Samacheer Kalvi Class 12 Maths Solutions Chapter 1 Applications of Matrices and Determinants Exercise 1.4

Question 1. Solve the following systems of linear equations by Cramer's rule:
(i) \( 5x - 2y + 16 = 0 \), \( x + 3y - 7 = 0 \)
(ii) \( \frac{3}{x} + 2y = 12 \), \( \frac{2}{x} + 3y = 13 \)
(iii) \( 3x + 3y - z = 11 \), \( 2x - y + 2z = 9 \), \( 4x + 3y + 2z = 25 \)
(iv) \( \frac{3}{x} - \frac{4}{y} - \frac{2}{z} - 1 = 0 \), \( \frac{1}{x} + \frac{2}{y} + \frac{1}{z} - 2 = 0 \), \( \frac{2}{x} - \frac{5}{y} - \frac{4}{z} + 1 = 0 \)
Answer:
(i) Given equations are: \( 5x - 2y + 16 = 0 \implies 5x - 2y = -16 \) \( x + 3y - 7 = 0 \implies x + 3y = 7 \) First, calculate the determinant \( \Delta \): \( \Delta = \left| \begin{matrix} 5 & -2 \\ 1 & 3 \end{matrix} \right| = (5 \times 3) - (-2 \times 1) = 15 - (-2) = 15 + 2 = 17 \) Since \( \Delta = 17 \neq 0 \), a unique solution exists. Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constants: \( \Delta_x = \left| \begin{matrix} -16 & -2 \\ 7 & 3 \end{matrix} \right| = (-16 \times 3) - (-2 \times 7) = -48 - (-14) = -48 + 14 = -34 \) Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constants: \( \Delta_y = \left| \begin{matrix} 5 & -16 \\ 1 & 7 \end{matrix} \right| = (5 \times 7) - (-16 \times 1) = 35 - (-16) = 35 + 16 = 51 \) Now, use Cramer's rule to find x and y: \( x = \frac{\Delta_x}{\Delta} = \frac{-34}{17} = -2 \) \( y = \frac{\Delta_y}{\Delta} = \frac{51}{17} = 3 \) Thus, the solution is \( x = -2, y = 3 \). Cramer's rule is especially useful for solving systems with a unique solution because it directly calculates each variable using determinants.
(ii) Given equations are: \( \frac{3}{x} + 2y = 12 \) \( \frac{2}{x} + 3y = 13 \) Let \( \frac{1}{x} = a \) and \( y = b \). The equations become: \( 3a + 2b = 12 \) \( 2a + 3b = 13 \) First, calculate the determinant \( \Delta \): \( \Delta = \left| \begin{matrix} 3 & 2 \\ 2 & 3 \end{matrix} \right| = (3 \times 3) - (2 \times 2) = 9 - 4 = 5 \) Since \( \Delta = 5 \neq 0 \), a unique solution exists. Next, calculate \( \Delta_a \) by replacing the a-coefficients with the constants: \( \Delta_a = \left| \begin{matrix} 12 & 2 \\ 13 & 3 \end{matrix} \right| = (12 \times 3) - (2 \times 13) = 36 - 26 = 10 \) Then, calculate \( \Delta_b \) by replacing the b-coefficients with the constants: \( \Delta_b = \left| \begin{matrix} 3 & 12 \\ 2 & 13 \end{matrix} \right| = (3 \times 13) - (12 \times 2) = 39 - 24 = 15 \) Now, use Cramer's rule to find a and b: \( a = \frac{\Delta_a}{\Delta} = \frac{10}{5} = 2 \) \( b = \frac{\Delta_b}{\Delta} = \frac{15}{5} = 3 \) Since \( a = \frac{1}{x} \), we have \( 2 = \frac{1}{x} \implies x = \frac{1}{2} \). And \( y = b \implies y = 3 \). Thus, the solution is \( x = \frac{1}{2}, y = 3 \). This method of substitution helps simplify equations that are not linear initially, making them solvable by standard methods like Cramer's rule.
