MCQ 11 Mark
If $x = -y$ and $y > 0$, which of the following is wrong?
AnswerCorrect option: D. $\frac{1}{\text{x}}-\frac{1}{\text{y}}=0$
Given:
$x = -y$ and $y > 0$
Now, we have:
$x^2y$
On substituting $x = -y$, we get:
$(-y)^2y = y^3 > 0$ $(\therefore\text{y}>0)$
This is true.
$x + y$
On substituting $x = -y$, we get:
$(-y) + y = 0$
This is also true.
$xy$
On substituting $x = -y,$ we get:
$(-y)y = -y^2 < 0 $$(\therefore\text{y}>0)$
This is again true.
$\frac{1}{\text{x}}-\frac{1}{\text{y}}=0$
$\Rightarrow\frac{\text{y}-\text{x}}{\text{xy}}=0$
On substituting $x = -y$, we get:
$\frac{\text{y}-(-\text{y})}{(-\text{y})\text{y}}=0$
$\Rightarrow\frac{2\text{y}}{-\text{y}^2}=0$
$\Rightarrow2\text{y}=0$
$\Rightarrow\text{y}=0$
Hence, from the above equation, we get $y = 0$, which is wrong.
View full question & answer→MCQ 21 Mark
If the lines given by $3x + 2ky = 2$ and $2x + 5y + 1 = 0$ are parallel then the value of $k$ is:
- A
$\frac{-5}{4}$
- B
$\frac{2}{5}$
- C
$\frac{3}{2}$
- ✓
$\frac{15}{4}$
AnswerCorrect option: D. $\frac{15}{4}$
$3x + 2ky = 2$ and $2x + 5y + 1 = 0$
$3x + 2ky - 2 = 0$ and $2x + 5y + 1 = 0$
If the lines are parallel, it means the system has an infinite number of solutions.
We know that,
the system of linear equations $a_1x + b_1x + c_1 = 0, a_2x + b_2y + c_2 = 0$ has a unique solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}.$
So, $\frac{3}{2}=\frac{2\text{k}}{5}=\frac{-2}{1}$
$\Rightarrow\frac{3}{2}=\frac{2\text{k}}{5}$
$\Rightarrow\text{k}=\frac{15}{4}$
View full question & answer→MCQ 31 Mark
If $\frac{2\text{x}}{3}-\frac{\text{y}}{2}+\frac{1}{6}=0$ and $\frac{\text{x}}{2}+\frac{2\text{y}}{3}=3$ then:
- ✓
$x = 2, y = 3$
- B
$x = -2, y = 3$
- C
$x = 2, y = -3$
- D
$x = -2, y = -3$
AnswerCorrect option: A. $x = 2, y = 3$
$\frac{2\text{x}}{3}-\frac{\text{y}}{2}+\frac{1}{6}=0$
Multiply by the $\text{LCM,} 6.$
$\Rightarrow 4x - 3y + 1 = 0$
$\Rightarrow 4x - 3y = -1 ....(i)$
$\frac{\text{x}}{2}+\frac{2\text{y}}{3}=3$
Multiply by the $\text{LCM,} 6.$
$3x + 4y = 18 ...(ii)$
Multiply equation $(i)$ and $(ii)$ by $4$ and $3$ respectively.
