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Jharkhand Board JAC Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.3

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
1. 2x² – 7x + 3 = 0
2. 2x² + x – 4 = 0
3. 4x² + 4\(\sqrt{3}\)x + 3 = 0
4. 2x² + x + 4 = 0
Solution:
1. 2x² – 7x + 3 = 0
Dividing throughout by 2, we get

Thus, the roots of the given quadratic equation are 3 and \(\frac{1}{2}\).

2. 2x² + x – 4 = 0
Dividing throughout by 2, we get

Thus, the roots of the given quadratic equation are \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\)

3. 4x² + 4\(\sqrt{2}\)x + 3 = 0
4x² + 4\(\sqrt{3}\)x + (\(\sqrt{3}\))² = 0
(2x + \(\sqrt{3}\))² = 0
(2x + \(\sqrt{3}\)) (2x + \(\sqrt{3}\)) = 0
2x + \(\sqrt{3}\) = 0 or 2x + \(\sqrt{3}\) =0
x = \(-\frac{\sqrt{3}}{2}\) or x = \(-\frac{\sqrt{3}}{2}\)
Thus, the roots of the given quadratic equation are \(-\frac{\sqrt{3}}{2}\) and \(-\frac{\sqrt{3}}{2}\)

4. 2x² + x + 4 = 0
Dividing throughout by 2, we get

But, the square of any real number cannot be negative.
Hence, the real roots of the given quadratic equation do not exist.

Question 2.
Find the roots of the quadratic equations given in Q.1 by applying the quadratic formula.
1. 2x² – 7x + 3 = 0
Here, a = 2, b = -7 and c = 3.
Then, b² – 4ac = (-7)² – 4(2)(3) = 25
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{7 \pm \sqrt{25}}{2(2)}\)
∴ x = \(\frac{7 \pm 5}{4}\)
∴ x = 3 or x = \(\frac{1}{2}\)
Thus, the roots of the given quadratic equation are 3 and \(\frac{1}{2}\).

2. 2x² + x – 4 = 0
Here, a = 2, b = 1 and c = -4.
Then, b² – 4ac = (1)² – 4(2)(-4) = 33
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{-1 \pm \sqrt{33}}{2(2)}\)
∴ x = \(\frac{-1 \pm \sqrt{33}}{4}\)
Thus, the roots of the given quadratic equation are \(\frac{-1 \pm \sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\)

3. 4x² + 4\(\sqrt{3}\)x + 3 = 0
Here, a = 4, b = 4\(\sqrt{3}\) and c = 3.
Then, b² – 4ac = (4\(\sqrt{3}\))² – 4(4)(3) = 0
Then, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{-4 \sqrt{3} \pm \sqrt{0}}{2(4)}\)
∴ x = \(-\frac{\sqrt{3}}{2}\) or x = \(-\frac{\sqrt{3}}{2}\)
Thus, the roots of the given quadratic equation are \(-\frac{\sqrt{3}}{2}\) and \(-\frac{\sqrt{3}}{2}\).

4. 2x² + x + 4 = 0
Here, a = 2, b = 1 and c = 4.
Then, b² – 4ac = (1)² – 4 (2)(4) = -31 < 0
Since b² – 4ac < 0 for the given quadratic equation, the real roots of the given quadratic equation do not exits.

Question 3.
Find the roots of the following equations:
1. x – \(\frac{1}{x}\) = 3, x ≠ 0
2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\); x ≠ -4, 7
Solution:
1. x – \(\frac{1}{x}\) = 3, x ≠ 0
∴ x² – 1 = 3x
∴ x² – 3x – 1 = 0
Here, a = 1, b = -3 and c = -1.
Then, b² – 4ac = (-3)² – 4(1)(-1) = 13
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
x = \(\frac{3 \pm \sqrt{13}}{2(1)}\)
x = \(\frac{3 \pm \sqrt{13}}{2}\)
Thus, the roots of the given equation are \(\frac{3+\sqrt{13}}{2}\) and \(\frac{3-\sqrt{13}}{2}\).

2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\); x ≠ -4, 7
∴ \(\frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}\)
∴ \(\frac{-11}{x^2-3 x-28}=\frac{11}{30}\)
∴ \(\frac{-1}{x^2-3 x-28}=\frac{1}{30}\)
∴ -30 = x² – 3x – 28
∴ x² – 3x + 2 = 0
Here, a = 1, b = -3 and c = 2.
Then, b² – 4ac = (-3)² – 4(1)(2) = 1
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{3 \pm \sqrt{1}}{2(1)}\)
∴ x = \(\frac{3 \pm 1}{2}\)
∴ x = 2 or x = 1
Thus, the roots of the given equation are 2 and 1.
Note: Here, the method of factorisation would turn out to be more easier.

Question 4.
The sum of the reciprocals of Rehman’s ages (in years), 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Solution:
Let the present age of Rehman be x years.
So, his age 3 years ago was (x – 3) years and his age 5 years hence will be (x + 5) years.
The sum of the reciprocals of these two ages (in years) is given to be \(\frac{1}{3}\).
∴ \(\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\)
∴ \(\frac{x+5+x-3}{(x-3)(x+5)}=\frac{1}{3}\)
∴ 3(2x + 2) = (x – 3)(x + 5)
∴ 6x + 6 = x² + 2x – 15
∴ x² – 4x – 21 = 0
∴ x² – 7x + 3x – 21 = 0
∴ x(x – 7) + 3(x – 7) = 0
∴ (x – 7)(x + 3) = 0
∴ x – 7 = 0 or x + 3 = 0
∴ x = 7 or x = -3
Now, since x represents the present age of Rehman, it cannot be negative, i,e., x ≠ -3.
∴ x = 7
Thus, the present age of Rehman is 7 years.

Question 5.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of those marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in Mathematics be x.
Then, her marks in English = 30 – x, as her total marks in Mathematics and English is 30.
Had she scored 2 marks more in Mathematics and 3 marks less in English, her score in Mathematics would be x + 2 and in English would be 30 – x – 3 = 27 – x.
∴ (x + 2) (27 – x) = 210
∴ 27x – x² + 54 – 2x = 210
∴ -x² + 25x + 54 – 210 = 0
∴ -x² + 25x – 156 = 0
∴ x² – 25x + 156 = 0
∴ x² – 13x – 12x + 156 = 0
∴ x(x – 13) – 12(x – 13) = 0
∴ (x – 13)(x – 12) = 0
∴ x – 13 = 0 or x – 12 = 0
∴ x = 13 or x = 12
Then, 30 – x = 30 – 13 = 17 or
30 – x = 30 – 12 = 18
Thus, Shefall’s marks in Mathematics and in English are 13 and 17 respectively or 12 and 18 respectively.

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:
Let the shorter side of the rectangular field be x m.
Then, it diagonal is (x + 60) m and the longer side is (x + 30) m.
In a rectangle, all the angles are right angles.
Hence, by Pythagoras theorem,
(Shorter side)² + (Longer side)² = (Diagonal)²
∴ x² + (x + 30)² = (x + 60)²
∴ x² + x² + 60x + 900 = x² + 120x + 3600
∴ x² – 60x – 2700 = 0
Here, a = 1, b = -60 and c = -2700.
Then, b² – 4ac = (-60)² – 4(1)(-2700)
= 3600 + 10800
= 14400
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{60 \pm \sqrt{14400}}{2(1)}\)
= \(\frac{60 \pm 120}{2}\)
∴ x = \(\frac{60+120}{2}\) or x = \(\frac{60-120}{2}\)
∴ x = 90 or x = -30
Since x denotes the shorter side of the rectangular field, x cannot be negative.
∴ x = 90
Then, x + 30 = 90 + 30 = 120
Thus, the shorter side (breadth) of the rectangular field is 90 m and the longer side (length) of the rectangular field is 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the smaller number be x.
Then, the larger number = \(\frac{x^2}{8}\)
Now, the difference of their squares is 180.
∴ \(\left(\frac{x^2}{8}\right)^2-(x)^2=180\)
∴ \(\frac{x^4}{64}-x^2=180\)
∴ x⁴ – 64x² – 11520 = 0
Let x² = y
∴ x⁴ = y²
∴ y² – 64y – 11520 = 0
Here, a = 1, b = -64 and c = -11520.
Then, b² – 4ac = (-64)² – 4(1)(-11520)
= 4096 + 46080
= 50176
Now, y = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
= \(\frac{64 \pm \sqrt{50176}}{2(1)}\)
= \(\frac{64 \pm 224}{2}\)
∴ y = \(\frac{64+224}{2}\) or y = \(\frac{64-224}{2}\)
∴ y = 144 or y = -80
∴ x² = 144 or x² = -80
But, x² = -80 is not possible.
∴ x² = 144
∴ x = 12 or x = -12
Then, \(\frac{x^2}{8}=\frac{144}{8}=18\)
Thus, the required numbers are 12 and 18 or -12 and 18.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let the usual uniform speed of the train be x km/h.
∴ Time taken to travel 360 km at the speed of x km/h = \(\frac{\text { distance }}{\text { speed }}\) = \(\frac{360}{x}\) hours.
If the speed had been 5 km/h more, the new speed would be (x + 5) km/h and the time taken to travel 360 km at this increased speed would be \(\frac{360}{x+5}\) hours.
Now, New time = Usual time – 1
∴ \(\frac{360}{x+5}=\frac{360}{x}-1\)
∴ 360x = 360x + 1800 – x(x + 5) (Multiplying by x(x + 5))
∴ 0 = 1800 – x² – 5x
∴ x² + 5x – 1800 = 0
∴ x² + 45x – 40x – 1800 = 0
∴ x(x + 45) – 40(x + 45) = 0
∴ (x + 45) (x – 40) = 0
∴ x + 45 = 0 or x – 40 = 0
∴ x = -45 or x = 40
As, x is the speed (in km/h) of the train, x = -45 is not possible.
∴ x = 40
Thus, the usual uniform speed of the train is 40 km/h.

Question 9.
Two water taps together can fill a tank in \(9 \frac{3}{8}\) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let the time taken by the tap with smaller diameter to fill the tank be x hours.
Then, the time taken by the tap with larger diameter = (x – 10) hours.
So, the part of the tank filled in one hour by the tap with smaller diameter = \(\frac{1}{x}\) and by the tap with larger diameter = \(\left(\frac{1}{x-10}\right)\)
So, the part of tank filled in one hour by both the taps together = \(\frac{1}{x}+\frac{1}{x-10}\)
Both the taps together can fill the tank in \(9 \frac{3}{8}\) hours, i.e., \(\frac{75}{8}\) hours.
∴ The part of tank filled in one hour by both the tank together = \(\frac{8}{75}\)
∴ \(\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}\).
∴ 75(x – 10) + 75x = 8x (x – 10) (Multiplying by 75x(x – 10))
∴ 75x – 750 + 75x = 8x² – 80x
∴ 8x² – 230x + 750 = 0
Here, a = 8, b = -230 and c = 750.
∴ b² – 4ac = (-230)² – 4 (8)(750)
= 52900 – 24000
= 28900
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{230 \pm \sqrt{28900}}{2(8)}\)
∴ x = \(\frac{230 \pm 170}{16}\)
∴ x = \(\frac{400}{16}\) or x = \(\frac{60}{16}\)
∴ x = 25 or x = 3.75
But, x ≠ 3.75, because for x = 3.75, x – 10 < 0.
∴ x = 25 and x – 10 = 15
Thus, the time taken by the tap with smaller diameter is 25 hours and that by the tap. with larger diameter is 15 hours.

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let the average speed of the passenger train be x km/h.
Then, the average speed of the express train is (x + 11) km/h.
∴ Time taken by passenger train to cover 132 km = \(\frac{132}{x}\) hours.
∴ Time taken by express train to cover 132 km = \(\frac{132}{x+11}\) hours.
Time taken by express train = Time taken by passenger train – 1
\(\frac{132}{x+11}=\frac{132}{x}-1\)
∴ 132x = 132(x + 11) – x(x + 11) (Multiplying by x(x + 11))
∴ 132x = 132x + 1452 – x² – 11x
∴ x² + 11x – 1452 = 0
Here, a = 1, b = 11 and c = -1452.
∴ b² – 4ac = (11)² – 4(1)(-1452)
= 121 + 5808 = 5929
Now, x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)
∴ x = \(\frac{-11 \pm \sqrt{5929}}{2(1)}\)
∴ x = \(\frac{-11 \pm 77}{2}\)
x = 33 or x = -44
x = -44 is inadmissible as x represents the speed of the passenger train.
∴ x = 33 and x + 11 = 44
Thus, the speed of the passenger train is 33 km/h and the speed of the express train is 44 km/h.

