ML Aggarwal Quadratic Equations Ch-Test Class 10 ICSE Maths Solutions

ML Aggarwal Quadratic Equations Ch-Test Class 10 ICSE Maths Solutions . We Provide Step by Step Answer of Ch-Test Questions for Quadratic Equations in One Variable as council prescribe guideline for upcoming board exam.  Visit official Website  CISCE  for detail information about ICSE Board Class-10.

ML Aggarwal Solutions of Quadratic Equations in one Variable Chapter-Test Questions for ICSE Class 10 Maths

Board	ICSE
Subject	Maths
Class	9th
Chapter-5	Quadratic Equations
Writer / Book	Understanding
Topics	Solutions of Ch-Test
Academic Session	2024-2025

 Quadratic Equations in one Variable Ch-Test

(Class 10 ICSE ML Aggarwal Maths Solutions)

Solve the following equations (1 to 4) by factorization :

Question -1

(i) x² + 6x – 16 = 0
(ii) 3x² + 11x + 10 = 0

Answer -1

(i) x² + 6x – 16 = 0

⇒ x² + 8x – 2x – 16 = 0
⇒ x (x + 8) – 2 (x + 8) = 0
⇒ (x + 8)(x – 2) = 0
Either x + 8 = 0,
then x = –8
or
x – 2 = 0,
then x = 2
Hence x = –8, 2.
(ii) 3x² + 11x + 10 = 0

⇒ 3x² + 6x + 5x + 10 = 0
3x(x + 2) + 5(x + 2) = 0
⇒ (x + 2)(3x + 5) = 0
Either x + 2 = 0,
then x = –2
or
3x + 5 = 0,
then 3x = –5

⇒ x = –5/3

Hence x = –2, –5/3.

Question -2

(i) 2x² + ax – a² = 0

(ii)√3 x² + 10 x +7√3=0

Answer- 2

(i) 2x² + ax – a² = 0
2x² + 2ax – ax – a² = 0

(i) x(x + 1) + (x + 2)(x + 3) = 42
(ii)6/x -2/(x-1)=1/x-2

Answer -3

(i) x(x + 1) + (x + 2)(x + 3) = 42
⇒ 2x² + 6x + 6 – 42 = 0
⇒ x2 + 3x – 18 = 0
⇒ x² + 3x – 18 = 0   ..(Dividing by 2)
⇒ x² + 6x – 3x – 18 = 0
⇒ x(x + 6) 3(x + 6) = 0
⇒ (x + 6)(x – 3) = 0
Either x + 6 = 0,
then x = –6
or
x – 3 = 0,
then x = 3
Hence x = –6, 3.
(ii) 6/x -2/(x-1)=1/x-2

⇒ (4x – 6)(x – 2) = x6 – x
⇒ 4x² – 8x – 6x + 12 = x² – x
⇒ 4x² – 14x + 12 – x² + x = 0
⇒ 3x² – 13x + 12 = 0
⇒ 3x² – 9x – 4x + 12 = 0
⇒ 3x(x – 3) -4(x – 3) = 0
⇒ (x – 3)(3x – 4) = 0
Either x – 3 = 0,
then x = 3
or
3x – 4 = 0,
then 3x = 4

⇒ x = 4/3

Hence x = 3, 4/3.

Question -4

(i)√(x+15) =x +3
(ii)√(3x²-2x-1)=2x-2

Answer -4

(i) √(x+15) =x +3
Squaring on both sides
x + 15 = (x + 3)²

⇒ x² + 6x + 9 – x – 15 = 0
⇒ x² + 5x – 6 = 0
⇒ x² + 6x – x – 6 = 0
⇒ x(x + 6) –1(x + 6) = 0
⇒ (x + 6)(x – 1) = 0
Either x + 6 = 0,
then x = -6
or
x – 1 = 0,
then x = 1
∴ x = –6, 1
Check :
(i) If x = –6 then
L.H.S. = √(x+15)
= √(-6+15)
=√ 9
= 3
R.H.S. = x + 3
= –6 + 3
= –3
∵ L.H.S. ≠ R.H.S.
∴ x = –6 is not a root

(ii)Squaring both sides
3x² – 2x – 1 = (2x – 2)²
⇒ 3x² – 2x – 1 = 4x² – 8x + 4
⇒ 4x² – 8x + 4 – 3x² + 2x + 1 = 0
⇒ x² – 6x + 5 = 0
⇒ x² – 5x – x + 5 = 0
⇒ x(x – 5) –1(x – 5) = 0
⇒ (x – 5)(x – 1) = 0
Either x – 5 = 0,
then x = 5
or
x – 1 = 0,
then x = 1
Check :
(i) If x = 5, then