$16x - 12y = -4 ...(iii)$
$9x + 12y = 54 ...(iv)$
Adding equations $(iii)$ and $(iv),$ we get
$25x = 50$
$\Rightarrow x = 2$
Substituting $x = 2$ in $(ii),$ we get $y = 3.$
View full question & answer→MCQ 41 Mark
The sum of the digits of a two$-$digit number is $15.$ The number obtained by interchanging the digits exceeds the given number by $9$. The number is:
AnswerLet the two-digit number be $xy.$
The given number $= 10x + y$
So, $x + y = 15 ...(i)$
The number obtained by interchanging the digits is $yx.$
$\Rightarrow 10y + x = 10x + y + 9$
$\Rightarrow -9x + 9y = 9$
$\Rightarrow -x + y = 1 ...(ii)$
Adding $(i)$ and $(ii),$ we get
$2y = 16$
$\Rightarrow y = 8$
Substituting in $(i),$ we get $x = 7.$
So, the given number is $xy = 78.$
View full question & answer→MCQ 51 Mark
If $2x - 3y = 7$ and $(a + b)x - (a + b - 3)y = 4a + b$ have an infinite number of solutions, then:
- A
$a = 5, b = 1$
- B
$a = -5, b = 1$
- C
$a = 5, b = -1$
- ✓
$a = -5, b = -1$
AnswerCorrect option: D. $a = -5, b = -1$
The given system equations can be written as follows:
$2x - 3y - 7 = 0$ and $(a + b) x - (a + b - 3) y - (4a + b) = 0$
The given equations are of the following form:
$a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$
Here, $a_1 = 2, b_1 = -3 c_1 = -7$ and $a_2 = (a + b), b_2 = - (a + b - 3)$ and $c_2 = - (4a + b)$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{2}{\text{(a+b)}},\frac{\text{b}_1}{\text{b}_2}=\frac{-3}{-(\text{a}+\text{b}-3)}=\frac{3}{-(\text{a}+\text{b}-3)}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{-7}{-(4\text{a}+\text{b)}}=\frac{7}{(4\text{a}+\text{b)}}$
For an infinite number of solution, we must have:
$\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\therefore\frac{2}{(\text{a}+\text{b)}}=\frac{3}{(\text{a}+\text{b}-3)}=\frac{7}{(4\text{a}+\text{b})}$
Now, we have:
$\frac{2}{\text({a}+\text{b})}=\frac{3}{(\text{a}+\text{b}-3)}$
$\Rightarrow\text{a}-6=\text{3a}+\text{3b}$
$\Rightarrow\text{a}+\text{b}+6=0\ ...(\text{i})$
Again, we have:
$\frac{3}{(\text{a}+\text{b}-3)}=\frac{7}{(4\text{a}+\text{b})}$
$\Rightarrow12\text{a}+3\text{b}=7\text{a}+7\text{b}-21$
$\Rightarrow5\text{a}-7\text{b}+21=0\ ....(\text{ii})$
On multiplying $(i)$ by $4$, we get:
$4a + 4b + 24 = 0$
On substituting $a = -5$ in, we get:
$\Rightarrow -5 + b + 6 = 0$
$\Rightarrow b = -1$
$\therefore a = -5$ and $b = -1$
View full question & answer→MCQ 61 Mark
In the given fraction, if $1$ is subtracted from the numerator and $2$ is added to the denominator, it become $\frac{1}{2}.$ If $7$ is subtracted from the numerator and $2$ is subtracted from the denominator, it become $\frac{1}{3}.$ The fraction is:
- A
$\frac{13}{24}$
- ✓
$\frac{15}{26}$
- C
$\frac{16}{27}$
- D
$\frac{16}{21}$
AnswerCorrect option: B. $\frac{15}{26}$
Let the fraction be $\frac{\text{x}}{\text{y}}.$
According to the first condition,
$\frac{\text{x}-1}{\text{y}+2}=\frac{1}{2}$
$\Rightarrow2\text{x}-2=\text{y}+2$
$\Rightarrow2\text{x}-\text{y}=4\ ...(\text{i})$
According to the second condition,
$\frac{\text{x}-7}{\text{y}-2}=\frac{1}{3}$
$\Rightarrow3\text{x}-21=\text{y}-2$
$\Rightarrow3\text{x}-\text{y}=19\ ...(\text{ii})$
Subtracting in $(i)$ from $(ii)$, we get
$\Rightarrow x = 15$
Substituting in $(i),$ we get $y = 26.$
So, the fraction is $\frac{15}{26}.$
View full question & answer→MCQ 71 Mark
The graphs of the equations $6x - 2y + 9 = 0$ and $3x - y + 12 = 0$ are two lines which are:
- A
- ✓
- C
Intersecting exactly at one point.