(iii) Given equations are: \( 3x + 3y - z = 11 \) \( 2x - y + 2z = 9 \) \( 4x + 3y + 2z = 25 \) First, calculate the determinant \( \Delta \): \( \Delta = \left| \begin{matrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{matrix} \right| \) \( = 3((-1 \times 2) - (2 \times 3)) - 3((2 \times 2) - (2 \times 4)) - 1((2 \times 3) - (-1 \times 4)) \) \( = 3(-2 - 6) - 3(4 - 8) - 1(6 + 4) \) \( = 3(-8) - 3(-4) - 1(10) \) \( = -24 + 12 - 10 = -22 \) Since \( \Delta = -22 \neq 0 \), a unique solution exists. Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constants: \( \Delta_x = \left| \begin{matrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{matrix} \right| \) \( = 11((-1 \times 2) - (2 \times 3)) - 3((9 \times 2) - (2 \times 25)) - 1((9 \times 3) - (-1 \times 25)) \) \( = 11(-2 - 6) - 3(18 - 50) - 1(27 + 25) \) \( = 11(-8) - 3(-32) - 1(52) \) \( = -88 + 96 - 52 = -44 \) Now, calculate \( \Delta_y \) by replacing the y-coefficients with the constants: \( \Delta_y = \left| \begin{matrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{matrix} \right| \) \( = 3((9 \times 2) - (2 \times 25)) - 11((2 \times 2) - (2 \times 4)) - 1((2 \times 25) - (9 \times 4)) \) \( = 3(18 - 50) - 11(4 - 8) - 1(50 - 36) \) \( = 3(-32) - 11(-4) - 1(14) \) \( = -96 + 44 - 14 = -66 \) Finally, calculate \( \Delta_z \) by replacing the z-coefficients with the constants: \( \Delta_z = \left| \begin{matrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{matrix} \right| \) \( = 3((-1 \times 25) - (9 \times 3)) - 3((2 \times 25) - (9 \times 4)) + 11((2 \times 3) - (-1 \times 4)) \) \( = 3(-25 - 27) - 3(50 - 36) + 11(6 + 4) \) \( = 3(-52) - 3(14) + 11(10) \) \( = -156 - 42 + 110 = -88 \) Use Cramer's rule to find x, y, and z: \( x = \frac{\Delta_x}{\Delta} = \frac{-44}{-22} = 2 \) \( y = \frac{\Delta_y}{\Delta} = \frac{-66}{-22} = 3 \) \( z = \frac{\Delta_z}{\Delta} = \frac{-88}{-22} = 4 \) Thus, the solution is \( x = 2, y = 3, z = 4 \). When solving systems with three variables, it's very important to calculate each determinant precisely to avoid errors in the final solution.
(iv) Given equations are: \( \frac{3}{x} - \frac{4}{y} - \frac{2}{z} - 1 = 0 \implies \frac{3}{x} - \frac{4}{y} - \frac{2}{z} = 1 \) \( \frac{1}{x} + \frac{2}{y} + \frac{1}{z} - 2 = 0 \implies \frac{1}{x} + \frac{2}{y} + \frac{1}{z} = 2 \) \( \frac{2}{x} - \frac{5}{y} - \frac{4}{z} + 1 = 0 \implies \frac{2}{x} - \frac{5}{y} - \frac{4}{z} = -1 \) Let \( a = \frac{1}{x} \), \( b = \frac{1}{y} \), \( c = \frac{1}{z} \). The system becomes: \( 3a - 4b - 2c = 1 \) (1) \( a + 2b + c = 2 \) (2) \( 2a - 5b - 4c = -1 \) (3) First, calculate the determinant \( \Delta \): \( \Delta = \left| \begin{matrix} 3 & -4 & -2 \\ 1 & 2 & 1 \\ 2 & -5 & -4 \end{matrix} \right| \) \( = 3((2 \times -4) - (1 \times -5)) - (-4)((1 \times -4) - (1 \times 2)) - 2((1 \times -5) - (2 \times 