Class 10

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i)2x² – 3x + 5 = 0
(ii) 3x² – 4√3x + 4 = o
(iii) 2x² – 6x + 3 = 0
Solution:
(i) Given quadratic equation is, 2x² – 3x + 5 = 0
Compare it with ax² + bx + c = 0
a = 2, b = -3, c = 5
D = b² – 4ac
= (-3)² 4 × 2 × 5
= 9 – 40 = -31 < 0
Hence, given quadratic equation has no real roots.

(ii) Given quadratic equation is, 3x² – 4√3x + 4 = 0
Compare it with ax² + bx + c = 0
a = 3, b = -4√3, c = 4
D = b² – 4ac
= (-4√3)² – 4 × 3 × 4
= 48 – 48 = 0
given equation has real and equal roots.
Now, x = \(\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}\)
= \(\frac{-(-4 \sqrt{3}) \pm \sqrt{0}}{2 \times 3}\) = \(\frac{2}{\sqrt{3}}\)
Hence, roots of given quadratic equation are \(\frac{2}{\sqrt{3}}\) and \(\frac{2}{\sqrt{3}}\).

(iii) Given quadratic equation is :
2x² – 6x + 3 = 0
Compare it with ax² + bx + c = 0
∴ a = 2, b = -6, c = 3
D = b² – 4ac
= (-6)² 4 × 2 × 3
= 36 – 24 = 12 > 0
∴ given equation has real and distinct roots.
Now, x = \(\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}\)

= \(\frac{-(-6) \pm \sqrt{12}}{2 \times 2}\)

= \(\frac{6 \pm 2 \sqrt{3}}{4}\)

= \(\frac{3 \pm \sqrt{3}}{2}\)

= \(\frac{3+\sqrt{3}}{2}\) and \(\frac{3-\sqrt{3}}{2}\)
Hence, roots of given quadratic equation are \(\frac{3+\sqrt{3}}{2}\) and \(\frac{3+\sqrt{3}}{2}\).

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
(ii) kx(x – 2) + 6 = 0
Solution:
(i) Given quadratic equation is : 2x² + kx + 3 = 0
Compare it with ax² + bx + c = 0
∴ a = 2, b = k, c = 3
∵ roots of the given quadratic equation are equal.
∴ D = 0
b² – 4ac = 0
Or(k)² – 4 × 2 × 3 = 0
Or k² – 24 = 0
Or k² = 24
Or k = ±√24
Or k = ±2√6.

(ii) Given quadratic equation is:
kx (x – 2) + 6 = 0
Or k – 2kx + 6 = 0
Compare it with ax² + bx + c = 0
∴ a = k, b = -2k, c = 6
∵ roots of the given quadratic equation are equal
∴ b² – 4ac = 0
Or(-2k)² – 4 × k × 6 = 0
Or 4k² – 24k = 0
Or 4k[k – 6]= 0
Either 4k = 0 Or k- 6 = 0
k = 0 Or k = 6
∴ k = 0, 6.

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find
its length and breadth.
Solution:
Let breadth of rectangular grove = x m
and length of rectangular grove = 2x m
Area of rectangular grove = length × breadth
= [x × 2x] m² = 2 × 2 m²
According to question
2x² = 800
x² = \(\frac{800}{2}\) = 400
x = ± √400
x = ± 20.
∵ length of rectangle cannot be negative.
So, we reject x = -20
∴ x = 20
∴ breadth of rectangular grove = 20 m
and length of rectangular grove = (2 × 20) m = 40 m.

Question 4.
Is the following situation possible?If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let age of one friend = x years
and age of 2nd friend = (20 – x) years
Four years ago,
Age of 1st friend = (x – 4) years
Age of 2nd friend = (20 – x – 4) years = (16 – x) years
∴ Their product = (x – 4) (16 – x)
= 16x – x2 – 64 + 4x
= – x² + 20x – 64
According to Question
– x² + 20x – 64 = 48
Or – x² + 20x – 64 – 48 = 0
Or – x² + 20x – 112 = 0
Or x² – 20x + 112 = 0 …………….(1)
Compare it with ax² + bx + c = 0
∴ a = 1, b = -20, c = 112
D = b² – 4ac
= (-20)² – 4× 1 × 112
= 400 – 448 = -48 < 0
∴ roots are not real
then no real value of x satisfies the quadratic equation (1).
Hence, given situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2 ? If so, find its length and breadth.
Solution:
Let length of rectangular park = x m
Breadth of rectangular park = y m
∴ Perimeter of rectangular park = 2 (x + y) m
and area of rectangular park = xy m²
According to 1st condition
2 (x + y) = 80
x + y = \(\frac{80}{2}\) = 40
y = 40 – x …………(1)
According to 2nd condition,
xy = 400
x (40 – x) = 400 [using (1)]
Or 40x – x² = 400
Or 40x – x² – 400 = 0
Or x² – 40x + 400 = 0
Compare it with ax² +bx + c = 0
a = 1, b = -40, c = 400
D = b² – 4ac
= (-40)² – 4 × 1 × 400
= 1600 – 1600 = 0
Now, x = \(\frac{-b \pm \sqrt{\mathrm{D}}}{2 a}\)

= \(\frac{-(-40) \pm \sqrt{0}}{2 \times 1}\)

= \(\frac{40}{2}\) = 20
When x = 20 then from (1)
y = 40 – 20 = 20
∴ Length and breadth of rectangular park are equal of measure 20 m.
Hence, given rectangular park exist and it is a square.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3

Question 1.
Find the roots of the following quadratic equations if they exist, by the method of completing the square:
(i) 2x² + 7x + 3
(ii) 2x² + x – 4 = 0
(ili) 4x² + 4√3x + 3 = 0
(iv) 2x² + x + 4 = 0
Solution:
(i) Given quadratic equation is
2x² – 7x + 3 = 0
Or 2x² – 7x = -3
Or x² – \(\frac{7}{2}\)x = –\(\frac{3}{2}\)
Or x² – \(\frac{7}{2}\)x + (\(\frac{-7}{4}\))² = \(\frac{-3}{2}+\left(\frac{-7}{4}\right)^{2}\)

Or \(\left(x-\frac{7}{4}\right)^{2}=\frac{-3}{2}+\frac{49}{16}\)

Or \(\left(x-\frac{7}{4}\right)^{2}=\frac{-24+49}{16}\)

Or \(\left(x-\frac{7}{4}\right)^{2}=\frac{25}{16}\)

Or x – \(\frac{7}{4}\) = \(\pm \sqrt{\frac{25}{16}}=\pm \frac{5}{4}\)

Case I:
When x – \(\frac{7}{4}\) = \(\frac{5}{4}\)
Or x = \(\frac{5}{4}+\frac{7}{4}=\frac{5+7}{4}\)
Or x = \(\frac{12}{4}\) = 3

Case II:
When x – \(\frac{7}{4}\) = \(\frac{-5}{4}\)
Or x = \(\frac{-5}{4}+\frac{7}{4}=\frac{-5+7}{4}\)
Or x = \(\frac{2}{4}=\frac{1}{2}\)
Hence, roots of given quadratic equation is 3, \(\frac{1}{2}\).

(ii) Given Quadratic Equation is
2x² + x – 4 = 0
Or 2x² + x = 4
Or x² + \(\frac{1}{2}\)x = \(\frac{4}{2}\)

Case I:
When x + \(\frac{1}{4}\) = \(\frac{-\sqrt{33}}{4}\)
Or x = \(-\frac{\sqrt{33}}{4}-\frac{1}{4}\)
Or x = \(\frac{-\sqrt{33}-1}{4}\)
Hence, roots of given quadratic equation are \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\).

(iii) Given quadratic equation is
4x² + 4√3x + 3 = 0
Or 4x² + 4√3x = -3
Or x² + \(\frac{4 \sqrt{3}}{4}\)x = \(\frac{-3}{4}\)
Or x² + √3x = \(\frac{-3}{4}\)
Or x² + √3x + \(\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{-3}{4}+\left(\frac{\sqrt{3}}{2}\right)^{2}\)
Or \(\left(x+\frac{\sqrt{3}}{2}\right)^{2}=\frac{-3}{4}+\frac{3}{4}\)
or (x + \(\frac{\sqrt{3}}{2}\))² = 0
(x + \(\frac{\sqrt{3}}{2}\)) (x + \(\frac{\sqrt{3}}{2}\)) = 0
Either x + \(\frac{\sqrt{3}}{2}\) = 0
x = –\(\frac{\sqrt{3}}{2}\)
Or x + \(\frac{\sqrt{3}}{2}\) = 0
Or x = –\(\frac{\sqrt{3}}{2}\)
Hence, roots of given quadratic equation are –\(\frac{\sqrt{3}}{2}\) and –\(\frac{\sqrt{3}}{2}\).

(iv) Given quadratic equation is
2x² + x + 4 = 0
2x² + x = -4
x² + \(\frac{1}{2}\)x = \(-\frac{4}{2}\)
Or x² + \(\frac{1}{2}\)x + (\(\frac{1}{4}\))² = -2 + (\(\frac{1}{4}\))²

Or \(\left(x+\frac{1}{4}\right)^{2}=-2+\frac{1}{16}\)
Or \(\left(x+\frac{1}{4}\right)^{2}=\frac{-32+1}{16}\)
Or \(\left(x+\frac{1}{4}\right)^{2}=\frac{-31}{16}<0\)

∴ square of any number cannot be negative. So, (x + \(\frac{1}{4}\))² cannot be negative for any real x.
∴ There is no real x whith satisfied the given quadratic equation.
Hence, given quadratic equation has no real roots.

Question 2.
Find the roots of the quadratic equations given in Q. 1 by applying the quadratic formula. Which of the above two
methods do you prefer, and why?
Solution:
(i) Given quadratic equation is
2x² – 7x + 3 = 0
Compare it with ax² + bx + c = 0
a = 2, b = -7, c = 3
Now, b² – 4ac = (-7)² 4 x 2 x 3
= 49 – 24
= 25 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-7) \pm \sqrt{25}}{2 \times 2}=\frac{7 \pm 5}{4}\)
= \(\frac{7+5}{4} \text { and } \frac{7-5}{4}\)
= \(\frac{12}{4} \text { and } \frac{2}{4}\)
= 3 and \(\frac{1}{2}\)
Hence, 3 and \(\frac{1}{2}\) are the roots of given quadratic equation.

(ii) Given quadratic equation is
2x² + x – 4 = 0
Compare it with ax² + bx + c = 0
∴ a = 2, b = 1, c = -4
Now,
b² – 4ac = (1)² – 4 × 2 × 4
= 1 + 32 = 33 > 0
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-1 \pm \sqrt{33}}{2 \times 2}=\frac{-1 \pm \sqrt{33}}{4}\)
= \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\)
Hence, \(\frac{-1+\sqrt{33}}{4}\) and \(\frac{-1-\sqrt{33}}{4}\) are the roots of given quadratic equation.

(iii) Given quadratic equation is
4x² + 4√3x + 3 = 0
Compare it with ax² + bx + c = 0
a = 4, b = 4√3, c = 3
b² – 4ac = (4√3)² – 4 × 4 × (3)
= 48 – 48 = 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-4 \sqrt{3} \pm \sqrt{0}}{2 \times 4}\)

= \(-\frac{4 \sqrt{3}}{8}\), \(-\frac{4 \sqrt{3}}{8}\)

= –\(\frac{\sqrt{3}}{2}\), –\(\frac{\sqrt{3}}{2}\)

Hence, –\(\frac{\sqrt{3}}{2}\), –\(\frac{\sqrt{3}}{2}\) are the roots of given quadratic equation.

(iv) Given quadratic equation is 2x² + x + 4 = 0
Compare it with ax² + bx + c = 0
∴ a = 2, b = 1, c = 4
Now, b² – 4ac = (1)² – 4 × 2 × 4
= 1 – 32 = -31 < 0
But
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
Since the square of a real number cannot be negative, therefore x will not have any real value.
Hence, there are no real roots for the given quadratic equation.

From above two questions, we used two methods to find the roots of the quadratic equations. It is very clear from above discussion that quadratic formula method is very convenient as compared to method of completing the square.

Question 3.
Find the roots of the following equations:

(i) x – \(\frac{1}{x}\) = 3, x ≠ 0
(ii) \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}\), x ≠ -4, 7
Solution:
(i) Given Equation is
x – \(\frac{1}{x}\) = 3
Or \(\frac{x^{2}-1}{x}\) = 3
Or x² – 1 = 3x
Or x² – 3x – 1 = 0
Compare it with ax² + bx + c = 0
∴ a = 1, b = -3, c = -1
Now, b² – 4ac = (-3)² – 4 . 1 . (-1)
= 9 + 4 = 13 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-3) \pm \sqrt{13}}{2 \times 1}\)
= \(\frac{3 \pm \sqrt{13}}{2 \times 1}\)
= \(\frac{3+\sqrt{13}}{2}\) and \(\frac{3-\sqrt{13}}{2}\)
Hence, \(\frac{3+\sqrt{13}}{2}\) and \(\frac{3-\sqrt{13}}{2}\) are the roots of given quadratic equation.

(ii) Given equation is

-11 × 30 = 11 (x2 – 3x – 28)
Or -30 = x² – 3x – 28
Or x² – 3x – 28 + 30 = 0
Or x² – 3x + 2 = 0
Compare it with ax² + bx + c = 0
a = 1, b = – 3, c = 2
Now, b² – 4ac = (-3)² – 4 × 1 ×2
= 9 – 8 = 1 > 0
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-3) \pm \sqrt{1}}{2 \times 1}=\frac{3 \pm 1}{2}\)
= \(\frac{3+1}{2}\) and \(\frac{3+1}{2}\)
= \(\frac{4}{2}\) and\(\frac{2}{2}\) and 1
Hence, 2 and 1 are the roots of given quadratic equation.

Question 4.
The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). F1nd his present age.
Solution:
Let Rehman’s present age = x years
3 years ago Rehman’s age (x – 3) years
5 years from now Rehman’s age =(x + 5) years
According to question,
\(\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\)

Or \(\frac{x+5+x-3}{(x-3)(x+5)}=\frac{1}{3}\)

Or \(\frac{2 x+2}{x^{2}+5 x-3 x-15}=\frac{1}{3}\)

Or \(\frac{2 x+2}{x^{2}+2 x-15}=\frac{1}{3}\)
Or 6x + 6 = x² + 2 -15
Or x² + 2x – 15 – 6x – 6 = 0
Or x² – 4x – 21 = 0, which is quadratic in x.
So compare it with ax² + bx + c =0
a = 1, b = -4, c = -21
Now, b² – 4ac = (- 4)² 4 × 1 × (-21)
= 16 + 84 = 100 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
x = \(\frac{-(-4) \pm \sqrt{100}}{2 \times 1}\)
= \(\frac{4 \pm 10}{2}\)
= \(\frac{4+10}{2}\) and \(\frac{4-10}{2}\)
\(\frac{14}{2}\) and \(\frac{-6}{2}\)
= 7 and -3
∵ age cannot be negative,
so, we reject x = – 3
∴ x = 7
Hence, Rehman’s present age = 7 years.

Question 5.
In a class test, the sum of Shefall’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less ¡n English, the product of their marks would have been 210. Find her marks In the two subjects.
Solution:
Let Shefali get marks in Mathematics = x
Shefali’s marks in English = 30 – x
According to 1st condition,
Shefali’s marks in Mathematics = x + 2
and Shefali’s marks in English = 30 – x – 3 = 27 – x
∴ Their product = (x + 2) (27 – x)
= 27x – x² + 54 – 2x
= x² + 25x + 54
According to 2nd condition,
-x²+ 25x+ 54 = 210
Or -x² + 25x + 54 – 210 = 0
Or -x² + 25x – 156 = 0
Or x² – 25x+ 156 = o
Compare it with ax² + bx + c = O
a = 1, b = -25, c = 156
Now, b² – 4ac = (-25)² – 4 × 1 × 156
= 625 – 624 = 1 > 0
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-25) \pm \sqrt{1}}{2 \times 1}\)
= \(\frac{25 \pm 1}{2}\)
= \(\frac{25+1}{2}\) and \(\frac{25-1}{2}\)
= \(\frac{26}{2}\) and \(\frac{24}{2}\)
= 13 and 12.

Case I:
When x = 13
then Shefaiis marks in Maths = 13
Shefali’s marks in English = 30 – 13 = 17.

Case II:
When x = 12
then Shefalis marks in Maths = 12
Shefali’s marks in English = 30 – 12 =18.
Hence, Shefalis marks in two subjects are 13 and 17 Or 12 and 18.