- D
Perpendicular to each other.
Answer$6x - 2y + 9 = 0$ and $3x - y + 12 = 0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{6}{3}=2$
$\frac{\text{b}_1}{\text{b}_2}=\frac{-2}{-1}=2$
$\frac{\text{c}_1}{\text{c}_2}=\frac{9}{12}=\frac{3}{4}$
Clearly, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}.$
We know that,
If in a system of linear equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$
We have $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2},$ then the given system has no solution.
So, the lines are parallel.
View full question & answer→MCQ 81 Mark
The graphs of the equations $5x - 15y = 8$ and $3\text{x}-9\text{y}=\frac{24}{5}$ are two lines which are:
- ✓
- B
- C
Intersecting exactly at one point.
- D
Perpendicular to each other.
View full question & answer→MCQ 91 Mark
$5$ years hence, the age of a man shall be $3$ times the age of his son. while $5$ years earlier the age of the man was $7$ times the age of his son. The present age of the man is:
- A
$45$ years.
- B
$50$ years.
- C
$47$ years.
- ✓
$40$ years.
AnswerCorrect option: D. $40$ years.
Let the present age of the man be $x$ years,
and his son's age be $y$ years.
According to the first condition,
$x + 5 = 3(y + 5)$
$\Rightarrow x + 5 = 3y + 15$
$\Rightarrow x - 3y = 10 ...(i)$
According to the second condition,
$x - 5 = 7(y - 5)$
$\Rightarrow x - 5 = 7y - 35$
$\Rightarrow x - 7y = -30 ...(ii)$
Subtracting $(ii)$ from $(i),$ we get
$4y = 40$
$\Rightarrow y = 10$
Substituting $y = 10$ in $(i),$ we get
$\Rightarrow x = 40.$
So, the present age of the man is $40$ years.
View full question & answer→MCQ 101 Mark
The pair of equations $2x + 3y = 5$ and $4x + 6y = 15$ has:
- A
- B
- C
Infinitely many solutions.
- ✓
Answer$2x + 3y = 5$ and $4x + 6y = 15$
$\Rightarrow 2x + 3y - 5 = 0$ and $4x + 6y - 15 = 0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2}$
$\frac{\text{c}_1}{\text{c}_2}=\frac{-5}{-15}=\frac{1}{3}$
Clearly, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
We know that,
The system of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ has no solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}.$
So, the pair of equations has no solution.
View full question & answer→MCQ 111 Mark
If $29x + 37y = 103$ and $37x + 29y = 95$ then:
Answer$29x + 37y = 103 ...(i)$
$37x + 29y = 95 ...(ii)$
Adding $(i)$ and $(ii),$ we get
$66x + 66y = 198$
$\Rightarrow x + y = 3 ...(iii)$
Subtract $(i)$ from $(ii),$ we get
$8x - 8y = 8$
$\Rightarrow x - y = 1 ....(iv)$
Adding $(iii)$ and $(iv)$, we get
$2x = 4$
$\Rightarrow x = 2$
Substituting $x = 2$ in $(iii),$ we get $y = 1.$
View full question & answer→MCQ 121 Mark
The graphs of the equations $2x + 3y - 2 = 0$ and $x - 2y - 8 = 0$ are two lines which are:
- A
- B
- ✓
Intersecting exactly at one point.
- D
Perpendicular to each other.
AnswerCorrect option: C. Intersecting exactly at one point.
$2x + 3y + 9 = 0$ and $x - 2y + 8 = 0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{2}{1}=2$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{-12}$
$\frac{\text{c}_1}{\text{c}_2}=\frac{9}{-8}$
Clearly, $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
We know that,
If in a system of linear equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$
We have $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$ then the system has a unique solution.
So, the pair of lines are intersecting exactly at one point.
View full question & answer→MCQ 131 Mark
In a $\triangle\text{ABC},\ \angle\text{C}=3\angle\text{B}=2(\angle\text{A}+\angle\text{B}),$ then $\angle\text{B}=?$
- A
$20^\circ$
- ✓
$40^\circ$
- C
$60^\circ$
- D
$80^\circ$
AnswerCorrect option: B. $40^\circ$
Give that in a $\triangle\text{ABC},$
$\angle\text{C}=3\angle\text{B}=2(\angle\text{A}+\angle\text{B})$
$\Rightarrow\angle\text{C}=3\angle\text{B}$ and $\angle\text{C}=2(\angle\text{A}+\angle\text{B})$
Consider, $\angle\text{C}=2(\angle\text{A}+\angle\text{B})$
$\Rightarrow3\angle\text{B}=2(\angle\text{A}+\angle\text{B})$
$\Rightarrow\angle\text{B}=2\angle\text{A}$
By the Angle Sum Property
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{A}+2\angle\text{A}+2(\angle\text{A}+2\angle\text{A})=180^\circ$
$\Rightarrow\angle\text{A}+2\angle\text{A}+2\angle\text{A}+4\angle\text{A}=180^\circ$
$\Rightarrow9\angle\text{A}=180^\circ$
$\Rightarrow\angle\text{A}=20^\circ$
So, $\angle\text{B}=2\angle\text{A}$
$\Rightarrow\angle\text{B}=40^\circ$
View full question & answer→MCQ 141 Mark
If $2^{\text{x+y}}=2^{\text{x}-\text{y}}=\sqrt8$ then the value of $y$ is:
- A
$\frac{1}{2}$
- B
$\frac{3}{2}$
- ✓
$0$
- D
Answer$\Rightarrow2\text{x} +\text{y}=2\text{x}-\text{y}=\sqrt{8}$
$\Rightarrow2\text{x}+\text{y}=\sqrt{8}\ ...(\text{i})$
$\Rightarrow2\text{x}-\text{y}=\sqrt{8}\ ...(\text{ii})$
Adding $(i)$ and $(ii),$ we get
$4\text{x}=2\sqrt8$
$\Rightarrow\text{x}=\frac{\sqrt8}{2}$
Substituting $\text{x}=\frac{\sqrt8}{2}$ in $(i),$ we get $y = 0.$
View full question & answer→MCQ 151 Mark
If $2x + 3y = 12$ and $3x - 2y = 5$ then:
- A
$x = 2, y = 3$
- B
$x = 2, y = -3$
- ✓
$x = 3, y = 2$
- D
$x = 3, y = -2$
AnswerCorrect option: C. $x = 3, y = 2$
$2x + 3y = 12 ....(i)$
$3x - 2y = 5 ...(ii)$
Multiplying equation $(i)$ and $(ii)$ by $2$ and $3$ respectively.
$4x + 6y = 24 ...(iii)$
$9x - 6y = 15 ....(iv)$
Adding equations $(iii)$ and $(iv),$ we get
$13x = 39$
$\Rightarrow x = 3$
Substituting $x = 3$ in $(ii),$ we get $y = 2.$
View full question & answer→MCQ 161 Mark
If $\frac{3}{\text{x+y}}+\frac{2}{\text{x}-\text{y}}=2$ and $\frac{9}{\text{x+y}}-\frac{4}{\text{x}-\text{y}}=1$ then:
- A
$\text{x}=\frac{1}{2},\ \text{y}=\frac{3}{2}$
- ✓
$\text{x}=\frac{5}{2},\ \text{y}=\frac{1}{2}$
- C
$\text{x}=\frac{3}{2},\ \text{y}=\frac{1}{2}$
- D
$\text{x}=\frac{1}{2},\ \text{y}=\frac{5}{2}$
AnswerCorrect option: B. $\text{x}=\frac{5}{2},\ \text{y}=\frac{1}{2}$
$\frac{3}{\text{x+y}}+\frac{2}{\text{x}-\text{y}}=2$
$\frac{9}{\text{x+y}}-\frac{4}{\text{x}-\text{y}}=1$
Put $\frac{1}{\text{x+y}}=\text{u}$ and $\frac{1}{\text{x}-\text{y}}=\text{v}$
So, we get
$3u + 2v = 2 ...(i)$
$9u - 4v = 1 ...(ii)$
Multiply $(i)$ by $2$ and add it to $(ii).$
$\Rightarrow 6u + 4v = 4$
$\Rightarrow 15u = 5$
$\Rightarrow\text{u}=\frac{1}{3}$
Substituting $\text{u}=\frac{1}{3}$ in $(i),$ we get $\text{v}=\frac{1}{2}.$
$\Rightarrow\frac{1}{\text{x+y}}=\frac{1}{3}$ and $\frac{1}{\text{x}-\text{y}}=\frac{1}{2}$
$\Rightarrow x + y = 3 ...(iii)$
$\Rightarrow x - y = 2 ...(iv)$
Adding $(iii)$ and $(iv),$ we get
$2\text{x}=5$
$\Rightarrow\text{x}=\frac{5}{2}$
Substituting $\text{x}=\frac{5}{2}$ in $(iii),$ we get $\text{y}=\frac{1}{2}.$
View full question & answer→MCQ 171 Mark
In a cyclic quadrilateral $\text{ABCD,}$ it is being given that $\angle\text{A}=(\text{x+y}+10)^\circ,$
$\angle\text{B}=(\text{y}+20)^\circ,$
$ \angle\text{C}=(\text{x+y}-30)^\circ$ and $\angle\text{D}=(\text{x+y}).$ then, $\angle\text{B}=?$
- A
$70^\circ$
- ✓
$80^\circ$
- C
$100^\circ$
- D
$110^\circ$
AnswerCorrect option: B. $80^\circ$
Given that in cydic quadrilateral $\text{ABCD,}$
$\angle\text{A}=(\text{x}+\text{y}+10)^\circ, \angle\text{B}=(\text{y+20})^\circ,$
$\angle\text{C}=(\text{x}+\text{y}+30)^\circ$ and $\angle\text{D}=(\text{x}+\text{y})^\circ$
We know that,
Opposite angles of a quadrilateral sum upto $180^\circ .$
$\Rightarrow\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow(\text{y}+20)^\circ+(\text{x+y})^\circ=180^\circ$
$\Rightarrow\text{x}+2\text{y}=160\ .....(\text{i})$
Similarly, $\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow(\text{x+y}+10)^\circ+(\text{x+y}-30)^\circ=180^\circ$
$\Rightarrow2\text{x}+2\text{y}=200$
$\Rightarrow\text{x+y}=100\ ...(\text{ii})$
Subtracting $(ii)$ from $(i),$ we get
$\text{y}=60$
$\angle\text{B}=(60+20)^\circ=80^\circ$
View full question & answer→MCQ 181 Mark
If $\frac{2\text{x}+\text{y}+2}{5}=\frac{3\text{x}-\text{y}+1}{3}=\frac{3\text{x}+2\text{y}+1}{6}$ then:
Answer$\frac{2\text{x}+\text{y}+2}{5}=\frac{3\text{x}-\text{y}+1}{3}=\frac{3\text{x}+2\text{y}+1}{6}$
$\Rightarrow\frac{2\text{x}+\text{y}+2}{5}=\frac{3\text{x}-\text{y}+1}{3}$ and $\frac{3\text{x}-\text{y}+1}{3}=\frac{3\text{x}+2\text{y}+1}{6}$
On solving we can obtain two equations in $x$ and $y.$
Multiplying each of the equation by the $\text{LCM}$ of the denominators, we get
$\Rightarrow 3(2x + y + 2) = 5(3x - y + 1)$ and $2(3x - y + 1) = 3x + 2y + 1$
$\Rightarrow 6x + 3y + 6 = 15x - 5y + 5$ and $6x - 2y + 2 = 3x + 2y + 1$
$\Rightarrow 9x - 8y = 1 ...(i)$
$\Rightarrow 3x - 4y = -1 ...(ii)$
Multiply $(ii)$ by $3,$ and subtract from $(i).$
$\Rightarrow 9x - 8y = 1$ and $9x - 12y = -3$
$\Rightarrow 4y = 4$
$\Rightarrow y = 1$
Substituting $y = 1$ in $(i),$ we get $x = 1.$
View full question & answer→MCQ 191 Mark
If a pair linear equations is inconsistent then their graph lines will be:
- ✓
- B
- C
- D
Intersecting or coincident.
AnswerA system of equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$ is said to be inconsistent if it has no solution.
If a pair of linear equation are parallel,
It has no solutions.
If a pair of linear equation are coincident,
If has infinite number of solutions.
If a pair of linear equations are intersecting,
It has a unique solution.
So, the pair of linear will be parallel.
View full question & answer→MCQ 201 Mark
The system $x - 2y = 3$ and $3x + ky = 1$ has a unique solution only when:
- A
$\text{k}=-6,$
- ✓
$\text{k}\neq-6$
- C
$\text{k}=0$
- D
$\text{k}\neq0$
AnswerCorrect option: B. $\text{k}\neq-6$
$x - 2y = 3$ and $3x + ky = 1$
We know that,
the system of linear equations $a_1x + b_1x + c_1 = 0, a_2x + b_2y + c_2 = 0$ has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
So, $\frac{1}{3}\neq\frac{-2}{\text{k}}$
$\Rightarrow\text{k}\neq-6.$
View full question & answer→MCQ 211 Mark
The pair of equations $2x + y = 5, 3x + 2y = 8$ has:
- ✓
- B
- C
- D
Infinitely many solutions.
AnswerThe given system of equations can be written can be follows:
$2x + y - 5 = 0$ and $3x + 2y - 8 = 0$
The given equations are of the following form:
$ax + by + c = 0$ and $ax + by + c = 0$
Here, $a = 2, b = 1, c = -5$ and $a = 3, b = 2$ and $c = -8$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{2}{3},\ \frac{\text{b}_1}{\text{b}_2}=\frac{1}{2}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{-5}{-8}=\frac{5}{8}$
$\therefore\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
The given system has a unique solution.
Hence, the lines intersect at one point.
View full question & answer→MCQ 221 Mark
The system $kx - y = 2$ and $6x - 2y = 3$ has a unique solution only when:
- A
$\text{k}=-6,$
- B
$\text{k}\neq-6$
- C
$\text{k}=0$
- ✓
$\text{k}\neq3$
AnswerCorrect option: D. $\text{k}\neq3$
$kx - y = 2$ and $6x - 2y = 3$
We know that,
the system of linear equations $a_1x + b_1x + c_1 = 0, a_2x + b_2y + c_2 = 0$ has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
So, $\frac{\text{k}}{6}\neq\frac{-1}{-2}$
$\Rightarrow\text{k}\neq3.$
View full question & answer→MCQ 231 Mark
If a pair of linear equations is consistent, then their graph lines will be:
- A
- B
- C
- ✓
Intersecting or coincident.
AnswerCorrect option: D. Intersecting or coincident.
A system of equations $a_1x + b_1y + c_1 = 0, a_2x + b_2y + c_2 = 0$ is said to be consistent if it has at least one solution.
If a pair of linear equations are parallel,
It has no solutions.
If a pair of linear equations are coincident,
It has infinite number of solutions.
If a pair of linear equation are intersecting,
It has a unique solution.
So, the pair of linear can be intersecting or coincident.
View full question & answer→MCQ 241 Mark
If $x - y = 2$ and $\frac{2}{\text{x+y}}=\frac{1}{5}$ then:
- A
$x = 4, y = 2$
- B
$x = 5, y = 3$
- ✓
$x = 6, y = 4$
- D
$x = 7, y = 5$
AnswerCorrect option: C. $x = 6, y = 4$
$x - y = 2 ....(i)$
$\frac{2}{\text{x+y}}=\frac{1}{5}$
$\Rightarrow 3x - 2y = 5 ...(ii)$
Adding equations $(i)$ and $(ii),$ we get
$2x = 12$
$\Rightarrow x = 6$
Substituting $x = 6$ in $(ii),$ we get $y = 4.$
View full question & answer→MCQ 251 Mark
The system$ x + 2y = 3$ and $5x + ky + 7 = 0$ has no solution, when:
- ✓
$\text{k}=10$
- B
$\text{k}\neq10$
- C
$\text{k}=\frac{-7}{3}$
- D
$\text{k}=-21$
AnswerCorrect option: A. $\text{k}=10$
$x + 2y = 3$ and $5x + ky + 7 = 0$
$x + 2y - 3 = 0$ and $5x + ky + 7 = 0$
We know that,
the system of linear equations $a_1x + b_1x + c_1 = 0, a_2x + b_2y + c_2 = 0$
has a unique solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}.$
So, $\frac{1}{5}=\frac{2}{\text{k}}\neq\frac{-3}{7}$
$\Rightarrow\frac{1}{5}=\frac{2}{\text{k}}$
$\Rightarrow\text{k}=0.$
View full question & answer→MCQ 261 Mark
For what value of k do the equations $kx - 2y = 3$ and $3x + y = 5$ represent two lines intersecting at a unique point?
AnswerCorrect option: D. All real values except $-6$
$kx - 2y = 3$ and $3x + y = 5$
We know that,
the system of linear equations $a_1x + b_1x + c_1 = 0, a_2x + b_2y + c_2 = 0$ has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
So, $\frac{\text{k}}{3}=\frac{-2}{1}$
$\Rightarrow\text{k}\neq-6.$
Thus, k can take any real values except $-6$.
View full question & answer→MCQ 271 Mark
The graphic representation of the equations $x + 2y = 3$ and $2x + 4y + 7 = 0$ gives a pair of:
AnswerThe given system of equations can be weitten as follows:
$x + 2y - 3 = 0$ and $2x + 4y + 7 = 0$
The given equations are of the following form:
$a_2x + b_1y + c_1 = 2, c_1 = 0$ and $a_2 = 2, b_2x + b_2y + c_2 = 0$
Here, $a_1=1, b_1=2, c_1=-3$ and $a_2=2, b_2=4$ and $c_2=7$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{1}{2},\frac{\text{b}_1}{\text{b}_2}=\frac{2}{4}=\frac{1}{2}$ and $\frac{\text{c}_1}{\text{c}_2}=\frac{-3}{7}$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{ c}_2}$
So, the given system has no solution.
Hence, the lines are parallel.
View full question & answer→MCQ 281 Mark
If $\frac{1}{\text{x}}+\frac{2}{\text{y}}=4$ and $\frac{3}{\text{y}}+\frac{1}{\text{x}}=11$ then:
- ✓
$\text{x}=2,\ \text{y}=3$
- B
$\text{x}=-2,\ \text{y}=3$
- C
$\text{x}=\frac{-1}{2},\ \text{y}=3$
- D
$\text{x}=\frac{-1}{2},\ \text{y}=\frac{1}{3}$
AnswerCorrect option: A. $\text{x}=2,\ \text{y}=3$
$\frac{2\text{x}}{3}-\frac{\text{y}}{2}+\frac{1}{6}=0$
Multiply by the $\text{LCM,} 6.$
$\Rightarrow 4x - 3y + 1 = 0$
$\Rightarrow 4x - 3y = -1 ....(i)$
$\frac{\text{x}}{2}+\frac{2\text{y}}{3}=3$
Multiply by the $\text{LCM}, 6.$
$3x + 4y = 18 ...(ii)$
Multiply equation $(i)$ and $(ii)$ by $4$ and $3$ respectively.