Question 11.
Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:
Let the side of the smaller square be x m.
Then, the perimeter of the smaller square = 4x m
and the area of the smaller square = x² m².
From the given, the perimeter of the bigger square = (4x + 24) m.
∴ Side of the bigger square = \(\frac{4 x+24}{4}\) = (x + 6) m and hence, the area of the bigger square = (x + 6)² m².
∴ x² + (x + 6)² = 468
∴ x² + x² + 12x + 36 – 468 = 0
∴ 2x² + 12x – 432 = 0
∴ x² + 6x – 216 = 0
∴ x² + 18x – 12x – 216 = 0
∴ x(x + 18) – 12(x + 18) = 0
∴ (x + 18) (x – 12) = 0
∴ x + 18 = 0 or x – 12 = 0
∴ x = -18 or x = 12
Here, x = -18 is not possible as x represents the side of a square.
∴ x = 12 and x + 6 = 18
Thus, the side of the smaller square is 12 m and the side of the bigger square is 18 m.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
1. x² – 3x – 10 = 0
2. 2x² + x – 6 = 0
3. \(\sqrt{2}\)x² + 7x + 5\(\sqrt{2}\) = 0
4. 2x² – x + \(\frac{1}{8}\) = 0
5. 100x² – 20x + 1 = 0
Solution:
1. x² – 3x – 10 = 0
∴ x² – 5x + 2x – 10 = 0
∴ x(x – 5) + 2(x – 5) = 0
∴ (x – 5)(x + 2) = 0
Hence, x – 5 = 0 or x + 2 = 0
∴ x = 5 or x = -2
Thus, the roots of the given equation are 5 and -2.
Verification:
For x = 5,
LHS = (5)² – 3(5) – 10
= 25 – 15 – 10
= 0
= RHS
For x = -2,
LHS = (2)² – 3(-2) – 10
= 4 + 6 – 10
= 0
= RHS
Hence, both the roots are verified.
Note that verification is not a part of the solution. It is meant only for your confirmation of receiving correct solution.

2. 2x² + x – 6 = 0
∴ 2x² + 4x – 3x – 6 = 0
∴ 2x (x + 2) – 3(x + 2) = 0
∴ (x + 2) (2x – 3) = 0
∴ x + 2 = 0 or 2x – 3 = 0
∴ x = -2 or x = \(\frac{3}{2}\)
Thus, the roots of the given equation are -2 and \(\frac{3}{2}\)

3. \(\sqrt{2}\)x² + 7x + 5\(\sqrt{2}\) = 0
∴ \(\sqrt{2}\)x² + 2x + 5x + 5\(\sqrt{2}\) = 0
∴ \(\sqrt{2}\)x(x + \(\sqrt{2}\)) + 5(x + \(\sqrt{2}\)) = 0
∴ (x + \(\sqrt{2}\))(\(\sqrt{2}\)x + 5) = 0
∴ x + \(\sqrt{2}\) = 0 or \(\sqrt{2}\)x + 5 = 0
∴ x = –\(\sqrt{2}\) or x = \(-\frac{5}{\sqrt{2}}\)
Thus, the roots of the given equation are –\(\sqrt{2}\) and \(-\frac{5}{\sqrt{2}}\)

4. 2x² – x + \(\frac{1}{8}\) = 0
∴ 16x² – 8x + 1 = 0 (Multiplying by 8)
∴ 16x² – 4x – 4x + 1 = 0
∴ 4x(4x – 1) -1 (4x – 1) = 0
∴ (4x – 1) (4x – 1) = 0
∴ 4x – 1 = 0 or 4x – 1 = 0
∴ x = \(\frac{1}{4}\) or x = \(\frac{1}{4}\)
Thus, the repeated roots of the given equation are \(\frac{1}{4}\) and \(\frac{1}{4}\)

5. 100x² – 20x + 1 = 0
100x² – 10x – 10x + 1 = 0
10x(10x – 1) -1 (10x – 1) = 0
(10x – 1)(10x – 1) = 0
10x – 1 = 0 or 10x – 1 = 0
x = \(\frac{1}{10}\) or x = \(\frac{1}{10}\)
Thus, the repeated roots of the given equation are \(\frac{1}{10}\) and \(\frac{1}{10}\).

Question 2.
Solve the problems given in Textual Examples:
1. John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find out how many marbles they had to start with.
2. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. Find out the number of toys. produced on that day.
Solution:
1. Let the number of marbles that John had be x.
Then, the number of marbles that Jivanti had is (45 – x).
After losing 5 marbles, the number of marbles left with John = x – 5.
After losing 5 marbles, the number of marbles left with Jivanti = (45 – x) – 5 = 40 – x.
Therefore, the product of marbles with them is (x – 5) (40 – x), which is given to be 124.
Hence, we get the following equation:
(x – 5)(40 – x) = 124.
∴ 40x – x² – 200 + 5x = 124
∴ -x² + 45x – 324 = 0
∴ x² – 45x + 324 = 0
∴ x² – 36x – 9x + 324 = 0
∴ x(x – 36) – 9(x – 36) = 0
∴ (x – 36)(x – 9) = 0
∴ x – 36 = 0 or x – 9 = 0
∴ x = 36 or x = 9
Here, both the answers are admissible.
∴ 45 – x = 45 – 36 = 9 or
45 – x = 45 – 9 = 36
Thus, the number of marbles with John and Jivanti to start with are 36 and 9 respectively or 9 and 36 respectively.

2. Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy on that day = 55 – x.
So, the total cost of production (in rupees) on that day = x (55 – x).
Hence, x(55 – x) = 750
∴ 55x – x² – 750 = 0
∴ x² – 55x + 750 = 0
∴ x² – 30x – 25x + 750 = 0
∴ x(x – 30) – 25(x – 30) = 0
∴ (x – 30)(x – 25) = 0
∴ x – 30 = 0 or x – 25 = 0
∴ x = 30 or x = 25
Here, both the answers are admissible.
Hence, the number of toys produced on that day is 30 or 25.

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let the first of the two numbers whose sum is 27 be x.
Then, the second number is 27 – x and the product of those two numbers is x (27 – x).
Their product is given to be 182.
∴ x (27 – x) = 182
∴ 27x – x² – 182 = 0
∴ x² – 27x + 182 = 0
∴ x² – 14x – 13x + 182 = 0
∴ x(x – 14) – 13(x – 14) = 0
∴ (x – 14)(x – 13) = 0
∴ x – 14 = 0 or x – 13 = 0
∴ x = 14 or x = 13
Here, both the answers are admissible.
Hence, if x = 14, it gives that the first number = x = 14 and the second number = 27 – x = 27 – 14 = 13.
And if x = 13, it gives that the first number = x = 13 and the second number = 27 – x = 27 – 13 = 14.
Thus, in either case, the required numbers are 13 and 14.

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let two consecutive positive integers be x and x + 1.
Then, the sum of their squares = (x)² + (x + 1)²
= x² + x² + 2x + 1
= 2x² + 2x + 1
This sum is given to be 365.
∴ 2x² + 2x + 1 = 365
∴ 2x² + 2x – 364 = 0
∴ x² + x – 182 = 0
∴ x² + 14x – 13x – 182 = 0
∴ x(x + 14) – 13(x + 14) = 0
∴ (x + 14)( x- 13) = 0
∴ x + 14 = 0 or x – 13 = 0
∴ x = -14 or x = 13
Since x is a positive integer, x = -14 is inadmissible.
∴ x = 13 and x + 1 = 13 + 1 = 14
Thus, the required consecutive positive integers are 13 and 14.

Question 5.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:
Let the base of the right triangle be x cm.
Then, its altitude is (x – 7) cm.
The hypotenuse of the right triangle is given to be 13 cm.
Now, by Pythagoras theorem.
(Base)² + (Altitude)² = (Hypotenuse)²
∴ (x)² + (x – 7)² = (13)²
∴ x² + x² – 14x + 49 = 169
∴ 2x² – 14x – 120 = 0
∴ x² – 7x – 60 = 0
∴ x² – 12x + 5x – 60 = 0
∴ x(x – 12) + 5(x – 12) = 0
∴ (x – 12)(x + 5) = 0
∴ x – 12 = 0 or x + 5 = 0
∴ x = 12 or x = -5
As the base of a triangle cannot be negative. x = -5 is inadmissible.
Hence, x = 12.
Then, the base of the triangle = x = 12 cm and the altitude of the triangle = x – 7 = 12 – 7
= 5 cm.
Thus, the base and the altitude of the given triangle are 12 cm and 5 cm respectively.

Question 6.
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was 90. find the number of articles produced and the cost of each article.
Solution:
Let the number of pottery articles produced on that day be x.
Then, according to the given, the cost of production (in rupees) of each article = 2x + 3.
Hence, total cost of production (in rupees) on that day = x(2x + 3) = 2x² + 3x.
This total cost of production is given to be ₹ 90.
∴ 2x² + 3x = 90
∴ 2x² + 3x – 90 = 0
∴ 2x² – 12x + 15x – 90 = 0
∴ 2x(x – 6) + 15(x – 6) = 0
∴ (x – 6) (2x + 15) = 0
∴ x – 6 = 0 or 2x + 15 = 0
∴ x = 6 or x = –\(\frac{15}{2}\)
Here, x = –\(\frac{15}{2}\) is inadmissible as x represents the number of articles produced.
Hence, x = 6 and 2x + 3 = 2(6) + 3 = 15.
Thus, the number of pottery articles produced on that day is 6 and the cost of production of each article is ₹ 15.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.1

Question 1.
Check whether the following are quadratic equations:
1. (x + 1)² = 2(x – 3)
2. x² – 2x = (-2)(3 – x)
3. (x – 2)(x + 1) = (x – 1)(x + 3)
4. (x – 3)(2x + 1) = x(x + 5)
5. (2x – 1)(x – 3) = (x + 5)(x – 1)
6. x² + 3x + 1 = (x – 2)²
7. (x + 2) = 2x(x² – 1)
8. x³ – 4x² – x + 1 = (x – 2)³
Solution:
1. Here, LHS = (x + 1)² = x² + 2x + 1 and
RHS = 2(x – 3) = 2x – 6.
Hence, (x + 1)² = 2(x – 3) can be rewritten as x² + 2x + 1 = 2x – 6
∴ x² + 2x + 1 – 2x + 6 = 0
∴ x² + 7 = 0
It is of the form ax² + bx + c = 0.
(a = 1, b = 0, c = 7)
Hence, the given equation is a quadratic equation.

2. Here, RHS = (-2)(3 – x) = -6 + 2x.
Hence, x² – 2x = (-2) (3 – x) can be rewritten as
x² – 2x = -6 + 2x
∴ x² – 4x + 6 = 0
It is of the form ax² + bx + c = 0.
(a = 1, b = -4, c = 6)
Hence, the given equation is a quadratic equation.

3. Here, LHS = (x – 2) (x + 1) = x² – x – 2 and
RHS = (x – 1)(x + 3) = x² + 2x – 3.
Hence, (x – 2)(x + 1)(x – 1)(x + 3) be rewritten as
x² – x – 2 = x² + 2x – 3
∴ -3x + 1 = 0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

4. Here, LHS = (x – 3) (2x + 1) = 2x² – 5x – 3
and RHS = x(x + 5) = x² + 5x.
Hence, (x – 3) (2x + 1) = x(x + 5) can be rewritten as
2x² – 5x – 3 = x² + 5x
∴ x² – 10x – 3 = 0
It is of the form ax² + bx + c = 0.
(a = 1, b = -10, c = -3)
Hence, the given equation is a quadratic equation.

5. Here, LHS = (2x – 1)(x – 3) = 2x² – 7x + 3 and
RHS = (x + 5) (x – 1) = x² + 4x – 5.
Hence, the given equation can be rewritten as
2x² – 7x + 3 = x² + 4x – 5
∴ x² – 11x + 8 = 0
It is of the form ax2 + bx + c = 0.
(a = 1, b = -11, c = 8)
Hence, the given equation is a quadratic equation.

6. Here, RHS = (x – 2)² = x² – 4x + 4
Hence, the given equation can be rewritten as
x² + 3x + 1 = x² – 4x + 4
∴ 7x – 3 = 0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

7. Here, LHS = (x + 2)³ = x³ + 6x² + 12x + 8 and
RHS = 2x (x² – 1) = 2x³ – 2x.
Hence, the given equation can be rewritten as
x³ + 6x² + 12x + 8 = 2x³ – 2x
∴ -x³ + 6x² + 14x + 8 = 0
It is not of the form ax² + bx + c = 0.
Hence, the given equation is not a quadratic equation.

8. Here, RHS = (x – 2)³ = x³ – 6x² + 12x – 8.
Hence, the given equation can be rewritten as
x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
2x² – 13x + 9 = 0
It is of the form ax² + bx + c = 0.
(a = 2, b = -13, c = 9)
Hence, the given equation is a quadratic equation.

Question 2.
Represent the following situations in the form of quadratic equations:
1. The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Solution:
Let the breadth (in metres) of the rectangular plot be x.
Then, the length (in metres) of the rectangular plot is 2x + 1.
Area of the rectangular plot = Length × Breadth
∴ 528 = (2x + 1) × x
(∵ Area is given to be 528 m²)
∴ 528 = 2x² + x
∴ 2x² + x – 528 = 0 is the required quadratic equation to find the length (2x + 1 m) and breadth (x m) of the rectangular plot.

2. The product of two consecutive positive integers is 306. We need to find the integers.
Solution:
Let the two consecutive positive integer be x and x + 1.
Then, their product = x(x + 1) = x² + x.
This product is given to be 306.
∴ x² + x = 306
∴ x² + x – 306 = 0 is the required quadratic equation to find the consecutive positive integers x and x + 1.