2)) \) \( = 3(-8 + 5) + 4(-4 - 2) - 2(-5 - 4) \) \( = 3(-3) + 4(-6) - 2(-9) \) \( = -9 - 24 + 18 = -15 \) Since \( \Delta = -15 \neq 0 \), a unique solution exists. Next, calculate \( \Delta_a \) by replacing the a-coefficients with the constants: \( \Delta_a = \left| \begin{matrix} 1 & -4 & -2 \\ 2 & 2 & 1 \\ -1 & -5 & -4 \end{matrix} \right| \) \( = 1((2 \times -4) - (1 \times -5)) - (-4)((2 \times -4) - (1 \times -1)) - 2((2 \times -5) - (2 \times -1)) \) \( = 1(-8 + 5) + 4(-8 + 1) - 2(-10 + 2) \) \( = 1(-3) + 4(-7) - 2(-8) \) \( = -3 - 28 + 16 = -15 \) Now, calculate \( \Delta_b \) by replacing the b-coefficients with the constants: \( \Delta_b = \left| \begin{matrix} 3 & 1 & -2 \\ 1 & 2 & 1 \\ 2 & -1 & -4 \end{matrix} \right| \) \( = 3((2 \times -4) - (1 \times -1)) - 1((1 \times -4) - (1 \times 2)) - 2((1 \times -1) - (2 \times 2)) \) \( = 3(-8 + 1) - 1(-4 - 2) - 2(-1 - 4) \) \( = 3(-7) - 1(-6) - 2(-5) \) \( = -21 + 6 + 10 = -5 \) Finally, calculate \( \Delta_c \) by replacing the c-coefficients with the constants: \( \Delta_c = \left| \begin{matrix} 3 & -4 & 1 \\ 1 & 2 & 2 \\ 2 & -5 & -1 \end{matrix} \right| \) \( = 3((2 \times -1) - (2 \times -5)) - (-4)((1 \times -1) - (2 \times 2)) + 1((1 \times -5) - (2 \times 2)) \) \( = 3(-2 + 10) + 4(-1 - 4) + 1(-5 - 4) \) \( = 3(8) + 4(-5) + 1(-9) \) \( = 24 - 20 - 9 = -5 \) Use Cramer's rule to find a, b, and c: \( a = \frac{\Delta_a}{\Delta} = \frac{-15}{-15} = 1 \) \( b = \frac{\Delta_b}{\Delta} = \frac{-5}{-15} = \frac{1}{3} \) \( c = \frac{\Delta_c}{\Delta} = \frac{-5}{-15} = \frac{1}{3} \) Substitute back to find x, y, and z: Since \( a = \frac{1}{x} \), \( 1 = \frac{1}{x} \implies x = 1 \). Since \( b = \frac{1}{y} \), \( \frac{1}{3} = \frac{1}{y} \implies y = 3 \). Since \( c = \frac{1}{z} \), \( \frac{1}{3} = \frac{1}{z} \implies z = 3 \). Thus, the solution is \( x = 1, y = 3, z = 3 \). Using substitutions like \( a = \frac{1}{x} \) transforms complex fractional equations into simpler linear systems that are easier to solve.In simple words: For each part, we wrote the equations clearly. Then, we used Cramer's rule, which involves calculating special numbers called determinants, to find the values of x, y, and z. For some problems, we first replaced complex terms like \( \frac{1}{x} \) with a simpler letter to make the equations easier to handle.

🎯 Exam Tip: Always check if Cramer's rule can be applied by ensuring the main determinant \( \Delta \) is not zero. If it is zero, the system either has no solution or infinitely many solutions, and Cramer's rule cannot directly solve it.

Question 2. In a competitive examination, one mark is awarded for every correct answer while \( \frac{1}{4} \) mark is deducted for every wrong answer. A student answered 100 questions and got 80 marks. How many questions did he answer correctly? (Use Cramer's rule to solve the problem).
