Pair of Linear Equations in Two Variables Test

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Pair of Linear Equations in Two Variables Test - 54

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Weekly Quiz Competition

Question 1
1 / -0

Using graphical method check whether the given equation is consistent:
$$3x- y = 4$$ and $$7x + 2y = 18$$

A
$$(2, 2)$$

B
$$(-2, -2)$$

C
$$(-2, 2)$$

D
$$(2, -2)$$

Solution

$$3x- y = 4$$ ----- (1)
$$7x + 2y = 18$$ ----- (2)
From equation (1) assume the value of $$x$$ and $$y$$ to satisfy the equation to zero.
$$3x- y- 4 = 0$$ ------ (3)
Put $$x = 0, y = -4$$ in equation (3)
$$3(0)- (-4)- 4 = 0$$
$$4- 4 = 0$$
$$0 = 0$$
Again put $$x = 2, y = 2$$ in equation (3)
$$3(2)- (2) -4 = 0$$
$$6- 2 - 4 = 0$$
$$6- 6 = 0$$
$$0 = 0$$
Now plotting $$(0, -4), (2, 2)$$ and joining them, we get a straight line.
From equation (2) assume the value of $$x$$ and $$y$$ to satisfy the equation to zero.
$$7x + 2y- 18 = 0$$ ----- (4)
Put $$x = 0, y = 9$$ in equation (4)
$$7(0) + 2(9) -18 = 0$$
$$18- 18 = 0$$
$$0 = 0$$
Again put $$x = 2, y = 2$$ in equation (4)
$$7(2) + 2(2)- 18 = 0$$
$$14 + 4- 18 = 0$$
$$18 -18 = 0$$
$$0 = 0$$
Plotting $$(0, 9), (2, 2)$$ and joining them, we get another straight line.
These lines intersect at the point $$(2, 2)$$ and therefore the solution of the equation is $$x = 2, y = 2$$.
Question 2
1 / -0

The equations $$4x - y = 1$$ and $$2x - 2y = 2$$ are __________.

A
Non-linear

B
Inconsistent

C
Consistent

D
Dependent

Solution

The general form for a pair of linear equations in two variables $$x$$ and $$y$$ is
$$a_{1}x + b_{1}y + c_{1}= 0$$ and
$$a_{2}x + b_{2}y + c_{2}= 0$$
$$4x - y - 1=0$$---$$(1)$$
$$2x - 2y -2=0$$---$$(2)$$
Comparing equations $$(1)$$ and $$(2)$$ with the general form of equations to consider their co-efficient,
$$a_{1}=4, a_{2} =2,b_{1}= -1, b_{2}= -2, c_{1}=-1, c_{2}=-2$$
We have, $$\dfrac{4}{2}\neq\dfrac{-1}{-2}$$
$$\Rightarrow \dfrac{a_{1}}{a_{2}}\neq\dfrac{b_{1}}{b_{2}}$$
Hence, the equations are consistent.
Question 3
1 / -0

The equation $$-4x - 4y = 2$$ and $$-2x - 2y = 1$$ have _________.

A
2 solutions

B
unique solution

C
no solutions

D
infinitely many solutions

Solution

The general form for a pair of linear equations in two variables $$x$$ and $$y$$ is
$$a_{1}x + b_{1}y + c_{1}= 0$$ and
$$a_{2}x + b_{2}y + c_{2}= 0$$
$$-4x -4y = 2$$---(1)
$$-2x - 2y = 1$$---(2)
Comparing equations (1) and (2) with the general form of equation to consider their co-efficient,
$$a_{1}=-4, a_{2} =-2,b_{1}= -4, b_{2}= -2, c_{1}=2, c_{2}=1$$
So, $$\dfrac{-4}{-2}=\dfrac{-4}{-2}=\dfrac{2}{1}$$
We know that $$\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}$$ is said to be dependent.
Hence, the equations are satisfying the above condition.
It is dependent and hence has infinitely many solutions
Question 4
1 / -0

The sum of unit place and tens place digits of a two digit number is $$12$$. If the new number formed by reversing those digits is less than the original number by $$18$$. Find the original number.