3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Solution:
Let Rohan’s present age (in years) be x.
Then, his mother’s present age (in years) = x + 26.
3 years from now, Rohan’s age (in years) will be x + 3 and his mother’s age (in years) will be x + 29.
The product of their ages (in years) 3 years from now is given to be 360.
Hence, (x + 3)(x + 29) = 360
∴ x² + 32x + 87 – 360 = 0
∴ x² + 32x – 273 = 0 is the required quadratic equation to find the present ages (in years) of Rohan (x) and his mother (x + 26).

4. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
Let the usual uniform speed of the train be x km/h.
Now, Time = \(\frac{\text { distance }}{\text { speed }}\)
∴ Time required to cover 480 km distance at usual speed = t₁ = \(\frac{480}{x}\) hours
If the speed is 8 km/hour less, the new speed would be (x – 8) km/hour.
∴ Time required to cover 480 km distance at new speed = t₂ = \(\frac{480}{x-8}\) hours
Now, the time required at new speed is 3 hours more than the usual time.
∴ t₂ = t₁ + 3
∴ \(\frac{480}{x-8}=\frac{480}{x}+3\)
∴ 480x = 480(x – 8) + 3x(x – 8)
(Multiplying by x(x – 8))
∴ 480x = 480x – 3840 + 3x² – 24x
∴ 0 = 3x² – 24x – 3840
∴ x² – 8x – 1280 = 0 is the required quadratic equation to find the usual speed (x km/h) of the train.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.7

Question 1.
The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution:
Let the age of Ani be x years and the age of Biju be y years.
Then, from the given,
x – y = 3 ………….(1)
or y – x = 3 ………(2)
Dharam is twice as old as Ani.
∴ Age of Dharam = 2x years
Biju is twice as old as his sister Cathy. Hence, the age of Cathy is half the age of Biju.
∴ Age of Cathy = \(\frac{y}{2}\) years
Naturally, Dharam is older than Cathy by 30 years.
2x – \(\frac{y}{2}\) = 30
∴ 4x – y = 60 ………..(3)
1. We solve equation (1) and equation (3).
Subtracting equation (1) from equation (3),
we get
(4x – y) – (x – y) = 60 – 3
∴ 3x = 57
∴ x = 19
Substituting x = 19 in equation (1), we get
19 – y = 3
∴ 19 – 3 = y
∴ y = 16
Hence, the age of Ani is 19 years and the age of Biju is 16 years.

2. We now solve equation (2) and equation (3). Adding equations (2) and (3), we get
(y – x) + (4x – y) = 3 + 60
∴ 3x = 63
∴ x = 21
Substituting x = 21 in equation (2), we get
∴ y – 21 = 3
∴ y = 24
Hence, the age of Ani is 21 years and the age of Biju is 24 years.
Thus, the ages of Ani and Biju are 19 years and 16 years respectively or 21 years and 24 years respectively.

Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their (respective) capital? (From the Bijaganita of Bhaskara II) [Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)].
Solution:
Let the amount with the first person (say A) be ₹ x and the amount with the second person (say B) be ₹ y.
If B gives ₹ 100 to A. then A will have ₹(x + 100) and B will have ₹(y – 100).
Then, from the first condition, we get
x + 100 = 2(y – 100)
∴ x + 100 = 2y – 200
∴ x – 2y = -300 ………..(1)
If A gives ₹ 10 to B, then A will have ₹(x – 10) and B will have ₹(y + 10).
Then, from the second condition, we get
y + 10 = 6(x – 10)
∴ y + 10 = 6x – 60
∴ 10 + 60 = 6x – y
∴ 6x – y = 70 ………..(2)
Multiplying equation (2) by 2, we get
12x – 2y = 140 ………..(3)
Subtracting equation (1) from equation (3),
we get
(12x – 2y) – (x – 2y) = 140 – (-300)
∴ 11x = 440
∴ x = 40
Substituting x = 40 in equation (1), we get
40 – 2y = -300
∴ 40 + 300 = 2y
∴ 2y = 340
∴ y = 170
Thus, the first person has got ₹ 40 and the second person has got ₹ 170.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have. been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:
Let the regular uniform speed of the train be x km/h and regular time taken by the train be y hours. Then, the total distance of the journey = speed × time = xy km.
Now, according to the first information, the new speed = (x + 10) km/h and new time = (y – 2) hours. Then, speed × time = distance gives
(x + 10) (y – 2) = xy
∴ xy – 2x + 10y – 20 = xy
∴ -2x + 10y = 20 …………..(1)
Again, according to the second condition. the new speed = (x – 10) km/h and new time = (y + 3) hours.
Then, (x – 10) (y + 3) = xy
∴ xy + 3x – 10y – 30 = xy
∴ 3x – 10y = 30 …………..(2)
Adding equations (1) and (2), we get
(-2x + 10y) + (3x – 10y) = 20 + 30
∴ x = 50
Substituting x = 50 in equation (1), we get
-2(50) + 10y = 20
∴ -100 + 10y = 20
∴ 10y = 120
∴ y = 12
Now, the total distance covered by the train = xy = 50 × 12 = 600 km.
Thus, the distance covered by the train is 600 km.

Question 4.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Solution:
Let the number of students in each row be x and the number of rows be y. Then total number of students = xy.
From the first condition, number of students in each row = x + 3 and the number of rows = y – 1.
∴ (x + 3)(y – 1) = xy
∴ xy – x + 3y – 3 = xy
∴ -x + 3y = 3 …………..(1)
From the second condition, number of students in each row = x – 3 and the number of rows = y + 2.
∴ (x – 3) (y + 2) = xy
∴ xy + 2x – 3y – 6 = xy
∴ 2x – 3y = 6 ……….. (2)
Adding equations (1) and (2), we get
(-x + 3y) + (2x – 3y) = 3 + 6
∴ x = 9
Substituting x = 9 in equation (1), we get
-9 + 3y = 3
∴ 3y = 12
∴ y = 4
Now, total number of students = xy = 9 × 4 = 36. Thus, the number of students in the class is 36.

Question 5.
In a ΔABC, ∠C = 3∠B = 2 (A + B). Find the three angles.
Solution:
In ΔABC, we have
∠A + ∠B + ∠C = 180°
∴ ∠A + ∠B + 3∠B = 180° (∵ ∠C = 3∠B)
∴ ZA + 4∠B = 180° …………..(1)
Again, ∠A + ∠B + ∠C = 180°
∠A + ∠B + 2(∠A + ∠B) = 180° (∵ ∠C = 2 (∠A + ∠B))
∴ 3(∠A + ∠B) = 180°
∴ ∠A + ∠B = 60° ……….(2)
Subtracting equation (2) from equation (1),
we get
(∠A + 4∠B) – (∠A + ∠B) = 180° – 60°
∴ 3∠B = 120°
∴ ∠B = 40°
Substituting ∠B = 40° in equation (2), we get
∠A + 40° = 60°
∴ ∠A = 20°
Substituting ∠B = 40° in ∠C = 3∠B, we get
∠C = 3(40°)
∴ ∠C = 120°
Thus, in ΔABC, ∠A = 20°; ∠B = 40°; ∠C = 120°
Note: Replacing ∠A + ∠B = \(\frac{\angle \mathrm{C}}{2}\) in ∠A + ∠B + ∠C = 180°, we get a simple equation in one variable is \(\frac{3}{2}\)∠C = 180°.
After solving it for ∠C, ∠B and ∠A can be found easily.

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y-axis.
Solution:
5x – y = 5 gives y = 5x – 5

x	1	2
y	0	5

3x – y = 3 gives y = 3x – 3

x	1	2
y	0	3

Now, we draw the graphs of both the equations.

From the graph, we observe that the co-ordinates of the vertices of the triangle formed by the graphs of 5x – y = 5 and 3x – y = 3 and the y-axis are (1, 0), (0, -3) and (0, -5).

Question 7.
Solve the following pair of linear equations:
1. px + qy = p – q
qx – py = p + q
2. ax + by = c
bx + ay = 1 + c
3. \(\frac{x}{a}-\frac{y}{b}=0\)
ax + by = a² + b²
4. (a – b)x + (a + b) y = a² – 2ab – b²
(a + b) (x + y) = a² + b²
5. 152x – 378y = -74
-378x + 152y = -604
Solution:
1. px + qy = p – q ……….(1)
qx – py = p + q ……….(2)
Multiplying equation (1) by p and equation (2) by q, we get
p²x + pqy = p² -pq ……….(3)
q²x – pqy = pq + q² ……….(4)
Adding equations (3) and (4), we get
(p²x + pqy) + (q²x – pqy) = (p² – pq) + (pq + q²)
∴ x(p² + q²) = p² + q²
∴ x = 1
Substituting x = 1 in equation (1), we get
p(1) + qy = p – q
∴ qy = -q
∴ y = -1
Thus, the solution of the given pair of linear equations is x = 1, y = -1.

2. ax + by = c ……….(1)
bx + ay = 1 + c ……….(2)
Multiplying equation (1) by a and equation (2) by b, we get
a²x + aby = ac ……….(3)
b²x + aby = b + bc ……….(4)
Subtracting equation (4) from equation (3), we get
(a²x + aby) – (b²x + aby) = ac – (b + bc)
x(a² – b²) = ac – b – bc
x = \(\frac{c(a-b)-b}{a^2-b^2}\)
Substituting x = \(\frac{c(a-b)-b}{a^2-b^2}\) in equation (1), we get

Thus, the solution of the given pair of linear equations is
x = \(\frac{c(a-b)-b}{a^2-b^2}\), y = \(\frac{c(a-b)+a}{a^2-b^2}\)

3. \(\frac{x}{a}-\frac{y}{b}=0\) ……….(1)
ax + by = a² + b² ……….(2)
From equation (1), we get
\(\frac{x}{a}=\frac{y}{b}\)
y = \(\frac{b}{a}\)x
Substituting y = \(\frac{b}{a}\)x in equation (2), we get

Substituting x = a in y = \(\frac{b}{a}\)x, we get
y = \(\frac{b}{a}\)(a)
∴ y = b
Thus, the solution of the given pair of linear equations is x = a, y = b.

4. (a – b)x + (a + b)y = a² – 2ab – b² ……….(1)
(a + b) (x + y) = a² + b²
∴ (a + b)x + (a + b)y = a² + b² ……….(2)
Subtracting equation (1) from equation (2), we get
[(a + b)x + (a + b)y] – [(a – b)x + (a + b)y] = (a² + b²) – (a² – 2ab – b²)
∴ x(a + b – a + b) = a² + b² – a² + 2ab + b²
∴ x (2b) = 2ab + 2b²
∴ x = \(\frac{2 b(a+b)}{2 b}\)
∴ x = a + b
Substituting x = a + b in equation (1), we get
(a – b)(a + b) + (a + b)y = a² – 2ab – b²
∴ a² – b² + (a + b) y = a² – 2ab – b²
∴ (a + b)y = -2ab
∴ y = \(-\frac{2 a b}{a+b}\)
Thus, the solution of the given pair of linear equations is x = a + b, y = \(-\frac{2 a b}{a+b}\)

5. 152x – 378y = -74 ……….(1)
-378x + 152y = -604 ……….(2)
Adding equations (1) and (2), we get
-226x – 226y = -678
∴ x + y = 3 (Dividing by -226) ……….(3)
Subtracting equation (2) from equation (1),
we get
(152x – 378y) – (-378x + 152y) = (-74) – (-604)
∴ 530x – 530y = 530
x – y = 1 (Dividing by 530) ……….(4)
Adding equations (3) and (4), we get
2x = 4
∴ x = 2
Substituting x = 2 in equation (3), we get 2 + y = 3
∴ y = 1
Thus, the solution of the given pair of linear equations is x = 2, y = 1.

Question 8.
ABCD is a cyclic quadrilateral (See the given figure). Find the angles of the cyclic quadrilateral.

