Find the values of $x$ and $y$ from the following pair of linear equations : $62x+43y = 167$ $43x + 62y = 148$
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$62 x + 43y = 167$ ...(i) $43 x + 62 y = 148$ ...(ii) Adding (i) and (ii) and simplifying, we get $x + y = 3$ ...(iii) Subtracting (ii) from (i) and simplifying, we get $x - y = 1$ ...(iv) Solving (iii) and (iv) to get $x = 2$ and $y = 1$
Assertion (A): The pair of linear equations $px + 3y + 59 = 0$ and $2x + 6y + 118 = 0$ will have infinitely many solutions if $p = 1$. Reason (R): If the pair of linear equations $px + 3y + 19 = 0$ and $2x + 6y + 157 = 0$ has a unique solution, then $p \neq 1$.
(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c)Assertion (A) is true, but Reason (R) is false.
(d)Assertion (A) is false, but Reason (R) is true.
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(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
Assertion (A): The system of linear equations $3x - 5y + 7 = 0$ and $-6x+10 y + 14 = 0$ is inconsistent. Reason (R) : When two linear equations don't have unique solution, they always represent parallel lines.
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(C) Assertion (A) is true, but Reason (R) is false. (1 Mark)
Assertion (A) : The value of p for which the system of equations $4x + py + 8 = 0$ and $2x + 2y + 2 = 0$ is consistent is $4$. Reason (R) : The system of equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ is consistent with infinitely many solutions, if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c)Assertion (A) is true, but Reason (R) is false.
(d)Assertion (A) is false, but Reason (R) is true.
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(D) Assertion (A) is false, but Reason (R) is true.
3 Marks Questions
243 Marks · 🔁 March 2023 & March 2025 · Basicopen ↗
A fraction becomes $\frac{1}{3}$, when $1$ is subtracted from the numerator and it becomes $\frac{1}{4}$, when $8$ is added to its denominator. Find the fraction.
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Let the fraction be $\frac{x}{y}$ [$\frac{1}{2}$ mark] $\frac{x - 1}{y} = \frac{1}{3} \Rightarrow 3x - y = 3 \dots (i)$ [$\frac{1}{2}$ mark] $\frac{x}{y + 8} = \frac{1}{4} \Rightarrow 4x - y = 8 \dots (ii)$ [$\frac{1}{2}$ mark] On solving the equations $(i)$ and $(ii)$, we get $x = 5, y = 12$ [$1$ mark] Required fraction is $\frac{5}{12}$ [$\frac{1}{2}$ mark]
Find the value of $k$ for which the following pair of linear equations will have infinitely many solutions : $kx + 3y - (k - 3) = 0$ and $12x + ky - k = 0$ Hence, find any two solutions of the given pair of equations.
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For infinitely many solutions: $\frac{k}{12} = \frac{3}{k} = \frac{k - 3}{k}$ [$1$ mark] $k^2 = 36$ and $k^2 - 3k = 3k$ $(k = \pm 6)$ and $(k = 6, 0)$ $\therefore k = 6$ [$1$ mark] For $k = 6$, equations are $6x + 3y = 3$ and $12x + 6y = 6$ any two correct solutions [$\frac{1}{2} + \frac{1}{2}$ mark]
Draw the graph of the following equations: $x + y = 5, x - y = 5$, and (i) find the solution of the equations from the graph. (ii) shade the triangular region formed by the lines and the $y$-axis.
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Correct graph of line for equation $x + y = 5$. Correct graph of line for equation $x - y = 5$. (i) $(5, 0)$ (ii) Correct shade the required triangular region.
An academy offering cricket coaching bought $10$ bats and $5$ balls for ₹32,500. Later, the academy bought $2$ bats and $8$ balls for ₹10,000. If there is no change in the cost of the bat and of the ball, find the cost of $1$ bat and $1$ ball.