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 shorter side of rectangular field = AD = x m

Longer side of rectangular field = AB = (x + 30) m
and diagonal of rectangular field = DB = (x + 60) m
In rectangle. the angle between the length and breadth is right angle.
∴ ∠DAB = 90°
Now, in right angled triangle DAB, using Pythagoras Theorem,
(DB)² = (AD)² + (AB)²
(x + 60)² = (x)² + (x + 30)²
Or x² + 3600 + 120x = x² + x² + 900 + 60x
Or x² + 3600 + 120x – 2x² – 900 – 60x = 0
Or -x² + 60x + 2700 = 0
Or x² – 60x – 2700 = 0
Compare it with ax² + bx + e = O
∴ a = 1, b = -60, c = -2700
and b² – 4ac = (-60)² – 4. 1 . (-2700)
= 3600 + 10800 = 14400 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-60) \pm \sqrt{14400}}{2 \times 1}\)

= \(\frac{60 \pm 120}{2}\)

= \(\frac{60+120}{2}\) and \(\frac{60-120}{2}\)
= \(\frac{180}{2}\) and \(\frac{-60}{2}\)
= 90 and – 30
∴ length of any side cannot be negative
So, we reject x = -30
∴ x = 90
Hence, shorter side of rectangular field = 90 m
Longer side of rectangular field = (90 + 30) m = 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 larger number = x .
Smaller number = y
According to 1st condition,
x² – y² = 180 ……………(1)
According to 2nd condition,
y² = 8x
From (1) and (2), we get
x² – 8x = 180
Or x² – 8x – 180 = 0
Compare it with ax² + bx + c = 0
∴ a = -1, b = -8, c = -180
and b² – 4ac = (-8)² – 4 × 1 × (-180)
= 64 + 720 = 784 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-(-8) \pm \sqrt{784}}{2 \times 1}\)
= \(\frac{8 \pm 28}{2}\)
= \(\frac{8+28}{2}\) and \(\frac{8-28}{2}\)
= \(\frac{36}{2}\) and \(\frac{-20}{2}\)
= 18 and -10
When x = – 10 then from (2),
y² = 8 (- 10) = – 80, which is impossible.
So, we reject x = – 10
When x = 18 then from (2).
y² = 8(18) = 144
Or y = ±√144
Or y = ± 12
Hence, required numbers are 18 and 12 Or 18 and -12.

Question 8.
A train travels 360 km ¡t 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 constant speed of the train = x km/hour
Distance covered by the train = 360 km
Time taken by the train = \(=\frac{\text { distance }}{\text { speed }}\)
(∵ speed = \(\frac{\text { Distance }}{\text { Time }}\))
= \(\frac{360}{x}\)
Increased speed of the train = (x + 5) km/hour
∴ Time taken by the train with increased speed = \(\frac{360}{x+5}\) hour
According to question

Or 1800 = x² + 5x
Or x² + 5x – 1800 = 0
Compare it with, ax² + bx + c = 0
a = 1, b = 5, c = – 1800
and b² – 4ac = (5)² 4 × 1 × (- 1800)
= 25 + 7200 = 7225 > 0

x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-5 \pm \sqrt{7225}}{2 \times 1}\)
= \(\frac{-5 \pm 85}{2}\)
= \(\frac{-5+85}{2}\) and \(\frac{-5-85}{2}\)
= \(\frac{80}{2}\) and \(\frac{-90}{2}\)
= 40 and – 45
∵ speed of any train cannot be negative.
So, we reject x = – 45
x = 40
Hence, speed of train = 40 km/hour.

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 time taken by larger tap to fill the tank = x hours.
Time taken by smaller tap to fill the tank = (x + 10) hours
In case of one hour:
Larger tap can fill the tank = \(\frac{1}{x}\)
Smaller tap can fill the tank = \(\frac{1}{x+10}\)
∴ Larger and smaller tap fill the tank = \(\frac{1}{x}\) + \(\frac{1}{x+10}\) ………….(1)
But, two taps together can fill the tank = 9\(\frac{3}{8}\)hour = \(\frac{75}{8}\) hour
Now, two taps together can fill the tank in one hour = \(\frac{8}{75}\) ……………..(2)
From (1) and (2), we get
\(\frac{1}{x}+\frac{1}{x+10}=\frac{8}{75} \)

Or \(\frac{x+10+x}{x(x+10)}=\frac{8}{75}\)

Or \(\frac{2 x+10}{x^{2}+10 x}=\frac{8}{75}\)

Or 75(2x + 10) = 8(x² + 10x)
Or 150x + 750 = 8x² + 80x
Or 8x² + 80x – 150x – 750 = 0
Or 8x² – 70x – 750 = 0
Or 4x² – 35x – 375 = 0
Compare it with ax² + bx + c = 0
∴ a = 4, b = -35, c = -375
and b² – 4ac = (35)² -4 × 4 × (-375)
= 1225 + 6000 = 7225 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-(-35) \pm \sqrt{7225}}{2 \times 4}\)

= \(\frac{35 \pm 85}{8}\)

= \(\frac{35+85}{8}\) and \(\frac{35-85}{8}\)

= \(\frac{120}{8}\) and \(\frac{-50}{8}\)

= 15 and \(\frac{-25}{4}\)

∵ time cannot be negative.
So,we reject x = \(\frac{-25}{4}\)
∴ x = 15
Hence, larger water tap fills the tank = 15 hours
and smaller water tap fills the tank = (15 + 10) hours = 25 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 to intermediate stations), if the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of the two trains.
Solution:
Let average speed of passenger train = x km/hour
Average speed of express train = (x+ 11) km/hour
Distance between Mysore and Bangalore = 132 km
Time taken by passenger train = \(\frac{132}{x}\) hour
[∵ Speed = \(=\frac{\text { Distance }}{\text { Time }}\) ]
Time taken by express train‚ = \(\frac{132}{x+11}\) hour
According to question,

Or 1452 = x² + 11x
Or x² + 11x – 1452 = 0
Compare it with ax² + bx + c = 0
∴ a = 1, b = 11, c = -1452
and b² – 4ac = (11)² – 4 × 1 × (- 1452)
= 121 + 5808 = 5929 > 0
∴ x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-11 \pm \sqrt{5929}}{2 \times 1}\)

= \(\frac{-11 \pm 77}{2}\)

= \(\frac{-11+77}{2}\) and \(\frac{-11-77}{2}\)

= \(\frac{66}{2}\) and \(\frac{-88}{2}\) = 33 and -44

∵ speed of any train cannot be negative
∴ x = 33
Hence, speed of passenger train = 33 km/hour
and speed of express train = (33 + 11) km/hour = 44 km/hour.

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:
In case of larger square
Let length of each side of square = x m
Area of square = x² m²
Perimeter of square = 4x m

In case of smaller square:

Let lenth of each side of square = y m
Area of square = y² m²
Perimeter of square = 4y m
According to 1st condition,
x² + y² = 468 …………….(1)
According to 2nd condition,
4x – 4y = 24
Or 4(x – y) = 24
Or x – y = 6
x = 6 + y
From (1) and (2), we get
(6 + y)² + y² = 468
Or 36 + y² + 12y + y² = 468
Or 2y² + 12y + 36 – 468 = 0
Or 2y² + 12y – 432 = 0
Or y² + 6y – 216 = 0
Compare it with ay² + by + c = 0
∴ a = 1, b = 6, c = -216
and b² – 4ac = (6)² – 4 × 1 × (- 216) = 36 + 864 = 900 > 0
∴ y = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

= \(\frac{-6 \pm \sqrt{900}}{2 \times 1}\)

= \(\frac{-6 \pm 30}{2}\)

= \(\frac{-6+30}{2}\) and \(\frac{-6-30}{2}\)

= \(\frac{24}{2}\) and \(\frac{-36}{2}\) = 12 and -18

∵ length of square cannot be negative
So, we reject y = – 18
∴ y = 12
From (2), x = 6 + 12 = 18
Hence, sides of two squares are 12 m and 18 m.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.2

Question 1.
Find the roots of the following quadratic equations by factorisation:
(i) x² – 3x – 10 = 0
(ii) 2x² + x – 6 = 0
(iii) √2x² + 7x + 5√2 = 0
(iv) 2 x² – x + \(\frac{1}{8}\) = 0
(v) 100x² – 20x + 1 = 0
Solution:
(i) Given quadratic
x² – 3x – 10 = 0
Or x² – 5x + 2x – 10 = 0
S = -3, p = -10
Or x (x – 5) + 2 (x – 5) = 0
Or (x – 5) (x + 2) = 0
Either x – 5 = 0 Or x + 2 = 0
x = 5 Or x = -2
Hence, 5 and -2 are roots of given Quadratic Equation.

(ii) Given quadratic equation
2x² + x – 6 = 0 =1
0r 2x² + 4x – 3x – 6 = 0
S = 1 P = -6 × 2 = -12
Or 2x (x + 2) -3 (x + 2) = 0
Or (x + 2) (2x – 3) = 0
Either x + 2 = 0 Or 2x – 3 = 0
x = -2 Or x = –\(\frac{3}{2}\)
Hence, – 2 and \(\frac{3}{2}\) are roots of given quadratic equation.

(iii) Given Quadratic Equation,
√2x² + 7x + 5√2 = 0
Or √2x² + 2x + 5x + 5√2 = 0
S = 7, P = √2 × 5√2 = 10
Or √2x (x + √2) + 5 (x + √2) = 0
Or (x + √2) (√2x + 5) = 0
Either x + √2 = 0 Or √2x + 5 = 0
x = -√2 Or x = –\(\frac{-5}{\sqrt{2}}\)
Hence, -√2 and \(\frac{-5}{\sqrt{2}}\) are roots of given quadratic equation.

(iv) Given quadratic equation
2x² – x + \(\frac{1}{8}\) = 0
Or \(\frac{16 x^{2}-8 x+1}{8}\) = 0
Or 16x² – 8x + 1 = 0
S = -8, P = 16 × 1 = 16
Or 16x² – 8x + 1 = 0
Or 16x² – 4x – 4x + 1 = 0
Or 4x(4x – 1) -1(4x – 1) = 0
Or (4x – 1) (4x – 1) = 0
Either 4x – 1 = 0
Or 4x – 1 = 0
x = \(\frac{1}{4}\) Or x = \(\frac{1}{4}\)
Hence, \(\frac{1}{4}\) and \(\frac{1}{4}\) are roots of given quadratic equation.

(v) Given quadratic equation,
100x² – 20x + 1 = 0
Or 100x² – 10x – 10x + 1 = 0
S = -20, P = 100 × 1 = 100
Or 10x(10x – 1) – 1 (10x – 1) = 0
Or (10x – 1)(10x – 1) = 0
Either 10x – 1 = 0 Or 10x – 1 = 0
x = \(\frac{1}{10}\) Or x = \(\frac{1}{10}\)
Hence, \(\frac{1}{10}\) and \(\frac{1}{10}\) are roots of given quadratic equation.

Question 2.
Solve the problems given in Example I. Statements of these problems are given below:
(i) John and Jivanti together have 45 marbles. Both of them lost S marbles each, and the product of the number of marbles they now have is 124. We would lfke to find out how many marbles they had to start with.

(ii) 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. We would like to find out the number of toys produced on that day.

Solution:
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x
The number of marbles Íeft withJohn, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5 = 40 – x
Therefore, their product = (x – 5) (40 – x)
= 40x – x² – 200 + 5x
= -x² + 45x – 200
According to question,
-x² + 45x – 200 = 124
Or -x² + 45x – 324 = 0
Or x² – 45x + 324 =0
Or x² – 36x – 9x + 324 = 0
S = -45, P = 324
Or x(x – 36) – 9(x – 36) = 0
Or (x – 36)(x – 9) = 0
Either x – 36 = 0, Or x – 9 = 0
x = 36 Or x = 9
∴ x = 36, 9
Hence, number of marbles they had to start with were 36 and 9 or 9 and 36.

(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x)
According to question.
x(55 – x) = 750
Or 55x – x² = 750
Or -x² + 55x – 750 = 0
Or x² – 55x – 750 = 0
Or x² – 30x – 25x + 750=0
S = -33, P = 750
Or x(x – 30) – 25(x – 30) = 0
Or (x – 30)(x – 25) = 0
Either x – 30 = 0 Or x – 25 = 0
x = 30 Or x = 25
∴ x = 30, 25
Hence, number of toys produced on that day were 30 and 25 or 25 and 30.

Question 3.
Find two numbers whose sum is 27 and product is 182.
Solution:
Let one number = x
2nd number = 27 – x
Their product = x (27 – x) = 27x – x²
According to question,
27x – x² = 182
Or – x² + 27x – 182 = 0
Or x² – 27x + 182 = 0
S = -27, P = 182
Or x² – 13x – 14x + 182 = 0
Or x(x – 13) – 14(x – 13) = 0
Or (x – 13) (x – 14) = 0
Either x – 13 = 0 Or x – 14 = 0
x = 13 Or x = 14
x = 13, 14
Hence, two numbers are 13 and 14 Or 14 and 13.

Question 4.
Find two consecutive positive integers, sum of whose squares is 365.
Solution:
Let one positive integer = x
2nd positive integer = x + 1
According to question,
(x)² + (x + 1)² = 365
Or x² + x² + 1 + 2x = 365
Or 2x² + 2x + 365 = 0
Or 2x² + 2x – 364 = 0
Or x² + x – 182 = 0
Or x² + 14x – 13x – 182 = 0
S = 1, P = -182
Or x(x + 14) – 13(x + 14) = 0
(x + 14)(x— 13) = O
Either x + 14 = 0
Or x = -14
Or
x – 13 = 0
x = 13
∵ We have positive integers.
So, we reject x = – 14.
∴ x = 13
∴ One positive integer = 13
and 2nd positive integer = 13 + 1 = 14
Hence, required consecutive positive integers are 13 and 14.

Question 5.
The altitude of a right triangle is 7 cm less than its base. 1f the hypotenuse is 13 cm, find the other two sides.
Solution:
Let base of right triangle = x cm
Altitude of right triangle = (x – 7) cm
and hypotenuse of right triangle = 13 cm (Given)
According to Pythagoras Theorem,
(Base)² + (Altitude)² = (Hypotenuse)²
(x)² ÷ (x – 7)² = (13)²
Or x² + x² + 49 – 14x = 169
Or 2x² – 14x + 49 – 169 = 0
Or 2x² – 14x – 120 = 0
Or 2[x² – 7x – 60] = 0
Or x² – 7x – 60 = 0
Or x² – 12x + 5x – 60 = 0
S = – 7 P = – 60
Or x(x – 12) + 5(x – 12) = 0
Or (x – 12) (x + 5) = 0
Either x – 12 = 0 Or x + 5 = 0
x = 12 Or x= – 5
∵ Length of any triangle cannot be negative.
So, we reject x = – 5
∴ x = 12
Hence, base of right triangle = 12 cm
Altitude of right triangle = (12 – 7) cm = 5 cm.

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, number of pottery articles produced by industry in one day = x
Cost of production of each article = ₹ (2x + 3)
∴ Total cost of production in panicular day = ₹ [x(2x + 3)] = ₹ (2x² + 3x)
According to question,
2x² + 3x = 90
2x² + 3x – 90 = 0
S = 3, P = 2 × -90 = -180
Or 2x² – 12x + 15x – 90 = 0
Or 2x (x – 6) + 15 (x – 6) = 0
Or (x – 6) (2x + 15) = 0
Either x – 6 = 0 Or 2x + 15 = 0
x = 6 Or x = \(\frac{-15}{2}\)
∵ number of articles cannot be negative
So, we reject x = 2
∴ x = 6
Hence, number of articles produced on certain day = 6
and cost of production of each article = ₹ [2 × 6 + 3] = ₹ 15.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1

Question 1.
Check whether the following are quadratic equations:
(i) (x + 1)² = 2(x – 3)
(ii) x² – 2x = (-2) (3 – x)
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
(iv) (x – 3)(2x + 1) = x (x + 5)
(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
(vi) x² + 3x + 1 = (x – 2)
(vii) (x + 2)³ = 2x(x² – 1)
(viii) x³ – 4x² – x + 1 = (x – 2)³

Solution:
(i) Given that
(x + 1)² = 2(x – 3)
Or x² + 1 + 2x = 2x – 6
Or x² + 1 + 2x – 2x + 6 = 0
Or x² + 7 = 0
Or x² + 0x + 7 = 0
which is in the formof ax² + bx + c = 0;
∴ It is a quadratic equation.

(ii) Given that
x² – 2x = (-2) (3 – x)
Or x² – 2x = -6 + 2x
Or x² – 2x + 6 – 2x = 0
Or x² – 4x + 6 = 0
which is the form of ax² + bx + c = 0; a ≠ 0
∴ It is the quadratic equation.

(iii) Given that ,
(x – 2) (x + 1) = (x – 1) (x + 3)
Or x² + x – 2x – 2 = x² + 3x – x – 3
Or x² – x – 2 = x² + 2x – 3
Or x² – x – 2 – x² -2x + 3 = 0
Or -3x + 1 = 0 which have no term of x².
So it is not a quadratic equation.

(iv) Given that
(x – 3)(2x + 1) = x(x + 5)
Or 2x² + x – 6x – 3 = x² + 5x
Or 2x² – 5x – 3 – x² – 5x = 0
Or x² – 10x – 3 = 0
which is a form of ax² + bx + c = 0; a ≠ 0
∴ It is a quadratic equation.

(v) Given that ,
(2x – 1) (x – 3) = (x + 5) (x – 1)
0r2x² – 6x – x + 3 = x² – x + 5x – 5
Or 2x² – 7x + 3 = x² + 4x – 5
Or 2x² – 7x + 3 – x² – 4x + 5 = 0
Or x² – 11x + 8 = 0
which is a form of ax² + bx + c = 0; a ≠ 0
∴ It is a quadratic equation.