A
$$70$$

B
$$75$$

C
$$80$$

D
$$85$$

Solution

Let the unit place digit $$= x$$
Ten's place digit $$= 12 - x$$
Original number $$=$$ $$10\times(12 - x) + x$$
$$= 120 - 10x + x $$
$$=120 - 9x$$
Reversing the digit:
Unit digit $$= 12 - x$$
Ten's digit $$= x$$
New number formed $$=$$ $$10\times x + 12 - x$$
$$= 9x + 12$$
As new number is less than the original number by $$18$$
New number $$=$$ original number $$- 18$$
$$9x + 12 = 120 - 9x - 18$$
$$18x = 120 - 30$$
$$18x =90$$
$$x = 5$$
Original number $$= 120 - 9x$$
$$= 120 - 9 \times 5$$
$$= 120 -45$$
$$= 75$$
Original number $$= 75.$$
Question 5
1 / -0

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.

A
$$100, 80$$

B
$$80, 100$$

C
$$110, 70$$

D
$$70, 110$$

E
None of these

Solution

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 $$(i)$$.
$$(i)-(ii) :$$
$$a- b -(a-2b) = 20-(-60)$$
$$\implies b =80$$
$$a =80+20 =100$$
Question 6
1 / -0

If the sum of $$2$$ real numbers is $$20$$ and their difference is $$6$$, find the value of their product.

A
$$51$$

B
$$64$$

C
$$75$$

D
$$84$$

E
$$91$$

Solution

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 $$y=$$ $$x\times y=13\times 7=91$$
Question 7
1 / -0

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

A
No

B
Only one

C
Only two

D
An infinite number of slutions

Solution

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.
Question 8
1 / -0

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.

A
$$-2$$

B
$$-\displaystyle\frac{1}{2}$$

C
$$2$$

D
$$6$$

Solution

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 $$a=6$$
Question 9
1 / -0

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

A
4

B
8

C
12

D
16

E
20

Solution
Question 10
1 / -0

If $$\left| x \right| +x+y=10$$ and $$x+\left| y \right| -y=12$$ then find the value of '$$x+y$$'.

A
$$\dfrac { 18 }{ 5 } $$

B
$$\dfrac { 17 }{ 5 } $$

C
$$\dfrac { 8 }{ 5 } $$

D
None of these

Solution

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)
$$ \dfrac{32}{5}-2y = 12$$
$$ 2y = \dfrac{32}{5}-12 $$
$$ y = \dfrac{-14}{5} $$
$$ \therefore x+y=\dfrac{32}{5}+\dfrac{-14}{5}$$
$$ \therefore x+y=\dfrac{18}{5}$$

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Pair of Linear Equations in Two Variables Test - 54

Result

Pair of Linear Equations in Two Variables Test - 54

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SHARING IS CARING

Question 1

$$(2, 2)$$

$$(-2, -2)$$

$$(-2, 2)$$

$$(2, -2)$$

Solution

Question 2

Non-linear

Inconsistent

Consistent

Dependent

Solution

Question 3

2 solutions

unique solution

no solutions

infinitely many solutions

Solution

Question 4

$$70$$

$$75$$

$$80$$

$$85$$

Solution

Question 5

$$100, 80$$

$$80, 100$$

$$110, 70$$

$$70, 110$$

None of these

Solution

Question 6

$$51$$

$$64$$

$$75$$

$$84$$

$$91$$

Solution

Question 7

No

Only one

Only two

An infinite number of slutions

Solution

Question 8

$$-2$$

$$-\displaystyle\frac{1}{2}$$

$$2$$

$$6$$

Solution

Question 9

4

8

12

16

20

Solution

Question 10

$$\dfrac { 18 }{ 5 } $$

$$\dfrac { 17 }{ 5 } $$

$$\dfrac { 8 }{ 5 } $$

None of these

Solution

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