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Let the cost of $1$ bat be $\text{Rs}x$ and the cost of $1$ ball be $\text{Rs}y$ A.T.Q. $10x + 5y = 32500$ or $2x + y = 6500$ -------(i) [1 mark] $2x + 8y = 10000$ or $2x + 8y = 10000$ ----(ii) [1 mark] Solving (i) and (ii) to get $x = 3000$ and $y = 500$ [1/2 + 1/2 mark] Cost of $1$ bat = ₹3,000 Cost of $1$ ball = ₹500
$x$ and $y$ are complementary angles such that $x : y = 1 : 2$. Express the given information as a system of linear equations in two variables and hence solve it.
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$x + y = 90^{\circ}$ $2x = y$ Solving to get $x = 30^{\circ}, y = 60^{\circ}$
Determine graphically, the coordinates of vertices of a triangle whose equations are $2x-3y+6=0$; $2x+3y-18=0$ and $x = 0$. Also, find the area of this triangle.
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Correct graph of the line $2x - 3y + 6 = 0$ (1 Mark) Correct graph of the line $2x + 3y - 18 = 0$ (1 Mark) Correct graph of the line $x = 0$ (1/2 Mark) Coordinates of vertices of the triangle are $A(0,2)$, $B(3,4)$ and $C(0,6)$ (1.5 Marks) Area of $\triangle ABC = \frac{1}{2} \times 4 \times 3 = 6$ sq. units (1 Mark)
Draw the graph of the pair of linear equations $x-y+2=0$ and $4x-y-4=0$. Calculate the area of the triangle formed by the lines so drawn and the x-axis.
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Correct graph of the equation $4x - y - 4 = 0$ (I) (2 Marks) Correct graph of the equation $x - y + 2 = 0$ (II) (2 Marks) Area of the required triangle $ABC = \frac{1}{2} \times 3 \times 4 = 6$ sq. units (III) (1 Mark)
3 chairs and 1 table cost ₹900; whereas 5 chairs and 3 tables cost ₹2,100. If the cost of 1 chair is $x$ and the cost of 1 table is $y$, then the situation can be represented algebraically as
A $2$-digit number is seven times the sum of its digits. The number formed by reversing the digits is $18$ less than the given number. Find the given number.
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Let unit's digit be $x$ and ten's digit be $y$. $\therefore$ Number $= 10y + x$ According to the first condition: $10y + x = 7(x + y)$ $10y + x = 7x + 7y$ $3y - 6x = 0$ $y = 2x$ (i) According to the second condition: Number formed by reversing digits is $10x + y$. $10x + y = (10y + x) - 18$ $10x + y - 10y - x = -18$ $9x - 9y = -18$ $x - y = -2$ $y - x = 2$ (ii) On solving (i) and (ii): Substitute (i) into (ii): $2x - x = 2 \Rightarrow x = 2$ Substitute $x=2$ into (i): $y = 2(2) = 4$ $\therefore$ required number is $10(4) + 2 = 42$
A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for $20$ days, she has to pay ₹1,500 as hostel charges while another student B, who takes food for $26$ days, pays ₹1,800. Find the fixed charges and the cost of food.
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Let the monthly fixed charges of hostel be $x$ and cost of food be $y$ per day. A.T.Q $x + 20y = 1500$ ---------(i) $x + 26y = 1800$ ---------(ii) Solving equations (i) & (ii) $x = 500, y = 50$ Hence, the monthly fixed charges of hostel be ₹500 and cost of food be ₹50 per day.
A coaching institute of Mathematics conducts classes in two batches I and II and fees for rich and poor children are different. In batch I, there are 20 poor and 5 rich children, whereas in batch II, there are 5 poor and 25 rich children. The total monthly collection of fees from batch I is ₹9000 and from batch II is ₹26,000. Assume that each poor child pays $x$ per month and each rich child pays $y$ per month. Based on the above information, answer the following questions : (i) Represent the information given above in terms of $x$ and $y$. (ii) Find the monthly fee paid by a poor child. OR Find the difference in the monthly fee paid by a poor child and a rich child. (iii) If there are 10 poor and 20 rich children in batch II, what is the total monthly collection of fees from batch II ?
