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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 6th Class Maths Book Solutions Chapter 7 Algebra Ex 7.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 7 Algebra Ex 7.3

1. Pick the algebraic expressions and the arithmetic expressions from the following:

Question (i)
(i) 2l – 3
(ii) 5 × 3 + 8
(iii) 6 – 3x
(iv) 51
(v) 2 × (21 – 18) + 9
(vi) \(\frac {6a}{5}\) + 2
(vii) 7 × 20 + 5 + 3
(viii) 8.
Solution:
Algebraic Expressions :
2l – 3, 6 – 3x, 51, \(\frac {6a}{5}\) + 2
Arithmetic Expressions :
5 × 3 + 8, 2 × (21 – 18) + 9, 7 × 20 + 5 + 3, 8.

2. Write the terms for the following expressions:

Question (i)
2y + 5z
Solution:
Terms of 2y + 5z = 2y, 5z

Question (ii)
6x – 3y + 8
Solution:
Terms of 6x – 3y + 8 = 6x, -3y, 8

Question (iii)
7a
Solution:
Terms of 7a = 7a

Question (iv)
3l – 5m + 2n
Solution:
Terms of 31 – 5m + 2n = 31, -5m, 2n

Question (v)
\(\frac {2l}{3}\) + x.
Solution:
Terms of = \(\frac {2l}{3}\) + x = \(\frac {2l}{3}\), x

3. Tell how the following expressions are formed.

Question (i)
a + 11
Solution:
a + 11 = a is increased by 11

Question (ii)
12 – x
Solution:
12 – x = x is subtracted from 12

Question (iii)
3z + 8
Solution:
3z + 8 = Three time of z is increased by 8

Question (iv)
6 – 5l
Solution:
6 – 5l = 5 times of l is subtracted from 6

Question (v)
\(\frac {5a}{4}\)
Solution:
\(\frac {5a}{4}\) = 5 times a is divided by 4.

4. Give expressions for the following:

Question (i)
10 is added to p
Solution:
10 is added to p = p + 10

Question (ii)
5 is subtracted from y
Solution:
S is subtracted from y = y – 5

Question (iii)
d is divided by 3
Solution:
d is divided by 3 = \(\frac {d}{3}\)

Question (iv)
l is multiplied by – 6
Solution:
l is multiplied by – 6 = – 6l

Question (v)
m is subtracted from l
Solution:
m is subtracted from 1 = 1 – m

Question (vi)
11 is added to 3x
Solution:
11 is added to 6x = 6x + 11

Question (vii)
y is divided by -2 and then 2 is added to the result
Solution:
y is multiplied by – 2 and then 2 is added to the result = – 2y + 2

Question (viii)
c is divided by 5 and then 7 is multiplied to the result
Solution:
c is divided by 5 and thus 7 is multiplied to the result = \(\frac {7c}{5}\)

Question (ix)
x is multiplied by 3 then subtracted this result from y
Solution:
x is multiplied by 3 then subtracted this result from y = y – 3x

Question (x)
a is added to b then c is multiplied with this result.
Solution:
a is added to b then c is multiplied by this result = (a + b) c

5. Write the number which is 15 less than y.
Solution:
The number which is 15 less than y = y – 15

6. Write the number which is 3 more than a.
Solution:
The number which is 3 more than a = a + 3

7. Find the number which is 1 more than twice of x.
Solution:
The number which is 1 more than twice of x = 2x + 1

8. Find the number which is 7 less than 5 times of y.
Solution:
The number which is 7 less than 5 times of y = 5y – 7

9. Somi’s present age is ‘a’ years. Express the following in algebraic form:

Question (i)
Her age after 15 years.
Solution:
Somi’s present age = ‘a’ years
Her age after 15 years = (a + 15) years

Question (ii)
Her age 2 years ago.
Solution:
Her age 2 years ago = (a – 2) years

Question (iii)
If Somi’s father’s age is 5 more than twice of her present age, express her father’ age.
Solution:
Somi’s father’s age is 5 more than twice of her present age
∴ Her father’s age = (2a + 5) years.

Question (iv)
If Somi’s sister is 4 years younger to her. Express her sister’s age.
Solution:
Somi’s sister is 4 years younger to her
∴ Her sister’s age = (a – 4) years

Question (v)
If Somi’s mother is 3 less than 3 times her present age. Express her mother’s age.
Solution:
Somi’s mother is 3 less than 3 times her present age
∴ Her mother’s age = (3a – 3) years.

10. The length of a floor is 10 more than two times of breadth what is the length if breadth is l meters?
Solution:
The breadth of floor = l metres
The length of the floor is 10 more than two times of its breadth
∴ Length of floor = (2l + 10) metres.

Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 7 Algebra Ex 7.2 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 6 Maths Chapter 7 Algebra Ex 7.2

1. Each side of equilateral triangle is denoted by ‘a’ then express the perimeter of the triangle using ‘a’.

Solution:
Each triangle of equilateral triangle = a
∴ Perimeter of equilateral triangle
= a + a + a = 3a

2. An isosceles triangle is shown. Express its perimeter in terms of ‘l’ and ‘b’.

Solution:
Perimeter of isosceles triangle = l + l + b
= 21 + b

3. Each side of regular hexagon is denoted by ‘S’ then express the perimeter of the regular hexagon using ‘S’.

Solution:
Each side of regular hexagon = S
Perimeter of regular hexagon
=S + S + S + S + S + S
= 6S

4. The cube has 6 faces and all of them are identify squares. If l is the length of an edge of a cube, find the total length of all edges of the cube in terms of ‘l’?

Solution:
Length of each edge of a cube = l
There are 12 edges of a cube
Total length of all edges of the cube
= 12 × l = 12l

5. Write commutative property of addition using variables x and y.
image
Solution:
According to commutative property of addition.
If the order of numbers, in addition, is changed it does not change their sum.
∴ x + y = y + x

6. Write associative property of multiplication using variables l, m and n.
Solution:
According to associative property of multiplication.
If three numbers can be multiplied in any order, it does not change their product.
∴ l × (m × n) = (l × m) × n

7. Write distributive property of multiplication over addition in terms of variables p, q and r respectively.
Solution:
According to Distributive property of multiplication over addition
p × (q + r) = p × q + p × r