Algebraic Equations: Solved Examples

Q 1. A man was asked about his sons and daughters. He replied, “Each of my sons has as many brothers as sisters and each of my daughters has twice as many brothers as sisters”. How many sons and daughters the man has?
Sol: Let the number of sons be x and daughters be y. So as per question, x – 1 = y........(1)
And 2(y – 1) = x........(2). Solving these simple equations, we get x = 4 and y = 3.
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Q 2. Solve the equation: P2 - 8P - 33 = 0
Sol: P = 11 or P = - 3. ⇒ We can factorize the above equation as (P – 11) (P + 3) = 0
Q 3. If one root of the quadratic equation P2 - 12P + Q = 0 is thrice the other root, then find the value of Q.
Sol. Let one root be 'a', then other root is 3a. Also sum of roots = a + 3a = 4a = 12 ⇒ a = 3.
Product of the roots = Q ⇒ a × 3a = 3a2 ⇒ 3 × 9 = 27
Q 4. A man hired a laborer on the condition that he would pay him a total of Rs. 10000 and also give him a cycle after service of two years. He served only for 1.5 years and received the cycle and Rs. 7000. Find the value of the cycle in rupees.
Sol: Let the value of the cycle be x rupees. So we can form the simple linear equation as 1.5/2 (10000 + x) = 7000 + x. Solving this linear equation in one variable, we get x = 2000.
Q 5. If the difference between the squares of two consecutive numbers is 17, then find the numbers.
Sol. Let numbers are P and P + 1. It is given that (P + 1)2 - (P)2 = 17
⇒ (P + 1 – P) (P + 1+ P) = 17 ⇒ (2P + 1) = 17 ⇒ 2P = 16 ⇒ P = 8.
So, the numbers are 8 and 9.
Q 6. 4 kg of sugar and 2 kg of flour are needed to be mixed to make 1 kg of ladoo, and 6 kg of sugar and 4 kg of flour are needed to be mixed to make 1 kg of barfi. How many kg of each type of sweet was manufactured if it is known that 260 kg of sugar and 160 kg of flour were used?
Sol: Let us take x to be the kg of ladoo and y to be the kg of barfi. Now the equations becomes 4x + 6y = 260 & 2x + 4y = 160.
Solve these linear equations and get the value of x as 20 and value of y as 30. So the required ingredients are 20 kg and 30 kg.
Q7. Solve: 2P2 - P - 28 = 0
Sol: Factorizing the given quadratic equation, we get 2P2 - 8P + 7P – 28 = 0
⇒ 2P (P – 4) + 7 (P – 4) = 0
⇒ (P – 4) (2P + 7) = 0 ⇒ P = 4 or P = -7.
Q8. There are some students and some benches in a class. If 3 students are seated on a bench, then 3 students are left standing. If 4 students are seated on a bench, then 3 benches are left vacant. Find the number of students and benches in the class.
Sol: Let the number of students be x and number of benches be y. So we can form two linear equations as 3y + 3 = x.......(1) and x/4 = y – 3.......(2). Solving these equations, we get x = 48 and y = 15. Hence there are 48 students and 15 benches in the class.
Q9. If the roots of P2 - 4P + Q = 0 are equal, then find the value of Q.
Sol. It is given that roots are equal. So value of Discriminant (D) should be zero. We know value of D is b2 – 4ac which is equal to 0. Hence (-4)2 - 4Q = 0 ⇒ 4Q = 16 ⇒ Q = 4
Q10. For what value of k, the linear equations 2x + 3y = k and 4x + 6y = 10 will have infinitely many solutions?
Sol: For infinitely many solutions, the given three ratios would be equal, i.e 2/4 = 3/6 = k/10. Solving this equation, we get the value of k as 5.
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