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Question 1: The pair of equations $y = 0$ and $y = -7$ has:
Solution: (D) No solution
Step 1: $y = 0$ is the x-axis.
Step 2: $y = -7$ is a line parallel to the x-axis.
Step 3: Since parallel lines never intersect, there is no solution. -
Question 2: The value of $k$ for which the system of equations $x + 2y = 3$ and $5x + ky + 7 = 0$ has no solution is:
Solution: (D) 10
Step 1: For no solution, $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$.
Step 2: Here $a_1=1, b_1=2, c_1=-3$ and $a_2=5, b_2=k, c_2=7$.
Step 3: $\frac{1}{5} = \frac{2}{k} \Rightarrow k = 10$. -
Question 3: If the lines given by $3x + 2ky = 2$ and $2x + 5y + 1 = 0$ are parallel, then the value of $k$ is:
Solution: (C) 15/4
Step 1: For parallel lines, $\frac{a_1}{a_2} = \frac{b_1}{b_2}$.
Step 2: $\frac{3}{2} = \frac{2k}{5}$.
Step 3: $4k = 15 \Rightarrow k = \frac{15}{4}$. -
Question 4: The pair of equations $x = a$ and $y = b$ graphically represents lines which are:
Solution: (D) Intersecting at (a, b)
Step 1: $x = a$ is a vertical line passing through $a$ on x-axis.
Step 2: $y = b$ is a horizontal line passing through $b$ on y-axis.
Step 3: They intersect at the point $(a, b)$. -
Question 5: The value of $k$ for which the pair of equations $2x + 3y = 5$ and $4x + ky = 10$ has infinitely many solutions is:
Solution: (C) 6
Step 1: Condition: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Step 2: $\frac{2}{4} = \frac{3}{k} = \frac{5}{10}$.
Step 3: $\frac{1}{2} = \frac{3}{k} \Rightarrow k = 6$. -
Question 6: One equation of a pair of dependent linear equations is $-5x + 7y = 2$. The second equation can be:
Solution: (D) $10x - 14y = -4$
Step 1: Dependent equations are coincident, so ratios of coefficients must be equal.
Step 2: Given $-5x + 7y - 2 = 0$. Multiply by -2: $10x - 14y + 4 = 0$.
Step 3: Rearranging gives $10x - 14y = -4$. -
Question 7: If $x = a, y = b$ is the solution of the equations $x - y = 2$ and $x + y = 4$, then the values of $a$ and $b$ are, respectively:
Solution: (C) 3 and 1
Step 1: Add the equations: $(x-y) + (x+y) = 2 + 4 \Rightarrow 2x = 6 \Rightarrow x = 3$.
Step 2: Substitute $x=3$ in $x+y=4 \Rightarrow 3+y=4 \Rightarrow y=1$.
Step 3: So $a=3, b=1$. -
Question 8: Aruna has only ₹1 and ₹2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹75, then the number of ₹1 and ₹2 coins are, respectively:
Solution: (D) 25 and 25
Step 1: Let $x$ be ₹1 coins and $y$ be ₹2 coins.
Step 2: $x + y = 50$ and $1x + 2y = 75$.
Step 3: Subtract eq 1 from eq 2: $(x+2y) - (x+y) = 75 - 50 \Rightarrow y = 25$.
Step 4: $x + 25 = 50 \Rightarrow x = 25$. -
Question 9: The father's age is six times his son's age. Four years hence, the age of the father will be four times his son's age. The present ages, in years, of the son and the father are, respectively:
Solution: (C) 6 and 36
Step 1: Let son's age $= S$, father's age $= F$. Given $F = 6S$.
Step 2: After 4 years: $F+4 = 4(S+4)$.
Step 3: Substitute $F=6S$: $6S+4 = 4S+16 \Rightarrow 2S = 12 \Rightarrow S=6$.
Step 4: $F = 6(6) = 36$. -
Question 10: If $29x + 37y = 103$ and $37x + 29y = 95$, then:
Solution: (A) $x=1, y=2$
Step 1: Add equations: $66x + 66y = 198 \Rightarrow x+y=3$.
Step 2: Subtract equations: $-8x + 8y = 8 \Rightarrow -x+y=1$.
Step 3: Add new equations: $2y = 4 \Rightarrow y=2$.
Step 4: $x + 2 = 3 \Rightarrow x=1$.