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(i) $20x + 5y = 9000$ $5x + 25y = 26000$ (ii) Solving the equations $x = 200$ Monthly fee paid by poor child = ₹200 OR (ii) getting $x=200$ and $y= 1000$ Difference in the fee = $1000 - 200 = \text{Rs}800$ (iii) $10x + 20y = 10(200) + 20(1000)$ $= \text{Rs}22000$
A school is organizing a grand cultural event to show the talent of its students. To accommodate the guests, the school plans to rent chairs and tables from a local supplier. It finds that rent for each chair is ₹50 and for each table is ₹200. The school spends ₹30,000 for renting the chairs and tables. Also, the total number of items (chairs and tables) rented are $300$. If the school rents 'x' chairs and 'y' tables, answer the following questions : (i) Write down the pair of linear equations representing the given information. (ii) (a) Find the number of chairs and number of tables rented by the school. OR (b) If the school wants to spend a maximum of ₹27,000 on $300$ items (tables and chairs), then find the number of chairs and tables it can rent. (iii) What is maximum number of tables that can be rented in ₹30,000 if no chairs are rented?
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(i) $x + y = 300$ and $50 x + 200 y = 30000$ or $x + 4y = 600$ (ii) (a) $x + y = 300$ and $x + 4y = 600$ Solving the equations, we get $x = 200$ and $y = 100$ $\therefore$ Number of chairs and tables rented by the school are $200$ and $100$ respectively. OR (b) $x + y = 300$ and $50x + 200y = 27000$ or $x + 4y = 540$ Solving the equations, we get $x = 220$ and $y = 80$ $\therefore$ Number of chairs and tables rented by the school are $220$ and $80$ respectively. (iii) Number of tables $= \frac{30000}{200} = 150$ $\therefore$ Maximum number of tables that can be rented is $150$ if no chairs are rented.
In Indian wedding ceremonies, grooms often wear garlands made with currency notes given by the relatives. One such garland (note mala) is made up of ₹200 and ₹500 notes. The total number of currency notes used in the garland is 40 and the total value of currency notes is ₹11,000. Express the given information algebraically as a system of linear equations in two variables. Hence, find the number of notes of each type used to make the garland.
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Let number of ₹200 notes be $x$ Let number of ₹500 notes be $y$ } (1/2 Mark) $x + y = 40$ --- (1) (1/2 Mark) $2x + 5y = 110$ --- (2) (1/2 Mark) Solving (1) and (2) we get $x = 30$ and $y = 10$ (1/2 Mark)
If the length of a rectangle is reduced by $5$ cm and its breadth is increased by $2$ cm, then the area of the rectangle is reduced by $80$ cm$^2$. However, if we increase the length by $10$ cm and decrease the breadth by $5$ cm, its area is increased by $50$ cm$^2$. Find the length and breadth of the rectangle.
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Let the length of rectangle be $x$ cm and the breadth be $y$ cm Area of rectangle = $xy$ ($\frac{1}{2}$) $(x-5) (y+2) = xy - 80 \Rightarrow 2x - 5y + 70 = 0$ (1$\frac{1}{2}$) $(x+10) (y-5) = xy + 50 \Rightarrow -5x + 10 y - 100 = 0$ (1$\frac{1}{2}$) Solving the two equations, we get $x = 40$ and $y = 30$ (1$\frac{1}{2}$) $\therefore$ Length of rectangle = $40$ cm and Breadth of rectangle = $30$ cm
If three times the greater of two numbers is divided by the smaller one, we get 4 as the quotient and 3 as the remainder. Also, if seven times the smaller number is divided by greater one, we get 5 as the quotient and 1 as the remainder. Find the numbers.
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Let the smaller number be $x$ and the greater number be $y$ $3y = 4x + 3$ ... (i) $7x = 5y + 1$ ... (ii) Solving (i) and (ii), we get $x = 18, y = 25$ $\therefore$ Smaller number is 18 and greater number is 25
A 2-digit number is obtained by either multiplying the sum of the digits by $7$ and then adding $3$ or by multiplying the difference of the digits by $19$ and then subtracting $1$. It is given that the digit at ten's place is greater than that of unit's place. Find the 2-digit number.