Answer: Let \( x \) be the number of questions answered correctly and \( y \) be the number of questions answered wrongly. According to the problem, the total number of questions answered is 100. So, \( x + y = 100 \) (1) For the marks obtained, one mark is awarded for each correct answer and \( \frac{1}{4} \) mark is deducted for each wrong answer. The student got 80 marks. So, \( 1x - \frac{1}{4}y = 80 \) To remove the fraction, multiply the second equation by 4: \( 4(x) - 4(\frac{1}{4}y) = 4(80) \implies 4x - y = 320 \) (2) Now we have a system of two linear equations: \( x + y = 100 \) \( 4x - y = 320 \) Using Cramer's rule: First, calculate the determinant \( \Delta \): \( \Delta = \left| \begin{matrix} 1 & 1 \\ 4 & -1 \end{matrix} \right| = (1 \times -1) - (1 \times 4) = -1 - 4 = -5 \) Since \( \Delta = -5 \neq 0 \), a unique solution exists. Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constants: \( \Delta_x = \left| \begin{matrix} 100 & 1 \\ 320 & -1 \end{matrix} \right| = (100 \times -1) - (1 \times 320) = -100 - 320 = -420 \) Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constants: \( \Delta_y = \left| \begin{matrix} 1 & 100 \\ 4 & 320 \end{matrix} \right| = (1 \times 320) - (100 \times 4) = 320 - 400 = -80 \) Now, use Cramer's rule to find x and y: \( x = \frac{\Delta_x}{\Delta} = \frac{-420}{-5} = 84 \) \( y = \frac{\Delta_y}{\Delta} = \frac{-80}{-5} = 16 \) So, the number of correct questions is 84, and the number of wrong questions is 16. The problem asked for the number of questions answered correctly. Therefore, the student answered 84 questions correctly. This type of problem, involving two unknowns and two conditions, is a classic example where systems of linear equations are very helpful.In simple words: We made two equations: one for the total number of questions and one for the total marks. Then, we used Cramer's rule to find how many questions were answered correctly and how many wrongly. The final answer for correct questions is 84.

🎯 Exam Tip: When setting up equations for word problems, ensure that each equation accurately represents the given conditions (e.g., total quantity, total value, total marks). A small error in setting up the equations will lead to an incorrect final answer.

Question 3. A chemist has one solution which is 50% acid and another solution which is 25% acid. How much should be mixed to make 10 litres of a 40% acid solution? (Use Cramer's rule to solve the problem).
Answer: Let \( x \) be the volume (in litres) of the 50% acid solution and \( y \) be the volume (in litres) of the 25% acid solution. The total volume of the mixture should be 10 litres. So, \( x + y = 10 \) (1) The total amount of acid in the mixture should be 40% of 10 litres, which is \( 0.40 \times 10 = 4 \) litres. The amount of acid from the 50% solution is \( 0.50x \). The amount of acid from the 25% solution is \( 0.25y \). So, \( 0.50x + 0.25y = 4 \) (2) To work with whole numbers, multiply equation (2) by 100: \( 50x + 25y = 400 \) Divide this equation by 25 to simplify: \( 2x + y = 16 \) (3) Now we have a system of two linear equations: \( x + y = 10 \) \( 2x + y = 16 \) Using Cramer's rule: First, calculate the determinant \( \Delta \): \( \Delta = \left| \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right| = (1 \times 1) - (1 \times 2) = 1 - 2 = -1 \) Since \( \Delta = -1 \neq 0 \), a unique solution exists. Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constants: \( \Delta_x = \left| \begin{matrix} 10 & 1 \\ 16 & 1 \end{matrix} \right| = (10 \times 1) - (1 \times 16) = 10 - 16 = -6 \) Then, calculate \( \Delta_y \) by replacing the y-coefficients with the constants: \( \Delta_y = \left| \begin{matrix} 1 & 10 \\ 2 & 16 \end{matrix} \right| = (1 \times 16) - (10 \times 2) = 16 - 20 = -4 \) Now, use Cramer's rule to find x and y: \( x = \frac{\Delta_x}{\Delta} = \frac{-6}{-1} = 6 \) \( y = \frac{\Delta_y}{\Delta} = \frac{-4}{-1} = 4 \) Therefore, 6 litres of the 50% acid solution and 4 litres of the 25% acid solution should be mixed to get 10 litres of a 40% acid solution. Mixture problems are common in chemistry and can often be solved efficiently by setting up simultaneous equations.In simple words: We set up two equations: one for the total amount of liquid and one for the total amount of acid. Then, using Cramer's rule, we found that 6 litres of the 50% acid and 4 litres of the 25% acid are needed.

🎯 Exam Tip: When dealing with percentages in mixture problems, always convert them to decimals before setting up equations to ensure accurate calculations.