(vi) Given that
x²+3x+1 = (x – 2)²
Or x² + 3x + 1 = x² + 4 – 4x
Or x² + 3x + 1 – x² – 4 + 4x = 0
Or 7x – 3 = 0
which have no term of x².
So it is not a quadratic equation.

(vii) Given that
(x + 2)³ = 2x(x² – 1)
Or x³ + (2)³ + 3 (x)² 2 + 3(x)(2)² = 2x³ – 2x
Or x³ + 8 + 6x² + 12x = 2x³ – 2x
Or x³ + 8 + 6x² + 12x – 2x³ + 2x = 0
Or -x³ + 6x² + 14x + 8 = 0
Here the highest degree of x is 3. which is a cubic equation.
∴ It is not a quadratic equation.

(viii) Given that
x³ – 4x² – x+ 1= (x – 2)³
Or x³ – 4x² – x + 1 = x³ – (2)³ + 3(x)² (-2) + 3 (x) (-2)²
Or x³ – 4x² – x + 1 = x³ – 8 – 6x² + 12x
Or x³ – 4x² – x + 1 – x³ + 8 + 6x² – 12x = 0
Or 2x² – 13x + 9 = 0
which is in the form of ax² + bx +c = 0; a ≠ 0
∴ It is a quadratic equation.

Question 2.
Represent the following situations in the form of quadratic equations:
(i) 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.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii) 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.

(iv) 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:
(i) Let Breadth of rectangular plot = x m
Length of rectangular plot= (2x + 1) m
∴ Area of rectangular plot = [x (2x + 1)] m² = (2x² + x) m²
According to question,
2x² + x = 528
S = 1
P = -528 × 2 = -1056
0r 2x² + x – 528 = 0
Or 2x² – 32x + 33x – 528 = 0
Or 2x(x – 16) + 33(x – 16) = 0
Or (x – 16) (2x + 33) = 0
Either x – 16 = 0 Or 2x + 33 = 0
x = 16 Or x = 2
∵ breadth of any rectangle cannot be negative, so we reject x = \(\frac{-33}{2}\), x = 16
Hence, breadth of rectangular plot = 16 m
Length of rectangular plot = (2 ×16 + 1)m = 33m
and given problem in the form of Quadratic Equation are 2x² + x – 528 = 0.

(ii) Let two consecutive positive integers are x and x + 1.
Product of Integers = x (x + 1) = x² + x
According to question,
Or x² + x – 306 = 0
S = 1, P = – 306
Or x² + 18x – 17x – 306 = 0
Or x(x + 18) -17 (x + 18) = 0
Or (x + 18) (x – 17) = 0
Either x + 18 = 0 Or x – 17 = 0
x = -18 Or x = 17
∵ We are to study about the positive integers, so we reject x = – 18.
x = 17
Hence, two consecutive positive integers are 17, 17 + 1 = 18
and given problem in the form of Quadratic Equation is x² + x – 306 = 0.

(iii) Let present age of Rohan = x years
Rohan’s mother’s age = (x + 26) years
After 3 years, Rohan’s age = (x + 3) years
Rohan’s mother’s age = (x + 26 + 3) years = (x + 29) years
∴ Their product = (x + 3) (x + 29)
= x² + 29x + 3x + 87
= x² + 32x + 87
According to question,
x² + 32x + 87 = 360
Or x² + 32x + 87 – 360 = 0
Or x² + 32x – 273 = 0
Or x² + 39x – 7x – 273 = 0
S = 32, P = – 273
Or x(x + 39) – 7(x + 39) = 0
Or (x + 39) (x – 7) =
Either x + 39 = Or x – 7 = 0
x = -39 Or x = 7
∵ age of any person cannot be negative so, we reject x = -39
∴ x = 7
Hence, Rohans present age = 7 years
and given problem in the form of Quadratic Equation is x² + 32x – 273 = 0.

(iv) Let u km/hour be the speed of train.
Distance covered by train = 480 km
Time taken by train = \(\frac{480}{u}\) hour
[ Using, Speed = \(\frac{\text { Distance }}{\text { Time }}\)
or Time = \(=\frac{\text { Distance }}{\text { Speed }}\) ]

If speed of train be decreased 8km/hr.
∴ New speed of train = (u – 8) km/hr.
and time taken by train = \(\frac{480}{u-8}\) hour
According to question.

or 3840 = 3 (u² – 8u)
or u² – 8u = 1280
or u² – 8u – 1280=0
or u² – 40u + 32u – 1280 = 0
S = -8, P = – 1280
or u(u – 40) + 32 (u – 40) = 0
or (u – 40)(u + 32) = 0
Either u – 40 = 0
or u + 32 = 0
u = 40 or u = -32
But, speed cannot be negative so we reject
u = – 32
∴ u = 40.
Hence speed of train is 40 km/hr Ans.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6

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

(i) \(\frac{1}{2 x}+\frac{1}{3 y}\) = 2
\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)

(ii) \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}\) = 2
\(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}\) = -1

(iii) \(\frac{4}{x}\) + 3y = 14
\(\frac{3}{x}\) – 4y = 23

(iv) \(\frac{5}{x-1}+\frac{1}{y-2}\) = 2
\(\frac{6}{x-1}-\frac{3}{y-2}\) = 1

(v) \(\frac{7 x-2 y}{x y}\) = 5
\(\frac{8 x+7 y}{x y}\) = 15

(vi) 6x + 3y = 6xy
2x + 4y = 5xy

(vii) \(\frac{10}{x+y}+\frac{2}{x-y}\) = 4
\(\frac{15}{x+y}-\frac{5}{x-y}\) = -2

(viii) \(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\)
\(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}\)

Solution:
(i) Given pair of linear equations are;
\(\frac{1}{2 x}+\frac{1}{3 y}\) = 2
\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)
putting \(\frac{1}{x}\) = u and \(\frac{1}{x}\) = v, then equations reduces to

Substitute this value of v in (1), we get:
3u + 2 (3) = 12
or 3u + 6 = 12
or 3u = 12 – 6 = 6
or u = \(\frac{6}{3}\) = 2
But \(\frac{1}{x}\) = u
or x = \(\frac{1}{u}\)
∴ x = \(\frac{1}{2}\)

and \(\frac{1}{y}\) = v
y = \(\frac{1}{v}\)
or y=—\(\frac{1}{3}\)
Hence x = \(\frac{1}{2}\) and y = \(\frac{1}{3}\)

(ii) Given pair of linear equation are
\(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}\) = 2
\(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}\) = -1
Putting \(\frac{1}{\sqrt{x}}\) = u and \(\frac{1}{\sqrt{y}}\) = v then equations reduces to
2u + 3v = 2 ……………(1)
and 4u – 9v = -1 ……………(2)
Multiplying (1) by 2, we get:
4u + 6v = 4
Now, (2) – (3) gives

Substitute this value of v in (1), we get:
2u + 3(\(\frac{1}{3}\)) = 2
or 2u + 1 = 2
or 2u = 2 – 1 = 1
or u = \(\frac{1}{2}\)

But \(\frac{1}{\sqrt{x}}\) = u
or (\(\frac{1}{\sqrt{x}}\))² = u²
or \(\frac{1}{x}\) = u²
or \(\frac{1}{x}\) = (\(\frac{1}{2}\))²
or x = 4

and \(\frac{1}{\sqrt{y}}\) = v
or (\(\frac{1}{\sqrt{y}}\))² = v²
or \(\frac{1}{y}\) = v²
or \(\frac{1}{y}\) = (\(\frac{1}{3}\))²
or y = 9
Hence x = 4 and y = 9.

(iii) Given pair of linear equations are
\(\frac{4}{x}\) + 3y = 14 and \(\frac{3}{x}\) – 4y = 23
putting \(\frac{1}{y}\) = v then equations reduces to
4v + 3y = 14 ………..(1)
and 3v – 4y = 23 ……………(2)
Multiplying (1) by 3 and (2) by 4, we get:
12v + 9y = 42 ………..(3)
and 12v – 16y = 92 ……………..(4)
Now, (4) – (3) gives

Substitute this value of y in (1), we get:
4v + 3(-2) = 14
or 4v – 6 = 14
or 4v = 14 + 6 = 20
or v = \(\frac{20}{4}\) = 5
But \(\frac{1}{x}\) = v
x = \(\frac{1}{v}\) = \(\frac{1}{5}\)
Hence x = \(\frac{1}{5}\) and y = -2.

(iv) Given pair of linear equation are
\(\frac{5}{x-1}+\frac{1}{y-2}\) = 2 and \(\frac{6}{x-1}-\frac{3}{y-2}\) = 12
Putting \(\frac{1}{x-1}\) = u and \(\frac{1}{y-2}\) = v then equations reduces to
5u + v = 2 ……………(1)
and 6u – 3v = 1 ……………(2)
Multiplying (1) by 3, we get:
15u + 3v = 6 ………..(3)
Now, (3) + (2) gives

u = \(\frac{7}{21}=\frac{1}{3}\)
Substitute this value of u in (1), we get:
5 × \(\frac{1}{3}\) + v = 2
or v = 2 – \(\frac{5}{3}\) = \(\frac{6-5}{3}\)
or v = \(\frac{1}{3}\)

But \(\frac{1}{x-1}\) = u
or \(\frac{1}{x-1}\) = \(\frac{1}{3}\)
or x – 1 = 3
or x = 3 + 1
or x = 4

and \(\frac{1}{y-2}\) = v
or \(\frac{1}{y-2}\) = \(\frac{1}{3}\)
or y – 2 = 3
or y = 3 + 2
or y = 5
Hence x = 4 or y = 5.

(v) Given pair of linear equation are
\(\frac{7 x-2 y}{x y}\) = 5
or \(\frac{7 x}{x y}-\frac{2 y}{x y}\) = 5
or \(\frac{7}{y}-\frac{2}{x}\) = 5
or \(-\frac{2}{x}+\frac{7}{y}\) = 5

and \(\frac{8 x+7 y}{x y}\) = 15
or \(\frac{8 x}{x y}+\frac{7 y}{x y}\) = 15
or \(\frac{8}{y}+\frac{7}{x}\) = 15
or \(\frac{7}{x}+\frac{8}{y}\) = 15

Putting \(\frac{1}{x}\) = u and \(\frac{1}{y}\) = v, then equations reduces to
-2u + 7v = 5 ………….(1)
and 7u + 8v = 15 ………….(2)
Multiplying (1) by 7 and (2) by 2, we get:
-14v + 49u = 35
and 14v + 16u = 30
Now, (3) + (4) gives,

Substitute thuis value of u in (1), we get:
-2(1) + 7v = 5
or 7v = 5 + 2
or 7v = 7
or v = \(\frac{7}{7}\) = 1

But \(\frac{1}{x}\) = u
or x = \(\frac{1}{u}\)
or x = \(\frac{1}{1}\)
or x = 1

and \(\frac{1}{y}\) = v
or y = \(\frac{1}{v}\)
or y = \(\frac{1}{1}\)
or y = 1
Hence x = 1 and y = 1.

(vi) Given pair of linear equations are
6x + 3y = 6xy
or \(\frac{6 x+3 y}{x y}=\frac{6 x y}{x y}\)
or \(\frac{6}{y}+\frac{3}{x}=6\)
or \(3\left[\frac{1}{x}+\frac{2}{y}\right]=6\)
or \(\frac{1}{x}+\frac{2}{y}=2\)

and 2x + 4y = 5xy
or \(\frac{2 x+4 y}{x y}=\frac{5 x y}{x y}\)
or \(\frac{2}{y}+\frac{4}{x}=5\)
or \(\frac{4}{x}+\frac{2}{y}=5\)
Putting \(\frac{1}{x}\) = u and \(\frac{1}{y}\) = v, then equations reduces to
u + 2v = 2 ……………(1)
and 4u + 2v = 5 ……….(2)
Now, (2) – (1) gives

or u = \(\frac{3}{3}\) = 1
Substitute this value of u in (1), we get:
1 + 2v = 2
or 2v = 2 – 1
or v = \(\frac{1}{2}\)

But \(\frac{1}{x}\) = u
or \(\frac{1}{x}\) = 1
or x = 1

and \(\frac{1}{y}\) = v
or \(\frac{1}{y}\) = \(\frac{1}{2}\)
or y = 2
Hence x = 1 and y = 2.

(vii) Given pair of linear equations are
\(\frac{10}{x+y}+\frac{2}{x-y}\) = 4
\(\frac{15}{x+y}-\frac{5}{x-y}\) = -2
Putting \(\frac{1}{x+y}\) = u and \(\frac{1}{x-y}\) = v, then equations reduces to
10u + 2v = 4 or
5u + v = 2 …………(1)
15u – 5v = -2 ………….(2)
Multiplying (1) by 5, we get
25u + 5v = 10 ………….(3)
Now, (3) + (2) gives

Substitute this value of u in (1), we get:
5(\(\frac{1}{5}\)) + v = 2
or 1 + v = 2
or v = 1
But \(\frac{1}{x+y}\) = u
or \(\frac{1}{x+y}\) = \(\frac{1}{5}\)
or x + y = 5 ……….(4)
and \(\frac{1}{x-y}\) = v
or \(\frac{1}{x-y}\) = 1
or x – y = 1 ………..(5)
Now, (4) + (5) gives

Substitute this value of x in (4), we get:
3 + y = 5
y = 5 – 3 = 2
Hence x = 3 and y = 2.

(viii) Given pair of linear equations are
\(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\) and \(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}\)
Putting \(\frac{1}{3 x+y}\) = u and \(\frac{1}{3 x-y}\) = y, then Equations reduces to
u + v = \(\frac{3}{4}\)
or 4u + 4v = 3
or 4u + 4v = 3 ………….(1)

and \(\frac{u}{2}-\frac{v}{2}=\frac{-1}{8}\)
or u – v = \(\frac{-1}{4}\)
or 4u – 4v = -1 ………………(2)
Now, (1) + (2) gives

Substitute this value of u in (1), we get:
4(\(\frac{1}{4}\)) + 4v = 3
or 4v = 2
or v = \(\frac{2}{4}=\frac{1}{2}\)
But \(\frac{1}{3 x+y}\) = \(\frac{1}{4}\)
or 3x + y = 4 …………(3)
and \(\frac{1}{3 x-y}\) = \(\frac{1}{2}\)
or 3x – y = 2 …………….(4)
Now, (3) + (4) gives

x = 1
Substitute this value of x in (3), we get:
3(1) + y = 4
or 3 + y = 4
or y = 4 – 3 = 1
Hence, x = 1 and y = 1.

Question 2.
Formulate the following problems as a pair of equations, and hence find their solutions.
(i) 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.

(ii) 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.

(iii) Roohi travels 300 km to her home party 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:
(i) Let the speed of Ritu in still water = x km/hour
and the speed of current = y km/hour
∴ speed in upstream = (x – y) km/hour
and speed in downstream = (x + y) km/hour
Distance covered by Ritu in downstream in 2 hours = Speed × Time
= (x + y) × 2 km
According to 1st condition
2(x + y) = 20
x + y = 10 ……………..(1)
Distance covered by Rim in upstream in 2 hours
= Speed × Time
= 2(x – y)km
According to 2nd condition,
2(x – y) =4
x – y = 2 ……………….(2)
Now, (1) + (2) gives

Substitute this value of x in (1), we get:
6 + y = 10
y = 10 – 6 = 4
Hence, Ritu’s speed in still water = 6 km/hour
and speed of current = 4 km/hour.

(ii) Let one woman can fmish the work = x days
One man can finish the work = y days
then one woman’s one day’s work = \(\frac{1}{x}\)
One man’s one day’s work = \(\frac{1}{y}\)
According to 1st condition,
\(\frac{2}{x}+\frac{5}{y}=\frac{1}{4}\) ……………(1)
According to 2nd equation
\(\frac{3}{x}+\frac{6}{y}=\frac{1}{3}\) ……………(2)
Putting \(\frac{1}{x}\) = u and \(\frac{1}{y}\) = v, then equations (1) and (2) reduces to
2u + 5v = \(\frac{1}{4}\)
8u + 20v = 1 …………….(3)
and 3u + 6v = \(\frac{1}{3}\)
9u + 18v = 1 ……………..(4)
Multiplying (3) by 9 and (4) by 8, we get:
72u + 180v = 9 ……………..(5)
and 72u + 144v = 8 …………..(6)
Now, (5) – (6) gives

or 9u + \(\frac{1}{2}\) = 1
or 9u = 1 – \(\frac{1}{2}\) = \(\frac{2-1}{2}\)
or 9u = \(\frac{1}{2}\)
or u = \(\frac{1}{2 \times 9}=\frac{1}{18}\)
But \(\frac{1}{x}\) = u
or \(\frac{1}{x}\) = \(\frac{1}{18}\)
or x = 18
and \(\frac{1}{y}\) = v
\(\frac{1}{y}\) = \(\frac{1}{36}\)
or y = 36
Hence, one woman and one man alone can finish work in 18 days and 36 days respectively.

