Q. The cost of 2 kg apples and 1 kg grapes was ₹ 160. After a month, the cost of 4 kg apples and 2 kg grapes was ₹ 300. Represent the situation algebraically.
- Sol.Let the cost of 1 kg of apples be ₹ a
- and the cost of 1 kg of grapes be ₹ g.
- Thus, 2a + g = 160 …(i)
- and 4a + 2g = 300
- or 2a + g = 150 …(ii)
- Thus, equations are 2a + g = 160
- and 2a + g = 150 Ans.
Q. Solve for x and y : 152x – 378y = – 74 and – 378x + 152y = – 604.
Sol. Given, 152x – 378y = – 74 …(i)
and – 378x + 152y = – 604 …(ii)
Adding (i) and (ii), we get
– 226x – 226y = – 678
⇒ x + y = 3 …(iii)
Subtracting (ii) from (i), we get
530x – 530y = 530
⇒ x – y = 1 …(iv)
Adding (iii) and (iv), we get
2x = 4
Substituting x = 2 in (iii), we get
2 + y = 3
⇒ y = 1
Thus, x = 2 and y = 1. Ans.
Q. Determine the value of k, for which the given system of equations has infinitely many solutions:
kx + 3y = k – 3 and 12x + ky = k.
Sol. Given,
kx + 3y – (k – 3) = 0 …(i)
and 12x + ky – k = 0 …(ii)
For this pair of equations to have infinite number of solutions, it must satisfy the condition of
k/12=3/k=(k-3)/k
⇒ k(k – 6) – 6(k – 6) = 0
⇒ k = 6
k = ± 6
Common value k = 6
Thus, the pair of equations will have infinite number of solutions for k = 6. Ans.
Q. Determine the value of k, for which the given system of equations has no solution: 3x + y = 1 and (2k – 1)x + (k – 1)y = 2k + 1.
- Sol. Given, 3x + y = 1 …(i)
- and (2k – 1)x + (k – 1)y = 2k + 1 …(ii)
- For this pair of equations not to hav e any solution, it must satisfy the condition of
- 3/(21-k)=1/(k-1)≠1/(2k=1)
- ⇒ 3(k – 1) = 2k – 1
- ⇒ 3k – 3 – 2k + 1 = 0
- ⇒ k – 2 = 0
- ⇒ k = 2
- Thus, the pair of equations will hav e no solution for k = 2. Ans.
Q. For what value of k will the system of equations 4x + ky + 8 = 0 and 2x + 2y + 2 = 0 has a unique solution?
- Sol. The system of equations will have a unique solution, if
- 4/2≠k/2
- or k≠4
Q. Determine the values of a and b for which the given system of equations has infinitely many solutions:
2x + 3y = 7 and (a – b)x + (a + b)y = 3a + b – 1.
- Sol. Given, 2x + 3y = 7 …(i)
- and (a – b)x + (a + b)y = 3a + b – 1 …(ii)
- For this pair of equations to have infinite number of solutions, it must satisfy the condition of
- 2/(a-b)=3/(a+b)=7/(3a+b-1)
- ⇒ 2(a + b) = 3(a – b)
- ⇒ 2a + 2b = 3a – 3b
- ⇒ a = 5b …(iii)
- and 2(3a + b – 1) = 7(a – b)
- ⇒ 6a + 2b – 2 = 7a – 7b
- ⇒ a – 9b = – 2 …(iv)
- Substituting a = 5b in equation (iv), we get
- 5b – 9b = – 2
- ⇒ – 4b = – 2
- ⇒ 2b = 1
- ⇒ b=1/2
- Thus, a = 5(1/2)=5/2
- Thus, the pair of equations will have infinite
- number of solutions for a =5/2 and b = 1/2. Ans.
Q. The path of a train A is given by the equation x + 2y – 4 = 0 and of B by the equation 2x + 4y – 12 = 0. Represent this situation graphically.
- Sol. Let XOX′ and YOY′ be the X-axis and Y-axis respectively.
- Now, x + 2y – 4 = 0
- ⇒ x = 4 – 2y
- If y = 2, x = 0
- If y = 1, x = 2
- If y = 0, x = 4
- Therefore,
x | 0 | 2 | 4 |
y | 2 | 1 | 0 |
- Now, 2x + 4y – 12 = 0
- ⇒ 4y = 12 – 2x
- ⇒ 2y = 6 – x
- ⇒ y = (6−x) / 2
- If x = 6, y = 0
- If x = 4, y = 1
- If x = 2, y = 2
- Therefore,
x | 6 | 4 | 2 |
y | 0 | 1 | 2 |
Thus, plotting the points S(6, 0), T(4, 1) and U(2, 2) on the graph paper, we get the graph of 2x + 4y = 12, which is represented by SU.
Q. Given the linear equation 2x + 3y – 8 = 0, write another equation in two variables such that the graphical representation of the pair so formed is
(i) intersecting,
(ii) parallel and
(iii) coincident.
- Ans. (i) 2x + 4y = 6
- (ii) 4x + 6y = 8
- (iii) 6x + 9y = 24
So, the system is consistent and thus the reason is also correct.