=8

R.H.S. = 2x – 2
= 2 x 5 – 2
= 10 – 2
= 8
∵ L.H.S. = R.H.S.
∴ x = 5 is a root
(ii) If x = 1, then

Question -5

(i) 2x² – 3x – 1 = 0
(ii) x(3x+1/2)=6

Answer -5

(i) 2x² – 3x – 1 = 0
Here a = 2, b = -3, c = -1
D = b² – 4ac
= (-3)² – 4 x 2 x (-1)
= 9 + 8
= 17

(ii)⇒ 6x² + x = 12
⇒ 6x² + x – 12 = 0
Here a = 6, b = 1, c = –12
D = b² – 4ac
= (1)² – 4 x 6 x (–12)
= 1 + 288
= 289

Question- 6

(i)( 2x+5)/(3x+4)=(x+1)/(x+3)
(ii) 2/(x+2)-1/(x+1)=4/(x+4)-3/(x+3)

Answer -6

(i) ( 2x+5)/(3x+4)=(x+1)/(x+3)
(2x + 5)(x + 3) = (x + 1)(3x + 4)

2x² + 6x + 5x + 15 = 3x² + 4x + 3x + 4
⇒ 3x² + 7x + 4 – 2x² – 11x – 15 = 0
⇒ x² – 4x – 11 = 0
Here a = 1, b = –4, c = –11
D = b² – 4ac
= (–4)² – 4 x 1 x (–11)
= 16 + 44

=60

(ii) 2/(x+2)-1/(x+1)=4/(x+4)-3/(x+3)

⇒ x² + 7x + 12 – x2 – 3x – 2 = 0
⇒ 4x + 10
⇒ 2x + 5 = 0
⇒ 2x = –5

⇒ x = -5/2
If x = 0, then

0/(x+2)(x+1)=0/(x+4)(x+3)

which is correct

hence x=0,-5/2

Question -7

(i)  (3x-4)/7+7/(3x-4)=5/2
(ii)( 4/x )-3 =5/(2x+3)

Answer -7

(i) (3x-4)/7+7/(3x-4)=5/2

let ( 3x–4)/7 = y, then

y+1/y=5/2

⇒ 2y² + 2 = 5y
⇒ 2y² – 5y + 2 = 0
⇒ 2y² – y – 4y + 2 = 0
⇒ y(2y – 1) –2(2y – 1) = 0
⇒ (2y – 1)(y – 2) = 0
Either 2y – 1 = 0,
then 2y = 1

⇒ y = 1/2
or
y – 2 = 0,
then y = 2

When y = 1/2,

then( 3x-4)/7=1/2
⇒ 6x – 8 = 7
⇒ 6x = 7 + 8
⇒ 6x = 15

⇒ x = 15/6=5/2
y = 2, then

(3x-4)/7=2/1
⇒ 3x – 4 = 14
⇒ 3x = 14 + 4 = 18

⇒ x = 18/3 = 6

∴ x = 6, 5/2.

(ii) ( 4/x )-3 =5/(2x+3)

⇒ (4 – 3x)(2x + 3) = 5x
⇒ 8x + 12 – 6x² – 9x – 5x = 0
⇒ –6x² – 6x + 12 = 0
⇒ x² + x – 2 = 0
⇒ x² + 2x – x – 2 = 0
⇒ x(x + 2) –1(x + 2) = 0
⇒ (x + 2)(x – 1) = 0
Either x + 2 = 0,
then x = –2
or
x – 1 = 0,
then x = 1
∴ x = 1, –2.

Question- 8

(i)  x² + (4 – 3a)x – 12a = 0
(ii)  10ax² – 6x + 15ax – 9 = 0,a≠0

Answer -8

(i)x² + (4 – 3a)x – 12a = 0
Here a = 1, b = 4 – 3a, c = -12a
∴ D = b² – 4ac
= (4 – 3a)² – 4 x 1 x (–12a)
= 16 – 24a + 9a² + 48a
= 16 + 24a + 9a² = (4 + 3a)

(ii)  10ax² – 6x + 15ax – 9 = 0,a≠0

Here a = 10a, b = –(6 – 15a), c = –9
D = b² – 4ac
= [–(6 – 15a)]² – 4 x 10a(–9)
= 36 – 180a + 225a² + 360a
= 36 + 180a + 225a² = (6 + 15a)²

= -30a/20a

= -3/2

Hence x = 3a/ 5

or -3/2.

