# Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Solutions Answers

Here you will find Quadratic Equations Class 10 Extra Questions Maths Chapter 4 with Answers Solutions, Extra Questions for Class 10 Maths are solved by experts and will guide students in the right direction.

## Extra Questions for Class 10 Maths Quadratic Equations with Answers Solutions

Extra Questions for Class 10 Maths Chapter 4 Quadratic Equations with Solutions Answers

### Quadratic Equations Class 10 Extra Questions Objective Type

Question 1.
The solution of quadratic equation x2 – x – 2 = 0 are:
(a) 1, -2
(b) -1, -2
(c) -1, 2
(d) 1, 2.
(c) -1, 2

Question 2.
The zero of the polynomial x2 + 2x – 3 are:
(a) 1, -3
(b) -1, 3
(c) -1, -3
(d) 1, 3
(a) 1, -3

Question 3.
The degree of the polynomial x3 – x + 7 is:
(a) 1
(b) 2
(c) 3
(d) none of these.
(c) 3

Question 4.
The zero of the polynomial p(x) = x2 + 1 are:
(a) real
(b) not real
(c) (a) and (b) both
(d) none of these.
(b) not real

Question 5.
(i) Every quadratic polynomial can have at the most:
(a) one zero
(b) two zeroes
(c) three zeroes
(d) none of these.
(b) two zeroes

(ii) From the equation 4√5 x2 + 7x – 3√5 = 0, the value of x will be:
(a)
(b)
(c)
(d)
(a)

Question 6.
If one root of quadratic equation x2 + kx + 3 = 0 is 1, then the value of k will be:
(a) 1
(b) -3
(c) -4
(d) -5.
(c) -4

Question 7.
If one root of quadratic equation 2x2 + px – 4 = 0 is 2, then the value of p will be:
(a) -3
(b) -2
(c) 2
(d) 3.
(b) -2

Question 8.
If one root of quadratic equation ax2 + bx + c = 0 is 1, then:
(a) a = 1
(b) b = 1
(c) c = 1
(d) a + b + c = 0
(d) a + b + c = 0

Question 9.
If 2x2 + 1 = 33, then the value of x will be:
(a) ±2
(b) ±3
(c) ±4
(d) ±1.
(c) ±4

Question 10.
One root of quadratic equation x2 + 3x – 10 = 0:
(a) -2
(b) +2
(c) 0
(d) 1
(b) +2

Question 11.
If = $$\frac {x}{6}$$ = $$\frac {6}{x}$$, then the value of x will be:
(a) 6
(b) -6
(c) 6
(d) none of these.
(c) 6

Question 12.
Discriminant of ax2 + bx + c = 0) is:
(a)
(b)
(c)
(d) none of these.
(d) none of these.

Question 13.
A quadratic equation ax2 + bx + c = 0, a ≠ 0 has equal roots if b2 – 4ac is:
(a) equal to 0
(b) ≥ 0
(c) ≤ 0
(d) none of these.
(a) equal to 0

Question 14.
The roots of ax2 + bx + c = 0, a 50 are real and unequal, if b2 – 4ac is:
(a) equal to 0
(b) ≥ 0
(c) ≤ 0
(d) none of these.
(b) ≥ 0

Question 15.
The roots of the equation 2x2 – 8x + c = 0 are eaual, the value of c is:
(a) 2
(b) 4
(c) 6
(d) 8
(d) 8

Question 16.
If the discriminant (D) of a quadratic equation ax2 + bx + c = 0, a ≠ 0 is greater than zero, the roots are:
(a) real and unequal
(b) real and equal
(c) not real
(d) none of these.
(a) real and unequal

Question 17.
An expression in α and β is called symmetrical expression if by interchanging α and β, the expression is:
(a) changed
(b) not changed
(c) may be (a) and (b)
(d) none of these.
(b) not changed