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Let the unit's place digit be $y$ and ten's digit be $x$. So, number be $10x + y$ Therefore, $10x + y = 7 (x + y) + 3$ $\Rightarrow x - 2y = 1$ --- (1) Also, $10x + y = 19 (x - y) - 1$ $\Rightarrow -9x + 20y = -1$ --- (2) Solving (1) and (2), we get $x = 9, y = 4$ $\therefore$ the required number is $94$.
A man lent a part of his money at $10\%$ p.a. and the rest at $15\%$ p.a. His income at the end of the year is ₹1,900. If he had interchanged the rate of interest on the two sums, he would have earned ₹200 more. Find the amount lent in both cases.
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Let amount lent for $10\%$ p. a. = $\text{Rs}x$ and amount lent for $15\%$ p. a. = $$\begin{aligned}& \text{Rs}y \\ & text{ATQ, } \frac{10x}{100} + \frac{15y}{100} = 1900 \\ & text{or } 2x + 3y = 38000 \\ & text{and } \frac{15x}{100} + \frac{10y}{100} = 2100 \\ & text{or } 3x + 2y = 42000 \\ & text{On solving these equations, we get} \\ & x = 10000 \text{ and } y = 6000 \\ & therefore \text{Amount lent for } 10\% \text{ p. a. = } \text{Rs}10000 \text{ \& money lent for } 15\% \text{ p. a. = } \text{Rs}6000\end{aligned}$$
Aarush bought $2$ pencils and $3$ chocolates for ₹ $11$ and Tanish bought $1$ pencil and $2$ chocolates for ₹ $7$ from the same shop. Represent this situation in the form of a pair of linear equations. Find the price of $1$ pencil and $1$ chocolate, graphically.
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Let the cost of $1$ pencil be $x$ and the cost of $1$ chocolate be $y$ $2x + 3y = 11$ (1 Mark) $x + 2y = 7$ (1 Mark) for correct graph of equation $2x + 3y = 11$ (1 Mark) for correct graph of equation $x + 2y = 7$ (1 Mark) $x=1, x=3$ (1/2+1/2 Mark) $\therefore$ Cost of $1$ pencil = ₹ $1$ and Cost of $1$ chocolate = ₹ $3$
A fraction becomes $\frac{5}{6}$ when $3$ is added to both the numerator and the denominator. If $2$ is added to both the numerator and the denominator, the fraction becomes $\frac{9}{11}$. Express the given information algebraically as a system of linear equations in two variables. Hence, find the original fraction.
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Let numerator be $x$ and denominator be $y$ $\therefore$ fraction is $\frac{x}{y}$ (1/2 Mark) $\frac{x+3}{y+3} = \frac{5}{6}$ (1 Mark) $\Rightarrow 6x – 5y = -3$ ---------- (i) (1/2 Mark) Also $\frac{x+2}{y+2} = \frac{9}{11}$ (1 Mark) $\Rightarrow 11x – 9y = -4$ ---------- (ii) (1/2 Mark) Solving (i) and (ii) we get $x = 7$ and $y = 9$ (1/2+1/2 Mark) Fraction is $\frac{7}{9}$ (1/2 Mark)
A bag contains some red and blue balls. Ten percent of the red balls, when added to twenty percent of the blue balls, give a total of $24$. If three times the number of red balls exceeds the number of blue balls by $20$, find the number of red and blue balls.
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Let number of red balls be $x$ & number of blue balls be $y$ A.T.Q. $\frac{10x}{100} + \frac{20y}{100} = 24$ or $x + 2y = 240$ .....(i) Also, $3x - y = 20$ ......(ii) Solving (i) and (ii), we get $x = 40, y = 100$ $\therefore$ Number of red balls = $40$ and Number of blue balls = $100$