Question- 9 Solve for x using the quadratic formula. Write your answer correct to two significant figures: (x – 1)² – 3x + 4 = 0. (2014)

Answer -9

(x – 1)² – 3x + 4 = 0
x² + 1 – 2x – 3x + 4 = 0

x² –  5x + 5 = 0
Here a = 1, b = – 5 and c = 5

= 3.618 or 1.382
∴ x = 3.618 ≈ 3.6

or 1.382 = 1.4(appr)

Question -10 Discuss the nature of the roots of the following equations:
(i) 3x² – 7x + 8 = 0
(ii) x² – x/2 – 4 = 0
(iii) 5x² – 6√5x + 9 = 0
(iv) √3x² – 2x – √3 = 0

Answer- 10

(i) 3x² – 7x + 8 = 0
Here a = 3, b = –7, c = 8
∴ D = b² – 4ac
= (–7)² – 4 x 3 x 8
= 49 – 96
= –47
∵ D < 0
∴ Roots are not real.

(ii) x² – x/2 – 4 = 0

Here a = 1, b = 12, c = –4

∴ D = b² – 4ac

= (1/2)²-4×1×(-4)

= (1/4)+16

= 65/4
∵ D > 0
∴ Roots are real and distinct.

(iii) 5x² – 6√5x + 9 = 0

Here a=5,b=-65,c=9
∴ D = b² – 4ac
= (-6√5)²-4×5×9
= 180 – 180
= 0
∴ D = 0
∴ Roots are real and equal.
(iv) √3x² – 2x – √3 = 0

Here a=√3,b=-2,c=-√3
∴ D = b² – 4ac
= (-2)²-4×√3×(-√3)
=  4 + 12
= 16
∵ D > 0
∴ Roots are real and distinct.

Question- 11 Find the values of k so that the quadratic equation (4 – k) x² + 2 (k + 2) x + (8k + 1) = 0 has equal roots.

Answer -11

Answer -12

3x² – 5x – 2m = 0
Here a = 3, b = -5, c = -2m

∴ D = b² – 4ac
= (5)² –4 x 3 x (–2m)
= 25 + 24m
∵ Roots are real and distinct
∴ D> 0
25 + 24m > 0
24m > –25

m > -25/24.

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Question -13 Find the value(s) of k for which each of the following quadratic equation has equal roots:
(i) 3kx² = 4 (kx – 1)
(ii)  (k + 4)x² + (k + 1)x + 1 =0
Also, find the roots for that value (s) of k in each case.

Answer -13

(i)3kx² = 4(kx – 1)
⇒ 3kx² = 4kx – 4
⇒ 3kx² – 4kx + 4 = 0
Here a = 3k, b = –4k, c = 4
∴ D = b² – 4ac
= (–4k)² – 4 x 3k x 4
= 16k² – 48k
∴ Roots are equal
∴ D = 0
⇒ 16k² – 48k = 0
⇒ k² – 3k = 0
⇒ k(k – 3) = 0
Either k = 0
or
k – 3 = 0
⇒ then k = 3
(k + 4)x² + (k + 1)x + 1 =0
Here a = k + 4, b = k + 1, c = 1
∴ D = b²  4ac
= (k + 1)² – 4 x (k + 4) x 1
= k² + 2k + 1 – 4k – 16
= k² – 2k – 15
∵ Root are equal
∴ k² – 2k – 15 = 0
⇒ k² – 5k + 3k – 15 = 0
⇒ k(k – 5) + 3(k – 5) = 0
⇒ (k – 5)(k + 3) = 0
Either k – 5 = 0,
then k = 5
or
k + 3 = 0,
then k = –3

Question -14 Find two natural numbers which differ by 3 and whose squares have the sum 117.

Answer- 14

Let first natural number = x
then second natural number = x + 3
According to the condition :
x² + (x + 3)² = 117

⇒ x² + x² + 6x + 9 = 117
⇒ 2x² + 6x + 9 – 1117 = 0
⇒ 2x² + 6x – 108 = 0
⇒ x² + 3x – 54 = 0   …(Dividing by 2)
⇒ x² + 9x – 6x – 54 = 0
⇒ x(x + 9) –6(x + 9)  0
⇒ (x + 9)(x – 6) = 0
Either x + 9 = 0,
then x = –9,
but it is not a natural number.
or
x – 6 = 0,
then x = 6
∴ First natural number = 6
and second numberr = 6 + 3 = 9

Question -15 Divide 16 into two parts such that the twice the square of the larger part exceeds the square of the smaller part by 164.

Answer -15

Let larger part = x
then smaller part = 16 – x
(∵ sum = 16)
According to the condition
2x² – (116 – x)² = 164
⇒ 2x² – (256 – 32x + x²) = 164
⇒ 2x² – 256 + 32x – x² = 164
⇒ x² + 32x – 256 – 164 = 0
⇒ x² + 32x – 420 = 0
⇒ x² + 42x – 10x – 420 = 0
⇒ x(x + 42) – 10(x + 42) = 0
⇒ (x + 42)(x – 10) = 0
Either x + 42 = 0,
then x = –42,
but it is not possible.
or
x – 10 = 0,
then x = 10
∴ Larger part = 10
and smaller part = 16 – 10 = 6.

Question- 16 Two natural numbers are in the ratio 3 : 4. Find the numbers if the difference between their squares is 175.

Answer- 16

Ratio in two natural numbers = 3 : 4
Let the numbers be 3x and 4x
According to the condition,

According to the condition,
(4x)² – (3x)² = 175
⇒ 16x² – 9x² = 175
⇒ 7x² = 175

⇒ x² = 175/7 = 25 = (±5)²
∴ x = 5, -5
But x – 5 is not a natural number
∴ x = 5
∴ Natural numbers are 3x, 4x
= 3 x 5, 4 x 5
= 15, 20.

Question -17 Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 sq. cm. Express this as an algebraic equation and solve it to find the sides of the squares.

Answer- 17

Side of first square = x cm .
and side of second square = (x + 4) cm
Now according to the condition,
(x)² + (x + 4)² = 656
⇒ x² – x² + 8x + 16 = 656
⇒ 2x² + 8x + 16 –  656 = 0
⇒ 2x² + 8x – 640 = 0
⇒ x² + 4x – 320 = 0  …(Dividing by 2)
⇒ x² + 20x – 16x – 320 = 0
⇒ x(x + 20) – 16(x + 20) = 0
⇒ (x + 20)(x – 16) = 0
EIther x + 20 = 0,
then x = –20,
but it not possible as it is in negative.
or
x – 16 = 0 then x = 16
Side of first square = 16 cm
and side of second square = 16 + 4 – 4 = 20 cm

Question- 18 The length of a rectangular garden is 12 m more than its breadth. The numerical value of its area is equal to 4 times the numerical value of its perimeter. Find the dimensions of the garden.

Answer -18

Let breadth = x m
then length = (x + 12) m
Area = l × b = x (x + 12) m²
and perimeter = 2 (l + b) = 2(x + 12 + x) = 2 (2x + 12) m
According to the condition.

Answer -19

Area of rectangular garden = 100 m²
Length of barbed wire = 30 m
Let the length of the side opposite to wall = x

and length of other each side = (30-x)/2
According to the condition,

x(30-x)/2 = 100

⇒ x(30 – x) = 200
⇒ 30x – x² = 200
⇒ x² – 30x + 200 = 0
⇒ x² – 20x – 10x + 200 = 0
⇒ x(x – 20) – 10(x – 20) = 0
⇒ (x – 20)(x – 10) = 0
Either x – 20 = 0,
then x = 20
or
x – 10 = 0,
then x = 10
(i) If x = 20,
then side opposite to the walI = 20m
and other side

= (30-20)/2

= 10/2
= 5m
(ii) If x = 10,
then side opposite to wall = 10m
and other side

= (30-10)/2

= 20/2
= 10m
∴ Sides are 20m, 5m or 10m.

Question- 20 The hypotenuse of a right-angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side, find the sides of the triangle.

Answer -20

Let the length of shortest side = x m
Length of hypotenuse = 2x – 1
and third side = x + 1
Now according to the condition,
(2x – 1)² = (x)² + (x + 1)²  …(By Pythagorus Theorem)
⇒ 4x² – 4x + 1 = x² + x² + 2x + 1
⇒ 4x² – 4x + 1 = 2x² – 2x – 1 = 0
⇒ 2x² – 6x = 0
⇒ x² – 3x = 0
⇒ x(x – 3) = 0  …(Dividing by 2)
Either x = 0,
but it is not possible
or
x – 3 = 0,
then x = 3
Shortest side = 3m
Hypotenuse = 2 x 3 – 1 = 6 – 1 – 5
Third side = x + 1 = 3 + 1 = 4
Hence sides are 3, 4, 5

Question -21 A wire ; 112 cm long is bent to form a right angled triangle. If the hypotenuse is 50 cm long, find the area of the triangle.