Question 18.
If x2 + 5bx + 16 = 0 has no real roots, then:
(a) b > $$\frac {8}{5}$$
(b) b < $$\frac {-8}{5}$$
(c) $$\frac {-8}{5}$$ < b < $$\frac {8}{5}$$
(d) none of these.
(c) $$\frac {-8}{5}$$ < b < $$\frac {8}{5}$$

Question 19.
If the roots of 5x2 – px + 1 = 0 are real and distinct, then:
(a) p > √15
(b) p > – 2√5
(c) -2√5 < p < 2√5
(d) p > 2√5 or p < – 2√5.
(d) p > 2√5 or p < – 2√5

Question 20.
The sum of the roots of the quadratic equation 3x2 + 4x = 0 is:
(a) 0
(b) –$$\frac {3}{4}$$
(c) –$$\frac {4}{3}$$
(d) $$\frac {4}{3}$$
(c) –$$\frac {4}{3}$$

Question 21.
The product of the roots of quadratic equation 3x2 – 4x = 0) is:
(a) 0
(b) –$$\frac {3}{4}$$
(c) –$$\frac {4}{3}$$
(d) –$$\frac {4}{3}$$
(a) 0

Question 22.
If one root of x2 + kx + 3 = 0 is 1, then the value of k will be:
(a) -4
(b) -3
(c) 1
(d) 5.
(b) -3

Question 23.
If the sum of the roots of the equation 3x2 + (2k + 1)x – (k + 5) = 0 be equal to their product, the value of k is:
(a) 5
(b) 4
(c) 3
(d) 2
(b) 4

Question 24.
The sum of the roots of quadratic equation 5 – 7x + 3x2 = 0 is:
(a) +$$\frac {7}{5}$$
(b) –$$\frac {7}{5}$$
(c) –$$\frac {7}{3}$$
(d) +$$\frac {7}{3}$$
(d) +$$\frac {7}{3}$$

Question 25.
If one root of quadratic equation x3 – 3x + 2 = 0 is 2, then the second root is:
(a) 3
(b) -1
(c) 1
(d) 2
(c) 1

Question 1.
If one root quadratic equation x2 + 2x – p = 0 is -2, then find the value of p.
Solution.
Given that – 2 is a root of given quadratic equation x2 + 2x – p = 0
∴ (-2)2 + 2 (-2) – p = 0
⇒ 4 – 4 – p = 0
∴ p = 0
Hence, p = 0

Question 2.
If x2 – 5 = 0, then find the value of x:
Solution.
x2 – $$\frac {1}{9}$$ = 0
x = ±$$\frac {1}{3}$$

Question 3.
If x = 12 and y = 5, then find the value of 9x2 + 49y2 – 42xy.
Solution.
We have:
9x2 + 49y2 – 42xy
= (3x)2 + (7y)2 – 2.3x.7y
= (3x – 7y)2
= (3.12 – 7.5)2
= (36 – 35)2 = 1

Question 4.
Solve the following quadratic equation for x:
4x2 + 46x – (a2 – 12) = 0
Solution.
4x2 + 4bx – a2 + b2 = 0
⇒ (2x)2 + 2.2xb + (b)2-(a)2 = 0
⇒ (2x + b)2-(a)2 = 0
⇒ (2x + b + a) (2x + b – a) = 0

Question 5.
Find the roots of the quadratic equation:
3x2 – 2 √6 x + 2 = 0
Solution.
3x2 – 2√6 x + 2 = 0
= (√3x)2 – 2√3x × √2 +(√2)2 = 0
(√3x – √2)2 = 0
∴ roots are $$\frac{\sqrt{2}}{\sqrt{3}}$$ or $$\frac{\sqrt{2}}{\sqrt{3}}$$

Question 6.
Find the equation whose roots are b – 2a and b + 2a.
Solution.
α = b – 2a
β = b + 2a
Sum of roots = α + β = 6-2a +b + 2a = 2b.
Product of roots = αß = (b – 2a)(b + 2a)
= b2 – (2a)2
= b2 – 4a2
∴ x2 – (sum of roots) x + product of roots = 0
∴ x2 – (2b)x + (b2 – 4a2) = 0
∴ x2 – 2bx + b2 – 4a2 = 0