(iii) Let speed of train = x km/hour
and speed of bus = y km/hour
Total distance = 300 km
Case I:
Time taken by train to cover 60 km = \(\frac{\text { Distance }}{\text { Speed }}\)
= \(\frac{60}{x}\) hours
Time taken by bus to cover 240 km = (300 – 60) = \(\frac{240}{y}\) hours
Total time = \(\left(\frac{60}{x}+\frac{240}{y}\right)\) hours
According to 1st condition,
\(\frac{60}{x}+\frac{240}{y}\) = 4
\(\frac{15}{x}+\frac{60}{y}\) = 1 ……………….(1)

Case II:
Time taken by train to cover 100 km = \(\frac{100}{x}\) hours
Time taken by bus to cover 200 km = 300 – 100 = \(\frac{200}{y}\) hours
∴ Total time = \(\left(\frac{100}{x}+\frac{200}{y}\right)\) hours
According to 2nd condition,
\(\left(\frac{100}{x}+\frac{200}{y}\right)\) = 4 hours 10 minutes
or \(\left(\frac{100}{x}+\frac{200}{y}\right)\) = \(\frac{25}{6}\)
\(\left(\frac{24}{x}+\frac{48}{y}\right)\) = 1 ……….(2)
Putting \(\frac{1}{x}\) = u and \(\frac{1}{y}\) = v in equations
(1) and (2) then equations reduces to
15u + 60v = 1
and 24u + 48v = 1
15u + 60v – 1 = 0
24u + 48v – 1 = 0

From I and III, we get:
\(\frac{u}{-12}=\frac{1}{-720}\)
⇒ u = \(\frac{12}{720}=\frac{1}{60}\)
From II and III, we get:
\(\frac{v}{-9}=\frac{1}{-720}\)
⇒ v = \(\frac{9}{720}=\frac{1}{80}\)
But \(\frac{1}{x}\) = u
or \(\frac{1}{x}\) = \(\frac{1}{u}\)
or x = 60
and \(\frac{1}{y}\) = v
or \(\frac{1}{y}\) = \(\frac{1}{80}\)
or y = 80
Hence, speed of train and bus are 60 km/hour and 80 km/hour respectively.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5

Question 1.
Which of the following pairs of linear equations has unique solution, no solution, or infinitély many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0
3x – 9y – 2 = 0

(ii) 2x + y = 5
3x + 2y = 8

(iii) 3x – 5y = 20
6x – 10y = 40

(iv) x – 3y – 7 = 0
3x – 3y – 15 = 0
Solution:
(i) Given pair of linear equation is:
x – 3y – 3 = 0
and 3x – 9y – 2 = 0

Here a₁ = 1, b₁ = -3, c₁ = -3
a₂ = 3, b₂ = -9, c₂ = -2

Now, \(\frac{a_{1}}{a_{2}}=\frac{1}{3}\);
\(\frac{b_{1}}{b_{2}}=\frac{-3}{-9}=\frac{1}{3}\);
\(\frac{c_{1}}{c_{2}}=\frac{-3}{-2}=\frac{3}{2}\)

∵ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given system of equations has no solution.

(ii) Given pair of linear equations
2x + y = 5
and 3x + 2y = 8
or 2x + y – 5 = 0
and 3x + 2y – 8=0
Here a₁ = 2, b₁ = 1, c₁ = -5
a₂ = 3,b₂ = 2, c₂ = 8
Now,
\(\frac{a_{1}}{a_{2}}=\frac{2}{3}\);
\(\frac{b_{1}}{b_{2}}=\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}=\frac{-5}{-8}=\frac{5}{8}\)

∵ \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
∴ given system of equation have unique solution.

or \(\frac{x}{2}=\frac{y}{1}=\frac{1}{1}\)
I          II         III
From I and III, we get:
\(\frac{x}{2}=\frac{1}{1}\)
⇒ x = 2

From I and III, we get:
\(\frac{y}{1}=\frac{1}{1}\)
⇒ y = 1
Hence, x = 2 and y = 1.

(iii) Given pair of linear equations is :
3x – 5y = 20
and 6x – 10y = 40
or 3x – 5y – 20 = 0
and 6x – 10y – 40 =0
Here a₁ = 3, b₁ = -5, c₁ = -20
a₂ = 6, b₂ = -10, c₂ = -40
Now,
\(\frac{a_{1}}{a_{2}}=\frac{3}{6}=\frac{1}{2}\);
\(\frac{b_{1}}{b_{2}}=\frac{-5}{-10}=\frac{1}{2}\);
\(\frac{c_{1}}{c_{2}}=\frac{-20}{-40}=\frac{1}{2}\)

∵ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
Hence, given system have infinite solution.

(iv) Given pair of linear equation is:
x – 3y – 7 = 0
and 3x – 3y – 15 = 0
Here a₁ = 1, b₁ = -3, c₁ = -7
a₂ = 3, b₂ = -3, c₂ = -15
Now,
\(\frac{a_{1}}{a_{2}}=\frac{1}{3}\);
\(\frac{b_{1}}{b_{2}}=\frac{-3}{-3}=1\);
\(\frac{c_{1}}{c_{2}}=\frac{-7}{-15}=\frac{7}{15}\)

∵ \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
∴ given system have unique solution.

From I and III, we get:
\(\frac{x}{24}=\frac{1}{6}\)
⇒ x = 4

From I and III, we get:
\(\frac{y}{-6}=\frac{1}{6}\)
⇒ y = -1
Hence, x = 4, y = -1.

Question 2.
(i) For which values of a and b does the following pair of linear eqU1tions have an infinite number of solutions?
2x + 3y = 7
(a – b)x ÷ (a + b)y = 3a + b – 2
(ii) For which value of k wifi the following pair of linear equations have no solution?
3x + y = 1
(2k – 1) x + (k – 1) y = 2k + 1
Solution:
(i) Given pair of linear equation are
2x + 3y = 7
and (a – b)x + (a + b)y = 3a + b – 2
or 2x + 3y – 7 = 0
and (a – b)x + (a + b)y – (3a + b – 2) = 0
Here a₁ = 2, b₁ = 3, c₁ = -7
a₂ = a – b, b₂ = a + b, c₂ = -(3a + b – 2)
∵ System of equation have an infinite number of solutions.

∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
\(\frac{2}{a-b}=\frac{3}{a+b}=\frac{-7}{-(3 a+b-2)}\)
Fron I and III, we get:
\(\frac{2}{a-b}=\frac{7}{3 a+b-2}\)
or 6a + 2b – 4 = 7a – 7b
or -a + 9b – 4 = 0
or a = 9b – 4 …………..(1)
From II and III. we get:
\(\frac{3}{a+b}=\frac{7}{3 a+b-2}\)
or 9a + 3b – 6 = 7a + 7b
or 2a – 4b – 6 = 0
or a – 2b – 3 = 0
Substitute the value of a from (1) in above, we get:
9b – 4 – 2b – 3 = 0
or 7b – 7 = 0
or 7b = 7
b = 1
Substitute this value of b in (1), we get
a = 9 × 1 – 4 = 9 – 4
a = 5
Hence a = 5 and b = 1

(ii) Given pair of linear equation are
3x + y = 1
and (2k – 1)x + (k – 1)y = 2k + 1
or 3x + y – 1 = 0
and(2k – 1)x + (k – 1)y – (2k + 1) = 0
Here a₁ = 3, b₁ = -1, c₁ = -1
a₂ = (2k – 1), b₂ = k – 1, c₂ = -(2k + 1)
∵ system of equations have no solution.
∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
\(\frac{3}{2 k-1}=\frac{1}{k-1} \neq \frac{-1}{-(2 k+1)}\)
I II III
From I and III, we get:
\(\frac{3}{2 k-1} \neq \frac{1}{(2 k+1)}\)
⇒ 6k + 3 ≠ 2k – 1
⇒ 4k ≠ -4
⇒ k ≠ –\(\frac{4}{4}\)
⇒ k ≠ -1
From II and III, we get:
\(\frac{3}{2 k-1}=\frac{1}{k-1}\)
⇒ 3k – 3 = 2k – 1
⇒ k = 2
Hence k = 2 and k ≠ -1.

Question 3.
Solve the following pair of linear equations by the substitution and cross- multiplication methods:
8x + 5y = 9
3x + 2y = 4
Solution:
Given pair of linear equation is:
8x + 5y = 9 ………….(1)
3x + 2y = 4 …………..(2)
Substitution Method:
From (2), 2y = 4 – 3x
y = \(\frac{4-3 x}{2}\) …………….(3)
Substitute this value of y in (1), we get:
8x + 5\(\frac{4-3 x}{2}\) = 9
or \(\frac{16 x+20-15 x}{2}\) = 9
or x + 20 = 18
or x = 18 – 20 = -2
Substitute this value of x in (3), we get:
y = \(\frac{4-3(-2)}{2}=\frac{4+6}{2}\)
= \(\frac{10}{2}\) = 5
Hence, x = -2 and y = 5.

Cross-multiplication Method:

Given pair of linear equation is:
8x + 5y – 9 = o
and 3x + 2y – 4= 0
Here a₁ = 8, b₁ = 5, c₁ = -9
a₂ = 3, b₂ = 2, c₂ = -4
Now,
\(\frac{a_{1}}{a_{2}}=\frac{8}{3}\);
\(\frac{b_{1}}{b_{2}}=\frac{5}{2}\);
\(\frac{c_{1}}{c_{2}}=\frac{-9}{-4}=\frac{9}{4}\)

∵ \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
∴ system have unique solution.

From I and III, we get:
\(\frac{x}{-2}=\frac{1}{1}\)
⇒ x = -2

From II and III, we get:
\(\frac{y}{5}=\frac{1}{1}\)
⇒ y = 5
Hence, x = -2 and y = 5.

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

(i) 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.

(ii) A fraction becomes when 1 is subtracted from the numerator and it becomes when 8 is added to its denominator. Find the fraction.

(iii) 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?

(iv) 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 differ it 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?

(v) 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 units. Find the dimensions of the rectangle.

Solution:
(i) Let monthly fixed hostel charges = ₹ x
and cost of food per day = ₹ y
According to 1st condition
x + 20y = 1000 ………….(1)
According to 2nd condition
x + 26y = 1180 …………..(2)

From I and III, we get:
\(\frac{x}{2400}=\frac{1}{6}\)
⇒ x = \(\frac{2400}{6}\) = 400

From II and III, we get:
\(\frac{y}{180}=\frac{1}{6}\)
⇒ y = \(\frac{180}{6}\) = 30

Hence, monthly fixed hostel charges and cost of food per day are 400 and 30 respectively.

(ii) Let numerator of fraction = x
Denominator of fraction = y
∴ required fraction = \(\frac{x}{y}\)
According to 1st condition,
\(\frac{x-1}{y}=\frac{1}{3}\)
or 3x – 3 = y
or 3x – y – 3 = 0 …………..(1)
According to 2nd condition,
\(\frac{x}{y+8}=\frac{1}{4}\)
or 4x = y + 8
or 4x – y – 8 = 0 …………….(2)

From I and III, we get:
\(\frac{x}{5}=\frac{1}{1}\)
⇒ x = 5

From II and III, we get:
\(\frac{y}{12}=\frac{1}{1}\)
⇒ y = 12
Hence, required fraction is \(\frac{5}{12}\).

(iii) Let, number of right questions attempted by Yash = x
and Number of wrong questions attempted by Yash = y
According to 1st condition,
3x – y = 40
or 3x – y – 40 = 0 …………….(1)
According to 2nd condition,
4x – 2y = 50
or 4x – 2y – 50 = 0 ……………(2)

\(\frac{x}{50-80}=\frac{y}{-160-(-150)}=\frac{1}{-6-(-4)}\)
or \(\frac{x}{-30}=\frac{y}{-10}=\frac{1}{-2}\)

From I and III, we get:
\(\frac{x}{-30}=\frac{1}{-2}\)
x = \(\frac{-30}{-2}\)
⇒ x = 15

From II and III, we get:
\(\frac{y}{-10}=\frac{1}{-2}\)
y = \(\frac{-10}{-2}\)
⇒ y = 5

∴ Number of right questions = 15
Number of wrong questions = 5
Hence, total number of questions = [No. of right questions] + [No. of wrong questions]
=15 + 5 = 20.

(iv) Let speed of car at place A = x km/hour
and speed of car at place B = y km/hour
Distance between places A and B = 100 km
In case of 5 hours
Distance covered by car A = 5x km
[∵ Distance = Speed × Time]
Distance covered by car B = 5y km
According to I st condition,
5x – 5y = 100
or x – y = 20
or x – y – 20 = 0
In case of one hour
Distance covered by car A = x km
[∵ Distance = Speed × Time]
Distance covered by car B = y km
According to 2nd condition,
x + y = 100
or x + y – 100 = 00 ……………….(2)

From I and III, we get:
\(\frac{x}{120}=\frac{1}{2}\)
x = \(\frac{1}{2}\) × 120
⇒ x = 60

From II and III, we get:
\(\frac{y}{171}=\frac{1}{19}\)
y = \(\frac{171}{19}\)
⇒ y = 9

Hence, length and breadth of rectangle are 17 units and 9 units respectively.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

Question 1.
Solve the following pair of equations by the elimination method and the substitution method.
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4= 0 and 9x = 2y + 7
(iv) \(\frac{x}{2}+\frac{2 y}{3}\) and x – \(\frac{y}{3}\) = 3
Solution:
(i) Given pair of linear equations
x + y = 5 …………..(1)
and 2x – 3y = 4 ………….(2)
Elimination Method
Multiplying (1) by 2, we get:
2x + 2y = 10 …………..(3)

Now, (3) – (2) gives

Substitute this value of y in (1), we get:
x + \(\frac{6}{5}\) = 5
or x = 5 – \(\frac{6}{5}\)
= \(\frac{25-6}{5}\) = \(\frac{19}{5}\)
Hence, x = \(\frac{19}{5}\) and \(\frac{6}{5}\)

Substitution Method:

From(2), 2x =4 + 3y
or x = \(\frac{4+3 y}{2}\) ……………..(4)
Substitute this value of x in (I), we get :
\(\frac{4+3 y}{2}\) + y = 5
Or \(\frac{4+3 y+2 y}{2}\)
Or 4 + 5y = 10
Or 5y = 10 – 4 = 6
Or y = \(\frac{6}{5}\)
Substitute this value of y in (4). we get:
x = \(\frac{4+\left(3 \times \frac{6}{5}\right)}{2}=\frac{4+\frac{18}{5}}{2}\)
= \(\frac{20+18}{5 \times 2}=\frac{38}{5 \times 2}\) = \(\frac{19}{5}\)
Hence x = \(\frac{19}{5}\) and y = \(\frac{6}{5}\)

(ii) Given pair of linear equation is :
3x + 4y = 10 …………….(1)
and 2x – 2y = 2 …………………(2)
Elimination Method
Multiplying equation (2) by 2, we get:
4x – 4y = 4 ……………(3)
Now, (3) + (1) gives

Substitute this value of x in (1), we get:
3(2) + 4y = 10
or 6 + 4y = 10
or 4y = 10 – 6
or 4y = 4
or y = \(\frac{4}{4}\) = 1
Hence x = 2 and y = 1.

Substitution method:

From (2),
2x = 2 + 2y
or x = y + 1 ………..(3)
Substitute this value of x in (1), we get :
3(y + 1) +4y = 10
or 3y + 3 + 4y = 10
or 7y = 10 – 3
or 7y = 7
or y = 1
Substitute this value of y in (3), we get :
x = 1 + 1 = 2
Hence, x = 2 and y = 1.

(iii) Given pair of linear equation is :
3x – 5y – 4 = 0 ……..(1)
and 9x = 2y + 7
or 9x – 2y – 7 = 0
Elimination Method:

Multiplying (1) by 3, we get:
9x – 15y – 12 = 0 ……………(3)
Now, (3) – (2) gives

Substitute this value of y in (1), we get:
3x – 5(-\(\frac{5}{13}\)) – 4 = 0
or 3x + \(\frac{25}{13}\) – 4 = 0
or 3x = 4 – \(\frac{25}{13}\)
or 3x = \(\frac{52-25}{13}\) = \(\frac{27}{13}\)
or x = \(\frac{27}{13} \times \frac{1}{3}\)
= \(\frac{9}{13}\)
Hence, x = \(\frac{9}{13}\) and y = – \(\frac{5}{13}\).

Substitution Method:

From(2), x = \(\frac{2 y+7}{9}\) …………..(4)
Substitute this value of x in (1), we get:
3\(\frac{2 y+7}{9}\) – 5y – 4 = 0
or \(\frac{2 y+7-15 y-12}{3}\) = 0
or – 13y – 5 = 0
or -13y = 5
or y = \(-\frac{5}{13}\)
Substitute this value of y in (4), we get:

Hence x = \(\frac{9}{13}\) and y = –\(\frac{5}{13}\)

(iv) Given pair of linear equation is:
\(\frac{x}{2}+\frac{2 y}{3}\) = -1
or \(\frac{3 x+4 y}{6}\) = -1
or 3x + 4y = -6 ……………(1)
x – \(\frac{y}{3}\) = 3
or \(\frac{3 x- y}{3}\) = 3
or 3x – y = 9 ……………(2)

Elimination Method:

Substitute this value of y in (1), we get :
3x + 4(-3) = -6
or 3x – 12 = -6
or 3x = -6 + 12
or 3x = 6
x = \(\frac{6}{3}\) = 2
Hence x = 2, y = – 3.