Solution:
ABCD is a cyclic quadrilateral.
∠A + ∠C = 180° and ∠B + ∠D = 180°
∠A + ∠C = 180° gives 4y + 20° – 4x = 180°
∴ 4y – 4x = 160
∴ y – x = 40° (Dividing by 4) …………..(1)
∠B + ∠D = 180° gives 3y – 5° – 7x + 5° = 180°
∴ 3y – 7x = 180° …………..(2)
From equation (1), we get y = x + 40°
Substituting y = x + 40° in equation (2), we get
3(x + 40°) – 7x = 180°
∴ 3x + 120° – 7x = 180°
∴ -4x = 60°
∴ x = -15°
Substituting x = -15° in equation (1), we get
y – (-15) = 40°
∴ y + 15° = 40°
∴ y = 25°
Now, ∠A = 4y + 20° = 4(25°) + 20° = 120°,
∠B = 3y – 5° = 3 (25°) – 5° = 70°,
∠C = -4x = -4(-15) = 60° and
∠D = -7x + 5° = -7(-15°) + 5° = 110°
Thus, in the given cyclic quadrilateral ABCD.
∠A = 120°, ∠B = 70°, ∠C = 60°, ∠D = 110°.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6

Question 1.
Sovle the following pairs of equations by reducing them to a pair of linear equations:

Solution:
1. The given pair of equations is

Taking \(\frac{1}{x}=a\) and \(\frac{1}{y}=b\) in both the equations, we get
\(\frac{1}{2}\)a + \(\frac{1}{3}\)b = 2 ……….(3)
\(\frac{1}{3}\)a + \(\frac{1}{2}\)b = \(\frac{13}{6}\) ………..(4)
Multiplying both the equations by 6, we get
3a + 2b = 12 ……(5)
2a + 3b = 13 ……(6)
Adding equations (5) and (6), we get
5a + 5b = 25
∴ a + b = 5 ……….(7)
Subtracting equation (6) from equation (5), we get
a – b = -1 ………..(8)
Equations (7) and (8) can be solved easily to get a = 2 and b = 3.
Then a = \(\frac{1}{x}\) = 2 and b = \(\frac{1}{y}\) = 3
∴ x = \(\frac{1}{2}\) and y = \(\frac{1}{3}\)
Thus, the solution of the given pair of equations is x = \(\frac{1}{2}\), y = \(\frac{1}{3}\)

2. The given pair of equations is
\(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2\) ……….(1)
\(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1\) ……….(2)
Taking \(\frac{1}{\sqrt{x}}=a\) and \(\frac{1}{\sqrt{y}}=b\) in both the equations, we get
2a + 3b = 2 ……….(3)
4a – 9b = -1 ……….(4)
Multiplying equation (3) by 3, we get
6a + 9b = 6 ……….(5)
Adding equations (4) and (5). we get
(4a – 9b) + (6a + 9b) = -1 + 6
∴ 10a = 5
∴ a = \(\frac{1}{2}\)
Substituting a = \(\frac{1}{2}\) in equation (3), we get
2(\(\frac{1}{2}\)) + 3b = 2
∴ 1 + 3b = 2
∴ 3b = 1
∴ b = \(\frac{1}{3}\)
Now, a = \(\frac{1}{\sqrt{x}}=\frac{1}{2}\)
∴ 2 = \(\sqrt{x}\)
∴ x = 4
Again b = \(\frac{1}{\sqrt{y}}=\frac{1}{3}\)
∴ 3 = \(\sqrt{y}\)
∴ y = 9
Thus, the solution of the given pair of equations is x = 4, y = 9,

3. The given pair of equations is
\(\frac{4}{x}\) + 3y = 14 ……….(1)
\(\frac{3}{x}\) – 4y = 23 ……….(2)
Taking \(\frac{1}{x}\) = a in both the equations, we get
4a + 3y = 14 ……….(3)
3a – 4y = 23 ……….(4)
Multiplying equation (3) by 4 and equation (4) by 3 and adding them, we get
4(4a + 3y) + 3(3a – 4y) = 4(14) + 3(23)
∴ 16a + 12y + 9a – 12y = 56 + 69
∴ 25a = 125
∴ a = 5
Now, a = \(\frac{1}{x}\) = 5
∴ x = \(\frac{1}{5}\)
Substituting x = \(\frac{1}{5}\) in equation (1), we get
\(\frac{4}{\frac{1}{5}}\) + 3y = 14
∴ 20 + 3y = 14
∴ 3y = -6
∴ y = -2
Thus, the solution of the given pair of equations is x = \(\frac{1}{5}\), y = -2.

4. The given pair of equations is
\(\frac{5}{x-1}+\frac{1}{y-2}=2\) ………..(1)
\(\frac{6}{x-1}-\frac{3}{y-2}=1\) …………(2)
Taking \(\frac{1}{x-1}=a\) and \(\frac{1}{y-2}=b\) in both the equations, we get
5a + b = 2 ………..(3)
6a – 3b = 1 ………..(4)
Multiplying equation (3) by 3 and then additing equation (4) to it, we get
3(5a + b) + (6a – 3b) = 3(2) + 1
∴ 15a + 3b + 6a – 3b = 6 + 1
∴ 21a = 7
a = \(\frac{1}{3}\)
Substituting a = \(\frac{1}{3}\) in equation (4), we get
6(\(\frac{1}{3}\)) – 3b = -1
∴ 2 – 3b = 1
∴ 1 = 3b
∴ b = \(\frac{1}{3}\)
Now, a = \(\frac{1}{x-1}=\frac{1}{3}\)
∴ x – 1 = 3
∴ x = 4
Again, b = \(\frac{1}{y-2}=\frac{1}{3}\)
∴ y – 2 = 3
∴ y = 5
Thus, the solution of the given pair of equations is x = 4, y = 5.

5. The given pair of equations is \(\frac{7 x-2 y}{x y}=5\) and \(\frac{8 x+7 y}{x y}=15\)
Hence, \(\frac{7}{y}-\frac{2}{x}=5\) ………..(1)
and \(\frac{8}{y}+\frac{7}{x}=15\) ………..(2)
Taking \(\frac{1}{y}\) = a and \(\frac{1}{y}\) = b in both the equations, we get
7a – 2b = 5 ………..(3)
8a + 7b = 15 ………..(4)
Expressing the equation in the standard form, we get
7a – 2b – 5 = 0 and 8a + 7b – 15 = 0

Now, a = \(\frac{1}{y}\) = 1 ∴ y = 1
and b = \(\frac{1}{x}\) = 1 ∴ x = 1
Thus, the solution of the given pair of equations is x = 1, y = 1.

6. The given pair of equations is 6x + 3y = 6xy and 2x + 4y = 5xy
Dividing both the equations by xy, we get
\(\frac{6}{y}+\frac{3}{x}=6\) ………..(1)
\(\frac{2}{y}+\frac{4}{x}=5\) ………..(2)
Taking \(\frac{1}{y}\) = a and \(\frac{1}{x}\) = b,
we get
6a + 3b = 6
i.e., 2a + b = 2 ……(3)
2a + 4b = 5 ……(4)
Subtracting equation (3) from equation (4),
we get
(2a + 4b) – (2a + b) = 5 – 2
∴ 3b = 3
∴ b = 1
Substituting b = 1 in equation (3), we get
2a + 1 = 2
∴ 2a = 1
∴ a = \(\frac{1}{2}\)
Now, a = \(\frac{1}{y}=\frac{1}{2}\) ∴ y = 2
and b = \(\frac{1}{x}\) = 1 ∴ x = 1
Thus, the solution of the given pair of equations is x = 1, y = 2.

7. The given pair of equations is
\(\frac{10}{x+y}+\frac{2}{x-y}=4\) ………..(1)
\(\frac{15}{x+y}-\frac{5}{x-y}=-2\) ………..(2)
Taking \(\frac{1}{x+y}=a\) and \(\frac{1}{x-y}=b\) in both the equations, we get
10a + 2b = 4
i.e., 5a + b = 2 ……(3)
15a – 5b = -2 ……(4)
Multiplying equation (3) by 5 and then adding equation (4) to it, we get
5(5a + b) + (15a – 5b) = 5(2) + (-2)
∴ 25a + 5b + 15a – 5b = 10 – 2
∴ 40a = 8
∴ a = \(\frac{1}{5}\)
Substituting a = \(\frac{1}{5}\) in equation (3), we get
5(\(\frac{1}{5}\)) + b = 2
∴ 1 + b = 2
∴ b = 1
Now, a = \(a=\frac{1}{x+y}=\frac{1}{5}\)
∴ x + y = 5 ……(5)
and b = \(\frac{1}{x-y}=1\)
∴ x – y = 1 ………(6)
Adding equations (5) and (6), we get
2x = 6
∴ x = 3
Substituting x = 3 in equations (5), we get
3 + y = 5
∴ y = 2
Thus, the solution of the given pair of equations is x = 3, y = 2.

8. The given pair of equations is
\(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\) ………(1)
\(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8}\) ………(2)
Taking \(\frac{1}{3 x+y}=a\) and \(\frac{1}{3 x-y}=b\) in both the equations, we get
a + b = \(\frac{3}{4}\) ………(3)
\(\frac{a}{2}-\frac{b}{2}=-\frac{1}{8}\)
i.e., a – b = –\(\frac{2}{8}\)
i.e., a – b = –\(\frac{1}{4}\) ………….(4)
Adding equations (3) and (4), we get
2a = \(\frac{2}{4}\)
∴ a = \(\frac{1}{4}\)
Substituting a = \(\frac{1}{4}\) in equation (3), we get
\(\frac{1}{4}+b=\frac{3}{4}\)
∴ b = \(\frac{1}{2}\)
Now, a = \(\frac{1}{3 x+y}=\frac{1}{4}\)
3x + y = 4 ………….(5)
and b = \(\frac{1}{3 x-y}=\frac{1}{2}\)
∴ 3x – y = 2 ………….(6)
Adding equations (5) and (6), we get
6x = 6
∴ x = 1
Substituting x = 1 in equations (5), we get
3(1) + y = 4
∴ y = 1
Thus, the solution of the given pair of equations is x = 1, y = 1.

2. Formulate the following problems as a pair of equations, and hence find their solutions:

Question 1.
Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
Solution:
Let Ritu’s speed of rowing in still water be x km/h and the speed of the current by y km/h.
Then, her net speed going down-stream = (x + y) km/h and her net speed going upstream = (x – y) km/h.
Also, time = \(\frac{\text { distance }}{\text { speed }}\)
Then, form the first condition, we get
2 = \(\frac{20}{x+y}\)
∴ x + y = 10 ……(1)
And, from the second condition, we get
2 = \(\frac{4}{x-y}\)
∴ x – y = 2 ………..(2)
Adding equations (1) and (2), we get
2x = 12
∴ x = 6
Substituting x = 6 in equation (1), we get
6 + y = 10
∴ y = 4
Thus, Ritu’s speed of rowing in still water is 6 km/h and the speed of the current is 4 km/h.

Question 2.
2 women and 5 men can together finish an embroidery work in 4 days. while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
Solution:
Suppose that 1 woman alone can finish the work in x days and 1 man alone can finish the work in y days.
∴ Work done by 1 woman in 1 day = \(\frac{1}{x}\) part and work done by 1 man in 1 day \(\frac{1}{y}\) part.
Then, work done by 2 women and 5 men together in 1 day = \(\left(\frac{2}{x}+\frac{5}{y}\right)\) part.
But, according to the first information given, 2 women and 5 men together finish the work in 4 days, i.e., in one day they can finish \(\frac{1}{4}\) part of the work.
Hence, we get the following equation:
\(\frac{2}{x}+\frac{5}{y}=\frac{1}{4}\) ……(1)
Similarly, from the second information given, we get
\(\frac{3}{x}+\frac{6}{y}=\frac{1}{3}\) ……(2)
Taking \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b in both the equations, we get
2a + 5b = \(\frac{1}{4}\) ……(3)
3a + 6b = \(\frac{1}{3}\) ……(4)
Multiplying equation (3) by 6 and equation (4) by 5, we get
12a + 30b = \(\frac{6}{4}\) ……(5)
15a + 30b = \(\frac{5}{3}\) ……(6)
Subtracting equation (5) from equation (6), we get
(15a + 30b) – (12a + 30b) = \(\frac{5}{3}-\frac{6}{4}\)
15a + 30b – 12a – 30b = \(\frac{20-18}{12}\)
∴ 3a = \(\frac{2}{12}\)
∴ a = \(\frac{1}{18}\)
Substituting a = \(\frac{1}{18}\) in equation (3), we get
2(\(\frac{1}{18}\)) + 5b = 4
∴ 5b = \(\frac{1}{4}-\frac{1}{9}\)
∴ 5b = \(\frac{5}{36}\)
∴ b = \(\frac{1}{36}\)
Now, a = \(\frac{1}{x}=\frac{1}{18}\) ∴ x = 18
and b = \(\frac{1}{y}=\frac{1}{36}\) ∴ y = 36
Thus, 1 woman alone can finish the work in 18 days and 1 man alone can finish the work in 36 days.

Question 3.
Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution:
Let the speed of the train be x km/h and the speed of the bus be y km/h.
Also, time = \(\frac{\text { distance }}{\text { speed }}\)
In the first case, she travels 60 km by train and the remaining 240 km (300 – 60) by bus.
∴ Time taken for journey by train = \(\frac{60}{x}\)h and time taken for journey by bus \(\frac{240}{y}\)h.
∴ Total time taken = \(\left(\frac{60}{x}+\frac{240}{y}\right)\)h
In the first case, total time taken = 4 h.
Hence, we get the following equation:
\(\frac{60}{x}+\frac{240}{y}=4\) ………..(1)
Similarly, in the second case, the distances she travels by train and bus are 100 km and 200 km respectively and time taken by those journies are \(\frac{100}{x}\)h and \(\frac{200}{y}\)h respectively.
In this case, the total time taken = 4 hours + 10 minutes = 4\(\frac{1}{6}\) hours.
Hence, we get the following equation:
\(\frac{100}{x}+\frac{200}{y}=4 \frac{1}{6}\)
∴ \(\frac{100}{x}+\frac{200}{y}=\frac{25}{6}\) ………..(2)
Taking \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b in both the equations, we get
60a + 240b = 4 ………..(3)
100a + 200b = \(\frac{25}{6}\) ………..(4)
Multiplying equation (3) by 5 and equation (4) by 6, we get
300a + 1200b = 20 ………..(5)
600a + 1200b = 25 ………..(6)
Subtracting equation (5) from equation (6), we get
(600a + 1200b) – (300a + 1200b) = 25 – 20
∴ 300a = 5
∴ a = \(\frac{1}{60}\)
Substituting a = \(\frac{1}{60}\) in (3), we get
60(\(\frac{1}{60}\)) + 240b = 4
∴ 1 + 240b = 4
∴ 240b = 3
∴ b = \(\frac{1}{80}\)
Now, a = \(\frac{1}{x}\) = \(\frac{1}{60}\)
∴ x = 60
and b = \(\frac{1}{y}\) = \(\frac{1}{80}\)
∴ y = 80
Thus, the speed of the train is 60 km/h and the speed of the bus is 80 km/h.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.5

Question 1.
Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross-multiplication method:
1. x – 3y – 3 = 0
3x – 9y – 2 = 0
2. 2x + y = 5
3x + 2y = 8
3. 3x – 5y = 20
6x – 10y = 40
4. x – 3y – 7 = 0
3x – 3y – 15 = 0
Solution:
1. x – 3y – 3 = 0 and 3x – 9y – 2 = 0
Here, a₁ = 1; a₂ = 3; b₁ = -3; b₂ = -9; c₁ = -3 and c₂ = -2.
Now, \(\frac{a_1}{a_2}=\frac{1}{3}, \quad \frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3}\) and \(\frac{c_1}{c_2}=\frac{-3}{-2}=\frac{3}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the given pair of linear equations has no solution.