Question 7.
If α and β are the roots of the equation x2 – x – 2 = 0, find that quadratic equation whose roots are (2α + 1) and (2β + 1).
Solution.
We have
x2 – x – 2 = 0

The required equation is is

Question 8.
If – 5 is a root of quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x) + k = 0 has equal roots. Find the value of k.
Solution.
Given – 5 is a root of quadratic equation 2x2 + px – 15 = 0
∴ 2(-5)2 + p(-5) – 15 = 0
50 – 5p – 15 = 0
5p = 35 ⇒ p = 7
Putting the value of p = 7 in given (ii) quadratic equation.
7x2 + 7x + k = 0
as the roots of this equation are equal
∴ b2 = 4ac
⇒ (7)2 = 4 × 7 × k
∴ k = $$\frac {7}{4}$$

Question 9.
If aand Bare the roots of the equation ax2 + bx + b = 0, then prove that:

Solution.
Since, a and ß are the roots of the equation
ax2 + bx + b = 0

Proved.

Question 10.
The length and breadth of a room are 15 m and 12 m. There is a verandah surrounding the room and area of this is 90m2. Find the width of verandah.
Solution.
Suppose the width of Verandah is x metre.
Length of verandah = (2x + 15) m.
Breadth of verandah = (2x + 12) m.
Area of verandah = (2x + 15) (2x + 12) – 15 × 12 = 90

∴ 4x2 + 24x + 30x + 180 – 180 – 90 = 0
∴ 4x2 + 54x – 90 = 0
∴ 2x2 + 27x – 45 = 0
∴ 2x2 + 30x – 3x – 45 = 0
∴ 2x(x + 15) – 3(x + 15) = 0
(x + 15) (2x – 3) = 0
x + 15 = 0
or 2x – 3 = 0
2x = 3
x = 3/2
The width of verandah = 3/2 m.

Question 11.
The square of a positive integer is greater than 11 times the integer by 26. Find the positive integer.
Solution.
Let the positive integer be x then according to the given condition,
x2 = 11x + 26
⇒ x2 – 11x – 26 = 0
⇒ x2 – 13x + 2x – 26 = 0
⇒ x(x – 13) + 2(x – 13) = 0
⇒ (x – 13)(x + 2) = 0
∴ x = -2 or x = 13
Neglecting the negative value of x, we have x = 13.
Hence, the integer is 13.

Question 12.
A dealer sells an article for ₹ 24 and gains as much per cent as the cost price of the article. Find the cost price of the article.
Solution.
Let cost price of article be ₹ x. Then, gain = x%.

⇒ x2 + 100x – 2400 = 0
⇒ (x + 120)(x – 20) = 0
⇒ x = -120 or x = 20
⇒  x = 20
[∴ Cost Price can never be negative]
Hence, the cost price of the article is ₹ 20.

Question 1.
The area of a right angle triangle is 30cm2. Find the lengths of the base if its height is 7 cm more then base.
Solution.
Let the length of base of triangle = x cm
∴ length of its height = (x + 7) cm

Area of right angle ∆ = $$\frac {1}{2}$$ × base × height
⇒ 30 = $$\frac {1}{2}$$x.(x + 7)
⇒ x2 + 7x – 60 = 0
⇒ x2 + (12 – 5)x – 60 = 0
⇒ x2 + 12x – 5x – 60 = 0
⇒ x(x +12) – 5(x + 12) = 0
⇒ (x + 12) (x – 5) = 0
∴ x = 5 or – 12
as length is never negative
∴ length of base = 5 cm.