Substitution Method:

From (2), y = 3x – 9 …………..(4)
Substitute this value of y in (1), we get :
3x + 4(3x – 9) = -6
or 3x + 12x – 36 = -6
or 15x = -6 + 36
or 15x = 30
or x = \(\frac{30}{15}\) = 2
Substitute this value of x in (4), we get :
y = 3 (2) – 9
= 6 – 9 = -3
Hence x = 2, y = -3.

Question 2.
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes if
we only add 1 to the denominator. What ¡s the fraction?

(ii) Five years ago, Nun was thrice as old as Sonu. Ten years later, Nun will be twice as old as Sonu. How old are Nun
and Sonti?

(iii) The sum of the digits of a two-digit nunaber ¡s 9. Also, nine times this number is twice the number obtained by reversing the order of the number. Find the number.

(iv) 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.

(v) 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:
(i) Let numerator of fraction = x
Denominator of fraction = y
Required fraction = \(\frac{x}{y}\)
According to 1st condition,
\(\frac{x+1}{y-1}\) = 1
or x + 1 = y – 1
or x – y + 2 = 0 …………….(1)
According to 2nd condition,
\(\frac{x}{y+1}=\frac{1}{2}\)
or 2x = y + 1
or 2x – y – 1 = 0 ……………..(2)
Now, (2) – (1) gives

or x = 3
Substitute this value of x in (2), we get :
2 × 3 – y – 1 = 0
or 6 – y – 1 = 0
or 5 – y = 0
or y = 5
Hence, required fraction is \(\frac{3}{5}\).

(ii) Let Nun’s present age = x years
Sonus present age = y years
Five years ago
Nun’s age = (x – 5) years
Sonus age = (y – 5) years
According to 1st condition,
x – 5 = 3(y – 5)
or x – 5 = 3y – 15
or x – 3y + 10 = 0 …………….(1)
Ten years later
Nun’s age = (x + 10) years
Sonu’s age = (y + 10) years
According to 2nd condition,
x + 10 = 2 (y + 10)
or x + 10 = 2y + 20
or x – 2y – 10 = 0
Now, (1) – (2) gives

or -y = -20
or y = 20
Substitute this value of y in (2), we get:
x – 2(20) – 10 = 0
or x – 40 – 10 = 0
or x = 50
Hence, Nun’s present age = 50 years
Sonu’s present age = 20 years.

(iii) Let unit’s digit = x
Ten’s digit = y
∴ Required Number = 10y + x
According to 1st condition,
x + y = 9 …………..(1)
On reversing
Unit’s digit = y
Ten’s digit = x
∴ Number = 10x + y
According to 2nd condition,
9[10y + x] = 2[10x + y]
or 90y + 9x = 20x + 2y
or 90y + 9x – 20x – 2y = 0
or -11x + 88y = 0
or x – 8y = 0 ………………(2)
Now, (2) – (1) gives

y = 1
Substitute this value of y in (2), we get :
x – 8 × 1 = 0
or x = 8
Hence, required number = 10y + x
= 10 × 1 + 8 = 18.

(iv) Let, Meena received number of Rs. 50 notes = x
also, Meena received number of Rs. 100 notes = y
According to 1st condition,
x + y = 25 ……………(1)
According to 2nd condition,
50x + 100y = 2000
or x + 2y = 40 ………………(2)
Now, (2) – (1) gives

Substitute this value of y in (1), we get:
x + 15 = 25
or x = 25 – 15 = 10
Hence, Meena received number of notes of
Rs. 50 and Rs. 100 are 10 and 15 respectively.

(v) Let fixed charges for first three days = ₹ x
An additional charge for each day thereafter = ₹ y
In case of Saritha,
x + 4y = 27 …………..(1)
In case of Susy,
x + 2y = 21
Now, (1) – (2) gives

or y = \(\frac{6}{2}\) = 3
Substitute this value of y in (2), we get:
x + 2(3) = 21
or x + 6 = 21
or x = 21 – 6 = 15
Hence, fixed charges for first three days and an additional charge for each day thereafter ₹ 15 and ₹ 3.

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically.

(i) 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.

(ii) 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:
(i) Let the number of boys in the Quiz = X
and the number of girls in the Quiz = y
Total number of students took part in Quiz = 10
x + y = 10
or x + y – 10 = 0
According to Question,
y = x + 4
or x = y – 4
Now, draw the graph of linear equations
x + y = 10
and x – y + 4 = 0
x + y = 10
or x = 10 – y
Putting y = 0 in (1), we get :
x = 10 – 0 = 10
Putting y = 7 in(1), we get:
x = 10 – 7 = 3
Putting y = 10 in (1) we get:
X = 10 – 10 = 0

Plotting the points A (10, 0), B (3, 7), C (0, 10) and drawing a line joining them we get the graph of the equation x + y = 10

Now x – y + 4 = 0
or x = y – 4
Putting y = 0 in (2), we get:
x = 0 – 4 = -4
Putting y = 7 in (2), we get:
x = 7 – 4 = 3
Putting y = 4 in (2), we get:
x = 4 – 4 = 0

Plotting the points D (-4, 0), B (3, 7), E (0, 4) and drawing a line joining them, we get the graph of the equation x – y + 4 = 0

From the graph it is clear that both the linear equations meets at a point B (3, 7).
∴ Point B (3, 7) is the graphic solution.
Hence, number of boys in the Quiz = 3
Number of girls in the Quiz = 7

(ii) Let cost of one pencil = ₹ x
and cost of one pen = ₹ y
According to 1st condition,
5x + 7y = 50
According to 2nd condition,
7x + 5y = 46
∴ Pairs of linear equations is
5x + 7y = 50
7x + 5y = 46
Now, draw the graph of these linear equations.
5x + 7y = 50
or 5x = 50 – 7y
x = \(\frac{50-7 y}{5}\) ………………..(1)
Putting y = 0 in (1), we get :
x = \(\frac{50-7 \times 0}{5}=\frac{50}{5}\)
Putting y = 5 in (1), we get:
x = \(\frac{50-7 \times 5}{5}=\frac{50-35}{5}\)
= \(\frac{15}{5}\) = 3
Putting y = 7 in (1), we get :
x = \(\frac{50-7 \times 7}{5}=\frac{50-49}{5}\)
= \(\frac{1}{5}\) = 0.2

Table:

Plotting the points A (10, 0), B (3, 5), C (0.2, 7) and drawing a line joining them, we get the graph of the equation
5x + 7y = 50
Now 7x + 5y = 46
0r 7x = 46 – 5y
or x = \(\frac{46-5 y}{7}\) …………….(2)
Putting y = 0 in (2), we get:

x = \(\frac{46-5 \times 0}{7}=\frac{46}{7}\) = 6.5
Putting y = 5 in (2), we get:
x = \(\frac{46-5 \times 5}{7}=\frac{46-25}{7}\)
= \(\frac{21}{7}\) = 3
Putting y = – 4 in (2), we get:
x = \(\frac{46-5 \times(-4)}{7}=\frac{46+20}{7}\)
= \(\frac{66}{7}\) = 9.5

Table:

Plotting the points E (6.5, 0), B (3, 5), F (9.5. -4) and drawing a line joining them, we get the graph of the equation.
7x + 5y = 46

From the graph, it is clear that both the linear equations meets at a point B (3, 5).
∴ point B (3, 5) is the graphic solution.
Hence, cost of one pencil = ₹ 3
Cost of one pen = ₹ 5

Question 2.
On comparing the ratios \(\frac{a_{1}}{a_{2}}\), \(\frac{b_{1}}{b_{2}}\) and \(\frac{c_{1}}{c_{2}}\) and find out whether the lines representing the following pairs of linear
equations intersect at point, are parallel or coincident :
(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0

(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0

(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Solution:
(i) Given pairs of linear equation
5x – 4y + 8 = 0
and 7x + 6y – 9 = 0
Here a₁ = 5, b₁ = – 4, c₁ = 8
a₂ = 7, b₂ = 6, c₂ = -9
Now,
\(\frac{a_{1}}{a_{2}}\) = \(\frac{5}{7}\);

\(\frac{b_{1}}{b_{2}}\) = \(-\frac{4}{6}\);

\(\frac{c_{1}}{c_{2}}\) = \(\frac{8}{-9}\)

∵ \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equation intersect at a point.

(ii) Given pairs of linear equation is
9x + 3y + 12 = 0
and 18x + 6y + 24 = 0
Here, a₁ = 9, b₁ = 3, c₁ = 12
a₂ = 18, b₂ = 6, c₂ = 24

Now \(\frac{a_{1}}{a_{2}}=\frac{9}{18}=\frac{1}{2}\);

\(\frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2}\);

\(\frac{c_{1}}{c_{2}}=\frac{12}{24}=\frac{1}{2}\)
Hence, given pairs of linear equation intersect at a point.

(iii) Given pairs of linear equation is
6x – 3y + 10 = 0
and 2x – y + 9 = 0
Here a₁ = 6, b₁ = – 3, c₁ = 10
a₂ = 2, b₂ = -1, c₂ = 9
Now,
\(\frac{a_{1}}{a_{2}}\) = \(\frac{6}{2}\) = 3;

\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{-1}\) = 3;

\(\frac{c_{1}}{c_{2}}\) = \(\frac{10}{9}\)

∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equations are parallel to each other.

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.
(i) 3x + 2y = 5; 2x – 3y = 7
(ii) 2x – 3y =8; 4x – 6y = 9
(iii) \(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7; 9x – 10y = 14
(iv) 5x – 3y = 11; -10x + 6y = -22
(v) \(\frac{4}{3}\)x + 2y = 8; 2x +3y = 12
Solution:
(i) Given pair of linear equation is
3x + 2y = 5
and 2x – 3y = 7
or 3x + 2y – 5 = 0
and 2x – 3y – 7 = 0
Here a₁ = 3, b₁ = 2, c₁ = -5
a₂ = 2, b₂ = -3, c₂ = -7
Now
\(\frac{a_{1}}{a_{2}}\) = \(\frac{3}{2}\);

\(\frac{b_{1}}{b_{2}}\) = \(\frac{2}{-3}\);

\(\frac{c_{1}}{c_{2}}\) = \(\frac{-5}{-7}\) = \(\frac{5}{7}\)
∴ \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equations are consistent.

(ii) Given pair of linear equation is:
2x – 3y = 8
and 4x – 6y = 9
Or 2x – 3y – 8 = 0
4x – 6y – 9 = 0
Here a₁ = 2, a₁ = -3, c₁ = -8
a₂ = 4, b₂ = -6, c ₂ = -9
Now
\(\frac{a_{1}}{a_{2}}\) = \(\frac{2}{4}\) = \(\frac{1}{2}\);

\(\frac{b_{1}}{b_{2}}\) = \(\frac{-3}{-6}\) = \(\frac{1}{2}\);

\(\frac{c_{1}}{c_{2}}\) = \(\frac{-8}{-9}\) = \(\frac{8}{9}\)

∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equations are inconsistent.

(iii) Given pair of linear equation is:
\(\frac{3}{2}\)x + \(\frac{5}{3}\)y = 7;
and 9x – 10y = 14

Or \(\frac{3}{2}\)x + \(\frac{5}{3}\)y – 7 = 0
and 9x – 10y – 14 = 0
Here a₁ = \(\frac{3}{2}\), b₁ = \(\frac{5}{3}\), c₁ = -7
a₂ = 9, b₂ = -10, c₂ = -14

Hence, given pairs of linear equations are consistent.

(iv) Given pair of linear equations is
5x – 3y= 11
and -10x + 6y = -22
Or 5x – 3y – 11 = 0
and -10x + 6y + 22 = 0
Here a₁ = 5, b₁ = -3, c₁ = -11
a₂ = -10, b₂ = 6, c₂ = 22
Hence given pair of linear equations is consistent.

(v) Given pair of linear equations is
\(\frac{4}{3}\)x + 2y = 8 and 2x + 3y = 12
or \(\frac{4}{3}\)x + 2y – 8 = 0
and 2x + 3y – 12 = 0
Here a₁ = \(\frac{4}{3}\), b₁ = 2, c₁ = -8
a₂ = 2, b₂ = 2, c₂ = -12
Now,

∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equations are consistent.

Question 4.
Which of the following pairs of linear equations are consistent / inconsistent? If consistent, obtain the solution graphically.
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0,4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0,4x – 4y – 5= 0
Solution:
(1) Given pair of linear equations is
x + y = 5
and 2x + 2y = 10
Or x + y – 5 = 0
2x + 2y – 10 = 0
Here a₁ = 1, b₁ = 1, c₁ = -5
a₂ = 2, b₂ = 2, c₂ = -10
Now

\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{2}\);

\(\frac{b_{1}}{b_{2}}\) = \(\frac{1}{2}\);

\(\frac{c_{1}}{c_{2}}\) = \(\frac{-5}{-10}\) = \(\frac{1}{2}\)

∴ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equations are consistent.
Draw the graph of these equations

x + y = 5
x = 5 – y ………….(1)
Putting y = 0 in (1), we get:
x = 5 – 0 = 5
Putting y = 3 in (1), we get
x = 5 – 3 = 2
Putting y = 5 in (1), we get
x = 5 – 5 = 0

Plotting the points A (5, 0), B (2, 3), C (0, 5) and drawing a line joining them, we ge the graph of the equation x + y = 5
2x + 2y = 10 Or 2 (x + y) = 10
Or x + y = 5
Or x = 5 – y
Putting y = 0 in (1), we get :
x = 5 – 0 = 5
Putting y = 2 in (2), we get :
x = 5 – 2 = 3
Putting y = 5 in (2), we get:
x = 5 – 5 = 0
Putting y = 2 in (2), we get :
x = 5 – 2 = 3
Putting y = 5 in (2), we get:
x = 5 – 5 = 0

Plotting the points A (5, 0), D (3, 2), C (0, 5) and drawing a line joining them, we get the graph of the equation 2x + 2y = 10

The graphs of two equations are coincident. Hence system of equations has infinitely many solutions i.e. consistent.

(ii) Given pair of linear equations is :
x – y = 8
and 3x – 3y = 16
Or
x – y – 8 = 0
and 3x – 3y – 16 = 0

Here a₁ = 1, b₁ = -1, c₁ = -8
a₂ = 3, b₂ = -3, c₂ = -16
Now,
\(\frac{a_{1}}{a_{2}}\) = \(\frac{1}{3}\);

\(\frac{b_{1}}{b_{2}}\) = \(\frac{-1}{-3}\); = \(\frac{1}{3}\)

\(\frac{c_{1}}{c_{2}}\) = \(\frac{-8}{-16}\) = \(\frac{1}{2}\)

∵ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given pairs of linear equations are inconsistent solution.

(iii) Given pailt of linear equations is :
2x + y – 6 = 0
and 4x – 2y – 4 = 0
Here a₁ = 2, b₁ = 1, c₁ = -6
a₂ = 4, b₂ = -2, c₂ = -4
Now
\(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2}\);

\(\frac{b_{1}}{b_{2}}=\frac{1}{-2}\);

\(\frac{c_{1}}{c_{2}}=\frac{-6}{-4}=\frac{3}{2}\)
∴ given pair of system is consistent.
Draw the graph of these linear equations
2x + y – 6 = 0
Or 2x = 6 – y
Or x = \(\frac{6-y}{2}\) ………….(1)
Putting y = 0 in (1), we get:
x = \(\frac{6-0}{2}=\frac{6}{2}\) = 3

putting y = 2 in (1), we get:
x = \(\frac{6-2}{2}=\frac{4}{2}\) = 2

Putting y = -2 in (1), we get:
x = \(\frac{6-(-2)}{2}=\frac{6+2}{2}=\frac{8}{2}\) = 4

Plotting the points A (3, 0), B (2, 2), C (4, -2) and drawing a line joining them, we get the graph of the equation.
2x + y – 6 = 0
Now 4x – 2y – 4= 0
or 2[2x – y – 2] = 0
or 2x – y – 2 = 0
or 2x = y + 2
or x = \(\frac{y+2}{2}\) …………..(2)
Putting y = 0 in (2), we get:
x = \(\frac{0+2}{2}=\frac{2}{2}\) = 1

Putting y = 2 in (2), we get :
\(\frac{2+2}{2}=\frac{4}{2}\) = 2

Putting y = – 2 in (2), we get:
x = \(\frac{-2+2}{2}=\frac{0}{2}\) = 0

Plotting the points D (1, 0), B (2, 2), E (0, -2) and drawing a line joining them, we get the graph of the equation
4x – 2y – 4 = 0

From the graph, it is clear that given system of equations meets at a point B (2, 2).
Hence, given pair of linear equations have unique solution.

(iv) Given pair of linear equations is :
2x – 2y – 2 = 0
and 4x – 4y – 5 = 0
Here a₁ = 2, b₁ = -2, c₁ = -2
a₂ = 4, b₂ = -4, c₂ = -5
Now,
\(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2}\);

\(\frac{b_{1}}{b_{2}}=\frac{-2}{-4}=\frac{1}{2}\);

\(\frac{c_{1}}{c_{2}}=\frac{-2}{-5}=\frac{2}{5}\)

∵ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
Hence, given pair of system have inconsistent solution.