2. 2x + y – 5 = 0 and 3x + 2y – 8 = 0
Here, a₁ = 2; a₂ = 3; b₁ = 1; b₂ = 2; c₁ = -5 and c₂ = -8.
Now, \(\frac{a_1}{a_2}=\frac{2}{3}, \quad \frac{b_1}{b_2}=\frac{1}{2}\) and \(\frac{c_1}{c_2}=\frac{-5}{-8}=\frac{5}{8}\)
Here, \(\frac{1}{2}\)
Hence, the given pair of linear equations has a unique solution.
Now,


Thus, the unique solution of the given pair of linear equations is x = 2, y = 1.

3. 3x – 5y – 20 = 0 and 6x – 10y – 40 = 0
Here, a₁ = 3, a₂ = 6, b₁ = -5, b₂ = -10, c₁ = -20 and c₂ = -40.
Now, \(\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}\), \(\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}\) and \(\frac{c_1}{c_2}=\frac{-20}{-40}=\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the given pair of linear equations has infinitely many solutions.

4. x – 3y – 7 = 0 and 3x – 3y – 15 = 0
Here, a₁ = 1, a₂ = 3, b₁ = -3, b₂ = -3, c₁ = -7 and c₂ = -15.
Now, \(\frac{a_1}{a_2}=\frac{1}{3}, \frac{b_1}{b_2}=\frac{-3}{-3}=1\) and \(\frac{c_1}{c_2}=\frac{-7}{-15}=\frac{7}{15}\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the given pair of linear equations has a unique solution.
Now,

Thus, the unique solution of the given pair of linear equations is x = 4, y = -1.

Question 2.
1. For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7
(a – b) x + (a + b) y = 3a + b – 2
2. For which value of k will the following pair of linear equations have no solution ?
3x + y = 1
(2k – 1)x + (k – 1) y = 2k + 1
Solution:
1. For the given pair of equations, we express them in the standard form as:
2x + 3y – 7 = 0 and
(a – b)x + (a + b)y – (3a + b – 2) = 0
Here, A₁ = 2; A₂ = a – b; B₁ = 3; B₂ = a + b; C₁ = -7 and C₂ = -(3a + b – 2)
Then, \(\frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{2}{a-b}, \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{3}{a+b}\) and \(\frac{C_1}{C_2}=\frac{-7}{-(3 a+b-2)}=\frac{7}{3 a+b-2}\)
For the pair of equations to have infinite number of solution, we must have

∴ 2(a + b) = 3(a – b)
∴ 2a + 2b = 3a – 3b
∴ 5b = a
∴ a = 5b ………..(1)
Again, \(\frac{3}{a+b}=\frac{7}{3 a+b-2}\)
∴ 3 (3a + b – 2) = 7(a + b)
∴ 9a + 3b – 6 = 7a + 7b
∴ 2a – 4b = 6
∴ 2 (5b) – 4b = 6 [from (1), a = 5b]
∴ 6b = 6
∴ b = 1
Now, a = 5b
∴ a = 5(1)
∴ a = 5
Thus, for a = 5 and b = 1, the given pair of linear equations will have an infinite number of solutions.

2. We express the given equations in the standard form as:
3x + y – 1 = 0 and
(2k – 1)x + (k – 1)y – (2k + 1) = 0.
Here, a₁ = 3; a₂ = 2k – 1; b₁ = 1; b₂ = k – 1; c₁ = -1 and c₂ = -(2k + 1).
Then, \(\frac{a_1}{a_2}=\frac{3}{2 k-1}, \quad \frac{b_1}{b_2}=\frac{1}{k-1}\) and \(\frac{c_1}{c_2}=\frac{-1}{-(2 k+1)}=\frac{1}{2 k+1}\)
For the pair of equations to have not solution, we must have
\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
∴ \(\frac{3}{2 k-1}=\frac{1}{k-1} \neq \frac{1}{2 k+1}\)
\(\frac{3}{2 k-1}=\frac{1}{k-1}\)
∴ 3(k – 1) = 2k – 1
∴ 3k – 3 = 2k – 1
∴ k = 2
For k = 2.
\(\frac{a_1}{a_2}=\frac{3}{2(2)-1}=1\), \(\frac{b_1}{b_2}=\frac{1}{2-1}=1\) and \(\frac{c_1}{c_2}=\frac{1}{2(2)+1}=\frac{1}{5}\)
Thus, k = 2 satisfies \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Thus, for k = 2, the given pair of linear equations has no solution.

Question 3.
Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4
Solution:
1. Substitution method:
8x + 5y = 9 ……….(1)
3x + 2y = 4 ……….(2)
From equation (2), we get y = \(\frac{4-3 x}{2}\)
Substituting y = \(\frac{4-3 x}{2}\) in equation (1).
we get
8x + 5(\(\frac{4-3 x}{2}\)) = 9
∴ 16x + 5(4 – 3x) = 18 (Multiplying by 2)
∴ 16x + 20 – 15x = 18
∴ x = -2
Substituting x = -2 in y = \(\frac{4-3 x}{2}\), we get
∴ y = \(\frac{4-3(-2)}{2}\)
∴ y = 5
Thus, the solution of the given pair of linear equations is x = -2, y = 5.

2. Cross-multiplication method:
We express both the equations in the standard form as:
8x + 5y – 9 = 0 and 3x + 2y – 4 = 0
Here, a₁ = 8; b₁ = 5; c₁ = -9; a₂ = 3; b₂ = 2 and c₂ = -4.
Now, we arrange the coefficients as required in cross-multiplication method as:

Thus, the solution of the given pair of linear equations is x = -2, y = 5.

4. Form the pair of linear equations in the following problems and find their solutions. (if they exist) by any algebraic method:

Question 1.
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 ₹ 1000 as hostel charges whereas a student B, who takes food for 26 days, pays ₹ 1180 as hostel charges. Find the fixed charges and the cost of food per day.
Solution:
Let the fixed monthly charge be ₹ x and the cost of food per day be ₹ y.
Then, from the give data, we get the following pair of linear equations:
x + 20y = 1000 ……….(1)
x + 26y = 1180 ………..(2)
These equations in standard form are:
x + 20y – 1000 = 0 ……….(3)
x + 26y – 1180 = 0 ……….(4)
Now, we solve the equation by cross-multiplication method.

Thus, monthly fixed charge is ₹ 400 and the cost of food per day is ₹ 30.
Note: Here, the elimination method would be much easter, but we have solved. it by cross-multiplication method to show more applications of that method.

Question 2.
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.
Solution:
Let the numerator and the denominator of the required fraction be x and y respectively.
Then, the required fraction = \(\frac{x}{y}\)
By the given data, we get
\(\frac{x-1}{y}=\frac{1}{3}\)
∴ 3x – 3 = y
∴ 3x – y = 3 ……….(1)
Also, \(\frac{x}{y+8}=\frac{1}{4}\)
∴ 4x = y +8
∴ 4x – y = 8 …………(2)
Subtracting equation (1) from equation (2),
we get
(4x – y) – (3x – y) = 8 – 3
∴ 4x – y – 3x + y = 5
∴ x = 5
Substituting x = 5 in equation (1),
we get
3(5) – y = 3
∴ 15 – 3 = y
∴ y = 12
Hence, the required fraction = \(\frac{x}{y}=\frac{5}{12}\)

Question 3.
Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Solution:
Suppose Yash gave x right answers and y wrong answers.
Then, from the given data, we get
3x – y = 40 ………(1)
and 4x – 2y = 50, i.e.. 2x – y = 25 ……….(2)
We express the equations in the standard form as
3x – y – 40 = 0 ………(3)
2x – y – 25 = 0 ………(4)
Then,

Then, the total number of questions in the test = x + y = 15 + 5 = 20.
Thus, there were 20 questions in the test.

Question 4.
Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Solution:
Let the speed of the car starting from A be x km/hour and the speed of the car starting from B be y km/hour such that x > y. If the cars travel in the same direction and meet, they should be travelling in the direction from A towards B as the car starting from A is faster than the car starting from B.

Suppose the cars meet at place P after 5 hours, when they travel in the same direction. Then, the distance travelled in 5 hours by the car starting from A= 5x km = AP (Distance = Speed × Time) Similarly, the distance travelled in 5 hours by the car starting from B = 5y km = BP.
Now, AB = 100 km
∴ AP – BP = 100
∴ 5x – 5y = 100
∴ x – y = 20 (Dividing by 5) ………(1)

Suppose the cars meet at place Q after 1 hour when they travel in opposite directions.
Then, distance travelled in 1 hour by the car starting from A = x km = AQ
Similarly, distance travelled in 1 hour by the car starting from B = y km = BQ.
Now, AB = 100 km
∴ AQ + BQ = 100
∴ x + y = 100 ………(2)
Adding equations (1) and (2), we get
(x – y) + (x + y) = 20 + 100
∴ 2x = 120
∴ x = 60
Substituting x = 60 in equation (2), we get
60 + y = 100
∴ y = 40
Thus, the speeds of the cars starting from A and B are 60 km/hour and 40 km/hour respectively under the assumption that x > y.

Question 5.
The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution:
Let the length of the rectangle be x units and its breadth be units.
Area of a rectangle = Length × Breadth
∴ Area of given rectangle = xy square units
According to the first condition given,
new reduced length = (x – 5) units,
new increased breadth = (y + 3) units and new reduced area = (xy – 9) square units.
Then, Length × Breadth = Area of a rectangle gives
(x – 5) (y + 3) = xy – 9
∴ xy + 3x – 5y – 15 = xy – 9
∴ 3x – 5y – 6 = 0 ……….(1)
Similarly, from the second condition given,
new increased length = (x + 3) units.
new increased breadth = (y + 2) units and
new increased area = (xy + 67) square units.
Again, Length × Breadth = Area of a rectangle gives
(x + 3)(y + 2) = xy + 67
∴ xy + 2x + 3y + 6 = xy + 67
∴ 2x + 3y – 61 = 0 ………..(2)
We solve these equations (1) and (2) by cross-multiplication method.


∴ x = \(\frac{323}{19}\) and y = \(\frac{171}{19}\)
∴ x = 17 and y = 9
Thus, the length and the breadth of the given rectangle are 17 units and 9 units respectively.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution method:
1. x + y = 5 and 2x – 3y = 4
2. 3x + 4y = 10 and 2x – 2y = 2
3. 3x – 5y – 4 = 0 and 9x = 2y + 7
4. \(\frac{x}{2}+\frac{2 y}{3}=-1\) and x – \(\frac{y}{3}\) = 3
Solution:
1. Elimination method:
x + y = 5 ………(1)
2x – 3y = 4 ……..(2)
We multiply equation (1) by 3 and equation (2) by 1 to get following equations:
3x + 3y = 15 ……(3)
2x – 3y = 4 ……(4)
Adding equations (3) and (4), we get
(3x + 3y) + (2x – 3y) = 15 + 4
∴ 5x = 19
∴ x = \(\frac{19}{5}\)
Substituting x = \(\frac{19}{5}\) in equation (1), we get
\(\frac{19}{5}\) + y = 5
∴ y = 5 – \(\frac{19}{5}\)
∴ y = \(\frac{6}{5}\)
Thus, the solution of the given pair of linear equations is x = \(\frac{19}{5}\), y = \(\frac{6}{5}\)
Substitution method:
x + y = 5 ……(1)
2x – 3y = 4 ……(2)
From equation (1), we get y = 5 – x.
Substituting y = 5 – x in equation (2), we get
2x – 3(5 – x) = 4
∴ 2x – 15 + 3x = 4
∴ 5x = 19
∴ x = \(\frac{19}{5}\)
Substituting x = \(\frac{19}{5}\) in y = 5 – x, we get
y = 5 – \(\frac{19}{5}\)
∴ y = \(\frac{6}{5}\)
Thus, the solution of the given pair of linear equations is x = \(\frac{19}{5}\), y = \(\frac{6}{5}\)

2. Elimination method:
3x + 4y = 10 ……(1)
2x – 2y = 2 ……(2)
We multiply equation (1) by 1 and equation (2) by 2 to get following equations:
3x + 4y = 10 ……(3)
4x – 4y = 4 ……(4)
Adding equations (3) and (4), we get
(3x + 4y) + (4x – 4y) = 10 +4
∴ 7x = 14
∴ x = 2
Substituting x = 2 in equation (1), we get
3(2) + 4y = 10
∴ 4y = 10 – 6
∴ 4y = 4
∴ y = 1
Thus, the solution of the given pair of linear equations is x = 2, y = 1.