Question 2.
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution.
Given that, hypotenuse of right triangle = 13 cm
Let the base of the right triangle = x
According to the question,
Altitude of the triangle = 7 cm
Less than its base = (x – 7) cm

By Pythagoras theorem, we have
In ∆ABC AC2 = BC2 + Ab2
⇒ (13)2 = x2 + (x – 7)2
⇒ 169 = x2 + x2 – 14x + 49
[∵ (a – b)2 = a2 – 2ab + b2]
⇒ 2x2 – 14x – 120 = 0
⇒ x2 – 7x – 60 = 0
x2 – (12x – 5x) – 60 = 0
⇒ x2 – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12) (x + 5) = 0
Now, x – 12 = 0 x = 12
and x + 5 = 0 = x = -5
Since, altitude of the triangle cannot be negative, hence x ≠ -5
Hence, base of the triangle = 12 cm
and altitude of the triangle = 12 – 7 = 5 cm.

Question 3.
Solve the equation

Solution.
Given,

⇒ (3x + 4)(x + 4) = 4(x + 1)(x + 2)
⇒ 3x2 + 16x + 16 = 4(x2 + 3x + 2)
⇒ 3x2 + 16x + 16 = 4x2 + 12x + 8
⇒ x2 – 4x – 8 = 0
⇒ Here a =1, b = -4, and c = -8.

Question 4.
A motor boat whose speed in still water is 24 km/h. takes 1 hour more to go 32 km up stream than to return down stream to the same spot. Find the speed of stream.
Solution.
Let the speed of stream be x km/h.
Given that the speed of boat in still water is 24 km/h
∴ Speed of boat in down stream = (24 + x) km/h
and speed of boat in up stream = (24 – x) km/h
Time taken by boat to travel 32 km in down

⇒ 32 × 2x = (24 – x) (24 + x)
⇒ 64x = 576 – x2
⇒ x2 + 64x – 576 = 0
⇒ x2 + (72 – 8)x – 576 = 0
⇒ x2 + 72x – 8x – 576 = 0
⇒ x(x + 72) – 8(x + 72) = 0 (x – 8)(x + 72) = 0
∴ x = 8 or -72
as speed is never negative
∴ x = 8
i.e., speed of stream = 8 km/h.

Question 5.

by factorization method
Solution.

⇒ x(a + b + x) = – ab
⇒ ax + bx + x2 + ab = 0
⇒ x2 + ax + bx + ab = 0
⇒ x(x + a) + b (x + a) = 0
⇒ (x + a) (x + b) = 0
∴ x = -a or -6.

Question 6.
If the equations x2 + kx + 64 = 0 and x2 – 8x + k = 0 have real roots. Find the positive value of k.
Solution.
x2 + kx + 64 = 0. …(i)
and x2 – 8x + k = 0…(ii)
If the equations have real roots, then D ≥ 0 Discriminant of (i) k2 – 256 and equation

(ii) 64 – 4k
= k2 – 256 ≥ 0 and 64 – 4k ≥ 0
⇒ k2 ≥ 256 and 64 ≥ 4k
⇒ k ≥ 16 and 16 ≥ k.
Hence, k = 16.

Question 7.
If the speed of a train is increased by 5 km/H then train takes 1 hour less time to cover a distance of 360 km. Find the speed of the train.
Solution.
Let the speed of train be x km/ Hr.
∴ Time taken by the train to cover 360
km = $$\frac {360}{x}$$ Hours
New speed of the train = (x + 5) km/Hr.
∴ New time taken to cover 360 km

⇒ 360 × 5 = x2 + 5x
⇒ x2 + 5x – 1800 = 0
⇒ x2 + 45x – 40x – 1800 = 0
⇒ x(x + 45) – 40(x + 45) = 0
⇒ (x + 45) (x – 40) = 0
∴ x = 40 or -45
as speed is never negative
∴ speed of train = 40 km/hour.