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 length of garden = x m
Width of garden =y m
Perimeter of garden = 2 [x + y] m
Half perimeter of garden = (x + y) m
According to 1st condition x = y +4
According to 2nd condition
x + y = 36
∴ Pair of linear equations is
x = y + 4
and x + y = 36
x = y + 4 ………………(1)
Putting y = 0 in (1), we get :
x = 0 + 4 = 4
Putting y = – 4 in (1), we get:
x = -4 + 4 = 0
Putting y = 16 in (1), we get :
x = 16 + 4 = 20

Plotting the points A (4, 0) B (0, -4), C (20, 16) and drawing a line joining them.
we get the graph of the equation.
x = y + 4
Now x + y = 36
x = 36 – y
Putting y = 12 in (2), we get:
x = 36 – 12 = 24
Putting y = 24 in (2), we get :
x = 36 – 24 = 12
Putting y = 16 in (2), we get:
x = 36 – 16 = 20

Plotting the points D (24, 12), E (12, 24) C(20, 16) and drawing a line joining them, we get the raph of the equation.
x + y = 36

From the graph. it is clear that pair of linear equations meet at a point C (20, 16).
∴ C (20, 16) i.e. x = 20 and y = 16 is the solution of linear equations.
Hence, length of garden = 20 m
Width of garden = 16 m

Another Method:

Let width of garden = x m
Lengh of garden = (x + 4) m
Perimeter of garden = 2 [Length + Width]
= 2 [x + x + 4] m
= 2[2x + 4]m
∴ Half perimeter of garden = (2x +4) m
According to Question,
2x + 4 = 36
or 2x = 36 – 4
or 2x = 32
or x = \(\frac{32}{16}\) = 2 m
Hence, width of garden = 16 m
and length of garden = (16 + 4)m = 20 m

Question 6.
Given the Linear equation 2x + 3y – 8 = 0, write another linear equadon in two variables such that the geometiical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Solution:
Case I. For Intersecting Unes
Given linear equation is:
2x + 3y – 8 = 0
There are many another linear equations in two variahes which satisfies the condition of intersecing lines i.e.
\(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

One of which is as follow:
3x – 2y – 6 = 0
Now, draw the graph of linear equations (1) and (2).
2x + 3y – 8 = 0
2x = 8 – 3y
Putting y = 0 in (1), we get;
x = \(\frac{8-3 \times 0}{2}=\frac{8}{2}\) = 4

Putting y = -2 in (1), we get:
x = \(\frac{8-3(-2)}{2}=\frac{14}{2}\) = 7

Piating y = 2 in (1) we get:
x = \(\frac{8-3 \times 2}{2}=\frac{8-6}{2}=\frac{2}{2}=1\)

Plotting the points A (4, 0), B (7, -2), C (1, 2) and drawing a line joining them, we get the graph of the equation
2x + 3y – 8 = 0
Now 3x – 2y – 6 = 0
3x = 6 + 2y
or x = \(\frac{6+2 y}{3}\)
Putting y = 0 in (1), we get:
x = \(\frac{6+2 \times 0}{3}=\frac{6}{3}\) = 2
Putting y = – 3 in (1), we get :
x = \(\frac{6+2(-3)}{3}\) = \(\frac{6-6}{3}\) = 0
Putting y = 3 in (2), we get :
x = \(\frac{6+2 \times 3}{3}=\frac{6+6}{3}=\frac{12}{3}\) = 4

From the graph it is clear that linear equations intersect at a point G.

Case II:
For Parallel Lines
Given linear equation is
2x + 3y – 8 = 0 ……………(1)
There are many other linear equation in two variables which satisfies the condition of parallel lines i.e.
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)
One of which is as follow :
2x + 3 – 5 = 0
Now, draw the graph of linear equations (1) and (3)
Table for linear equation 2x + 3y – 8 = 0 is follow:

Consider, 2x + 3y – 5 = 0
or 2x = 5 – 3y
or x = \(\frac{5-3 y}{2}\) …………..(3)
Putting y = 0 in (3), we get :
x =\(\frac{5-3 \times 0}{2}=\frac{5}{2}\) = 2.5

Putting y = 3 in (3), we get:
x = \(\frac{5-3 \times 3}{2}=\frac{5-9}{2}=\frac{-4}{2}\) = -2

Putting y = 3 in (3), we get:
x = \(\frac{5-3(-3)}{2}=\frac{5+9}{2}=\frac{14}{2}\) = 7

Plotting the points G (2.5, 0), H (-2, 3), I (7, -3) and drawing a line joining them,
we get the graph of the equation
2 + 3y – 5 = 0.

Case III. For Coincident Lines
Given linear equation is
2x + 3y – 8 = 0
There are many other linear equations ¡n two variables which satisfies the condition of coincident lines i.e.
\(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
One of which is as follow :
6x+ 9y – 24 = 0
Now, draw the graph of linear equations (1) and (4).
Consider linear equation (4)
6x + 9y – 24 = o
or 3[2x + 3 – 8] = 0
or 2x + 3y – 8 = 0
∴ The points of both are same and line of both equations are same.

Question 7.
Drab tht graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed b these lines and the x-axis and shade the triangular region. (Pb. 2018 Set I, II, III)
Solution:
Consider the pair of linear equation
x – y + 1 =0
and 3x + 2y – 12 = 0
x – y + 1 = 0
or x = y – 1
Putting y = 0 in (1), we get:
x = 0 – 1 = -1
Puning y = 3 in (1) we get:
x = 3 – 1 = 2
Putting y = 1 in (1) , we get:
x = 1 – 1 = 0

PloningthepointsA(-1, 0), B(2, 3), C(0, 1) and drawing a line ioining them. we get the graph of the equation x – y + 1 = 0
3x + 2y – 12 = 0
or 3x = 12 – 2y
or x = \(\frac{12-2 y}{3}\) …………..(2)
Putting y = 0 in (2), we get:
x = \(\frac{12-2 \times 0}{3}=\frac{12}{3}\) = 4
Putting y = 3 in (2), we get:
x = \(\frac{12-2 \times 3}{3}=\frac{12-6}{3}=\frac{6}{3}\) = 2
Putting y = 6 in (2), we get:
x = \(\frac{12-2 \times 6}{3}=\frac{12-12}{3}=0\)

Table:

Plotting the points D(4, 0) B (2, 3), E (0, 6) and drawing a line oinin them, we get the graph of the equation 3x – 2y – 12 = 0

The vertices of the triangle formed by pair of linear equations and the x-axis are shaded in the graph. The triangle so formed is ∆ABD.Coordinates of the vertices of ∆ABD are
A(-1, 0), B(2, 3) and D(4, 0).
Now, length of Base AD = AO + OD = 1 + 4 = 5 units
Length of perpendicular BF = 3 units
∴ Area of ∆ABD = \(\frac{1}{2}\) × Base × altitude
= \(\frac{1}{2}\) × AD × BF
= (\(\frac{1}{2}\) × 5 × 3) sq. units
= \(\frac{15}{2}\) = 7.5 sq. units

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Question 1.
Solve the following pair of linear equations by the substitution method:
(i) x + y = 14
x – y = 4

(ii) s – t = 3
\(\frac{s}{3}+\frac{t}{2}\) = 6

(iii) 3x – y = 3
9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3

(v) √2x + √3y = 0
√3x – √8y = 0

(vi) \(\frac{3 x}{2}-\frac{5 y}{3}=-2\)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)

Solution:
(i) Given pair of linear equati3ns
x + y = 14 …………(1)
and x – y = 4
From (2) x = 4 ………….(3)
Substitute this value of x in Equation we get: .
4 + y + y = 14
2y = 14 – 4
2y = 10
y = \(\frac{10}{2}\) = 5
Substitute this value of y in equation (3), we get:
x = 4 + 5 = 9
Hence x = 9 and y = 5

(ii) Given pair of linear equations
s – t = 3 …………….(1)
and
\(\frac{s}{3}+\frac{t}{2}\) = 6
\(\frac{2 s+3 t}{6}\) = 6
2s + 3t = 36 …………….(2)
From (1), s = t + 3 ……………….(3)
Substitute this value of s in equation (1), we get:
2(3 + t) + 3t = 36
Or 6 + 2t + 3t = 36
Or 6 + 5t = 36
Or 5t = 36 – 6
Or 5t = 30
Or t = \(\frac{30}{5}\) = 6
Substitute this value of r in equation (3), we get:
s = 3 + 6 = 9
Hence, s = 9 and t = 6

(iii) Given pair of linear equation is:
3x – y = 3 .
and 9x – 3y = 9
From (1),
3x – 3 = y
Or y = 3x – 3 ………….(3)
Substitute this value of y in equation (2), we get :
9x – 3(3x – 3) = 9
Or 9x – 9x + 9 = 9
Or 9 = 9
This statement is true for all values of x. However, we do not get a specific value of x as a solution. Therefore we cannot obtain a specific value of y. This situation has arises because both the given equations are same. Therefore, equations (1) and (2) have infinitely many solutions.

(iv) Given pair of linear equation is:
0.2x + 0.3y = 1.3
or \(\frac{2}{10} x+\frac{3}{10} y=\frac{13}{10}\)
or 2x + 3y = 13 ……………..(1)
0.4x + 0.5y = 2.3
or \(\frac{4}{10} x+\frac{5}{10} y=\frac{23}{10}\)
Or 4x + 5y = 23 ……………(2)
From (1),
2x = 13 – 3y
x = \(\frac{13-3 y}{2}\) …………..(2)
Substitrne this value of x in (2), we get:
4[latex]\frac{13-3 y}{2}[/latex] + 5y = 23
26 – 6y + 5y = 23
-y = 23 – 26 = -3
y = 3
Substitute this value of y in (3), we get:
x = \(\frac{13-3 \times 3}{2}\)
= \(\frac{13-9}{2}=\frac{4}{2}\) = 2
Hence, x = 2 and y = 3.

(v) Given pair of linear equation is:
√2x + √3y = 0 ……………..(1)
√3x – √8y = 0 …………..(2)
From (2), √3x = √8y
or x = \(\frac{\sqrt{8}}{\sqrt{3}}\)y …………..(3)
Substitute this value of x in (1), we get
√2(\(\frac{\sqrt{8}}{\sqrt{3}}\)y) + √3y = 0
or [\(\frac{4}{\sqrt{3}}\) + √3]y = 0
y = 0
Substitute this value of y in (3), we get:
x = \(\frac{\sqrt{8}}{\sqrt{3}}\) × 0 = 0
Hence x = 0 and y = 0

(vi) Given pair of linear equation is:
\(\frac{3 x}{2}-\frac{5 y}{3}=-2\)

or \(\frac{9 x-10 y}{6}\) = -2
or 9x – 10y = -12 …………..(1)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)
\(\frac{2 x+3 y}{6}=\frac{13}{6}\)
or 2x + 3y = \(\frac{13}{6}\) × 6
or 2x + 3y = 13 …………..(2)
From (1), 9x = 10y – 12
or x = \(\frac{10 y-12}{9}\) …………….(3)
2[latex]\frac{10 y-12}{9}[/latex] + 3y = 13

or \(\frac{20 y-24}{9}\) + 3y = 13

or \(\frac{20 y-24+27 y}{9}\) = 13

or 47y – 24 = 13 × 9 = 117
47y = 117 + 24 = 141
or y = \(\frac{141}{47}\) = 3
substitute this value of y in (3), we get
x = \(\frac{10 \times 3-12}{9}=\frac{30-12}{9}\)

= \(\frac{18}{9}\) = 2
Hence, x = 2 and 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:
Given pair of linear equations is:
2x + 3y = 11
and 2x – 4y = -24 …………(2)
From (2),
2x = 4y – 24
2x = 2 [2y – 12]
Or x = 2y – 12 …………..(3)
Substitute this value of x in (1), we get:
2 (2y – 12) + 3y = 11
Or 4y – 24 + 3y = 11
Or 7y = 11 + 24
Or 7y = 35
y = \(\frac{35}{7}\) = 5
Substitute this value of y in (3), we get:
x = 2(5) – 12 = 10 – 12 = -2
Now, consider y = mx + 3
Substitute the value of x = -2, y = 5, we get:
5 = m(-2) + 3
Or 5 – 3 = -2m
Or 2 = – 2m
Or -2m = 2
Or m = -1
Hence, x = -2, y = 5 and m = -1

Question 3.
Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) 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.

(iv) 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 kilometre? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes \(\frac{9}{11}\), if 2 is added to both the numerator and the denominator. If 3 ¡s added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.

(vi) 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:
(I) Let two number be x and y.
According to 1st condition,
x – y=26 ……….(1)
According to 2nd condition,
x = 3y …………(2)
Substitute this value of x in (1), we get :
3y – y = 26
Or 2y = 26
y = \(\frac{26}{2}\) = 13
Substitute this value of y in (2), we get:
x = 3 × 13 = 39
Hence, two numbers are 39, 13.

(ii) Let, required two supplementary angles are x, y and x > y
According to 1st condition,
x + y = 180 ………..(1)
According to 2nd condition,
x = y + 18 …………..(2)
Substitute this value of x in (1), we get:
y + 18 + y = 180
Or 2y = 180 – 18
or 2y =162
Or y = \(\frac{162}{2}\) = 81
Substitute this value of y in (2), we get:
x = 81 + 18 = 99
Hence, required angles are 99, 81.

(iii) Let cost of one bat = ₹ x
and cost of one ball = ₹ y
According to 1st condition,
7x + 6y = ₹ 3800 ……………(1)
According to 2nd condition,
3x + 5y = ₹ 1750 …………….(2)
From(1), 7x = 3800 – 6y
Or x =\(\frac{3800-6 y}{7}\) ………….(3)
substitute this value of x in (2), we get:
3[latex]\frac{3800-6 y}{7}[/latex] + 5y = 1750
Or \(\frac{11400-18 y+35 y}{7}\) = 1750
Or 11400 + 17y = 1750 × 7
0r 11400 + 17y = 12250
Or 17y = 12250 – 11400
Or 17y = 850
or y = \(\frac{850}{17}\) = 50
Substitute this value of y in (3), we get:
x = \(\frac{3800-6 \times 50}{7}\)
= \(\frac{3800-300}{7}=\frac{3500}{7}\)
x = 500
Hence cost of one bat = ₹ 500
and cost of one ball = ₹ 50.

(iv) Let the fixed charges for the taxi = ₹ x
and charges for travelling one km = ₹ y
According to 1st condition,
x + 10y = 105 …………..(1)
According to 2nd condition,
x + 15y = 155 ……………(2)
From (1),
x = 105 – 10y …………..(3)
Substitute the value of x in (2), we get
105 – 10y + 15y = 155
Or 5y = 155 – 105
Or 5y = 50
Or y = \(\frac{50}{5}\) = 10
Substitute the value of y in (3), we get:
x = 105 – 10 × 10
= 105 – 100 = 5
Hence, fixed charges for the taxi = ₹ 5
and charges for travelling one km = ₹ 10
Also, charges for travelling 25 km = ₹(10 × 25) + ₹ 5
= ₹[250 + 5] = ₹ 255

(v) Let numerator of given fraction = x
Denominator of given fraction = y
∴ Required fraction = \(\frac{x}{y}\)
Acccrding to 1st condition,
\(\frac{x+2}{y+2}=\frac{9}{11}\)
Or 11(x + 2) = 9(y + 2)
Or 11x + 22 = 9y + 18
Or 11x = 9y + 18 – 22
Or 11x = 9y – 4
Or x = \(\frac{9 y-4}{11}\) ……………..(1)
Acccrding to 2nd equation
\(\frac{x+3}{y+3}=\frac{5}{6}\)
Or 6 (x + 3) = 5(y + 3)
Or 6x + 18 = 5y + 15
Or 6x – 5y = 15 – 18
Or 6x – 5y = -3
Putting the value of x from (1). we get:
6[latex]\left[\frac{9 y-4}{11}\right][/latex] – 5y = -3
Or \(\frac{54 y-24}{11}\) – 5y = -3
Or \(\frac{54 y-24-55 y}{11}\) = -3
Or -y – 24 = -3 × 11
Or -y = -33 + 24
Or -y = -9
Or y = 9
Substitute the value of y in (1), we get:
x = \(\frac{9 \times 9-4}{11}=\frac{81-4}{11}\)
= \(\frac{77}{11}\) = 7
Hence, required fraction is \(\frac{7}{9}\).

(vi) Let Jacob’s present age = x years
and Jacob son’s present age = y years
Five years hence
Jacob’s age = (x + 5) years
His son’s age = (y + 5)years
According to 1st condition.
x + 5 = 3(y + 5)
Or x + 5 = 3y + 15
Or x = 3y + 15 – 5
Or x = 3y + 10 ……………(1)
Five years ago
Jacobs age = (x – 5) years
His son’s age = (y – 5) years
According to 2nd condition.
x – 5 = 7(y – 5)
Or x – 5 = 7y – 35
Or x – 7y = -35 + 5
Or x – 7y = -30
Substitute the value of x from (1), we get:
3y + 10 – 7y = -30
– 4y = – 30 – 10
-4y = -40
y = 10
Substitute tins value of y in (1), we get:
x = 3 (10) + 10
= 30 + 10 = 40
Hence. Jacob and his son’s ages are 40 years and 10 years respecùveiy.

Punjab State Board PSEB 10th Class Hindi Book Solutions Hindi Grammar prapatra poorti प्रपत्र पूर्ति Exercise Questions and Answers, Notes.