Question 4. A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself? (Use Cramer's rule to solve the problems).
Answer: Let \( x \) be the time (in minutes) pump A takes to fill the tank alone, and \( y \) be the time (in minutes) pump B takes to fill the tank alone. In 1 minute, pump A fills \( \frac{1}{x} \) of the tank. In 1 minute, pump B fills \( \frac{1}{y} \) of the tank. When both pumps A and B work together to fill the tank, they complete it in 10 minutes. So their combined filling rate is \( \frac{1}{10} \) tank per minute. \( \frac{1}{x} + \frac{1}{y} = \frac{1}{10} \) (1) When pump B runs in reverse, it empties the tank. So its rate becomes negative, \( -\frac{1}{y} \). With pump A filling and pump B emptying, the tank is filled in 30 minutes. So the combined rate is \( \frac{1}{30} \) tank per minute. \( \frac{1}{x} - \frac{1}{y} = \frac{1}{30} \) (2) Let \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \). The equations become: \( a + b = \frac{1}{10} \) \( a - b = \frac{1}{30} \) Using Cramer's rule: First, calculate the determinant \( \Delta \): \( \Delta = \left| \begin{matrix} 1 & 1 \\ 1 & -1 \end{matrix} \right| = (1 \times -1) - (1 \times 1) = -1 - 1 = -2 \) Since \( \Delta = -2 \neq 0 \), a unique solution exists. Next, calculate \( \Delta_a \) by replacing the a-coefficients with the constants: \( \Delta_a = \left| \begin{matrix} \frac{1}{10} & 1 \\ \frac{1}{30} & -1 \end{matrix} \right| = (\frac{1}{10} \times -1) - (1 \times \frac{1}{30}) = -\frac{1}{10} - \frac{1}{30} = -\frac{3}{30} - \frac{1}{30} = -\frac{4}{30} = -\frac{2}{15} \) Then, calculate \( \Delta_b \) by replacing the b-coefficients with the constants: \( \Delta_b = \left| \begin{matrix} 1 & \frac{1}{10} \\ 1 & \frac{1}{30} \end{matrix} \right| = (1 \times \frac{1}{30}) - (\frac{1}{10} \times 1) = \frac{1}{30} - \frac{1}{10} = \frac{1}{30} - \frac{3}{30} = -\frac{2}{30} = -\frac{1}{15} \) Now, use Cramer's rule to find a and b: \( a = \frac{\Delta_a}{\Delta} = \frac{-\frac{2}{15}}{-2} = \frac{2}{15} \times \frac{1}{2} = \frac{1}{15} \) \( b = \frac{\Delta_b}{\Delta} = \frac{-\frac{1}{15}}{-2} = \frac{1}{15} \times \frac{1}{2} = \frac{1}{30} \) Substitute back to find x and y: Since \( a = \frac{1}{x} \), \( \frac{1}{15} = \frac{1}{x} \implies x = 15 \). Since \( b = \frac{1}{y} \), \( \frac{1}{30} = \frac{1}{y} \implies y = 30 \). Therefore, pump A takes 15 minutes to fill the tank alone, and pump B takes 30 minutes to fill the tank alone. Work-rate problems often become simpler when you think about the fraction of work completed per unit of time, which helps in forming linear equations.In simple words: We used the idea of "work per minute" for each pump. We made two equations, one for when both pumps fill together, and one for when one fills and the other empties. Solving these equations with Cramer's rule showed that pump A fills the tank in 15 minutes and pump B fills it in 30 minutes on its own.

🎯 Exam Tip: In work-rate problems, remember that if a machine or person is emptying or working against the main task, their rate should be subtracted, not added, in the combined rate equation.

Question 5. A family of 3 people went out for dinner in a restaurant. The cost of two dosai, three idlies and two vadais is Rs 150. The cost of the two dosai, two idlies and four vadais is Rs 200. The cost of five dosai, four idlies and two vadais is T 250. The family has Rs 350 in hand and they ate 3 dosai and six idlies and six vadais. Will they be able to manage to pay the bill within the amount they had?