Question 8.
Solve the following equation

Solution.
Let = $$\frac{x-2}{x+2}$$ = y
∴ Equation reduced to
y + $$\frac {3}{y}$$ – 4 = 0
⇒ y2 + 3 – 4y = 0
⇒ y2 – 4y + 3 = 0
⇒ y2 – 3y – y + 3 = 0
⇒ y(y – 3)-1 (y – 3) = 0
⇒ (y – 3)(y – 1) = 0
∴ y = 3 or 1.
Where
y = 3 = $$\frac{x-2}{x+2}$$
3x + 6 = x – 2
⇒ 2x = -8
∴ x = -4
Where
y = 1 = $$\frac{x-2}{x+2}$$
x + 2 x + 2 = x – 2 which is not true
∴ x = -4.

Question 9.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is a. Find his present age.
Solution.
Let the present age of Rehman – x year
∴ Rehman’s age, 3 yr ago = (x – 3) yr
Rehman’s age, after 5 yr = (x + 5) yr
According to the question,

∴ 6x + 6 = x2 + 2x – 15
⇒ x2 + 2x – 6x – 15 – 6 = 0
x2 – 4x − 21 = 0
x2 – (7x – 3x) – 21 = 0
⇒ x2 – 7x + 3x – 21 = 0
(By factorization method)
⇒ x(x – 7) + 3(x – 7) = 0
⇒ (x – 7)(x + 3) = 0
∴ x – 7 = 0 ⇒ x = 7 or x + 3 = 0 ⇒ x = -3
But x +-3 age cannot be negative.
So, present age of Rehman = 7 yr.

Question 10.
The difference of two numbers is 2 and the sum of their square is 34. Find the number.
Solution.
Let Numbers be x and y, such x > y
∴ x – y = 2 …..(i)
and x2 + y2 = 34 …(ii)
from (i) x = 2 + y
⇒ (2 + y)2 + y2 = 34
⇒ 2y2 + 4y – 30 = 0
⇒ y2 + 2y – 15 = 0
⇒ (y + 5) (y – 3) = 0
∴ y = 3 or – 5
where y = 3, x = 2 + 3 = 5
∴ Numbers are 5 and 3.

Question 11.
The speed of boat in still water is 15km/H. The boat completes 30 km distance in going downstream and 30 km up stream in total 4 Hr 30 minute. Find the speed of stream.
Solution.
Let the speed of stream = x km/ Hr.
∴ speed of boat in down stream = (15 + x) km/H
and speed of boat in up stream = (15 – x) km/H
Time taken 30 km in down stream = $$\frac{30}{(15+x)}$$ Hrs

⇒ 2 × 30 × 30 = 9(225 – x2)
⇒ 200 = 225 – x2
⇒ x2 = 225 – 200
= 25
x = √25 = ± 5
as speed is never negative:. speed to stream = 5 km/Hr.

Question 12.
Determine whether the quadratic equation 9x2 + 7x – 2 = 0 has real roots. If so, find roots of quadratic equation.
Solution.
9x2 + 7x – 2 = 0
Here a = 9, b = 7 and c= -2
D = b2 – 4ac .
= (7)2 – 4 × 9 × -2
= 49 + 72 = 121
∴ roots of quadratic equation are real.
∴ roots are

Question 13.
By solving the equation

get a quadratic equation. Find the nature of roots. Using formula, solve the quadratic equation.
Solution.
Given,

Nature of root, D = b2 – 4ac = (4)2 – 4 × 1 × 0 = 16 > 0
∴ root of the equation are real.

Question 14.
Some vessels are manufactured in a day in a small industry. On a particular day it is observed that the cost of manufacturing each item was 3 more than twice the number of vessels manufactured on that day. If the total manufacturing cost on that day was 90, then find the number of manufactured vessels and manufacturing cost of each item.
Solution.
Let Number of vessel manufacture in a day = x
∴ Cost of manufacturing = (2x + 3).
According to Question,
x(2x +3) = 90
⇒ 2x2 + 3x – 90 = 0
⇒ 2x2 + 15x – 12x – 90 = 0
⇒ x( 2x + 15)-6 (2x + 15) = 0
⇒ (x – 6) (2x + 15) = 0
∴ x = 6 or – $$\frac {15}{2}$$
as number of vessels are not negative
∴ x = 6
Also the manufacturing cost = 2 × 6 + 3 = ₹ 15.

Question 15.
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 of the two friends = x yr
Then, age of other friend = (20 – x) yr
(∵ The sum of the ages of two friends is 20 yr)
4 yr ago age of one of the two friends = (x – 4) yr
4 yr ago age of the other friend
= (20 – x – 4) yr = (16 – x) yr
According to the question,
(x – 4)(16 – x) = 48
⇒ 16x – x2 – 64 + 4x = 48
⇒ x2 – 20x + 112=0
On comparing the above equation with ax2 + bx + c = 0, we get
a = 1, b = -20 and c = 112
∴ Discriminant, D = b2 – 4ac
= (-20)2 – 4 × 1 × 112
= 400 – 448 = 48 < 0
which implies that the real roots are not possible because this condition represents imaginary roots. So, the solution does not exist.

Question 16.
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 the breadth of the park = x metre
Then, according to the question, Perimeter of a rectangular park = 80 m
= Length + Breadth = 40 m
Length = (40 – x) m
∴ Area of a rectangular park = Length × Breadth = (40 – x)x m2
But according to the question,
Area of the rectangular park is 400 m2.
∴ (40 – x)x = 400
⇒ x2 – 40x + 400 = 0
⇒ x2 – 2x × 20 + (20)2 = 0
⇒ (x – 20)2 = 0
[∵ a2 – 2ab + b2 = (a – b)2]
⇒ x = 20
Thus, breadth of the park = 20 m
and length of the park = (40 – 20) m = 20 m
So, it is possible to design the rectangular park of perimeter 80 m and area 400 m2. But this rectangle will be a square of side 20 m.

Question 1.
Solve:

x ≠ 0.
Solution.
The given equation is:

Therefore, the given equation can be written as

Now, substituting y = x + $$\frac {1}{x}$$, we get
4y2 + 8y – 45 = 0
⇒ 4y2 + 18y – 10y – 45 = 0
⇒ 2y (2y + 9) – 5(2y +9) = 0
⇒ (2y + 9) (2y – 5) = 0

Now, from (ii), we get,
2(x2 + 1) = 5x
⇒ 2x2 – 5x + 2 = 0
⇒ 2x2 – 4x – x + 2 = 0
⇒ 2x(x – 2) – 1(x – 2) = 0
⇒ (x – 2)(2x – 1) = 0
x = 2 and x = $$\frac {1}{2}$$
Hence, the solutions of the given equation are

Question 2.
Solve: 2x4 – x3 – 11x2 + x + 2 = 0.
Solution.
Given equation is
2x4 – x3 – 11x2 – x + 2 = 0
Divide both sides by x2,

Now substitute x + $$\frac {1}{x}$$ = y
∴ 2y2 – y – 15 = 0
⇒ 2y2 – 6y + 5y – 15 = 0
⇒ 2y(y – 3) + 5(y – 3) = 0
⇒ (y – 3)(2y + 5) = 0
∴ y = 3 or y = –$$\frac{5}{2}$$
y = 3 = $$\frac{x^{2}+1}{x}$$
When
⇒ x2 + 1 = 3x
⇒ x2 – 3x + 1 = 0
Here, a = 1, b = -3 and c = 1

⇒ 2x2 + 5x + 2 = 0
⇒ 2x2 + 4x + x + 2 = 0
⇒ 2x(x + 2) + 1(x + 2) = 0
⇒ (x + 2)(2x + 1) = 0
x = -2, – $$\frac{1}{2}$$
Hence, the solutions are -2, –$$\frac{1}{2}$$, $$\frac{3 \pm \sqrt{5}}{2}$$.

Question 3.
Solve $$\sqrt{3 x+10}$$ + $$\sqrt{6-x}$$ = 6.
Solution.
Given equation is:
$$\sqrt{3 x+10}$$ + $$\sqrt{6-x}$$ = 6.
We must look for solutions which satisfy
3x + 10 ≥ 0 and 6 – x ≥ 0
x ≥ –$$\frac {10}{3}$$ and x ≤ 6
i.e., –$$\frac {10}{3}$$ ≤ x ≤ 6.
Now, we first transform one of the radicals co the R.H.S.
We have,
$$\sqrt{3 x+10}$$ = 6 – $$\sqrt{6-x}$$
Squaring both sides, we get
3x + 10 = 36 + 6 – x – 12$$\sqrt{6-x}$$
or 4x – 32 = -127$$\sqrt{6-x}$$
or 8 – x = 3$$\sqrt{6-x}$$
Again, squaring both sides, we get
64 + x2 – 16x = 9 (6 – x)
or 64 + x2 – 16x – 54 + 9x = 0
or x2 – 7x + 10 = 0
or (x − 2)(x – 5) = 0
∴ x = 2 and x = 5.
which satisfy $$\frac{-10}{3}$$ ≤ x ≤ 6.
Hence, the solutions are x = 2 and x = 5.

Question 4.
One person purchased some cloth for ₹ 2250. Had the price of cloth been ₹ 50 per metre higher, then he would have been able to purchase the cloth less by 1.5 metre. Find the measure of the cloth purchased and price per metre.
Solution.
Let the price per metre of cloth be ₹ x
Amount spend = ₹ 2250
∴ Length of cloth purchased
$$\frac{2250}{x}$$ = metre.
Now, new price of cloth per metre become = ₹ (x + 50)
∴ Length of cloth purchase
= $$\frac{2250}{x + 50}$$ metre.
According to the question,

⇒ 2250 × 50 × 2 = 3(x2 + 50x)
⇒ x2 + 50x – 75000 = 0
⇒ x2 + 300x – 250x – 75000 = 0
⇒ x(x + 300) – 250(x + 300) = 0
⇒ (x + 300)(x – 250) = 0
∴ x = -300 or x = 250
Neglecting the -ve sign value i.e.,
x = -300
Hence, cost of cloth per metre = ₹ 250.
and length of cloth $$\frac{2250}{250}$$ = 9 metre.

Question 5.
A two-digit number is such that the product of the digits is 12, when 36 is added to the number, the digit interchange their places. Find the number.
Solution.
Suppose two digit number is 10x + y.
Whose y = unit’s place and x = ten’s place digit.
According two question
xy = 12 …..(i)
and 10x + y + 36 = 10y + x.
∴ 9x – 9y = -36
∴ x – y = -4
x = y – 4
Put x = y – 4 in (i)
(y – 4)y = 12
∴ y2 – 4y – 12 = 0
∴ y2 – 6y + 2y – 12 = 0
∴ y(y – 6) + 2(y – 6) = 0
∴ (y – 6) (y + 2) = 0
y – 6 = 0 or y + 2 = 0
y = 6 or y = -2
(not acceptable).
∴ Put y = 6 in (i) we get
x (6) = 12
x = 2
The two digit number is
10x + y = 10(2) + 6 = 26.

Question 6.
The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side, find the sides of the field.
Solution.
Let PQRS be the rectangular field. Let the shorter side QR of the rectangle = x m.

According to the question,
Diagonal of the rectangle PR = 60 m more than the shorter side = (x + 60) m
Side of the rectangle PQ = 30 m more than the shorter side = (x + 30) m
∴ By Pythagoras theorem,
In ∆PQR, PR2 = PQ2 + QR2
(∵ In rectangle every adjacent side makes an angle 90° to each other)
⇒ (x + 60)2 = (x + 30)2 + x2
⇒ x2 + 120x + 3600 = x2 + 60x + 900 + x2
[∵ (a + b)2 = a2 + 2ab + b2]
⇒ (2x2 – x2) + (60x – 120x) + (900 – 3600) = 0
x2 – 60x – 2700 = 0
x2 – (90x – 30x) – 2700 = 0
x2 – 90x + 30x – 2700 = 0
(By factorization method)
⇒ x(x – 90) + 30(x – 90) = 0
⇒ (x – 90)(x + 30) = 0
∴ Either x – 90 = 0 ⇒ x = 90 or x + 30 = 0
⇒ x = – 30 which is not possible because side cannot be negative.
∴ x = 90
So, breadth of the rectangle = 90 m and length of the rectangle = 90 + 30 = 120 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution.
Let the required numbers be x and y, where x > y
Difference of squares of two numbers = 180
Given, x2 – y2 = 180 …(i)
Square of smaller number = 8 × Larger number
⇒ y2 = 8x …..(ii)
From Eqs. (i) and (ii), we have
x2 – 8x = 180
⇒ x2 – 8x – 180 = 0
x2 – (18x – 10x) – 180 = 0
⇒ x2 – 18x + 10x – 180 = 0
(By factorization method)
⇒ x(x – 18) + 10(x – 18) = 0
⇒ (x – 18)(x + 10) = 0
∴ x – 18 = 0 or x + 10 = 0
∴ x = 18 or x = – 10
Now, x = 18
⇒ y2 = 8 × 18 = 144
⇒ y = ± 12
⇒ y = 12 or -12 [From EQuestion (ii)]
Again, x = -10
⇒ y2 = [8 (-10)] = -80
which is not possible i.e., imaginary value.
Hence, the numbers are (18 and 12) or (18 and -12).

Question 8.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bengaluru (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 kmh-1 more than that of the passenger train, find the average speed of the two trains.
Solution.
Let the average speed of the passenger train = x kmh-1.
Then, the average speed of the express train = (x + 11) kmh-1
Time taken by passenger train to cover

⇒ x (x + 11) = 132 × 11
⇒ x2 + 11x – 1452 = 0
x2 + (44 – 33)x – 1452 = 0
⇒ x2 + 44x – 33x – 1452 = 0
(By fatorization method)
= x(x + 44) – 33(x + 44) = 0
⇒ (x + 44)(x – 33) = 0
∴ x + 44 = 0
⇒ x = -44 or x – 33 = 0
⇒ x = 33
Since, x ≠ -44 because speed can’t be negative.
Hence, the average speed of the passenger train = 33 km/h
and the average speed of the express train = (33 + 11) km/h = 44 km/h.

Question 9.
Out of a certain number of Saras birds, one-fourth of the number are moving about in lotus plants, $$\frac {1}{9}$$th coupled with $$\frac {1}{4}$$th as well as 7 times the square root of the number move on a hill, 56 birds remain in vakula tree. What is the total number of birds ?
Solution.
Let the total number of birds be x.
∴ Number of birds moving about in lotus plants = $$\frac {x}{4}$$
and the number of birds moving on a hill
⇒ $$\frac {x}{9}$$ + $$\frac {x}{4}$$ + 7√x
Also, the number of birds in vakula tree = 56
∴ According to the given information, we have

9x + 2x + 126 √x – 18x + 1008 = 0
⇒ -7x + 126 √x + 1008 = 0
⇒ x – 18√x – 144 = 0
Putting √x = y, we get
y2 – 18y – 144 = 0
⇒ y2 – 24y + 6y – 144 = 0
⇒ y(y – 24) + 6(y – 24) = 0
⇒ (y – 24)(y + 6) = 0
⇒ y = 24 or y = -6.
But y = -6, since √x = y is positive
∴ y = 24
⇒ √x = 24
∴ x = 576.
Hence, the total number of birds is 576.

Question 10.
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 a rectangular mango grove = x metre
∴ Length of a rectangular mango grove = 2x metre(By given condition)
∴ According to the question,
Area of rectangular mango grove = 800 m2
⇒ Length × Breadth = 2x (x) = 800
⇒ 2x2 = 800
⇒ x2 = 400
⇒ x = ± 20
But (∵ breadth never be negative)
∴ Length 2x = 40 m and Breadth = 20 m.

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