PSEB 10th Class Hindi Grammar प्रपत्र पूर्ति

प्रपत्र पाठ्यक्रम-बैंक, डाकघर तथा रेलवे से संबंधित प्रपत्र पूर्ति
‘प्रपत्र’ अंग्रेज़ी शब्द फार्म का अनुवाद है। आम बोलचाल की भाषा में प्रपत्र को फार्म भी कहते हैं। प्रपत्र शब्द पत्र के आगे ‘प्र’ उपसर्ग लगाने से बना है। बालक के जन्म से लेकर मृत्यु तक के प्रपत्र भरकर बालक का जन्म प्रमाण-पत्र तथा मृतक का मृत्यु प्रमाण-पत्र प्राप्त करना होता है। इसके अतिरिक्त स्कूल, कॉलेज में प्रवेश के लिए, परीक्षा देने के लिए, नौकरी के लिए, प्रतियोगी परीक्षाओं के लिए, राशन कार्ड, आधार कार्ड, बिजली-पानीटेलीफोन-मोबाइल, गैस कनेक्शन लेने के लिए, पासपोर्ट बनवाने, बैंक/डाकखाने में पैसे जमा कराने या निकालने के लिए, रेल-हवाई यात्रा के समय आरक्षण कराने के लिए, विवाह के पंजीकरण, वाहन पंजीकरण, व्यवसाय पंजीकरण आदि अनेक कार्यों के लिए भी प्रपत्र भरने होते हैं। कोई भी प्रपत्र भरते समय निम्नलिखित तथ्यों का ध्यान रखना चाहिए-

(क) प्रपत्र के साथ दिए गए निर्देशों को अच्छी प्रकार से पढ़ कर ही प्रपत्र भरना चाहिए।
(ख) प्रपत्र में दिए गए निर्देश के अनुसार ही प्रपत्र को पैन/बॉलपैन/पेंसिल से भरना चाहिए।
(ग) प्रपत्र में जिन स्तंभों (कॉलमों) को अंग्रेज़ी के बड़े (कैपिटल) अक्षरों में भरना हो, उन्हें बड़े अक्षरों में भरना चाहिए।
(घ) प्रपत्र पर अपना छायाचित्र (फोटो) उचित स्थान पर चिपकाएं या पिन से नत्थी कीजिए। यदि फ़ोटो को सत्यापित (अटैस्ट) कराना है अथवा अपने हस्ताक्षर करने हैं, तो वह भी कीजिए।
(ङ) प्रपत्र के साथ निर्देशानुसार स्वहस्ताक्षरित अथवा राजपत्रित अधिकारी द्वारा सत्यापित (अटैस्टिड) दस्तावेज़ लगाएँ।
(च) प्रपत्र का कोई भी स्तम्भ (कॉलम) रिक्त न छोड़ें। यदि वह स्तम्भ आप पर लागू नहीं होता तो वहां लागू नहीं लिख दीजिए।
(छ) प्रपत्र में दिए गए स्थान पर अपने हस्ताक्षर कीजिए तथा स्थान और दिनांक लिखिए।
(ज) प्रपत्र साफ़, स्पष्ट, सुंदर तथा पढ़ा जा सके-ऐसे अक्षरों में भरा जाना चाहिए।
(झ) प्रपत्र भरकर व्यक्तिगत रूप से जमा कराते समय उसकी जमा करने वाले कार्यालय से रसीद ले लें। डाक से रजिस्ट्री अथवा स्पीड-पोस्ट से प्रपत्र भरकर भेज सकते हैं। यहाँ कुछ प्रपत्रों को भरने के उदाहरण दिए जा रहे हैं।

1. उदाहरण-किसी खेल में चयन के लिए दिए गए प्रपत्र को भरना-

2. उदाहरण-नौकरी के लिए दिए गए प्रपत्र को भरना
(क) व्यक्तिगत तथा पारिवारिक विवरण

(ग) व्यावसायिक अनुभव (यदि कोई हो)

दसवीं के पाठ्यक्रम में बैंक, डाकघर तथा रेलवे से संबंधित प्रपत्रों की पूर्ति निर्धारित है। इसलिए यहाँ इन्हीं तीनों विभागों से संबंधित प्रपत्रों पर विस्तार से चर्चा की जा रही है-

(क) बैंक से संबंधित प्रपत्र
आधुनिक युग में बैंकों का हमारे दैनिक जीवन में बहुत महत्त्व है। इससे हमारे लेन-देन तथा व्यवसाय में बहुत सहायता मिलती है। बैंक में खाता खुलवाने के लिए कुछ प्रमाण-पत्रों की आवश्यकता होती है, जो निम्नलिखित हैं-

1. आवासीय पते का प्रमाण-पत्र; जैसे-राशन कार्ड, आधार कार्ड, ड्राइविंग लाइसेंस, पासपोर्ट, वरिष्ठ नागरिक प्रमाण-पत्र आदि।
2. जिस बैंक में खाता खुलवाना हो वहाँ के किसी खाताधारी से गवाही।
3. फ़ोटो।
4. पैनकार्ड की प्रतिलिपि।
5. जन-धन योजना में शून्य राशि से खाता खुल सकता है।
6. विद्यार्थियों को अपने विद्यालय के पहचान-पत्र तथा प्राचार्य के हस्ताक्षरित खाता खोलने के फार्म के आधार पर बैंक में शून्य राशि से खाता खोलने की सुविधा है।
7. बैंक में खाता खोलने वाले को बैंक में अपने नमूने के हस्ताक्षर भी देने पड़ते हैं, जिससे कोई अन्य व्यक्ति उसके खाते से रुपए नहीं निकाल सके।
8. बैंक में खाता खोलने के बाद मिलने वाले ए०टी०एम० कार्ड से खाता धारक किसी भी ए०टी०एम० बूथ से कभी भी पैसे निकलवा सकता है तथा खरीददारी भी कर सकता है।

बैंक में खातों के प्रकार-बैंक में खातों के अनेक प्रकार होते हैं; जैसे-बचत खाता, चालू खाता, आवर्ती खाता, सावधि जमा खाता आदि। विद्यार्थियों को बचत खाता खुलवाना चाहिए, जिसमें उनकी छात्रवृत्ति, अपनी बचत आदि जमा हो सकती है। बचत खाता, आवर्ती खाता तथा सावधि खाता पर बैंक जमा राशि पर नियमानुसार ब्याज भी देते हैं, जो खाताधारक के खाते में जमा होता रहता है। बैंक के लेन-देन में मुख्य रूप से निम्नलिखित प्रपत्र प्रयोग में आते हैं-

1. बैंक में खाता खोलने का प्रपत्र
2. रुपए नकद जमा करने का प्रपत्र
3. राशि चैक द्वारा जमा करने का प्रपत्र
4. रुपए निकलवाने का प्रपत्र/चैक
5. ड्राफ्ट बनवाने का प्रपत्र।

यहां प्रमुख प्रपत्रों की रूपरेखा तथा उनके भरने के उदाहरण दिए जा रहे हैं-
1. बैंक में खाता खोलने का प्रपत्र

बैंक में खाता खोलने के प्रपत्र के प्रत्येक कॉलम को ध्यान से पढ़कर साफ़-साफ़ शब्दों में भरिए। खाता एक व्यक्ति अकेले या दो-तीन व्यक्ति मिलकर भी खोल सकते हैं तथा इस संबंध में परिचालन का विकल्प चुन सकते हैं। किसी खाता धारक से परिचय भी दिया जाता है तथा नामांकन का पूरा विवरण भी भरिए। नामिती नाबालिग हो तो उसके संरक्षक का नाम भी लिखना होता है। खाता खोलने हेतु आवेदन फार्म का एक उदाहरण यहां दिया जा रहा है-
1. खाता खोलने का प्रपत्र

प्रश्न 1.
विद्यार्थियों को बैंक में खाता खुलवाने के लिए किन-किन दस्तावेज़ों की आवश्यकता होती है?
उत्तर:
आधुनिक युग में बैंकों का हमारे दैनिक जीवन में बहुत महत्त्व है। इससे हमारे लेन-देन तथा व्यवसाय में बहुत सहायता मिलती है। बैंक में खाता खुलवाने के लिए कुछ प्रमाण-पत्रों की आवश्यकता होती है, जो अग्रलिखित हैं-

1. आवासीय पते का प्रमाण-पत्र ; जैसे-राशन कार्ड, आधार कार्ड, ड्राइविंग लाइसेंस, पासपोर्ट वरिष्ठ नागरिक प्रमाण-पत्र आदि।
2. जिस बैंक में खाता खुलवाना हो वहाँ के किसी खाताधारी से गवाही।
3. फ़ोटो।
4. पैनकार्ड की प्रतिलिपि।
5. जन-धन योजना में शून्यराशि से खाता खुल सकता है।
6. विद्यार्थियों को अपने विद्यालय के पहचान-पत्र तथा प्राचार्य के हस्ताक्षरित खाता खोलने के फार्म के आधार पर बैंक में शून्य राशि से खाता खोलने की सुविधा है।
7. बैंक में खाता खोलने वाले को बैंक में अपने नमूने के हस्ताक्षर भी देने पड़ते हैं, जिससे कोई अन्य व्यक्ति उसके ख़ाते से रुपए नहीं निकाल सके।

2. बैंक में पैसे जमा कराने का प्रपत्र
बैंक में नकद राशि जमा करने के लिए जो प्रपत्र भरना होता है, उसके दो हिस्से होते हैं। एक बैंक की प्रति तथा दूसरी ग्राहक की प्रति होती है। इन दोनों हिस्सों को भरना होता है, जिसमें दिनांक, खाता संख्या, खाताधारी का नाम, जमा राशि तथा जमा राशि का विवरण कि किस-किस राशि के कितने नोट हैं लिखना होता है। दिए गए स्थान पर मोबाइल नंबर लिखकर जमाकर्ता के हस्ताक्षर होते हैं। यहाँ एक उदाहरण दिया जा रहा है जिसमें विजय सिंह और सुरजीत कौर के खाते में एक हज़ार रुपए नकद जमा करने के लिए प्रपत्र भरा गया है-

3. बैंक में चैक द्वारा राशि जमा कराने का प्रपत्र बैंक में चैक जमा कराने का प्रपत्र भी दो भागों में होता है। एक प्रति बैंक की तथा दूसरी ग्राहक की होती है। इसमें दिनांक, खाते का प्रकार तथा संख्या, खातेदार का नाम, चैक संख्या, चैक जारी करने वाले बैंक तथा शाखा का नाम, दिनांक तथा राशि भरनी होती है। यहाँ एक उदाहरण दिया जा रहा है जिसमें मनमोहन सिंह अपने खाते में दस हज़ार रुपए का चैक जमा कर रहे हैं-

4. बैंक से रुपए निकलवाने का प्रपत्र/चैक बैंक से रुपए निकलवाने के लिए बैंक के भगतान प्रपत्र तथा बैंक द्वारा दिए गए चैक का प्रयोग किया जाता है। चैक द्वारा किसी संस्था अथवा व्यक्ति को भी भुगतान कर सकते हैं। बैंक से भुगतान प्रपत्र पर रुपए निकलवाने के लिए पासबुक साथ लगानी पड़ती है। इसमें खातेदार अपने खाते की संख्या, निकाले जाने वाली राशि, दिनांक तथा अपने हस्ताक्षर कर के राशि निकलवा सकता है। प्रीतम कौर अपने खाते से दस हज़ार रुपए भुगतान प्रपत्र पर निकाल रही है। इसका उदाहरण यहाँ दिया जा रहा है-

चैक द्वारा किसी को भुगतान करने अथवा स्वयं रुपए निकालने के लिए चैक में दिनांक, खाता संख्या, किसे भुगतान करना है, राशि तथा खाताधारी के हस्ताक्षर करना आवश्यक होता है। यहाँ एक चैक राजनाथ सिंह द्वारा भारतीय जीवन बीमा निगम को आठ हजार नौ सौ बीस रुपए का जारी किया गया है उसका प्रारूप यहाँ दिया जा रहा है। चैक को रेखांकित कर देना चाहिए इस चैक को कोई अन्य नहीं भुना सकता।

5. बैंक ड्राफ्ट बनवाने का प्रपत्र
बैंक ड्राफ्ट एक मांग-प्रपत्र होता है, जिसे कोई भी व्यक्ति बैंक को निर्धारित शुल्क देकर बनवा सकता है। यह बैंक की एक शाखा द्वारा अपनी उस शाखा के नाम जारी किया जाता है जहाँ ग्राहक ने अपनी राशि भेजनी है। बैंक ड्राफ्ट पाने वाला व्यक्ति उस शाखा से ड्राफ्ट की राशि नकद अथवा अपने बैंक खाते में जमा करवा कर प्राप्त कर सकता है। बैंक ड्राफ्ट के प्रपत्र में दिनांक, आवेदन का नाम पता, किसके पक्ष में, कहाँ भेजना है, ड्राफ्ट की राशि, शुल्क आवेदक के हस्ताक्षर, मोबाइल नंबर भरने होते हैं। यहाँ राजेन्द्र सिंह द्वारा श्री रघुराज भालेराव पाठक को नागपुर पाँच हज़ार रुपए का ड्राफ्ट बनवाने का प्रपत्र उदाहरण के रूप में भरकर दिया जा रहा है। ड्राफ्ट के लिए राजेन्द्र सिंह ने दिए पाँच हजार तीस रुपए हैं परन्तु उसे ड्राफ्ट पांच हज़ार रुपए का ही मिलेगा। तीस रुपए ड्राफ्ट बनवाने का शुल्क है। इसके भी दो भाग होते हैं। एक ग्राहक के लिए तथा दूसरा बैंक के लिए-

बोर्ड परीक्षाओं में पूछे गए प्रपत्र-पूर्ति सम्बन्धी प्रश्नोत्तर

प्रश्न 1.
मान लो आपका नाम निर्मला देवी है। आपको सेविंग्स बैंक खाता नं0 79684 में से दस हज़ार रुपये निकलवाने हैं। अत: आप नीचे दिए गए प्रपत्र की रूपरेखा को अपनी उत्तर-पुस्तिका पर उतार कर भरें-

उत्तर:

प्रश्न 2.
मान लो आपका नाम तमन्ना है। आपका मुख्य डाकघर, अमृतसर में बचत खाता नं0 764329 है। आपके इस खाते में ₹4,500/- हैं। आपको इस खाते में ₹ 1,500/- जमा करवाने हैं। इसलिए नीचे दिए गए प्रपत्र के प्रारुप को अपनी उत्तर-पुस्तिका पर उतार कर भरें-

उत्तर:

प्रश्न 3.
मान लीजिए आपका नाम विनोद कुमार है। आप पटियाला में रहते हैं। आपको अपने बचत खाता नं० 17321986 में र दो हज़ार जमा करवाने हैं। अत: आप नीचे दिए गए प्रपत्र की रूपरेखा को अपनी उत्तरपुस्तिका पर उतार कर भरें-

उत्तर:

(ख) डाकघर से संबंधित प्रपत्र

डाकघर में भी बैंक की तरह अनेक प्रकार से लेन-देन होता है तथा यहाँ भी रुपए जमा करने, निकालने, मनीआर्डर से रुपए भेजने आदि से संबंधित अनेक कार्य होते हैं, जिनके लिए अनेक प्रपत्रों का प्रयोग करना पड़ता है। डाकखाने में प्रयोग में आने वाले मुख्य प्रपत्र अग्रलिखित हैं-

1. खाता खोलने का प्रपत्र
2. रुपए जमा कराने का प्रपत्र
3. रुपए निकलवाने का प्रपत्र
4. मनीआर्डर से रुपए भेजने का प्रपत्र

1. डाकघर में खाता खोलने का प्रपत्र
डाकघर में खाता खोलने के लिए दिए गए प्रपत्र को ध्यानपूर्वक पढ़कर सभी कॉलम साफ़-साफ़ शब्दों में भरने चाहिए। खाता एक व्यक्ति अथवा संयुक्त रूप से दो-तीन व्यक्ति भी खोल सकते हैं। परिचायक का नाम, पता तथा खातेदारों के हस्ताक्षर करने होते हैं। खाते में नामांकन भी किया जा सकता है। उदाहरण के लिए एक भरा हुआ प्रपंत्र यहाँ दिया जा रहा है-
डाकघर बचत-बैंक
खाता खोलने के लिए आवेदन पत्र

4. मैंहम किसी भी समय सुसंगत नियम में विनिर्दिष्ट सीमा के भीतर अपने सभी एकल या संयुक्त बचत बैंक/सा० स० ज० खातों में अतिशेष बनाए रखने का और डाकघर बचत बैंक से मांग किए जाने पर ऐसे सभी खातों की विशिष्टियां भी देने का वचन देता हूँ/देते हैं।

5. मैं/हम केंद्रीय सरकार द्वारा बनाए गए ऐसे नियमों का पालन करने के लिए सहमत हूँ/हैं जो समय-समय पर खाते को लागू हों।

6. मैं हम सरकारी बचत बैंक अधिनियम, 1873 (1873 का 5) की धारा 4 के अधीन नीचे व्यक्ति (व्यक्तियों) को अपनी मृत्यु हो जाने की दशा में खाते में जमा रकम के लिए एकमात्र प्राप्तकर्ता (प्राप्तकर्ताओं) के रूप में नाम निर्दिष्ट करता हूँ करते हैं।