Substitution method:
3x + 4y = 10 ……(1)
2x – 2y = 2 ……(2)
From equation (2), we get 2x = 2y + 2.
i.e., x = y + 1.
Substituting x = y + 1 in equation (1), we get
3(y + 1) + 4y = 10
∴ 3y + 3 + 4y = 10
∴ 7y = 7
∴ y = 1
Substituting y = 1 in x = y + 1, we get
x = 1 + 1
x = 2
Thus, the solution of the given pair of linear equations is x = 2, y = 1.

3. Elimination method:
3x – 5y – 4 = 0 ……(1)
9x = 2y + 7 ……(2)
i.e., 3x – 5y = 4 ……(3)
9x – 2y = 7 ……(4)
We multiply equation (3) by 3 and equation (4) by 1 to get following equations:
9x – 15y = 12 ……(5)
9x – 2y = 7 ……(6)
Subtracting equation (5) from equation (6),
we get
(9x – 2y) – (9x – 15y) = 7 – 12
∴ 9x – 2y – 9x + 15y = -5
∴ 13y = -5
∴ y = –\(\frac{5}{13}\)
Substituting y = –\(\frac{5}{13}\) in equation (5), we get
9x – 15(-\(\frac{5}{13}\)) = 12
∴ 9x + \(\frac{75}{13}\) = 12
∴ 9x = 12 – \(\frac{75}{13}\)
∴ 9x = \(\frac{75}{13}\)
∴ x = \(\frac{9}{13}\)
Thus, the solution of the given pair of linear equations is x = \(\frac{9}{13}\), y = –\(\frac{5}{13}\)

Substitution method:
3x – 5y – 4 = 0 ………(1)
9x = 2y + 7 ………(2)
From equation (2), we get x = \(\frac{2 y+7}{9}\).
Substituting x = \(\frac{2 y+7}{9}\) in equation (1).
we get
3(\(\frac{2 y+7}{9}\)) – 5y – 4 = 0
∴ \(\frac{2 y+7}{9}\) – 5y – 4 = 0
∴ 2y + 7 – 15y – 12 = 0 (Multiplying by 3)
∴ -13y – 5 = 0
∴ -13y = 5
∴ y = –\(\frac{5}{13}\)
Substituting y = –\(\frac{5}{13}\) in x = \(\frac{2 y+7}{9}\)
we get,

Thus, the solution of the given pair of linear equations is x = \(\frac{9}{13}\), y = –\(\frac{5}{13}\).

4. Elimination method:
\(\frac{x}{2}+\frac{2 y}{3}=-1\) ………(1)
\(x-\frac{y}{3}=3\) ………(2)
We multiply equation (1) by 3 and equation (2) by 6 to get following equations:
\(\frac{3}{2}\)x + 2y = -3 ………(3)
6x – 2y = 18 ………(4)
Adding equations (3) and (4), we get
(\(\frac{3}{2}\)x + 2y) + (6x – 2y) = -3 + 18
∴ \(\frac{15}{2}\)x = 15
∴ x = 2
Substituting x = 2 in equation (3), we get
\(\frac{3}{2}\)(2) + 2y = -3
∴ 3 + 2y = -3
∴ 2y = -6
∴ y = -3
Thus, the solution of the given pair of linear equations is x = 2, y = -3.

Substitution method:
\(\frac{x}{2}+\frac{2 y}{3}=-1\) ………(1)
\(x-\frac{y}{3}=3\) ………(2)
From equation (2), we get x = \(\frac{y}{3}+3\)
Substituting x = \(\frac{y}{3}+3\) in equation (1).
we get

Substituting y = -3 in x = \(\frac{y}{3}+3\), we get
x = \(\frac{-3}{3}+3\)
∴ x = -1 + 3
∴ x = 2
Thus, the solution of the given pair of linear equations is x = 2, y = -3.

Form the pair of linear equations in the following problems and find their solutions (if they exist) by the elimination method:

Question 1.
If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction?
Solution:
Let the numerator of the required fraction be x and the denominator be y.
Then, the required fraction is \(\frac{x}{y}\)
From the first condition given, we get
\(\frac{x+1}{y-1}=1\)
∴ x + 1 = y – 1
∴ x – y = -2 …..(1)
From the second condition, we get
\(\frac{x}{y+1}=\frac{1}{2}\)
∴ 2x = y + 1
∴ 2x – y = 1 ………(2)
Now, subtracting equation (1) from equation (2), we get
(2x – y) – (x – y) = 1 – (-2)
∴ 2x – y – x + y = 3
∴ x = 3
Substituting x = 3 in equation (1), we get
3 – y = -2
∴ -y = – 2 – 3
∴ -y = -5
∴ y = 5
So, the fraction \(\frac{x}{y}=\frac{3}{5}\)
Thus, the pair of linear equations formed is x – y = -2 and 2x – y = 1; and the required fraction is \(\frac{3}{5}\)

Question 2.
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?
Solution:
Let the present age of Nuri be x years. and the present age of Sonu be y years.
Five years ago, the age of Nuri was (x – 5) years and the age of Sonu was (y – 5) years.
Then, from the first condition given, we get
(x – 5) = 3(y – 5)
∴ x – 5 = 3y – 15
∴ x – 3y = -10 ……….(1)
Ten years later the ages of Nuri and Sonu will be (x + 10) years and (y + 10) years respectively.
Then, from the second condition given, we get
(x + 10) = 2(y + 10)
∴ x + 10 = 2y + 20
∴ x – 2y = 10 ………..(2)
Subtracting equation (1) from equation (2).
we get
(x – 2y) – (x – 3y) = 10 – (-10)
∴ x – 2y – x + 3y = 10 + 10
∴ y = 20
Substituting y = 20 in equation (2).
we get
x – 2 (20) = 10
∴ x – 40 = 10
∴ x = 50
So, the present ages of Nuri and Sonu are 50 years and 10 years.
Thus, the pair of linear equations formed is x – 3y = -10 and x – 2y = 10 and the present ages of Nuri and Sonu are 50 years and 20 years respectively.

Question 3.
The sum of the digits of a two digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Solution:
Let the digit at tens place be x and the digit at units place be y in the original number.
Then, the original number = 10x + y.
From the first condition given, we get x + y = 9 ………..(1)
If the digits are reversed, in the new number the digit at tens place is y and the digit at units place is x.
Then, the new number = 10y + x.
From the second condition given, we get
9(10x + y) = 2(10y + x)
∴ 90x + 9y = 20y + 2x
∴ 88x – 11y = 0
∴ 8x – y = 0 (Dividing by 11) ……….(2)
Adding equations (1) and (2), we get
(x + y) + (8x − y) = 9 + 0
∴ 9x = 9
∴ x = 1
Substituting x = 1 in equation (1), we get
1 + y = 9
∴ y = 8
So, the original number = 10x + y
= 10(1) + 8 = 18
Thus, the pair of linear equations formed is x + y = 9 and 8x – y = 0; and the original number is 18.

Question 4.
Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
Solution:
Suppose Meena received x notes of ₹ 50 and y notes of ₹ 100.
So, the total amount received by her = ₹ (50x + 100y)
From the first condition given, the total amount is ₹ 2000. So, we get
50x + 100y = 2000
∴ x + 2y = 40 (Dividing by 50) ……(1)
From the second condition given, we get
x + y = 25 ……(2)
Subtracting equation (2) from equation (1),
we get
(x + 2y) – (x + y) = 40 – 25
∴ y = 15
Substituting y = 15 in equation (2), we get
x + 15 = 25
∴ x = 10
Hence, Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.
Thus, the pair of linear equations formed is x + 2y = 40 and x + y = 25, and Meena received 10 notes of ₹ 50 and 15 notes of ₹ 100.

Question 5.
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
Let the fixed charge for first three days be ₹ x and the additional charge for each day exceeding the first three days be ₹ y. Saritha kept the book for 7 days.
So, she has to pay the fixed charge plus the additional charge for 4(7 – 3) days. Hence, we get the following equation for Saritha:
x + 4y = 27 …………(1)
Similarly, Susy has to pay the fixed charge plus the addition charge for 2 (5 – 3) days.
Hence, we get the following equation for Susy:
x + 2y = 21 ……….(2)
Subtracting equation (2) from equation (1),
we get
(x + 4y) – (x + 2y) = 27 – 21
∴ 2y = 6
∴ y = 3
Substituting y = 3 in equation (1), we get
x + 4(3) = 27
∴ x + 12 = 27
∴ x = 15
Hence, the fixed charge for first three days is ₹ 15 and the addition charge for each day thereafter is ₹ 3.
Thus, the pair of linear equations formed is x + 4y = 27 and x + 2y = 21; and the fixed charge and the additional charge per day are ₹ 15 and ₹ 3 respectively.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

Question 1.
Solve the following pair of linear equations by the substitution method:
1. x + y = 14
x – y = 4
2. s – t = 3
\(\frac{s}{3}+\frac{t}{2}=6\)
3. 3x – y = 3
9x – 3y = 9
4. 0.2x + 0.3y = 1.3
0.4x + 0.5 y = 2.3
5. \(\sqrt{2}\)x + \(\sqrt{3}\)3y = 0
\(\sqrt{3}\)x – \(\sqrt{8}\)y = 0
6. \(\frac{3 x}{2}-\frac{5 y}{3}=-2\)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)
Solution:
1. x + y = 14 ……….(1)
x – y = 4 ……..(2)
From equation (1), we get y = 14 – x.
Substituting y = 14 – x in equation (2).
we get
x – (14 – x) = 4
∴ x – 14 + x = 4
∴ 2x = 4 + 14
∴ 2x = 18
∴ x = 9
Substituting x = 9 in equation (1), we get
9 + y = 14
∴ y = 5
Thus, the solution of the given pair of linear equations is x = 9, y = 5.
Verification: x + y = 9 + 5 = 14 and x – y = 9 – 5 = 4.
Hence, the solution is verified.

2. s – t = 3 ……….(1)
\(\frac{s}{3}+\frac{t}{2}=6\) ……..(2)
From equation (1), we get s = t + 3.
Substituting s = t + 3 in equation (2),
we get
\(\frac{t+3}{3}+\frac{t}{2}=6\)
∴ 2 (t + 3) + 3t = 36 (Multiplying by 6)
∴ 2t + 6 + 3t = 36
∴ 5t = 30
∴ t = 6
Substituting t = 6 in equation (1), we get s – 6 = 3
∴ s = 9
Thus, the solution of the given pair of linear equations is s = 9, t = 6.
Verification: s – t = 9 – 6 = 3 and
\(\frac{s}{3}+\frac{t}{2}=\frac{9}{3}+\frac{6}{2}=6\)
Hence, the solution is verified.

3. 3x – y = 3 ……….(1)
9x – 3y = 9 ……….(2)
From equation (1), we get y = 3x – 3.
Substituting y = 3x – 3 in equation (2).
we get
9x – 3(3x – 3) = 9
∴ 9x – 9x + 9 = 9
∴ 9 = 9
Here, we do not get the value of x, but we get a true statement 9 = 9.
Hence, the given pair of linear equations has infinitely many solutions given by y = 3x – 3, where x is any real number.

4. 0.2x + 0.3y = 1.3 ……….(1)
0.4x + 0.5y = 2.3 ……….(2)
Not mandatory, but for convenience we multiply both the equations by 10 and get equations with integer coefficients as:
2x + 3y = 13 ……(3)
4x + 5y = 23 ……..(4)
From equation (3), we get x = \(\frac{13-3 y}{2}\).
Substituting x = \(\frac{13-3 y}{2}\) in equation (4),
we get
4(\(\frac{13-3 y}{2}\)) + 5y = 23
∴ 26 – 6y + 5y = 23
∴ -y = -3
∴ y = 3
Substituting y = 3 in x = \(\frac{13-3 y}{2}\)
x = \(\frac{13-3(3)}{2}\)
∴ x = \(\frac{4}{2}\)
∴ x = 2
Thus, the solution of the given pair of linear equations is x = 2, y = 3.
Verification:
0.2x + 0.3y = (0.2) (2) + (0.3) (3) = 1.3 and
0.4x + 0.5y = (0.4) (2) = (0.5) (3) = 2.3.
Hence, the solution is verified.

5. \(\sqrt{2}\)x + \(\sqrt{3}\)y = 0 ……….(1)
\(\sqrt{3}\)x – \(\sqrt{8}\)y = 0 ……….(2)
From equation (2), we get x = \(\frac{\sqrt{8}}{\sqrt{3}}\)y.
Substituting x = \(\frac{\sqrt{8}}{\sqrt{3}}\)y in equation (1),
we get
\(\sqrt{2}\left(\frac{\sqrt{8}}{\sqrt{3}} y\right)+\sqrt{3} y=0\)
∴ \(\frac{4}{\sqrt{3}} y+\sqrt{3} y=0\)
∴ \(y\left(\frac{4}{\sqrt{3}}+\sqrt{3}\right)=0\)
∴ y = 0
Substituting y = 0 in x = \(\frac{\sqrt{8}}{\sqrt{3}}\)y, we get
x = \(\frac{\sqrt{8}}{\sqrt{3}}\)(0)
∴ x = 0
Thus, the solution of the given pair of linear equations is x = 0, y = 0.

6. \(\frac{3 x}{2}-\frac{5 y}{3}=-2\) ……….(1)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\) ……….(2)
Not mandatory, but for convenience we multiply both the equations by 6 and get
9x – 10y = – 12 ……….(3)
2x + 3y = 13 ……….(4)
From equation (3), we get x = \(\frac{10 y-12}{9}\).
Substituting x = \(\frac{10 y-12}{9}\) in equation (4).
we get
2(\(\frac{10 y-12}{9}\)) + 3y = 13
∴ 2(10y – 12) + 27y = 117 (Multiplying by 9)
∴ 20y – 24 + 27y = 117
∴ 47y = 141
∴ y = 3
Substituting y = 3 in x = \(\frac{10 y-12}{9}\), we get
x = \(\frac{10(3)-12}{9}\)
∴ x = \(\frac{18}{9}\)
∴ x = 2
Thus, the solution of the given pair of linear equations is x = 2, y = 3.

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx +3.
Solution:
2x + 3y = 11 …..(1)
2x – 4y = -24 …..(2)
From equation (2), we get x = \(\frac{4 y-24}{2}\) = 2y – 12.
Substituting x = 2y – 12 in equation (1), we get
2 (2y – 12) + 3y = 11
∴ 4y – 24 + 3y = 11
∴ 7y = 35
∴ y = 5
Substituting y = 5 in x = 2y – 12, we get
x = 2(5) – 12
∴ x = 10 – 12
∴ x = -2
Now, for x = -2 and y = 5, y = mx + 3 gives
5 = m(-2) + 3
∴ 5 = -2m + 3
∴ 2m = 3 – 5
∴ 2m = -2
∴ m = -1
Thus, the solution of the given pair of equations is x = -2, y = 5 and m = -1 satisfies y = mx +3.

Form the pair of linear equations for the following problems and find their solution by substitution method:

Question 1.
The difference between two numbers is 26 and one number is three times the other. Find them.
Solution:
Let the greater number be x and the smaller number be y
Then, from the given information, we get the following pair of linear equations:
x – y = 26 ……….(1)
x = 3y ………….(2)
Substituting x = 3y in equation (1).
we get
3y – y = 26
∴ 2y = 26
∴ y = 13
Then, x = 3y gives x = 3 × 13 = 39.
Thus, the required numbers are 39 and 13.
Verification: The difference of numbers = 39 – 13 = 26 and 39 = three times 13.

Question 2.
The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution:
Let the measure (in degrees) of the greater angle be x and that of the smaller angle be y.
Then, from the given data, we get the following pair of linear equations:
x + y = 180 ……….(1)
x – y = 18 ………….(2)
From equation (2), we get x = y + 18.
Substituting x = y + 18 in equation (1).
we get
y + 18 + y = 180
∴ 2y = 162
∴ y = 81
Substituting y = 81 in equation (2), we get
x – 81 = 18
∴ x = 99
Thus, the measures (in degrees) of the given angles are 99 and 81.
Verification: Larger angle-Smaller angle = 99° – 81° = 18° and Larger angle + Smaller angle = 99° + 81° = 180°, i.e., the angles are supplementary angles.

Question 3.
The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
Solution:
Let the cost of each bat be ₹ x and the cost of each ball be ₹ y.
Then, from the given data, we get the following pair of linear equations:
7x + 6y = 3800 ……….(1)
3x + 5y = 1750 ……….(2)
From equation (2), we get x = \(\frac{1750-5 y}{3}\)
Substituting x = \(\frac{1750-5 y}{3}\) in equation (1) we get
7(\(\frac{1750-5 y}{3}\)) + 6y = 3800
∴ 7(1750 – 5y) + 18y = 11400 (Multiplying by 3)
∴ 12250 – 35y + 18y = 11400
∴ -17y = 11400 – 12250
∴ -17y = -850
∴ 17y = 850
∴ y = 50
Substituting y = 50 in x = \(\frac{1750-5 y}{3}\), we get
x = \(\frac{1750-5(50)}{3}\)
∴ x = \(\frac{1500}{3}\)
∴ x = 500
Thus, the cost of each bat is ₹ 500 and the cost of each ball is ₹ 50.
Verification:
Cost of 7 bats and 6 balls = 7 × 500 + 6 × 50 = ₹ 3800
Cost of 3 bats and 5 balls = 3 × 500 + 5 × 50 = ₹ 1750

Question 4.
The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km. the charge paid is ₹ 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Solution:
Let the fixed charge be ₹ x and the charge for the distance covered be ₹ y per km.
Then, from the given data, we get the following pair of linear equations:
x + 10y = 105 ……….(1)
x + 15y = 155 ……….(2)
From equation (1), we get x = 105 – 10y.
Substituting x = 105 – 10y in equation (2), we get
(105 – 10y) + 15y = 155
∴ 105 + 5y = 155
∴ 5y = 50
∴ y = 10
Substituting y = 10 in x = 105 – 10y.
we get
x = 105 – 10(10)
∴ x = 5
Thus, the fixed charge is ₹ 5 and the charge for the distance covered is ₹ 10 per km.
So, the total charge a person has to pay for travelling d km, is given by
Total charge = ₹(5 + 10d)
∴ Total charge to be paid for travelling 25 km = ₹ (5 + 10 × 25) = ₹ 255.

Question 5.
A fraction becomes \(\frac{9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.
Solution:
Let the numerator of the fraction be x and the denominator of the fraction be y.
Then, the required fraction is \(\frac{x}{y}\)
Then, from the given data, we get the following pair of equations:
\(\frac{x+2}{y+2}=\frac{9}{11}\)
∴ 11(x + 2) = 9(y + 2)
∴ 11x + 22 = 9y + 18
∴ 11x – 9y = -4 is the first linear equation derived from the data.
Similarly, \(\frac{x+3}{y+3}=\frac{5}{6}\)
∴ 6x + 18 = 5y + 15
∴ 6x – 5y = -3 is the second linear equation derived from the data.
Hence, required pair of linear equations is as follows:
11x – 9y = -4 ……….(1)
6x – 5y = -3 ……….(2)
From equation (2), we get x = \(\frac{5 y-3}{6}\)
Substituting x = \(\frac{5 y-3}{6}\) in equation (1),
we get
11(\(\frac{5 y-3}{6}\)) – 9y = -4
∴ 11 (5y – 3) – 54y = -24 (Multiplying by 6)
∴ 55y – 33 – 54y = -24
∴ y = 9
Substituting y = 9 in x = \(\frac{5 y-3}{6}\), we get
x = \(\frac{5(9)-3}{6}\)
∴ x = \(\frac{42}{6}\)
∴ x = 7
Thus, the required fraction is \(\frac{7}{9}\)
Verification:
\(\frac{7+2}{9+2}=\frac{9}{11}\) and \(\frac{7+3}{9+3}=\frac{10}{12}=\frac{5}{6}\)

Question 6.
Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution:
Let the present age of Jacob be x years and the present age of his son be y years.
∴ Five years hence, the age of Jacob will be (x + 5) years and the age of his son will be (y + 5) years.
Then, from the given data,
(x + 5) = 3(y + 5)
∴ x + 5 = 3y + 15
∴ x – 3y = 10
Again, five years ago, the age of Jacob was (x – 5) years and the age of his son was (y – 5) years.
Then, from the given data,
(x – 5) = 7(y – 5)
∴ x – 5 = 7y – 35
∴ x – 7y = -30
Hence, the required pair of linear equations is as follows:
x – 3y = 10 ……….(1)
x – 7y = -30 ……….(2)
From equation (2), we get x = 7y – 30.
Substituting x = 7y – 30 in equation (1).
we get
7y – 30 – 3y = 10
∴ 4y = 40
∴ y = 10
Substituting y = 10 in x = 7y – 30, we get
x = 7(10) – 30
∴ x = 40
Thus, the present ages of Jacob and his son are 40 years and 10 years respectively.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2

Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically:
1. 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
2. 5 pencils and 7 pens together cost 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.
Solution:
1. Let the number of boys be x and the number of girls be y.
Then, the equations formed as follows:
x + y = 10 ……. (1)
and y = x + 4,
i.e., y – x = 4 …… (2)
To draw the graphs of these equations,
we find two solutions for each equation.
For equation (1), x + y = 10 gives y = 10 – x.

x	0	5
y	10	5

For equation (2), y – x = 4 gives y = x + 4.

x	0	2
y	4	6

Two lines intersect at point (3, 7). Hence, x = 3 and y = 7 is the required solution of the pair of linear equations.
Thus, 3 boys and 7 girls took part in the quiz.
Verification : x = 3 and y = 7 satisfy both the equations x + y = 10 and y – x = 4.

2. Let the cost of each pencil be ₹ x and the cost of each pen be ₹ y.
Then, from the given information, we receive the following equations:
5x + 7y = 50 ……… (1)
7x + 5y = 46 ……….. (2)
To draw the graphs of these equations, we find two solutions for each equation.
For equation (1), 5x + 7y = 50 gives
y = \(\frac{50-5 x}{7}\)

x	3	10
y	5	0

For equation (2), 7x + 5y = 46 gives
y = \(\frac{46-7 x}{5}\)

x	3	8
y	5	-2

Two lines intersect at point (3, 5). Hence, x = 3 and y = 5 is the required solution of the pair of linear equations.
Thus, the cost of each pencil is ₹ 3 and the cost of each pen is ₹ 5.
Verification : x = 3 and y = 5 satisfy both the equations 5x + 7y = 50 and 7x + 5y = 46.

Question 2.
On comparing the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}\) and \(\frac{c_1}{c_2}\), find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
1. 5x – 4y + 8 = 0; 7x + 6y – 9 = 0
2. 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
3. 6x – 3y + 10 = 0; 2x – y + 9 = 0
Solution:
5x – 4y + 8 = 0; 7x + 6y – 9 = 0
For the given pair of linear equations,
a₁ = 5, b₁ = -4, c₁ = 8, a₂ = 7, b₂ = 6 and c₂ = -9
Now, \(\frac{a_1}{a_2}=\frac{5}{7}, \quad \frac{b_1}{b_2}=\frac{-4}{6}=-\frac{2}{3}\)
and \(\frac{c_1}{c_2}=\frac{8}{-9}=-\frac{8}{9}\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the lines representing the given pair of linear equations intersect at a point.

2. 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
For the given pair of linear equations, a₁ = 9, b₁ = 3, c₁ = 12, a₂ = 18, b₂ = 16 and c₂ = 24.
Now, \(\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the lines representing the given pair of linear equations are coincident lines.

3. 6x – 3y + 10 = 0; 2x – y + 9 = 0
For the given pair of linear equations, a₁ = 6, b₁ = -3, c₁ = 10, a₂ = 2, b₂ = -1 and c₂ = 9.
Now, \(\frac{a_1}{a_2}=\frac{6}{2}=3, \quad \frac{b_1}{b_2}=\frac{-3}{-1}=3\)
and \(\frac{c_1}{c_2}=\frac{10}{9}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the lines representing the given pair of linear equations are parallel lines.

Question 3.
On comparing the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}\) and \(\frac{c_1}{c_2}\), find out whether the following pair of linear equations are consistent or inconsistent:
1. 3x + 2y = 5; 2x – 3y = 7
2. 2x – 3y = 8; 4x – 6y = 9
3. \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7; 9x – 10y = 14
4. 5x – 3y = 11; -10x + 6y = -22
5. \(\frac{4}{3}\)x + 2y = 8; 2x + 3y = 12
Solution:
1. 3x + 2y = 5; 2x – 3y = 7
For the given pair of linear equations, a₁ = 3, b₁ = 2, c₁ = -5, a₂ = 2, b₂ = -3 and c₂ = -7.
Now, \(\frac{a_1}{a_2}=\frac{3}{2}, \frac{b_1}{b_2}=\frac{2}{-3}=-\frac{2}{3}\)
and \(\frac{c_1}{c_2}=\frac{-5}{-7}=\frac{5}{7}\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the given pair of linear equations is consistent.

2. 2x – 3y = 8: 4x – 6y = 9
For the given pair of linear equations,
a₁ = 2, b₁ = -3, c₁ = -8, a₂ = 4, b₂ =-6 and c₂ = -9.
Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-8}{-9}=\frac{8}{9}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the given pair of linear equations is inconsistent.

3. \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7; 9x – 10y = 14
Multiplying the first equation by 6 and expressing both the equations in the standard form, we get following equations:
9x + 10y – 42 = 0; 9x – 10y – 14 = 0
For the given pair of linear equations, a₁ = 9, b₁ = 10, c₁ = -42, a₂ = 9, b₂ = -10 and c₂ = -14.
Now, \(\frac{a_1}{a_2}=\frac{9}{9}=1, \frac{b_1}{b_2}=\frac{10}{-10}=-1\)
and \(\frac{c_1}{c_2}=\frac{-42}{-14}=3\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the given pair of linear equations is consistent.

4. 5x – 3y = 11; -10x + 6y = -22
For the given pair of linear equations, a₁ = 5, b₁ =-3, c₁ = -11, a₂ = -10, b₂ = 6 and c₂ = 22.
Now, \(\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2}, \frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-11}{22}=-\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the given pair of linear equations is consistent and dependent.

5. \(\frac{4}{3}\)x + 2y = 8; 2x + 3y = 12
For the given pair of linear equations, a₁ = \(\frac{4}{3}\), b₁ = 2, c₁ = -8, a₂ = 2, b₂ = 3 and c₂ = -12.
Now, \(\frac{a_1}{a_2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}, \quad \frac{b_1}{b_2}=\frac{2}{3}\)
and \(\frac{c_1}{c_2}=\frac{-8}{-12}=\frac{2}{3}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the given pair of linear equations is consistent and dependent.

Question 4.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
1. x + y = 5; 2x + 2y = 10
2. x – y = 8; 3x – 3y = 16
3. 2x + y – 6 = 0; 4x – 2y – 4 = 0
4. 2x – 2y – 2 = 0; 4x – 4y – 5 = 0
Solution:
1. x + y = 5; 2x + 2y = 10
For the given pair of linear equations,
a₁ = 1, b₁ = 1, c₁ = -5, a₂ = 2, b₂ = 2 and c₂ = -10.
Now, \(\frac{a_1}{a_2}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-5}{-10}=\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Hence, the given pair of linear equations is consistent and dependent.
Now, we draw the graphs of both the equations.
x + y = 5 gives y = 5 – x.

x	0	5
y	5	0

2x + 2y = 10 gives y = \(\frac{10-2 x}{2}\)

x	1	3
y	4	2

Here, lines representing both the equations. are coincident. Hence, any point on the line gives a solution. In general, y = 5 – x, where x is any real number is a solution of the given pair of linear equations.

2. x – y = 8; 3x – 3y = 16
For the given pair of linear equations, a₁ = 1, b₁ = -1, c₁ = -8, a₂ = 3, b₂ = -3 and c₂ = -16.
Now, \(\frac{a_1}{a_2}=\frac{1}{3}, \quad \frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3}\)
and \(\frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the given pair of linear equations is inconsistent.

3. 2x + y – 6 = 0; 4x – 2y – 4 = 0
For the given pair of linear equations, a₁ = 2, b₁ = 1, c₁ = -6, a₂ = 4, b₂ = -2 and c₂ = -4.
Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-6}{-4}=\frac{3}{2}\)
Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
Hence, the given pair of linear equations is consistent.
Now, we draw the graphs of both the equations.
2x + y – 6 = 0 gives y = 6 – 2x.

x	0	3
y	6	0

4x – 2y – 4 = 0 gives y = \(\frac{4 x-4}{2}\) = 2x – 2.

x	0	1
y	-2	0

Here, the lines intersect at point (2, 2). Hence, x = 2 and y = 2 is the unique solution of the given pair of linear equations.

4. 2x – 2y – 2 = 0; 4x – 4y – 5 = 0
For the given pair of linear equations, a₁ = 2, b₁ = -2, c₁ = -2, a₂ = 4, b₂ = -4 and c₂ = -5.
Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}\)
and \(\frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}\)
Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
Hence, the given pair of linear equations is inconsistent.

Question 5.
Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:
Let the length and breadth of the rectangular garden be x m and y m respectively.
Then, from the given data, x = y + 4 and 36 = \(\frac{1}{2}\)[2(x + y)] i.e., x + y = 36 as the perimeter of a rectangle = 2 (length + breadth).
To draw the graphs, we find two solutions of each equation.
x = y + 4 gives y = x – 4

x	8	24
y	4	20

x + y = 36 gives y = 36 – x

x	12	24
y	24	12

Here, the lines intersect at point (20, 16). Hence, x = 20 and y = 16 is the unique solution of the pair of linear equations.
Thus, for the rectangular garden, length = 20 m and breadth = 16 m.

Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
1. intersecting lines
2. parallel lines
3. coincident lines
Solution:
1. For intersecting lines, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\). The given equation is 2x + 3y – 8 = 0. We can give another equation as 3x + 4y – 24 = 0. Here, \(\frac{a_1}{a_2}=\frac{2}{3}\) and \(\frac{b_1}{b_2}=\frac{3}{4}\) satisfying \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

2. For parallel lines, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\). The given equation is 2x + 3y – 8 = 0. We can give another equation as 6x + 9y – 10 = 0.
Here, \(\frac{a_1}{a_2}=\frac{1}{3}, \quad \frac{b_1}{b_2}=\frac{1}{3}\) and \(\frac{c_1}{c_2}=\frac{4}{5}\) satisfying \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).

3. For coincident lines, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). The given equation is 2x + 3y – 8 = 0.
We can give another equation as 10x + 15y – 40 = 0. Here, \(\frac{a_1}{a_2}=\frac{1}{5}\), \(\frac{b_1}{b_2}=\frac{1}{5}\) and \(\frac{c_1}{c_2}=\frac{1}{5}\) satisfying \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Question 7.
Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:
x – y + 1 = 0 gives y = x + 1

x	-1	2
y	0	3

3x + 2y – 12 = 0 gives y = \(\frac{12-3 x}{2}\)

x	0	4
y	6	0

The vertices of the triangle formed by the given lines and the x-axis are (-1, 0), (4, 0) and (2, 3).

Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1

Question 1.
Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting ?) Represent this situation algebraically and graphically.
Solution:
Let the present age of Aftab be x years and the present age of his daughter be y years. Then, seven years ago, the age of Aftab was x – 7 years and the age of his daughter was y – 7 years.
So, from the given data.
x – 7 = 7(y – 7)
∴ x – 7 = 7y – 49
x – 7y + 42 = 0 …….. (1)
Similarly, three years from now, the age of Aftab will be x + 3 years and the age of his daughter will be y + 3 year.
So, according to the given data,
x + 3 = 3(y + 3)
∴ x + 3 = 3y + 9
∴ x – 3y – 6 = 0 ….. (2)
Thus, the equations x – 7y + 42 = 0 and x – 3y – 6 = 0 represent the given situation algebraically.
To represent the given situation graphically. we draw the graphs of both the equations.
x – 7y + 42 = 0
∴ y = \(\frac{42+x}{7}\)

x	0	35
y	6	11

x – 3y – 6 = 0
∴ y = \(\frac{x-6}{3}\)

x	0	30
y	-2	8

The above graph represents the situation graphically.
We observe that the lines intersect at point (42, 12).

Question 2.
The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹ 1300. Represent this situation algebraically and geometrically.
Solution:
Let the cost of one bat be ₹ x and the cost of one ball be ₹ y.
Then, the total cost of 3 bats is ₹ 3x and that of 6 balls is ₹ 6y. From the data, the total cost is ₹ 3900.
∴ 3x + 6y = 3900
∴ x + 2y = 1300
Similarly, the cost of 1 bat is ₹ x and the total cost of 3 balls is ₹ 3y. From the data, the total cost is ₹ 1300.
∴ x + 3y = 1300
Thus, the equations x + 2y = 1300 and x + 3y = 1300 represent the given situation algebraically.
To represent the given situation geometrically. we draw the graphs of both the equations.
x + 2y = 1300

x	100	1300
y	600	0

x + 3y = 1300

x	100	1300
y	400	0

The above graph represents the situation geometrically.
We observe that the lines intersect at point (1300, 0).

Question 3.
The cost of 2 kg of apples and 1 kg of grapes on a day was found to be After a month, the cost of 4 kg of apples and 2 kg of grapes is 300. Represent the situation algebraically and geometrically.
Solution:
Let the cost of 1 kg of apples be ₹ x and the cost of 1 kg of grapes be ₹ y.
Then, from the given data, 2x + y = 160 and 4x + 2y = 300.
Thus, the equations 2x + y = 160 and 4x + 2y = 300 represent the given situation algebraically.
To represent the given situation geometrically. we draw the graphs of both the equations.
2x + y = 160

x	0	80
y	160	0

4x + 2y = 300

x	0	75
y	150	0

The above graph represents the situation geometrically.
We observe that the lines are parallel.

Jharkhand Board JAC Class 10 Maths Solutions Chapter 2 Polynomials Ex 2.4 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 2 Polynomials Exercise 2.4

Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
1. 2x³ + x² – 5x + 2; \(\frac{1}{2}\), 1, -2
2. x³ – 4x² + 5x – 2; 2, 1, 1
Solution:
1. Let p(x) = 2x³ + x² – 5x + 2

Hence, \(\frac{1}{2}\) is a zero of
p(x) = 2x³ + x² – 5x + 2.
Again,
p(1) = 2(1)³ + (1)² – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
Hence, 1 is a zero of p(x) = 2x³ + x² – 5x + 2.
Again,
p(-2) = 2(-2)² + (-2)² – 5(-2)+2
= -16 + 4 + 10 + 2 = 0
Hence, -2 is a zero of
p(x) = 2x³ + x² – 5x + 2.
Now, for p(x) = 2x³ + x² – 5x + 2;
a = 2, b = 1, c = -5 and d = 2.
The zeroes of p (x) are α = \(\frac{1}{2}\), β = 1 and γ = -2.
Now, α + β + γ = \(\frac{1}{2}\) + 1 + (-2)
= \(-\frac{1}{2}=\frac{-(1)}{2}=\frac{-b}{a}\)
αβ + βγ + γα = (\(\frac{1}{2}\)) (1) + (1) (-2) + (-2) (\(\frac{1}{2}\))
= \(\frac{1}{2}\) – 2 – 1 = \(\frac{-5}{2}=\frac{c}{a}\) and
αβγ = (\(\frac{1}{2}\))(1)(-2) = -1 = \(\frac{-(2)}{(2)}=\frac{-d}{a}\)

2. Let p (x) = x³ – 4x² + 5x – 2.
Then, p (2) = (2)³ – 4(2)² + 5(2) – 2
= 8 – 16 + 10 – 2 = 0
Hence, 2 is a zero of p (x) = x³ – 4x² + 5x – 2.
Again,
p(1) = (1)³ – 4(1)² + 5(1) – 2
= 1 – 4 + 5 – 2 = 0
Hence, 1 is a repeated zero of p(x) = x³ – 4x² + 5x – 2.
Now, for p(x) = x³ – 4x² + 5x – 2, a = 1, b = -4, c = 5 and d = -2.
The zeroes of p (x) are α = 2, β = 1 and γ = 1.
Now,
α + β + γ = 2 + 1 + 1 = 4 = \(\frac{-(-4)}{1}=\frac{-b}{a}\)
αβ + βγ + γα = (2) (1) + (1) (1) + (1) (2)
= 2 + 1 + 2 = 5 = \(\frac{5}{1}=\frac{c}{a}\) and
αβγ = (2) (1) (1) = 2 = \(\frac{-(-2)}{1}=\frac{-d}{a}\)

Question 2.
Find a cubic polynomial with the sum of its zeroes, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, 7, 14 respectively.
Solution:
Let p (x) = ax³ + bx² + cx + d, a ≠ 0, be the required cubic polynomial and its zeroes be α, β and γ
Then, as given in the data,
α + β + γ = 2 ∴ \(\frac{-b}{a}\) = 2
αβ + βγ + γα = -7 ∴ \(\frac{c}{a}\) = -7
αβγ = \(\frac{-d}{a}\) = -14
So, if a = 1, then b = -2, c = -7 and d = 14.
Hence, the required polynomial is p(x) = x³ – 2x² – 7x + 14.

Question 3.
If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.
Solution:
For the given polynomial x³ – 3x² + x + 1.
A = 1, B = -3, C = 1 and D = 1.
Its zeroes are given to be a – b, a and a + b.
Now, sum of zeroes = (a – b) + a + (a + b) = 3a
From the polynomial,
Sum of zeroes = \(\frac{-B}{A}=\frac{-(-3)}{1}=3\)
Hence, 3a = 3 ∴ a = 1
Product of zeroes = (a – b) × a × (a + b)
= a(a² – b²)
From the polynomial.
Product of zeroes = \(\frac{-\mathrm{D}}{\mathrm{A}}=\frac{-1}{1}=-1\)
Hence, a (a² – b²) = -1
1(1² – b²) = -1 ∴ 1 – b² = -1
∴ 1 + 1 = b² ∴ b² = 2
∴ b = ±\(\sqrt{2}\)
Thus, a = 1 and b = ±\(\sqrt{2}\).

Question 4.
It two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2±\(\sqrt{3}\), find other zeroes.
Solution:
Let p (x) = x⁴ – 6x³ – 26x² + 138x – 35
2 + \(\sqrt{3}\) and 2 – \(\sqrt{3}\) are zeroes of p (x).
∴ (x – 2 – \(\sqrt{3}\))(x – 2 + \(\sqrt{3}\)) = (x – 2)² – (\(\sqrt{3}\))²
= x² – 4x + 4 – 3
= x² – 4x + 1
is a factor of p(x).

Then, Quotient = x² – 2x – 35
= x² – 7x + 5x – 35
= x(x – 7) + 5(x – 7)
=(x – 7)(x + 5)
x – 7 = 0 gives x = 7 and x + 5 = 0 gives x = -5.
Hence, the other zeroes of the given polynomial are 7 and -5.

Question 5.
If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Solution:

But, the remainder is given to be x + a.
Hence, comparing the coefficients of x and the constant term, we get
2k – 9 = 1 and k² – 8k + 10 = a
Now, 2k – 9 = 1
∴ 2k = 10
∴ k = 5
and a = k² – 8k + 10
∴ a = (5)² – 8(5) + 10
∴ a = 25 – 40 + 10
∴ a = -5
Thus, k = 5 and a = -5.