\Rightarrow 15y = 155-x$

$\Rightarrow y = \dfrac{155-x}{15}...(3)$

Substituting y from eq3 in eq1

$\Rightarrow x+\dfrac{10\left ( 155-x \right )}{15} = 105$

$\Rightarrow 15x+1550-10x =1575$

$\Rightarrow 5x = 1575-1550$

$\Rightarrow 5x = 25$

$\Rightarrow x = 5$

Substituting x in eq2

$\Rightarrow 5+15y = 155$

$\Rightarrow 15y = 155-5$

$\Rightarrow 15y = 150$

$\Rightarrow y = 10$

Hence

$x= 5$ and

$y = 10$

Fixed charge is Rs.

55 and the charge per kilometer is Rs.

1010

For

2525 km person have to pay =

$5+10 \times 25= 255 Rs.$

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,

$x = 39$ ,

$y = 13$

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

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

$\Rightarrow 4 +3y = 13$

$\Rightarrow 3y = 13-4$

$\Rightarrow 3y = 9$

$\Rightarrow y = 3$

$\Rightarrow x = 2 , y = 3$

Solve the following pair of linear equations by the substitution method

$s-t=3, \dfrac {s}{3}+\dfrac {t}{2}=6$

0%
$s = 2, t = 7$
0%
$s = 9, t = 6$
0%
$s = 3, t = 1$
0%
$s = 6, t = -9$

Explanation

$s-t=3$ .....(i)

$\frac {s}{3}+\frac {t}{2}=6$ .....(ii)
From (i)

$s = t + 3$ ...(iii)
Substituting s from (iii) in (ii), we get

$\frac {t+3}{3}+\frac {t}{2}=6$

$\Rightarrow 2(t + 3) + 3t = 36$

$\Rightarrow 5t + 6 = 36$

$\Rightarrow t = 6$
From (iii),

$s =6 +3 = 9$ ,
Hence,

$s = 9, t = 6$

A fraction becomes

$\dfrac {9}{11}$ , if

$2$ is added to both the numerator and the denominator. If

$3$ is added to both the numerator and the denominator it becomes

$\dfrac {5}{6}$ . Find the fraction.

0%
$\dfrac{7}{9}$
0%
$\dfrac{5}{7}$
0%
$\dfrac{9}{11}$
0%
None of these

Explanation

Let $\dfrac {x}{y}$ be the fraction, where $x$ and $y$ are positive integers.

$\Rightarrow \dfrac {x+2}{y+2}=\dfrac {9}{11}$ ..(1)
$\Rightarrow \dfrac {x+3}{y+3}=\dfrac {5}{6}$ ..(2)

From (1) we get
$\Rightarrow 11x+22 = 9y+18$
$\Rightarrow 11x-9y = -4$ ...(3)

From (2) we get
$\Rightarrow 6x +18 = 5y +15$
$\Rightarrow 6x-5y = -3$ ...(4)

From (4) will get
$y = \dfrac{3+6x}{5}$

Now substituting $y$ in (3),

$\Rightarrow 11x-\dfrac{9\left ( 3+6x \right )}{5} = -4$

$\Rightarrow 55x-27-54x = -20$

$\Rightarrow 55x-54x = -20+27$

$\Rightarrow x = 7$

Now substituting $x$ in (4)

$\Rightarrow 6\times 7 -5y = -3$

$\Rightarrow 42-5y = -3$

$\Rightarrow -5y = -3 - 42$

$\Rightarrow -5y = -45$

$\Rightarrow y = 9$

Hence, $x = 7$ and $y = 9$ .

And fraction $\dfrac{x}{y}=\dfrac{7}{9}$

Five years hence, the age of jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages?

0%
Present age of Jacob: $50$ years and Present age of Jacob's son: $15$ years
0%
Present age of Jacob: $50$ . years and Present age of Jacob's son: $10$ years
0%
Present age of Jacob: $40$ years and Present age of Jacob's son: $10$ years
0%
Present age of Jacob: $35$ years and Present age of Jacob's son: $15$ years

Explanation

Let

$x$ be the present age of Jacob's son and

$y$ be the present age of Jacob. Then as per the question

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

$(y- 5) = 7 (x - 5)\ ......(2)$

From equation

$(1)$ ,

$\Rightarrow y +5 = 3x +15$

$\Rightarrow 3x-y = -10\ ......(3)$

From equation

$(2)$ ,

$\Rightarrow y-5 = 7x-35$

$\Rightarrow 7x-y = 30\ .......(4)$

from equation

$(3)$ we get

$\Rightarrow -y = -10 -3x$

$\Rightarrow y = 10 +3x$

Now Substitute

$y$ in

$(4)$

$\Rightarrow 7x-\left ( 10+3x \right ) = 30$

$\Rightarrow 7x -10-3x = 30$

$\Rightarrow 4x = 30+10$

$\Rightarrow 4x = 40$

$\Rightarrow x = 10$

Now substituting

$x$ in (1) will give

$\Rightarrow 3\times 30-y = -10$

$\Rightarrow 30 - y = -10$

$\Rightarrow -y = -40$

$\Rightarrow y = 40$

Hence, Jacob's son's current age

$=$

$10$ and Jacob's Current age

$=$

$40$ .

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

The larger of two supplementary angles exceeds the smaller by

$18$ degrees.

0%
$x + y = 100$ and $x - y = 18$ , $x=90$ and $\,y=16$
0%
$x + y = 60$ and $x - y = 18$ , $x=95$ and $\,\,y=70$
0%
$x + y = 180$ and $x - y = 18$ , $x=99$ and $\,\,y=81$
0%
None of these

Explanation

Let the supplementary angles be

$x$ and

$y$ ,

$(x> y)$

$\therefore$ Sum of both angles is

$180^0$

$\Rightarrow x + y = 180$ ...

$(1)$

The larger angle exceeds the smaller by

$18^0$

$\Rightarrow x - y = 18$ ...

$(2)$

$\Rightarrow y=x -18$

Put value of

$y$ in (1), we get

$2x=198$

$\Rightarrow x = 99$

Now substituting the value of

$x$ in equation

$(1)$

$\Rightarrow 99+y = 180$

$\Rightarrow y = 180-99$

$\Rightarrow y = 81$

Hence,

$x = 99^0$ and

$y = 81^0$

The pair of linear equations for the given problem are

$x + y = 180$ and

$x - y = 18$ .

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

Five years ago Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

0%
Nuri is $40$ years old and Sonu is $10$ years old.
0%
Nuri is $50$ years old and Sonu is $20$ years old.
0%
Nuri is $70$ years old and Sonu is $30$ years old.
0%
Nuri is $60$ years old and Sonu is $10$ years old.

Explanation

Let present age of Nuri=

$x$ years
Let present age of sonu=

$y$ years
Five years ago
Nuri was

$\left(x-5 \right) years$
Sonu was

$\left(y-5 \right) years$
According to the first condition

$\Rightarrow \left(x-5 \right)=3\left(y-5 \right)$

$\Rightarrow x-5=3y-15 \Rightarrow x-3y=-10.........eq1$
Ten years later
Nuri will be

$\left(x+10 \right) years$
Sonu will be

$\left(y+10 \right)$

$\Rightarrow \left(x+10 \right)=2\left(y+10 \right)$

$\Rightarrow x+10=2y+20 \Rightarrow x+2y=10.....eq2$
Subtracting

$eq 1 and eq2$

$\Rightarrow -y=-2- \Rightarrow y=20$
put y=20 in

$eq1$

$x-3y=10 \Rightarrow x=10-60 \Rightarrow x=50$
Hence, present age of Nuri= 50 years
Let present age of sonu=20 years

Solve the following pair of equations by reducing them to a pair of linear equations:

$\dfrac {1}{(x-1)}+\dfrac {1}{(y-2)}=2, \ \dfrac {6}{(x-1)}-\dfrac {2}{(y-2)}=1$

0%
$x=\dfrac{11}{5},\,\,y=\dfrac{13}{11}$
0%
$x=\dfrac{13}{5},\,\,y=\dfrac{30}{11}$
0%
$x=\dfrac{4}{5},\,\,y=\dfrac{20}{11}$
0%
None of these

Explanation

The given equations are

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

$\dfrac { 6 }{ x-1 } -\dfrac { 2 }{ y-2 } =1$

Let $\dfrac { 1 }{ x-1 } =u$ and $\dfrac { 1 }{ y-2 } =v$

$u+v=2$ .......(i) and

$6u-2v=1$ ......(ii)

Multiply (i) by $6$

$\Rightarrow 6u+6v=12$ ......(iii)

Subtract (ii) from (iii)

$\Rightarrow 8v=11$

$\Rightarrow v=\dfrac { 11 }{ 8 }$

Therefore $\dfrac { 1 }{ y-2 } =\dfrac { 11 }{ 8 }$

$\Rightarrow y-2=\dfrac { 8 }{ 11 }$

$\Rightarrow y=\dfrac { 30 }{ 11 }$

Putting $v=\dfrac { 11 }{ 8 }$ in (i), we get

$u+\dfrac { 11 }{ 8 } =2$

$\Rightarrow u=\dfrac { 5 }{ 8 }$

$\Rightarrow \dfrac { 1 }{ x-1 } =\dfrac { 5 }{ 8 }$

$\Rightarrow x-1=\dfrac { 8 }{ 5 }$

$\Rightarrow x=\dfrac { 13 }{ 5 }$

Therefore, $\left( x,y \right) =\left( \dfrac { 13 }{ 5 } ,\dfrac { 30 }{ 11 } \right)$

Solve the following pairs of equations by reducing them to a pair of linear equations:

$\displaystyle \dfrac {7x-2y}{xy}=5, \dfrac {8x+7y}{xy}=15$

0%
$x=0,\,\,y=2$
0%
$x=2,\,\,y=7$
0%
$x=13,\,\,y=0$
0%
$x=1,\,\,y=1$

Explanation

The given equations are

$\displaystyle \frac { 7x-2y }{ xy } =5\quad \& \quad \frac { 8x+7y }{ xy } =15.$

On simplifying, we get

$\displaystyle \frac { 7 }{ y } -\frac { 2 }{ x } =5$ and

$\dfrac { 8 }{ y } +\dfrac { 7 }{ x } =15\\$

Let

$\displaystyle \frac { 1 }{ x } =u$ and

$\dfrac { 1 }{ y }$

Then the equations reduce to

$7v-2u=5$ ...(i) and

$8v+7u=15$ ...(ii)

Multiplying (i) by

$8$ and (ii) by

$7$ , we get

$56v-16u=40$ ...(iii) and

$56v+49u=105$ ...(iv)

Subtracting (iv) from (iii),

$-65u=-65$

$\Rightarrow u=1$ .

$\therefore \dfrac { 1 }{ x } =1\Rightarrow x=1$

Putting

$u=1$ in

$(i)$ ,

$7v-2\times 1=5\\ \Rightarrow v=1\\ \Rightarrow \dfrac { 1 }{ y } =1\\ \Rightarrow y=1$

$(x,y)=( 1,1)$

The real numbers

$x$ and

$y$ are such that

$\displaystyle x+\frac{2}{y}=\frac{8}{3}$ and

$\displaystyle y+\frac{2}{x}=3$ .

The value of

$xy$ , is

0%
$\displaystyle \frac{4}{3}$
0%
$\displaystyle \frac{16}{9}$
0%
$2$
0%
$4$

Explanation

$\displaystyle x+\frac{2}{y}=\frac{8}{3}$ ...(i)

$\displaystyle y+\frac{2}{x}=3\Rightarrow y=\frac{3x-2}{x}$ ...(ii)

From (i)

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

$\Rightarrow \displaystyle 3(3x^2-2x+2x)=8(3x-2)$

$\Rightarrow \displaystyle 9x^2-24x+16=0$

$\Rightarrow \displaystyle (3x-4)^2=0$

$\Rightarrow \displaystyle x=\frac43$

From (ii), we get

$y=\dfrac{3\times\frac43-2}{\frac43}=\dfrac32$
Now,

$xy=\dfrac43\times\dfrac32=2$
Hence, option C is correct.

Solve the following pair of equations by reducing them to a pair of linear equations:

$\displaystyle \frac {2}{\sqrt x}+\frac {3}{\sqrt y}=2, \frac {4}{\sqrt x}-\frac {9}{\sqrt y}=-1$

0%
$x=1,\,\,y=4$
0%
$x=3,\,\,y=2$
0%
$x=4,\,\,y=9$
0%
$x=16,\,\,y=25$

Explanation

Let us put

$\displaystyle \frac { 1 }{ \sqrt { x } } =u\quad \& \quad \frac { 1 }{ \sqrt { y } } =v$ .

Then the equations become

$2u+3v=2$ ......(i) and

$4u-9v=-1$ .....(ii)

Multiplying (i) by

$2$ and subtracting the result from (ii),

$-15v=-5$

$\Rightarrow v= \dfrac { 1 }{ 3 }$

So,

$\dfrac { 1 }{ \sqrt { y } } =\dfrac { 1 }{ 3 }$

$\Rightarrow y=9$

Again, putting

$v= \dfrac { 1 }{ 3 }$ in

$(i)$ ,

$2u+3\times \dfrac { 1 }{ 3 } =2$

$\Rightarrow u=\dfrac { 1 }{ 2 }$ i.e.

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

$\Rightarrow x=4.\\$

So,

$(x,y)=(4,9)$

Hence, option C is correct.

For which values of

$a$ and

$b$ does the following pair of linear equations have an infinite number of solutions?

$2x + 3y = 7$

$(a - b)x + (a + b) y = 3 a + b - 2$

0%
$a=1$ and $b=7$
0%
$a=13$ and $b=6$
0%
$a=2$ and $b=10$
0%
$a=5$ and $b=1$

Explanation

Let

$2x + 3y - 7 = 0$ ........(i)
and

$(a- b)x + (a + b)y - (3 a + b - 2) = 0$ .......(ii)
For infinite number of solutions, we have

$\dfrac {a-b}{2}=\dfrac {a+b}{3}=\dfrac {3a+b-2}{7}$
For first and second, we have

$\dfrac {a-b}{2}=\dfrac {a+b}{3}$ or

$3a - 3b = 2a + 2b$

$a = 5b$ ........(i)
From second and third, we have

$\dfrac {a+b}{3}=\dfrac {3a+b-2}{7}$

$7a + 7b = 9a + 3b - 6 \ or \ 4b = 2a - 6$

$2b = a - 3$ ........(ii)
From (i) and (ii), eliminating

$a$ ,

$2b = 5b - 3 \Rightarrow b = 1$
Substituting

$b= 1$ in (i), we get

$a = 5$

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

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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