2. डाकघर में पैसे जमा कराने का प्रपत्र
डाकघर में धनराशि जमा करवाने के प्रपत्र में खाताधारी को डाकघर का पता, दिनांक, खातेधारी का नाम, खाता संख्या, जमा कराने की राशि शब्दों और अंकों में, जमा के बाद कुल राशि लिख कर अपने हस्ताक्षर करने होते हैं। इसके साथ डाकघर बचत बैंक की पास बुक भी लगानी होती है, जिसमें बचत बैंक सहायक जमा की गई राशि लिखकर, डाकघर की मोहर लगाकर तथा अपने हस्ताक्षर कर जमाकर्ता को वापिस कर देता है। यहां जालंधर के डाकघर में सतीश कुमार शर्मा के खाता संख्या 1901 में एक हज़ार रुपए जमा करने के प्रपत्र को भरकर उदाहरण के रूप में दिया जा रहा है।

3. डाकघर बचत बैंक से पैसे निकलवाने का प्रपत्र
डाकघर बचत बैंक से पैसे निकलवाने के लिए प्रपत्र में जमाकर्ता को डाकघर का नाम, दिनांक, खाता संख्या, निकाली गई राशि, शेष राशि, यदि आवश्यकता हो तो संदेशवाहक का नाम, हस्ताक्षर देने होते हैं। अदायगी आदेश में राशि मिलने के हस्ताक्षर खातेदार अथवा संदेशवाहक द्वारा किए जाते हैं। यहां एक उदाहरण के रूप में भरा हुआ प्रपत्र दिया जा रहा है-

प्रश्न 3.
मान लीजिए आपका नाम नरेश कुमार है। आपको लोकेश कुमार को दिनांक 10.04.2016 को ₹10,000/- का स्वहस्ताक्षरित रेखांकित किया हुआ चेक लिखकर देना है। इस अनुसार निम्नलिखित चेक के प्रपत्र को भरें-

उत्तर:

4. मनीऑर्डर से रुपये भेजने का प्रपत्र
मनीऑर्डर से रुपए भेजने का काम डाकघर में वर्षों से हो रहा है। डाकघर से मनीऑर्डर प्रपत्र लेकर भरना होता है। इसके 6 भाग होते हैं। पहला भाग जिसे रुपये भेजने हैं रुपयों की संख्या शब्दों और अंकों में प्राप्तकर्ता का नाम, पता, पिनकोड, दिनांक लिखकर भेजने वाला अपने हस्ताक्षरकर्ता है। चौथा भाग में भेजने वाले का नाम पता तथा पिन कोड लिखना होता है। छठे भाग में भेजने वाला संदेश लिख सकता है। दूसरा, तीसरा और पांचवां भाग डाक विभाग ने भरना होता है। डाविभाग भेजने वाली राशि पर निर्धारित शुल्क भी लेता है। यहाँ मनीऑर्डर प्रपत्र को भर कर प्रस्तुत किया जा रहा है-

(संदेश के लिए स्थान)
जालन्धर शहर प्रिय, सुजान
02-12-2000 तुम्हारे पत्र के अनुसार 800 सौ रुपए भेज रहे हैं । मिलने पर सूचना देना और रुपयों की ज़रूरत हो तो लिखना। सब को सत श्री अकाल कहना।
तुम्हारा मित्र,
सुखदेव सिंह

5. इलेक्ट्रानिक मनीआर्डर (ई०एम०ओ०) आधुनिक युग में इंटरनेट की प्रगति के साथ-साथ अपनी धनराशि किसी दूसरे को भेजने का इलेक्ट्रानिक मनीऑर्डर जल्दी से जल्दी और सरलता से भेजने का डाक विभाग द्वारा चलाया गया मनीऑर्डर का नया रूप है। इससे आम मनीआर्डर भेजने की अपेक्षा कम समय लगता है। जिस दिन ई०एम०ओ० किया जाता है, वह उसी दिन प्राप्तकर्ता को मिल जाता है। इस पर भी डाक विभाग निर्धारित शुल्क लेता है। इसके प्रपत्र की रूपरेखा यहाँ प्रस्तुत की जा रही है-

6. मोबाइल मनीट्रांसफर सर्विस (इंसटेंट मनीऑर्डर) मोबाइल मनीट्रांसफर सर्विस धन भेजने का सबसे तेज़, सुविधाजनक, विश्वसनीय और सरल माध्यम है। इसे इंसटेट मनीऑर्डर अथवा तत्काल मनीआर्डर भी कहते हैं। इस माध्यम से कुछ ही मिनटों में धनराशि प्राप्तकर्ता को मिल जाती है। डाकघर से प्राप्त प्रपत्र को भर कर अधिकारी को देने पर प्रेषक को कम्प्यूटर द्वारा सोलह अंकों का सीलबंद गुप्त नम्बर दिया जाता है, जिसे प्रेषक प्राप्तकर्ता को फोन, एस०एम०एस० अथवा ई-मेल द्वारा बता देता है, जो इंसटेंट मनीऑर्डर के अपने नगर के केंद्र पर जाकर वहाँ के अधिकारी को बताता है तथा अपनी पहचान का फोटो पहचानपत्र दिखाकर प्रेषक द्वारा उसे भेजी गई राशि प्राप्त कर सकता है। इस सेवा में उन्नीस हज़ार रुपए तक की राशि नकद प्राप्त की जा सकती है जबकि बीस हज़ार रुपए अथवा उससे अधिक राशि का भुगतान चैक द्वारा किया जाता है। इस सेवा को प्राप्त करने का प्रपत्र अग्रलिखित है-

प्रश्न 1.
इन्सटेंट मनीऑडर्र (आई०एम०ओ) किसे कहते हैं?
उत्तर:
मोबाइल मनीट्रांसफर सर्विस धन भेजने का सबसे तेज़, सुविधाजनक, विश्वसनीय और सरल माध्यम है। इसे इंसटेट मनीऑर्डर अथवा तत्काल मनीआर्डर भी कहते हैं। इस माध्यम से कुछ ही मिनटों में धनराशि प्राप्तकर्ता को मिल जाती है। डाकघर से प्राप्त प्रपत्र को भरकर अधिकारी को देने पर प्रेषक को कम्प्यूटर द्वारा सोलह अंकों का सीलबंद गुप्त नम्बर दिया जाता है, जिसे प्रेषक प्राप्तकर्ता को फोन, एस०एम०एस० अथवा ई०-मेल द्वारा बता देता है, जो इंसटेंट मनीऑर्डर के अपने नगर के केंद्र पर जाकर वहाँ के अधिकारी को बताता है तथा अपनी पहचान का फोटो पहचानपत्र दिखाकर प्रेषक द्वारा उसे भेजी गई राशि प्राप्त कर सकता है। इस सेवा में उन्नीस हजार रुपए तक की राशि नकद प्राप्त की जा सकती है जबकि बीस हज़ार रुपए अथवा उससे अधिक राशि का भुगतान चैक द्वारा किया जाता है।

7. अंतर्राष्ट्रीय धन अंतरण
डाकघर में विदेशों से धन मंगवाने अथवा भेजने के लिए मनीऑर्डर सेवाओं के अन्तर्गत अंतर्राष्ट्रीय धन अंतरण सेवा, मनीग्राम, इलेक्ट्रानिक क्लियरेंस सेवाएं (ई०सी०एस०) आदि भी समुचित मूल्य पर उपलब्ध हैं।

8. इंडियन पोस्टल ऑर्डर
इंडियन पोस्टल ऑर्डर डाकघर से निश्चित राशि के प्राप्त किये जा सकते हैं। इनके द्वारा एक व्यक्ति किसी दूसरे व्यक्ति अथवा संस्था को धनराशि भेज सकता है। ये भारत के सभी डाकघरों से प्राप्त किए जा सकते हैं। विभिन्न प्रतियोगी परीक्षाओं, नौकरियों के आवेदन-पत्रों आदि का शुल्क इंडियन पोस्टल ऑर्डर द्वारा मांगा जाता है। इन्हें रेखांकित भी कर सकते हैं। इसकी खरीद तिथि से वैधता दो वर्षों की होती है। इसे प्रयोग नहीं करने पर निश्चित अवधि में वापस करने पर खरीदी हुई राशि वापस भी ली जा सकती है।

इसके दो भाग फॉइल और काउंटर फॉइल होते हैं जिसके नाम इंडियन पोस्टल ऑर्डर भेजा जाता है उसे फॉइल कहते हैं और जो हिस्सा भेजने वाले के पास रहता है उसे काउंटर फॉइल कहते हैं। पोस्टल आर्डर में राशि प्राप्तकर्ता का नाम, पता, शहर, प्रेषक का नाम पता, दिनांक तथा जिस डाकघर से प्राप्तकर्ता को धनराशि प्राप्त करनी है उसका नाम लिखा जाता है। यदि इसे रेखाकिंत करना है तो वह भी किया जाता है। यहाँ इंडियन पोस्टल ऑर्डर का एक उदाहरण दिया जा रहा है, जिसमें गुरनाम सिंह, फिरोजपुर, पंजाब लोक सेवा आयोग, लुधियाना को एक रुपए का पोस्टल ऑर्डर भेज रहा है-

9. रजिस्ट्री-पत्र की पावती
डाकघर से किसी को रजिस्ट्री-पत्र भेजते समय उसके साथ रजिस्ट्री-पत्र की पावती भी लगानी होती है। यह पावती रजिस्ट्री-पत्र प्राप्तकर्ता हस्ताक्षर कर प्रेषक के पास भेज देता है। इसमें भेजने वाले को अपना तथा प्राप्तकर्ता का पूरा पता लिखना होता है। इसके आगे-पीछे दो भाग होते हैं। इसका प्रारूप निम्नलिखित है-
रजिस्ट्री-पावती प्रपत्र

(ग) रेलवे आरक्षण प्रपत्र
रेलवे में आरक्षण करवाने अथवा आरक्षण रद्द करवाने के लिए प्रपत्र भरना होता है। दोनों ही स्थितियों में एक प्रकार का ही प्रपत्र प्रयोग में आता है। भरने से पहले उस पर लिखना होता है कि आरक्षण लेने के लिए प्रपत्र भर रहे हैं । अथवा आरक्षण रद्द करवाने के लिए। इस प्रपत्र में तीन भाग होते हैं । प्रथम भाग में आरक्षण लेने अथवा रद्द की जाने वाली रेलगाड़ी का नाम, क्रमांक, यात्रा करने की तिथि, श्रेणी, कहां से कहाँ तक, यात्रियों के नाम, आयु, लिंग, आवेदक का नाम, पता, मोबाइल नम्बर तथा हस्ताक्षर होते हैं। दूसरा भाग में वापसी यात्रा का विवरण तथा तीसरा भाग रेलवे के कर्मचारी भरते हैं। इसका एक उदाहरण यहाँ दिया जा रहा है-

रेलवे आरक्षण कराने के लिए प्रपत्र का नमूना
उत्तर:

आगे की यात्रा/वापसी यात्रा का विवरण

टिप्पणी : 1. अधिकतम अनुपेय यात्रियों की संख्या प्रति मांग पत्र 6 व्यक्ति।
2. एक बार एक व्यक्ति से केवल एक ही मांग पत्र स्वीकार किया जाएगा।
3. कृपया खिड़की छोड़ने से पहले अपने टिकट और शेष राशि की जांच कर लें।
4. ठीक ढंग से न भरे हुए तथा अपठनीय फार्म स्वीकार नहीं किए जाएंगे।
5. विशेष मांग पर उपलब्धता के आधार पर विचार किया जाएगा।

अब रेलवे आरक्षण/आरक्षण रद्द करने की सुविधा इंटरनेट/मोबाइल फोन द्वारा भी उपलब्ध है, जिसकी पूरी जानकारी आई०आर०सी०टी०सी० की बेबसाइट पर उपलब्ध है।

1. मान लीजिए आपका नाम तमन्ना शर्मा है।आपका गुड बैंक, शाखा गंगानगर में एक बचत खाता है। आपका खाता नम्बर-45632789 है। आपका मोबाइल नं0 1009321432 है। आपको दिनांक 22.10.2015 को अपने इस खाते में 12,000 रुपये नकद जमा करवाने हैं। आपके पास जमा करवाने के लिए एक हज़ार के 8 नोट, पांच सौ के नोट तथा पचास के 20 नोट हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:
उत्तर:
गुड बैंक, गंगानगर
बैंक में रुपये जमा करवाने के लिए प्रपत्र ( बैंक प्रति)

गुड बैंक, गंगानगर
बैंक में रुपये जमा करवाने के लिए प्रपत्र (ग्राहक प्रति)

2. मान लीजिए आपका नाम रत्नलीन कौर है। आपका मोबाइल नम्बर 1890876535 है। आपका हिमालय बैंक, शाखा सोलन में एक बचत खाता नम्बर 96237180 है। आपको दिनांक 11.08.2015 को अपने इस खाते में से 7,000 रुपये निकालवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें
उत्तर:

3. मान लीजिए आपका नाम प्रकाश सिंह है। आपका भारतीय डाक के सेक्टर 15 चंडीगढ़ में 344565 नम्बर एक बचत खाता है। आपके इस खाते में अब तक 45,000 रुपये जमा हैं। आपको दिनांक 02.01.2016 को अपने इस खाते में से 4,000 रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:
उत्तर:

4. मान लीजिए आपका नाम तुषार कपूर है। आप मकान नं० 987, पुरानी दिल्ली रोड, गुड़गाँव (हरियाणा)-122001 में रहते हैं। आपका बेटा मकान नं० 456 सिविल लाइन्स बरेली (उत्तर प्रदेश) 243001 में रहता है। आप अपने बेटे को भारतीय डाक के माध्यम से दिनांक 05.11.2015 को मनीआर्डर के द्वारा 5,000/- रुपये भेजना चाहते हैं। इस प्रपत्र में उसे संदेश भी लिखें ‘अपनी सेहत का ध्यान रखना’। इस अनुसार निम्नलिखित प्रपत्र भरें।

5. मान लीजिए आपका नाम शिखर कुमार है। आपको लोकेश कुमार को दिनांक 6.12.15 को 10,000/- रु० का स्वहस्ताक्षरित रेखांकित किया हुआ चेक लिखकर देना है। इस अनुसार निम्नलिखित चैक के प्रपत्र को भरें:

बोर्ड परीक्षा में पूछे गए प्रश्न

निम्नलिखित में से किसी एक प्रपत्र को अपनी उत्तर-पुस्तिका पर उतारकर भरें
1. (क) मान लीजिए आपका नाम सुलोचना कुमारी है। आपका हिमालय बैंक, शाखा-शिमला में एक बचत खाता है, जिसका नम्बर, 18954326781 है। आपको अपने इस खाते में से 2500 रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें-

उत्तर:

(ख) मान लीजिए आपका नाम रूप सिंह है। आपका भारतीय डाक के सेक्टर 15, करनाल में एक बचत खाता है, जिसका नम्बर 128947654431 है। आपको अपने इस खाते में से 4000/- रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:

उत्तर:

2. (क) मान लीजिए आपका नाम पीयूष सिन्हा है। आपका कल्याण बैंक, शाखा-भुवनेश्वर में एक बचत खाता है, जिसका नम्बर 1150000234567 है। आपको अपने इस खाते में 18000/- रुपये जमा करवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:

उत्तर:

(ख) मान लीजिए आपका नाम दीदार सिंह है। आपका भारतीय डाक के सेक्टर-25, देहरादून में एक बचत खाता है, जिसका नम्बर 18634956744 है। आपको अपने इस खाते में से 10000/- रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें:

उत्तर:

3. (क) मान लीजिए आपका नाम निर्मला कौर है। आपका भारतीय डाक के सेक्टर-42, जम्मू में एक बचत खाता है, जिसका नम्बर 62745367823 है। आपको अपने इस खाते से 8200/- रुपये निकलवाने हैं। इस अनुसार निम्नलिखित प्रपत्र भरें-

उत्तर:

(ख) मान लीजिए आपका नाम दिनेश कुमार है। आपको मुनीश कुमार को 12000/- रु० का स्वहस्ताक्षरित चेक लिखकर देना है। इस अनुसार निम्नलिखित चेक के प्रपत्र को भरें:

उत्तर:

4. (क) निम्नलिखित में से किसी एक प्रपत्र को अपनी उत्तर-पुस्तिका पर उतारकर भरें-
(i) मान लीजिए आपका नाम पूनम कुमारी है। आपका भविष्य बैंक, शाखा-शिमला में एक बचत खाता है, जिसका नम्बर 373748488 है। आपको अपने इस खाते में से ₹ 5500/- निकलवाने हैं। इस अनुसार
निम्नलिखित प्रपत्र की रूपरेखा अपनी उत्तर पुस्तिका पर उतार कर भरें-

उत्तर:

(ख) मान लीजिए आपका नाम सुप्रिया देवी है। आपका भारतीय डाक के सेक्टर-15 नोएडा में एक बचत खाता है, जिसका नम्बर 834747993 है। आपको अपने इस खाते में से दिनांक 23.05.2018 को ₹4500/- निकलवाने हैं। इस अनुसार अग्रलिखित प्रपत्र की रूपरेखा अपनी उत्तर-पुस्तिका पर उतार कर भरें:

उत्तर: