Answer: (a) 96 Solution: Suppose the tens and the units digits of the required no. be $\mathrm{x}$ and $\mathrm{y}$, respectively. The required number $=(10 x+y)$ As per the question, we have:...

Answer: (a) coincident Solution: The correct option is (a). The given system of equations can be written as follows: $5 x-15 y-8=0$ and $3 x-9 y-\frac{24}{5}=0$ Given equations are of the following...

Answer: Solution: The given system of equations are as follows: $2 x+3 y-2=0$ and $x-2 y-8=0$ They are of the following form: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ Here, $a_{1}=2,...

Answer: (b) parallel Solution: The given system of equations are as follows: $6 x-2 y+9=0$ and $3 x-y+12=0$ They are of the following form: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$...

Answer: (c) Solution: The correct answer is option (C). It is clear that, Reason (R) is false. Upon solving $x+y=8$ and $x-y=2$, we obtain: $x=5$ and $y=3$ Therefore, the given system has a unique...

Answer: (d) 40 years Solution: Suppose the present age of the man be $\mathrm{x}$ years. And his son's present age be $y$ years. 5 years later: $\begin{array}{l} (x+5)=3(y+5) \\ \Rightarrow x+5=3...

Answer: (b) $80^{\circ}$ Solution: Correct option is (b). In a cyclic quadrilateral $\mathrm{ABCD}$ : $\begin{array}{l} \angle A=(x+y+10)^{0} \\ \angle B=(y+20)^{0} \\ \angle C=(x+y-30)^{0} \\...

Answer: (a) parallel Solution: If a pair of linear equations in two variables is inconsistent, then no solution exists as they have no common point. And, since there is no common solution, their...

Answer: (d) intersecting or coincident Solution: If a pair of linear equations is consistent, then the two graph lines either intersect at a point or coincidence.

Answer: (d) no solution Here, $a_{1}=3, b_{1}=2 k, c_{1}=-2, a_{2}=2, b_{2}=5$ and $c_{2}=1$ $\therefore \frac{a_{1}}{a_{4}}=\frac{3}{2}, \frac{b_{1}}{b_{L}}=\frac{2 k}{5}$ and...

Answer: (d) no solution Solution: We can write the given system of equations as: $x+2 y+5=0$ and $-3 x-6 y+1=0$ Given equations are of the following form: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2}...

Answer: (d) all real values except -6 Solution: We can write the given system of equations as follows: $\mathrm{kx}-2 y-3=0$ and $3 \mathrm{x}+\mathrm{y}-5=0$ Given equations are of the following...

Answer: (d) $\frac{15^{2}}{4}$ Solution: We can write the given system of equations as follows: $3 x+2 k y-2=0$ and $2 x+5 y+1=0$ Given equations are of the following form: $a_{1} x+b_{1} y+c_{1}=0$...

Answer: (a) $\mathrm{k}=10$ We can write the given system of equations as follows: $x+2 y-3=0$ and $5 x+k y+7=0$ The given equations are of the following form: $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2}...

Answer: (b) $\mathrm{k} \neq-6$ Solution: The correct option is (b). We can write the given system of equations as follows: $x-2 y-3=0$ and $3 x+k y-1=0$ given equations are of the following form:...

Answer: (d) $k \neq 3$. Solution: The given system of equations are $\begin{array}{l} \mathrm{kx}-\mathrm{y}-2=0\dots \dots(i) \\ 6 \mathrm{x}-2 \mathrm{y}-3=0\dots \dots(ii) \end{array}$ Here,...

Answer: (b) $x=\frac{2}{3}, y=1$ Solution: The given system of equations are $\begin{array}{l} \frac{2}{x}+\frac{3}{y}=6\dots (i) \\ \frac{1}{x}+\frac{1}{2 y}=2\dots (ii) \end{array}$ Multiplying...

Answer: (e) 0 Solution: $\begin{array}{l} \because 2^{x+y}=2^{x-y}=\sqrt{8} \\ \therefore \mathrm{x}+\mathrm{y}=\mathrm{x}-\mathrm{y} \\ \Rightarrow \mathrm{y}=0 \end{array}$

Answer: $(a) x=1, y=2$ Solution: The given system is $29 x+37 y=103\dots \dots(i)$ $37 x+29 y=95\dots \dots(ii)$ Adding equation(i) and equation(ii), we get $66 x+66 y=198$ $\Rightarrow x+y=3\dots...

Answer: (c) $x=3, y=4$ Solution: The given system of equations are $\begin{array}{l} 4 x+6 y=3 x y\dots (i) \\ 8 x+9 y=5 x y\dots (ii) \end{array}$ Dividing equation(i) and equation(ii) by $x y$, we...

Answer: (b) $x=\frac{5}{2}, y=\frac{1}{2}$ Solution: The given system of equations are $\begin{array}{l} \frac{3}{x+y}+\frac{2}{x-y}=2\dots \dots(i) \\ \frac{9}{x+y}-\frac{4}{x-y}=1\dots \dots(ii)...

Answer: $($ a) $x=1, y=1$ Solution: Considering $\frac{2 x+y+2}{5}=\frac{3 x-y+1}{3}$ and $\frac{3 x-y+1}{3}=\frac{3 x+2 y+1}{3}$. Now, on simplifying these equations, we obtain $\begin{array}{l}...

Answer: (d) $x=\frac{-1}{2}, y=\frac{1}{3}$ Solution: The given system is $\begin{array}{l} \frac{1}{x}+\frac{2}{y}=4\dots \dots(i) \\ \frac{3}{y}-\frac{1}{x}=11\dots \dots(ii) \end{array}$ Adding...

Answer: $($ a) $x=2, y=3$ Solution: The given system is $\begin{array}{l} \frac{2 x}{3}-\frac{y}{2}=-\frac{1}{6}\dots \dots(i) \\ \frac{x}{2}+\frac{2 y}{3}=3\dots \dots(ii) \end{array}$ Multiplying...

Answer: (c) $x=6, y=4$ Solution: The given system is $\begin{array}{l} x-y=2\dots (i) \\ x+y=10\dots (ii) \end{array}$ Adding equation(i) and equation(ii), we get $2 x=12 \Rightarrow x=6$ Now,...

Answer: (c) $x=3, y=2$ Solution: The given system is $\begin{array}{l} 2 x+3 y=12\dots \dots(i) \\ 3 x-2 y=5\dots \dots(ii) \end{array}$ Multiplying equation(i) by 2 and equation(ii) by 3 and then...