Answer -21

Perimeter of a right angled triangle = 112 cm
Hypotenuse = 50 cm
∴ Sum of other two sides = 112 – 50 = 62 cm
Let the length of first side = x
and length of other side = 62 – x
According to the condition
(x)² + (62 – x)² = (50)²  …(By Pythagorus Theorem)
⇒ x² + 3844 – 124x + x² = 2500
⇒ 2x² – 124x + 3844 – 2500 = 0
⇒ 2x² – 124 + 1344 = 0
⇒ x² – 62x + 672 = 0   …(Dividing by 2)
⇒ x² – 48x – 14x + 672 = 0
⇒ x(x –  48) –14(x – 48) = 0
⇒ (x – 48)(x – 14) = 0
Either x – 48 = 0,
then x = 48
or
x – 14 = 0,
then x = 14
(i) If x = 48,
then one side = 48cm
and other side = 62 – 48 = 14cm
(ii) If x = 14,
then one side = 14cm
and other side = 62 – 14 = 48
Hence sides are 14cm, 48cm.
Question -22 The speed of a boat in still water is 11 km/ hr. It can go 12 km up-stream and return downstream to the original point in 2 hours 45 minutes. Find the speed of the stream

Answer -22

Speed of a boat in still water = 11 km/hr
Let the speed of stream = x km/hr.
Distance covered = 12 km.
Time taken = 2 hours 45 minutes
now according to the condition

⇒ 1331 – 11x² = 4 x 12 x 22 = 1056
⇒ 1331 – 11x² = 1056
⇒ 1331 – 1056 – 11x² = 0
⇒ -11x² + 275 = 0
⇒ x² – 25 = 0  …(Dividing by -11)
⇒ (x + 5)(x – 5) = 0
Either x + 5 = 0,
then x = –5,
but it is not possible as it is in negative.
or
x – 5 = 0,
then x = 5
Hence speed of stream = 5km/hr.

Question- 23 A man spent Rs. 2800 on buying a number of plants priced at Rs x each. Because of the number involved, the supplier reduced the price of each plant by Rupee 1.The man finally paid Rs. 2730 and received 10 more plants. Find x.

Answer- 23

Amount spent = Rs. 2800
Price of each plant = Rs. x
Reduced price = Rs. (x – 1)

No. of plants in first case = 2800/x

No. of plants received in second case = (2800/x)+10
Amount paid = Rs. 2730
According to the condition,

(2800/x +10)(x-1)=2730

(2800 +10 x)(x-1)=2730x

⇒ (2800 + 10)(x – 1) = 2730x
⇒ 2800x – 2800 + 10x² – 10x – 2730 = 0
⇒ 10x² + 2800x – 10x – 2730x – 2800 = 0
⇒ 10x² + 60x – 2800 = 0
⇒ x² + 60x – 280 = 0  …(Dividing by 10)
⇒ x² + 20x – 14x – 280 = 0
⇒ x(x + 20) – 14(x + 20) = 0
⇒ (x + 20)(x – 14) = 0
Either x + 20 = 0,
then x = –20,
but it is not possible as it is in negative.
or
x – 14 = 0,
then x = 14.

Question- 24 Forty years hence, Mr. Pratap’s age will be the square of what it was 32 years ago. Find his present age.

Answer -24

Let Partap’s present age = x years

40 years hence his age = x + 40
and 32 years ago his age = x – 32
According to the condition
x + 40 = (x – 32)²
⇒ x + 40 = x² – 64x + 1024
⇒ x² – 64x + 1024 – x – 40 = 0
⇒ x² – 65x + 1024 – x – 40 = 0
⇒ x² – 65x + 984 = 0
⇒ x² – 24x – 41x + 984 = 0
⇒ x(x – 24) – 41(x – 24) = 0
⇒ (x – 24)(x – 41) = 0
Either x – 24 = 0,
then x = 24
but it is not possible as it is less than 32
or
x – 41 = 0,
then x = 41
Hence present age = 41 years.

— : End of ML Aggarwal Solutions Quadratic Equations Ch-Test Class 10 ICSE Maths Solutions :–

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