$16x - 12y = -4 ...(iii)$
$9x + 12y = 54 ...(iv)$
Adding equations $(iii)$ and $(iv),$ we get
$25x = 50$
$\Rightarrow x = 2$
Substituting $x = 2$ in $(ii),$ we get $y = 3.$
View full question & answer→MCQ 291 Mark
If $\frac{2}{\text{x}}+\frac{3}{\text{y}}=6$ and $\frac{1}{\text{x}}+\frac{1}{2\text{y}}=2$ then:
- A
$\text{x}=1,\ \text{y}=\frac{2}{3}$
- ✓
$\text{x}=\frac{2}{3},\ \text{y}=1$
- C
$\text{x}=1,\ \text{y}=\frac{3}{2}$
- D
$\text{x}=\frac{3}{2},\ \text{y}=1$
AnswerCorrect option: B. $\text{x}=\frac{2}{3},\ \text{y}=1$
$\frac{2}{\text{x}}+\frac{3}{\text{y}}=6$ and $\frac{1}{\text{x}}+\frac{1}{2\text{y}}=2$
Multiplying $\frac{1}{\text{x}}+\frac{1}{2\text{y}}=2$ by $2$,
we get $\frac{2}{\text{x}}+\frac{1}{\text{y}}=4$
So, we have equations, $\frac{2}{\text{x}}+\frac{3}{\text{y}}=6$ and $\frac{2}{\text{x}}+\frac{1}{\text{y}}=4$
Put $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}$
So, we get
$2u + 3v = 6 ...(i)$
$2u + v = 4 ....(ii)$
Substituting $v = 1$ in $(ii),$ we get $\text{u}=\frac{3}{2}$
$\Rightarrow\frac{1}{\text{x}}=\frac{3}{2}$ and $\frac{1}{\text{y}}=1$
$\Rightarrow\text{x}=\frac{2}{3}$ and $\text{y}=1.$
View full question & answer→MCQ 301 Mark
If $4x + 6y = 3xy$ and $8x + 9y = 5xy$ then:
Answer$4x + 6y = 3xy$ and $8x + 9y = 5xy$
Dividing through out by $xy,$ we get
$\frac{4}{\text{y}}+\frac{6}{\text{x}}=3$ and $\frac{8}{\text{y}}+\frac{9}{\text{x}}=5$
That is, $\frac{6}{\text{x}}+\frac{4}{\text{y}}=3$ and $\frac{9}{\text{x}}+\frac{8}{\text{y}}=5$
Put $\frac{1}{\text{x}}=\text{u}$ and $\frac{1}{\text{y}}=\text{v}$
So, we get
$6u + 4v = 3 ...(i)$
$9u + 8v = 5 ...(ii)$
Multiply $(i)$ by $2$ and subtract $(ii)$ from the resultant.
$\Rightarrow 12u + 8v = 6$ and $9u + 8v = 5$
$\Rightarrow 3u = 1$
$\Rightarrow\text{u}=\frac{1}{3}$
Substituting $\text{u}=\frac{1}{3}$ in $(i),$ we get $\text{v}=\frac{1}{4}.$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{3}$ and $\frac{1}{\text{y}}=\frac{1}{4}$
$\Rightarrow x = 3$ and $y = 4.$
View full question & answer→MCQ 311 Mark
The pair of equations $x + 2y + 5 = 0$ and $-3x - 6y + 1 = 0$ has:
- A
- B
- C
Infinitely many solutions.
- ✓
Answer$x + 2y + 5 = 0$ and $-3x - 6y + 1 = 0$
$\frac{\text{a}_1}{\text{a}_2}=\frac{1}{-3}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{2}{-6}=\frac{1}{-3}$
$\frac{\text{c}_1}{\text{c}_2}=5$
Clearly, $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
We know that,
The system of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ has no solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}.$
So, the pair of equations has no solution.
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