Answer: Let \( x \) be the cost of one dosai, \( y \) be the cost of one idly, and \( z \) be the cost of one vadai. From the given information, we can form the following equations: \( 2x + 3y + 2z = 150 \) (1) \( 2x + 2y + 4z = 200 \) (2) \( 5x + 4y + 2z = 250 \) (3) First, calculate the determinant \( \Delta \): \( \Delta = \left| \begin{matrix} 2 & 3 & 2 \\ 2 & 2 & 4 \\ 5 & 4 & 2 \end{matrix} \right| \) \( = 2((2 \times 2) - (4 \times 4)) - 3((2 \times 2) - (4 \times 5)) + 2((2 \times 4) - (2 \times 5)) \) \( = 2(4 - 16) - 3(4 - 20) + 2(8 - 10) \) \( = 2(-12) - 3(-16) + 2(-2) \) \( = -24 + 48 - 4 = 20 \) Since \( \Delta = 20 \neq 0 \), a unique solution exists. Next, calculate \( \Delta_x \) by replacing the x-coefficients with the constants: \( \Delta_x = \left| \begin{matrix} 150 & 3 & 2 \\ 200 & 2 & 4 \\ 250 & 4 & 2 \end{matrix} \right| \) \( = 150((2 \times 2) - (4 \times 4)) - 3((200 \times 2) - (4 \times 250)) + 2((200 \times 4) - (2 \times 250)) \) \( = 150(4 - 16) - 3(400 - 1000) + 2(800 - 500) \) \( = 150(-12) - 3(-600) + 2(300) \) \( = -1800 + 1800 + 600 = 600 \) Now, calculate \( \Delta_y \) by replacing the y-coefficients with the constants: \( \Delta_y = \left| \begin{matrix} 2 & 150 & 2 \\ 2 & 200 & 4 \\ 5 & 250 & 2 \end{matrix} \right| \) \( = 2((200 \times 2) - (4 \times 250)) - 150((2 \times 2) - (4 \times 5)) + 2((2 \times 250) - (200 \times 5)) \) \( = 2(400 - 1000) - 150(4 - 20) + 2(500 - 1000) \) \( = 2(-600) - 150(-16) + 2(-500) \) \( = -1200 + 2400 - 1000 = 200 \) Finally, calculate \( \Delta_z \) by replacing the z-coefficients with the constants: \( \Delta_z = \left| \begin{matrix} 2 & 3 & 150 \\ 2 & 2 & 200 \\ 5 & 4 & 250 \end{matrix} \right| \) \( = 2((2 \times 250) - (200 \times 4)) - 3((2 \times 250) - (200 \times 5)) + 150((2 \times 4) - (2 \times 5)) \) \( = 2(500 - 800) - 3(500 - 1000) + 150(8 - 10) \) \( = 2(-300) - 3(-500) + 150(-2) \) \( = -600 + 1500 - 300 = 600 \) Use Cramer's rule to find x, y, and z: \( x = \frac{\Delta_x}{\Delta} = \frac{600}{20} = 30 \) \( y = \frac{\Delta_y}{\Delta} = \frac{200}{20} = 10 \) \( z = \frac{\Delta_z}{\Delta} = \frac{600}{20} = 30 \) So, the cost of one dosai is Rs 30, one idly is Rs 10, and one vadai is Rs 30. Now, calculate the total cost for the food they ate (3 dosai, 6 idlies, and 6 vadais): Total cost = \( 3x + 6y + 6z \) Total cost = \( 3(30) + 6(10) + 6(30) \) Total cost = \( 90 + 60 + 180 \) Total cost = Rs 330 The family has Rs 350 in hand. Since the total bill is Rs 330, and they have Rs 350, they will be able to manage to pay the bill. Solving for individual prices first is a crucial step in budget problems, allowing you to accurately calculate total costs for different combinations of items.In simple words: First, we figured out the cost of one dosai, one idly, and one vadai using Cramer's rule from the given information. Then, we calculated the total cost of what the family ate. Their bill was Rs 330, and they had Rs 350, so they could pay for their meal.

🎯 Exam Tip: For problems with multiple conditions and a final question, break it down into smaller, manageable steps. First, calculate the unknown unit costs, then use those to answer the final part of the question.